Field Arithmetic, Third edition (Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge A Series of Modern Surveys in Mathematics)

16.11 Abelian Extensions of Hilbertian Fields We give here some results about Galois extensions of Hilbertian fields which involve small groups: Proposition 16.11.1: Let N be a Galois extension of a Hilbertian field K. Suppose Gal(N/K) is small. Then each separable Hilbert subset H of N r contains a separable Hilbert subset of K r . In particular, N is Hilbertian. Proof: By definition H = HK (f1 , . . . , fk ; g), where fi ∈ N (T1 , . . . , Tr )[X] is irreducible and separable, i = 1, . . . , k, and g ∈ N [T], g 6= 0. Let n = max(degX (f1 ), . . . , degX (fk )). Choose a finite extension L of K in N that contains all coefficients of f1 , . . . , fk , g. Put m = [L : K]. Denote the composite of all extensions of K in N of degree at most mn by M . By assumption, [M : K] < ∞. By Corollary 12.2.3, HM (f1 , . . . , fk ; g) contains a separable Hilbert subset HK of K r . Let a ∈ HK . Consider i between 1 and k. Then g(a) 6= 0 and fi (a, X) is irreducible over M . Let b be a zero in fi (a, X). Then L(b) is linearly disjoint from M over L. But, [N ∩ L(b) : K] ≤ mn. Hence, N ∩ L(b) ⊆ M ∩ L(b) = L.  Thus, fi (a, X) is irreducible over N . Consequently, a ∈ H. Theorem 16.11.2: Let K be a Hilbertian field and A a finite Abelian group. ˆ × A occurs over K. Then Z Proof: For each prime number p Corollary 16.6.7 gives a Zp -extension Lp of K. The sequence of all these extensions is linearly disjoint over K. Hence, ˆ of K (Lemma 1.4.5). the compositum L of all Lp is a Z-extension

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By Proposition 16.3.5 and Remark 16.2.2, there is an X-stable polynomial f ∈ K[T, X] which is Galois in X with Gal(f (T, X), K(T )) ∼ = A. In (resp. H ) be the set of all a∈K particular, Gal(f (T, X), L(T )) ∼ A. Let H = 1 2 ∼ ∼ ˆ with Gal(f (a, X), K) = A (resp. Gal(f (a, X), L) = A). Since Z is small, L is Hilbertian (Proposition 16.11.1). Moreover, each Hilbert subset of L contains a Hilbert subset of K. By Proposition 13.1.1, H1 (resp. H2 ) contains a Hilbert subset of K (resp. L). Hence, H1 ∩ H2 contains a Hilbert subset H of K. Choose a ∈ H. Let M be the splitting field of f (a, X) over K. Put N = ˆ × A. LM . Then Gal(M/K) ∼ = Gal(N/L) ∼ = A. Therefore, Gal(N/K) ∼ =Z  Theorem 16.11.3 ([Kuyk2, p. 113]): Every Abelian extension N of a Hilbertian field K is Hilbertian. Proof ([Weissauer, Satz 9.8]): We may assume that N 6= K and choose σ ∈ Gal(N/K), σ 6= 1. The fixed field L of σ in N is a proper subfield of N . Hence, it has a finite proper extension M in N . Let [M : L] = m. Since Gal(N/L) is a closed subgroup of an Abelian group, L/K is Galois. By Theorem 13.9.1(b), M is a Hilbertian field. Now observe that σ m generates Gal(N/M ). Therefore, by Proposition 16.11.1, N is Hilbertian.  Remark 16.11.4: We cannot draw the conclusion, in Theorem 16.11.3, that every separable Hilbert subset of N contains a separable Hilbert subset of K. Indeed, consider N = Kab , the maximal Abelian extension of K. Suppose char(K) 6= 2. Then there exists no a ∈ K with X 2 − a irreducible over Kab , even though X 2 − T is absolutely irreducible. When char(K) = 2, there is no a ∈ K with X 2 − X − a irreducible over Kab although X 2 − X − T is absolutely irreducible and separable in X.  Lemma 16.11.5: Let K be a Hilbertian field. Then Gal(K) is neither prosolvable nor small. Proof: By Corollary 16.2.7(a), K has a Galois extension L with Galois group S5 . Hence, Gal(K) is not prosolvable. Again, by Corollary 16.2.7(b),(c), K has infinitely many quadratic extensions. Therefore, Gal(K) is not small.  Proposition 16.11.6: Let N be a Galois extension of a Hilbertian field K. Suppose N 6= Ks . Then Gal(N ) is neither prosolvable ([F.K.Schmidt], [Kuyk2, Thm. 2]) nor it is contained in a closed small subgroup of Gal(K). Proof: By assumption, N has a proper finite separable extension N 0 . By Theorem 13.9.1(b), N 0 is Hilbertian. Hence, by Lemma 16.11.5, Gal(N 0 ) is not prosolvable. Consequently, Gal(N ) is not prosolvable. Now consider an extension M of K in N . Let M 0 be a finite extension of M with N M 0 = N 0 . In particular, M 0 6⊆ N . Hence, by Theorem 13.9.1(a), M 0 is Hilbertian. So, by Lemma 16.11.5, Gal(M 0 ) is not small. It follows from Remark 16.10.3(b) that Gal(M ) is not small. 

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Example 16.11.7: A Hilbertian field which is not a proper finite extension of any field. For each p, Q has a Galois extension N with Gal(N/Q) ∼ = Zp (Lemma 16.6.4 or Corollary 16.6.7). By Proposition 16.11.1, N is Hilbertian. Since Zp has no nontrivial finite subgroups (Lemma 1.4.2(c)), N is not a proper finite extension of any field. 

16.12 Regularity of Finite Groups over Complete Discrete Valued Fields The results of this section depends on a theorem of Harbater whose proof is unfortunately outside the scope of this book. Proposition 16.12.1: Suppose K is a complete field under a discrete valuation and G is a finite group. Then G is regular over K. On the proof: Harbater’s proof of the proposition uses the language of formal schemes [Harbater1, Thm. 2.3]. Liu [Liu1] and Serre [Serre8, Thm. 8.4.6] translate the proof into the language of rigid analytic geometry. Both approaches rely on general GAGA theorems relating formal (resp. rigid analytic) geometry to algebraic geometry. A short cut proof which is more algebraic than the former ones appears in [Haran-V¨ olklein] and in [V¨olklein, p. 239]. Each of these proofs actually provides a Galois extension F of K(t) (with t transcendental over K), regular over K, with Gal(F/K(t)) ∼  = G. Proposition 16.12.2: Let K be a PAC field and G a finite group. Then G is regular over K. ˆ = K((z)) of formal power series in z is complete under a Proof: The field K discrete valuation (Example 3.5.1). By Proposition 16.12.1, G is regular over ˆ By Proposition 16.2.8, there exists an absolutely irreducible polynomial K. ˆ ˆ )) ∼ f ∈ K[T, X], monic and Galois in X, with Gal(f (T, X), K(T = G. Choose ˆ u1 , . . . , un ∈ K such that K[u] contains the coefficients of f and f is Galois over K(u, T ) with Gal(f (u, T, X), K(u)) ∼ = G. Write f (T, X) = g(u, T, X) with g a polynomial with coefficients in K. ˆ is regular Denote the K-variety that u generates in An by V . Since K/K (Example 3.5.1), so is K(u)/K (Corollary 2.6.5(b)). Hence, V is absolutely irreducible (Corollary 10.2.2(a)). Proposition 16.1.4 gives a nonempty Zariski open subset V0 of V satisfying: For each a ∈ V0 (K), the polynomial g(a, T, X) is absolutely irreducible and Galois over K(T ). Moreover, Gal(g(a, T, X), K(T )) ∼ = Gal(g(u, T, X), K(u, T )) ∼ = G. Since K is PAC, there exists an a ∈ V0 (K). By the preceding paragraph, g(a, T, X) is absolutely irreducible, monic and Galois in X, and Gal(g(a, T, X), K(T )) ∼ = G. Consequently, G is regular over K.



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Remark 16.12.3: Ample fields. The assumption “K is PAC” in the proof of Proposition 16.12.2 is used only to find a K-rational point of V0 . In many instances the condition “K is PAC” can be replace by a weaker one: We say that a field K is ample if K is existentially closed in K((t)) (cf. Proposition 11.3.5). For example, PAC fields and Henselian fields are ample. For more details see [Haran-Jarden6, §6].  Proposition 16.12.4: Let K be a field and G a finite group. (a) Then G is regular over a finite Galois extension L of K. (b) Moreover, suppose K is Hilbertian. Then L has a finite Galois extension with Gal(N/L) ∼ = G. Proof of (a): Ks is PAC (Section 11.1). By Proposition 16.12.2, G is regular over Ks . Hence, there is an irreducible Galois polynomial f ∈ Ks [T, X] in X with T = (T1 , . . . , Tr ) and Gal(f (T, X), Ks (T)) ∼ = G. By Lemma 16.2.4(a),(b), there is a finite Galois extension L of K such that f (T, X) is Galois over L(T) and Gal(f (T, X), L(T)) ∼ = G. Consequently, by Lemma 16.2.4(c), G is regular over L. Proof of (b): By Corollary 12.2.3, L is Hilbertian. Proposition 16.1.5 gives a ∈ Lr with f (a, X) Galois over L and Gal(f (a, X), L) ∼ = G. Let N be the splitting field of f (a, X) over L. Then Gal(N/L) ∼ G.  = Remark 16.12.5: An evaluation of Proposition 16.12.4(b). The proof of Proposition 16.12.4(b) relies on the deep result, Proposition 16.12.1. If we ask L/K only to be finite and separable, the proof becomes quite elementary. To this end embed G in Sn , say, for n = |G|. Then use the general polynomial of degree n to find a Galois extension N of K with Gal(N/K) ∼ = Sn . Denote the fixed field of G in N by L. Then Gal(N/L) ∼ = G. However, unless G is  trivial, G ∼ = Z/2Z, or G = Sn itself, L will not be Galois over K.

Exercises 1. Let K be a Zp extension of Q and l 6= p a prime number. Use Lemma 16.6.9 to prove that l is unramified in K. 2. Prove that Zp × Zp does not occur over Q. Hint: Use Exercise 1 and Kronecker-Weber: Every finite Abelian extension of Q is contained in Q(ζn ) for some positive integer n. Alternatively, use Lemma 16.6.4 to prove that ˆ where G is a product of finite groups. Then use Gal(Qab /Q) ∼ = G×Z Kronecker-Weber. 3. Suppose each of the direct powers Gn of a finite group G can be realized over the field K. Prove there exists a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions of K with Gal(Li /K) ∼ = G, i = 1, 2, 3, . . . . 4. Use Corollary 16.2.7, to prove that if K is a Hilbertian field, then the group Gal(Kab /K) is not finitely generated.

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5. Strengthen Theorem 16.11.6 to prove that if L is a separable algebraic extension of a Hilbertian field K and Gal(L) is a prosolvable group, then the Galois closure of L/K is Ks . 6. Let G be a profinite group, N be a closed characteristic subgroup of G, and α an automorphism of G. Suppose the centralizer of N in G is trivial. Prove that the map α 7→ α|N is an embedding of Aut(G) into Aut(N ). Hint: Use the identity α(n−1 gn) = α(n)−1 α(g)α(n) for g ∈ G and n ∈ N . 7. Let G be a finitely generated group and H an open subgroup. Prove that H is S finitely generated. Hint: Let x1 , . . . , xe be generators of G and n write G = · i=1 Hgi with g1 = 1. Prove that H is generated by all elements −1 gi x±1 j gk which belong to H.

Notes ˜ = Let α: G → Gal(L/K) be an embedding problem over a field K. Put G ˜ . We call a homomorphism γ: G → Gal(K), A = Gal(L/K), and ϕ = resK/L ˜ G satisfying α ◦ γ = ϕ a solution of the embedding problem if γ is surjective, because we are exclusively looking for solutions of embedding problems over fields. However, in the framework of profinite groups, it is of great interest ˜ → A satisfying α ◦ γ = ϕ which are to consider also homomorphisms γ: G not necessarily surjective. In Chapter 22 we call such γ’s weak solutions of ˜ to be projective if every finite the embedding problem (ϕ, α) and define G ˜ embedding problem for G has a weak solution. Other authors (e.g. [MalleMatzat, p. 296]) prefer the terminology “solution” and “proper solution” rather than “weak solution” and “solution”. Likewise, what we call a “regular solution” of the embedding problem α: G → Gal(L(t)/K(t)) in Definition 16.4.1, is called a “proper parametric solution” in [Malle-Matzat, p. 296]. The use of the group epimorphism A wr G → G n A (Lemma 16.4.3) replaces Uchida’s proof of Proposition 16.4.4 as presented in [Fried-Jarden3, Lemma 24.46]. We have borrowed this approach from [Malle-Matzat, Section IV.2.2] [Serre8, Thm. 1.2.1] gives a cohomological proof of Proposition 16.5.1. Our proof of Whaple’s theorem (Theorem 16.6.6) is an elaboration of [Kuyk-Lenstra]. The special case of Proposition 16.6.10 when K = Q appears in [Serre8, p. 36, Exer. 1]. The article [Geyer-Jensen1] studies the number of linearly disjoint Zp -extensions of a given field K. Theorem 2.7 of that article says this number does not grow in a finitely generated regular extension of a finitely generated extension of Q. In characteristic 0 this is essentially Proposition 16.6.10. The realization of symmetric and alternating groups over Q using Hilbert irreducibility theorem goes back to [Hilbert]. Our treatment of this subject in Section 16.7 is a workout of [Serre10, pp. 42-43].

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The concept of GAR-realization and its application to solve embedding problems with kernels having trivial centers go back to [Matzat]. See also [Malle-Matzat, Chap. IV.3]. The concept of small profinite groups appears in [Klingen1]. Theorem 22.9.7 partially generalizes Theorem 16.11.3 to pronilpotent extensions of K. Geyer (private communication) uses wreath products to prove that if K is a Hilbertian field, then Gal(K) has no nontrivial finitely generated normal closed subgroups. Proposition 16.11.6 generalizes this result. Example 16.11.7 settles a question of Sonn. The proof of Proposition 16.12.2 essentially depends on the following result: (1) For a field K and a finite group G there is an absolutely irreducible variety V over K satisfying this: If V (K) 6= ∅, then G is regular over K. [Fried-V¨olklein1, Thm. 2] proves (1) in characteristic 0 as an application of the parametrization of Galois extensions of C(t) by moduli spaces. The proof of Proposition 16.12.2 assumes G is regular over a regular extension of K (e.g. over K((t)) ) and construct V satisfying the conclusion of (1). Pop observed that the condition “K is existentially closed in K((t))” is common to PAC fields and to Henselian fields as well as to other families of fields. He then observed that “G is regular over K” is an elementary property of K and deduced Proposition 16.12.2 for ample fields. This is included in a letter from P. Roquette to W.-D. Geyer in February, 1991. Pop’s observation replaces the use of Bertini-Noether in the proof of Proposition 16.12.2. Pop actually uses the terminology “large fields” instead of “ample fields” [Pop, p. 4]. We prefer the latter terminology because the adjective “large” in the naive sense has been attached to algebraic extensions of Hilbertian fields in several works (e.g. in [Frey-Jarden, Geyer-Jarden1, Jarden2, Jarden4, Jarden7, Jarden12, Jacobson-Jarden, Lubotzky-v.d.Dries]).

Chapter 17. Free Profinite Groups We continue the discussion on profinite groups of Chapters 1 and 16. Central to this chapter is a discussion on free profinite groups. In particular, we prove that an open subgroup of a free profinite group is free (Proposition 17.6.2).

17.1 The Rank of a Profinite Group The rank of a finitely generated profinite group is defined to be the minimal number of generators of the sets (Section 16.10). The definition of the rank of arbitrary profinite group G puts a topological condition on the minimal set of generators of G. Once this condition is satisfied, the cardinality m of that set depends only on the group. Indeed, if G is not finitely generated, then m is the cardinality of the set of all open subgroups of G (Proposition 17.1.2) and rank(G) is defined to be m. The above mentioned topological condition generalizes convergence of a sequence. A subset X of a profinite group G is said to converge to 1, if X r N is a finite set for every open normal subgroup N of G. Proposition 17.1.1 ([Douady]): Every profinite group G has a system of generators that converges to 1. Proof: Without loss assume G is infinite. Well order the set N of open normal subgroups of G: N =T {Nα | α < m}, where m is an infinite cardinal number. For β ≤ m let Mβ = α<β Nα . Choose a set I of cardinality greater than m. We break the rest of the proof into parts to do a transfinite induction on β ≤ m and define Xβ = {xβi | i ∈ I} ⊆ G with these properties: (1a) Xβ generates G modulo Mβ . (1b) For each open normal subgroup N of G that contains Mβ the set {i ∈ / N } is finite. I | xβi ∈ (1c) xαi ≡ xβi mod Mα , for all α < β and i ∈ I. In particular, Xm is a system of generators of G that converges to 1. Part A: Induction for a successor ordinal. Let γ = β + 1 < m and suppose we have defined Xα for each α ≤ β. The set Iβ = {i ∈ I | xβi 6∈ Mβ } = S α<β {i ∈ I | xβi 6∈ Nα } is a union of no more than |β| finite sets. Hence, |Iβ | ≤ ℵ0 · |β| ≤ m < |I|. Let y1 , . . . , yk be generators of Mβ modulo Mγ . Choose k distinct elements i1 , . . . , ik in I r Iβ . Define xγi as follows: For the finite number of i’s with xβi ∈ G r Mβ Nβ , let xγi = xβi . For xβi ∈ Mβ Nβ r Mβ , choose xγi ∈ Nβ such that xγi ≡ xβi mod Mβ . For i = ij and j ∈ {1, 2, . . . , k} choose xγi = yj . Finally, for all other i, define xγi = 1. Then Xγ satisfies (1) with γ replacing β. For example, in order to prove (1b), consider an open normal subgroup N of G that contains Mγ . By definition, Mβ ∩ Nβ = Mγ . Hence, by Lemma 1.2.5, Mβ Nβ has an open

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normal subgroup U containing Mβ such that U ∩ Nβ ≤ N . For all but finitely many i ∈ I we have xβi ∈ U r Mβ or xγi = 1. In the former case xγi ∈ Nβ and xγi ≡ xβi mod Mβ , so xγi ∈ U ∩ Nβ ≤ N . Thus, all but finitely many i ∈ I belong to N . Part B: Induction for a limit ordinal. Assume γ is a limit ordinal ≤ m and Xα has been defined for each α < γ. Let i ∈ I. By (1c), the collection {xαi Mα | α < γ} of closed sets satisfies the finite intersection property. Therefore, there is an xγi ∈ G with xγi ≡ xαi mod Mα for each α < γ. Suppose N is an open normal subgroup of G that contains Mγ . Then there is an α < γ with Mα ⊆ N (Lemma 1.2.2(a)). Hence, xγi ∈ N for all but finitely many i ∈ I. Finally, for each α < γ, the set Xγ generates G Lemma 1.2.2(b) and by induction, modulo Mα .TPut H = hxT γi | i ∈ Ii. By T HMγ = H β<γ Mβ = β<γ HMβ = β<γ G = G. This concludes the transfinite induction.  We define the rank of a nonfinitely generated profinite group G as the cardinality of a system of generators of G that converges to 1. The following result proves this definition to be independent of the particular system of generators: Proposition 17.1.2: Let G be a nonfinitely generated profinite group. Denote the family of all open (resp. open normal) subgroups of G by M (resp. N ). Suppose X is a system of generators of G that converges to 1. Then |X| = |M| = |N |. Proof: First note that G has only finitely many open subgroups containing each N ∈ N . Hence, |M|S= |N |. It remains to prove that |X| = |N |. By definition, X = N ∈N (X r N ) is a union of |N | finite sets. Hence, r |X| ≤ |N |. For each finite subset A of X, let N (A) S = {N ∈ N | X A ⊆ N }. If we show that N (A) is countable, then N = A N (A) would be a union of no more than |X| countable sets. Thus, |N | ≤ |X|, and we are done. Let K(A) be the minimal closed normal subgroup of G that contains X r A. Then G/K(A) is finitely generated and there exists a bijective correspondence between N (A) and the family of open normal subgroups of G/K(A). By Lemma 16.10.2, both are therefore countable.  Remark 17.1.3: On the definition of the rank. If G is an infinite finitely generated profinite group, then G has infinitely many open normal subgroups. Thus, Proposition 17.1.2, does not hold in this case. Moreover, {2, 3} is a ˆ although rank(Z) ˆ = 1. Thus, the definitions minimal set of generators of Z, we gave to the rank of finitely generated and nonfinitely generated groups differ from each other. A unified definition of rank(G) in both cases could be taken as the minimal cardinality of a system of generators that converges to 1.  Corollary 17.1.4: Let ϕ: G → A be an epimorphism of profinite groups. Then rank(A) ≤ rank(G).

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Proof: First suppose rank(G) < ∞. Then ϕ maps each finite set of generators of G onto a set of generators of A. Hence, rank(A) ≤ rank(G). Now suppose both ranks are infinite. Then the map N 7→ ϕ−1 (N ) maps the set of all open normal subgroups of A injectively into the set of all open normal subgroups of G. Therefore, by Proposition 17.1.2, rank(A) ≤ rank(G).  Corollary 17.1.5: Let G be a profinite group of infinite rank and H a closed subgroup. Then rank(H) ≤ rank(G). If H is open, then rank(H) = rank(G). Proof: The map G0 7→ G0 ∩ H surjects the set of all open subgroups of G onto the set of all open subgroups of H (Lemma 1.2.5). Hence, by Proposition 17.1.2, rank(H) ≤ rank(G). If H is open, the fibers of that map are finite. Hence, by Proposition 17.1.2, rank(H) ≥ rank(G). Consequently, in this case, rank(H) = rank(G).  In Section 17.6 we prove rank(H) to be finite if rank(G) is finite. Moreover, we estimate rank(H) in terms of rank(G) and (G : H). Corollary 17.1.6: Let G be a profinite group, m an infinite cardinal number, and I a set of cardinality at most m. (a) For T each i ∈ I let Ni be a closed normal subgroup of G. Suppose i∈I Ni = 1 and Q rank(G/Ni ) ≤ m for each i ∈ I. Then rank(G) ≤ m. (b) Suppose G = i∈I Gi with rank(Gi ) ≤ m. Then rank(G) ≤ m. Moreover, rank(G) = m if rank(Gj ) = m for at least one j ∈ I or each Gi is nontrivial and |I| = m. Proof of (a): Denote the set of all open normal subgroups of G which contains Ni by Ni . Each open normal subgroup of G is in Ni S for some i ∈ I (Lemma 1.2.2(a)). Hence, by Proposition 17.1.2, rank(G) ≤ | i∈I Ni | ≤ m. Proof of (b): For each i ∈ I, let Ni be the kernel of the projection G → Gi . By (a), rank(G) ≤ m. If rank(Gj ) = m for at least one j ∈ I, then rank(G) ≥ m (Corollary 17.1.5). The same conclusion holds if |I| = m. Thus, rank(G) = m.  Example 17.1.7: Countable rank groups. (a) The rank of a profinite group G is ≤ ℵ0 if and only if G has a descending sequence G = G0 ≥ G1 ≥ G2 ≥ . . . of open normal subgroups with a trivial intersection. (b) If K is a countable field, then it has at most countably many finite  Galois extensions. Hence, rank(Gal(K)) ≤ ℵ0 .

17.2 Profinite Completions of Groups We have already seen that profinite groups appear as Galois groups of Galois extensions (Corollary 1.3.4). In this section we study profinite groups that arise as completions of abstract groups.

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Consider a directed family N of normal subgroups of finite index in a group G. Thus, Tr (1) for all N1 , . . . , Nr ∈ N there is an N ∈ N with N ≤ i=1 Ni . To each pair N1 ≤ N2 of groups in N associate the quotient map G/N1 → G/N2 to obtain a compatible system of maps that defines a profinite group ˆ = lim G/N . We call G ˆ the profinite completion of G with respect G ←− to N . If N is the family of all normal subgroups of G of finite index, then ˆ is the profinite completion of G. G ˆ whose The map g 7→ (gN )N ∈N is a canonical homomorphism θ: G → G kernel N∞ is the intersection of all N ∈ N . For each N ∈
N denote the ˆ onto G/N by πN . Thus, πN (xN 0 N 0 )N 0 ∈N = xN N and projection of G πN (θ(g)) = gN for each g ∈ G. Note that πN is surjective. Finally, let M be the family of all subgroups M of G which contain some N ∈ N . For each ˆ be the closure of θ(M ) in G. ˆ The ambiguity in the definition M ∈ M let M ˆ is settled in Part (a4) of the following lemma: of G Lemma 17.2.1: (a) The following statements hold for each N ∈ N : ˆ = Ker(πN ). (a1) N ˆ N ˆ → G/N . ¯N : G/ (a2) πN induces an isomorphism π ˆ (a3) The set {N | N ∈ N } is a basis for the open neighborhoods of 1 in ˆ G. ˆ (a4) θ(G) is dense in G. ˆ (b) The map M 7→ M maps M bijectively onto the family of open subgroups ˆ ) = M. ˆ For each M ∈ M we have θ−1 (M of G. (c) The following statements hold for all M1 , M2 ∈ M: ˆ1 ≤ M ˆ 2. (c1) M1 ≤ M2 if and only if M ˆ2 : M ˆ 1 ). (c2) If M1 ≤ M2 , then (M2 : M1 ) = (M ˆ ˆ (c3) M1 / M2 if and only if M1 / M2 . ˆ 1. ˆ 2 /M (c4) If M1 / M2 , then M2 /M1 ∼ =M ˆ ∼ (d) Let M ∈ M. Put NM = {N ∈ N | N ≤ M }. Then M = lim M/N , ←− where N ranges over NM . Proof: Consider N ∈ N . ˆ = Ker(πN ). Since θ(N ) is contained in Ker(πN ) and Part A: We prove: N ˆ Ker(πN ) is closed, N ≤ Ker(πN ). Conversely, consider x ∈ Ker(πN ). Every open neighborhood of x contains a coset of the form x·Ker(πN 0 ) with N 0 ∈ N . Choose N 00 ∈ N with N 00 ≤ N ∩ N 0 and let πN 00 (x) = yN 00 with y ∈ G. Then yN = πN (x) = N and yN 0 = πN 0 (x). Hence, θ(y) ∈ θ(N ) ∩ x · Ker(πN 0 ). ˆ meets θ(N ). It follows that x Thus, every open neighborhood of x in G ˆ ˆ belongs to the closure N of θ(N ) in G. Applying the arguments of the latter paragraph to G, we conclude that ˆ G is dense in G. Statement (a2) follows from statement (a1) and (a3) holds because ˆ {Ker(πN ) | N ∈ N } is a basis for the open neighborhoods of 1 in G.

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ˆ N ˆ . By definition, Ker(θ) ≤ N . Hence, Part B: The map θN : G/N → G/ the map gN 7→ θ(g)θ(N ) is an isomorphism of G/N onto θ(G)/θ(N ). ˆ . Then, by Part A, gN = πN (θ(g)) = N . Let g ∈ G with θ(g) ∈ N ˆ = θ(N ). Hence, g ∈ N . Therefore, θ(G) ∩ N ˆ there is a g ∈ G with πN (x) = gN = πN (θ(g)). Hence, For each x ∈ G ˆ . Therefore, G ˆ = θ(G)N ˆ. x ∈ θ(g)N ˆ is an isomorphism of θ(G)/θ(N ) onto Thus, the map θ(g)θ(N ) 7→ θ(g)N ˆ N ˆ defined by θN (gN ) = θ(g)N ˆ ˆ N ˆ . It follows that the map θN : G/N → G/ G/ ¯N . is an isomorphism. By definition, θN is the inverse of π / ˆ xG x x πN x x xx   |xx θN / ˆ ˆ G/N o G/N G

θ

π ¯N

Part C: A bijection of families of subgroups. Let M be a subgroup of G ˆ is an open subgroup of G, ˆ hence containing some N ∈ N . Then θ(M )N ˆ ≤ θ(M )N ˆ . On the other hand, from closed, which contains θ(M ). Hence, M ˆ ≤M ˆ , so M ˆ = θ(M )N ˆ. N ≤ M follows θ(M )N G

/ θ(G)

ˆ G

M

/ θ(M )

ˆ M

N

/ θ(N )

ˆ N

ˆ ) = θ−1 (θ(M )) = M . It follows Finally, Ker(θ) ≤ N ≤ M . Hence, θ−1 (M ˆ from Part B that the map M 7→ M is a bijection of all subgroups of G which ˆ which contain N ˆ . In particular, (c1)-(c4) contain N onto all subgroups of G hold when N ≤ M1 , M2 . Part D: Conclusion of the proof. Consider open subgroups M10 and M20 of ˆ By (a3), there exist N ∈ N with N ˆ ≤ M 0 ∩ M 0 . By Part C, there are G. 1 2 ˆ i , i = 1, 2. open subgroups M1 , M2 of G which contain N such that Mi0 = M This concludes the proof of (b) and (c). ˆ Finally, consider M ∈ M. By (a3), T {N | ˆN ∈ NM } is a directed family ˆ of open normal subgroups of M and N ∈NM N = 1. Hence, by Lemma 1.2.4, ˆ ∼ ˆ ∼ ˆ /N ˆ , where N ranges over NM . By (c4), M M = lim M/N , where = lim M ←− ←− N ranges over NM .  The completion of a group is universal in the sense phrased by the following result:

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Lemma 17.2.2: Let G be a group, N a directed family of normal subgroups of G of finite index, H a profinite group, and α: G → H a homomorphism. Suppose α−1 (H0 ) contains some N ∈ N for each open normal subgroup H0 ˆ = lim G/N be the profinite completion of G with respect to of H. Let G ←− ˆ the canonical map of G into G. ˆ Then, there is a unique N and θ: G → G ˆ homomorphism α ˆ : G → H satisfying α ˆ ◦ θ = α. ˆ We prove Proof: The uniqueness of α ˆ follows from the density of θ(G) in G. the existence of α. ˆ Consider an open normal subgroup H0 of H. Choose N ∈ N with ˆ → H/H0 by αH (ˆ N ≤ α−1 (H0 ). Define a homomorphism αH0 : G 0 g) = ˆ g ))H0 , gˆ ∈ G. In particular, αH0 (θ(g)) = α(g)H0 for each g ∈ G. α(πN (ˆ The definition of αH0 is independent of N . Indeed, if N 0 ∈ N and 0 g )N = πN (ˆ g )N , hence πN 0 (ˆ g )α−1 (H0 ) = πN (ˆ g )α−1 (H0 ), N ≤ N , then πN 0 (ˆ g ))H0 = α(πN (ˆ g ))H0 . so α(πN 0 (ˆ Consider now an additional open normal subgroup H1 of H which is contained in H0 . Let πH1 ,H0 : H/H1 → H/H0 be the quotient map. Choose N ∈ N with N ≤ α−1 (H1 ) and use N to define αH1 . Then πH1 ,H0 ◦ αH1 = αH0 . ˆ → H with αH (ˆ The latter conclusion yields a homomorphism α ˆ: G g) = 0 ˆ Conseα ˆ (ˆ g )H0 for each open normal subgroup H0 of H and every gˆ ∈ G. quently, α ˆ (θ(g)) = α(g) for each g ∈ G.  As a corollary of Lemma 17.2.2 we prove that the completion of a group is unique: Lemma 17.2.3: Let N and N 0 be directed families of normal subgroups of finite index of a group G. Suppose N and N 0 are cofinite in each other; that is, each N ∈ N contains some N 0 ∈ N 0 and each N 0 ∈ N 0 contains ˆ = lim G/N (resp. G ˆ 0 = lim G/N 0 ) be the completion some N ∈ N . Let G ←− ←− ˆ (resp. θ0 : G → G ˆ0) of G with respect to N (resp. N 0 ) and let θ: G → G be the corresponding canonical homomorphism. Then there exists a unique ˆ→G ˆ 0 such that α ◦ θ = θ0 . isomorphism α: G ˆ 0 is the closure N ˆ0 Proof: By Lemma 17.2.1, each open normal subgroup of G ˆ 0 of θ0 (N 0 ) for some N 0 ∈ N 0 . Moreover, N 0 = (θ0 )−1 (N ˆ 0 ). Lemma 17.2.2 in G ˆ → G ˆ 0 with α ◦ θ = θ0 . By symmetry, gives a unique homomorphism α: G ˆ0 → G ˆ such that α0 ◦ θ0 = θ. Thus, there exists a unique homomorphism α0 : G 0 0 0 α ◦α◦θ = α ◦θ = θ = idGˆ ◦θ. The uniqueness part of Lemma 17.2.2 applied ˆ implies that α0 ◦ α = id ˆ . By symmetry, α ◦ α0 = id ˆ 0 . to the map θ: G → G G G Consequently, α is an isomorphism.  Remark 17.2.4: The group Zp is the completion of Z with respect to the ˆ is the completion of Z with respect family {pi Z | i ∈ N}, while the group Z to the family {nZ | n ∈ N}. 

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17.3 Formations of Finite Groups We classify profinite groups according to their ranks and their finite quotients. To this end consider a family C of finite groups containing the trivial group. Each group in C is called a C-group. We call C a formation if it is closed under taking quotients and fiber products (Section 22.2). In other words, C satisfies the following conditions: ¯ is a homomorphic image of G, then G ¯ ∈ C. (1a) If G ∈ C and G (1b) Let G be a finite group and N1 , N2 / G. Suppose G/N1 , G/N2 ∈ C and N1 ∩ N2 = 1. Then G ∈ C. We call C a Melnikov formation if it is closed under taking quotients, normal subgroups and extensions. This means that C satisfies this condition: ¯ → 1 be a short exact sequence of finite groups. (1c) Let 1 → N → G → G ¯ ∈ C. Then G ∈ C if and only if N, G In particular, C satisfies (1a) and (1b). Indeed, under the assumptions of (1b), N2 is isomorphic to N1 N2 /N1 and N1 N2 /N1 is a normal subgroup of G/N1 , so N2 ∈ C. Since also G/N2 ∈ C, we have G ∈ C. It follows that every Melnikov formation is a formation. We say C is a full formation if it is closed under taking quotients, subgroups, and extensions. In other words, C is a Melnikov formation which satisfies the following condition: (1d) If G ∈ C and H ≤ G, then H ∈ C. A pro-C group is an inverse limit G = lim Gi of C-groups for which ←− the connecting homomorphisms Gj → Gi are epimorphisms. Thus, G has a collection N of open normal subgroups with G/N ∈ C for each N ∈ N and T N ∈N N = 1. Lemma 17.3.1: Let C be a family of finite groups satisfying (1a). Consider a profinite group G. Then this holds: (a) If G is pro-C, then each homomorphic image of G is pro-C. (b) Suppose C is closed under taking normal subgroups and G is pro-C. Then every closed normal subgroup N of G is pro-C. (c) Suppose C is closed under taking subgroups and G is pro-C. Then every closed subgroup H of G is pro-C. (d) Suppose (1b) holds. Let N1 , N2 be closed normal subgroups of G. Suppose G/N1 , G/N2 are pro-C. Then G/N1 ∩ N2 is pro-C. (e) Suppose (1c) holds. Let N be a closed normal subgroup of G. Suppose both N and G/N are pro-C. Then G is pro-C. (f) Suppose C is closed under taking normal subgroups. Then inverse limits of pro-C groups are pro-C groups. Proof of (a): Let N be a closed normal subgroup of G. First suppose N is open. Then G has an open normal subgroup M in N with G/M ∈ C. Since G/N is a homomorphic image of G/M , it belongs to C.

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T In the general case N = j∈J Nj with Nj normal and open in G. By the preceding paragraph, G/Nj ∈ C for each j ∈ J. Thus, G/N = lim G/Nj ←− is pro-C. Proof of (b): Let N be a closed normal subgroup of G. Then N ∼ = lim N/H ∩ N with H ranging over all open normal subgroups H of ←− G. Consider an open normal subgroup H of G. By (a), G/H ∈ C. Also, HN/H / G/H. Hence, N/H ∩ N ∼ = HN/H ∈ C. Consequently, N is pro-C. Proof of (c): As the proof of (b). Proof of (d): Let H be an open normal subgroup of G. By (a), G/Ni H ∈ C, i = 1, 2. Hence, by (1b), G/N1 H ∩ N2 H ∈ C. By Lemma 1.2.2(b), T H N1 H ∩ N2 H = N1 ∩ N2 . Hence, G/N1 ∩ N2 is a pro-C group. Proof of (e): Let H be an open normal subgroup of G. Then G/HN is a quotient of G/N . Hence, G/HN ∈ C. Also, HN/H ∼ = N/H ∩ N ∈ C. Hence, by (1c), G/H ∈ C. Consequently, G is a pro-C group. Proof of (f): It suffices to prove that if G is a profinite group and N is a directed family of closed normal subgroups such that G/N is pro-C for each N ∈ N and the intersection of all N ∈ N is trivial, then G is pro-C. Indeed, G∼ = lim G/H, where H ranges over all open normal subgroups of G which ←− contain N for some N ∈ N . By (a), G/H ∈ C for each such H. Hence, G is pro-C.  In particular, the consequences of Lemma 17.3.1 except (c) hold if C is a formation and all consequences of Lemma 17.3.1 hold if C is full. Definition 17.3.2: Maximal pro-C quotients. Let C be a formation of finite groups and G a profinite group. Denote the family of all closed normal subgroups N of G with G/N ∈ T C by N . By Lemma 17.3.1(d), N is closed under intersections. Put NC = N ∈N N . Then G/NC is pro-C. Moreover, ¯ is a quotient of G and G ¯ ∈ C. Then G ¯ is a quotient of G/NC . We suppose G call G/NC the maximal pro-C quotient of G.  Example 17.3.3: Standard families. (a) C contains all finite groups: pro-C groups are just profinite groups. (b) C consists of all finite p-groups, p a prime: pro-C groups are then called pro-p groups. (c) C consist of all solvable groups: pro-C groups are prosolvable groups. Each of the families (a), (b), and (c) is full. (d) C consists of all finite Abelian (resp. nilpotent) groups: pro-C groups are Abelian profinite (resp. pronilpotent). In this case (1a) and (1b) hold. Thus, C is a formation. But (1c) is false. For example, in the short exact sequence 1 → A3 → S3 → C2 → 1 both A3 and C2 are Abelian but S3 is not nilpotent. Therefore, C is not Melnikov. 

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Remark 17.3.4: Melnikov formations. Let D be a nonempty family of finite simple groups. Denote the family of all finite groups whose composition ¯ → 1 is a short exact sequence factors belong to D by C. If 1 → N → G → G ¯ belong to D, of finite groups and the composition factors of both N and G then each composition factor of G belongs to D. Thus, C is a Melnikov formation. We call C the formation generated by D. Conversely, let C be a Melnikov formation. Denote the family of all composition factors of groups belonging to C by D. Then D generates C. We call D the underlying set of simple groups of C. By the preceding paragraph, the consequences of Lemma 17.3.1(a),(b),(d),(e),(f) hold for pro-C groups. If D consists of a single simple group S, we call a pro-C group also a proS group. In particular, if S = Z/pZ, pro-C groups are just pro-p groups. If S is a simple non-Abelian group, no nontrivial subgroup is an S-group. Hence, C is not full. When D is the set of all finite groups, pro-C groups are just profinite groups. 

17.4 Free pro-C Groups Free pro-C groups arise as completions of free abstract groups. Like the latter, they are defined by a universal property in the category of pro-C groups: Definition 17.4.1: Let X be a set and G a profinite group. A map ϕ: X → G is convergent to 1 if X r ϕ−1 (H) is a finite set for each open normal subgroup H of G. Let C be a formation of finite groups and X a set. A free pro-C group on X is a pro-C group Fˆ with a map ι: X → Fˆ satisfying: (1a) ι converges to 1 and ι(X) generates Fˆ . (1b) For each map ϕ of X into a pro-C group G which converges to 1 and satisfies G = hϕ(X)i there exists a unique epimorphism ϕ: ˆ Fˆ → G with ϕˆ ◦ ι = ϕ.  When C is full, each closed subgroup of G is pro-C. Hence, the condition on ϕ(X) in (1b) to generate G is redundant. We construct free pro-C groups from free abstract groups. Recall that an abstract group F is free on a subset X if X generates F and every map ϕ0 of X into an abstract group G uniquely extends to a homomorphism ϕ: F → G. We call X a basis of F or also a free set of generators of F . If X is infinite, then |X| = |F |. If X is finite, say of cardinality n, then there is an epimorphism of F onto (Z/2Z)n . Since the latter group is not generated by less than n elements, n is the minimal number of generators of F . We call |X| the rank of F . Proposition 17.4.2: Let C be a formation of finite groups and X a set. Then there exists a free pro-C group Fˆ on X. More precisely: Let F be the free abstract group on X. Denote the family of normal subgroups N of F such that F/N ∈ C and X r N is finite

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347

by N . Let Fˆ = lim F/N be the profinite completion of F with respect to ←− N , θ: F → Fˆ the canonical homomorphism, and ι: X → Fˆ the restriction of θ to X. Then Fˆ is a free pro-C group on X with ι the associated map. Proof: By (1) of 17.3, N is a directed family (Section 17.2). Hence, Fˆ is a pro-C group. ˆ is an open normal subgroup of Fˆ , By Lemma 17.2.1, Fˆ = hι(X)i. If H −1 ˆ −1 ˆ then θ (H) ∈ N , so X r ι (H) is finite. Therefore, ι converges to 1. Next consider a map ϕ of X into a pro-C group G which converges to 1 such that ϕ(X) generates G. Extend ϕ in the unique possible way to a homomorphism ϕ: F → G of abstract groups. Let H be an open normal subgroup of G. Then ϕ−1 (H) is a normal subgroup of F and ϕ(F )H = G, because G is the closure of ϕ(F ). Hence, F/ϕ−1 (H) ∼ = G/H ∈ C. Also, X r ϕ−1 (H) −1 is finite. Therefore, ϕ (H) ∈ N . Lemma 17.2.2 gives a homomorphism ϕ: ˆ Fˆ → G satisfying ϕˆ ◦ θ = ϕ, so ϕˆ ◦ ι = ϕ. Consequently, Fˆ is a free pro-C group on X.  The universal property of a free pro-C group on a set X implies its uniqueness: Lemma 17.4.3: Let C be a formation of finite groups. Consider a free pro-C group Fˆ on a set X and let ι: X → Fˆ be the associated map. Let κ: X → X 0 be a bijection of sets. (a) Then Fˆ is a free pro-C group on X 0 with the associated map ι◦κ−1 : X 0 → Fˆ . (b) Let F 0 be a free pro-C group on X 0 with an associated map ι0 : X 0 → F 0 . Then there exists a unique isomorphism κ ˆ : Fˆ → F 0 with ι0 ◦ κ = κ ˆ ◦ ι. Proof: We prove only (b), leaving (a) to the reader. By definition, ι0 (κ(X)) = ι0 (X 0 ) generates F 0 . Also, the map ι0 ◦ κ: X → F 0 converges to 1. Indeed, if H 0 is an open subgroup of F 0 , then X 0 r(ι0 )−1 (H 0 ) is finite. Since κ is bijective, X r(ι0 ◦ κ)−1 (H 0 ) = κ−1 (X 0 r(ι0 )−1 (H 0 )) is finite. It ˆ ◦ ι = ι0 ◦ κ. follows that there exists a unique epimorphism κ ˆ : Fˆ → F 0 with κ 0 0 0 0 Similarly, there exists an epimorphism κ : F → Fˆ with κ ◦ ι = ι ◦ κ−1 . The ˆ and κ ˆ ◦ κ0 are the identity uniqueness part of (1b) implies that both κ0 ◦ κ maps. Consequently, κ ˆ is an isomorphism.  Following Proposition 17.4.2 and Lemma 17.4.3, we denote the unique (up to isomorphism) free pro-C group on a set X by FˆX (C). If X is empty, then FˆX (C) = 1. The next result shows that in general the map ι of (1) is injective: Lemma 17.4.4: Let C be a formation of finite groups, X a set, and ι: X → FˆX (C) be the associated map into the free pro-C group on X. Suppose there is a nontrivial group C ∈ C with rank(C) ≤ |X|. Then ι is injective. Moreover, ι(x) 6= 1 for each x ∈ X. Proof: Consider distinct elements x and y of X. Put r = rank(C). Let c1 , . . . , cr be generators of C. Then c1 , . . . , cr are distinct and ci 6= 1 for

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each i. When r = 1, define ϕ(x) = c1 and ϕ(z) = 1 for all z ∈ X r{x}. When r ≥ 2, choose x3 , . . . , xr ∈ X r{x, y}. Then let ϕ(x) = c1 , ϕ(y) = c2 , ϕ(xi ) = ci for i = 3, . . . , r, and ϕ(z) = 1 for z ∈ X r{x, y, x3 , . . . , xr }. In each case hϕ(X)i = C and ϕ: X → C converges to 1. Hence, there is an epimorphism ϕ: ˆ Fˆ → C with ϕˆ ◦ ι = ϕ. By definition, ϕ(x) 6= ϕ(y). Hence, ι(x) 6= ι(y). Moreover, ϕ(x) 6= 1, hence ι(x) 6= 1.  Assuming C, X, ι, and C are as in Lemma 17.4.4, we may identify X with a subset of FˆX (C) which does not contain 1 and ι: X → FˆX (C) with the inclusion map. Then FˆX (C) becomes a free pro-C group with basis X in the following sense: Definition 17.4.5: Let C be a formation of finite groups, Fˆ a pro-C group, and X a subset of Fˆ which does not contain 1. We say that Fˆ is a free pro-C group with basis X if the following hold: (2a) X generates Fˆ and converges to 1. (2b) Each map of X into a pro-C group G which converges to 1 and satisfies G = hϕ(X)i extends uniquely to an epimorphism ϕ: ˆ Fˆ → G.  Lemma 17.4.6: Let C be a formation of finite groups and Fˆ a free pro-C group with a basis X. Suppose C contains a nontrivial group of rank at most |X|. Then: (a) rank(Fˆ ) = |X|. (b) Suppose e = |X| < ∞. Then every set of generators of Fˆ of e elements is a basis of Fˆ . (c) Let F be the free abstract group on X and N (X) the set of all normal subgroups N of F with F/N ∈ C and X r N finite. Then Fˆ is the profinite completion of F with respect to N (X) and the canonical map θ: F → Fˆ maps each x ∈ X onto itself. Proof of (a) when X is finite: Suppose X = {x1 , . . . , xe } with x1 , . . . , xe distinct. Let {s1 , . . . , sd } be a system of generators of Fˆ , with d ≤ e. Put sd+1 = · · · = se = 1. Then the map xi 7→ si , i = 1, . . . , e extends to an epimorphism ϕ of Fˆ onto itself. By Proposition 16.10.6, ϕ is an automorphism. Since xi 6= 1 for all i, we have d = e, so {s1 , . . . , se } is a basis of Fˆ . This proves (a) for finite X and proves (b). Proof of (a) when X is infinite: Suppose X is infinite. By definition, X converges to 1. Hence, |X| = rank(Fˆ ) (Section 17.1). This concludes the proof of (a). Proof of (c): By Propositions 17.4.2 and 17.4.4, the free pro-C group F 0 with basis X is the completion of F with respect to N (X). Hence, by Lemma 17.4.3, the identity map of X extends to an isomorphism F 0 → Fˆ . This proves the claim.  Remark 17.4.7: Let C be a formation of finite groups and m ≥ 1 a cardinal number. Suppose C contains a group of rank at most m (e.g. when C is full). Then, Propositions 17.4.2 and 17.4.4 give a free pro-C group FˆX (C) with a

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basis X of cardinality m. By Lemma 17.4.3, FˆX (C) is uniquely determined up to an isomorphism by m and C. We denote FˆX (C) by Fˆm (C) and call it the free pro-C group of rank m. In particular, we write Fˆe (C) when m is a positive integer e and Fˆω (C) for the free pro-C group of rank ℵ0 . For ˆ C the family of all finite groups, we simplify Fˆm (C) to Fˆm (e.g. Fˆ1 = Z). ˆ When C is the family of all p-groups (resp. solvable groups), we use Fm (p) (resp. Fˆm (solv)) for Fˆm (C) (e.g. Fˆ1 (p) = Zp ). Now assume rank(G) > m for all nontrivial G ∈ C (so m is finite). Let F be a free abstract group on a set X of m elements. Then F has no proper normal subgroup N with F/N ∈ C. Thus, FˆX (C) is trivial and Fˆm (C) is undefined. To avoid this situation, whenever we say “let Fˆm (C) be a free pro-C group of rank m”, we tacitly assume C has nontrivial groups of rank at most m. For example, consider the case where C is a Melnikov formation which contains only non-Abelian groups. In this case, Fˆ1 (C) is undefined. By the classification of finite simple groups, each finite simple group is generated by two elements [Aschbacher-Guralnick, Thm. B]. Hence, for each cardinal number m ≥ 2 there is a free pro-C group on a set of cardinality m. Nevertheless, we will not use this application of the Classification in this book.  Proposition 17.4.8: Let C be a formation of finite groups. Then every pro-C group G is a quotient of a free pro-C group; if rank(G) = m, then G is a quotient of Fˆm (C). Proof: Use Douady (Proposition 17.1.1).



Next we show that every free pro-C group of rank ≤ m appears both as a closed subgroup and as a quotient of Fˆm (C): Lemma 17.4.9: Let C be a formation of finite groups, F a free pro-C group with a basis X, C a nontrivial group in C. (a) Let A be a subset of X of cardinality at least rank(C). Then hAi is a free pro-C group with basis A. Moreover, denote the minimal closed normal subgroup of F containing X r A by N . Then F = hAi n N . Thus, F/N is isomorphic to the free pro-C group on A. (b) F is the inverse limit of free pro-C groups on A where A ranges over all finite subsets of X of cardinality at least rank(C). Proof of (a): Since X r A ⊆ N , N · hAi = F . Define a map ϕ0 : X → FˆA (C) by ϕ0 (x) = x if x ∈ A and ϕ0 (x) = 1 if x ∈ X r A. By (1), ϕ0 extends to an epimorphism ϕA : F → FˆA (C) which is the identity on hAi. Since X r A ≤ Ker(ϕA ), N ≤ Ker(ϕA ). Hence, N ∩ hAi = 1. Conversely, let k ∈ Ker(ϕA ). Then, k = zn, with n ∈ N and z ∈ hAi. Apply ϕA to obtain 1 = z. Thus, k ∈ N . It follows that N = Ker(ϕA ). Consequently, F = hAi n N and hAi ∼ = F/N ∼ = FˆA (C). Proof of (b): Suppose A ⊆ B are finite subsets of X with rank(C) ≤ |A|. By (a), there is an epimorphism ϕB,A : FˆB (C) → FˆA (C) with ϕB,A (x) = x for x ∈ A and ϕB,A (x) = 1 for x ∈ B r A. Hence, ϕB,A ◦ ϕB = ϕA .

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Put F 0 = lim FˆA (C) and let πA : F 0 → FˆA (C) be the canonical projection. ←− The maps ϕA : F → FˆA (C) of (a) define an epimorphism ϕ: F → F 0 . For each A we have ϕA = πA ◦ ϕ. Hence, ϕ is injective on hAi. But the groups hAi generate F . Consequently, ϕ is an isomorphism.  Lemma 17.4.10: Let B ⊆ C be formations of finite groups and m a cardinal number. Suppose there is a B ∈ B with rank(B) ≤ m. Then Fˆm (B) is the maximal pro-B quotient of Fˆm (C). In other words, let N be the intersection of all open normal subgroups H of Fˆm (C) with Fˆm (C)/H ∈ B. Then Fˆm (C)/N ∼ = Fˆm (B). If X is a basis of Fˆm (C), then X/N is a basis of Fˆm (B). Suppose in addition, B is Melnikov. Then N has no proper open normal subgroup M with N/M ∈ B. Proof: Choose bases X and Y of cardinality m for Fˆm (B) and Fˆm (C), respectively. Let ϕ0 : Y → X be a bijective map. Since each pro-B group is a pro-C group, ϕ0 extends to an epimorphism ϕ: Fˆm (C) → Fˆm (B). Conversely, let G be a pro-B group and γ: Fˆm (C) → G an epimorphism. −1 Then γ ◦ ϕ−1 0 (X) converges to 1 and generates G. Hence, γ ◦ ϕ0 extends to an epimorphism ψ: Fˆm (B) → G such that ψ ◦ ϕ = γ. According to Definition 17.3.2, Fˆm (B) is the maximal pro-B quotient of Fˆm (C). Hence, N = Ker(ϕ). Finally, suppose B is Melnikov and M is a proper open normal subgroup of N with N/M ∈ B. Then M = H ∩ N for S some open subgroup H T of Fˆm (C) n n (Lemma 1.2.5) and M / H. Write Fˆm (C) = i=1 Hzi . Then M0 = i=1 M zi is an open subgroup of N which is normal in Fˆm (C). Moreover, N/M0 ∈ B. Hence, by Lemma 17.3.1(e), Fˆm (C)/M0 is a pro-B group. This contradicts the minimality of N .  Lemma 17.4.11: Let C be a formation of finite groups, e a positive integer, and G a pro-C group of rank at most e. Then each epimorphism β: G → Fˆe (C) is an isomorphism. Proof: By assumption, there is an epimorphism α: Fˆe (C) → G. By Proposition 16.10.6(b), β is an isomorphism. 

17.5 Subgroups of Free Discrete Groups Free profinite groups are introduced in Proposition 17.4.2 as inverse limits of finite quotients of free abstract groups. It is therefore reassuring that the study of closed subgroups of free profinite groups depends on the classical result that subgroups of free abstract groups are free. Consider a free abstract group F on a set X. With each x ∈ X associate a new letter x−1 and put X −1 = {x−1 | x ∈ X}. Each element f ∈ F can be presented as a word in the letters of X ∪ X −1 : (1)

f = s1 s2 · · · sn , with s1 , . . . , sn ∈ X ∪ X −1

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Among all possible words that express f there is a unique one in which n is minimal. In this case si 6= s−1 i+1 for each i, 1 ≤ i ≤ n − 1, and the word s1 s2 · · · sn is said to be reduced. We say that f starts with s1 and ends with sn . Define lengthX (f ) to be n. Now let H be a subgroup of F and define the length of the right coset Hf for f ∈ F : lengthX (Hf ) = min{lengthX (hf ) | h ∈ H}. Definition 17.5.1: Representation of right cosets. Let R be a S system of representatives of the right cosets of F modulo H. Thus, F = · r∈R Hr. For each f ∈ F let r = ρR (f ) be the unique element of R satisfying Hf = Hr. The function ρ = ρR : F → R satisfies the following conditions: (2a) ρ(f ) ∈ Hf . (2b) ρ(hf ) = ρ(f ) for all h ∈ H and f ∈ F . (2c) R = ρ(F ). These conditions imply that (3a) ρ(ρ(f )g) = ρ(f g), for all f, g ∈ F ; and (3b) ρ(r) = r for each r ∈ R. Conversely, if a function ρ: F → R satisfies (2), then R is a system of  representatives of the right cosets of F modulo H and ρR = ρ. Lemma 17.5.2: Let t be an element of X ∪ X −1 not in H. Then there exists a system of representatives R for the right cosets of F modulo H with the following properties: (4a) lengthX (ρR (f )) = lengthX (Hf ) for each f ∈ F . (4b) If ρR (f ) = s1 s2 · · · sn is a reduced presentation of ρR (f ), then s1 s2 · · · si ∈ R for each i between 1 and n. (4c) 1, t ∈ R. Proof: It suffices to define a function ρ: F → R satisfying (2) and (4). We define ρ(1) = 1 and ρ(t) = t. Then, lengthX (ρ(f )) = 1 = lengthX (Hf ) for f = 1 and f = t. Assume by induction, that in addition to the right cosets H and Ht, we have chosen representatives for all right cosets of length smaller than a positive integer k. Let Hf be a right coset of length k such that Hf 6= Ht. Without loss suppose that lengthX (f ) = k and let f = s1 s2 · · · sk be a reduced presentation of f . Then for f 0 = s1 s2 · · · sk−1 , lengthX (Hf 0 ) ≤ k − 1. Thus, ρ(f 0 ) is already defined and it has a reduced presentation ρ(f 0 ) = s01 s02 · · · s0l , with l ≤ k − 1, such that s01 s02 · · · s0i ∈ R for each i, 1 ≤ i ≤ l. In particular, there exists h ∈ H such that f 0 = hρ(f 0 ). Hence, f = f 0 sk = hs01 · · · s0l sk . The assumption lengthX (f ) = lengthX (Hf ) implies that l = k − 1 and that s01 · · · s0l sk is a reduced word. Define ρ(f ) = s01 · · · s0l sk and observe that conditions (4a) and (4b) are satisfied. This completes the induction. 

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Notation 17.5.3: The Schreier construction [Kurosh, §36]. Let R be a system of representatives of the right cosets of F modulo H and t an element of X ∪ X −1 r H. We call R a Schreier system (with respect to X, H, t) if it satisfies Condition (4) appearing in Lemma 17.5.2. In this case we put ρ = ρR and define a map ϕR : R × X → F by ϕR (r, x) = rxρ(rx)−1 ,

r ∈ R, x ∈ X.

By (3b), ϕR (r, x) = 1 if and only if rx ∈ R. Put Y = YR = {ϕR (r, x) | r ∈ R, x ∈ X} r{1}. Since Hrx = Hρ(rx), Y is a subset of H. We show below that H is free and Y is a basis of H. We call Y a Schreier basis of H (with respect to X, t).  Qn Use the convention i=1 fi = f1 f2 · · · fn for products of elements of a group. Lemma 17.5.4: The following hold for the objects F, X, H, t, R, Y constructed above: (a) The set Y generates H. (b) lengthY (h) ≤ lengthX (h), for each h ∈ H. (c) If the reduced presentation of an element h ∈ H starts with t, then lengthY (h) < lengthX (h). (d) X ∩ H ⊆ Y . (e) Let N be a subgroup of H and x ∈ X ∩ N . Suppose N is normal in F . Then ϕR (r, x) = rxr−1 for each r ∈ R. (f) Suppose in addition to (e) that (F : H) < ∞ and X r N is finite. Then Y r N is finite. Proof of (a): Put ρ = ρR and ϕ = ϕR . Let h = s1 s2 · · · sn

(5)

be a presentation of an element h ∈ H as a (not necessarily reduced) word. For 0 ≤ i ≤ n put gi = s1 s2 · · · si . Then g0 = 1 and gn = h. In particular, ρ(g0 ) = ρ(gn ) = 1. Therefore (6)

h=

n Y

ρ(gi−1 )si ρ(gi )−1 .

i=1

∈ X. In the first case, by For each i, 1 ≤ i ≤ n, either si ∈ X or s−1 i (3), the ith factor in the right hand side of (6) is ρ(gi−1 )si ρ(gi )−1 = ρ(gi−1 )si ρ(gi−1 si )−1 = ρ(gi−1 )si ρ(ρ(gi−1 )si )−1 = ϕ(ρ(gi−1 ), si ).

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In the second case, −1 −1 = ρ(ρ(gi )s−1 ρ(gi−1 )si ρ(gi )−1 = ρ(gi s−1 i )si ρ(gi ) i )si ρ(gi )   −1
−1 −1 −1 = ρ(gi )s−1 = ϕ(ρ(gi ), s−1 . i ρ ρ(gi )si i )

Thus, we may rewrite (6) as (7)

h=

n Y

ϕ(ri , xi )εi ,

i=1

where  (8)

(ri , xi , εi ) =

(ρ(gi−1 ), si , 1) if si ∈ X −1 (ρ(gi ), s−1 . i , −1) if si ∈ X

This is a presentation of h as a product of elements of Y ∪ Y −1 . Proof of (b): Suppose now that the presentation (5) of h is reduced. Then (7) implies that lengthY (h) ≤ n = lengthX (h). Proof of (c): If s1 = t in (5), then (4c) gives ρ(g0 )s1 ρ(g1 )−1 = t · t−1 = 1. Hence, lengthY (h) ≤ n − 1, as asserted. Proof of (d): If x ∈ X ∩ H, then ρ(x) = 1, and x = ϕ(1, x) ∈ Y . Proof of (e): If x ∈ X ∩N , then Hρ(rx) = Hrx = H(rxr−1 )r = Hr. Hence, ρ(rx) = r and ϕ(r, x) = rxr−1 . Proof of (f): Here |R| = (F : H) < ∞. By (e), ϕ(r, x) = rxr −1 ∈ N for all r ∈ R and x ∈ X ∩ N . Therefore, Y r N is contained in the finite set {ϕ(r, x) | r ∈ R, x ∈ X r N }.  Lemma 17.5.5: H is a free group on Y . Proof: Put ρ = ρR and ϕ = ϕR . To each pair (r, x) ∈ R×X with ϕ(r, x) 6= 1 attach a letter ψ(r, x). For (r, x) ∈ R × X with ϕ(r, x) = 1 put ψ(r, x) = 1. Then let Y ∗ be the set of all ψ(r, x) with (r, x) ∈ R × X and ϕ(r, x) 6= 1. Denote the free abstract group on Y ∗ by H ∗ . Map H into H ∗ by mapping h ∈ H, presented by (5), onto (9)

h∗ =

n Y

ψ(ri , xi )εi , with ri , xi , εi defined by (8), i = 1, . . . , n.

i=1

We prove that the map h 7→ h∗ is a homomorphism of H onto H ∗ mapping Y bijectively onto Y ∗ . It will follow that this map is an isomorphism.

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Part A: The map h 7→ h∗ is well defined. If si+1 = s−1 i , then gi+1 = gi−1 si si+1 = gi−1 . Therefore, if si ∈ X, then (ri , xi , εi ) = (ρ(gi−1 ), si , 1) and (ri+1 , xi+1 , εi+1 ) = (ρ(gi+1 ), s−1 i+1 , −1). Thus, (ri , xi ) = (ri+1 , xi+1 ), εi+1 = −εi , and ψ(ri+1 , xi+1 ) = ψ(ri , xi ). Similarly, the latter two equalities hold if si ∈ X −1 . Two presentations of h as words in X ∪ X −1 are obtained from each other by inserting or canceling finitely many factors of the form ss−1 , with s ∈ X ∪ X −1 . Thus, the preceding paragraph shows that the definition of h∗ is independent of the presentation (5) of h. Part B: The map h 7→ h∗ is a homomorphism. Let h0 = s01 s02 · · · s0n0 be another element of H with s01 , s02 , . . . , s0n0 ∈ X ∪ X −1 . By (2b), ρ(s1 · · · sn s01 · · · s0j ) = ρ(hs01 · · · s0j ) = ρ(s01 · · · s0j ). Therefore, the presentation (6) for hh0 is the product of the presentation (6) for h followed by the presentation (6) for h0 . This implies (hh0 )∗ = h∗ (h0 )∗ . In other words, the map h 7→ h∗ is a homomorphism of H into H ∗ . Part C: We prove that ψ(r, x) = ϕ(r, x)∗ for each (r, x) ∈ R × X with ϕ(r, x) 6= 1. Let r = u1 u2 · · · uk and ρ(rx) = v1 v2 · · · vl be the reduced presentations of r and ρ(rx) as words in the letters of X ∪ X −1 . Apply (5) and (6) to h = ϕ(r, x): −1

ϕ(r, x) = rxρ(rx)

=

k Y

ui · x ·

i=1

=

k Y

l−1 Y

−1 vl−j

j=0

ρ(u1 · · · ui−1 )ui ρ(u1 · · · ui−1 ui )−1

i=1

· u1 · · · uk xρ(rx)−1 ·

l−1 Y

−1 −1 −1 −1 −1 ρ(rxvl−1 · · · vl−j+1 )vl−j ρ(rxvl−1 · · · vl−j+1 vl−j ) .

j=0 −1 as an empty product with the value 1. Here, for j = 0, regard vl−1 · · · vl−j+1

Also, for j = l − 1 we have ρ(rxvl−1 · · · v1−1 ) = ρ(h) = 1. Use (3) to define appropriate ri , xi , εi and rj0 , x0j , ε0j as in(8) such that ϕ(r, x) =

k Y

l−1 Y

ϕ(ri , xi )εi · ϕ(r, x) ·

i=1

0

ϕ(rj0 , x0j )εj .

j=0

From definition (9) ϕ(r, x)∗ =

k Y i=1

ψ(ri , xi )εi · ψ(r, x) ·

l−1 Y j=0

0

ψ(rj0 , x0j )εj .

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355

For each i, 1 ≤ i ≤ k, compute from (3a), (3b), and (4b) that ϕ(ri , xi )εi = ρ(u1 · · · ui−1 )ui ρ(u1 · · · ui−1 ui )−1 = (u1 · · · ui−1 )ui (u1 · · · ui−1 ui )−1 = 1, so ψ(ri , xi ) = 1. Similarly, for each j, 0 ≤ j ≤ l − 1, −1 −1 ) = ρ(ρ(rx)vl−1 · · · vl−j+1 ) ρ(rxvl−1 · · · vl−j+1

= ρ(v1 v2 · · · vl−j ) = v1 v2 · · · vl−j . Hence 0

−1 −1 −1 −1 −1 ϕ(rj0 , x0j )εj = ρ(rxvl−1 · · · vl−j+1 )vl−j ρ(rxvl−1 · · · vl−j+1 vl−j ) −1 = (v1 · · · vl−j )vl−j (v1 · · · vl−j−1 )−1 = 1.

Thus, ψ(rj0 , x0j ) = 1. Consequently, ϕ(r, x)∗ = ψ(r, x). This concludes the proof of the lemma.  We summarize the information achieved in this section up to this point: Proposition 17.5.6 (Schreier): Let F be a free abstract group on a set X and H a subgroup of F . Then H is free. Moreover, for each t ∈ X ∪X −1 r H there exist a Schreier system of representatives R for the right cosets of F modulo H and a Schreier basis Y = YR for X, H, t. By definition, R and Y satisfy all assertions of Lemma 17.5.4: (a) lengthY (h) ≤ lengthX (h), for each h ∈ H. (b) If the reduced presentation of an element h ∈ H starts with t, then lengthY (h) < lengthX (h). (c) X ∩ H ⊆ Y . (d) Let N be a subgroup of H and x ∈ X ∩ N . Suppose N is normal in F . Then ϕR (r, x) = rxr−1 for each r ∈ R. (e) Moreover, suppose (F : H) < ∞ and X r N is finite. Then Y r N is finite. Proposition 17.5.7 (Nielsen-Schreier Formula): Let F be a free abstract group of rank e and let H be a subgroup of index n. Then H is free of rank 1 + n(e − 1). Proof: Continue to use the notation of the previous lemmas. In this case the basis X of F contains e elements and the system of representatives R of F modulo H has n elements. Therefore, the number of elements in the basis Y of H (Lemma 17.5.6) equals ne minus the number of elements of the set Γ = {(r, x) ∈ R × X | rx = ρ(rx)}. If we prove that |Γ| = n − 1, then rank(H) = ne − (n − 1) = 1 + n(e − 1),

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and we are done. To this end, define two subsets of R for each positive integer k: Rk = {r ∈ R | lengthX (r) = k and r ends with some x ∈ X}; and Rk0 = {r ∈ R | lengthX (r) = k and r ends with some x−1 ∈ X −1 }. S In addition, let R00 = ∅ and R0 = · k≥0 Rk0 ∪· Rk+1 . Then R = {1} ∪· R0 . Thus, |R0 | = n − 1. On the other hand, for each k ≥ 0 define two subsets of Γ: Γk = {(r, x) ∈ Γ | lengthX (r) = k and lengthX (rx) = k + 1}; and Γ0k = {(r, x) ∈ Γ | lengthX (r) = k and lengthX (rx) = k − 1}. S Also let Γ00 = ∅. Then Γ = · k≥0 Γ0k ∪· Γk . Now define a bijective map τ : Γ → R0 . Suppose k > 0 and (r, x) ∈ Γ0k . Then r ends with x−1 . Define τ (r, x) to be r. Conversely, if r ∈ Rk0 , then r ends with some x−1 ∈ X −1 . Hence, lengthX (rx) = k − 1. Therefore, by (4b), rx ∈ R, so ρ(rx) = rx and (r, x) ∈ Γ0k . Consequently, τ maps Γ0k bijectively onto Rk0 . Finally, suppose k ≥ 0 and (r, x) ∈ Γk . Define τ (r, x) = rx and note that rx ∈ Rk+1 . Conversely, if r0 ∈ Rk+1 , then, by (4b), r0 = rx, where x ∈ X and r ∈ R is of length k. By (3b), rx = ρ(rx). Hence, (r, x) ∈ Γk . Consequently, τ maps Γk bijectively onto Rk+1 . The definition of τ is now complete. We conclude that |Γ| = |R0 | = n−1, as desired.  We define the rank of an abstract group G as the minimal cardinality of a set of generators of G. If G is finite, than the rank of G as an abstract group is equal to the rank of G as a profinite group. However, if G is infinite, then the rank of G as a profinite group may be less than the rank of G as ˆ as a profinite group is 1 while an abstract group. For example, the rank of Z ℵ0 its rank as an abstract group is 2 . Nevertheless, it should be always clear from the context in which sense we use rank(G). Corollary 17.5.8: Let G0 be a subgroup of index n of an abstract group G of rank e. Then rank(G0 ) ≤ 1 + n(e − 1). Proof: There is an epimorphism α: F → G, where F is the free abstract group on e generators. Put F0 = α−1 (G0 ). Then (F : F0 ) = n. By Proposition 17.5.7, rank(F0 ) = 1 + n(e − 1). Therefore, rank(G0 ) ≤ 1 + n(e − 1).  Remark 17.5.9: Residually finite groups. Let G be an abstract group and C a family of finite groups. Then G is residually-C if the intersection of all normal subgroups N of G with G/N ∈ C is 1. When C consists of all finite groups, we use residually finite for residually-C.

17.5 Subgroups of Free Discrete Groups

357

Suppose C is a formation of finite groups. Denote the family of all N / G with G/N ∈ C by N . Then N is directed. Let θ be the canonical map of G ˆ with respect to N . Then G is residually C if and only into its completion G if θ is injective.  Let x1 , . . . , xe be free generators of a free abstract group F and w a nonempty reduced word in {x1 , . . . , xe } of length n. Then there exists a homomorphism ϕ: F → Sn+1 with ϕ(w) 6= 1 [Kurosh, p. 42]. Thus, F is residually finite. Proposition 17.5.11 below generalizes this conclusion considerably. Lemma 17.5.10 (Levi): Let F be a free abstract group and F > F1 > F2 > · · · a decreasing sequence T∞of subgroups. Suppose each Fi+1 is a characteristic subgroup of Fi . Then i=1 Fi = 1. T∞ Proof (Lubotzky): Assume i=1 Fi contains a nontrivial element w. By Proposition 17.5.6, each Fi is free. Let lengthi (w) = min(lengthX (w)) where X ranges on all bases of Fi . If we prove that lengthi+1 (w) < lengthi (w), we will draw a contradiction. Indeed, let X be a basis of Fi with lengthX (w) = lengthi (w). Let w = s1 s2 · · · sn be a reduced word in X ∪ X −1 . Assume s1 ∈ Fi+1 . Replacing s1 with its inverse, if necessary, we may assume s1 ∈ X. Let x ∈ X. The bijective map ϕ0 : X → X which permutes x and s1 and fixes all other elements of X extends to an automorphism ϕ: Fi → Fi . Since Fi+1 is a characteristic subgroup of Fi , ϕ maps Fi+1 onto itself. In particular, x ∈ / Fi+1 . Fi+1 . It follows that Fi+1 = Fi . This contradiction proves that s1 ∈ Proposition 17.5.6 gives a basis Y of Fi+1 such that length TY∞(w) < lengthX (w). Hence, lengthi+1 (w) < lengthi (w). We conclude that i=1 Fi = 1.  Proposition 17.5.11 ([Ribes-Zalesskii, Prop. 3.3.15]): Let C be a Melnikov formation of finite groups, m a cardinal number, F the free abstract group of rank m, and Fˆ = Fˆm (C) the free pro-C group of rank m. Then F is residually-C and the canonical map θ: F → Fˆ is injective. Proof: The assumption on the existence of Fˆ yields G ∈ C with rank(G) ≤ m. Let F1 be the intersection of all normal subgroups N of F with F/N ∼ = G. Then F1 is a characteristic subgroup of F . By Proposition 17.5.6, F1 is free. To estimate the rank of F1 we choose a prime number p such that Z/pZ is not a composition factor of G. Then F has a normal subgroup N with F/N ∼ = F/N . Hence, rank(F1 ) ≥ m. = (Z/pZ)m and F1 /N ∩ F1 ∼ We may therefore define F2 to be the intersection of all normal subgroups N of F1 with F1 /N ∼ = G. Then F2 is characteristic in F1 , hence in F , and rank(F2 ) ≥ m. Continue by induction to construct an descending sequence F > F1 > F2 > · · · of characteristic T∞ subgroups of F with rank(Fi ) ≥ m. By Levi’s result (Lemma 17.5.10), i=1 Fi = 1. Therefore, F is residually-C. 

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Remark 17.5.12: Let F be the free abstract group of rank m ≥ 2 and let C be an infinite family of finite non-Abelian simple groups. By [Weigel1,2,3] or [Dixon-Pyber-Seress-Shalev], F is residually-C. Note that the assumption of the groups in C to be non-Abelian is essential. If C consists of all groups Z/pZ, then the intersection of all normal subgroups N of F with F/N ∈ C contains the commutator subgroup [F, F ] of F which is not trivial. 

17.6 Open Subgroups of Free Profinite Groups We prove in this section that an open subgroup of a free profinite group is free and satisfies the Nielsen-Schreier formula for the ranks: Remark 17.6.1: Identifying F as a subgroup of Fˆ . Let C be a Melnikov formation of finite group and F the free abstract group on a set X. Suppose there is a nontrivial group C ∈ C with rank(C) ≤ |X|. Let FˆX (C) be the corresponding profinite completion. We use Proposition 17.5.11 to identify F with a dense abstract subgroup of FˆX (C) and to identify the canonical map θ: F → FˆX (C) with the inclusion map. Conversely, let Fˆ be a free pro-C group with a basis X. Lemma 17.4.3 extends the identity map X → X to an isomorphism FˆX (C) → Fˆ . The latter maps F onto the abstract subgroup F0 of Fˆ generated by X. It follows that F0 is free on X.  Proposition 17.6.2: Let C be a Melnikov formation of finite groups, Fˆ a ˆ an open subgroup of Fˆ . Suppose H ˆ is pro-C (e.g. C free pro-C-group, and H ˆ / Fˆ ). Then H ˆ is a free pro-C group. Moreover, if e = rank(Fˆ ) is is full or H ˆ = 1 + (Fˆ : H)(e ˆ finite, then rank(H) − 1). If m = rank(Fˆ ) is infinite, then ˆ = m. rank(H) Proof: Let X be a basis for Fˆ and F the free abstract group on X. Consider the family N (X) of all normal subgroups N of F with X r N finite and F/N ∈ C. By Lemma 17.4.6(c), Fˆ is the profinite completion of F with respect to N (X). By Proposition 17.5.11, we may assume that F is contained in Fˆ and θ: F → Fˆ is the inclusion map. Lemma 17.2.1 gives a subgroup H ˆ is the closure of H in Fˆ . of F and a M ∈ N (X) such that M ≤ H and H ˆ is finite and H ˆ = lim H/N , where N ranges Moreover, (F : H) = (Fˆ : H) ←− over N0 (X) = {N ∈ N (X) | N ≤ H}. Let Y , R, and ϕ be as in Proposition 17.5.6. Then R is finite, H is free, Y is a basis of H, and Y r N is finite for each N ∈ N0 (X). By Proposition 17.4.2, the free pro-C-group on Y is isomorphic to the inverse limit lim H/N , ←− where N ranges over the family N (Y ) of all open normal subgroups N of H such that Y r N is finite and H/N ∈ C. We show that N0 (X) and N (Y ) are ˆ is the free cofinite in each other. By Lemma 17.2.3, this will prove that H pro-C-group on Y . ˆ . By Lemma Indeed, let N ∈ N0 (X). Denote the closure of N in Fˆ by N ∼ ˆ ˆ ˆ ˆ ˆ 17.2.1, N / H and H/N = H/N . Since H is pro-C, we have H/N ∈ C, so

17.6 Open Subgroups of Free Profinite Groups

359

N ∈ N (Y ). r Conversely, let N ∈ N T T (Y ). Then H/N ∈ C and Y N is finite. Let N0 = r∈R N r and H0 = r∈R H r . Then N0 and H0 are normal subgroups of F , M ≤ H0 , and N0 ≤ H0 . Since N H0 /N is normal in H/N , it belong to C. Since H0 /(N ∩ H0 ) is isomorphic to N H0 /N , it belongs toTC. Hence, H0 /(N r ∩H0 ) ∈ C for each r ∈ R. Since R is finite, H0 /N0 = H0 / r∈R (N r ∩ H0 ) ∈ C (Condition (1b) of Section 17.3)). In addition, F/H0 ∈ C because F/M ∈ C and F/H0 is a quotient of F/M . Finally, since C is Melnikov, F/N0 ∈ C. F

N0

N

N H0

N ∩ H0

H0

H

M If we prove that X r N0 is finite, it will follow that N0 ∈ N0 (X). This will complete the proof of the cofiniteness of N (Y ) and N0 (X) in each other. Indeed, X r M , Y r N , and R are finite sets. Thus, almost all x ∈ X belong to M . For each x ∈ X ∩ M and r ∈ R we have ϕ(r, x) = rxr −1 (Proposition 17.5.6(d)). For each r ∈ R the map x 7→ rxr−1 of X ∩ M into Y is injective. Hence, only finitely many x ∈ X ∩ M satisfy rxr−1 ∈ Y r N . Therefore, X0 = {x ∈ X ∩M | ϕ(r, x) = rxr−1 and rxr−1 ∈ N for all r ∈ R} is cofinite in X. It follows that almost every y ∈ Y can be written as y = vxv −1 with x ∈ X0 and v ∈ R. For each u ∈ R there exist r ∈ R and h ∈ H with uv = hr. Therefore, uyu−1 = uvxv −1 u−1 = hrxr−1 h−1 ∈ hN h−1 = N , so, y ∈ N u . It follows that almost every y ∈ Y belongs to N0 . By Lemma 17.5.4(d), X r N0 = (X r H) ∪ (X ∩ H r N0 ) ⊆ (X r H) ∪ (Y r N0 ) is finite, as claimed. The rank formula for the finitely generated case now follows from Proposition 17.5.7. When m is infinite, use Corollary 17.1.5.  Corollary 17.6.3: Let G be a profinite group of rank at most e and H an open subgroup of G of index n. Then rank(H) ≤ 1 + n(e − 1). Proof: There exists an epimorphism θ: Fˆe → G and (Fˆe : θ−1 (H)) = n. By Proposition 17.6.2, rank(θ−1 (H)) = 1 + n(e − 1). Hence, by Corollary 17.1.4, rank(H) ≤ 1 + n(e − 1).  The following result is used in Section 25.7 to prove that a closed norˆ of a free pro-C group is free pro-C if N ˆ contains nontrivial mal subgroup N elements of the underlying abstract free group. Proposition 17.6.4: Let C be a Melnikov formation, Fˆ a free pro-C group of rank m ≥ 2, X a basis of Fˆ , F the abstract group of Fˆ generated by X, ˆ a closed normal subgroup of Fˆ . Suppose N = F ∩ N ˆ 6= 1. Then: and N

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Chapter 17. Free Profinite Groups

ˆ containing N ˆ with a Schreier basis Y (a) Fˆ has an open normal subgroup E with respect to X such that Y ∩ N 6= ∅. ˆ containing N ˆ with (b) If m < ∞ than Fˆ has an open normal subgroup E ˆ ˆ ˆ ˆ rank(E/N ) < 1 + (F : E)(m − 1). ˆ , t 6= 1. Denote the set of all open normal Proof of (a): Let t ∈ F ∩ N ˆ by H. For each E ˆ ∈ H put E = F ∩ E. ˆ subgroups of Fˆ which contain N ˆ ∈ H, a Schreier basis Y of E with respect to X (Notation 17.5.3), and Find E a presentation t = s1 s2 · · · sn as a word in Y with a minimal length among all presentations of this type. We show that s1 ∈ N . ˆ is the intersection of all H ˆ ∈ H. Assume s1 ∈ / N . By Lemma 1.2.3, N ˆ ˆ ˆ ˆ / H. By Proposition Hence, there is an H ∈ H such that H ≤ E and s1 ∈ 17.5.6(b), H has a Schreier basis Z with respect to Y with lengthZ (t) < lengthY (t), a contradiction. ˆ Proof of (b): Continuing the proof of (a), we conclude from s1 ∈ Y ∩ N ˆ ˆ ˆ ˆ ˆ and from Nielsen-Schreier that rank(E/N ) < rank(E) = 1 + (F : E)(m − 1).  Corollary 17.6.5: Let C be a Melnikov formation and Fˆ a free pro-C-group of positive rank. Then Fˆ is infinite. Proof: Assume Fˆ is finite. Then the trivial group is a free pro-C-group of positive rank (Proposition 17.6.2). This is a contradiction.  Example 26.1.10 shows that Corollary 17.6.5 does not hold for arbitrary formations.

17.7 An Embedding Property Here is a characterization of free pro-C groups of finite rank in terms of their finite quotients: Lemma 17.7.1: Let C be a formation of finite groups, e a positive integer, and F a pro-C group. Then F ∼ = Fˆe (C) if and only if F satisfies the following conditions: (a) Every finite homomorphic image of F is generated by e elements. (b) Every C-group of rank at most e is a homomorphic image of F . Proof: A compactness argument shows that (a) holds if and only if rank(F ) ≤ e. Now apply Proposition 16.10.7(b).  We conclude this section by establishing an “embedding property” of Fˆe (C). It depends on a surprising lemma that allows us to “lift generators” of a homomorphic image: Lemma 17.7.2 (Gasch¨ utz): Let π: G → H be an epimorphism of profinite groups with rank(G) ≤ e. Let h1 , . . . , he be a system of generators of H.

Exercises

361

Then there exists a system of generators g1 , . . . , ge of G such that π(gi ) = hi , i = 1, . . . , e. Proof (Roquette): We start with the crucial case: G is a finite group. For each subgroup C of G satisfying π(C) = H and all systems of generators a1 , . . . , ae of H denote the number of e-tuples c ∈ C e that generate C and satisfy π(c) = a by ϕC (a). We prove by induction on |C| that ϕC (a) is independent of a. Assume ϕB (a) is independent of a for every proper subgroup B of C |C| . Then there are exactly me elements satisfying π(B) = H. Let m = |H| b ∈ C e with π(b) = a. Each such b generates a subgroup B of C satisfying π(B) = H. Hence, (1)

me = ϕC (a) +

X0 B
ϕB (a),

P0 indicates a sum over groups with π(B) = H. By assumption, the where P0 is independent of a. Therefore, so is ϕC (a). Now choose a system of generators g10 , . . . , ge0 for G. Then π(g0 ) = h0 generates H. By the preceding paragraph, ϕG (h) = ϕG (h0 ) ≥ 1. Consequently, G has a system of generators g1 , . . . , ge such that π(g) = h. For G general present π: G → H as an inverse limit of epimorphisms of finite groups πi : Gi → Hi , i ∈ I. Specifically, if j ≥ i, then there are epimorphisms ξji : Gj → Gi and ηji : Hj → Hi such that πi ◦ ξji = ηji ◦ πj . In addition, there are epimorphisms ξi : G → Gi and ηi : H → Hi such that πi ◦ ξi = ηi ◦ π. For each i ∈ I denote the set of e-tuples x ∈ Gei that generate Gi and satisfy πi (x) = ηi (h) by Ai . By the case where G is finite, Ai is nonempty. In addition, Ai is finite. If j ≥ i and y ∈ Aj , then ξji (y) ∈ Ai . By Corollary 1.1.4, the inverse limit A of the inverse system hAi , ξji ii,j∈I is nonempty. Each element in A defines a system of generators g1 , . . . , ge of G with π(g) = h.  Proposition 17.7.3: Let C be a formation of finite groups, e a positive integer, and (ϕ: Fˆe (C) → H, α: G → H) a pair of epimorphisms, where G is a pro-C group and rank(G) ≤ e. Then there exists an epimorphism γ: Fˆe (C) → G such that α ◦ γ = ϕ. Proof: Let x1 , . . . , xe be a basis of Fˆe (C). Then ϕ(x1 ), . . . , ϕ(xe ) generate H. By Lemma 17.7.2, there are g1 , . . . , ge which generate G such that α(gi ) = ϕ(xi ), i = 1, . . . , e. The map xi 7→ gi , i = 1, . . . , e, extends to an epimorphism  ϕ: Fˆe (C) → G such that α ◦ γ = ϕ.

Exercises Let θ: G → H be an epimorphism of profinite groups. (a) Prove that θ maps subsets of G that converge to 1 onto subsets of H that converge to 1.

1.

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Chapter 17. Free Profinite Groups

(b) Let {hi | i ∈ I} be a subset of H that converges to 1. Imitate the proof of Proposition 17.1.1 to prove that G contains a subset {gi | i ∈ I} which converges to 1 with θ(gi ) = hi for each i ∈ I. 2. Let G be a profinite group, H an open subgroup, and G0 a dense subset of G. Prove that G0 ∩ H is dense in H. 3. Let D be a family of finite simple groups and C the Melnikov formation generated by C (Remark 17.3.4). Suppose every subgroup of each S ∈ D belongs to C. Prove that C is full. Hint: Let G ∈ C. Choose a composition series of G and apply induction on its length. 4. Use Lemma 17.4.10 to prove that the quotient of Fˆe modulo its commuˆ e. tator subgroup (Fˆe )0 is isomorphic to Z ˆ be a profinite completion of a finitely generated group G. Prove 5. Let G ˆ ≤ rank(G). that rank(G) 6. For f ≥ e > 1 use Nielsen-Schreier to prove for each x1 , . . . , xf ∈ Fˆe −1 that the inequality (Fˆe : hxi) < ∞ implies (Fˆe : hxi) ≤ fe−1 . Conclude for x1 , . . . , xe ∈ Fˆe that either hxi = Fˆe or (Fˆe : hxi) = ∞.

Notes Ershov [Ershov2, Lemma 1, and Exercise 9 of Chapter 21] suggests an alternative proof of Douady’s theorem (Proposition 17.1.1). The open subgroup theorem (Proposition 17.6.2) for a free profinite group is a special case of the subgroup theorem for free products of profinite groups appearing in [Binz-Neukirch-Wenzel]. The latter, is based on the deeper Kurosh subgroup theorem for free products of abstract groups [Specht, p. 189, Satz 8] while our proof of Theorem 17.6.2 is based on the simpler Schreier construction of Section 17.5. Gasch¨ utz’s proof of Lemma 17.7.2 [Gasch¨ utz] provides a (complicated) formula for ϕG (a). The proof of Levi’s Lemma 17.5.10 depends on Schreier’s construction of a basis of a subgroup of a free abstract group rather than on Nielsen’s construction as in [Lyndon-Schupp, p. 14, Prop. 3.3]. The latter proof considers a nontrivial element w in Fi (notation as in Lemma 17.5.10) and a basis X of F and proves that lengthX (w) ≥ i. Our proof constructs a basis Y of Fi such that lengthY (w) ≤ max(0, lengthX (w) − i).

Chapter 18. The Haar Measure It is well known that every locally compact group admits a (one sided) translation invariant Haar measure. Applications of the Haar measure in algebraic number theory to local fields and adelic groups appear in [Cassels-Fr¨ohlich, Chap. II] and [Weil6]. Here we use it to investigate absolute Galois groups of fields. Since these groups are compact, the Haar measure is a two sided invariant. We provide a simple direct proof of the existence and uniqueness of the Haar measure of profinite groups (Sections 18.1 and 18.2). We normalize the Haar measure µ of a compact group G so that µ(G) = 1. Thus, µ is a probability measure. When G = Gal(K) is the absolute Galois group of a field K, µ-independence of sets is related to linear disjointness of fields (Section 18.5). For K a Hilbertian field, we prove that hσ1 , . . . , σe i ∼ = Fˆe for almost all σ ∈ Gal(K)e . In addition, if K is countable, then the fixed fields Ks (σ1 , . . . , σe ) are PAC for almost all σ ∈ Gal(K)e (Section 18.6). In the uncountable case we provide examples where the set of σ ∈ Gal(K)e such that Ks (σ1 , . . . , σe ) is PAC, is nonmeasurable. The complete proof that every field is stable (Definition 18.9.1) lies unfortunately outside the scope of this book. Section 18.9 outlines the main ingredients and steps of the proof. Once this is done, we are able to construct for each countable Hilbertian field an abundance of Galois extensions which are PAC (Theorem 18.10.2).

18.1 The Haar Measure of a Profinite Group Let G be a profinite group. We define the completed Haar measure of G and prove its uniqueness. In the next section we prove the existence of the Haar measure. Consider a collection A of subsets of G. The σ-algebra generated by A is the smallest collection A0 of subsets of G which contains A and is closed under taking complements and countable unions. When A is a Boolean algebra, A0 is also the smallest collection of subsets of G which contains A and is monotone. That is, A0 is closed under countable increasing unions and countable decreasing intersections [Halmos, p. 27, Thm. B] The Borel field of G is the σ-algebra generated by all closed (= compact) subsets of G. We denote it by B or also by B(G) if reference to G is needed. Consider a function µ: B → R satisfying: (1a) 0 ≤ µ(B) ≤ 1 for each B ∈ B. (1b) µ(∅) = 0 and µ(G) = 1. S∞ (1c) P Let B1 , B2 , B3 , . . . be pairwise disjoint Borel sets. Then µ( i=1 Bi ) = ∞ i=1 µ(Bi ) (σ- additivity).

364

Chapter 18. The Haar Measure

(1d) If B ∈ B and g ∈ G, then µ(gB) = µ(Bg) = µ(B) (translation invariance). (1e) For all B ∈ B and each ε > 0 there are an open set U and a closed set C satisfying C ⊆ B ⊆ U and µ(U r C) < ε (regularity). Condition (1) has the following immediate consequences: (2a) Let A ⊆ B be Borel sets. Then µ(B r A) = µ(B) − µ(A). In particular, µ(G r A) = 1 − µ(A). S∞ P∞ (2b) Let B1 , B2 , B3 , . . . be Borel sets. Then µ( i=1 Bi ) ≤ i=1 µ(Bi ). Sn−1 0 r (Write Bn = Bn i=1 Bi and apply (1c).) , B , . . . be Borel sets satisfying µ(Bi ) = 0, i = 1, 2, 3, . . . . (2c) Let B1 , B 2 3 S∞ Then µ( i=1 Bi ) = 0 (use 2b). (2d) Let B1 , B T2∞, B3 , . . . be Borel sets satisfying µ(Bi ) = 1, i = 1, 2, 3, . . . . Then µ( i=1 Bi ) = 1 (use (2a) and (2c)). (2e) SupposeSA1 ⊆ A2 ⊆ A3 ⊆ · · · is an increasing sequence of Borel sets. ∞ Then µ( i=1 Ai ) = limi→∞ µ(Ai ) (use (1c)). (2f) Suppose TA1 ⊇ A2 ⊇ A3 ⊇ · · · is a decreasing sequence of Borel sets. ∞ Then µ( i=1 Ai ) = limi→∞ µ(Ai ) (take complements and use (2a) and (2e)). 1 (2g) Suppose S n H is an open subgroup of G of index n. Then µ(H) = n (write G = · i=1 gi H and apply (1c) and (1d)).  Lemma 18.1.1: 1 . (a) If H is a closed subgroup of G, then µ(H) = (G:H) ¯ (b) Let N be an open normal subgroup of G, A a subset of G/N , and A = ¯ ¯ Then µ(A) = |A| {g ∈ G | gN ∈ A}. . (G:N )

(c) Finally, if U is a nonempty open subset of G, then µ(U ) > 0. Sn Proof of (a): First P suppose (G : H) = n is finite and write G = · i=1 gi H. n By (1), 1 = µ(G) = i=1 µ(gi H) = nµ(H). Hence, µ(H) = n1 , as claimed. Now assume H has infinite index. Then H is contained in the intersection of a decreasing sequence of open subgroups, H1 > H2 > . . . . Thus, 1 µ(H) ≤ limi→∞ (G:H = 0. i) Proof of (b): Use (a) and (1d). Proof of (c): By definition, U contains a coset gH, where H is an open 1 subgroup of G. By (a) and (1d), µ(U ) ≥ µ(gH) = µ(H) = (G:H) > 0.  Definition 18.1.2: Zero sets. Call a subset A of G a zero set (with respect to (B, µ)) if A ⊆ B ∈ B with µ(B) = 0.  Let Bˆ be the σ-algebra of G generated by B and the zero sets. Write Bˆµ , ˆ ˆ of Bˆ is the B(G), or Bˆµ (G) if a reference to µ or G is needed. Each set B union of a set B ∈ B and a zero set. Extend µ to a function µ ˆ: Bˆ → R by ˆ defining µ ˆ(B) to be µ(B). The extended measure µ ˆ has properties (1) and (2) with B replaced by Bˆ and also the following:

18.1 The Haar Measure of a Profinite Group

365

ˆµ (3a) Bˆ contains all zero sets with respect to (B, ˆ) (ˆ µ is complete). (3b) For each B ∈ Bˆ there are B1S, B2 ∈ B with B1 ⊆TB ⊆ B2 and ∞ ∞ µ(B2 r B1 ) = 0 (Take B1 = n=1 B1,n and B2 = n=1 B2,n with B1,n , B2,n ∈ B satisfying B1,n ⊆ B1,n+1 ⊆ B ⊆ B2,n+1 ⊆ B2,n and µ(B2,n r B1,n ) < n1 .) The following result shows that µ ˆ is determined by its values on openclosed sets. A compactness argument shows that each open-closed set is a union of finitely many left cosets of open normal subgroups of G. The latter union can be made disjoint. Hence, µ ˆ is unique. Proposition 18.1.3: Let B0 be the Boolean algebra of open-closed sets in G and B1 the σ-algebra generated by B0 . Then: (a) For every U ∈ Bˆµ there exist A, B ∈ B1 with A ⊆ U ⊆ B and µ(B r A) = 0. (b) Suppose µ and ν are functions from B to R satisfying (1). Then (Bˆµ , µ ˆ) = (Bˆν , νˆ). S Proof of (a): It suffices to prove (a) for U open. We write U = i∈I xi Mi with openSnormal subgroups Mi of G and xi ∈ G and let α be the supremum
of all µ i∈I 0 xi Mi where I 0 is a countable subset of I. For each n we
S choose a countable subset Jn of I such that α − µ i∈Jn xi Mi < n1 . Let S∞ S J = n=1 Jn . Then A = j∈J xj Mj ⊆ U , A ∈ B1 , andTµ(A) = α. Now we consider the closed normal subgroup N = j∈J Mj of G. Then G/N has a countable basis for its topology. Let π: G → G/N be the quotient map. The sets π(xi M ) = xi Mi N/N are open in G/N and their union is π(U ). By a lemma of Lindel¨ of [Hocking-Young, p. 66], I has a countable S subset K such that π(U ) = k∈K π(xk Mk ). In addition, π −1 (π(A)) = A and π(U r A) = π(U ) r π(A). Hence, U r A ⊆ π −1 (π(U r A)) =

[

(xk Mk N r A).

k∈K

The right hand side of this equality, which we denote by B0 , belongs to B1 . If we prove that µ(B0 ) = 0, then B = A ∪ B0 ∈ B1 and µ(B r A) = 0, as needed. It suffices to prove that µ(xk Mk N r A) = 0 for each k ∈ K. r Sr k Mk N A) > 0. Write N = S r Assume there exists k ∈ K with µ(x · ρ=1 (N ∩ Mk )nρ . Then xk Mk N r A = ρ=1 (xk Mk nρ r A), so there exists ρ with µ(xk Mk nρ r A) > 0. Note that Anρ = A, because nρ ∈ Mj for each j ∈ J. Hence, µ(xk Mk r A) = µ((xk Mk r A)nρ ) = µ(xk Mk nρ r A) > 0. S  It follows that µ x M ∪ x M = µ(A) + µ(xk Mk r A) > α. This j j k k j∈J contradiction to the definition of α concludes the proof of (a). Proof of (b): Denote the collection of all Borel sets on which µ and ν coincide by B2 . Lemma 18.1.1 implies S each open subgroup of G belongs to B2 . Hence, r µ and ν coincide on all sets · i=1 gi N with N open normal in G. In other

366

Chapter 18. The Haar Measure

words, µ and ν coincide on B0 . By (2e) and (2f), B2 is monotone. It follows that µ and ν coincide on B1 . Consider now a closed subset C of G. The proof of (a) gives a set D ∈ B1 with C ⊆ D and µ(C) = µ(D). By the preceding paragraph, ν(D) = µ(D). Thus, ν(C) ≤ ν(D) = µ(C). By symmetry, µ(C) ≤ ν(C). Hence, µ(C) = ν(C), so C ∈ B2 . It follows that B = B2 . In particular, µ(B) = 0 if and only if ν(B) = 0 for each B ∈ B. Hence, a subset A of G is a zero set with respect to (B, µ) if and only if A is a zero set with respect to ˆ = νˆ.  (B, ν). Therefore, Bˆµ = Bˆν and µ Call each set in Bˆ measurable. Example 18.1.4: Haar measure on finite groups. Let G be a finite group equipped with the discrete topology. Then each subset B of G is open and  µ(B) = |B| |G| is the unique Haar measure of G. Example 18.1.5: Infinite profinite groups have nonmeasurable subsets. If G is an infinite profinite group, it has a countable abstract subgroup H. Let R of G modulo H. Then G = S be a set of representatives for the left cosetsP · h∈H hR. Assume R is measurable. Then 1 = h∈H µ(hR). Since µ(hR) = µ(R) for each h ∈ H, this is impossible. Therefore, R is nonmeasurable. 

18.2 Existence of the Haar Measure Proposition 18.1.3 gives the uniqueness of the Haar measure on a profinite group. The definition of µ on the Boolean algebra B0 of open-closed sets in G is a straightforward application of the invariance property of µ. A theorem of Caratheodory [Halmos, p. 42, Thm. A] then extends µ to the σ-algebra Bˆ0 generated by B0 . If rank(G) ≤ ℵ0 , Bˆ0 contains every open set and therefore every Borel set of G. In the general case one has to work harder. The proof that we give here for arbitrary G avoids Caratheodory’s theorem. Proposition 18.2.1: Every profinite group G has a unique Haar measure. Proof: Proposition 18.1.3 asserts the uniqueness of µ. The proof of the existence of µ divides into five parts. It uses the notation of Proposition 18.1.3. Sm Part A: A measure on B0 . Represent B ∈ B0 as B = · i=1 gi N with N m an open normal subgroup. Define µ0 (B) to be (G:N ) . We have to show that this is independent of the choice of N . Suppose N 0 is an open S n normal subgroup of G contained in N . Put n = (N : N 0 ). Then N = · j=1 hj N 0 Sm Sn and B = · i=1 · j=1 gi hj N 0 . The computation of µ0 (B) with respect to N 0 mn m gives (G:N 0 ) = (G:N ) . Thus, µ0 (B) is well defined. By definition, µ0 (G) = 1. Finite additivity and translation invariance of µ0 on B0 are easy exercises.

18.2 Existence of the Haar Measure

367

Part B: Extension of µ0 to open sets.

For each open subset U of G let

µ1 (U ) = sup(µ0 (A) | A ∈ B0 and A ⊆ U ). The function µ1 extends µ0 and has these properties: (1a) U1 and U2 open =⇒ µ1 (U1 ∪ U2 ) ≤ µ1 (U1 ) + µ1 (U2 ). (1b) U1 and U2 open disjoint =⇒ µ1 (U1 ∪ U2 ) = µ1 (U1 ) + µ1 (U2 ). ). (1c) U1 ⊆ U2 open =⇒ µ1 (U1 ) ≤ µ1 (U2S P∞ ∞ (1d) Ui is open, i = 1, 2, 3, . . . =⇒ µ1 ( i=1 Ui ) ≤ i=1 µ1 (Ui ). (1e) U open and x ∈ G =⇒ µ1 (xU ) = µ1 (U x) = µ1 (U ). To prove (1a), denote U1 ∪ U2 by U . For ε > 0 let A ∈ B0 with A ⊆ U and µ1 (U ) < µ0 (A) + ε. The open set Ui is a union of sets in B0 , i = 1, 2. Since A is compact, A is contained in a union B1 ∪ · · · ∪ Bn such that for each i, Bi ∈ B0 and Bi ⊆ U1 or Bi ∈ B0 and Bi ⊆ U2 . Let C1 be the union of all Bi which are contained in U1 . Let C2 be the union of all Bi which are contained in U2 . Then C1 ∪ C2 = B1 ∪ · · · ∪ Bn and µ1 (U ) − ε < µ0 (A) ≤ µ0 (C1 ∪ C2 ) ≤ µ0 (C1 ) + µ0 (C2 ) ≤ µ1 (U1 ) + µ1 (U2 ). Since this inequality holds for each ε > 0, µ1 (U ) ≤ µ1 (U1 ) + µ1 (U2 ). To prove (1b), let ε > 0. Choose Ai ∈ B0 , Ai ⊆ Ui such that µ1 (Ui ) < µ0 (Ai ) + 2ε , i = 1, 2. Since µ0 is additive, µ1 (U1 ) + µ1 (U2 ) < µ0 (A1 ) + µ0 (A2 ) + ε = µ0 (A1 ∪ A2 ) + ε ≤ µ1 (U1 ∪ U2 ) + ε. Hence, µ1 (U1 ) + µ1 (U2 ) ≤ µ1 (U1 ∪ U2 ). Combining this with (1a) gives (1b). For (1c), let ε > 0 and choose A ∈ B0 such that A ⊆ U1 and µ1 (U1 ) < µ0 (A) + ε. Since A ⊆ U2 , µ1 (U 1 ) < µ1 (U2 ) + ε. Hence, µ1 (U1 ) ≤ µ1 (U2 ). S∞ To prove (1d), let U = i=1 Ui and let ε > 0. Choose A ∈ B0 with A ⊆S U and µ1 (U ) < µ0 (A) + ε. Since A is compact, there exists n with n A ⊆ i=1 Ui . Hence, by (1c) and (1a), µ1 (U ) ≤ µ1 (

n [

Ui ) + ε ≤

i=1

n X

µ1 (Ui ) + ε ≤

i=1

∞ X

µ1 (Ui ) + ε,

i=1

and (1d) follows. Finally, multiplication from the left (resp. right) with an element x ∈ G is a homeomorphism of G onto itself. Therefore, (1e) follows from the translation invariance of µ0 . Part C: An outer measure.

For each E ⊆ G let

µ2 (E) = inf(µ1 (U ) | U open and E ⊆ U ). The function µ2 extends µ1 and has these properties: (2a) E1 ⊆ E2 =⇒ µ2 (E1 ) ≤ µ2 (E2 ).

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Chapter 18. The Haar Measure

S∞ P∞ (2b) µ2 ( i=1 Ei ) ≤ i=1 µ2 (Ei ). (2c) µ2 (xE) = µ2 (Ex) = µ2 (E), x ∈ G. Both (2a) and (2c) are easy exercises. For (2b), let ε > 0. Choose Ui open such that Ei ⊆ Ui and µ1 (Ui ) ≤ µ2 (Ei ) + 2εi . By (1d) and (2a) µ2 (

∞ [

i=1

Ei ) ≤ µ1 (

∞ [

Ui ) ≤

i=1

∞ X

µ1 (Ui ) ≤

i=1

∞ X

µ2 (Ei ) + ε,

i=1

and (2b) follows. Part D: Measurable sets. Call a subset E of G measurable if for each A ⊆ G we have µ2 (A) = µ2 (A ∩ E) + µ2 (A r E). The following rules hold: (3a) E is measurable =⇒ G r E is measurable. (3b) E1 and E2 are measurable =⇒ E1 ∪ E2 is measurable. (3c) µ2 (E) = 0 =⇒ E is measurable. Sn (3d) P E1 , . . . , En are measurable and mutually disjoint =⇒ µ2 (A∩ i=1 Ei ) = n i=1 µ2 (A ∩ Ei ) for each A ⊆ G. S∞ and mutually disjoint =⇒ E = i=1 Ei (3e) E1 , E2 , E3 , . . . are measurable P∞ is measurable and µ2 (E) = i=1 µ2 (Ei ). (3f) Each B ∈ B0 is measurable. (3g) Each open set is measurable. (3h) E is measurable and x ∈ G =⇒ xE and Ex are measurable. Rule (3a) follows from the definition. To prove (3b), let A ⊆ G. By (2b) µ2 (A) ≤ µ2 (A ∩ (E1 ∪ E2 )) + µ2 (A r(E1 ∪ E2 )) = µ2 ((A ∩ E1 ) ∪ ((A r E1 ) ∩ E2 )) + µ2 ((A r E1 ) r E2 ) ≤ µ2 (A ∩ E1 ) + µ2 ((A r E1 ) ∩ E2 ) + µ2 ((A r E1 ) r E2 ) = µ2 (A ∩ E1 ) + µ2 (A r E1 ) = µ2 (A). Thus, these inequalities are in fact equalities. Consequently, E1 ∪ E2 is measurable. To prove (3c), let A ⊆ G. By (2a) and (2b), µ2 (A) ≤ µ2 (A ∩ E) + µ2 (A r E) ≤ µ2 (E)+µ2 (A) = µ2 (A). Hence, µ2 (A) = µ2 (A∩E)+µ2 (A r E). Thus, E is measurable. The proof of (3d) is done by induction. It is trivial for n = 1. Assume it holds for n − 1. Then n n n     [ [ [


µ2 A ∩ Ei = µ2 A ∩ Ei ∩ En + µ2 A ∩ Ei r E n i=1

i=1

i=1

= µ2 (A ∩ En ) + µ2 A ∩

n−1 [

Ei




i=1

= µ2 (A ∩ En ) +

n−1 X i=1

µ2 (A ∩ Ei ).

18.2 Existence of the Haar Measure

For (3e), (3b) implies that Fn = Hence, for each A ⊆ G, (3d) implies

369

Sn

i=1

Ei is measurable, n = 1, 2, 3, . . . .

µ2 (A) = µ2 (A ∩ Fn ) + µ2 (A r Fn ) ≥ µ2 (A ∩ Fn ) + µ2 (A r E) =

n X

µ2 (A ∩ Ei ) + µ2 (A r E).

i=1

Therefore, by (2b), µ2 (A) ≥

∞ X

µ2 (A ∩ Ei ) + µ2 (A r E) ≥ µ2 (A ∩ E) + µ2 (A r E) ≥ µ2 (A).

i=1

It follows that E is measurable. Sn Pn P∞ By (3d), µ2 (E) ≥ µ2 ( i=1 Ei ) = i=1 µ2 (Ei ). Hence, by (2b), µ2 (E) = i=1 µ2 (Ei ). To prove (3f), let A ⊆ G and U be an open set that contains A. Then U ∩ B and U r B are open disjoint sets. By (2a) and (1b) µ2 (A ∩ B) + µ2 (A r B) ≤ µ2 (U ∩ B) + µ2 (U r B) = µ1 (U ). Take the infimum on all U to conclude that µ2 (A) ≤ µ2 (A∩B)+µ2 (A r B) ≤ µ2 (A). Hence, B is measurable. Now we prove (3g): Let U be an open set. Consider A ⊆ G and ε > 0. Since µ2 (U ) = µ1 (U ) (Part C), there is a B ∈ B0 with B ⊆ U and µ2 (U ) − µ2 (B) < ε. Since B and U r B are disjoint open sets, (1b) implies µ2 (U ) = µ2 (B) + µ2 (U r B). Hence, µ2 (U r B) < ε. By (3f), µ2 (A) = µ2 (A ∩ B) + µ2 (A r B). Therefore, µ2 (A) ≤ µ2 (A ∩ U ) + µ2 (A r U ) ≤ µ2 (A ∩ (U r B)) + µ2 (A ∩ B) + µ2 (A r B) ≤ µ2 (U r B) + µ2 (A) < ε + µ2 (A) It follows that µ2 (A) = µ2 (A ∩ U ) + µ2 (A r U ). Thus, U is measurable. ˆ Let Bˆ be the collection of all measurable sets. Part E: The Baire field B. By (3a), (3b) and (3e), Bˆ is a σ-algebra. Since Bˆ contains each open set, it contains each Borel set. Denote the restriction of µ2 to Bˆ by µ. By (3e), (3h), and (2c), µ is a σ-additive invariant measure. For each E ∈ Bˆ and every ε > 0 there is an open set U with E ⊆ U and µ2 (U r E) < ε. Apply this to G r E to conclude the existence of a closed set C ⊆ E such that µ(E r C) < ε. T Thus, µ is regular. In particular, each E ∈ Bˆ is contained in a ∞ Borel set F = i=1 Ui with Ui open, i = 1, 2, 3, . . . such that µ(F r E) = 0. ˆ µ) is Combining this with (3c), we conclude that µ is complete on Bˆ and (B, the completion of the Borel field B with respect to the restriction of µ to B. Thus, µ is the desired Haar measure of G. 

370

Chapter 18. The Haar Measure

Proposition 18.2.2: Let π: G → H be an epimorphism of profinite groups and µG , µH the corresponding Haar measures. Then µH (B) = µG (π −1 (B)) for each measurable subset B of H. Proof: The map B 7→ π −1 (B) maps closed subgroups of H onto closed subgroups of G. In addition, it commutes with complements and unions. Hence, it maps B(H) into B(G). The function µG ◦ π −1 satisfies Condition (1) of Section 18.1 for B ∈ B(H). This is clear for (1a)-(1d) of Section 18.1. To prove (1e) of Section 18.1, consider B ∈ B(H) and ε > 0. Then G has a closed subset C with C ⊆ π −1 (B) and µG (π −1 (B) r C) < ε. Then π(C) is closed in H and π(C) ⊆ B. Thus, π −1 (π(C)) ⊆ π −1 (B) and µG (π −1 (B r π(C))) ≤ µG (π −1 (B) r C) < ε. Applying this result to H r B and taking complements, we get an open subset U of H with B ⊆ U and µG ◦ π −1 (U r B) < ε. By Proposition 18.1.3, the completion of (B(H), µG ◦π −1 ) coincides with ˆ (B(H), µH ). ˆ If It remains to prove that µH (B) = µG (π −1 (B)) for each B ∈ B(H). A is a µH -zero set, then A ⊆ A1 for some A1 ∈ B(H) with µH (A1 ) = 0. Then µG (π −1 (A)) ≤ µG (π −1 (A1 )) = µH (A1 ) = 0, so µG (π −1 (A)) = 0. An ˆ arbitrary B ∈ B(H) differs from a set in B(H) by a µH -zero set. Hence, −1  µH (B) = µG (π (B)) holds for B. Example 18.2.3: Finite quotients. Let π be an epimorphism of a profinite group G onto a finite group H. Then, by Example 18.1.4 and Proposition |B| for each subset B of H.  18.2.2, µG (π −1 (B)) = |H| Proposition 18.2.4: Let H be an open subgroup of a profinite group G. ˆ Then µH (B) = (G : H)µG (B) for each B ∈ B(H). ˆ is a Haar measure of H, so it Proof: The restriction of (G : H)µG to B(H)  coincides with µH .

18.3 Independence Now that we have a unique normalized Haar measure µ on G, we may regard G as a probability space. Recall thatT a family {A Qi | i ∈ I} of measurable subsets of G is µ-independent if µ( i∈J Ai ) = i∈J µ(Ai ) for each finite subset J of I. LemmaS18.3.1: The hold Pnfollowing P for measurable subsets A1 , . . . , An of G: n k−1 (a) µ i=1 Ai ) = k=1 (−1) 1≤i1 <···
µ

n [ i=1

Ai ) =

n X k=1

(−1)k−1

X 1≤i1 <···
µ(Ai1 ) · · · µ(Aik )

18.3 Independence

371

and G r A1 , . . . , G r An are µ-independent. Proof of (a): The principle is trivial for n = 1. For n = 2 it follows from the identity A1 ∪ A2 = (A1 r A1 ∩ A2 ) ∪· A2 . Suppose it holds for n − 1. Then

µ(

n [

n−1 [

Ai ) = µ(

i=1

Ai ∪ An )

i=1 n−1 [

= µ(

n−1 [

Ai ) + µ(An ) − µ(

i=1

=

n−1 X

X

(−1)k−1

µ(Ai1 ∩ · · · ∩ Aik ) + µ(An )

1≤i1 <···
k=1

−

n−1 X

X

(−1)k−1

n X k=1

X

(−1)k−1

µ(Ai1 ∩ · · · ∩ Aik ∩ An )

1≤i1 <···
k=1

=

Ai ∩ An )

i=1

µ(Ai1 ∩ · · · ∩ Aik )

1≤i1 <···
Proof of (b): Since µ(Ai1 ∩ · · · ∩ Aik ) = µ(Ai1 ) · · · µ(Aik ), (1) follows from (a). It remains to prove that G r A1 , . . . , G r An are independent. By symmetry, it suffices to prove the independence condition for G r A1 , . . . , G r Am , for all m ≤ n. Indeed,

µ(

m \

G r Ai ) = µ(G r

i=1

m [

Ai ) = 1 − µ(

i=1

=

m X k=0

=

m Y i=1

(−1)k

n [

Ai )

i=1

X

µ(Ai1 ) · · · µ(Aik )

1≤i1 <···
(1 − µ(Ai )) =

m Y

µ(G r Ai ).

i=1

Consequently, G r A1 , . . . , G r Am are µ-independent.



Q∞ a3 , . . . be a sequence of nonzero real numbers. Define i=1 ai Let a1 , a2 ,Q n to be limn→∞ i=1 ai , if the limit exists. Q∞ Lemma 18.3.2: P∞Suppose 0 < ai < 1, i = 1, 2, 3, . . . . Then i=1 (1 − ai ) = 0 if and only if i=1 ai = ∞. Proof: Consider 0 < x < P 1. Then log(1 − x) = ∞ x . other hand, log(1 − x) ≥ − i=1 xi = − 1−x

P∞

i=1

i

− xi < −x. On the

372

Chapter 18. The Haar Measure

Now suppose that log

n Y

P∞

i=1

ai = ∞. Then

(1 − ai ) =

i=1

n X

log(1 − ai ) < −

i=1

n X

ai .

i=1

Qn Qn Hence, log limn→∞ i=1 (1 − ai ) = limn→∞ log i=1 (1 − ai ) = −∞. ThereQ∞ fore, i=1 (1 − ai ) = 0. P ∞ Conversely, suppose i=1 ai < ∞. Then limi→∞ ai = 0. Hence, there ai ≥ − aci , is a c > 0 with 1 − ai > c, i = 1, 2, . . . . Thus, log(1 − ai ) ≥ − 1−a i P∞ Q∞ 1 iQ= 1, 2 . . . . Therefore, log i=1 (1 − ai ) ≥ − c i=1 ai > −∞. Consequently, ∞  i=1 (1 − ai ) > 0. Example 18.3.3: Zeta function. The most prominent example of infinite products is the Euler product for the Riemann zeta function: ζ(s) =

∞ X Y 1 −1 1 1 − = , ns ps p n=1

s>1

P with p ranging over all prime numbers. Here p p1s < ∞ for s > 1. Hence, P1 Q by Lemma 18.3.2, p (1 − p1s ) > 0. For s = 1 we have p = ∞ [LeVeque, Q 1  Thm. 6-13] and (1 − p ) = 0. Lemma 18.3.4: Let A1 , A2 , A3 , . . . bePa µ-independent sequence S∞of measur∞ able subsets of a profinite group G. If i=1 µ(Ai ) = ∞, then µ( i=1 Ai ) = 1. Q∞ By Lemma 18.3.1, Proof: By Lemma 18.3.2, i=1 (1 − µ(Ai )) = 0. G r A1 , G r A2 , . . . are µ-independent. Hence, µ(G r

∞ [

Ai ) = µ(

i=1

∞ \

i=1

= lim

n→∞

Thus, µ(

S∞

i=1

Ai ) = 1.

(G r Ai )) = lim µ n Y i=1

n→∞

(1 − µ(Ai )) =

n \

G r Ai




i=1 ∞ Y

(1 − µ(Ai )) = 0.

i=1



Of fundamental importance is the following strengthening of Lemma 18.3.4: Lemma 18.3.5 (Borel-Cantelli): Let A1 , A2 , A3 , . . . be a sequence of measurable subsets of a profinite group G. Put A= A0 =

∞ ∞ [ \ n=1 i=n ∞ ∞ \ [ n=1 i=n

Ai = {x ∈ G | x belongs to infinitely many Ai ’s} Ai = {x ∈ G | x belongs to all but finitely many Ai ’s}.

18.3 Independence

373

ThenPthe following hold: ∞ (a) i=1 µ(Ai ) < ∞ =⇒ µ(A) = 0. P∞ (b) Suppose A1 , A2 , A3 , . . . are µ-independent and i=1 µ(Ai ) = ∞. Then µ(A) = 1. Q ∞ (c) i=1 µ(Ai ) > 0 =⇒ µ(A0 ) = 1. Q∞ (d) Suppose A1 , A2 , A3 , . . . are µ-independent, µ(Ai ) > 0, and i=1 µ(Ai ) = 0. Then µ(A0 ) = 0. S∞ P∞ Proof: For each positive n, µ(A) ≤ µ( i=n Ai ) ≤ If i=n µ(Ai ). P ∞ i=1 µ(Ai ) < ∞, the right hand side approaches 0 as n tends to infinity. Thus, µ(A) = 0 and (a) follows. By Lemma 18.3.4, the hypotheses of (b) S∞ = 1 for every positive n.QTherefore, µ(A) = 1.
implies µ( i=n Ai )Q ∞ ∞ r Now suppose P∞ i=1 µ(Ai ) > 0. Then i=1 1 − µ(G Ai ) > 0. By r Ai ) < ∞. Hence, by (a), almost no x ∈ G Lemma 18.3.2, i=1 µ(G belongs to infinitely many (G r Ai )’s. Therefore, almost all x ∈ G belong to all but finitely many Ai ’s. This means, µ(A0 ) = 1. Finally, suppose A1 , A2 , A3T , . . . are µ-independent, µ(Ai ) > 0 for each i, Q∞ Q∞ ∞ and i=1 µ(Ai ) = 0. Then µ( i=n Ai ) = i=n µ(Ai ) = 0 for each n. So,  µ(A0 ) = 0. Remark 18.3.6: (a) Chebyshev’s inequality [R´enyi, p. 391] sharpens (b) of Lemma 18.3.5 by allowing one to assume only that the sets A1 , A2 , . . . are pairwise µindependent. (b) The most frequently used cases of 18.3.5 are when PLemma ∞ (b1) µ(Ai ) is a positive constant, where i=1 µ(Ai ) obviouslyP diverges, or 1 (b2) µ(Ap ) = p1e where p ranges over all prime numbers; then pe diverges for e = 1 (Example 18.3.3) and converges for e ≥ 2.  Sequences A1 , A2 , . . . with µ(Ai ) = 1 (resp. µ(Ai ) = 0), i = 1, 2, . . . are clearly µ-independent. The next lemma ties independence to the group structure on G: Lemma 18.3.7: Open subgroups H1 , . . . , Hn of a profinite group G are µindependent if and only if (2)

(G :

n \

Hi ) =

i=1

n Y

(G : Hi ).

i=1

Suppose H1 , . . . , Hn are µ-independent open subgroups of G. For each i let A¯i be a set of left cosets of Hi in G. Put Ai = {g ∈ G | gHi ∈ A¯i }. Then A1 , . . . , An are µ-independent and (3)

µ(

n \

i=1

Ai ) =

n Y

|A¯i | . (G : Hi ) i=1

374

Chapter 18. The Haar Measure

Proof: Suppose (2) holds and let I be a subsetT of {1, . . . , n}. WithTno loss m n assume I = {1, . . . , m} with m ≤ n. Let K = i=1 Hi and H = i=1 Hi . ) embeds the coset space G/K into The canonical map gK 7→ (gH1 , . . . , gHmQ m G/H1 × · · · × G/Hm . Hence, (G : K) ≤ i=1 (G : Hi ). Similarly, (K : H) ≤

n Y i=m+1

(K : K ∩ Hi ) ≤

n Y

(G : Hi ).

i=m+1

Combining these with the Qm(G : H) = (G : K)(K : H) and Tm(2), we conclude Qm equality (G : K) = i=1 (G : Hi ). Thus, (2) gives µ( i=1 Hi ) = i=1 µ(Hi ), as desired. Necessity of (2) for µ-independence is clear. Now suppose H1 , . . . , Hn are µ-independent. Put A = A1 ∩ · · · ∩ An ¯ then gH = aH for some and A¯ = {aH | a ∈ A}. If g ∈ G and gH ∈ A, ¯ a ∈ A. Hence, gHi ∈ Ai , so g ∈ Ai , i = 1, . . . , n and g ∈ A. It follows ¯ List the elements of A¯ as a1 H, . . . , am H. Then that S A = {gH | gH ∈ A}. m ¯ A = · i=1 ai H and µ(A) = |A|µ(H). Next define a map α: G/H → G/H1 × · · · × G/Hn by α(gH) = (gH1 , . . . , gHn ). By (2), α is bijective. In particular, α maps A¯ bijectively onto A¯1 × · · · × ¯ = |A¯1 | · · · |A¯n |. In addition, µ(H) = µ(H1 ) · · · µ(Hn ). Hence, A¯n , so |A| Qn Qn ¯ µ(A) = |A|µ(H) = i=1 |A¯i |µ(Hi ) = i=1 µ(Ai ). Thus, A1 , . . . , An are µ-independent and (3) holds.  Example 18.3.8: Relatively prime indices. Let H1 , . . . , Hn be open subgroups of a profinite group G. Suppose (G : H1 ), . . . , (G : Hn ) are relatively primes in pairs. Then (2) holds. Hence, by Lemma 18.3.7, H1 , . . . , Hn are µ-independent.  Part (b) of the following lemma generalizes Lemma 16.8.3(a). Q Lemma 18.3.9: Let G = i∈I Si be a direct product of finite non-Abelian simple groups Si and N a closed normal subgroup of G. For each i ∈ I let πi : G → Si be the projection on the ith factor. Then the following hold: = Sj if and only if Sj ≤ N . (a) πj (N )Q (b) N = i∈I0Q Si , where I0 = {i ∈ I | Si ≤ N }. Moreover, G = N × N 0 0 with N = i∈I r I0 Si . (c) Let α be an automorphism of G with α(N ) = N . Then α(N 0 ) = N 0 . (d) Suppose G is a normal subgroup of a profinite group F and N / F . Then N0 / F. Proof of (a): First suppose Sj ≤ N . Since πj is the identity map on Sj , we have πj (N ) = Sj . Now suppose Sj 6≤ N . Since Sj is simple, Sj ∩ N = 1. Hence, [n, s] = 1 for all n ∈ N and s ∈ Sj . Therefore, [πj (n), s] = 1. Since Sj is non-Abelian, its center is trivial. Consequently, πj (n) = 1.

18.3 Independence

375

Q Proof of (b): By definition, i∈I0 Si ≤ N . QConversely, let n ∈ N . If j ∈ I r I0 , then πj (n) = 1, by (a). Hence, n ∈ i∈I0 Si . Q Proof of (c): Si ≤ N if and only if α(Si ) ≤ N , so α(N 0 ) = Si 6≤N α(Si ) = Q 0 Si 6≤N Si = N . Proof of (d): Apply (c) to conjugation of G by elements of F .  Qr Lemma 18.3.10: Let G = i=1 Si be a direct product of r finite simple nonAbelian groups Si . Then G has exactly r normal subgroups N with G/N simple. Proof: Let N be a normal subgroup of G such that G/N is simple. By Q Lemma Q 18.3.9, N = i∈I0 Si , where I0 is a subset of {1, 2, . . . , r}. Then Since G/N is simple, I r I0 = {j} for some j between G/N = i∈I r I0 Si . Q 1 and r. Thus, N = i6=j Si . Consequently, there are exactly r possibilities for N .  Lemma 18.3.11: Let G be a profinite group and λ an ordinal number. For each α < λ let Nα be an open normal subgroup of G. SupposeTG/Nα is a simple non-Abelian group and Nα 6= Nα0 if α 6= α0 . Then G/ α<λ Nα ∼ = Q are µ-independent. Moreover, if N is an open α<λ G/Nα and the Nα ’sT subgroup of G containing α<λ Nα and G/N is simple, then N = Nα for some α < λ. T Proof: For each γ ≤ λ let Mγ = α<γ Nα . Suppose by transfinite induction Q for each β < γ there exists an isomorphism ϕβ : G/Mβ → α<β G/Nα such that the following diagram commutes for all β < β 0 < γ: (4)

G/Mβ 0  G/Mβ

ϕβ 0

ϕβ

/

/

Q

G/Mα

α<β 0

Q

 α<β

G/Mα .

Here the left vertical map is the quotient map and the right vertical map is the projection on the first β coordinates. If γ Q is a limit ordinal, T Qthen Mγ = ∼ ∼ ∼ M . Hence, G/M G/M lim G/M lim = = = β γ β α β<γ α<γ G/Mα . ←− ←− α<β 0 Moreover, Diagram (4) is commutative with γ replacing β . Now suppose γ = β + 1. Let N be an open normal subgroup of G containing Mβ such that G/N Tr is simple. By Lemma 1.2.2(a), there are . . . , αr ≤ βQsuch that i=1 Nαi ≤ N . By the induction hypothesis, α1 , T r r 18.3.10, Nα1 , . . . , Nαr are all G/ i=1 Nαi ∼ = i=1 G/Nαi . Hence, by Lemma Tr open normal subgroups of G containing i=1 Nαi with a simple coquotient. Thus, N = Nαi for some i between 1 and r. It follows that Mβ 6≤ Nβ . ∼ Therefore, Q Mβ Nβ = G (because G/Nβ is simple) and G/Mγ = G/Mβ × ∼ G/Nβ = α<γ G/Mα , as claimed.  The µ-independence of the Nα ’s follows now from Lemma 18.3.7.

376

Chapter 18. The Haar Measure

18.4 Cartesian Product of Haar Measures The direct product G×H of profinite groups G and H is a profinite group. By Propositions 18.2.1, G × H has a unique Haar measure µG×H . Consider also the product measure µG × µH of G × H [Halmos, Chapter VII]. It is first defined on sets of the form A × B with A a µG -measurable set and B a µH measurable set (i.e. measurable rectangles) by the rule (µG ×µH )(A×B) = µG (A)µH (B). Then µG × µH extends to a σ-additive measure on the σalgebra generated by these rectangles [Halmos, p. 144, Thm. B]. Finally, one completes µG × µH by adding all zero sets. We prove that µG×H coincides with µG × µH . Given families A and B of subsets of G and H, we denote the set of all pairs (A, B) with A ∈ A and B ∈ B, as usual, by A × B. Following [Neveu, p. 71], we denote the σ-algebra of G × H generated by all of the sets A × B with (A, B) ∈ A × B by A ⊗ B. The latter notation is partially justified by the following result. Lemma 18.4.1: Let G and H be sets, A a collection of subsets of G, and B a collection of subsets of H. Let A0 be the σ-algebra of G generated by A, B 0 the σ-algebra of H generated of B, and C a σ-algebra of G × H. Suppose A × H and G × B are in C for all (A, B) ∈ A × B. Then A0 ⊗ B 0 ⊆ C. Proof: Let A1 = {A ⊆ G | A × H ∈ C}. Then A1 contains A and is closed under countable unions and taking complements. Hence, A0 ⊆ A1 . In other words, A × H ∈ C for each A ∈ A0 . Similarly, G × B ∈ C for each B ∈ B 0 . It follows that A × B = (A × H) ∩ (G × B) ∈ C for all (A, B) ∈ A0 × B 0 .  Consequently, A0 ⊗ B 0 ⊆ C. Proposition 18.4.2: The Haar measure on G × H coincides with µG × µH . Proof: Let B0 (G) be the Boolean algebra of all open-closed subsets of G, ˆ B1 (G) the σ-algebra generated by B0 (G), B(G) the Borel field of G, and B(G) the family of all measurable subsets of G. Use the corresponding notation for H and for G × H. In addition let R0 be the Boolean algebra of G × H generated by all of the sets A × B with (A, B) ∈ B0 (G) × B0 (H), R1 = ˆ be the completion of R with respect B0 (G)⊗B0 (H), R = B(G)⊗B(H), and R ˆ = B(G ˆ × H) and (µG × µH )(C) = to µG × µH . In four parts we prove that R ˆ µG×H (C) for each C ∈ R. Part A: R0 = B0 (G × H), R1 = B1 (G × H), and B1 (G) ⊗ B1 (H) ⊆ R1 . If G0 is an open subgroup of G and H0 is an open subgroup of H, then G0 × H0 is an open subgroup of G × H. Therefore, R0 ⊆ B0 (G × H). Conversely, let M be an open subgroup of G × H. Then M ∩ G is an open subgroup of G, M ∩ H is an open subgroup of H, and L = (M ∩ G) × (M ∩ H) ≤ M . Every union of cosets of M is a union of cosets of L. Thus, B0 (G × H) ⊆ R0 . The combination of the preceding two paragraphs proves that R0 = B0 (G × H). Closing under countable unions and taking complements, we find that R1 = B1 (G × H).

18.4 Cartesian Product of Haar Measures

377

Finally, by Lemma 18.4.1, B1 (G) ⊗ B1 (H) ⊆ R1 . Part B: µG × µH and µG×H coincide on R1 . Let A ∈ B0 (G) and B ∈ B0 (H). Then, there exist open S s M and N of G and H, S r normal subgroups respectively, such that A = · i=1 gi M and B = · j=1 hj N . Hence, [ rs µG×H (A × B) = µG×H ( · (gi , hj )M × N ) = (G : M )(H : N ) i,j r s [ [ = µG ( · gi M ) · µH ( · hj N ) = (µG × µH )(A × B). i=1

j=1

It follows that A × B belongs to the set ˆ ∩ B(G ˆ × H) | (µG × µH )(C) = µG×H (C)}. S = {C ∈ R Observe that S is closed under countable unions and taking complements. Therefore, R1 ⊆ S. In addition, if E0 ⊆ E1 , E1 ∈ S, and (µG × µH )(E1 ) = 0, then E0 ∈ S. Part C: S contains R. Let A be a closed subset of G and B a closed subsets of H. By Proposition 18.1.3, there exists an A1 ∈ B1 (G) with A ⊆ A1 and µG (A) = µG (A1 ). Similarly, there exists a B1 ∈ B1 (H) with B ⊆ B1 and µH (B) = µH (B1 ). By Part A, A1 ×B1 ∈ R1 . By Part B, A1 ×B1 ∈ S. Hence, µG×H (A × B) ≤ µG×H (A1 × B1 ) = (µG × µH )(A1 × B1 ) = (µG × µH )(A × B). Conversely, since A × B is closed in G × H, Proposition 18.1.3 gives C1 ∈ B1 (G × H) with A × B ⊆ C1 and µG×H (A × B) = µG×H (C1 ). Hence, by Parts A and B, (µG × µH )(A × B) ≤ (µG × µH )(C1 ) = µG×H (C1 ) = µG×H (A × B). It follows that (µG × µH )(A × B) = µG×H (A × B). In other words, A × B ∈ S. By Lemma 18.4.1, R = B(G) ⊗ B(H) ⊆ S. ˆ can be written as C = Part D: Conclusion of the proof. Each C ∈ R C0 ∪ C 0 with C0 ⊆ C1 , C1 ∈ R, (µG × µH )(C1 ) = 0, and C 0 ∈ R. By Part ˆ ⊆ S. C, both C1 and C 0 are in S. It follows that C ∈ S. Therefore, R ˆ By Proposition 18.1.3, each D ∈ B(G×H) can be written as D = D0 ∪D0 with D0 ⊆ D1 , D1 ∈ B1 (G × H), µG×H (D1 ) = 0, and D0 ∈ B1 (G × H). By Part A, B1 (G × H) = R1 , so, by Part B, B1 (G × H) ⊆ S. Therefore, D ∈ S. ˆ × H) ⊆ S. Thus, B(G ˆ = S = B(G ˆ × H), as claimed. It follows that R  We apply Proposition 18.4.2 to the product Ge of e copies of G. For simplicity, we also denote the Haar measure on Ge by µ. It has this property: for A1 , . . . , Ae measurable subsets of G, (1)

µ(A1 × · · · × Ae ) = µ(A1 ) · · · µ(Ae ),

where µ on the right (resp. left) denotes Haar measure on G (resp. Ge ). As an immediate consequence of (1) we find: sequence Q of µ-independent subsets of G, (2) Suppose A1i , A2i , AQ 3i , . . . is aQ e e e i = 1, . . . , e. Then i=1 A1i , i=1 A2i , i=1 A3i , . . . is a µ-independent sequence of subsets of Ge .

378

Chapter 18. The Haar Measure

18.5 The Haar Measure of the Absolute Galois Group We consider the Haar measure µ on the absolute Galois group, Gal(K), of a fixed field K. If L is a finite separable extension of K, then (Gal(K) : Gal(L)) = [L : K]. This enables us to rephrase the first part of Lemma 18.3.7 as follows: Lemma 18.5.1: Let L1 , . . . , Ln be finite separable extensions of K. Then the following conditions are equivalent: (a) Gal(L1 ), . . . , Gal(L Qnn) are µ-independent. (b) [L1 · · · Ln : K] = i=1 [Li : K]. (c) L1 , . . . , Ln are linearly disjoint over K. Now use Lemma 18.5.1 to rephrase Lemma 18.3.5 and the second part of Lemma 18.3.7 in terms of Galois theory: Lemma 18.5.2: Let L1 , L2 , L3 , . . . be a linearly disjoint sequence of finite separable extensions of K. For each i ≥ 1 let A¯i be a set of left cosets of Gal(Li )e in Gal(K)e and Ai = {σ ∈ Gal(K)e | σGal(Li )e ∈ A¯i }. Then P∞ |A¯i | the sequence A1 , A2 , A3 , . . . is µ-independent. If i=1 [Li :K]e = ∞, then S∞ µ( i=1 Ai ) = 1. Proof: By Lemma 18.5.1, Gal(L1 ), Gal(L2 ), Gal(L3 ), . . . are µ-independent. Hence, by (2) of Section 18.4, so are Gal(L1 )e , Gal(L2 )e , Gal(L3 )e , . . . . By Lemma 18.3.7, applied to G = Gal(K)e , A1 , A2 , A3 , . . . are µ-independent. P∞ |A¯i | ¯i | |A Now suppose i=1 [Li :K]e = ∞. Then, in view of µ(Ai ) = [Li :K]e , S∞  Lemma 18.3.4 implies µ( i=1 Ai ) = 1. Lemma 18.3.5 has a similar interpretation: Lemma 18.5.3: Let L1 , L2 , L3 , . . . be finite separable extensions of K. For each i ≥ 1 let A¯i be a set of left cosets of Gal(Li )e in Gal(K)e and Ai = {σ ∈ Gal(K)e | σGal(Li )e ∈ A¯i }. Let A be the set of all σ ∈ Gal(K)e which belong to infinitely many Ai ’s. P∞ ¯i | (a) If i=1 [L|iA:K] e < ∞, then µ(A) = 0. P∞ ¯i | (b) Suppose L1 , L2 , L3 , . . . are linearly disjoint over K and i=1 [L|iA:K] e = ∞, then µ(A) = 1. Remark 18.5.4: If Li is a Galois extension of K, replace Gal(K)e /Gal(Li )e in the foregoing lemmas by Gal(Li /K)e and Ai by {σ ∈ Gal(K)e | resLi σ ∈  A¯i }. If K is a Hilbertian field, we can construct linearly disjoint sequences L1 , L2 , . . . of finite Galois extensions of K with Gal(Li /K) ∼ = Sn , i = 1, 2, . . . (Corollary 16.2.7). We interpret this in terms of µ-independence, abbreviating “for all but a set of measure 0” by “almost all”:

18.5 The Haar Measure of the Absolute Galois Group

379

Lemma 18.5.5: Suppose K is Hilbertian. Let π1 , . . . , πe be e elements of the symmetric group Sn . Then for almost all σ ∈ Gal(K)e there exists a continuous homomorphism ρ: Gal(K) → Sn with ρ(σj ) = πj , j = 1, . . . , e. Proof: Corollary 16.2.7 gives a linearly disjoint sequence, L1 , L2 , L3 , . . . of Galois extensions of K with Gal(Li /K) ∼ = Sn , i = 1, 2, 3, . . . . For each i choose an isomorphism ρi : Gal(Li /K) → Sn and σi1 , . . . , σie ∈ Gal(Li /K) with ρi (σij ) = πj , j = 1, . . . , e. Put σi = (σi1 , . . . , σie ), A¯i = {σi }, and Ai = {σ ∈ Gal(K)e | resLi (σ) = σi }. For each i and each σ ∈ Ai , ρi ◦ resLi is a homomorphism from Gal(K) into Sn that maps σij onto πj , j = 1, . . . , e. The assumption of Lemma 18.5.2 holds here trivially: P∞ S∞ P∞ |A¯i | 1  i=1 [Li :K]e = i=1 (n!)e = ∞. Consequently, µ( i=1 Ai ) = 1. Theorem 18.5.6 (The Free Generators Theorem [Jarden3, Thm. 5.1]): Let K be a Hilbertian field and e a positive integer. Then hσ1 , . . . , σe i ∼ = Fˆe for e almost all (σ1 , . . . , σe ) ∈ Gal(K) . Proof: Let G be a finite group generated by e elements. Embed G in Sn for some positive integer n. Choose π1 , . . . , πe ∈ Sn which generate the image of G in Sn . For almost all σ ∈ Gal(K)e Lemma 18.5.5 gives a homomorphism ρ: Gal(K) → Sn with ρ(σj ) = πj , j = 1, . . . , e. Thus, S(G) = {σ ∈ Gal(K)e | G is a homomorphic image of hσi} has measure 1. Since there are only countably many finite groups, the intersection of all S(G)0 s has measure 1. In other words, for almost all σ ∈ Gal(K)e , each group with e generators is a quotient of hσi. By Lemma 17.7.1, hσi ∼ = Fˆe .  The group generated by σ1 = · · · = σe = 1, is certainly not isomorphic to Fˆe . The same holds when σ1 , . . . , σe equal to the same involution of Gal(K). Thus, we cannot remove the phrase “almost all” from Theorem 18.5.6. We continue now with more results that indicate the behavior of e-tuples σ excluding a subset of Gal(K)e of measure zero.  Definition 18.5.7: A field M is called e-free if Gal(M ) ∼ = Fˆe . For a field K and for a σ ∈ Gal(K)e denote the fixed field in Ks of the entries of σ by Ks (σ). From Theorem 18.5.6, if K is Hilbertian, then Ks (σ) is e-free for almost all σ ∈ Gal(K)e . Let K be a finite field. Although K is not Hilbertian, Theorem 18.5.6 holds for e = 1. For e ≥ 2 we obtain an entirely different result: Lemma 18.5.8: ˆ we have hzi ∼ ˆ and (Z ˆ : hzi) = ∞. (a) For almost all z ∈ Z =Z ˆ and ˆ e , we have hzi ∼ (b) Suppose e ≥ 2. Then, for almost all z ∈ Z = Z ˆ (Z : hzi) < ∞. T∞ ˆ Q Then Z0l ∼ Proof of (a): For each prime number l let Z0l = i=1 li Z. = Zp ˆ 0 ∼ where p ranges over all prime numbers excluding l, and Z/Z l = Zl . Hence, T 0 0 ˆ r µ(Zl ) = 0. Therefore, µ( l (Z Zl )) = 1.

380

Chapter 18. The Haar Measure

T ˆ r Z0l ). Then for each prime l, Z0l is a proper closed Suppose z ∈ l (Z 0 subgroup of hz, Zl i. By Lemma 1.4.2(e), hzi/hzi ∩ Z0l ∼ = hz, Z0l i/Z0l ∼ = Zl . i Hence, Z/l Z is a homomorphic image of hzi for each positive integer i. By the Chinese remainder theorem the same holds for Z/nZ for each positive ˆ integer n. We conclude from Lemma 17.7.1 that hzi ∼ =Z ˆ : hzi) = ∞ for almost all z ∈ Z, ˆ recall that the sum To prove (Z P1 P ˆ l µ(lZ) = l over all primes l diverges [LeVeque, p. 100]. By Example ˆ are µ-independent. By Borel-Cantelli (Lemma 18.3.5(b) 18.3.8, the groups lZ ˆ with Ai = li Z where li is the ith prime) the set of z that belongs to infinitely ˆ has measure 1. Since the intersection of infinitely many of the groups lZ ˆ ˆ : hzi) = ∞ for each z in that set. many groups lZ has infinite index, (Z P∞ ˆ e Proof of (b): Under the assumption e ≥ 2 we have, n=1 µ((nZ) ) = P∞ −e e ˆ < ∞. By Lemma 18.3.5(a), almost all e-tuples z ∈ Z belong to n=1 n e ˆ : hzi) < ∞. By Lemma ˆ (nZ) for only finitely many n. For each such z, (Z ˆ is isomorphic to Z. ˆ Thus, hzi ∼ ˆ for almost 1.4.4, every open subgroup of Z =Z e ˆ  all z ∈ Z . ˆ We reformulate Lemma 18.5.8 for For K a finite field, Gal(K) ∼ = Z. Gal(K): Corollary 18.5.9 ([Jarden4, p. 122]): Let K be a finite field. For almost ˜ all σ ∈ Gal(K) the field K(σ) is 1-free and infinite. If e ≥ 2, then for almost ˜ is finite (and therefore 1-free). all σ ∈ Gal(K)e , the field K(σ) ˆn Remark 18.5.10: Chapter 26 generalizes Lemma 18.5.8 to the groups Z and Fˆn . 

18.6 The PAC Nullstellensatz In this section we use measure theory to produce an abundance of algebraic PAC fields. They arise as the fixed fields, Ks (σ), of σ ∈ Gal(K)e for K a countable Hilbertian field: Theorem 18.6.1 (PAC Nullstellensatz [Jarden2, p. 76]): Let K be a countable Hilbertian field and e a positive integer. Then Ks (σ) is a PAC field for almost all σ ∈ Gal(K)e . Proof: The proof has two parts. We use countability of K only in the second. Part A: Zeros of one polynomial. Let f ∈ K[T, X] be a nonconstant absolutely irreducible polynomial. Denote the set of all σ ∈ Gal(K)e such that f has a zero in Ks (σ) by S(f ). Our goal is to prove that µ(S(f )) = 1. Without loss assume f is separable in X and let d = degX (f ). We use induction to construct a linearly disjoint sequence L1 , L2 , L3 , . . . of separable extensions of K of degree d such that f has a zero in each of the Li ’s.

18.6 The PAC Nullstellensatz

381

Assume L1 , . . . , Ln are already defined. Then L = L1 · · · Ln is a finite separable extension of K and f is irreducible over L. By Corollary 12.2.3, there exists a ∈ K such that f (a, X) is an irreducible separable polynomial of degree d over L. For b a zero of f (a, X), the field Ln+1 = K(b) is a separable extension of K of degree d. It contains a zero of f and it is linearly disjoint from L over K. This completes our induction. By construction, S∞S(f ) contains S ∞ e e Gal(L ) . Lemma 18.5.2, with A = Gal(L ) , gives µ( i i i i=1 i=1 Gal(Li )) = 1. Hence, µ(S(f )) = 1. Part B: Countability. There are countably many nonconstant absolutely irreducible polynomials f ∈ K[T, X]. Thus, the intersection S of all the sets S(f ) has measure 1. By Theorem 11.2.3, the fixed field Ks (σ) of each etuple σ ∈ S is a PAC field. Therefore, Ks (σ) is a PAC field for almost all  σ ∈ Gal(K)e . Remark 18.6.2: We cannot drop the phrase “almost all” from Theorem ˜ induced by complex 18.6.1. For example, if σ is the automorphism of Q ˜ conjugation, Q(σ) is a real closed field and not PAC (Theorem 11.5.1). The ˜ given by the prime next example [Ax2, p. 269] arises from the valuation of Q p. Let Qp,ur be the maximal unramified extension of Qp . It is well known ˆ [Cassels-Fr¨ ohlich, p. 28]. Choose a generator σ ¯ of that Gal(Qp,ur /Qp ) ∼ =Z ˜ p . Then the field Gal(Qp,ur /Qp ) and extend σ ¯ to an automorphism σ of Q ˜ p (σ) is a totally ramified extension of Qp . It therefore has Fp as its E =Q ˆ (Lemma 17.4.11). It is also well known residue field. Also, Gal(E) = hσi ∼ =Z ˜ p (a corollary of Krasner’s Lemma [Ribenboim, p. 190]). ˜ · Qp = Q that Q ˆ Hence, Q(σ ˜ 0 ) is a Thus, with σ0 = resQ˜ (σ), we have that hσ0 i ∼ = hσi ∼ = Z. 1-free field that admits a valuation with a finite residue field. By Corollary ˜ 0 ) is not PAC. 11.5.5 (or Exercise 7 of Chapter 11), Q(σ ˜ 0 ) and L 6= Q, ˜ then More generally, if L is an algebraic extension of Q(σ L is Henselian with respect to a rank-1 valuation. By Theorem 11.5.5, L is ˜ τ1 , . . . , σ τe ) is an e-free field not PAC. By [Jarden14, Thm. 21.3], M = Q(σ 0 0 e for almost all (τ1 , . . . , τe ) ∈ Gal(Q) . Nevertheless M is not a PAC field,  since the residue field of M with respect to the p-adic valuation is Fp . ˜ Problem 18.6.3: Let σ ∈ Gal(Q) for which Q(σ) is neither PAC nor for˜ mally real. Does Q(σ) admit a valuation with a non-algebraically closed completion? Finite fields are not Hilbertian. Still, Theorem 18.6.1 holds for them with e = 1: ˜ Proposition 18.6.4: Let K be a finite field. Then K(σ) is a PAC field for almost all σ ∈ Gal(K). ˜ Proof: For almost all σ ∈ Gal(K) the field K(σ) is an infinite extension ˜ of K (Corollary 18.5.9). Hence, by Corollary 11.2.4, K(σ) is a PAC field. 

382

Chapter 18. The Haar Measure

Remark 18.6.5: K finite and e ≥ 2. The analog of Proposition 18.6.4 for e ≥ 2 is false. Indeed, by Corollary 18.5.9, for almost all σ ∈ Gal(K)e , the ˜ ˜ field K(σ) is finite. Hence, by Proposition 11.1.1, K(σ) is not PAC.  Remark 18.6.6: Let K be a countable infinite field such that Ks (σ) is a PAC e-free field for all positive integers e and almost all σ ∈ Gal(K)e . One may ask whether K is necessarily Hilbertian? Example 26.1.11 gives an example where this is not the case. 

18.7 The Bottom Theorem We supplement the free generators theorem for a Hilbertian field K. Not only is it true that a σ selected at random in Gal(K)e generates a group isomorphic to Fˆe , but also, the field Ks (σ) has no proper subfields of finite codegree (Theorem 18.7.7). Our first concern is to eliminate the possibility that Ks (σ) has a proper subfield of finite codegree which is the fixed field of some involution of Gal(K) acting on Ks (σ) (Proposition 18.7.5): Lemma 18.7.1: Let K be a field of characteristic 0 and f ∈ K(T1 , . . . , Tr )[X1 , . . . , Xn ] an irreducible polynomial. Then the polynomial 3 3 X X 2 2 f( Y1j ,..., Yrj , X 1 , . . . , Xn )

(1)

j=1

j=1

is defined and irreducible in K(Y)[X]. ∂f Proof: Assume without loss that T1 , . . . , Tr occur in F . Then ∂T 6= 0 for i 2 2 2 i = 1, . . . , r. Take gi (Yi ) = Yi1 +Yi2 +Yi3 in Corollary 10.3.2. Then gi (Yi )+c ˜ Therefore, the polynomial in (1) is is absolutely irreducible for all c ∈ K. irreducible in K(Y)[X]. 

Lemma 18.7.2: Let K be a formally real Hilbertian field, H a Hilbert subset of K r and a1 < b1 , . . . , ar < br rational numbers. Then there is a (z1 , . . . , zr ) ∈ H with ai < zi < bi in each ordering of K, i = 1, . . . , r . Proof: We prove the lemma only for the case r = 1. The general case is analogous. Write H as HK (f1 , . . . , fl ) with fj ∈ K(T )[X1 , . . . , Xn ], irreducible, j = 1 1 . Then fj (a + c+T , X) 1, . . . , l. Let a < b be rational numbers. Put c = b−a 1 is irreducible. Hence, so is fj (a + c+Y 2 +Y 2 +Y 2 , X) (Lemma 18.7.1). Since 1

2

3

K is Hilbertian there exist y1 , y2 , y3 ∈ K with y1 6= 0 such that fj (a + 1 , X) is defined and irreducible in K[X], j = 1, . . . , l . Let z = c+y 2 +y 2 +y 2 1

a+

2

3

1 c+y12 +y22 +y32

. Then a < z < b in every ordering of K and fj (z, X) is

defined and irreducible in K[X], j = 1, . . . , l.



18.7 The Bottom Theorem

383

The following result strengthens Corollary 16.2.7 in the case where char(K) = 0: Lemma 18.7.3: Let K be a formally real Hilbertian field and m ≥ 2 a positive integer. Then there exists a linearly disjoint sequence {Ki /K}∞ i=1 of Galois extensions such that Gal(Ki /K) ∼ = Sm and Ki /K has a totally imaginary quadratic subextension Ei /K (i.e. Ei is contained in no real closure of K), i = 1, 2, 3, . . . . Proof: Suppose by induction K1 , . . . , Kn and E1 , . . . , En have already been constructed. Put L = K1 · · · Kn . Consider the general polynomial of degree m, m Y (X − xi ). f (T, X) = X m + T1 X m−1 + · · · + Tm = i=1

Put ∆(T) =

Q

i<j (xi

(X 2 + 1)

2

− xj ) . Now let c1 , . . . , cm ∈ R with

m−2 Y

(X − i) = X m + c1 X m−1 + · · · + cm = f (c, X).

i=1

Then ci is an integer, i = 1, . . . , m, and ∆(c) < 0. Since ∆(T) is a polynomial with integral coefficients, there exist rational numbers ai , bi with ai < ci < bi , i = 1, . . . , m, such that for all t1 , . . . , tm ∈ R (2)

ai < ti < bi , i = 1, . . . , m,

implies ∆(t) < 0.

Since each real closure of K is an elementary equivalent to R [Prestel2, p. 51], (2) holds for all t1 , . . . , tm ∈ K and for every ordering < of K. By Lemma 13.1.1 and Corollary 12.2.3 there exists a Hilbert subset H of K m such that for each t ∈ H, Gal(f (t, X), K) ∼ = Gal(f (t, X), L) ∼ = Sm . Choose t ∈ H such that ai < ti < bi , i = 1, . . . , m, in each ordering < of K (Lemma 18.7.2). Let Kn+1 be the splitting field of f (t, X) over K. Then Kn+1 pis linearly disjoint S and E = K( ∆(t)) is a totally from L over K, Gal(Kn+1 /K) ∼ = m n+1 imaginary quadratic extension of K contained in Kn+1 . This completes the induction.  Lemma 18.7.4: Let p be an odd prime, γ be the cycle (12 . . . p) in Sp , N the normalizer of hγi in Sp , and π an element of N of order 2. Then π ∈ Ap if and only if p ≡ 1 mod 4. p+3 Proof (Cherlin): The element τ = (1 p)(2 p−1) · · · ( p−1 2 2 ) of Sp is a prodp−1 uct of 2 transpositions. Hence, τ ∈ Ap if and only if p ≡ 1 mod 4. It satisfies τ −1 γτ = γ −1 . Hence, τ ∈ N . Since p is odd, γ is in Ap . It suffices therefore to check that each element π ∈ N of order 2 lies in the coset hγiτ . Observe that hγi is its own centralizer in G. Indeed, suppose σ commutes with γ. Find i with γ i (1) = σ −1 (1). Then prove inductively that γ i (k) = σ −1 (k) for all k. Therefore, σ = γ −i .

384

Chapter 18. The Haar Measure

Consider π ∈ N of order 2. Then π −1 γπ = γ j for some j ∈ N with j 2 ≡ 1 mod p. Hence, j ≡ ±1 mod p. If j ≡ 1 mod p, then π commutes with γ. By the preceding paragraph π ∈ hγi. Since p is odd, this is a contradiction. Thus, π −1 γπ = γ −1 . Hence, (πτ −1 )−1 γ(πτ −1 ) = γ. Consequently, πτ −1 ∈ hγi, so π ∈ hγiτ , as required.  Proposition 18.7.5 ([Jarden3, p. 299]): Let K be a Hilbertian field. Then for almost all σ ∈ Gal(K)e there exists no τ ∈ Gal(K), τ 6= 1, of finite order such that [Ks (σ) : Ks (σ, τ )] < ∞. Proof: By Artin-Schreier we only have to prove the proposition for K a formally real field, τ 2 = 1, and τ 6= 1 [Lang7, p. 299, Cor. 9.3]. Moreover, it suffices to prove for each n ∈ N, that for almost all σ ∈ Gal(K)e there exists ˜ ˜ : K(σ, τ )] = n. no τ ∈ Gal(K) with τ 2 = 1, τ 6= 1, and [K(σ) Choose a prime p ≡ 1 mod 4 with p ≥ n. Consider the sequence {Ki /K}∞ i=1 of fields constructed in Lemma 18.7.3 for m = p. In particular, there is an isomorphism θi : Sp → Gal(Ki /K), i = 1, 2, 3, . . . . Let γi = θi ((1 2 . . . p)). Denote the set of all σ ∈ Gal(K)e for which there exists i with resKi (σj ) = γi , j = 1, . . . , e by S. By Lemma 18.5.2, µ(S) = 1. We prove that each σ ∈ S has the desired property. ˜ : Assume there is a τ ∈ Gal(K) with τ 2 = 1, τ 6= 1, and [K(σ) ˜ ˜ K(σ, τ )] = n. Let L be the smallest normal extension of K(σ, τ ) that con˜ ˜ ˜ tains K(σ). Then [L : K(σ, τ )] divides n!, so [L : K(σ)] divides (n − ˜ 1)!. Hence, p does not divide [L : K(σ)]. In addition, there is an i with ˜ = Ki (γi ) and [Ki ∩ L : resKi (σj ) = γi , j = 1, . . . , e. Then Ki ∩ K(σ) ˜ ˜ Ki (γi )] = [(Ki ∩L)K(σ) : K(σ)]. Thus, [Ki ∩L : Ki (γi )] divides the relatively ˜ Hence, Ki ∩ L = Ki (γi ). prime numbers [Ki : Ki (γi )] = p and [L : K(σ)]. It follows for τ¯ = resKi (τ ) that Ki (γi )/Ki (γi , τ¯) is a Galois extension. That is, τ¯ belongs to the normalizer of hγi i in Gal(Ki /K). Since τ¯2 = 1 and p ≡ 1 mod 4, Lemma 18.7.4 implies that τ¯ belongs to θi (Ap ), a subgroup of Gal(Ki /K) of index 2. Since Sp has only one subgroup of index 2, ˜ ), a real closed field θi (Ap ) = Gal(Ki /Ei ). Thus, Ei is contained in K(τ [Lang7, p. 299, Cor. 9.3 and p. 452, Prop. 2.4]. This contradicts the choice made for Ki /K from Lemma 18.7.3 that Ei /K is a totally imaginary extension. Consequently, a τ as above does not exist.  Lemma 18.7.6: Let K be a Hilbertian field. Write Fˆe as the inverse limit π3 π2 π1 H3 −→ H2 −→ H1 of epimorphisms of finite groups. Then there exists · · · −→ π3 π2 π1 a sequence · · · −→ C3 −→ C2 −→ C1 of finite groups such that for each k ∈ N the following hold: (3a) Hk is a subgroup of Ck , and πk : Ck+1 → Ck extends πk : Hk+1 → Hk . (3b) There exists a subgroup Bk of Ck such that Ck = Bk Hk = {bh | b ∈ Bk , h ∈ Hk } and Bk ∩ Hk = 1. (3c) For almost all σ ∈ Gal(K)e there exists a homomorphism ρ: Gal(K) → Ck such that ρ(hσi) = Hk .

18.7 The Bottom Theorem

385

Proof: For each k ∈ N let nk = |Hk |. Multiplication from the right gives an embedding αk of Hk into the symmetric group Snk satisfying: (4) For all 1 ≤ i, j ≤ nk there is a unique h ∈ Hk such that αk (h)(i) = j. Let Ck = Sn1 × · · · × Snk and define λk : Hk → Ck by λk (h) = ((α1 ◦ π1 ◦ · · · ◦ πk−1 )(h), (α2 ◦ π2 ◦ · · · ◦ πk−1 )(h), . . . , αk (h)). This is an embedding, so identify Hk with its image in Ck . Let πk : Ck+1 → Ck be projection on the first k coordinates. Then (3a) holds. Assertion (3b) follows from (4) for Bk = {(τ1 , . . . , τk ) ∈ Ck | τk (1) = 1}. ¯e ∈ Hk with Hk = h¯ σ1 , . . . , σ ¯e i. For To prove (3c), choose σ ¯1 , . . . , σ each i, 1 ≤ i ≤ e, denote the jth component of σ ¯i , viewed as an element ¯ij . For almost all σ ∈ Gal(K)e and each of Ck = Sn1 × · · · × Snk , by σ j, 1 ≤ j ≤ k, there is a continuous homomorphism ρj : Gal(K) → Snj with ρj (σ1 , . . . , σe ) = (¯ σ1j , . . . , σ ¯ej ) (Lemma 18.5.5). Combine ρ1 , . . . , ρk to a homomorphism ρ: Gal(K) → Ck with ρ(σi ) = (¯ σi1 , . . . , σ ¯ik ) = σ ¯i for  i = 1, . . . , e. Thus, ρ(hσi) = Hk . Theorem 18.7.7 (The Bottom Theorem [Haran2]): Let K be a Hilbertian field and e ∈ N. Then for almost all σ ∈ Gal(K)e , Ks (σ) is a finite extension of no proper subfield that contains K. Proof: Let S be the set of all σ ∈ Gal(K)e satisfying this: (5a) hσi ∼ = Fˆe . (5b) There is no τ ∈ Gal(K), τ 6= 1, of finite order with (hσ, τ i : hσi) < ∞. (5c) For each positive integer k there is a homomorphism ρ: Gal(K) → Ck with ρ(hσi) = Hk . Here we use the notation of Lemma 18.7.6. Theorem 18.5.6, Proposition 18.7.5, and Lemma 18.7.6 imply that S is the intersection of three sets of measure 1. Thus, µ(S) = 1. We show that each σ ∈ S satisfies the conclusion of the theorem. Put H = hσi. Assume H is an open subgroup of index n of a closed subgroup G of Gal(K). By (5b), G is torsion free. For each positive integer k denote the set of all homomorphisms ρ: G → Ck with ρ(H) = Hk by Rk . By (5c), Rk is nonempty. Since (G : H) < ∞, G is finitely generated. Hence, Rk is finite. The homomorphisms πk of (3a) induce an inverse system π

π

π

3 2 1 · · · −→ R3 −→ R2 −→ R1

whose inverse limit is nonempty. By Corollary 1.1.4, there are ρk ∈ Rk with πk ◦ ρk+1 = ρk , k = 1, 2, 3, . . . . Let Gk = ρk (G), k = 1, 2, 3, . . . . The ρk ’s define an epimorphism ρ: G → lim Gk with ρ(H) = lim Hk . From ←− ←− the conditions of Lemma 18.7.6 and Condition (5a), ρ(H) = Fˆe ∼ = H. By Lemma 17.4.11, H ∩ Ker(ρ) = 1. But then |Ker(ρ)| = (H · Ker(ρ) : H) ≤ (G : H) < ∞. Since G is torsion free, Ker(ρ) = 1. Hence, ρ maps G (resp. H) isomorphically onto lim Gk (resp. lim Hk ). This gives k0 with ←− ←−

386

Chapter 18. The Haar Measure

(6) (Gk : Hk ) = (G : H) = n for all k ≥ k0 . Assume without loss that k0 = 1. For each k ∈ N let Bk be the set of all subgroups Bk0 of Gk satisfying (7) Bk0 Hk = Gk and Bk0 ∩ Hk = 1. The condition Bk Hk = Ck (Lemma 18.7.6(3b)) implies (Bk ∩ Ck )Hk = Gk . Thus, Bk is nonempty and finite. By (6) and (7), the order of each Bk0 ∈ Bk is n. Moreover, πk−1 (Bk0 )Hk−1 = πk−1 (Bk0 )πk−1 (Hk ) = πk−1 (Gk ) = Gk−1 . Hence, n = (Gk−1 : Hk−1 ) = (πk−1 (Bk0 )Hk−1 : Hk−1 ) =

|πk−1 (Bk0 )| . |πk−1 (Bk0 ) ∩ Hk−1 |

Since |πk−1 (Bk0 )| ≤ |Bk0 | = n, this implies |πk−1 (Bk0 ) ∩ Hk−1 | = 1. Therefore, π2 π1 π3 B3 −→ B2 −→ B1 , is an inverse system with πk−1 (Bk0 ) ∈ Bk−1 . Thus, · · · −→ a nonempty inverse limit. In other words, there is a Bk0 ≤ Gk of order n with 0 , k = 2, 3, . . . . The inverse limit B 0 = lim Bk0 is a subgroup πk−1 (Bk0 ) = Bk−1 ←− of G of order n. But G is torsion free. Consequently, n = 1.  Problem 18.7.8: Is the following generalization of the bottom theorem true: Let K be a Hilbertian field and e a positive integer. Then for almost all σ ∈ Gal(K)e , Ks (σ) is a finite extension of no proper subfield.

18.8 PAC Fields over Uncountable Hilbertian Fields In contrast to Theorem 18.6.1, if K is an uncountable Hilbertian field, it is possible that the set (1)

Se (K) = {σ ∈ Gal(K)e | Ks (σ) is a PAC field}

is not even measurable. The main theorem of this section says this happens if K is a rational function field in uncountably many variables over a field E0 or if K = C(t). Lemma 18.8.1: Let G be a profinite group and S a subset of G. Denote the Boolean algebra of all open-closed subsets of G by B0 and the σ-algebra generated by B0 by B1 . Suppose for each B ∈ B1 there exists an epimorphism r of G onto a profinite group H such that (2)

B = r−1 (r(B)) and µH (r(S)) = 1.

Then G r S contains no subset of positive measure. Proof: Let B 0 be a measurable subset of G r S. Proposition 18.1.3 gives a set B ∈ B1 with B ⊆ B 0 and µ(B 0 r B) = 0. Let r: G → H be an epimorphism satisfying (2). Since B ⊆ G r S, we have S ⊆ G r B and µH (r(G r B)) = 1. In addition, G r B = r−1 (r(G r B)). By Proposition 18.2.2, µ(G r B) = 1.  Hence, µ(B 0 ) = 0.

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Lemma 18.8.2: Let E be a field, A an infinite subset of E, and F ⊆ E[X1 , . . . , Xn ] a set of nonzero polynomials with |F| < |A|. Then the cardinality of the set C = {a ∈ An |f (a) 6= 0 for each f ∈ F} is equal to |A|. In particular, if {Ui | i ∈ I} isTa family of nonempty Zariski E-open sets in An and |I| < |A|, then |An ∩ ( i∈I Ui )| = |A|. Proof: Our assertion is true for n = 1, because every polynomial f ∈ F has only finitely many zeros. Assume, by induction, the assertion is true for n−1, where n ≥ 2. Each f ∈ F has a nonzero coefficient g ∈ E[X1 , . . . , Xn−1 ]. Hence, by the induction hypothesis, applied to the set of those g’s, the cardinality of the set B = {(a1 , . . . , an−1 ) ∈ An−1 | f (a1 , . . . , an−1 , Xn ) 6= 0 for every f ∈ F} is equal to that of A. For each (a1 , . . . , an−1 ) ∈ B the case n = 1 gives an element an ∈ A such that f (a1 , . . . , an ) 6= 0 for each f ∈ F. This gives an embedding of B into C. Consequently, |C| = |A|. Since each Zariski open set is defined by finitely many polynomial inequalities, the last assertion follows from the above.  A finite separable extension has only finitely many subextensions. Define the rank of an infinite separable algebraic extension of fields as the cardinality of the family of all finite subextensions. Thus, if L is the compositum of m finite separable extensions of a field K and m is an infinite cardinal number, then rank(L/K) = m. If L/K is a Galois extension and Gal(L/K) is not finitely generated, then rank(L/K) = rank(Gal(L/K)) (Proposition 17.1.2). Lemma 18.8.3: Let K = E(t) be a rational function field in one variable over an infinite field E. Consider a separable extension L of K with rank(L/K) < |E|. Let f ∈ K[X, Y ] be an irreducible polynomial in L[X, Y ], separable in Y . Then there exists an x ∈ K such that f (x, Y ) is irreducible and separable in L[Y ]. Proof: Let {Ki | i ∈ I} be the family of all finite extensions of K contained in L. By assumption, |I| < |E|. By Corollary 12.2.3, every separable Hilbert subset of Ki contains a separable Hilbert subset of K. Therefore, by Proposition 13.2.1, there exists a nonempty Zariski E-open set Ui ⊆ A2 with f (a + bt, Y ) irreducible and separable in Ki [Y ] for each (a, b) ∈ Ui (E). By T Lemma 18.8.2, i∈I Ui (E) is nonempty. If (a, b) lies in this intersection and x = a + bt, then f (x, Y ) is irreducible and separable over every Ki , hence also over L.  Lemma 18.8.4: Let K be as in Lemma 18.8.3 and N a Galois extension of K with rank(N/K) < |K|. Then each σ = (σ1 , . . . , σe ) ∈ Gal(N/K)e extends to τ = (τ1 , . . . , τe ) ∈ Gal(K)e such that Ks (τ ) is a PAC field. Proof: Wellorder the absolutely irreducible polynomials of K[X, Y ] which are separable in Y in a transfinite sequence {fα | α < m}, where m = |K|.

388

Chapter 18. The Haar Measure

For each α < m we produce a finite separable extension Kα of K containing a zero of fα and satisfying this: For each β < m the set of fields {N }∪{Kα | α < β} is linearly disjoint over K. Indeed, let β < m. Assume that Kα has been defined for each α < β. Let L be the compositum of N and of all fields Kα with α < β. Then L is a separable extension of K with rank(L/K) < m. Lemma 18.8.3 gives an x ∈ K with fβ (x, Y ) irreducible and separable in L[Y ]. Choose y ∈ Ks with f (x, y) = 0. Then Kβ = K(y) is linearly disjoint from L over K. The compositum M of all Kα is a separable algebraic extension of K which is linearly disjoint from N . Hence, res: Gal(M ) → Gal(N/K) is an epimorphism. By Theorem 11.2.3, M is PAC. Extend σ1 , . . . , σe to automorphisms τ1 , . . . , τe ∈ Gal(M ). Their fixed field Ks (τ ) is an algebraic extension  of M . From Corollary 11.2.5, Ks (τ ) is a PAC field. The next result is a weak form of Theorem 18.6.1 for K = E(t). Proposition 18.8.5 ([Jarden-Shelah]): Let K = E(t) be a rational function field in one variable over an uncountable field E. For each e ≥ 1, the complement of Se (K) in Gal(K)e contains no subset of positive measure. Proof: Use Lemma 18.8.1 with G = Gal(K)e . Note first that every openclosed subset U of Gal(K)e is a union of finitely many products of the form U1 × · · · × Ue where each Ui is the union of translates of Gal(L) for L (independent of i) a finite Galois extension of K. For B ∈ B1 , let N be the compositum of all fields L associated with those products U1 × · · · × Ue that are required to express B as an element of B1 . Since there are at most countable such L’s, rank(N/K) ≤ ℵ0 . Moreover, r−1 (r(B)) = B, where r: Gal(K)e → Gal(N/K)e is the restriction map. By Lemma 18.8.4, r(Se (K)) = Gal(N/K)e . Thus, Lemma 18.8.1 implies  that Gal(K)e r Se (K) contains no set of positive measure. When K is a rational function field in uncountably many variables over a field E0 , we are nearly ready to conclude that Se (K) is nonmeasurable. Lemma 18.8.6: (a) Let L/K0 be a field extension with L 6= Ls and v a Henselian val¯ v = K0 . Then each uation of L which is trivial on K0 such that L σ ∈ Gal(K0 )e r{1} extends to ρ ∈ Gal(L)e with Ls (ρ) non-PAC. (b) Let K = E(t) be a rational function field in one variable over a field E. Consider a subfield K0 of K and σ ∈ Gal(K0 )e . Then σ extends to ρ ∈ Gal(K)e with Ks (ρ) non-PAC in each of the following cases: (b1) K0 = E. ^ (b2) K0 = E0 (t) with E0 algebraically closed and E = E 0 (u) with u transcendental over E0 . Proof of (a): By Lemma 2.6.9, L is a regular extension of K0 . Hence, each σ ∈ Gal(K0 )e r{1} extends to ρ ∈ Gal(L)e with Ls (ρ) 6= Ls . Then Ls (ρ) is

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Henselian (Section 11.5) but not separably closed. It follows from Corollary 11.5.6 that Ls (ρ) is not PAC. Proof of (b): By (a), it suffices to construct an algebraic extension L of K with L 6= Ls and a Henselian valuation v of L which is trivial on K0 such ¯ v = K0 . that L First suppose K0 = E. Let v be the valuation of K/E with v(t) = 1. ¯ v = E and L 6= Ls , as Choose a Henselian closure (L, v) of (K, v). Then L desired. Now suppose K0 and E are as in (b2). Let v0 be the valuation of E0 (u) over E0 with v0 (u) = 1. Extend v0 to a valuation v˜0 of E. As E0 is ¯v˜ . Next extend v˜0 to a valuation v of K/K0 algebraically closed, E0 = E 0 ¯ with Kv = K0 (Example 2.3.3). Then let (L, v) be the Henselian closure of ¯ v = K0 is not separably closed. Hence, L 6= Ls , as (K, v). In particular, L desired.  Lemma 18.8.7: Let F be a field, e a positive integer, and B1 the σ-algebra generated by all open-closed subsets of Gal(F )e . Let F = {Fi | i ∈ I} be a family of subfields of F . For each i ∈ I, let ri : Gal(F )e → Gal(Fi )e be the restriction map. Suppose Fi is algebraically closed in F . Suppose also, S and for each subset I0 of I with |I0 | ≤ ℵ0 there exists j ∈ I F = i∈I F i S such that i∈I0 Fi ⊆ Fj . Then, for each B ∈ B1 there is an i ∈ I with B = ri−1 (ri (B)). Proof: First suppose B is open-closed. Then there is a finite Galois extension ¯ of Gal(F 0 /F )e . F 0 of F such that B is the lifting to Gal(F )e of some subset B 0 Then there is an i ∈ I and a Galois extension Fi of Fi with Fi0 F = F 0 . By assumption, Fi0 ∩ F = Fi . Hence, res: Gal(F 0 /F )e → Gal(Fi0 /Fi ) is an isomorphism. Therefore, B = ri−1 (ri (B)). Next let B10 be the union of all σ-algebras generated by countable collections of open-closed subsets of Gal(K)e . Then B10 is a sub-σ-algebra of B1 which contains all open-closed subsets. Therefore, B10 = B1 . Now consider B ∈ B1 . The preceding paragraph gives open-closed sets B1 , B2 , B3 , . . . which generate a σ-algebra containing B. For each k the first (rik (Bk )). By assumption, there paragraph gives ik ∈ I such that Bk = ri−1 k S∞ is an i ∈ I with k=1 Fik = Fi . For each k we have ri−1 (ri (Bk )) = Bk . The family of all subsets A of Gal(F )e satisfying ri−1 (ri (A)) = A is closed under countable unions and taking complements. Hence, it contains B, as contended.  Theorem 18.8.8: Let K be one of the following types of fields: (a) K = E0 (T ), where E0 is a field and T is an uncountable set of algebraically independent elements over E0 . (b) K = E(t), where E is an uncountable algebraically closed field and t is a transcendental element over E. Then, for each positive integer e neither Se (K) nor Gal(K)e r Se (K) contains a set of positive measure.

390

Chapter 18. The Haar Measure

Proof: In both cases K = E(t) with t transcendental over an uncountable field E. Thus, from Proposition 18.8.5 we need only to prove that Se (K) contains no set of positive measure. This will be achieved by applying Lemma 18.8.1 to G = Gal(K)e and S = Gal(K)e r Se (K). So, consider B ∈ B1 . First suppose that (a) holds. Then K is the union of all E0 (T0 ) where T0 ranges over all countable subsets of T . By Lemma 18.8.7, there is a countable subset T1 of T such that B = r−1 (r(B)) where r: Gal(K)e → Gal(E0 (T1 ))e is the restriction map. Since T is uncountable there is a t ∈ T r T1 . Let E = E0 (T r{t}) and let r1 : Gal(K)e → Gal(E)e be the restriction map. By Lemma 18.8.6, r1 (Gal(K)e r Se (K)) ⊇ Gal(E)e r{1}. Hence, r(Gal(K)e r Se (K)) ⊇ Gal(E0 (T1 ))e r{1}. Now suppose (b) holds. Then K is the union of all fields E0 (t) with E0 a countable algebraically closed subfields of E. Lemma 18.8.7 gives a countable algebraically closed subfield E1 of E such that B = r1−1 (r1 (B)), where r1 : Gal(K)e → Gal(E1 (t))e is the restriction map. Since E is uncountable, it has a nonempty transcendence basis U over E1 . Choose u ∈ U and let E2 be the algebraic closure of E1 (U r{u}). Then E is the algebraic closure of E2 (u). Moreover, with r: Gal(K)e → Gal(E2 (t))e the restriction map, r−1 (r(B)) = B. By Lemma 18.8.6(b2), r(Gal(K)e r Se (K)) ⊇ Gal(E2 (t))e r{1}. In both cases we conclude from Lemma 18.8.1 that Se (K) contains no set of positive measure.  Corollary 18.8.9: Let K be a field of type (a) or (b), as in Theorem 18.8.8. Then, for each finite separable extension K 0 of K, the set Se (K 0 ) is nonmeasurable. Proof: Note that Se (K 0 ) = Se (K)∩Gal(K 0 )e . Assume Se (K 0 ) is measurable. If µ(Se (K 0 )) > 0, then µ(Se (K)) > 0. Otherwise, µ(Se (K 0 )) = 0. Hence, µ(Gal(K)e r Se (K)) ≥ µ(Gal(K 0 )e r Se (K 0 )) = [K 01:K]e > 0. In both cases we get a contradiction to Theorem 18.8.8.  Proposition 13.4.6 produces an uncountable Hilbertian PAC field K. Since every algebraic extension of K is again a PAC field, Se (K) = Gal(K)e for each positive integer e. Thus, Theorem 18.8.8 is false for arbitrary uncountable Hilbertian fields. It is still open if case (b) of Theorem 18.8.8 holds without the assumption that E is algebraically closed. Problem 18.8.10: Let E(t) be a rational function field in one variable over an uncountable field E. Is Se (E(t)) nonmeasurable?

18.9 On the Stability of Fields We defined stable polynomials in Section 16.2 in order to realize certain groups over Hilbertian fields. Here we interpret stability in terms of function field properties in order to construct special PAC fields:

18.9 On the Stability of Fields

391

Definition 18.9.1: Let F/K be a field extension. Call F/K a function field of several variables if F/K is finitely generated and regular. We say F/K is stable if it has a separating transcendence base t1 , . . . , tr such that the Galois closure, Fˆ , of the (separable) extension F/K(t) is regular over K. ˜ K(t)). ˜ This is equivalent to Gal(Fˆ /K(t)) ∼ Refer to t1 , . . . , tr as a = Gal(Fˆ K/ stabilizing base for F/K. Call K stable if every finitely generated regular extension F/K is stable.  The following sufficient condition for stability is an analog of Remark 16.2.2: Lemma 18.9.2: Let t1 , . . . , tr be a separating transcendence base of a finitely generated regular extension F/K. Let n = [F : K(t)]. Denote the Galois ∼ ˜ K(t)) ˜ closure of F/K(t) by Fˆ . If Gal(Fˆ K/ = Sn , then t1 , . . . , tr stabilizes F/K. Thus, F/K is stable. ∼ ˜ K(t)) ˜ ˜ Proof: By Galois theory Gal(Fˆ K/ Since [F : = Gal(Fˆ /Fˆ ∩ K(t)). ˆ ˆ ˜ ˆ ˜ K(t)] = n, n! = [F : F ∩ K(t)] ≤ [F : K(t)] ≤ n!. Hence, Fˆ ∩ K(t) = K(t) ∼ ˜ K(t)) ˜ and Gal(Fˆ K/ = Gal(Fˆ /K(t)). Thus, t1 , . . . , tr is a stabilizing base for F/K.  Refer to the transcendence base t1 , . . . , tr of Lemma 18.9.2 as a symmetrically stabilizing base and to n as its degree. Theorem 18.9.3 ([Fried-Jarden1,2, Geyer-Jarden3, Madan-Madden, Geyer2, Neumann]): Every field is stable. Moreover, every function field of several variables has a symmetrically stabilizing base of arbitrarily high degree. Outline of proof: As the list of creditors indicates, the proof was developed over a long period of time, each new step covering more fields and having an increased level of difficulty. Each of the proofs considers a function field F of one variable over K and produces a symmetrically stabilizing element t. Then it applies Lemma 18.9.2 to conclude F/K is stable. The production of a symmetrically stabilizing base for arbitrary function fields is reduced to the case r = 1. We survey the various proofs and indicate the new ingredients that enter in each of them: Part A: Reduction to function fields of one variable. Suppose F/K is a function field of transcendence degree r > 1. If K is infinite, MatsusakaZariski (Proposition 10.5.2) gives x, y ∈ F and c ∈ K with F/K(x + cy) regular of transcendence degree r − 1. In the general case there are x, y ∈ F and a positive integer n with F/K(x + y n ) regular [Neumann, Lemma 2.1, due to Geyer]. Let t1 be x + cy in the former case or x + y n in the general case. Induction gives a symmetrically stabilizing base t2 , . . . , tr for F/K(t1 ) of large degree n. Then t1 , t2 , . . . , tr form a symmetrically stabilizing base of degree n for F/K. Assume from now on F/K is a function field of one variable.

392

Chapter 18. The Haar Measure

Part B: K is PAC [Fried-Jarden1]. Since K is PAC, F has a prime divisor p of degree 1. Let g be the genus of F/K and choose a prime number l > 2g − 1 with char(K) - l(l − 2). Lemma 3.2.3 gives an element x ∈ F with find a K-rational prime div∞ (x) = (l − 2)p. Now use that K is infinite Pto m divisor q of K(x)/K unramified in F . Let q = i=1 qi be its decomposition Pm in F . Then i=1 deg(qi ) = [F : K(x)] = deg(div∞ (x)) = l − 2. The divisor d = q1 + · · · + qm + 2p of F/K satisfies deg(d) = l > 2g − 1. By Exercise 3 of Chapter 3 there exists t ∈ F with div∞ (t) = d. Hence, [F : K(t)] = l. ˜ The pole divisor of t in Now extend the field of constants from K to K. ˜ ˜ decomposes in F K ˜ as p1 + . . . + pl−2 + 2p0 where p1 , . . . , pl−2 , p0 are K(t)/ K ˜ K ˜ that lies over p. Consider distinct and p0 is the unique prime divisor of KF/ ˜ of F K/ ˜ K(t). ˜ ˜ Fˆ that the Galois closure Fˆ K Let ˆ p be a prime divisor of K ˜ Fˆ /K(t)) ˜ acts faithfully on the cosets extends p0 . The Galois group Gal(K ˜ Fˆ /KF ˜ ) and the inertia group of ˆ of Gal(K p contains a transposition [Fried˜ : K(t)] ˜ Jarden1, p. 777]. Since [KF = l is a prime, it is an elementary lemma ∼ ˜ Fˆ /K(t)) ˜ of group theory that Gal(K = Sl [Waerden1, p. 201], as desired. Part C: char(K) = 0 [Fried-Jarden2]. In this case F has a projective plane curve model C defined over K having only nodes (i.e. singular points of multiplicity 2) as singularities [Hartshorne, p. 313]. Like any other irreducible ˜ plane curve in characteristic 0, C has these properties (over K): (1a) C has only finitely many inflection points (i.e. points at which the tangent cuts C with multiplicity at least 3). (1b) C has only finitely many double tangents (i.e. tangents in at least two distinct points). ˜ (1c) Only finitely many tangents to C pass through each point of P2 (K) (i.e. C has no strange points). Let n = deg(C). Since K is infinite, there is an o ∈ P2 (K) r C(K) outside the finitely many lines involved in (1). Each such o satisfies this: (2) Every line that passes through o cuts C in at least n − 1 points, and all but finitely many lines that pass through o cut C in n points. Use o to project C onto P1 . In field theoretic terms this means there exists t ∈ F with [F : K(t)] = n such that all inertial groups of places of ˜ K(t) ˜ ˜ K(t)) ˜ Fˆ K/ are generated by 2-cycles. As in Part B, consider Gal(Fˆ K/ ˜ KF ˜ ). The as a transitive permutation group on the set of cosets of Gal(Fˆ K/ ˜ of the group generated by the inertial groups of the extension fixed field in Fˆ K ˜ K(t) ˜ ˜ Fˆ K/ is an unramified extension of K(t). Thus, by Remark 3.6.2(b), ˜ ˜ K(t)) ˜ this fixed field is K(t). Hence, Gal(Fˆ K/ is generated by its inertial groups. Another elementary lemma of group theory [Fried-Jarden2, Lemma 1.4] implies that this group is Sn , as desired. Part D: K is infinite and perfect [Geyer-Jarden1]. Not all plane curves in positive characteristic satisfy Condition (1). For example, if char(K) = p 6= 0 and C is defined by X0p−1 X2 + X1p = 0, then all tangents to C at finite points pass through (0:1:0). Thus, (0:1:0) is a strange point of C. The construction of a projective plane node model for F/K satisfying (1) must

18.9 On the Stability of Fields

393

therefore be done with care. It is carried out in few steps. For simplicity assume genus(F/K) > 0 and char(K) 6= 2. The reduction step from arbitrary finitely generated regular extensions F/K to function fields of one variable forces us to prove the latter case not only for K perfect but for an arbitrary infinite field K and for F/K ˜ K). ˜ conservative. That is, genus(F/K) = genus(F K/ (3a) Choose x, y ∈ F with F = K(x, y) separable over K(x). Then choose u ∈ K such that the affine curve Γ generated over K by (x, y + ux2 ) has only finitely many inflection points. Then Γ has only finitely many double tangents and no strange points. (3b) Let (x0 :x1 :x2 ) be a generic homogeneous point of Γ. For i = 0, 1, 2 let yi = xi . Then use the condition “F/K is conservative” to choose y3 , . . . , yn ∈ F with n ≥ 4 large such that y = (y0 :y1 : · · · :yn ) generates a smooth curve ∆ in Pn over K. The projection on the first three coordinates is a rational morphism which fails to be regular only at finitely many points. Use this projection to prove that ∆ has only finitely many inflection points, finitely many double tangents and no strange points. ˜ such that the tangent through (3c) Choose a noninflection point a ∈ ∆(K) a is not double and it intersects only finitely many tangents to ∆. (3d) Choose a point o ∈ Pn (K) not on ∆ such that the projection π from o into Pn−1 maps ∆ onto a model Λ of F/K in Pn−1 such that π(a) is a simple noninflection point of Λ with a nondouble tangent. (3e) If n ≥ 4, choose o such that, in addition, Λ is smooth and the tangent to Λ at π(a) intersects only finitely many tangents of Λ. Then repeat step (3d). (3f) If n = 3, choose o such that, in addition, the only singular points of Λ are nodes and the tangent at π(a) goes through no singular point. Consequently, C = Λ satisfies (1). The condition “F/K is conservative” is equivalent to “F/K has a nor˜ in F K ˜ mal projective model Γ”. That is, the local ring of each point of Γ(K) is integrally closed [Rosenlicht, Thm. 12]. Suppose now F/K is a function field of several variables. Define F/K to be conservative if it satisfies the above condition. This is always the case when K is perfect. However, in the reduction process of Part A, K(t1 ) is no longer perfect, so one constructs t1 such that F/K(t1 ) is conservative. Then proceed by induction. Part E: K is finite [Madan-Madden, Geyer2]. Following [Geyer2] choose large prime numbers k, l with l − k ≥ 3. Write l = k + 2m. The Riemann Hypothesis for function fields (Theorem 4.5.2) gives t ∈ F r K with F/K(t) ˜ factors as Pm 2pi + Pk qi with separable whose pole divisor (over K) i=1 i=1 ˜ K. ˜ Let Fˆ be the Galois p1 , . . . , pm , q1 , . . . , qk distinct prime divisors of F K/ ˜ K). ˜ Then [F : K(t)] = [F K ˜ : K(t)] ˜ closure of F/K and G = Gal(Fˆ K/ = l. Hence, G may be considered as a primitive subgroup of Sl . As such G contains

394

Chapter 18. The Haar Measure

a cycle of length k. By a theorem of Jordan [Wielandt, Thm. 13.9], Al ≤ G. Also, G contains an odd permutation. Therefore, G = Sl . Part F: K is infinite [Neumann]. In this case, F/K need not be conservative. Hence, F/K need not have a projective plane node model (as in Part D) [Rosenlicht, Thm. 12]. Instead, [Neumann, Prop. 2.11] considers a large prime number l and constructs a plane model C for F/K with n = deg(C) large satisfying, in addition to (1), the following conditions: (4a) All singular points of C are either nodes or cusps of multiplicity at most l. (4b) C has at least one cusp of multiplicity l. ˜ is contained Here a cusp is a singular point of C whose local ring in F K in a unique valuation ring. As in Part C, use a point o ∈ P2 (K) outside C and the finitely many tangents involved in (1). Projection from o onto P1 produces an element t ∈ F r K satisfying this: ˜ ˜ Then p = Pn−k pi + kq, where (5) Let p be a prime divisor of K(t)/ K. i=1 p1 , . . . , pn−k , q are distinct prime divisors of F/K. The coefficient k takes the value 0 for all but finitely many p, it is at most l, and it takes each of the values 2 and l at least once. ˜ K) ˜ ∼ This guarantees that Gal(Fˆ K/ = Sn [Neumann, Prop. 2.7]. The construction of C has the same pattern as the construction of Part D, although it is more refined due to the possible lack of smooth model of F/K: (6a) Construct a projective plane model Γ1 for F/K satisfying (1) with a cusp p1 of multiplicity l [Neumann, Lemma 2.7]. (6b) Construct a projective model Γ2 over K in Pm for some m with a point p2 whose local ring (over K) coincides with that of p1 . The local rings (over K) of all other points are integrally closed [Rosenlicht, p. 174, Thm. 5]. (6c) Let ϕ: Γ1 → Γ2 be a birational map over K which maps p1 biregularly onto p2 . Denote the graph of ϕ by Γ. It is a projective curve in some Pm which satisfies (1) and has a cusp p of multiplicity l. Moreover, all ˜ are cusps of multiplicity at most l. This singular points of Γ over K construction replaces the construction in (3b) which assumes that F/K is conservative. (6d) Project Γ from a suitable point of Pm (K) into Pm−1 , being careful to preserve all properties of Γ except the following: When m = 3, the projected curve may have also nodes as singularities. Indeed, it must have at least one node. Infinitely many such projections produce the desired curve C. 

18.10 PAC Galois Extensions of Hilbertian Fields Let K be a countable Hilbertian field. We know Ks (σ) is PAC for almost all σ ∈ Gal(K)e (Theorem 18.6.1). Hence, each algebraic extension of those

18.10 PAC Galois Extensions of Hilbertian Fields

395

Ks (σ)’s is PAC (Corollary 11.2.5). However, since Gal(Ks (σ)) = hσi is finitely generated, none of these extensions except Ks is Galois over K (Proposition 16.11.6). In this section we use the stability of fields to construct an abundance of Galois extensions of K which are PAC Lemma 18.10.1: Let K be a field and M an algebraic extension of K. Suppose each X-stable polynomial f ∈ K[T, X] with Gal(f (T, X), K(T )) ∼ = Sn for some positive integer n has infinitely many zeros in M . Then M is a PAC field. Proof: Apply Theorem 11.2.3. We need to prove that every nonconstant absolutely irreducible polynomial g ∈ K[X, Y ] has an M -zero. Let (x, y) be a generic point of V (g). Then F = K(x, y) is a regular extension of K. Theorem 18.9.3 gives a symmetrically stabilizing element t for F/K. That is, with Fˆ being the Galois closure of F/K(t) we have Gal(Fˆ /K(t)) ∼ = Sn for some positive integer n. Let z be a primitive element for Fˆ /K(t). Choose z to be integral over K[t]. Put f (t, Z) = irr(z, K(t)). Then f is Z-stable h2 (t,z) over K, x = hh10(t,z) (t) , and y = h0 (t) for suitable polynomials h0 , h1 , h2 with coefficients in K such that h0 (t) 6= 0. By assumption, f has a zero (a, b) in M with h0 (a) 6= 0. Then

h1 (a,b) h2 (a,b) h0 (a) , h0 (a)

is an M -zero of g.



Let K be a field and σ1 , . . . , σe ∈ Gal(K). Denote the maximal Galois extension of K in Ks (σ) by Ks [σ]. Note that Gal(Ks [σ]) = hσ τ | τ ∈ Gal(K)i. Thus, Gal(Ks [σ]) is the smallest closed normal subgroup of Gal(K) containing hσi. We denote that group by [σ]. Theorem 18.10.2: Let K be a countable Hilbertian field and e a positive integer. Then Ks [σ] is PAC for almost all σ ∈ Gal(K)e . Proof: Denote the set of all polynomials f ∈ K[T, X] which are X-stable by F. For each f ∈ F, Lemma 16.2.6 gives a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions of K and a sequence a1 , a2 , a3 , . . . of distinct zeros of f such that for each i, ai is Li -rational and Gal(Li /K) ∼ = Gal(f (T, X), K(T )). Denote the set of σ ∈ Gal(K)e which are contained in infinitely many of the sets Gal(Li )e T by S(f ). By Borel-Cantelli (Lemma 18.5.3(b)), µ(S(f )) = 1. Hence, S = f ∈F S(f ) also has measure 1. If σ ∈ S, then there is an infinite set I of positive integers such that Li ⊆ Ks (σ). By definition, Li ⊆ Ks [σ]. Thus, ai is a Ks [σ]-rational point of f . It follows from Lemma  18.10.1 that Ks [σ] is PAC. ˜ In Section 25.10 we prove that almost all K[σ] mentioned in Theorem 18.10.2 are Hilbertian. Here we give an example of a PAC Galois extension of K which is Hilbertian and has a special Galois group. Theorem 18.10.3 ([Fried-Jarden2, Thm. 4.4 and Fried-V¨olklein2, p. 475, Cor. 1]): Let K be a countable Hilbertian field K. Then KQhas a Galois ∞ extension N which is Hilbertian and PAC with Gal(N/K) ∼ = k=1 Sk .

396

Chapter 18. The Haar Measure

Proof: List all plane curves over K in a sequence C1 , C2 , C3 , . . . . By induction construct a sequence k1 < k2 < k3 < · · · of positive integers, a sequence p1 , p2 , p3 , . . . of points in A2 (Ks ), and a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions of K satisfying this: (1a) Gal(Lk /K) ∼ = Sk , k = 1, 2, 3, . . . . (1b) pi ∈ Ci (Lki ). (1c) The points p1 , p2 , p3 , . . . are distinct. Suppose k1 , . . . , kn−1 , p1 , . . . , pn−1 , L1 , . . . , Ln−1 , and Lk1 , . . . , Lkn−1 have been constructed with the above mentioned properties. Choose a generic point (x, y) for Cn over K. Then F = K(x, y) is a function field of one variable over K. Theorem 18.9.3 gives a symmetrically stabilizing element t for F/K of degree kn greater than kn−1 . Denote the Galois closure of F/K(t) by Fˆ . Then Fˆ /K is regular and Gal(Fˆ /K(t)) ∼ = Skn . Now use Lemma 13.1.1: Specialize t to an element a ∈ K and extend the specialization to a K-place ϕ of Fˆ which is finite at x, y such that the residue field Lkn of F is Galois over K, linearly disjoint from L1 · · · Ln−1 Lk1 · · · Lkn−1 , and Gal(Lkn /K) ∼ = Skn . Then pn = (ϕ(x), ϕ(y)) is an Lkn -rational point of Cn . A careful choice of a will place pn outside the set {p1 , . . . , pn−1 }. If n belongs to {k1 , . . . , kn }, then Ln is already defined. Otherwise, use Corollary 16.2.7 to construct a Galois extension Ln of K which is linearly disjoint from L1 · · · Ln−1 Lk1 · · · Lkn over K and with Gal(Ln /K) ∼ = Sn . This completes the induction. Q∞ ∼ Q∞ Put N = k=1 Lk . Then N is a Galois extension and Gal(N/K) = S . Moreover, N is a finite proper extension of the Galois extension k Qk=1 ∞ k=2 Lk of K. By Weissauer (Theorem 13.9.1(b)), N is Hilbertian. Finally, each plane curve over K has an N -rational zero. It follows from Lemma 11.2.3 that N is a PAC field.  We generalize that part of Theorem 18.10.3 which states that N is Hilbertian. For each field K we denote the compositum of all Galois extensions L of K with Gal(L/K) ∼ = Sn for some n by Ksymm . Similarly Kalt denotes the compositum of all Galois extensions L of K with Gal(L/K) ∼ = An for some n. Theorem 18.10.4: Let K be a Hilbertian field. Then Ksymm is PAC and Hilbertian. Proof: In order to prove that Ksymm is PAC, it suffices to prove that every absolutely plane curve C defined over K has a Ksymm -rational point (Theorem 11.2.3). Indeed, let (x, y) be a generic point over K and let F = K(x, y) be the corresponding function field of C over K. By Theorem 18.9.3, F/K has a symmetrically stabilizing element t. Denote the Galois closure of F/K(t) by Fˆ . Then Gal(Fˆ /K) ∼ = Sn with n = [F : K(t)]. By Lemma 13.1.1, Fˆ has a K-place ϕ such that a = ϕ(t) ∈ K, b = ϕ(x) 6= ∞, c = ϕ(y) 6= ∞, and the

18.11 Algebraic Groups

397

residue field of Fˆ at ϕ is Galois over K with Galois group Sn . It follows that (b, c) ∈ C(Ksymm ), as desired. Next let K 0 be the compositum of all finite Galois extensions L of K with Gal(L/K) ∼ = S2n+1 for some positive integer n. Let K 00 be the compositum of all finite Galois extensions L of K with Gal(L/K) ∼ = S2n for some positive integer n. Then both K 0 and K 00 are Galois extensions of K and Ksymm = K 0 K 00 . By Corollary 16.2.7(a), K has a finite Galois extension L in K 0 with Gal(L/K) ∼ = S5 . If K 0 ⊆ K 00 , then the simple group A5 is a composition factor of Gal(K 00 /K), hence also of S2n for some positive integer n. This means that A5 = A2n , so 5 = 2n. This contradiction implies that K 0 6⊆ K 00 . Exchanging the roles of K 0 and K 00 and replacing 5 by 6 in the above argument proves that K 00 6⊆ K 0 . It follows from Corollary 13.8.4 that Ksymm is Hilbertian.  A similar argument as in the last paragraph of the proof of Theorem 18.10.4 that applies Proposition 16.7.8 to the groups An rather than to the groups Sn proves that Ksymm is Hilbertian if K is. However, in contrast to Ksymm , we do not know if Kalt is PAC. Problem 18.10.6 ([Fried-V¨ olklein2, p. 476]): Let K be a Hilbertian field. Is Kalt PAC? ˜ . Then M Definition 18.10.6: Let M be a field and let w be a valuation of M is said to have the density property with respect to w if V (M ) is w-dense ˜ ) for every variety V defined over M . in V (M  As an application of Lemma 18.9.3 (using only the case where the ground field is PAC) [Fried-Jarden1, p. 785] proves the following strong density prop˜ erty for almost all fields K(σ): Theorem 18.10.7: Let K be a countable Hilbertian field equipped with a ˜ by W and let e be a valuation v. Denote the set of all extensions of v to K e positive integer. Then for almost all σ ∈ Gal(K) , the field Ks (σ) has the density property with respect to each w ∈ W . ˜, If a field M has the density property with respect to a valuation w of M then M is PAC, so Theorem 18.10.4 strengthens Theorem 18.6.1. Problem 11.5.4 asks whether every PAC field has the density property with respect to ˜. every valuation w of M

18.11 Algebraic Groups In this section, K denotes a field finitely generated over its prime field. In particular, if K is infinite, then K is Hilbertian (Theorem 13.4.2). We list some results about algebraic groups over the fields Ks (σ) partially extending the PAC Nullstellensatz. In this survey we assume familiarity with the concepts of an algebraic group and of an Abelian variety. For an algebraic

398

Chapter 18. The Haar Measure

group G defined over K, we let Gtor (K) be the set of all p ∈ G(K) of fi(K) = {p ∈ Gal(K) | pn = 1}, and for each prime number nite order, GnS ∞ l, Gl∞ (K) = i=1 Gli (K). In particular, if A is an Abelian variety defined over an extension L of K, then A(L) is an Abelian group and rank(A(L)) is its rank as a Z-module. Theorem 18.11.1 ([Frey-Jarden]): Suppose K is infinite. Let e be a positive integer. Then almost all σ ∈ Gal(K)e have this property: For each Abelian variety A of positive dimension defined over Ks (σ), rank(A(Ks (σ))) = ∞. Thus, A(Ks (σ)) contains points a1 , a2 , a3 , . . . which are linearly independent over Z. The celebrated Mordell-Weil Theorem asserts that A(K) is a finitely generated Abelian group [Lang3, p. 71, Thm. 1]. In particular, rank(A(K)) is finite and Ator (K) is a finite group. By Theorem 18.11.1, Ks (σ) is a “large algebraic extension of K” as far as the ranks of Abelian groups are concerned. In contrast, the following conjecture draws a line between 1 and the integers ˜ greater than 1. Whereas, K(σ) are “large” with respect to torsion if e = 1, they seem to be “small” if e ≥ 2. In what follows we let l vary on the set of prime numbers. Conjecture 18.11.2 ([Geyer-Jarden1, p. 620]): For almost all ˜ σ ∈ Gal(K)e each Abelian variety A with dim(A) ≥ 1 defined over K(σ) has these properties: ˜ 6= 0. (a) If e = 1, there exist infinitely many l’s with Al (K(σ)) ˜ (b) If e ≥ 2, then there exist only finitely l’s with Al (K(σ)) 6= 0. ˜ is finite for every l. (c) If e = 1, then Al∞ (K(σ)) Remark 18.11.3: Partial results. [Geyer-Jarden1] proves Conjecture 18.11.2 for elliptic curves, that is, when dim(A) = 1. In the general case the conjecture is true if K is finite [Jacobson-Jarden1, p. 114]. The article [JacobsonJarden2] proves part (c) of the conjecture in each case and part (b) for char(K) = 0. [Geyer-Jarden6] proves a weak version of (a) when K is a number field and A is defined over K: There exists a finite Galois extension L over K such that almost all σ ∈ Gal(L)e satisfy (a), (b), and (c) of Conjecture 18.11.2. Moreover, if End(A) = Z and dim(A) is either odd or 2, or 6, L may be taken to be K. All other cases of the conjecture are still open.  Analogous results hold for linear algebraic groups: Theorem 18.11.4 ([Jarden7, Thms. B and E]): For almost all σ ∈ Gal(K)e ˜ each linear algebraic group G defined over K(σ) has the following properties: ˜ is not bounded, then there (a) If e = 1 and the order of the torsion of G(K) ˜ 6= 1. are infinitely many l’s with Gl (K(σ)) ˜ (b) If e ≥ 2, then the order of the torsion of G(K(σ)) is bounded. For a later use we prove Theorem 18.11.4 in the case where G is the multiplicative group Gm of a field. In this case the torsion part is the group

18.11 Algebraic Groups

399

U of roots of unity. Lemma 18.11.5: Let K be a finitely generated extension of Q and e a positive integer. Then the following statements hold for almost all σ ∈ Gal(K)e : ˜ (a) If e = 1, then there are infinitely many l’s with ζl ∈ K(σ). ˜ (b) If e ≥ 2, there are only finitely many l’s with ζl ∈ K(σ). (c) If e ≥ 1, then for each l there are only finitely many positive integers i ˜ with ζli ∈ K(σ). Proof of (a) and (b): Let Se (K) be the set of all σ ∈ Gal(K)e for which ˜ For K = Q the series there are infinitely many l with ζl ∈ K(σ). X l≥3

X 1 1 = [Q(ζl ) : Q]e (l − 1)e l≥3

diverges if e = 1 and converges if e ≥ 2. In addition, the fields Q(ζl ), with l ranging over all prime numbers, are linearly disjoint over Q (Example 2.5.9). Hence, by Lemma 18.5.3, almost all σ ∈ Gal(K) belong to infinitely many Gal(Q(ζl )) and almost no σ ∈ Gal(K)e belong to infinitely many Gal(Q(ζl ))e if e ≥ 2. Thus, µQ (S1 (Q)) = 1 and µQ (Se (Q)) = 0 if e ≥ 2. If K is a finite extension of Q, then Se (K) = Se (Q) ∩ Gal(K)e . Hence, µK (S1 (K)) = 1 and µK (Se (K)) = 0 if e ≥ 2. Finally, in the general case, ˜ ∩ K is a finite extension of Q. Let ρ: Gal(K)e → Gal(K0 )e be the K0 = Q ˜ if and only restriction map. If σ ∈ Gal(K)e and σ0 = ρ(σ), then ζl ∈ K(σ) −1 ˜ if ζl ∈ K0 (σ0 ). Hence, Se (K) = ρ (Se (K0 )). By Proposition 18.2.2 and the preceding paragraph, µK (S1 (K)) = 1 and µK (Se (K)) = 0 if e ≥ 2. Proof of (c): Since the intersection of countably many sets of measure 1 is a set of measure 1, it suffices to fix l and to prove that for almost all ˜ σ ∈ Gal(K)e there are S∞only finitely many i’s with ζli ∈ K(σ). Put K(ζl∞ ) = i=1 K(ζli ). Since [Q(ζli ) : Q] is (l − 1)li−1 if l 6= 2 and i−2 if l = 2 and i ≥ 2, the field Q(ζl∞ ) is an infinite extension of Q. Since 2 ˜ ∩ K is a finite extension of Q, the field K(ζl∞ ) is an infinite extension Q of K. Hence, µK (Gal(K(ζl∞ ))e ) = 0 (Lemma 18.1.1). If σ ∈ Gal(K)e and ˜ ˜ then ζli ∈ K(σ) for all i, so there are infinitely many i’s with ζli ∈ K(σ), e σ ∈ Gal(K(ζl∞ )) . Consequently, for almost all σ ∈ Gal(K)e there are only ˜  finitely many i’s with ζli ∈ K(σ). We denote the group of roots of unity of a field F by U (F ). Lemma 18.11.6: Let F be a field. Suppose ζl ∈ F for only finitely many l’s. For each l suppose in addition that F contains ζli for only finitely many i’s. Then U (F ) is finite. Proof: Denote the set of all positive integers n with ζn ∈ F by N . If n ∈ N and li |n, then ζli ∈ F . Hence, by assumption, only finitely many prime numbers l divide integers n belonging to N . Moreover, the powers of those l’s are bounded. Therefore, N is finite. 

400

Chapter 18. The Haar Measure

˜ Theorem 18.11.7: For almost all σ ∈ Gal(K)e , U (K(σ)) is infinite if e = 1 and finite if e ≥ 2. Proof: If K is a finitely generated extension of Q, the theorem is a consequence of Lemmas 18.11.5 and 18.11.6. Suppose K is a finitely gener˜ p ∩ K is a finite field. For almost all ated extension of Fp . Then K0 = F e ˜ σ ∈ Gal(K0 ) the field K0 (σ) is infinite if e = 1 and finite if e ≥ 2 (Corollary ˜ 0 (σ)× . Applying Proposition 18.2.2 to the map ˜ 0 (σ)) = K 18.5.9). Also, U (K e e res: Gal(K) → Gal(K0 ) , we conclude that for almost all σ ∈ Gal(K)e , ˜ U (K(σ)) is infinite if e = 1 and finite if e ≥ 2. 

Exercises 1. Let B1 , B2 , B3 , . . . be measurableSsubsets of aPprofinite group G. Suppose ∞ ∞ µ(Bi ∩ Bj ) = 0 for i 6= j. Prove: µ i=1 Bi ) = i=1 µ(Bi ). 2. Let µ be the Haar measure on a profinite group G. Prove that µ(A−1 ) = µ(A) for every measurable subset A of G. Hint: Use the uniqueness of µ. 3. Let L/K be a finite Galois extension and σ0 ∈ Gal(L/K)e . Consider a linearly disjoint sequence of finite extensions P∞M1 , M2 , M3 , . . . of L satisfying this: Mi /K is Galois, i = 1, 2, 3, . . . and i=1 [Mi : K]−e = ∞. Let σi ∈ e e Gal(Mi /K) S∞with resL (σi ) = σ0 . Put Ai = {σ ∈ Gal(K) | resMi (σ) = σi }. Prove: µ( i=1 Ai ) = [L : K]−e . Hint: Extend σ0 to ρ ∈ Gal(K)e and consider the sets ρ−1 Ai , i = 1, 2, 3, . . . . 4. Follow the following outline to give an alternative proof of Theorem 18.6.1 based on Exercise 3 of Chapter 11 rather than on the stronger Theorem 11.2.3. (a) Consider a finite separable extension L of K, an absolutely irreducible polynomial f ∈ L[T1 , . . . , Tr , X], separable in X, and a nonzero g ∈ L[T1 , . . . , Tr ]. Denote the set of σ ∈ Gal(K)e for which there are a1 , . . . , ar , b in Ks (σ) with f (a, b) = 0 and g(a) 6= 0 by S(L, f, g). Imitate the construction of Part A of the proof of Theorem 18.6.1 to prove that S(L, f, g) has measure 1 in Gal(L)e . Deduce from Proposition 18.2.4 that Gal(K)e r S(L, f, g) is a measure zero subset of Gal(K)e . (b) Conclude the proof of Theorem 18.6.1 by considering all possible L, f and g. 5. For K a countable Hilbertian field, consider a finite proper separable extension L0 of K. Order the nonconstant absolutely irreducible polynomials f ∈ K[T, X] in a sequence. Use it to construct a separable algebraic extension L of K which is PAC and linearly disjoint from L0 . In particular, L 6= Ks . 6. Let K be a countable Hilbertian PAC field with char(K) = 0. Let F be as in the proof of Theorem 18.10.2. Order the elements of F in two sequences such that each polynomial f ∈ F appears infinitely many times in both of them. Use these sequences to construct two linearly disjoint Galois extensions N1 and N2 of K such that each of them is Hilbertian and PAC [Dries-Smith].

Notes

401

7. Let K and F be as in Exercise 6. List all finite extensions of K as K1 , K2 , K3 , . . . . Use Lemma 2.5.6 to select suitable subsequences of F and to construct by induction a descending sequence N1 ⊇ N2 ⊇ N3 ⊇ · · · of Galois extensions of K which are Hilbertian T∞ and PAC and such that Ki ∩ Ni = K for i = 1, 2, 3, . . . . Observe that i=1 Ni = K [Dries-Smith]. ˜ 8. Prove: The set of all σ ∈ Gal(Q) with Q(σ) non-PAC is dense in Gal(Q). Hint: Use the arguments of Remark 18.6.2 9. Let B be a subset of positive Haar measure of a profinite group G. Prove: ¯ with rank(G) ¯ ≤ ℵ0 There is an epimorphism π of G onto a profinite group G and µG¯ (π(B)) = µG (B). 10. Let K be a Hilbertian field. Use Corollary 16.3.6 and the proof of the free generators theorem (Theorem 18.5.6) to prove for each positive integer ˆ e for almost all σ ∈ Gal(Kab /K)e . e that hσi ∼ =Z

Notes The proof of the existence of the Haar measure for a profinite group (Proposition 18.2.1) is easier than the proof of the existence of the Haar measure for arbitrary locally compact groups [Halmos, Chap. XI]. It was worked out by Comay. Let G and H be profinite groups and A and B famailies of subsets of G and H, respectively. Following [Halmos1, p. 140] but in contradiction to the usual convention of set theory, [Fried-Jarden5, Sec. 18.4] uses A × B for the σ-algebra generated by all of the sets A × B with A ∈ A and B ∈ B. Section 18.4 uses now the more appropriate notation A ⊗ B. [Ax1, p. 177] proves the free generators theorem (Theorem 18.5.6) in the special case K = Q and e = 1 . His method of proof is restricted to this case and is more in the nature of the proof of Lemma 18.5.8. Theorem 18.6.1 (the PAC Nullstellensatz) was originally proved in [Jarden1] in respond to a problem of [Ax2, p. 269] to produce a proper subfield ˜ which is PAC. of Q The bottom theorem (Theorem 18.7.7) was conjectured in [Jarden3, p. 300] and was proved there for e = 1. A proof for e ≤ 5 is given in [Jarden6]. [Chatzidakis2] proves the bottom theorem for countable Hilbertian fields. A pro-p theoretic analog to the bottom theorem is Serre’s theorem [Serre1]: If a torsion free finitely generated pro-p group G contains an open free pro-p group F , then G is pro-p free. [Stallings] proves an analogous theorem for abstract groups. It is shown in [Jarden3, p. 300] that the profinite analog of the Serre-Stallings theorem would imply the bottom theorem. In [Jarden11] it is proved that if G is a torsion-free profinite group which contains a free profinite subgroup F of rank e and 2 ≤ e < ∞, then (G : F ) divides e − 1. In particular, for e = 2, G = F . So the profinite analog of the Serre-Stalling theorem holds for e = 2.

402

Chapter 18. The Haar Measure

For e 6= 2 the analog is false. Brandis [Jarden11, p. 32] gives a counter example for e = 1. [Haran1] and Melnikov (private communication) give counter examples for e ≥ 3. Theorem 18.8.8(a) was originally proved in [Jarden-Shelah]. Theorem 18.8.8(b) was left in [Jarden-Shelah] as an open problem. It was settled in [Fried-Jarden3] and later also in [Ershov6]. We consider the stability of fields (Theorem 18.9.3) as one of the cornerstones of Field Arithmetic. The proof of this theorem which was started by [Fried-Jarden1] in 1976, and continued in [Fried-Jarden2], [Geyer-Jarden3], [Madan-Madden], and [Geyer2] was brought to a successful end by [Neumann] in 1998.

Chapter 19. Effective Field Theory and Algebraic Geometry Present fashion in field theory and algebraic geometry is to replace classical constructive proofs by elegant existence proofs. For example, it is rare for students to see an actual procedure for factoring polynomials in Q[X1 , . . . , Xn ] in the course of finding out that it is a unique factorization domain. But constructive factorization is the essential backbone of constructive demonstrations that every K-closed set is the union of finitely many K-varieties, (i.e. Hilbert’s basis theorem) and that every K-variety can be normalized. Since one of the goals of this book is to present decision procedures for theories of PAC fields, we require a preliminary concept, “a presented field with elimination theory”, in order to display the essential (for our purposes), basic explicit operations of field theory and algebraic geometry. In particular, all fields finitely generated over their prime fields, and their algebraic closures, are fields with elimination theory (Corollary 19.2.10).

19.1 Presented Rings and Fields The basic structures are the prime fields Q, Fp , finite algebraic extensions of these fields, the ring Z and rings of polynomials over these rings. We consider elements of these rings as recognizable and we may perform computations with them effectively. To make this precise we now employ the concept of primitive recursive functions (Sections 8.3 and 8.4). Consider a sequence (ξ1 , ξ2 , ξ3 , . . .) of symbols. Define polynomial words inductively: Each element of Z and each ξi is a polynomial word. If t1 and t2 are polynomial words and n ∈ Z, then n · t1 , (t1 + t2 ), and (t1 · t2 ) are polynomial words. Denote the set of all formal quotients of polynomial words by Λ. For example, ((ξ1 + ξ2 ) + ξ3 ) and (ξ1 + (ξ2 + ξ3 )) are two distinct polynomial words and ((5 · ξ1 ) + (ξ2 · ξ2 ))/(ξ2 + (−1 · ξ2 )) is an element of Λ. Writing each n ∈ N in its decimal form and ξi as ξ[i], we view Λ as a subset of the set Λ0 of all finite strings in the finite alphabet (α1 , . . . , α19 ) = (0, 1, . . . , 9, ξ, /, +, ·, −, (, ), [, ]). A G¨ odel numbering (or a primitive recursive indexing [Rabin]) on Λ0 (resp. Λ) is given by the injective map i0 : Λ0 → N (resp. its restriction i to Λ) defined by (1)

j(1) j(2) p2

i0 (αj(1) αj(2) · · · αj(n) ) = p1

· · · pj(n) n ,

where 2 = p1 < p2 < p3 < · · · is the sequence of prime numbers. Then i(Λ) is a primitive recursive subset of N. For each n ∈ N, let i(n) : Λn → Nn be the nth power if i.

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To each function f : Λn → Λ there corresponds a unique function f 0 : Nn → N such that f 0 ◦ i(n) = i ◦ f and f 0 is identically 1 on Nn r i(n) (Λ(n) ). Call f primitive recursive if f 0 is primitive recursive. Similarly, a subset ∆ of Λn is primitive recursive if i(n) (∆) is primitive recursive. The following sets and functions over Λ are primitive recursive: (2a) N = {1, 2, . . .} and its primitive recursive functions and sets. (2b) Z = {0, ±1, ±2, . . .}. (2c) All sets {ξj | j ∈ S} with S a primitive recursive subset of N. (2d) The set Π of all polynomial words. (2e) For each n ∈ N, the set Πn of all polynomial words in ξ1 , . . . , ξn . (2f) The function from Π to N that maps π ∈ Π to the smallest n with π ∈ Πn . P ai ξ1i1 ξ2i2 · · · ξnin having a (2g) The subset Π0n of Πn of all polynomial words canonical form, where ai ∈ Z, the i1 , i2 , . . . , in are nonnegative integers, and the monomials are ordered, say, lexicographically. (2h) The function that maps π ∈ Πn onto its canonical form in Π0n . (2i) The various degree functions on Π0n . (2j) The various coefficient functions on Πn . (2k) Addition and multiplication on the union of the sets Π0n , n = 1, 2, . . . . In short, the usual information on polynomials is available by means of primitive recursive functions. Refer to all above mentioned functions and sets, as well as the constant functions and projections from Λn onto Λ, as basic functions and sets. These give the minimal framework for field theoretic operations. Definition 19.1.1: A ring R is said to be presented if there exists an injective map j: R → Λ such that j(R) is a primitive recursive subset of Λ and addition and multiplication are primitive recursive functions over R (via j). In addition the primitive recursive construction of this data from basic functions should be “given explicitly” (i.e. in a naive sense suitable for practical purposes).  A field F is presented if, in addition, the characteristic of F is “given explicitly”, and the inverse function on F × is a primitive recursive function, “explicitly given” in terms of basic functions. An integral domain R is presented in its quotient field F if both R and F are presented with R a primitive recursive subset of F . Note that a presented ring is countable and that Z, Q, and Fp can be presented. In order to work with polynomials over presented rings, introduce the set Σ of all polynomial words in X1 , X2 , X3 , . . . with coefficients in Λ. Define primitive recursive functions as above. In particular, Λ should be a primitive recursive subset of Σ. For R a presented ring with a presentation j: R → Λ, extend j to an embedding of S = R[X1 , X2 , X3 , . . .] into Σ by mapping each polynomial in S to its canonical form in Σ. Thus, we may refer to primitive recursive functions on S.

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An effective algorithm over a presented ring R is an explicitly given (in terms of basic functions) primitive recursive map λ: A → B, where A and B are explicitly given primitive recursive subsets of S n and S m , respectively. Definition 19.1.2: A presented field K is said to have the splitting algorithm if K has an effective algorithm for factoring each elements of K[X] into a product of irreducible factors. Lemma 19.1.3 shows that Q and each  of the fields Fp have a splitting algorithm. Of course, sole reliance on this definition would result in tedious descriptions of algorithms. In practice we work in terms of previously defined algorithms by using basic operations such as composition, induction, case distinction, etc. but not the unbounded µ operator (Definition 8.5.1). We accept without proof the effectiveness of the following well known algorithms: (3a) Finding the remainder and quotient of the division of two elements of N. (3b) Writing an element in N as a product of prime powers. (3c) The Euclidean algorithm to find the greatest common divisor of polynomials over a presented field. (3d) The Gauss elimination procedure to solve systems of linear equations over a presented field. (3e) For K a presented infinite field and f ∈ K[X1 , . . . , Xn ] an explicitly given nonzero polynomial, finding a1 , . . . , an with f (a1 , . . . , an ) 6= 0. The following lemma of Kronecker gives the foundational effectiveness results in algebra: Lemma 19.1.3: The following algorithms are effective: (a) Factoring an element of Z[X] into a product of irreducible factors. (b) Factoring an element of Q[X] into a product of irreducible factors. (c) Factoring an element of K[X1 , . . . , Xn ] into a product of irreducible factors, with K a presented field with a splitting algorithm. Proof of (a): Given a polynomial f ∈ Z[X] of degree n, factor out the greatest common divisor of its coefficients to assume f is primitive. It suffices to decide for each 0 < m < n if f has a factor g ∈ Z[X] of degree m and in the affirmative case to compute g. To this end, compute all factors of f (0), f (1), . . . , f (m) (include ±1). Each (m + 1)-tuple of possibilities for g(0), g(1), . . . , g(m) arises by choosing a factor of f (i), i = 0, . . . , m, for g(i). Each such possibility yields a unique polynomial b0 + b1 X + · · · + bm X m = g1 (X) for which g1 (i) = g(i), i = 0, . . . , m. Indeed, this gives a system of linear equations for b0 , . . . , bm b0 + b1 i + · · · + bm im = g(i),

i = 0, . . . , m,

whose coefficients form a Vandermonde matrix. If b0 , b1 , . . . , bm happen to be in Z, check if g1 (X) divides f (X).

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If none of the above cases leads to a factor of f (X), then f (X) has no factor of degree m. Proof of (b): Write each polynomial in Q[X] as r · f (X), where r ∈ Q and f (X) is a primitive element of Z[X]. By Gauss’ Lemma [Lang7, p. 181, Thm. 2.1] each decomposition of f (X) into irreducible factors in Q[X] corresponds to a decomposition of f (X) in Z[X]. Now apply (a). Proof of (c): The Kronecker substitution Sd (Section 11.3) applied to f ∈ K[X1 , . . . , Xn ] with d > degXi (f ), i = 1, . . . , n, gives Sd (f ) ∈ K[Y ]. If Sd (f ) has a nontrivial monic irreducible factor p ∈ K[Y ], then there is a unique polynomial g ∈ K[X1 , . . . , Xn ] with Sd (g) = p. Check if g divides f . If g fails to divide f for all possible monic p, then f is irreducible. 

19.2 Extensions of Presented Fields We generalize Lemma 19.1.3(b) to fields finitely generated over their prime fields. Let F be a field extension of a presented field K. Call an element x of F presented over K if either x is algebraic over K and irr(x, K) is explicitly given or it is known that x is transcendental over K. Lemma 19.2.1: If an element x is presented over a presented field K, then K(x) is also a presented field. Proof: Let j: K → Λ be the presentation of K. Apply the transformation ξi 7→ ξi+1 , i = 1, 2, 3, . . . , to assume that ξ1 6∈ j(K). First suppose x is algebraic over K of degree n. Use the Euclidean algorithm to assign to each element y of K[x] a polynomial gy in ξ1 of degree at most n − 1 with coefficients in j(K) for which j −1 (gy )(x) = y. This presents K(x) as a ring. If g ∈ K[X] is a nonzero polynomial of degree at most n − 1, then we can write the inverse of g(x) as cn−1 xn−1 + · · · + c0 , with c0 , . . . , cn−1 ∈ K, as follows: Write g(x)(cn−1 xn−1 +· · ·+c0 ) as a polynomial h(x) of degree at most n − 1. Then, with h(x) set equal to 1, solve the system of n linear equations in the unknowns c0 , . . . , cn−1 that results from equating the coefficients of the powers of x on both sides. Therefore, K(x) is a presented field. If x is transcendental over K, then assign to each element in K(x) the corresponding function p(ξ1 )/q(ξ1 ), where gcd(p, q) = 1 and q is monic.  The construction of the presentation of K(x) in Lemma 19.2.1 requires a change in the presentation of K. The essential point, however, is that K is a primitive recursive subset of K(x). We say K(x) is presented over K. Let x be a presented algebraic element over K with [K(x) : K] = n. Each y ∈ K(x) can be presented (with respect to x) as y = a0 + a1 x + · · · + an−1 xn−1 , with a0 , a1 , . . . , an−1 presented elements of K. Then we can effectively find the norm, NK(x)/K (y), as the determinant of the matrix, relative to the basis 1, x, . . . , xn−1 , of the linear operator given by multiplication

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407

by y. Thus, the norm function normK(x)/K is primitive recursive. To find irr(y, K), check which of 1, y, y 2 , . . . , y n−1 is the first that linearly depends (over K) on the previous elements. An n-tuple (x1 , x2 , . . . , xn ) is said to be presented over K if xi is presented over K(x1 , . . . , xi−1 ), i = 1, 2, . . . , n. A sequence x1 , x2 , x3 , . . . of elements of an extension F of K is said to be presented over K if the function n 7→ irr(xn , K(x1 , . . . , xn−1 )) is primitive recursive. In both cases K(x1 , x2 , . . .) is said to be presented over K. In this context, for x transcendental over K, define irr(x, K) to be the zero polynomial. In all cases K(x1 , x2 , . . .) is presented over K(x1 , . . . , xm ) for m = 0, 1, 2, . . . . For the case of the infinite sequence, change, if necessary, the presentation j: K 7→ Λ so that infinitely many of ξi do not appear in j(K), say, by using the transformation ξi 7→ ξ2i . Lemma 19.2.2: If K is a field with a splitting algorithm and x is separable (this includes the transcendental case) and presented over K, then K(x) also has a splitting algorithm. Proof: If x is transcendental over K, then Lemma 19.1.3(c) provides an algorithm for factorization in the ring K[x, X]. By Gauss’ Lemma [Lang7, p. 181, Thm. 2.1] this algorithm extends to K(x)[X]. For the remaining three parts of the proof assume x is separably algebraic over K. Part A: Adjoining two more variables. Let f ∈ K(x)[Z] be a monic polynomial of degree m. Consider the following polynomial of K(x)[Y, Z]: (1)

g(Y, Z) = f (Y − xZ) = (Y − xZ)m + g ∗ (Y, Z)

where g ∗ ∈ K(x)[Y, Z] is of total degree < m and g is monic of degree m as a polynomial in Y over K(x, Z). The norm, normK(x)/K (g(Y, Z)) = h(Y, Z), of g is a monic polynomial in Y over K(Z) of degree m · [K(x) : K] of the form h(Y, Z) = normK(x)/K (Y − xZ)m + h∗ (Y, Z), where h∗ ∈ K[Y, Z] is of total degree less than m · [K(x) : K] = deg(h). Decompose h (Lemma 19.1.3(c)) into irreducible factors in K[Y, Z]: h(Y, Z) = h1 (Y, Z) · · · hr (Y, Z). By (1), the highest homogeneous part in Y, Z of each of the factors hi , is a factor of normK(x)/K (Y −xZ)m . Since each such monomial is defined over K, it must be a power of normK(x)/K (Y −xZ). Thus (2)

hi (Y, Z) = normK(x)/K (Y − xZ)mi + hi ∗ (Y, Z)

where 1 ≤ mi ≤ m and hi ∗ ∈ K[Y, Z] has degree smaller than mi · [K(x) : K] = deg(hi ), i = 1, . . . , r. For each i apply the Euclidean algorithm to find the greatest common divisor of g and hi in K(x, Z)[Y ]. Since both g and hi are monic in Y , all

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polynomials involved in the procedure belong to K(x)[Y, Z]. Since g divides the product h of the hi ’s, at least one of the resulting polynomials has positive degree. Thus, for some i, either g divides hi or 0 < deg(gcd(g, hi )) < deg(g). Part B: The case where g(Y, Z)q(Y, Z) = hi (Y, Z) with q ∈ K(x)[Y, Z]. Compare the highest homogeneous parts in Y and Z on each side of this equation. By (1) and (2): (3)

(Y − xZ)m q ∗ (Y, Z) = normK(x)/K (Y − xZ)mi

with q ∗ ∈ K(x)[Y, Z] a homogeneous polynomial. From the separability of x over K conclude that the factor Y − xZ appears on the right hand side of (3) with exact multiplicity mi . Hence, m ≤ mi . Therefore, m = mi and h = hi is an irreducible polynomial in K[Y, Z]. It follows that g is an irreducible element of K(x)[Y, Z]. Consequently, f is an irreducible element of K(x)[X]. Part C: There is an i with 0 < deg(gcd(g, hi )) < deg(g). Then g(Y, Z) = g1 (Y, Z)g2 (Y, Z) with g1 , g2 ∈ K(x)[Z, Y ] monic in Y , and degY (g1 ), degY (g2 ) > 1. Substitute Z = 0 and Y = X to obtain a nontrivial factorization f (X) = g1 (X, 0)g2 (X, 0) for f in K(x)[X]. Continue by induction on the degree of f to complete the factorization of f into irreducible factors in K(x)[X].  From the next lemma we derive a simple conclusion about a presented ntuple (x1 , . . . , xn ) over a field K: If each xi is separable over K(x1 , . . . , xi−1 ), then the order of the entries of (x1 , . . . , xn ) is unimportant. Lemma 19.2.3: Let (x, y) be a presented pair over a field K with a splitting algorithm. Then we may effectively present y over K. If K(y) has a splitting algorithm, then we may effectively present x over K(y). Proof: First assume x is transcendental over K. If irr(y, K(x)) ∈ K[X], then irr(y, K) = irr(y, K(x)) and x is transcendental over K(y). Otherwise, y is transcendental over K and irr(y, K(x)) involves x. Write irr(y, K(x)) as a monic polynomial in Y with coefficients in K(x). Multiply irr(y, K(x)) with the least common multiple of the denominators of the coefficients to obtain an irreducible polynomial f (x, Y ) in K[x, Y ]. Now replace x by X and Y by y and rewrite f (X, y) as a polynomial in X with coefficients in K[y]. Finally, divide f (X, y) by the leading coefficient to obtain irr(x, K(y)). Now assume x is algebraic over K. If y is transcendental over K(x), then y is transcendental over K, so assume y is algebraic over K(x). Then irr(y, K)|NK(x)/K (irr(y, K(x)). Decompose NK(x)/K (irr(y, K(x)) into irreducible factors over K and check which of the factors is zero when evaluated at y. This factor will be irr(y, K). Also, irr(x, K(y)) divides irr(x, K). If K(y) has a splitting algorithm, we can factor irr(x, K) over K(y) and check which of the factors is annihilated by x. This factor will be irr(x, K(y)). 

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It may, however, happen that for a presented n-tuple (x1 , . . . , xn ) over a field K the field extension K(x1 , . . . , xn )/K is separable but there exists i with xi inseparable over K(x1 , . . . , xi−1 ). The next lemma handles this inconvenient situation: Lemma 19.2.4: Let x = (x1 , . . . , xn ) be a presented n-tuple over a field K with a splitting algorithm such that K(x)/K is separable. Then: (a) For each explicitly given permutation π of {1, 2, . . . , n} we can present (xπ(1) , . . . , xπ(n) ) over K. (b) We can effectively find π and r ≤ n such that {xπ(1) , . . . , xπ(r) } is a separating transcendence base for K(x)/K. (c) K(x) has a splitting algorithm. Proof: Statement (c) follows from (a) and (b) by Lemma 19.2.2. To prove (a) and (b), we do an induction on n and let π be a permutation of {1, 2, . . . , n}. Since π is a product of permutations of two successive numbers we may assume π permutes k and k + 1, and leaves all other i unchanged. By assumption, (xk , xk+1 ) is presented over K(x1 , . . . , xk−1 ). By induction, K(x1 , . . . , xk−1 ) has a splitting algorithm. Hence, by Lemma 19.2.3, we can effectively present xk+1 over K(x1 , . . . , xk−1 ). Again, by induction, K(x1 , . . . , xk−1 , xk+1 ) has a splitting algorithm. Therefore, by Lemma 19.2.3, we can effectively present xk over K(x1 , . . . , xk−1 , xk+1 ). Consequently, (xπ(1) , . . . , xπ(n) ) can be presented over K. To construct a separating transcendence base for K(x)/K find a maximal sequence of integers τ1 < · · · < τr between 1 and n such that xτi is transcendental over K(x1 , . . . , xτi −1 ), i = 1, . . . , r. Then {xτ1 , . . . , xτr } is a transcendence base for K(x)/K. By the preceding paragraph, we may reorder the xi ’s to assume x1 , . . . , xr is a transcendence base for K(x)/K. If r = n or if char(K) = 0, we are done, so assume char(K) = p and r < f is a polyn. Find f ∈ K[X1 , . . . , Xn ] of minimal degree with f (x) = 0. PIf m nomial in X1p , . . . , Xnp , then f (x) = 0 is a nontrivial relation i=1 ci Mip = 0, with M1 , . . . , Mm monomials in x1 , . . . , xn and c1 , . . . , cm ∈ K. Since K(x) and K 1/p are linearly disjoint over K, the linear dependence of M1 , . . . , Mm over K 1/p gives a linear dependence of M1 , . . . , Mm over K. This gives an equation g(x) = 0 with g ∈ K[X1 , . . . , Xn ] of lower degree than f . This is a contradiction. It follows that there is some Xi , say X1 , that appears in f to a power not divisible by p. The element x1 is separably algebraic over K(x2 , . . . , xn ), and we may find irr(x1 , K(x2 , . . . , xn )). Now, use the induction hypothesis to find a permutation π of {2, . . . , n} such that (xπ(2) , . . . , xπ(r+1) ) is a separating transcendence base for  K(x2 , . . . , xn )/K. Lemma 19.2.5: Let E and F be algebraically independent extensions of a field K. Suppose E/K is separable. Then: (a) EF/F is separable.

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(b) If in addition F/K is separable, then EF/K is separable. Proof: Statement (b) follows from (a) by Corollary 2.6.2(a). In order to prove (a), we may assume E/K is finitely generated. Let t1 , . . . , tr be a separating transcendence base for E/K. Then t1 , . . . , tr are algebraically independent over F and EF/F (t) is separable algebraic. Therefore, EF/F is separable.  Lemma 19.2.6: Let x = (x1 , . . . , xm ) and y = (y1 , . . . , yn ) be presented tuples over a field K with a splitting algorithm. Suppose K(x) and K(y) are separable over K and algebraically independent. Then (x, y) can be presented over K and K(x, y) has a splitting algorithm. Proof: Suppose we have already presented (x, y) over K. By Lemma 19.2.5, K(x, y) is separable over K. Hence, by Lemma 19.2.4, K(x, y) has a splitting algorithm. Use induction on n to present (x, y) over K. First suppose, n = 1. If y1 is transcendental over K, then y1 is also transcendental over K(x). Suppose y1 is algebraic over K. By Lemma 19.2.4, K(x) has a splitting algorithm. Decompose irr(y1 , K(x)) into irreducible factors over K(x) and check which has y1 as a root. This will be irr(y1 , K(x)). Now suppose n > 1 and there is a presentation of (x, y1 , . . . , yn−1 ) over K. Lemma 19.2.4 presents (x, y1 , . . . , yn−1 ) over K(y1 , . . . , yn−1 ). Since K(y) is algebraically independent from K(x) over K, it is algebraically independent from K(x, y1 , . . . , yn−1 ) over K(y1 , . . . , yn−1 ) (Section 2.6, discussion preceding Lemma 2.6.5). The case n = 1 gives a presentation of (x, y) over K(y), hence also over K.  Every field extension of a perfect field is separable. This yields another application of Lemma 19.2.4: Lemma 19.2.7: Let K be a perfect field with a splitting algorithm. If an ntuple (x1 , . . . , xn ) is presented over K, then L = K(x1 , . . . , xn ) has a splitting algorithm. Thus, every finitely generated presented extension of K has a splitting algorithm. This applies, in particular, to the case where K is a finite field. Definition 19.2.8: A presented field K is said to have an elimination theory if every finitely generated presented extension F of K has a splitting algorithm. Consequently, F itself has an elimination theory.  Lemma 19.2.9: Let (x1 , . . . , xn ) be a presented n-tuple over a field K with an elimination theory and π a permutation of (1, 2, . . . , n). Then we can present (xπ(1) , . . . , xπ(n) ) over K. Moreover, we can find π such that xπ(1) , . . . , xπ(r) is a transcendence base of K(x)/K. Proof: As in the proof of Lemma 19.2.4, we may assume that π permutes k and k + 1 and fixes all other i. By assumption, (x1 , . . . , xk−1 ) is presented

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over K. Since K has elimination theory, K(x1 , . . . , xk−1 ) has a splitting algorithm. By assumption, (xk , xk+1 ) is presented over K(x1 , . . . , xk−1 ). Hence, by Lemma 19.2.3, xk can be presented over K(x1 , . . . , xk−1 , xk+1 ). Again, by assumption, (xk+2 , . . . , xn ) is presented over K(x1 , . . . , xn+1 ). Hence,  (x1 , . . . , xk−1 , xk+1 , xk , xk+2 , . . . , xn ) is presented over K. Corollary 19.2.10: Every presented perfect field K with a splitting algorithm has an elimination theory. Hence, every presented field K, finitely generated over its prime field, has an elimination theory. Remark 19.2.11: The phrase “elimination theory” derives from the observation that there are effective versions of the classical elimination theory algorithms of algebraic geometry if and only if K satisfies the condition of Definition 19.2.8. Note, that in contrast to Lemma 19.2.7, there are fields with a splitting algorithm which do not have an elimination theory [Fr¨ohlich-Shepherdson, Thm. 7.27]. Such fields have finite purely inseparable extensions that do not have a splitting algorithm. In addition, a presented infinite algebraic extension of Q need not have a splitting algorithm [Fr¨ohlich-Shepherdson, Thm. 7.12]. 

19.3 Galois Extensions of Presented Fields Among the infinite extensions of a presented field which can be presented are the algebraic closure and the separable closure. First we give a constructive version of the primitive element theorem: Lemma 19.3.1 (Kronecker): Let L = K(x1 , . . . , xn ) be a separable algebraic presented extension of a field K with a splitting algorithm. Then one can effectively find z ∈ L such that K(z) = L. Proof: It suffices to consider the case where K is infinite: Let T1 , . . . , Tn be indeterminates and consider the element x = T1 x1 + · · · + Tn xn from the rational function field L(T). Compute a nonzero polynomial f ∈ K[T, X] of ∂f (T, x) 6= lowest degree with f (T, x) = 0. Since x is separable over K(T), ∂X ∂f 0. Since K is infinite, we can find a1 , . . . , an ∈ K with ∂X (a, a1 x1 + · · · + an xn ) 6= 0. With z = a1 x1 + · · · + an xn , express x1 , . . . , xn as elements of K(z). Indeed, the partial derivative with respect to Ti of the identity f (T, T1 x1 + · · · + Tn xn ) = 0 gives ∂f ∂f (T, T1 x1 + · · · + Tn xn ) = 0, (T, T1 x1 + · · · + Tn xn ) + xi · ∂Ti ∂X i = 1, . . . , n. Substituting a1 , . . . , an for T1 , . . . , Tn , respectively, we conclude ∂f ∂f (a, z)/ ∂X (a, z), i = 1, . . . , n. Thus, K(z) = L.  that xi = − ∂T i

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Lemma 19.3.2: Given a separable polynomial f (X) of degree n over a field K with a splitting algorithm, we may present the n-tuple (x1 , . . . , xn ) of all roots of f (X), and therefore the splitting field L of f (X) over K. Moreover, we can effectively compute the Galois group G = Gal(f, K), through its action on L, as a group of permutations of {x1 , . . . , xn }. If H is a presented subgroup of G, then we can compute the fixed field L(H) of H acting on L. Note: An effective computation of Gal(f, K) is a primitive recursive map f 7→ Gal(f, K) from the above polynomials in K[X] of degree n into subgroups of Sn . Proof: Let u be transcendental over K. Then Gal(f, K) = Gal(f, K(u)). Replacing K by K(u) if necessary, we may assume K is infinite. Let g ∈ K[X] be an irreducible factor of f (X) having x1 as a zero. Then (Lemma 19.2.2) K(x1 ) has a splitting algorithm. Proceeding by induction we may present the splitting field of f (X)/(X − x1 ) over K(x1 ), hence the splitting field of f over K. To compute Gal(f, K) we may assume f has no multiple roots. Consider the element x = x1 T1 + . . . + xn Tn which appears in the proof of Lemma 19.3.1. Let E = K(T), F = L(T). Consider each element σ of Gal(L/K) as an element of Sn through the formula σ(xi ) = xσ(i) , i = 1, . . . , n. On the other hand, σ uniquely extends to an element of Gal(F/E). Embed Sn into Aut(F/L) through the formula (Ti )π = Tπ−1 (i) for π ∈ Sn and i = 1, . . . , n. Thus, for σ ∈ Gal(L/K), σ(x) = xσ(1) T1 + · · · + xσ(n) Tn = x1 Tσ−1 (1) + · · · + xn Tσ−1 (n) = xσ . Compute the polynomial h(X) = irr(x, E). Act on h(X) by Sn from the right through the action on the indeterminates. We show that Gal(f, K) = {π ∈ Sn | hπ = h}. Indeed, if hπ = h for some π ∈ Sn , then x and xπ are conjugate over E. Hence, there exists σ ∈ Gal(f, K) with xσ = σ(x) = xπ . Thus, Tσ−1 (i) = Tπ−1 (i) , i = 1, . . . , n. Therefore, σ = π. Conversely, if σ ∈ Gal(f, K), then x and σ(x) are conjugate over E. Hence, hσ = irr(x, E) = h. Finally, let H be a subgroup of G. To find L(H), we may assume f is . . , xm } be an irreducible in K[X] and L = K(x1 ) (Lemma 19.3.1). Let {x1 , .Q m H-orbit in the action of H on {x1 , . . . , xn }. Compute p(X) = i=1 (X − xi ) and adjoin its coefficients to K. The field K 0 so obtained is in L(H). In addition, L = K 0 (x1 ). Hence, [L : K 0 ] ≤ m = |H|. Consequently, K 0 = L(H). 

19.4 The Algebraic and Separable Closures of Presented Fields We show how to present the algebraic and separable closures of fields with elimination theory:

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Lemma 19.4.1: If K is a field with a splitting algorithm, then we can effectively find a sequence x1 , x2 , x3 , . . . of elements of Ks that presents Ks over K as a field with a splitting algorithm. If K has an elimination theory, then ˜ has one too. K Proof: Order the set of all nonconstant separable polynomials of K[X] in a primitive recursive sequence p1 , p2 , p3 , . . . . Inductively construct a sequence x1 , x2 , x3 , . . . of separable algebraic elements over K, a sequence n1 ≤ n2 ≤ n3 ≤ · · · of positive integers, and a sequence K ⊆ K1 ⊆ K2 ⊆ · · · of field extensions such that Kk = K(x1 , x2 , . . . , xnk ) is the splitting field of p1 . . . pk . Then K(x1 , x2 , x3 , . . .) is the separable closure of K and it is presented by the sequence x1 , x2 , x3 , . . . . We must still show how to factor polynomials over Ks into irreducible factors. Let f ∈ Ks [X] be a nonconstant polynomial. Then f has coefficients for some j. Compute g(X) = normL/K (f (X)). Factor in L = K(x1 , . . . , xj ),Q r g over K, g(X) = i=1 gi (X qi ), with gi ∈ K[X] separable, irreducible, and qi a power of char(K) (or qi = 1 if char(K) = 0), i = 1, . . . , r. Then find k ≥ j with g1 , . . . , gr ∈ {p1 , . . . , pk }. The factorization of f (X) into irreducible polynomials over Kk is the desired factorization of f over Ks . If K is a field with elimination theory, an analogous construction, with p1 , p2 , p3 , . . . now all nonconstant irreducible polynomials of K[X], proves ˜ has a splitting algorithm. Since K ˜ is perfect, K ˜ is a field with that K elimination theory.  Relative algebraic closures also occur in several constructions: Lemma 19.4.2: Let K be a field with a splitting algorithm and let F = K(x1 , . . . , xn ) be a separable presented extension of K. Then one can effectively find the separable algebraic closure L of K in F . Proof: Use Lemma 19.2.4 to rearrange (x1 , . . . , xn ) so that (x1 , . . . , xr ) is a separating transcendence base for the extension F/K. By Lemma 19.3.1 we may assume n = r + 1. Then f (X) = irr(xr+1 , L(x1 , . . . , xr )) is also irr(xr+1 , Ks (x1 , . . . , xr )). Apply Lemma 19.4.1 to compute f (X) effectively. Write its coefficients (x1 ,...,xr ) , where pi , qi are relatively prime polynomials as rational functions pqii(x 1 ,...,xr ) in Ks [X1 , . . . , Xr ] and one of the coefficients of qi is 1, i = 0, . . . , k − 1. Then the coefficients b1 , . . . , bm of p0 , q0 , . . . , pk−1 , qk−1 , belong to L. Since F is linearly disjoint from Ks over K(b1 , . . . , bm ), we have L = K(b1 , . . . , bm ). 

19.5 Constructive Algebraic Geometry Throughout this section we fix a field K with elimination theory. We show how to carry out the basic operations of algebraic geometry over K:

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Lemma 19.5.1 (Noether Normalization Theorem): Let (x1 , . . . , xn ) be a presented n-tuple over K and assume that K is infinite. Then we can effectively compute the transcendence degree r of K(x) over K. Moreover, we can effectively find linear forms (1)

ti =

n X

aij xj ,

i = 1, . . . , n

j=1

with aij ∈ K and det(aij ) 6= 0 such that t1 , . . . , tr is a transcendence base for K(x)/K, and t1 , . . . , tn (as well as x1 , . . . , xn ) are integral over K[t1 , . . . , tr ] and presented over K(t1 , . . . , tr ). Proof: If each xi is transcendental over K(x1 , . . . , xi−1 ), then take r = n and ti = xi , i = 1, . . . , n. Otherwise, assume xn is algebraic over K(x1 , . . . , xn−1 ) (Lemma 19.2.9). Find an irreducible polynomial f ∈ K[X] such that f (x) = 0 in which Xn Pd occurs. Write f as the sum of its homogeneous parts, f (X) = j=0 fj (X), where fd (X) 6= 0 (so that fd (X1 , . . . , Xn−1 , 1) 6= 0). Find c1 , . . . , cn−1 ∈ K with fd (c1 , . . . , cn−1 , 1) 6= 0 (K is infinite). Then put ti = xi − ci xn , i = 1, . . . , n − 1, and tn = xn . We claim tn is integral over K[t1 , . . . , tn−1 ] and presented over K(t1 , . . . , tn−1 ). PHence,k1so are kxn1 , . . . , xn . Indeed, write fd (X) = k bk X1 · · · Xn , where k1 + · · · + kn = d for each k = (k1 , . . . , kn ). Then the coefficient of Xnd in the transformed polynomial f (T1 + c1 Xn , . . . , Tn−1 + cn−1 Xn , Xn ) = g(T1 , . . . , Tn−1 , Xn ) is P kn−1 = fd (c1 , . . . , cn−1 , 1) 6= 0 and degXn (g) = d. Since bk ck11 · · · cn−1 (t1 , . . . , tn−1 , xn ) is a zero of g(T1 , . . . , Tn−1 , Xn ), we may compute irr(xn , K(t1 , . . . , tn−1 )). Thus, xn = tn and xi = ti + ci xn , i = 1, . . . , n − 1, are integral over K[t1 , . . . , tn−1 ]. Now apply the induction hypothesis to (t1 , . . . , tn−1 ) and use transitivity of integral dependence to conclude the lemma.  Lemma 19.5.2: Let (x1 , . . . , xn ) be a presented n-tuple over K. Then we can effectively find a finite set of polynomials gi ∈ K[X1 , . . . , Xn ], i ∈ I, such that V (gi | i ∈ I) is the K-variety generated by x. Proof: The first two parts of the proof handle the case where K is infinite. The third part reduces the finite case to the infinite case. Part A: Finding the gi when K is infinite. Apply Lemma 19.5.1 to compute the transcendence degree r of K(x)/K. Assume, after applying an invertible linear transformation, that x1 , . . . , xr form a transcendence base for K(x)/K and that x1 , . . . , xn are integral over K[xP 1 , . . . , xr ]. Let t1 , . . . , tn n be algebraically independent over K(x). Then z = i=1 xi ti is integral over K[t, x1 , . . . , xr ]. Thus, we can find a polynomial f (T, X1 , . . . , Xr , Z), irreducible in K[T, X1 , . . . , Xr ,P Z] and monic inP Z, such that f (t, x1 , . . . , xr , z) = n i1 in 0. Write f (T, X1 , . . . , Xr , i=1 Xi Ti ) = i∈I gi (X)T1 · · · Tn , with gi ∈ K[X], i ∈ I.

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Part B: V (gi | i ∈ P I) is the K-variety V generated by x. By definition, f (t, x1 , . . . , xr , z) = i∈I gi (x)ti11 · · · tinn = 0. As t1 , . . . , tn are algebraically independent over K(x), gi (x) = 0 for each i ∈ I. Hence, V ⊆ V (gi | i ∈ I). ˜ n be a common zero of the gi . We show x → x0 Conversely, let x0 ∈ K is a K-specialization. First, extend the K-specialization (t, x1 , . . . , xr ) → (t, x01 , . . . , x0r ) to a K-specialization (2)

(t, x1 , . . . , xr , z) → (t, x01 , . . . , x0r , z 0 )

Pn 0 where z 0 = i=1 xi ti . Indeed, f is irreducible and monic. Hence, if h ∈ K[T, X1 , . . . , Xr , Z] is a polynomial with h(t, x1 , . . . , xr , z) = 0, then h is a multiple of f in K[T, P X1 , . . . , Xr , Z] (by Gauss’ Lemma). In addition, f (t, x01 , . . . , x0r , z 0 ) = i∈I g(x0 )ti11 · · · tinn = 0. Hence, h(t, x0n , . . . , x0r , z 0 ) = 0. Therefore, (2) holds. Now use that xr+1 , . . . , xn are integral over K[x1 , . . . , xr ] to extend (2) to a specialization (3)

(t, x1 , . . . , xr , xr+1 , . . . , xn , z) → (t, x01 , . . . , x0r , x00r+1 , . . . , x00n , z 0 ),

˜ In particular, with x00r+1 , . . . , x00n ∈ K. z 0 = x01 t1 + · · · + x0r tr + x00r+1 tr+1 + · · · + x00n tn . Pn Combining this with z 0 = i=1 x0i ti we conclude that x00i = x0i , i = r+1, . . . , n. This gives the K-specialization x → x0 . Part C: K is finite. Extend K by an element u, transcendental over K(x). By Parts A and B, we can effectively find gi ∈ K(u)[X], for i ∈ I, such that V (gi | i ∈ I)Pis the K(u)-variety generated by x. Without loss write the gi as gi (X) = j∈Ji gij (X)uj , with gij ∈ K[X]. Then V (gij | i ∈ I, j ∈ Ji ) is the K-variety generated by x.  We now present the main theorem of classical elimination theory: Lemma 19.5.3: Let f1 (X, Y ), . . . , fs (X, Y ) be explicitly given nonzero polynomials in K[X1 , . . . , Xn , Y ]. Then we can effectively compute a collection g1 (X), . . . , gr (X) of nonzero polynomials in K[X], called a resultant system of f1 , . . . , fs with respect to Y satisfying: For each x ∈ An , g1 (x) = · · · = gr (x) = 0 if and only if (4a) f1 (x, Y ), . . . , fs (x, Y ) have a common zero in the algebraic closure of K(x); or (4b) the leading coefficient of fi (as a polynomial in Y ) vanishes at x, i = 1, . . . , s. Proof: Let f (X) = a0 X k + · · · + ak and g(X) = b0 X l + · · · + bl be polynomials with coefficients in an integral domain R. Then Resultant(f, g) is the determinant of an explicit (k + l) × (k + l) matrix whose entries are the

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ai ’s, bj ’s and zeros [Lang7, p. 200]. Moreover, Resultant(f, g) = 0 if and only if f and g have a common root (in the algebraic closure of Quot(R)) or a0 = b0 = 0. Returning to the original fi ’s, let di = degY (fi ) and let ai (X) be the leading coefficient of fi (X, Y ) as a polynomial in Y , i = 1, . . . , s. Choose d ≥ di , i = 1, . . . , s and let u1 , . . . , us , v1 , . . . , vs be algebraically independent elements over the universal domain containing K. Compute the resultant of the polynomials

(5)

p(Y ) = u1 Y d−d1 f1 (X, Y ) + · · · + us Y d−ds fs (X, Y ) q(Y ) = v1 (Y − 1)d−d1 f1 (X, Y ) + · · · + vs (Y − 1)d−ds fs (X, Y )

with respect P to Y . It is a polynomial g ∈ K[u, v, X]. Write it in the form r g(u, v, X) = j=1 gj (X)mj (u, v) where the mj (u, v) are distinct monomials Ps in i=1 ui ai (X) is the leading coefficient of p(Y ) while Psu, v. Note that v a (X) is the leading coefficient of q(Y ). i=1 i i Claim: g1 (X), . . . , gr (X) is a resultant system of f1 (X, Y ), . . . , fs (X, Y ) with respect to Y .P Indeed, let x ∈ An . Suppose g1 (x) = · · · = gr (x) = 0. r Then g(u, v, x) = j=1 gj (x)mj (u, v) = 0. Hence, there is a y with (6)

s X i=1

ui y d−di fi (x, y) = p(y) = 0 and

s X

vi (y − 1)d−di fi (x, y) = q(y) = 0

i=1

Ps Ps or i=1 ui ai (x) = 0 and i=1 vi ai (x) = 0. In the latter case a1 (x) = · · · = If the latter case does not as (x) = 0. P Ps occur, there is an i with ai (x) 6= 0. s u a (x) = 6 0 and Therefore, i i i=1 i=1 vi ai (x) 6= 0. Hence, by (6), y ∈ ^ ^ ] K(x, u) ∩ K(x, v) = K(x). Since y 6= 0 or y 6= 1, (6) implies f1 (x, y) = · · · = fs (x, y) = 0. Pr Conversely, suppose (4) holds. Then j=1 gj (x)mj (u, v) = g(u, v, x) = 0. Hence, g1 (x) = · · · = gr (x) = 0.  Remark 19.5.4: If r = 0, then (4a) holds for each x ∈ An for which not all leading coefficients of f1 , . . . , fs vanish. Thus, if r = 0 and x1 , . . . , xn are algebraically independent over K, then f1 (x, Y ), . . . , fs (x, Y ) have a common factor d(x, Y ) in K(x)[Y ] of positive degree. Without loss (Gauss’ lemma) we may take d(x, Y ) to be a primitive nontrivial common factor of  f1 (x, Y ), . . . , fs (x, Y ) in K[x, Y ]. Remark 19.5.5: Universal bounds on degrees. Suppose K in Lemma 19.5.3 is an arbitrary field and f1 , . . . , fs are nonzero polynomials in K[X, Y ]. The proof of Lemma 19.5.3 then proves the existence of the resultant system g1 , . . . , gr ∈ K[X] of f1 , . . . , fs . Moreover, the proof gives a bound on r and deg(g1 ), . . . , deg(gr ) which depends on deg(f1 ), . . . , deg(fs ) but neither on

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417

K nor on the coefficients of f1 , . . . , fs . (We will henceforth say that r and deg(g1 ), . . . , deg(gs ) are bounded.) Indeed, let p and q be as in (5). Put k = deg(p) and l = deg(q). The resultant g(u, v, X) of the polynomials (5) is a (k+l)×(k+l) determinant in the coefficients of f1 (X, Y ), . . . , fs (X, Y ) viewed as polynomials Pr in Y . Therefore, deg(g(u, v, X)) is bounded. By definition, g(u, v, X) = j=1 gj (X)mj (u, v).  Consequently, r and deg(g1 ), . . . , deg(gr ) are bounded. Proposition 19.5.6: Let f1 , . . . , fs be explicitly given nonzero polynomials in K[X1 , . . . , Xn ]. Then we can effectively decompose the K-closed set A = V (f1 , . . . , fs ) into a union A = V1 ∪ · · · ∪ Vm of K-varieties. Furthermore, we can explicitly give a generic point and the dimension of Vi , i = 1, . . . , m. Proof: Adjoin n elements t1 , . . . , tn , algebraically independent over the universal domain containing K, to K. We construct polynomials over K(t) in X1 , . . . , Xn and then eliminate X1 , . . . , Xn , one by one. The components Vi are defined by polynomials that appear in Part D and Part E of this proof. As in the proof of Lemma 19.5.2 we may assume that K is infinite. Part A: Elimination of Xn . Apply a linear transformation to X1 , . . . , Xn to assume (Lemma 19.5.1) that the coefficient of Xn in f1 is 1. Consider this collection of polynomials in K[t, X, Z]: (7a) (7b)

e(t, X, Z) = Z − t1 X1 − · · · − tn Xn ;

and

f1 (X), . . . , fs (X). Compute (Lemma 19.5.3) a resultant system of (7) with respect to Xn :

(8)

g1,j (t, X1 , . . . , Xn−1 , Z),

j = 1, . . . , r1 .

Since the polynomials of (7) have no common factor in K[t, X, Z], Remark 19.5.4 implies r1 ≥ 1. If r1 = 1, let h1 = g1,1 and h2 = · · · = hn = 1. Then the elimination procedure stops here and we skip to the second paragraph of Part D. Otherwise, r1 ≥ 2. Apply (3c) of Section 19.1 and Lemma 19.1.3 in K[t, X1 , . . . , Xn−1 , Z] to find h1 (t, X1 , . . . , Xn−1 , Z) = gcd(g1,1 , . . . , g1,r1 ) and e1,1 , . . . , e1,r1 ∈ K[t, X1 , . . . , Xn−1 , Z] such that g1,j = h1 e1,j , j = 1, . . . , r1 . If Z occurs in e1,j for some j, say in e1,1 , then, with another linear change of variables (not affecting Xn ), we may assume that the leading coefficient of Z in e1,1 is 1. Compute a resultant system of e1,1 , . . . , e1,r1 with respect to Z: (9)

d1j (t, X1 , . . . , Xn−1 ),

j = 1, . . . , m1 .

Consider the collection (9) as polynomials in t and display their nonzero coefficients (in K[X1 , . . . , Xn−1 ]): (10)

f1j (X1 , . . . , Xn−1 ),

j = 1, . . . , s1 .

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Part B: Properties of the collection (10): (11a) If (x, z) is a common zero of the system (7), then (x, z) is a zero of h1 , or of all e1,1 , . . . , e1,r1 , f1,1 , . . . , f1,s1 (Lemma 19.5.3); (11b) If (x1 , . . . , xn−1 , z) is a presented zero of h1 over K(t), then we can present an algebraic zero xn of Pf1 (x1 , . . . , xn−1 , Xn ) n over K(x1 , . . . , xn−1 ), such that x ∈ A and z = i=1 ti xi . (11c) If (x1 , . . . , xn−1 , z) is a presented common zero of all e1,j and f1,j , then we can present an algebraic zero xn ofPf1 (x1 , . . . , xn−1 , Xn ) over n K(x1 , . . . , xn−1 ), such that x ∈ A and z = i=1 ti xi . (11d) The collection (10) is nonempty. Apply Remark 19.5.4 to see (11c). Otherwise, e1,1 , . . . , e1,r1 have a common linear factor in K[t, X1 , . . . , Xn−1 , Z], contrary to their definition. Part C: Elimination of Xn−1 .

Apply Part A to

{e1,1 , . . . , e1,r1 ; f1,1 , . . . , f1,s1 } as it is applied to (7). That is, make a linear transformation so that the leading coefficient of Xn−1 in f1,1 is 1. Then display a resultant system of {e1,1 , . . . , e1,r1 ; f1,1 , . . . , f1,s1 } with respect to Xn−1 : g2j (t, X1 , . . . , Xn−2 , Z),

j = 1, . . . , r2 .

Since e1,1 , . . . , e1,r1 have no common factor, r2 ≥ 1. If r2 = 1, let h2 = g2,1 , h3 = · · · = hn = 1, and skip to Part D. Otherwise, find h2 (t, X1 , . . . , Xn−2 , Z) = gcd(g2,1 , . . . , g2,r2 ) and e2,j (t, X1 , . . . , Xn−2 , Z) such that g2,j = h2 e2,j , j = 1, . . . , r2 . After a suitable linear transformation which makes the leading coefficient of Z in e2,1 equal 1, construct a resultant system, d2,j (t, X1 , . . . , Xn−2 ),

j = 1, . . . , m2 ,

of e2,1 , . . . , e2,r2 with respect to Z. Display the nonzero coefficients of d2,j , j = 1, . . . , m2 , as (12)

f2j (X1 , . . . , Xn−2 ),

j = 1, . . . , s2

(again s2 ≥ 1).

Then the following holds: (13a) If (x, z) is a common zero of the system (7), then (x, z) is a zero of h1 , or of h2 , or of all e2j and f2j . (13b) If (x1 , . . . , xn−2 , z) is a presented zero of h2 over K(t), then we can present algebraic Pnelements xn−1 , xn over K(x1 , . . . , xn−2 ), such that x ∈ A and z = i=1 ti xi . (13c) If (x1 , . . . , xn−2 , z) is a presented common zero of all e1,j and f1,j , then we can present algebraic Pn elements xn−1 , xn over K(x1 , . . . , xn−2 ) such that x ∈ A and z = i=1 ti xi .

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Part D: The induction step. Proceed by induction to find m between 1 and n and for each i between m and n polynomials hi (t, X1 , . . . , Xn−i , Z), eij (t, X1 , . . . , Xn−i ), j = 1, . . . , ri , and fij (X1 , . . . , Xn−i ), j = 1, . . . , si , as in Parts A, B, and C such that rn , . . . , rm+1 ≥ 2 and rm = 1. Let h1 = . . . = hm−1 = 1. These polynomials have the following properties: (14a) If (x, z) is a common zero of the system (7), then (x, z) is a zero of one of the hi ’s. (If m = n, then fn1 is a nonzero constant, so fn1 (x) 6= 0.) (14b) Let 1 ≤ i ≤ n. If (x1 , . . . , xn−i , z) is a presented zero of hi over K(t), then we can present algebraic elements Pn xn−i+1 , . . . , xn over K(x1 , . . . , xn−i ), such that x ∈ A and z = i=1 ti xi . (14c) If (x1 , . . . , xn−i , z) is a presented common zero of all ei,j and fi,j , then we can present algebraic elements xn−i+1 , . . . , xn over K(x1 , . . . , xn−i ) Pn such that x ∈ A and z = i=1 ti xi . Decompose hi into irreducible factors in K[t, X1 , . . . , Xn−i , Z];

(15)

hi (t, X1 , . . . , Xn−i , Z) =

bi Y

hij (t, X1 , . . . , Xn−i , Z)aj ,

j=1

i = 1, . . . , n. For each i and j, display the coefficients of hij (t, X1 , . . . , Xn−i , t1 X1 + . . . + tn Xn ) (as a polynomial in t1 , . . . , tn ): (16)

hijk (X1 , . . . , Xn ),

k = 1, . . . , lij .

Now consider the K-closed set set Vij = V (hij,1 , . . . , hij,lij ) ). If hi happens to be 1, then Vij is empty for all j. Sn Sbi Vij . Indeed, let x ∈ A and z = t1 x1 + · · · + tn xn . Part E: A = i=1 j=1 Then (x, z) is a zero of (7). By (14a) there is an i with hi (t, x, z) = 0. By (15) there exists a j with hij (t, x1 , . . . , xn−i , z) = 0. Thus, x is a common zero of the hijk , k = 1, . . . , lij , and x ∈ Vij . Conversely, let (x1 , . . . , xn ) ∈ Vij . Then hi (t, x1 , . . . , xn−i , z) = 0, where z = t1 x1 + · · · + tn xn . By (14b), there exist x0n−i+1 , . . . , x0n algebraic over K(x1 , . . . , xn−i ) with (x1 , . . . , xn−i , x0n−i+1 , . . . , x0n ) ∈ A and z = t1 x1 + · · · + tn−i xn−i + tn−i+1 x0n−i+1 + · · · + tn x0n . But then x0j = xj , j = n − i + 1, . . . , n. Hence, x ∈ A.

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Part F: Vij is either empty or it is a K-variety of dimension n − i. Suppose hij is nonconstant. Then Z actually occurs in hij . Otherwise, we may start with a zero (x1 , . . . , xn−i ) of hij (t, X1 , . . . , Xn−i ) and conclude for two distinct elements z and z 0 the existence of xn−i+1 , . . . , xn , algebraic over K(x1 , . . . , xn−i ) such that both (x, z) and (x, z 0 ) are zeros of (7a). Then z = z 0 , a contradiction. From this we produce a generic point of Vij : Starting from elements x1 , . . . , xn−i , algebraically independent over K, choose z with hij (t, x1 , . . . , xn−i , z) = 0. Then hi (t, x1 , . . . , xn−i , z) = 0. By (14b), thereP are xn−i+1 , . . . , xn in the algebraic closure of K(x1 , . . . , xi−1 ) such that n z = i=1 ti xi . It follows that x belongs to Vij and has maximal transcendence degree over K. Therefore, dim(Vij ) = n − i. Moreover, x is a generic point of Vij , so Vij is a K-variety. Indeed, (t, x1 , . . . , xn−i , z) is a generic point of the K-variety V (hij ). Consider x0 ∈ Vij . Let z 0 = t1 x01 + · · · + tn x0n . Then hij (t, x01 , . . . , x0n−i , z 0 ) = 0. Hence, (t, x01 , . . . , x0n−i , z 0 ) is a K-specialization of (t, x1 , . . . , xn−i , z). As in Part E, this specialization extends to a K-specialization (t, x, z) → (t, x01 , . . . , x0n−i , x00n−i+1 , . . . , x00n , z 0 ) with z 0 = t1 x01 + · · · + tn−i x0n−i + tn−i+1 x00n−i+1 + · · · + tn x00n . Hence, x0i = x00i , i = n − i + 1, . . . , n. It follows that x0 is a K-specialization of x. This proves our claim and concludes the proof of the proposition.  Remark 19.5.7: (a) A K-closed set A in An is said to be presented if polynomials f1 , . . . , fs ∈ K[X1 , . . . , Xn ] are given explicitly such that A = V (f1 , . . . , fs ). (b) If A and B are two given K-closed sets in An , then Proposition 19.5.6 allows us to effectively decompose A into a union of K-varieties and to compute generic points of each of them. Substituting these points in the polynomials that define B, we can decide if A ⊆ B. (c) The decomposition of A described in Proposition 19.5.6 may not be into K-irreducible components, because there might be inclusions among the Vij ’s. This, however, can be checked using (b). Throw away the unnecessary K-varieties from the union to effectively obtain a decomposition of A into K-irreducible components.  Remark 19.5.8: Universal bounds on degrees. Suppose K in Proposition 19.5.6 is an arbitrary field and f1 , . . . , fs are nonzero polynomials in K[X]. Then the proof of Proposition 19.5.6 proves the existence of a decomposition of A = V (f1 , . . . , fs ) into a union V1 ∪ · · · ∪ Vm of K-varieties. Moreover, it gives polynomials in K[X] which define Vj with degrees bounded by a number e. Both m and e depend only on the degrees of f1 , . . . , fs but neither on K nor on the coefficients of f1 , . . . , fs : First, by Remark 19.5.5, g1,j in (8) have bounded degrees. Then both h1 and e1,j , as factors of g1,j , have bounded degrees. Again, by Remark 19.5.5,

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the polynomials d1j in (9) have bounded degrees. Hence, the degrees of their coefficients f1j are bounded. The proof of Proposition 19.5.6 proceeds in Part C and Part D by induction on n. After at most n steps it produces polynomials hi ∈ K(t1 , . . . , tn , X1 , . . . , Xn−i , Z), i = 1, . . . , n, with bounded degrees. Let hij be the irreducible factors of hi , as in (15). Then hijk (X1 , . . . , Xn ), k = 1, . . . , lij , are the coefficients of hij (t, X1 , . . . , Xn−i , t1 X1 + · · · + tn Xn ). Both the degrees of hij and their number are bounded. Moreover, A = Sn S bi j=1 Vij , where Vij is defined by hijk , k = 1, . . . , lij . Part F of the i=1 proof of Proposition 19.5.6 then proves that each of the Vij is either empty or it is a K-variety. This completes the proof of our claim. Note that each Vij can be defined by a bounded number of polynomials. Indeed, suppose Vij = V (g1 , . . . , gq ) where each gk is a polynomial in K[X1 , . . . , Xn ] of degree at most e. There are only finitely many mono
— mials in X1 , . . . , Xn , say r, of degree at most e (indeed, r = n+e e see Exercise 3). Hence, the dimension of the space of all polynomials in K[X1 , . . . , Xn ] of degree at most e is r. Renumbering g1 , . . . , gq , we Pqmay assume g1 , . . . , gr0 , with r0 ≤ min(q, r), generate the K-vector space i=1 Kgi . Then, Vij = V (g1 , . . . , gr0 ), as claimed.  As a corollary of Remark 19.5.8 we prove that the concept of an absolutely irreducible variety is elementary. We have been able to circumvent this result in this book and instead use the easier result that absolute irreducibility of polynomials is elementary. For example, we have proved that pseudo algebraic closeness of a field can be checked by absolutely irreducible polynomials and deduce that “K is PAC” is an elementary statement on K (Section 11.3). However, there are occasions where one has to use the elementary nature of absolute irreducibility of varieties directly. We therefore put this result here as a convenient reference. Let I be the set of all n-tuples (i1 , . . . , in ) of nonnegative integers with i1 + · · · + in ≤ d. Put r = |I|. Choose a bijective map j: I → {1, . . . , r}. Then the general polynomial in X1 , . . . , Xn of degree d can be written as P f (T, X1 , . . . , Xn ) = T X1i1 · · · Xnin , with T = (T1 , . . . , Tr ). Every j(i) i∈I polynomial in K[X1 , . . . , Xn ] of degree at most d can then be written as f (a, X1 , . . . , Xn ) with a ∈ K r . Proposition 19.5.9: For all positive integers d, m, n there is a formula θ(T1 , . . . , Tm ) in L(ring) satisfying this: Let K be a field and f1 , . . . , fm be polynomials in K[X1 , . . . , Xn ] of degree at most d with vectors of coefficients a1 , . . . , am , respectively. Then the Zariski K-closed subset of An defined by the system of equations f (ai , X1 , . . . , Xn ) = 0, i = 1, . . . , m, is absolutely irreducible if and only if θ(a1 , . . . , am ) holds in K. Proof: Let K be a field and V a Zariski closed subset of An defined by polynomials in K[X1 , . . . , Xn ] of degrees at most d. Remark 19.5.8 gives positive integers e, s, and k depending only on n and d such that ”V is ˜ is equivalent to the following statement: irreducible over K”

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˜ 1 , . . . , Xn ], i = 1, . . . , k, j = (17) There exist no polynomials gij ∈ K[X 1, . . . , s, of degree at most e and with k ≥ 2 such that V =

k [

V (gi1 , . . . , gis )

i=1

and no V (gi1 , . . . , gis ) is contained in another. For all distinct i, i0 the statement V (gi1 , . . . , gis ) 6⊆ V (gi0 1 , . . . , gi0 s ) is ˜ to “There exist x with gi1 (x) = · · · = gir (x) = 0 and equivalent over K 0 gi j (x) 6= 0 for at least one j.” Statement (17) is therefore equivalent to a ˜ 1 , . . . , Tm ) of L(ring) with the following property: formula θ(T ˜ r . Then ˜ be an algebraically closed field and a1 , . . . , am ∈ K (18) Let K ˜ V (f (a1 , X), . . . , f (am , X)) is irreducible over K if and only if ˜ 1 , . . . , am ) is true in K. ˜ θ(a Elimination of quantifiers (Theorem 9.2.1) gives a quantifier free formula ˜ 1 , . . . , Tm ) over every algebraically θ(T1 , . . . , Tm ) which is equivalent to θ(T closed field. Observe that a quantifier free sentence with parameters in K is ˜ It follows that for every field K and true in K if and only if it is true in K. r all a1 , . . . , am ∈ K the following chain of equivalencies holds: θ(a1 , . . . , am ) is true in K ˜ ⇐⇒ θ(a1 , . . . , am ) is true in K ˜ 1 , . . . , am ) is true in K ˜ ⇐⇒ θ(a ˜ ⇐⇒ V (f (a1 , X), . . . , f (am , X)) is irreducible over K ⇐⇒ V (f (a1 , X), . . . , f (am , X)) is absolutely irreducible. This completes the proof of the proposition.



19.6 Presented Rings and Constructible Sets Let R be an integral domain presented in its quotient field K. Assume K has elimination theory. The finitely generated integral domains over K are exactly the set of coordinate rings of K-varieties. Later we use the following discussion to consider reduction of these varieties modulo prime ideals of R. Let S = R[x], where x = (x1 , . . . , xn ) is a presented n-tuple over K. Assume without loss that x1 , . . . , xr is a transcendence base for K(x)/K (Lemma 19.2.9). For each r < i ≤ n write irr(xi , K(x1 , . . . , xi−1 )) in the 1 ,...,xi−1 ,Xi ) form fi (xh(x with fi ∈ R[X1 , . . . , Xi ] and h ∈ R[X1 , . . . , Xr ]. Each 1 ,...,xr ) element of S 0 = R[x1 , . . . , xn , h(x1 , . . . , xr )−1 ] has a unique presentation as (1)

X gi (x1 , . . . , xr ) i · x r+1 . . . xinn , h(x1 , . . . , xr )d(i) r+1

19.6 Presented Rings and Constructible Sets

423

where the sum ranges over values of ir+1 , . . . , in from 0 to degXr+1 (fr+1 ) − 1, . . . , degXn (fn ) − 1, respectively, and for each i = (ir+1 , . . . , in ) the polynomial gi (x1 , . . . , xr ) of R[x1 , . . . , xr ] is not divisible in K[x1 , . . . , xr ] by h(x1 , . . . , xr ). An arbitrary element y of K(x) may be presented in the form (2)

X gi (x1 , . . . , xr ) i · x r+1 . . . xinn , hi (x1 , . . . , xr ) r+1

where the exponents ir+1 , . . . , in satisfy the same condition as above and gi , hi are relatively prime in K[x1 , . . . , xr ]. A necessary condition for y to belong to S 0 is that each hi (x1 , . . . , xr ) divides a power of h(x1 , . . . , xr ) in fi (x1 ,...,xr ) K[x1 , . . . , xr ]. In this case, rewrite the ith coefficient as ci ·h(x d(i) with 1 ,...,xr ) fi ∈ R[x1 , . . . , xr ] and ci ∈ R. Then y belongs to S 0 if and only if c1i fi ∈ R[x1 , . . . , xr ] for each i. We can check this effectively. We sum up this discussion: Lemma 19.6.1: In the notation above, S 0 is also a primitive recursive subset of K(x). Definition 19.6.2: Let h1 , . . . , hm ∈ K[X1 , . . . , Xn ] and P a Boolean polynomial in the symbols ∪, ∩, and 0 (i.e. taking complements). Call A = set in An . The dimension of A P (V (h1 ), . . . , V (hm )) a K-constructible
is max trans.deg(K(x)/K) | x ∈ A . If L is a field containing K, denote the set of points of A with coordinates in L by A(L). Call A presented if h1 , . . . , hm and P are presented. Note that A can be presented in many ways. In particular, if K is the quotient field of a presented integral domain R, then we may assume that h1 , . . . , hm have coefficients in R.  Lemma 19.6.3 (Chevalley): Let f1 , . . . , fm ∈ K[X1 , . . . , Xn ]. Define a Kmorphism ϕ: An → Am by ϕ(x) = (f1 (x), . . . , fm (x)). If A ⊆ An is a presented K-constructible set, then we can effectively describe B = ϕ(A) as a presented K-constructible set. Proof: Let Ω be a universal domain containing K. Then B = {(y1 , . . . , ym ) ∈ Ωm | θ(y1 , . . . , ym ) is true in Ω}, where θ(Y1 , . . . , Ym ) is the formula (∃X1 ) · · · (∃Xn )[f1 (X) = Y1 ∧ · · · ∧ fm (X) = Ym ∧ X ∈ A]. Theorem 9.3.1(a) effectively gives a quantifier-free formula θ0 (Y1 , . . . , Ym ) (in the language L(ring, K)) equivalent to θ(Y1 , . . . , Ym ) over Ω. This formula  gives B as a constructible subset of Am .

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Call a K-constructible subset A of An K-basic if A = V r V (g), where V = V (f1 , . . . , fm ) with f1 , . . . , fm ∈ K[X1 , . . . , Xn ] is a K-variety and g ∈ K[X1 , . . . , Xn ] does not vanish on V . If x is a generic point of V , then call K[A] = K[x, g(x)−1 ] the coordinate ring of A and K(A) = K(x) the function field of A. Then the dimension of A is the transcendence degree of K(A)/K. Furthermore, the K-basic set A is normal if K[A] is an integrally closed domain, and A is presented if the polynomials f1 , . . . , fm , g and the ring K[A] are presented. Remark 19.6.4: Let A = V r V (g) be a K-basic set, x a generic point of V , and u ∈ K(x). By Lemma 2.4.4, u is in K[A] = K[x, g(x)−1 ] if and only if for each a ∈ A there are h1 , h2 ∈ K[X] with and h2 (a) 6= 0 and u = hh12 (x) (x) . In 0r 0 V (h) of A, with x a generic particular, consider a K-basic subset B = V point of V 0 . Then g(b) 6= 0 for each b ∈ B. Thus, K[x0 , g(x0 )−1 ] ⊆ K[B]. In other words, the K-specialization extends to a K-homomorphism of K[A] into K[B].  Definition 19.6.5: Let P be a property of constructible sets (e.g. basic, normal, nonsingular, etc.). A P -stratification of a constructible set A is a finite collection S {Ai | i ∈ I} of disjoint constructible sets having property P such that A = · i∈I Ai . Refer to Ai as a P -set, i ∈ I.  Lemma 19.6.6 (The Stratification Lemma): Let P be a property of constructible sets. Suppose for each presented nonempty K-basic set A we can effectively compute a nonempty K-basic P -set B, K-open in A. Then we can effectively produce a P -stratification of each presented nonempty constructible set. Proof: Start with a presented nonempty K-constructible set A in An and do an induction on dim(A). S r T si [V (fij ) r V (gij )], with fij , gij ∈ Without loss assume A = i=1 j=1 K[X1 , . . . , Xn ]. Use the identity si \

[V (fij ) r V (gij )] = V (fi1 , . . . , fi,si ) r V (gi1 · · · gi,si )

j=1

to rewrite A in the form (3)

A=

r [

(Vi r Ui ),

i=1

where Vi and Ui are presented Zariski K-closed sets. Apply Proposition 19.5.6 to write each Vi as a union of K-irreducible components and thus assume that each Vi is K-irreducible. If Vi = Vj , replace (Vi r Ui )∪(Vj r Uj ) by Vi r Ui ∩Uj . Thus, assume the Vi ’s are distinct. Replace Vi r Ui by Vi r Vi ∩ Ui to assume Ui ⊆ Vi for each

19.7 Basic Normal Stratification

425

i. Then assume Ui ⊂ Vi . Finally, reorder the indices such that dim(Ur ) ≤ dim(Ui ) for i = 1, . . . , r − 1 and rewrite A as a disjoint union: A1 ∪· A2 ∪· A3 Sr−1 Sr−1 with A1 = Vr r Ur , A2 = i=1 [Vi r(Vr ∪ Ui )] and A3 = i=1 [(Vi ∩ Ur ) r Ui ]. Hence, dim(A3 ) ≤ dim(Ur ) < dim(Vr ) ≤ dim(A). Apply the induction hypothesis to effectively produce a P -stratification of A3 . The A2 term has less than r components. Use induction on r to assume that the union is disjoint. Thus, assume that (4)

A = V rU

with U = V (g1 , . . . , gk ).

Since A 6= ∅, there is an i with V (gi ) ∩ V ⊂ V . Then A = [V r V (gi )] ∪· [V ∩ V (gi ) r U ] and the second term in the union has dimension less than dim(A). By the induction hypothesis, it suffices to stratify the first term. Hence, we may assume that A is K-basic. By assumption, we can effectively find a nonempty P -set B, K-open in A with A = B ∪· (A r B). Since dim(A r B) < dim(A), we can effectively P -stratify A. 

19.7 Basic Normal Stratification We first give a constructive version of the normalization procedure. The next lemma circumvents difficulties that arise when K is imperfect: Lemma 19.7.1: Let (x1 , . . . , xn ) be a presented n-tuple over K. Then we can effectively compute a finite purely inseparable extension K 0 of K with K 0 (x)/K 0 separable. Moreover, if char(K) 6= 0, we can effectively find a basis w1 , . . . , wl of K 0 /K and a power q of char(K) with w1 = 1 and wq ∈ K for all w ∈ K 0 . Proof: Without loss char(K) = p > 0. Assume x1 , . . . , xr is a transcendence base for K(x)/K. If r = n, take K 0 = K. Otherwise, compute a nonzero polynomial f ∈ K[X1 , . . . , Xr+1 ] of minimal degree with (1)

f (x1 , . . . , xr+1 ) =

X

i

r+1 ai xi11 · · · xr+1 = 0,

with ai ∈ K.

i∈I

Let q be the maximal power of p that divides all exponents of (1). Put 1/q K1 = K(ai | i ∈ I). Then x1 , . . . , xr+1 satisfy the following irreducible relation over K1 : X 1/q i /q ir+1 /q ai x11 · · · xr+1 = 0. i∈I

One of the variables, say x1 , appears in this relation in a monomial which is not a pth power. Hence, x1 is separably algebraic over K1 (x2 , . . . , xn ). Now use induction on n to compute a finite purely inseparable extension K 0 of K1 such that K 0 (x2 , . . . , xn )/K 0 is separable. Then K 0 (x)/K 0 is also separable. In addition, find a basis w100 , . . . , wl0000 for K 0 /K1 .

426

Chapter 19. Effective Field Theory and Algebraic Geometry 1/q

Now order the ai ’s in a sequence a1 , . . . , ar . Examine which of the aj 1/q

1/q

K-linearly depends on a1 , . . . , aj−1 to find a basis w10 , . . . , wl00 for K1 /K. Then {wi0 wj00 | 1 ≤ i ≤ l0 , 1 ≤ j ≤ l00 } is a basis for K 0 /K which can be w ordered as w1 , . . . , wl . Finally, replace wj , if necessary, by w1j to assure  w1 = 1. Lemma 19.7.2: Let (x1 , . . . , xn , z) be a presented (n + 1)-tuple over K with z separably algebraic over K(x). Then we can effectively find a polynomial g ∈ K[X] with these properties: (2) g(x) 6= 0, and K[x, g(x)−1 ] is integrally closed and presented in its function field. (3) The ring K[x, g(x)−1 , z] is a presented cover of the ring K[x, g(x)−1 ] with z a primitive element. Proof: Suppose we have found g ∈ K[X] satisfying (2). Present irr(z, K(x)) as a quotient of a polynomial in K[x, Z] by a polynomial in K[x]. Then multiply g by the product of the denominator and the discriminant of irr(z, K(x)). By the remarks leading up to Lemma 19.6.1, the ring K[x, g(x)−1 , z] is presented over K. Also (Definition 6.1.3), z is a primitive element for the cover K[x, g(x)−1 , z]/K[x, g(x)−1 ]. The argument that finds g satisfying (2) breaks into two cases. Case A: K(x)/K is separable. Reorder x1 , . . . , xn according to Lemma 19.2.4 to assume xi is separable over K(x1 , . . . , xi−1 ), i = 1, . . . , n. Use induction on n to effectively find a polynomial g1 ∈ K[X1 , . . . , Xn−1 ] with g1 (x) 6= 0 and K[x1 , . . . , xn−1 , g1 (x)−1 ] integrally closed. If xn is transcendental over K(x1 , . . . , xn−1 ), then K[x1 , . . . , xn , g1 (x)−1 ] is also integrally closed [Zariski-Samuel2, p. 85 or p. 126]. Otherwise, xn is separable algebraic over K(x1 , . . . , xn−1 ) and from the above, with z = xn , we can effectively find a multiple g ∈ K[X1 , . . . , Xn−1 ] of g1 such that K[x1 , . . . , xn , g(x)−1 ] is a ring cover of K[x1 , . . . , xn−1 , g(x)−1 ]. In particular K[x, g(x)−1 ] is integrally closed. Case B: K(x)/K is general. Assume without loss char(K) 6= 0. Apply Lemma 19.7.1 to find a finite purely inseparable extension K 0 of K with K 0 (x)/K 0 separable. Lemma 19.7.1 gives a linear basis w1 , . . . , wl with w1 = 1 for K 0 /K and a power q of char(K) with wq ∈ K for each w ∈ K 0 . The wi ’s need not be linearly independent over K(x). Find among them a basis, say w1 , . . . , wk , for K 0 (x)/K(x) and compute a polynomial a ∈ K[X] with a(x) 6= 0 and wi ∈ K[x, a(x)−1 , w1 , . . . , wk ], i = 1, . . . , l. Note: K(x) ∩ K 0 [x] ⊆ K[x, a(x)−1 ]. Use Case A to find h ∈ K 0 [X] with h(x) 6= 0 and K 0 [x, h(x)−1 ] integrally closed. Now let g = ahq . Then K(x) ∩ K 0 [x, h(x)−1 ] is contained in the ring  K[x, g(x)−1 ], which is therefore integrally closed. Combine Lemma 19.7.2 with the stratification lemma:

Exercises

427

Proposition 19.7.3: There is an effective procedure for producing a basic normal stratification of a given constructible set.

Exercises 1. (a) Use the inequality pn+1 ≤ (p1 p2 · · · pn ) − 1 (Euclid) to show for each n n ∈ N that p1 p2 · · · pn ≤ 22 , where p1 < p2 < · · · is the sequence of primes. (b) Use (a) to show for each r ∈ N that {pj11 · · · pjnn | n ∈ N, 1 ≤ j1 , . . . , jn ≤ r} is a primitive recursive subset of N. 2. The racetrack problem. Consider the real points C1 (R) in the ellipse X 2 + 2Y 2 = 1 as the inside rail of a racetrack. Let C2 (R) be the locus of points (U, V ) traced out by the points on the outside of the ellipse that are 1 unit distance from the ellipse along a perpendicular to the ellipse. Let W ⊆ A4 with coordinates (X, Y, U, V ) be given by the equations f1 = X 2 + 2Y 2 − 1 = 0, f2 = XV − 2Y U + XY = 0, f3 = (U − X)2 + (V − Y )2 − 1 = 0. Let π: A4 → A2 with π(X, Y, U, V ) = (U, V ) be the projection onto the last coordinates. (a) Show that C2 (R) = π(W )(R). (b) Show that π(W ) = V (h), where h is a polynomial of degree 8 (= deg(f1 ) deg(f2 ) deg(f3 )) that generates the ideal Q[U, V ] ∩ I(f1 , f2 , f3 ) of Q[U, V ]: (b1) First eliminate X to show that Y, U, V satisfy: g1 = (1 − 2Y 2 )(V + Y )2 − 4(Y U )2 = 0,
g2 = (U V − Y U )2 + (V − Y )2 − 1 (V + Y )2 = 0. Notice that both g1 and g2 are polynomials of degree 4 in Y with coefficients in Z[U, V ] and with leading coefficients −2 and 1, respectively. (b2) Now apply the proof of Theorem 9.2.1 (expression (1)) to g1 and g2 to eliminate Y . Alternatively, apply the resultant to g1 and g2 . 3. Prove
that the number of monomials in X1 , . . . , Xn of degree at most d is n+d d . Hint: Given integers i1 , . . . , in between 0 and d define j1 = i1 + 1, j2 = i1 + i2 + 2, . . ., jn = i1 + i2 + · · · + in + n. Prove that the map (i1 , . . . , in ) 7→ (j1 , . . . , jn ) defines a bijective map between the set of all monomials in X1 , . . . , Xn of degree at most d and the set of all subsets with n elements of the set {1, 2, . . . , n + d}.

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Chapter 19. Effective Field Theory and Algebraic Geometry

Notes This Chapter is an elaboration of [Fried-Haran-Jarden, §2]. Most of it is due to Kronecker. The common feature is the use of the “method of indeterminates.” Our sources include [Waerden3] for the proofs of Lemmas 19.1.3, 19.2.2, and 19.3.2; [Waerden1] for the proof of Lemma 19.5.3; a model for the proof of Lemma 19.2.4 from [Lang4]; [Zariski-Samuel2] for the proofs of Lemmas 19.3.1 and 19.5.1; and an elaboration of [Waerden2] for the proof of Proposition 19.5.6. Kronecker’s algorithm is no longer the state of the art for a deterministic procedure for factoring polynomials with coefficients in Z (Lemma 19.1.3). Indeed, [Lenstra-Lenstra-Lov´ asz] gives a bound on the time of factorization which is a polynomial in the height of the polynomial and [Lenstra] extends this to polynomials in several variables. No one has yet tested the effect of these procedures on such practical applications as the algorithm of Proposition 19.5.6. But this would be a key ingredient in producing a computer program that actually accomplishes the Galois stratification procedure and its corollaries of Chapters 30 and 31. L. v. d. Dries and K. Schmidt [v.d.Dries-Schmidt] use nonstandard approach and the notion of faithful flatness to prove a stronger result than that achieved in Remark 19.5.8: Given positive integers d and n, there is a positive integer e such that for every field K and all polynomials f1 , . . . , fm ∈ K[X1 , . . . , Xn ] of degree at most d, the ideal I generated by f1 , . . . , fm is prime if and only if for all g, h ∈ K[X1 , . . . , Xn ] of degree at most e, the relation gh ∈ I implies g ∈ I or h ∈ I. The idea of checking absolute irreducibility over the algebraic closure of a field K first and then using elimination of quantifiers to descend to K was communicated to the authors by v. d. Dries.

Chapter 20. The Elementary Theory of e-Free PAC Fields This chapter presents one of the highlights of this book, the study of the elementary theory of e-free PAC fields. We apply the elementary equivalence theorem for arbitrary PAC fields (Theorem 20.3.3) to the theory of perfect e-free PAC fields containing a fixed countable base field K. If K is finite and e = 1 or K is Hilbertian and e ≥ 1, then this theory coincides with ˜ the theory of all sentences with coefficients in K that are true in K(σ), for e almost all σ ∈ G(K) (Section 20.5). In particular, if K is explicitly given with elimination theory, then this theory is recursively decidable. In the special case where K is a global field and e = 1 we prove a transfer theorem ˜ (Theorem 20.9.3): A sentence θ of L(ring, OK ) is true among the fields K(σ) with probability equal to the probability that θ is true among the residue fields of K. Finally, we prove that the elementary theory of finite fields is recursively decidable (Theorem 20.10.6).

20.1 ℵ1 -Saturated PAC Fields We start with a result that strengthens the PAC property of a field which is also ℵ1 -saturated (Section 7.7): Lemma 20.1.1: Let K be an ℵ1 -saturated PAC field and R an integral domain which is countably generated over K. (a) Suppose R is contained in a field F regular over K. Then there exists a K-homomorphism ϕ: R → K. (b) Suppose in addition, char(K) = p > 0. Let S be a subset of R, pindependent over F p with |S| ≤ min(ℵ0 , [K : K p ]). Then ϕ can be chosen such that ϕ(S) is p-independent over K p . Proof: By assumption, R = K[x1 , x2 , x3 , . . .]. Denote the ideal of all polynomials in K[X1 , . . . , Xn ] that vanish at (x1 , . . . , xn ) by In . Let fn,1 , . . . , fn,rn be a system of generators for In . Then I1 ⊆ I2 ⊆ I3 · · ·. Since K[X1 , . . . , Xn ]/In ∼ =K K[x1 , . . . , xn ] and K(x1 , . . . , xn ), as a subfield of F , is regular over K, V (In ) is a variety defined over K (Corollary 10.2.2(a)). Since K is PAC, there exist elements a1 , . . . , an ∈ K with fnj (a1 , . . . , an ) = 0, j = 1, . . . , rn . Each fij with 1 ≤ i ≤ n and 1 ≤ j ≤ ri is a linear combination of fn1 , . . . , fnrn . Hence, (1)

ri n ^ ^ i=1 j=1

fij (a1 , . . . , ai ) = 0.

430

Chapter 20. The Elementary Theory of e-Free PAC Fields

The saturation property of K gives b1 , b2 , b3 , . . . in K with fnj (b1 , . . . , bn ) = 0 for all n and j. The map xi 7→ bi , i = 1, 2, 3, . . . extends to a Khomomorphism ϕ: R → K. Now assume char(F ) = p > 0 and S is as in (b). The essential case occurs when [K : K p ] ≥ ℵ0 and S = {s1 , s2 , s3 , . . .} is infinite. Write sn = gn (x1 , . . . , xkn ) with gn ∈ K[X1 , . . . , Xkn ]. Without loss assume that n ≤ kn ≤ kn+1 for all n. For each n Proposition 11.4.1 gives a1 , . . . , akn in K such that, in addition to (1), gi (a1 , . . . , aki ), i = 1, . . . , n, are p-independent  over K p . Then proceed as before.

20.2 The Elementary Equivalence Theorem of ℵ1 -Saturated PAC Fields Conditions for the elementary equivalence of two PAC fields E and F are central to the model theory of PAC fields. If E and F are elementarily equivalent, then E and F have the same characteristic and thus the same prime field K. Since a polynomial f ∈ K[X] has a zero in E if and only if ˜ ∼ ˜ (Lemma 20.6.3). Assume it has a zero in F , it follows that E ∩ K =F ∩K ˜ ˜ therefore that L = E ∩ K also equals F ∩ K. More generally (Section 23.4), if L0 is a finite Galois extension of L and E 0 is a finite Galois extension of E containing L0 , then there exists a finite Galois extension F 0 of F , containing L0 , and a commutative diagram (1)

Gal(F 0 /F ) Gal(E 0 /E) o II II vv II vv v I v res II v res $ zvv 0 Gal(L /L) ϕ

with ϕ an isomorphism. If E and F are countable, then an inverse limit gives a similar diagram with absolute Galois groups replacing corresponding relative Galois groups. The next lemma shows this last condition to be essentially sufficient for the elementary equivalence of E and F . The basic element is Lemma 20.2.2 whose main ingredients, the field crossing argument and the use of existence of rational points on varieties, have already appeared in the proof of the Chebotarev density theorem and will appear again in Proposition 24.1.1. Let L, M be fields and Φ: Ls → Ms be an embedding with Φ(L) ⊆ M . Then Φ induces a homomorphism ϕ: Gal(M ) → Gal(L) with ϕ(σ)x = Φ−1 (σΦ(x)) for all σ ∈ Gal(M ) and x ∈ Ls . If Φ is an isomorphism with Φ(L) = M , then ϕ is an isomorphism. Lemma 20.2.1: Let E/L be a regular extension of fields. Then Es /Ls is regular. ˜ over L, so ELs is linearly Proof: By assumption, E is linearly disjoint from L ˜ ˜ disjoint from L over Ls . Since Es /ELs is separable and E L/EL s is purely

20.2 The Elementary Equivalence Theorem of ℵ1 -Saturated PAC Fields

431

˜ over ELs . It follows from the inseparable, Es is linearly disjoint from E L ˜ over tower property of linear disjointness that Es is linearly disjoint from L  Ls . In other words, Es /Ls is regular. Lemma 20.2.2 (Embedding Lemma [Jarden-Kiehne, p. 279]): Let E/L and F/M be separable field extensions satisfying: E is countable and F is PAC and ℵ1 -saturated. In addition, suppose there are an isomorphism Φ0 : Ls → Ms with Φ0 (L) = M and a commutative diagram (2)

Gal(E) o

ϕ

res

 Gal(L) o

Gal(F ) res

ϕ0

 Gal(M )

where ϕ0 is the isomorphism induced by Φ0 and ϕ is a homomorphism. If char(L) = p > 0, add the assumption that [E : E p ] ≤ [F : F p ]. Then there exists an extension of Φ0 to an embedding Φ: Es → Fs which induces ϕ with F/Φ(E) separable. ˜ is a separable extension, E ∩ L ˜ = E ∩Ls . Similarly, F ∩ Proof: Since E ∩ L/L ˜ and resM (Gal(F )) = ˜ M = F ∩ Ms . In addition, resLs (Gal(E)) = Gal(E ∩ L) s ˜ ˜ ˜ . Therefore, replace Gal(F ∩ M ). The hypotheses thus give Φ0 (E∩ L) = F ∩ M ˜ and F ∩ M ˜ to assume that E/L and F/M L and M , respectively, by E ∩ L are regular extensions. By Lemma 20.2.1, Es /Ls and Fs /Ms are regular. Without loss assume L = M , Φ0 and ϕ0 are the identity isomorphisms, and E is algebraically independent from F over L. It follows from Corollary 2.6.8 that (3) EF is a separable (and regular) extension of both E and F . Also, by Lemma 2.6.7, Es is linearly disjoint from Fs over Ls . We divide the rest of the proof into three parts to separate out the use of the PAC property. Part A: The field crossing argument. From (2), ϕ(σ)x = σx for each σ ∈ Gal(F ) and x ∈ Ls . Thus, by Lemma 2.5.5, each σ ∈ Gal(F ) extends uniquely to a σ ˜ ∈ Gal(Es Fs /EF ) with  σ ˜x =

ϕ(σ)x if x ∈ Es σx if x ∈ Fs .

˜ = σ. The map σ 7→ σ ˜ embeds Gal(F ) into Gal(Es Fs /EF ) and resFs Es /Fs σ Let D be the fixed field of the image of Gal(F ). Then res: Gal(Es Fs /D) → Gal(F ) is an isomorphism, so D ∩ Fs = F and DFs = Es Fs . Since D/EF is separable, (3) implies that D is a separable extension of both E and F . Hence, by Lemma 2.6.4, (4) D is a regular extension of F .

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Part B: Use of the PAC property. Note that Es Fs is an algebraic extension choose yi ∈ Fs of D. Hence, Es ⊆ Es Fs = D[Fs ] = Fs [D]. For each x ∈ Es P and di ∈ D, where i ranges over a finite set Ix such that x = i∈Ix yi di . Let D0 = E ∪ {di | x ∈ Es , i ∈ Ix }. Then, Es ⊆ Fs [D0 ] and since E is countable, so is D0 . We illustrate the relations among all rings and fields mentioned so far in the following diagram. (5) E

L

Es   

| |||

Ls

Fs [D0 ] GGG r r GG r r

E s Fs yy y y F [D0 ] D E KK EEE KK E KK KK Fs KK KK KK wwwww F

If p = char(L) > 0, let S be a p-basis of E over E p . Since F (D0 )/E is separable, S is p-independent over F (D0 )p . Also, |S| ≤ [E : E p ] ≤ [F : F p ]. Applying Lemma 20.1.1 using the hypothesis and (4), we conclude that there exists an F -homomorphism Ψ: F [D0 ] → F with Ψ(S) p-independent over F p (if p > 0). Since E is a field, Ψ is injective on E, so Ψ(S) is a p-basis of Ψ(E) over Ψ(E)p . Therefore, F/Ψ(E) is separable. Since D is linearly ˜ Fs [D0 ] → Fs disjoint from Fs over F , Ψ extends to an Fs -homomorphism Ψ: (Lemma 2.5.5). Part C: Conclusion of the proof. (6)

Check that

˜ σ x) = σ Ψ(x) ˜ Ψ(˜ for each σ ∈ Gal(F )

and each x ∈ Fs ∪ D0 . Thus, (6) holds for each x ∈ Fs [D0 ]. Since Es ⊆ ˜ to Es by Φ. Fs [D0 ], (6) holds for each x ∈ Es . Denote the restriction of Ψ Then Φ is an Ls -embedding of Es into Fs which satisfies the conclusion of the Lemma.  Recall that if K is a field, then L(ring, K) denotes the first order language of the theory of rings augmented with constant symbols for the elements of K (Example 7.3.1). We apply Skolem-L¨ owenheim (Proposition 7.4.2) and the Cantor back and forth argument to improve Lemma 20.2.2: Lemma 20.2.3: Let E/L and F/M be separable field extensions with both L and M countable and containing a given field K. Assume E and F have the same imperfect degree, they are PAC and ℵ1 -saturated, and there exists a K-isomorphism Φ0 : Ls → Ms with Φ0 (L) = M . Assume in addition that (2) is a commutative diagram with ϕ an isomorphism. Then E is K-elementarily equivalent to F . Proof: Skolem-L¨ owenheim gives a countable K-elementary subfield M1 of F that contains M . Since M1 /M is separable, we may apply Lemma 20.2.2

20.3 Elementary Equivalence of PAC Fields

433

to the diagram Gal(E)

res◦ϕ−1

/ Gal(M1 )

res

 Gal(L)

res

ϕ−1 0

 / Gal(M )

to conclude that there is an extension of Φ−1 0 to an embedding Ψ1 : M1,s → Es with these properties: E/Ψ1 (M1 ) is separable and the diagram Gal(E)

res◦ϕ−1

/ Gal(M1 )

res

 Gal(M10 )

res

ψ1

 / Gal(M1 )

is commutative, where M10 = Ψ1 (M1 ) and ψ1 is the isomorphism induced by Ψ1 . Now reverse the roles of E and F to find a countable K-elementary subfield L1 of E that contains M10 and an embedding Φ1 : L1,s → Fs that extends Ψ−1 1 appropriately. Proceed by induction to construct two towers of countable fields L ⊆ L1 ⊆ L2 ⊆ · · · ⊆ E

and M ⊆ M1 ⊆ M2 ⊆ · · · ⊆ F

and embeddings Φi : Li → Mi+1 , and Ψi : Mi → Li satisfying: Li (resp. Mi ) is a K-elementary subfieldSof E (resp. F ), theSmap Φi extends Ψ−1 i , and Ψi ∞ ∞ extends Φ−1 i−1 . Let L∞ = i=1 Li and M∞ = i=1 Mi . By Lemma 7.4.1(b), L∞ and M∞ are K-elementary subfields of E and F , respectively. Moreover, the Φi combine to give a K-isomorphism Φ∞ : L∞ → M∞ . Consequently,  E ≡K F . Remark 20.2.4: Suppose E and F have the same cardinality, say m and they are m+ -saturated. Then with m+ -saturated replacing ℵ1 -saturated in Lemma 20.2.3, we may conclude that E ∼ =K F . The introduction of the m+ saturated concept in this approach usually forces one to use the continuum hypothesis 2ℵ0 = ℵ1 . Of course, a general principle of set theory asserts that arithmetical theorems proved using the continuum hypothesis hold even without assuming it. Our exposition achieves the same result directly. 

20.3 Elementary Equivalence of PAC Fields We wish to remove the restriction of saturation put on the fields E and F from Lemma 20.2.2. This requires, however, a strengthening of (2) of Section 20.2 to an ultraproduct statement:

Chapter 20. The Elementary Theory of e-Free PAC Fields

434

Lemma 20.3.1: Let {Ei | i ∈ I} and {Fi | i ∈ I} be families of fields. For ) → Gal(Ei ) be each i ∈ I let ϕi : Gal(FiQ Q an isomorphism. Q Let D be an Fi /D, and ϕ∗ = ϕi /D. Then: ultrafilter of I. Put E ∗ = QEi /D, F ∗ = Q Gal(Fi )/D → Gal(Ei )/D induces an isomor(a) The isomorphism ϕ∗ : phism ϕ: Gal(F ∗ ) → Gal(E ∗ ). (b) Suppose in addition, there exists a field K which is contained in all Ei and Fi . For each i, let Li be a finite Galois extension of K such that the diagram (1)

Gal(Fi ) Gal(Ei ) o FF FF xx x F x res FFF xxres # {xx Gal(Li /K) ϕi

is commutative. Assume {i ∈ I | L ⊆ Li } ∈ D for each finite extension L of K. Then the diagram (2)

Gal(E ∗ ) o Gal(F ∗ ) CC CC {{ CC {{ { res CC { res ! }{{ Gal(K) ϕ

is commutative. Proof of (a): of F ∗ of degree n. Then there Q Let N be a Galois extension ∗ exists y ∈ Fi,s /D such that N = F (y). Choose a representative {yi | i ∈ I} of y. Then there exists D ∈ D such that N Qi = Fi (yi ) is a Galois extension of Fi of degree n for all i ∈ D, and N = Ni /D. Denote the fixed field the isomorphism of Gal(Ni /Fi ) in Ei,s of ϕi (Gal(Ni )) by Mi . Let ϕN,i be Q Mi /D is a Galois extension of onto Gal(Mi /EQ i ) induced by ϕi . Then M = E ∗ and ϕN = ϕN,i /D is an isomorphism of Gal(N/F ∗ ) onto Gal(M/E ∗ ) which is induced by ϕ∗ . If N 0 is a finite Galois extension of F ∗ that contains N , then the extension M 0 of E ∗ corresponding to N 0 contains M and ϕN 0 induces ϕN on the group Gal(N/F ∗ ). Thus, the inverse limit of the ϕN ’s defines an isomorphism ϕ: Gal(F ∗ ) → Gal(E ∗ ) which is induced by ϕ∗ . Proof of (b): Consider y ∈ Ks . Let L be a finite Galois extension of K which contains y. By assumption, Q D = {i ∈ I | L ⊆ Li } ∈ D. For each Ni /D. For each i ∈ D and each σi ∈ i ∈ I let Ni = LFi and N = Gal(Ni /Fi ) the commutativity of (1) implies ϕN,i (σi )(y) = σi (y). Hence, for each σ ∈ Gal(N/F ∗ ), ϕN (σ)(y) = σ(y). Therefore, ϕ(σ)(y) = σ(y) for each σ ∈ Gal(F ∗ ). Thus, (2) is commutative.  In the special case where Ei = E, Fi = F and ϕi = ϕ0 for each i ∈ I, E and F are canonically embedded in E ∗ and F ∗ , respectively, and ϕ∗ induces ϕ0 on Gal(F ):

20.3 Elementary Equivalence of PAC Fields

435

Corollary 20.3.2: Let E and F be fields, ϕ: Gal(F ) → Gal(E) an isomorphism, and D an ultrafilter of I. Put E ∗ = E I /D and F ∗ = F I /D. Then there exists a commutative diagram (3)

Gal(E ∗ ) o

ϕ∗

res

 Gal(E) o

Gal(F ∗ ) res

ϕ

 Gal(F )

where ϕ∗ is an isomorphism. Theorem 20.3.3 (Elementarily Equivalence Theorem): Let E/L and F/M be separable field extensions with both L and M containing a field K. Suppose E and F are PAC fields having the same imperfect degree. In addition, suppose there exist a K-isomorphism Φ0 : Ls → Ms with Φ0 (L) = M and a commutative diagram (4)

Gal(E) o

ϕ

res

 Gal(L) o

Gal(F ) res

ϕ0

 Gal(M ) ,

where ϕ is an isomorphism and ϕ0 is the isomorphism induced by Φ0 . Then E is K-elementarily equivalent to F . Proof: Assume first L is countable. Choose a countable set I and a nonprincipal ultrafilter D of I. Then, in the notation of Corollary 20.3.2, combine the diagrams (3) and (4) to get a commutative diagram (5)

Gal(E ∗ ) o

ϕ∗

res

 Gal(L) o

Gal(F ∗ ) res

ϕ0

 Gal(M )

where ϕ∗ is an isomorphism. The fields E ∗ and F ∗ are PAC (Corollary 11.3.3) and ℵ1 -saturated (Lemma 7.7.4). By Lemma 20.2.3, E ∗ ≡K F ∗ . Therefore, E ≡K F . For the general case, let θ be a sentence of L(ring, K) which is true in E. There are only finitely many elements of K, say x1 , . . . , xn , that occur in θ. Let K0 be a countable subfield of K that contains x1 , . . . , xn . By Skolem-L¨owenheim (Proposition 7.4.2), L contains an elementary countable subfield L0 that contains K0 . Let M0 = Φ0 (L0 ). Then L/L0 and M/M0

436

Chapter 20. The Elementary Theory of e-Free PAC Fields

are separable extensions and θ is a sentence of L(ring, K0 ). Also, there is a commutative diagram Gal(L) o

ϕ0

 Gal(L0 ) o

ϕ0

Gal(M )  Gal(M0 )

where both horizontal arrows are the isomorphisms induced by Φ0 . The first  part of the proof gives E ≡K0 F . Hence, θ is true in F . A special case of Theorem 20.3.3 is useful in investigating model completeness of PAC fields: Corollary 20.3.4: Let F/K be a separable extension of PAC fields such that F and K have the same imperfect degree. Suppose res: Gal(F ) → Gal(K) is an isomorphism. Then F is an elementary extension of K. The elementary equivalence theorem for separably closed fields is another special case of Theorem 20.3.3. Corollary 20.3.5 ([Ershov1, Prop.]): Let E and F be separably closed fields of characteristic p > 0 with the same imperfect degree. Then E and F are elementarily equivalent. ˜ p in Theorem 20.3.3.  Proof: Set L = M = F

20.4 On e-Free PAC Fields In Theorem 20.3.3 we may assume without loss that E/L and F/M are regular extensions (e.g. as in the proof of Lemma 20.2.2). It is the relation between the groups Gal(E) (resp. Gal(F )) and Gal(L) (resp. Gal(M )) via restriction that complicates applications. When Gal(E) and Gal(F ) are isoutz’ lemma morphic to the free profinite group, Fˆe , on e generators, Gasch¨ (Lemma 17.7.2) comes to our aid: Proposition 20.4.1: Let E and F be e-free PAC fields with the same imper˜ ∩ E and F/K ˜ ∩F fect degree and with a common subfield K. Suppose E/K ∼ ˜ ˜ are separable extensions and K ∩ E =K K ∩ F . Then E ≡K F . ˜ ∩ E. By Proposition 17.7.3, there exists an isomorphism Proof: Put L = K ϕ: Gal(F ) → Gal(E) satisfying resEs /Ls ◦ ϕ = resFs /Ls . Hence, by Theorem  20.3.3, E ≡K F . ˜ ∩ E is also a perfect field and therefore E/K ˜ ∩E If E is perfect, then K is a separable extension. This simplifies Proposition 20.4.1: Corollary 20.4.2: Let K be a subfield of perfect e-free PAC fields E and ˜ ∩E ∼ ˜ ∩ F , then E ≡K F . F . If K =K K

20.4 On e-Free PAC Fields

437

Corollary 20.4.3: Let E and F be e-free PAC fields with the same imperfect degree. If F is a regular extension of E, then F is an elementary extension of E. We axiomatize the concept of e-free PAC fields for model-theoretic applications: Proposition 20.4.4: Let K be a field and e a positive integer. Then there exists a set Ax(K, e) of axioms in the language L(ring, K) such that a field extension F of K satisfies the axioms if and only if it is perfect, PAC and e-free. The axioms are sentences that interpret the field axioms, perfectness axioms, [p 6= 0] ∨ (∀X)(∃Y )[Y p = X], as p ranges over primes, the positive diagram of K (Example 7.3.1) and the following axioms: (a) PAC axioms: Every absolutely irreducible polynomial f (X, Y ) of degree d has a zero, d = 1, 2, . . . . (b) e-free axioms: The finite groups which appear as Galois groups over F are exactly the groups of rank bounded by e. Proof: Section 11.3 translates the PAC axioms into elementary statements. Thus, it suffices to translate the e-free axioms into elementary statements. For this, consider a polynomial f (X) = X n + u1 X n−1 + · · · + un with indeterminate coefficients u1 , . . . , un and a subgroup G of Sn which is given by its action on {1, 2, . . . , n}. Suppose the following assertion is an elementary statement on u1 , . . . , un : “The polynomial f is irreducible, separable, normal and has G as a Galois group.” Then consider all subgroups G1 , . . . , Gr of Sn which are generated by e elements. Restate axiom (b): “For each n and for every irreducible, separable, and normal polynomial f of degree ≤ n, the Galois group of f is isomorphic to one of the groups G1 , . . . , Gr ; and for each group G of order ≤ n and of rank ≤ e, there exists an irreducible, separable, normal polynomial of degree at most n with Galois group isomorphic to G .” The normality condition on f means that a root z of f (X) (in the algebraic closure), gives all other roots as polynomials in z of degree at most n − 1 with coefficients in F . To eliminate the reference to F˜ , use congruences modulo f (X) as follows: There exist polynomials p1 (Z) = Z, p2 (Z), . . . , pn (Z) of degree at most n − 1 with (1)

f (X) ≡

n Y

(X − pi (Z)) mod f (Z).

i=1

Of course, (1) is actually n congruence conditions on the coefficients of the powers of X on both sides. For example, equating the free coefficients on both sides gives the condition un ≡ (−1)n p1 (Z) · · · pn (Z) mod f (Z),

Chapter 20. The Elementary Theory of e-Free PAC Fields

438

which is equivalent to the existence of a polynomial g of degree at most n(n − 1) with un = (−1)n p1 (Z) · · · pn (Z) + g(Z)f (Z). Thus, the normality condition on f is elementary. The condition “Gal(f, F ) is isomorphic to G as a permutation group” may be shown to be elementary by considering the action on the roots of f . Then eliminate the reference to F˜ as before. Indeed, suppose f is monic, irreducible, separable, normal, and p1 (z) = z = z1 , p2 (z) = z2 , . . ., pn (z) = zn are the roots of f with pi ∈ F [X] a polynomial of degree at most |Gal(f, F )| − 1. Suppose σ ∈ Gal(f, F ). Then, for each i, σzi = pi (σz). Hence, zσ(i) = pi (zσ(1) ). Hence, pσ(i) (z) = pi (zσ(1) ). Conversely, suppose σ ∈ Sn satisfies pσ(i) (z) = pi (zσ(1) ) for i = 1, . . . , n. Let τ be the unique element of Gal(f, F ) with τ (1) = σ(1). Then pσ(i) (z) = pi (zσ(1) ) = pi (zτ (1) ) = pτ (i) (z). Hence, σ(i) = τ (i) for each i. Therefore, σ = τ ∈ Gal(f, F ). Consequently, “Gal(f, F ) ∼ = G00 is equivalent to n ^ ^ σ∈G i=1

[pσ(i) (z) = pi (zσ(1) )] ∧

^

n _

[pσ(i) (z) 6= pi (zσ(1) )].



σ∈Sn rG i=1

Remark 20.4.5: (a) If, in an application, K is the quotient field of a distinguished subring R, we may replace the positive diagram of K by the positive diagram of R. (b) When desired, axioms indicating that the imperfect exponent of F is m (0 ≤ m ≤ ∞) may replace the perfect axioms. (c) If K is presented with elimination theory, then Ax(K, e) can be effectively presented. (d) Let K be a field and G a group of order n. Then “G occurs as a Galois group over K” is an elementary statement on K. Indeed, the proof of Proposition 20.4.4 presents the equivalent statement “there is a monic Galois polynomial f ∈ K[X] with Gal(f, K) ∼ = G” as an elementary statement.  Proposition 20.4.6 ([Klingen1]): Let K and L be elementarily equivalent fields. Suppose Gal(K) is a small profinite group. Then Gal(K) ∼ = Gal(L). Proof: By Remark 20.4.5, a finite group G occurs as a Galois group over K if and only if G occurs as a Galois group over L. Thus, Gal(K) and Gal(L) have the same finite quotients. It follows from Proposition 16.10.7 that Gal(K) ∼  = Gal(L).

20.5 The Elementary Theory of Perfect e-Free PAC Fields We interpret the elementary theory of perfect e-free PAC fields that contain a field K in the following cases: (1a) K is finite and e = 1.

20.5 The Elementary Theory of Perfect e-Free PAC Fields

439

(1b) K is countable and Hilbertian, and e ≥ 1. In each of these cases (K, e) is called a Hilbertian pair. ˜ For σ = (σ1 , . . . , σe ) in Gal(K)e , let K(σ) = Ks (σ)ins be the maximal ˜ of the unique purely inseparable extension of Ks (σ). It is the fixed field in K ˜ ˜ have the extension of σ to automorphisms of K. The fields Ks (σ) and K(σ) ˜ ˜ is a same absolute Galois group. If Ks (σ) is PAC, so is K(σ). But K(σ) perfect field. Apply Corollary 18.5.9 and Proposition 18.6.4 to case (1a) and Theorems 18.5.6 and 18.6.1 to case (1b): ˜ Theorem 20.5.1: Suppose (K, e) is a Hilbertian pair. Then K(σ) is a e perfect e-free PAC field for almost all σ ∈ Gal(K) . Recall the regular ultrafilters of Section 7.6 when the index set S is Gal(K)e and “small sets” are the subsets of Gal(K)e of measure zero. In particular, a regular ultrafilter of Gal(K)e contains all subsets of Gal(K)e of measure 1. We will compare an arbitrary e-free PAC field with a regular ˜ ultraproduct of the fields K(σ). ˜ Denote the theory of all sentences of L(ring, K) which are true in K(σ) e for almost all σ ∈ Gal(K) by Almost(K, e). Lemma 20.5.2: Let (K, e) be a Hilbertian pair. (a) A field F is a model of Almost(K, e) if and only if it is K-elementarily ˜ equivalent to a regular ultraproduct of the fields K(σ). ˜ (b) Every regular ultraproduct of the fields K(σ) is a perfect e-free PAC field. Proof: Statement (a) is a special case of Proposition 7.8.1(b). Statement (b) follows from Proposition 20.4.4 and Theorem 20.5.1.  We define the corank of a field K as the rank of Gal(K). Lemma 20.5.3: Let K be a field and e a positive integer. Then, for every perfect field F of corank at most e that contains K, there exists a regular ˜ ˜ ∩E ∼ ˜ ∩ F. ultraproduct E of the K(σ)’s with K =K K Proof: Let τ1 , . . . , τe be generators of Gal(F ). For each finite Galois extension L of K the set S(L) = {σ ∈ Gal(K)e | resL (σ) = resL (τ )} has a positive Haar measure, so S(L) is not small. If L is contained in a larger finite Galois extension L0 of K, then S(L0 ) ⊆ S(L). By Lemma 7.6.1 there exists a regular ultrafilter D of Gal(K)e which contains the sets S(L) as L runs over all finite Galois extensions of K. Q ˜ Let E = K(σ)/D. Then L∩E = L∩F for every finite Galois extension ˜ ∩ E and K ˜ ∩ F are perfect, L of K. Therefore, Ks ∩ E = Ks ∩ F . Since K they are equal.  Theorem 20.5.4: If (K, e) is a Hilbertian pair, then Ax(K, e) (Proposition 20.4.4) is a set of axioms for Almost(K, e). Specifically, a field F is perfect,

440

Chapter 20. The Elementary Theory of e-Free PAC Fields

e-free, PAC, and contains K if and only if it satisfies each sentence θ which ˜ is true in K(σ) for almost all σ ∈ Gal(K)e . Proof: Suppose that F |= Ax(K, e). By Lemma 20.5.3 there exists a regular ˜ ˜ ∩E ∼ ˜ ∩ F . By Lemma ultraproduct E of the K(σ)’s such that K =K K 20.5.2(b), E |= Ax(K, e). Corollary 20.4.2 now gives E ≡K F . From Lemma 20.5.2(a), F |= Almost(K, e). Theorem 20.5.1 gives the converse. 

20.6 The Probable Truth of a Sentence Let K be a field and let e be a positive integer. For a sentence θ of L(ring, K), consider the truth set of θ: (1)

˜ |= θ}. S(K, e, θ) = {σ ∈ G(K)e | K(σ)

Refer to the case where K has elimination theory (Definition 19.2.8) and e and θ are explicitly given, as the explicit case. Regard the measure of ˜ S(K, e, θ) (if it exists) as the probability that θ is true among the K(σ)’s. Call a sentence λ of the form (2)

P ((∃X)[f1 (X) = 0], . . . , (∃X)[fm (X) = 0])

with f1 , . . . , fm ∈ K[X] separable polynomials and P a Boolean polynomial (Section 7.6), a test sentence. In this case it is fairly easy to describe the set S(K, e, λ). Indeed, the splitting field L of the polynomial f1 · · · fm is a finite Galois extension of K. Denote the set of all τ ∈ Gal(L/K)e with L(τ ) |= λ by S0 . Then (3)

S(K, e, λ) = {σ ∈ G(K)e | resL (σ) ∈ S0 }.

Indeed, if λ has the form (∃X)[f1 (X) = 0], then S0 consists exactly of the τ ∈ Gal(L/K)e for which L(τ ) contains at least one root of f1 (X). Since L contains all roots of f1 (X), this gives (3). An induction on the structure of P gives (3) in general. From (3) (4)

µ(S(K, e, λ)) =

|S0 | . [L : K]e

In the explicit case the right hand side of (4) can be computed effectively from Lemma 19.3.2. Lemma 20.6.1: Let K be a field, e be a positive integer, and λ a test sentence. Then µ(S(K, e, λ)) is a rational number which, in the explicit case, can be effectively computed. The reduction of arbitrary sentences to test sentences depends on a general result of field theory:

20.6 The Probable Truth of a Sentence

441

Lemma 20.6.2: Let K ⊆ L ⊆ L0 be a tower of fields. Suppose L0 /K is algebraic and L ∼ =K L0 . Then L = L0 . Proof: Consider x ∈ L0 . Let f = irr(x, K). Denote the set of all zeros of f in L (resp. L0 ) by Z (resp. Z 0 ). Then Z ⊆ Z 0 and |Z| = |Z 0 |, so Z = Z 0 . Therefore, x ∈ L.  Lemma 20.6.3: Let E and F be fields having a common subfield K. (a) Suppose each irreducible polynomial f ∈ K[X] which has a root in E ˜ ∩ E into K ˜ ∩ F. has a root in F . Then there exists a K-embedding of K (b) Suppose an irreducible polynomial f ∈ K[X] has a root in E if and only ˜ ∩ F. ˜ ∩E ∼ if it has a root in F . Then K =K K Proof: Statement (b) follows from (a) by Lemma 20.6.2. Thus, it suffices to prove (a). Assume without loss E and F are algebraic over K. If K is finite and x ∈ E, then K(x) is a Galois extension of K. By assumption, irr(x, K) has a root in F . Hence, K(x) ⊆ F . Therefore, E ⊆ F . Now assume K is infinite. Let L be a finite extension of K in E. Choose ˆ ∩ F . List the ˆ of K which contains L. Put L0 = L a finite normal extension L ˜ as σ1 , . . . , σn . Put Li = σi (L0 ), i = 1, . . . , n. By K-isomorphisms of L0 into K assumption, for each x ∈ L the polynomial f = irr(x, K) has a root x0 ∈ L0 . ˆ Then τ −1 |L0 = σi for Extend the map x 7→ x0 to a K-automorphism τ of L. some i, so x = τ −1 (x0 ) = σi (x0 ) ∈ Li . It follows that L ⊆ L1 ∪ · · · ∪ Ln . Now consider each of the fields L, L1 , . . . , Ln as a subspace of the finite dimensional ˆ over K. Since K is infinite, there exists j with L ⊆ Lj . vector space L Therefore, σj−1 (L) ⊆ L0 ⊆ F . Denote the finite nonempty set of all K-embeddings of L into F by I(L). If L is contained in another finite extension L1 of K, contained in E, then restriction defines a canonical map of I(L1 ) into I(L). Take the inverse limit of the I(L)’s to establish the existence of a K-embedding of E into F (Corollary 1.1.4), as desired.  Corollary 20.6.4: Let E and F be perfect fields with a common subfield K. (a) Suppose each separable irreducible polynomial f ∈ K[X] which has a ˜ ∩ E can be K-embedded into K ˜ ∩ F. root in E has a root in F . Then K (b) Suppose a separable irreducible polynomial f ∈ K[X] has a root in E if ˜ ∩ F. ˜ ∩E ∼ and only if it has a root in F . Then K =K K Proof: Lemma 20.6.3 covers the case where char(K) = 0, so we assume char(K) = p > 0. Let f ∈ K[X] be an irreducible polynomial. Then there exists a separable irreducible polynomial g ∈ K[X] and a power q of p such that f (X) = g(X q ). Since E (resp. F ) is perfect, f has a root in E (resp. F ) if and only if g has one, as well. Therefore, we may apply Lemma 20.6.3 to conclude both (a) and (b).  We combine Corollary 20.4.2 with Corollary 20.6.4:

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Lemma 20.6.5: Let (K, e) be a Hilbertian pair and E and F models of Almost(K, e). Then E ≡K F is and only if E and F satisfy exactly the same test sentences. Proof: Suppose E and F satisfy the same test sentences. Then a separable polynomial f ∈ K[X] has a root in E if and only if f has a root in F . Since E and F are models of Almost(K, e), they are perfect. Hence, by Corollary ˜ ∩E ∼ ˜ ∩ F . Thus, by Corollary 20.4.2, E ≡K F . 20.6.4, K  =K K Proposition 20.6.6: Let (K, e) be a Hilbertian pair. For each sentence θ of L(ring, K) there exists a test sentence λ satisfying: (a) The sets S(K, e, θ) and S(K, e, λ) differ only by a zero set; the sentence θ ↔ λ belongs to Almost(K, e). (b) There is a formal proof (δ1 , . . . , δn ) of θ ↔ λ from the set of axioms Ax(K, e); both λ and (δ1 , . . . , δn ) can be found in the explicit case in a recursive way by checking all proofs from Ax(K, e). Proof: Proposition 7.8.2 gives a test sentence λ satisfying (a). By Theorem 20.5.4, Ax(K, e) is a set of axioms for Almost(K, e). Therefore, by Corollary 8.2.6, Ax(K, e) ` θ ↔ λ, which is the first part of (b). The second part of (b) follows from Proposition 8.7.2.  We use Proposition 20.6.6 to generalize Lemma 20.6.1 to arbitrary sentences: Theorem 20.6.7 ([Jarden-Kiehne]): Let (K, e) be a Hilbertian pair and θ a sentence of L(ring, K). Then µ(S(K, e, θ)) is a rational number which, in the explicit case, can be recursively computed. In particular, the theory Almost(K, e) is recursively decidable. Proof: Indeed, µ(S(K, e, θ)) = 1 if and only if the sentence θ belongs to Almost(K, e).  Remark 20.6.8: Chapter 31 uses algebraic geometry to prove that Almost(K, e) is primitive recursive. 

20.7 Change of Base Field Suppose θ is a sentence of L(ring, K) and K 0 is a field containing K. Then θ is also a sentence of L(ring, K 0 ). It is therefore natural, to relate S(K, e, θ) and S(K 0 , e, θ). More generally, we relate S(K, e, θ) and S(K 0 , e, θ) when θ is an “infinite sentence”. To define the latter concept, we adjoin the symbol W∞ i=1 to the language and define the set of infinite sentences to be the smallest set of strings that satisfy the following rules: (1a) Every sentence of L(ring, K) is an infinite sentence. (1b) If θ is an infinite sentence, then so is ¬θ. W∞ (1c) For θ1 , θ2 , θ3 , . . . a sequence of infinite sentences, i=1 θi is an infinite sentence.

20.7 Change of Base Field

443

The interpretation of infinite sentences is given by the following rule: W∞ For F an extension field of K, F |= i=1 θi if F |= θi for some integer i. If θ is an infinite sentence, use (1) of Section 20.6 to define its truth set, S(K, e, θ). Observe that the operator S(K, e, ∗) commutes with infinite disjunctions. Also, if fields E and F containing K are K-elementarily equivalent, then the same infinite sentences are true in both of them. Consider a regular extension K 0 of K. Let ρ: Gal(K 0 )e → Gal(K)e be the ˜ ˜ ˜ ∩K ˜ 0 (σ) = K(ρ(σ)) = K(σ) for each σ ∈ Gal(K 0 )e . restriction map. Then K 0 e 0 Denote the measure of Gal(K ) by µ and use the rule µ0 (ρ−1 (A)) = µ(A) for each measurable subset A of Gal(K)e (Proposition 18.2.2). Theorem 20.7.1 ([Jacobson-Jarden1, Thm. 1.1]): Let (K, e) and (K 0 , e) be Hilbertian pairs such that K 0 is a regular extension of K. Then: ˜ ˜ 0 (σ). (a) For almost all σ ∈ Gal(K 0 )e , K(σ) ≺K (b) For each infinite sentence θ of L(ring, K), S(K 0 , e, θ) and ρ−1 (S(K, e, θ)) differ only by a zero set. (c) µ0 (S(K 0 , e, θ)) = µ(S(K, e, θ)). ˜ eProof: Denote the set of all σ in Gal(K)e (resp. in Gal(K 0 )e ) with K(σ) free and PAC by S (resp. S 0 ). By Theorem 20.5.1, µ(S) = 1 and µ0 (S 0 ) = 1. ˜ is an elementary Hence, µ0 (ρ−1 (S) ∩ S 0 ) = 1. By Corollary 20.4.3, K(σ) ˜ 0 (σ) for every σ ∈ ρ−1 (S) ∩ S 0 . This completes the proof of (a). subfield of K Statement (b) follows from (a); and (c) follows from (b).  Consider the special case where K 0 = K(t1 , . . . , tr ) is the field of rational functions over K in the variables t1 , . . . , tr . Let R = K[t1 , . . . , tr ] be the corresponding ring of polynomials. Regard a sentence of L(ring, R) as a formula θ(t1 , . . . , tr ) of L(ring, K) involving the variables t1 , . . . , tr . If a1 , . . . , ar ∈ K, then θ(a) is a sentence of L(ring, K). The next theorem generalizes Hilbert’s irreducibility theorem (Exercise 4): Theorem 20.7.2: Let K be a countable Hilbertian field, e a positive integer, θ(t1 , . . . , tr ) a sentence of L(ring, R). Then there exists a separable Hilbert subset H of K r such that (2)

µ0 (S(K 0 , e, θ(t))) = µ(S(K, e, θ(a)))

for each a ∈ H. Proof: Using test sentences divides the proof into two parts. Part A: Reduction to test sentences. Proposition 20.6.6 gives a test sentence λ(t) for θ(t) of the form P (∃X)[f1 (t, X) = 0], . . . , (∃X)[fm (t, X) =
0] with f1 , . . . , fm ∈ K(t)[X] separable polynomials and P a boolean polynomial. Moreover, there exists a formal proof (δ1 (t), . . . , δn (t)) of θ(t) ↔ λ(t) from Ax(K 0 , e). The axioms of the positive diagram of K 0 (Example 7.3.1) involved in this proof have the form r1 (t)+r2 (t) = r3 (t) or r1 (t)·r2 (t) = r3 (t)

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Chapter 20. The Elementary Theory of e-Free PAC Fields

where r1 , r2 , r3 ∈ K(t). Define U to be the K-Zariski open set of all a ∈ Ar at which none of the denominators of the fi ’s and the rj ’s vanishes. Thus, ˜ K(σ) |= δi (a) for each a ∈ U (K), i = 1, . . . , n. Therefore, for each a ∈ U (K) ˜ ˜ e-free and PAC, K(σ) |= θ(a) ↔ λ(a). and for each σ ∈ Gal(K)e with K(σ) 0 Now apply Theorem 20.5.1 to both K and K to obtain (3a) (3b)

µ0 (S(K 0 , e, θ(t))) = µ0 (S(K 0 , e, λ(t)) µ(S(K, e, θ(a))) = µ(S(K, e, λ(a)))

and for each a ∈ U (K).

Part B: Test sentences. Let L0 be the splitting field of f1 (t, X) · · · fm (t, X) over K 0 . Take a primitive element z for L0 /K 0 which is integral over R and let g(t, X) = irr(z, K 0 ). Put h = gf1 · · · gm . Denote the set of all a ∈ U (K) such that the discriminants (and therefore also the leading coefficients) of f1 , . . . , fm , g remain nonzero under the specialization t → a by H 0 . Make H 0 smaller, if necessary, to assume that the specialization t → a induces an isomorphism of Gal(L0 /K 0 ) onto Gal(L/K), where L is the splitting field of h(a, X) over K, that preserves the operation on the roots of f1 , . . . , fn (Lemma 13.1.1(a)). For a ∈ H 0 , the number of σ 0 ∈ Gal(L0 /K 0 )e with L0 (σ 0 ) |= λ(t) is equal to the number of σ ∈ Gal(L/K)e such that L(σ) |= λ(a). Hence µ0 (S(K 0 , e, λ(t))) = µ(S(K, e, λ(a))). Therefore, (2) follows from (3). By Lemma 13.1.1, H 0 contains a Hilbert subset of K r . The theorem follows.  Remark: With K fixed, Section 30.6 analyzes the effect of a change in e on µ(S(K, e, θ)). 

20.8 The Fields Ks (σ1 , . . . , σe ) The free generators theorem (Theorem 18.5.6) and the PAC Nullstellensatz (Theorem 18.6.1) establish properties satisfied by almost all fields Ks (σ). As applications, however, the last sections developed a theory of properties ˜ shared by almost all fields K(σ). The next section explains this shift. Comparison of the theory of these fields to the theory of finite (and therefore ˜ perfect) fields forces us to consider the perfect fields K(σ) rather than the imperfect fields Ks (σ). ˜ Nevertheless, if we replace K(σ) by Ks (σ) and make some obvious changes, the results in Sections 20.4, 20.5, and 20.6 as well as their proofs remain valid. First, the analog of Corollary 20.4.2: Proposition 20.8.1: Let E and F be separable extensions of a field K with the same imperfect degree. Suppose E and F are e-free, PAC, Ks ∩ E ∼ =K Ks ∩ F . Then E ≡K F . Now replace Ax(K, e) by a set of axioms Ax0 (K, e). A field extension F of K satisfies Ax0 (K, e) if and only if F is PAC, e-free, [F : F p ] = [K : K p ],

20.8 The Fields Ks (σ1 , . . . , σe )

445

and F/K is separable. To express the separability of F/K by sentences of L(ring, K), choose a p-basis B of K over K p . Then “F/K is separable” if and only if “B0 is p-independent over F p ” for all finite subsets B0 of B. Having done this, we write the analog of Theorem 20.5.1: Theorem 20.8.2: Suppose (K, e) is an Hilbertian pair. Then, for almost all σ ∈ Gal(K)e , the field Ks (σ) is e-free, PAC, separable over K, and [Ks (σ) : Ks (σ)p ] = [K : K p ]. We denote the theory of all sentences of L(ring, K) which are true in Ks (σ) for almost all σ ∈ Gal(K)e by Almost0 (K, e). Then the analog of Lemma 20.5.2 holds: Lemma 20.8.3: Let (K, e) be an Hilbertian pair. (a) A field F is a model of Almost0 (K, E) if and only if it is K-elementarily equivalent to a regular ultraproduct of the fields Ks (σ). (b) Every regular ultraproduct of the fields Ks (σ) is e-free and PAC, has the same imperfect degree as K, and is separable over K. Now we present the analog of Lemma 20.5.3: Lemma 20.8.4: Let K be a field, e a positive integer, and F a field of corank at most e which is separable over K. Then there exists a regular ultraproduct E of the Ks (σ)’s with Ks ∩ E ∼ =K Ks ∩ F . This gives the analog of Theorem 20.5.4: Theorem 20.8.5: Let (K, e) be an Hilbertian pair. Then Ax0 (K, e) is a set of axioms for Almost0 (K, e). Finally, for a sentence θ of L(ring, K) let S 0 (K, e, θ) = {σ ∈ Gal(K)e | Ks (σ) |= θ}. Then the following analog of Theorem 20.6.7 holds: Theorem 20.8.6: Let (K, e) be an Hilbertian pair and θ a sentence of L(ring, K). Then µ(S 0 (K, e, θ)) is a rational number which, in the explicit case, can be recursively computed. In particular, the theory Almost0 (K, e) is recursively decidable. The results of Section 20.7 are in general false for the fields Ks (σ). Suppose for example, K = Fp (t) and K 0 = Fp (t, u) with t, u algebraically independent elements over Fp . Then the separable extensions of K have imperfect exponent 1 while the separable extensions of K 0 have imperfect exponent 2. Therefore, no Ks0 (σ) is elementarily equivalent to Ks (σ).

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Chapter 20. The Elementary Theory of e-Free PAC Fields

20.9 The Transfer Theorem This section connects the elementary theory of finite fields with the elemen˜ tary theory of the fields K(σ). Let K be a global field and OK the ring of integers of K. Denote the set of nonzero prime ideals of OK by P (K). It is equipped with the Dirichlet density δ (Section 6.3). We consider models of the language L(ring, OK ) ¯ p , for that are field extensions either of K or of one of the residue fields K p ∈ P (K). ¯ p for almost By Proposition 7.8.1, a sentence θ of L(ring, OK ) is true in K all p ∈ P (K) if and only if θ is true in every nonprincipal ultraproduct of ¯ p /D is one of these ultraproducts, then, by Proposition ¯ p ’s. If F = Q K the K 7.9.1 and Corollary 11.3.4, F is a perfect, 1-free, PAC field that contains K. Since K is a global field, it is Hilbertian (Theorem 13.4.2). Hence, by Theorem 20.5.4, F is a model of Almost(K, 1). We have therefore proved: ˜ Lemma 20.9.1: If a sentence θ of L(ring, OK ) is true in K(σ) for almost all ¯ σ ∈ Gal(K), then θ is true in Kp for almost all p ∈ P (K). For a given sentence θ of L(ring, OK ) we compare the sets ˜ S(θ) = S(K, 1, θ) = {σ ∈ Gal(K) | K(σ) |= θ} and ¯ A(θ) = A(K, θ) = {p ∈ P (K) | Kp |= θ}, using the Dirichlet density δ of P (K) and the Haar measure µ of Gal(K). Lemma 20.9.2: Let λ be the test sentence (1)


p (∃X)[f1 (X) = 0], . . . , (∃X)[fm (X) = 0] ,

where f1 , . . . , fm ∈ K[X] are separable polynomials and p is a Boolean polynomial. Let B be the set of all p ∈ P (K) such that all coefficients of fi are p-integral and the leading coefficient and the discriminant of the fi ’s are p-units, i = 1, . . . , m. Denote the splitting field of f1 · · · fm over K by L. Then: (a) For each p ∈ B, every P ∈ P (L) over p, every σ ∈ DP , and every field ¯ p satisfying L ¯P ∩ F = L ¯ P (¯ extension F of K σ ) (where σ ¯ is the image of ¯ ¯ σ under the map DP → Gal(LP /Kp ) induced by P) we have L(σ) |= λ ⇐⇒ F |= λ. ¯ ¯ (b) Let S(λ) = {σ ∈ Gal(L/K) | L(σ) |= λ}. Then δ(A(λ)) = |S(λ)| . [L:K]

Proof of (a): First note that B is a cofinite set, because each fi is separable. Suppose p ∈ B. Then p is unramified in L (Section 6.2). If λ is (∃X)[fi (X) = 0], statement (a) is a reinterpretation of Lemma 6.1.8(a). The general case follows by induction on the structure of λ. Proof of (b): Use (a) and the Chebotarev density theorem (Theorem 6.3.1). 

20.9 The Transfer Theorem

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Theorem 20.9.3 (The Transfer Theorem): Let OK be the ring of integers of a global field K and θ a sentence of L(ring, OK ). Then S(θ) is measurable, A(θ) has a Dirichlet density, and µ(S(θ)) = δ(A(θ)). Proof: Proposition 20.6.6 provides a test sentence λ of the form (1) with ˜ θ ↔ λ true in K(σ) for almost all σ ∈ Gal(K). Without loss assume the coefficients of the polynomials f1 , . . . , fm belong to OK . Thus (Lemma 20.9.1), ¯ p , for almost all p ∈ P (K). Hence, S(θ) ≈ S(λ) (i.e. S(θ) θ ↔ λ is true in K and S(λ) differ by a set of measure zero) and A(θ) ≈ A(λ) (i.e. A(θ) and A(λ) differ by a finite set). Therefore, it suffices to prove the theorem for λ, rather than for θ. Let L be the splitting field of the polynomial f1 . . . fm . Then L is a ¯ finite Galois extension of K, S(λ) = {τ ∈ Gal(L/K) | L(τ ) |= λ} is a union ¯ of conjugacy classes of Gal(L/K), and S(λ) = {σ ∈ Gal(K) | resL σ ∈ S(λ)}. Hence, µ(S(λ)) =

¯ |S(λ)| [L:K] .

By Lemma 20.9.2, 

(2)

and δ(A(λ)) =

A(λ) ≈ {p ∈ P (K) | ¯ |S(λ)| [L:K] .

L/K p



¯ ⊆ S(λ)}

Consequently, µ(S(λ)) = δ(A(λ)).



¯ If δ(A(λ)) = 0, then µ(S(λ)) = 0 and S(λ) is empty. Therefore, by (2), A(λ), hence A(θ), are finite sets. Therefore, Theorem 20.6.7 gives the following supplement to Theorem 20.9.3: Theorem 20.9.4 ([Ax1, p. 161, Cor.]): Let θ be a sentence of L(ring, OK ). Then δ(A(θ)) is a rational number which is positive if A(θ) is infinite. In the explicit case δ(A(θ)) can be recursively computed. The special case K = Q gives the decidability of the theory of sentences which are true in Fp for almost all p. The next result represents this theory in two more ways. Consider the set Q of all powers of prime numbers. Call a subset B of Q small if only finitely many prime numbers divide the elements of B. Proposition 20.9.5: The following three statements about a sentence θ of L(ring) are equivalent: ˜ (a) Q(σ) |= θ for almost all σ ∈ Gal(Q). (b) Fp |= θ for almost all p ∈ P (Q). (c) Fq |= θ for almost all q ∈ Q (i.e. for all but a small set). Proof: The equivalence “(a) ⇐⇒ (b)” is a special case of the transfer theorem. The implication “(c) =⇒ (b)” follows from the definitions. Finally, the implication “(a) Q =⇒ (c)” follows from Lemma 20.5.2(a), because every regular ultraproduct Fq /D, where D contains all complements of small sets of Q, is a 1-free PAC field of characteristic zero (Corollary 11.3.4). 

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Chapter 20. The Elementary Theory of e-Free PAC Fields

Corollary 20.9.6 ([Ax2, p. 265]): Let K be a given global field. Then the ¯ p for almost all theory of all sentences in L(ring, OK ) which are true in K p ∈ P (K) is recursively decidable. Proof: By Theorem 20.9.4, the set of all sentences θ of L(ring, K) with δ(A(θ)) = 1 is recursive.  If δ(A(θ)) = 1, then there are only finitely many p ∈ P (K) for which θ ¯ p . The proof of the next theorem gives a recursive procedure may be false in K for displaying these primes: Theorem 20.9.7 ([Ax2, p. 264]): Let K be a given global field. Then the ¯ p for all p ∈ P (K) is recursively decidable. theory of all sentences true in K Proof: We follow the pattern of proof of Theorem 20.9.3. Part A: Finding a test sentence. Let θ ∈ L(ring, OK ). Proposition 20.6.4 recursively gives a test sentence λ of the form (2) of Section 20.6 and a formal proof (δ1 , . . . , δn ) from Ax(K, 1) (in the language L(ring, K)) of θ ↔ λ. Let A0 be the set of all p ∈ P (K) that divide one of the denominators of the elements of K involved in δ1 , . . . , δn . If a field F of characteristic not in A0 contains a homomorphic image of OK and if the axioms among the δi are true in F , then (δ1 , . . . , δn ) is a valid proof in F . In particular, θ ↔ λ is true in F . Part B: Reduction modulo p of the PAC axioms. Let 1 ≤ i ≤ n. If δi is the axiom “each absolutely irreducible polynomial f (X, Y ) of degree d has ¯ p | ≤ (d − 1)4 . By a zero”, define Ai to be the set of all p ∈ P (K) with |K ¯ Corollary 5.4.2, if p ∈ P (K) r Ai , then Kp |= δi . Otherwise, let Ai = A0 . ¯ p is perfect and 1-free, K ¯ p |= θ ↔ λ for Let B1 = A0 ∪ A1 ∪ · · · ∪ An . Since K each p ∈ P (K) r B1 . Part C: Exceptional primes for λ. Construct the splitting field L of the product of the polynomials f1 , . . . , fm appearing in λ and check if there exists ¯p τ ∈ Gal(L/K) with L(τ ) 6|= λ. In this case, λ (and therefore θ) is false in K L/K
for almost all p ∈ P (K) with τ ∈ p (Lemma 20.9.2). By Chebotarev, infinitely many primes p satisfy the latter condition. Assume therefore that L(τ ) |= λ for each τ ∈ Gal(L/K). Compute a finite subset B2 of P (K) including all p which are either ramified in L or divide one of the discriminants ¯ p |= λ (or the leading coefficients) of f1 , . . . , fm . Let B = B1 ∪ B2 . Then K ¯ r (Lemma 20.9.2), so Kp |= θ for each p ∈ P (K) B. ¯ p |= θ for each of the Complete the procedure by checking whether K finitely many exceptional primes p ∈ B. 

20.10 The Elementary Theory of Finite Fields We conclude with a discussion of the theory of finite fields and its decidability. ˆ (i.e. 1-free), Call a field F pseudo finite if F is perfect, Gal(F ) ∼ =Z and PAC.

20.10 The Elementary Theory of Finite Fields

449

Lemma 20.10.1: Every nonprincipal ultraproduct of distinct finite fields is pseudo finite. Proof: Let F1 , F2 , F3 , . . . be distinct finite fields. Consider a nonprincipal Q ultrafilter D of N and let F = Fi /D. Then F is perfect, 1-free, and PAC (Proposition 20.4.4). Thus, F is pseudo finite.  The following result is a special case of Corollary 20.4.2 in the case e = 1: Proposition 20.10.2: Let E and F be pseudo finite fields. ˜ ∼ ˜ Then (a) Suppose E and F contain a common field K and E ∩ K =K F ∩ K. E ≡K F . (b) Suppose F is a regular extension of E. Then E ≺ F . Lemma 20.10.3: Let K be a finite field and L an algebraic extension of K. For each positive integer n denote the unique extension ofQK of degree n by Kn . Then there exists a nonprincipal ultraproduct D = Kn /D such that ˜ = L. D∩K Proof: For each positive integer d the set Ad = {n ∈ N | Kn ∩ Kd = L ∩ Kd } is infinite. If d|d0 , then Ad0 ⊆ Ad . Thus, given d1 , . . . , dr , we put d = lcm(d1 , . . . , dr ) and conclude from the relation Ad ⊆ Ad1 ∩ · · · ∩ Adr that the latter intersection is an infinite set. By Lemma 7.5.4, there Q exists a nonprincipal ultrafilter D on N which contains each Ad . Let D = Kn /D be the corresponding ultraproduct. ˜ = L: Consider a positive integer d ∈ D, let x be We prove that D ∩ K a primitive element for Kd over K, and put f = irr(x, K). Then f is Galois over K. If Kd ⊆ L, then Kd ⊆ Kn for each n ∈ Ad , so f has a root in Kn for each n ∈ Ad . By Loˇs (Proposition 7.7.1), f has a root in D, so D ⊆ L. If Kd 6⊆ L, then L ∩ Kd ⊂ Kd , so f has no root in Kn for each n ∈ Ad . By ˜ = L. Loˇs, f has no root in D, so Kd 6⊆ D. It follows that D ∩ K  Proposition 20.10.4: Let θ be a sentence of L(ring). Then θ is true in almost all (i.e. all but finitely many) finite fields if and only if θ is true in every pseudo finite field. Proof: First suppose θ is true in almost all finite fields. Consider a pseudo finiteQ field F . By Lemma 20.10.3, there exists a nonprincipal ultraproduct ˜p = D ∩ F ˜ p . By Lemma 20.10.1, D is pseudo D = Fpn /D such that F ∩ F finite. Hence, by Proposition 20.10.2, F ≡K D. By Loˇs (Proposition 7.7.1), θ is true in D. Therefore, θ is true in F . . of distinct finite Conversely, suppose θ is false in a sequence F1 , F2 , F3 , . .Q fields. Choose a nonprincipal ultrapower D of N. Put F = Fi /D. By Loˇs, θ is false in F . By Lemma 20.10.1, F is pseudo finite.  Corollary 20.10.5 ([Ax2, p. 240]): A field F is pseudo finite if and only if F is an infinite model of the theory of finite fields. Proof: First suppose F is pseudo finite. Then, by Proposition 20.10.4, each sentence θ which holds in every finite field holds in F .

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Chapter 20. The Elementary Theory of e-Free PAC Fields

Conversely, suppose F is an infinite model of the theory of finite fields. ˆ (Proposition 20.4.4). For each positive Then F is perfect and Gal(F ) ∼ =Z integer d, let θd be the sentence “there are at most (d − 1)4 distinct elements or every absolutely irreducible polynomial in the variables X, Y of degree at most d has a zero”. By Corollary 5.4.2(b), θd holds in each finite field. Hence, θd holds in F . But F is infinite. Hence, every absolutely irreducible polynomial of degree d has a zero in F . Consequently, F is PAC and therefore pseudo finite.  Like the theory of all residue fields of a given global field, the theory of finite fields is decidable. This was a problem raised by Tarski and solved by Ax: Theorem 20.10.6 ([Ax2, p. 264]): The theory of all sentences of L(ring) which are true in every finite field is recursively decidable. Proof: Consider θ ∈ L(ring). Follow the proof of Theorem 20.9.7 in the case K = Q to check if θ is true in Fp for all p ∈ P (Q). In the affirmative case, choose an integer n greater than (d − 1)4 for all d’s that appear in Part B and greater than of the primes belonging to Part C of that proof. Then conclude from Lemma 20.9.2 at the end of Part C of Theorem 20.9.7 that Fpi |= θ for each p ≥ n and each i. We therefore only need to check if, for a given prime p, the sentence θ is true in Fpi for all i ∈ N. This is equivalent to checking if θ is true in Fp (t)p for each p ∈ P (Fp (t)). This, again is a special case of Theorem 20.9.7. Now, the proof is complete.  Example 20.10.7: Pseudo finite fields. (a) If a sentence θ of L(ring) holds in infinitely many finite fields, then it holds in every nonprincipal ultraproduct of those fields. Hence, by Lemma 20.10.1, θ holds in every pseudo finite field. Similarly, let K be a global field and θ a sentence of L(ring, OK ). Sup¯ p with p ranging over an infinite set A of prime divisors of pose θ holds in K ¯ p ’s. Reduction K. Then θ holds in each nonprincipal ultraproduct F of the K modulo p embeds OK in F . Thus, F extends K. ˜ (b) Let K be a global field. Then K(σ) is pseudo finite for almost all σ ∈ Gal(K) (Theorem 20.5.1 for e = 1). Consider a sentence θ of L(ring, OK ) ¯ p for infinitely many prime p of K. By Theorem 20.9.4, which holds in K δ(A(θ)) > 0. Hence, by the Transfer Theorem, µ(S(θ)) > 0. In particular, ˜ ˜ there is a σ ∈ Gal(K) such that K(σ) is pseudo finite and θ holds in K(σ). (c) Let F be an infinite algebraic extension of Q Fp . Then F is perfect and PAC (Corollary 11.2.4). Moreover, Gal(F ) ∼ = l∈S Zl for some set S of prime numbers (Exercise 8 of Chapter 1). Suppose F has an extension of ˆ Hence, F is a pseudo degree l for every prime number l. Then Gal(F ) ∼ = Z. finite field. 

Exercises

451

Theorem 20.10.8: K be a global field. Denote the set of all nonprinciQ Let ¯ p /D, where p ranges over P (K), by F. Let T be the pal ultraproducts K set of all sentences θ ∈ L(ring, OK ) which hold in almost all residue fields ¯ p . Then: K (a) Each F ∈ F is pseudo finite. (b) A sentence θ of L(ring, OK ) belongs to T if and only if θ holds in every F ∈ F. (c) Every model of T is elementarily equivalent in L(ring, OK ) to some F ∈ F. ˜ ∼ ˜ (d) For every σ ∈ Gal(K) there is an F ∈ F with F ∩ K =K K(σ). Proof: Statement (a) is a special case of Lemma 20.10.1. Statement (b) is a special case of Proposition 7.8.1(a). Statement (c) is a special case of Proposition 7.8.1(b). Statement (d) can be proved directly from the Chebotarev density theorem, but we deduce it here from Lemma 20.5.3. By that lemma, ˜ ˜ )’s with E ∩ K ˜ ∼ By there is a regular ultraproduct E of the K(τ =K K(σ). Lemma 20.5.2, E is a pseudo finite field. By (c), there is an F ∈ F which is elementarily equivalent in L(ring, OK ) to E. It follows from 20.6.3, that ˜ ∼ ˜ Consequently, F ∩ K ˜ ∼ ˜ E∩K  =K F ∩ K. =K K(σ).

Exercises 1. Give an example showing that the hypothesis of Corollary 20.3.4 can hold nontrivially, by taking K to be a PAC field for which Gal(K) is finitely generated. Then, let F be a nonprincipal ultraproduct of countably many copies of K. 2. Let λ be the test sentence (Section 20.6) (∃X)[f1 (X) = 0 ∨ f2 (X) = 0 ∨ f3 (X) = 0] with fi (X) = X 2 − ai and ai ∈ Z for i = 1, 2, 3, nonzero integers. What is the exact condition on a1 , a2 , a3 such that µ(S(K, 1, λ)) = 1? Now answer the same question for µ(S(K, 2, λ)) = 1. 3. [Jarden8, p. 149] It is a consequence of Proposition 20.6.6(a) that if (K, e) is a Hilbertian pair and θ is a sentence of L(ring, K), then S(K, e, θ) is a measurable set. We outline an alternative proof, valid for every countable field K and every positive integer e: ˜ define Let ϕ(X1 , . . . , Xn ) be a formula of L(ring, K). For x1 , . . . , xn ∈ K ˜ ˜ and K(σ) |= ϕ(x)} S(K, e, ϕ(x)) = {σ ∈ Gal(K)e | x1 , . . . , xn ∈ K(σ) (a) Use induction on structure to show that S S(K, e, ϕ(x)) is a Borel set. Hint: Use the identity S(K, e, (∃Y )ϕ(x, Y )) = y∈K˜ S(K, e, ϕ(x, y)). (b) Deduce that S(K, e, θ) is a Borel set for each infinite sentence of L(ring, K).

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4. Prove the converse of Theorem 20.7.2: Let K be a field and t an indeterminate. Consider the ring R = K[t] and the field K 0 = K(t). Suppose for each positive integer e and each sentence θ(t) of L(ring, R) there exists a nonempty subset H of K with µ0 (S(K 0 , e, θ(t))) = µ(S(K, e, θ(a))) for each a ∈ H. Prove that K is Hilbertian. Hint: Let f ∈ K[X] be separable. Then f is irreducible over K if and only if for all large e there exists S ⊆ Gal(K)e of positive measure such that ˜ f is irreducible over K(σ) for all σ ∈ S. 5. ([Ax2, p. 260]) Let K be a global field τ ∈ Gal(K). Prove that there Q and ˜ ∩F ∼ ˜ ). ¯ p /D with K exists a nonprincipal ultraproduct F = K = K(τ Hint: Either reproduce Ax’s direct application of the Chebotarev density theorem, or combine the Transfer theorem, Lemma 20.5.3 and Proposition 7.8.1. 6. Give an example of an infinite sentence of L(ring) which is true in each ˜ field Q(σ) but is false in each field Fp . Thus, the Transfer theorem does not generalize to infinite sentences. Indeed, the Haar measure is σ-additive while the Dirichlet density is only finitely-additive. 7. Verify the Transfer theorem over Q for the test sentence (∃X)[X 2 = a], where a is an integer, by using quadratic reciprocity and the Dirichlet density theorem (Corollary 6.3.2) for primes in arithmetic progressions. 8. Let K be a global field and f ∈ K[X] be separable and irreducible with deg(f ) = n > 1. Denote the set of prime numbers p such that f has no zero mod p by A. Put G = Gal(f, K) and let S be the set of all σ ∈ G which fix no zero of f . |S| (a) Apply the Transfer theorem to prove: δ(A) = |G| . Alternatively, use the Chebotarev density theorem to prove the equality directly. 1 1 1 (b) Suppose Gal(f, K) ∼ − 3! + · · · + (−1)n n! . = Sn . Prove: δ(A) = 2! 9. Let f ∈ Z[X, Y ] be an absolutely irreducible polynomial. Prove that for almost all primes p there exists a ∈ Fp such that f (a, Y ) decomposes into linear factors in Fp [Y ]. Hint: Let F = Q(x, y) be the function field of V (f ). Choose a stabilizing element t for the extension F/Q (Theorem 18.9.3). Let Fˆ be the Galois hull of F/Q(t). One way to complete the proof is to specialize t to elements of Q infinitely often so that x, y are integral over the corresponding local ring and such that the residue fields of Fˆ form a linearly disjoint sequence of Galois extensions of Q (as in the proof of Theorem 18.10.3). Use BorelCantelli (Lemma 18.3.5) to prove that for almost all σ ∈ Gal(K) there exists ˜ a ∈ Q(σ) such that f (a, Y ) decomposes into linear factors. Now use the transfer principle. Alternatively, let z be a primitive element for the extension Fˆ /Q(t). Let g ∈ Z[T, Z] be an absolutely irreducible polynomial with g(t, z) = 0. Now find an h ∈ Z[X], h 6= 0 such that for almost all p and for each b ∈ Fp

Notes

453

with h(b) 6= 0 and g(b, T ) has zeros in Fp the specialization t → b extends to a specilization (t, x) → (b, a) and the polynomial f (a, Y ) decomposes into linear factors. Thus, the result follows directly from Corollary 5.4.2. 10. Show that the small sets of Q in Section 20.9 cannot be replaced by finite sets, by giving a sentence θ ∈ L(ring) for which θ is true for all p ∈ P (Q) but the q ∈ Q for which Fq 6|= θ is infinite. 11. Let K be a number field. Denote the set of all prime ideals of OK by P (K). For each finite extension L of K let Splt(L/K) be the set of all p ∈ P (K) which split completely in L. (a) Prove that there exists an ultrafilter L of P (K) which contains Splt(L/K) for each finite extension L of K and each subset of P (K) of Dirichlet density 1. Q ˜ ⊆ F. ¯ p /L. Prove that Q (b) Let L be as in (a). Put F = p∈P (K) K 0 (c) Let LQbe another ultrafilter of P (K) satisfying the conditions of (a). Put ¯ p /L0 . Prove that F 0 ≡ F . F0 = K

Notes Ax initiated the investigation of free PAC fields in [Ax1] and [Ax2] by connecting the theory of finite fields with the theory of pseudo finite fields. This connection is based on the special case of the embedding lemma (Lemma 20.2.2) where E and F are 1-free and Gal(L) is procyclic (Ax’s proof [Ax2, p. 248] is very complicated) and on Exercise 5. The transfer theorem [Jarden2] strengthens this connection. The treatment of perfect e-free PAC fields can be found in [JardenKiehne]. [Cherlin-v.d.Dries-Macintyre] introduces the treatment of imperfect fields that are e-free and PAC. Finally, we follow [Jacobson-Jarden1] and include finite fields, with e = 1 as base fields, in addition to the countable Hilbertian fields. In [Ax1, Lemma 5], Ax proves Lemma 20.6.3(b) essentially for the case ˜ ∩E/K and K ˜ ∩F/K are separable extensions. His reduction of where both K the general case to the separable case is vague. The argument which appears in our proof and which uses vector spaces is due independently to B. Poizat [Poizat] and to H. W. Lenstra.

Chapter 21. Problems of Arithmetical Geometry We apply the model theory - measure theory technique of Chapter 20 to concrete field arithmetic problems. The transfer principles between properties ˜ of finite fields and properties of the fields K(σ) can often be accomplished through direct application of the Chebotarev density theorem. Usually, however, application of the transfer theorem avoids a repetition of arguments. This chapter includes the theory of Ci -fields, Kronecker conjugacy of global field extensions, Davenport’s problem on value sets of polynomials over finite fields, a solution of Schur’s conjecture on permutation polynomials, and a solution of the generalized Carlitz’s conjecture on the degree of a permutation polynomial in characteristic p. Each of these concrete problems focuses our attention on rich historically motivated concepts that could be overlooked in an abstract model theoretic viewpoint.

21.1 The Decomposition-Intersection Procedure The classical diophantine concern is the description of the Q-rational (resp. Zrational) points of a Zariski Q-closed set A. The decomposition-intersection procedure reduces this concern to the study of a union A∗ of subvarieties (Section 10.2) of A defined over Q. We start with an arbitrary base field K. To each nonempty Zariski K-closed set A in affine space, An , or in the projective space, Pn , there corresponds a canonical K-closed subset A0 defined as follows: First, decompose S ˜ A into its K-components, A = i Vi . Then decompose each Vi into its KS components, Vi = j Wij . For a fixed i, {Wij }j is a complete set of conjugate ˜ (Section 10.2). Thus, the intersection Ui = T Wij varieties, defined over K j S is invariant under the action of Gal(K): Ui is a K-closed set. Denote i Ui by A0 . Continue the procedure to obtain a descending sequence A ⊇ A(1) ⊇ (2) A ⊇ · · · ⊇ A(m) ⊇ A(m+1) ⊇ · · · of K-closed sets, with A(m+1) = (A(m) )0 . Hilbert’s basis theorem (Lemma 10.1.1) gives an integer m with A(r) = A(m) for all r ≥ m. Denote A(m) by A∗ . ˜ of A, A(1) , . . . , A(m) . The Let W1 , . . . , Ws be a list of all K-components 0 compositum, L , of the fields of definition of W1 , . . . , Ws is a finite normal extension of K. The maximal separable extension L of K in L0 is the smallest Galois extension of K such that each of W1 , . . . , Ws is L-closed. Call L the Galois splitting field of A over K. Lemma 21.1.1: Let M be a field extension of K with L ∩ M = K. Then: (a) A∗ contains all M -closed subvarieties of A. In particular, A∗ (M ) = A(M ).

21.2 Ci -Fields and Weakly Ci -Fields

455

(b) A∗ is nonempty if and only if A contains an M -closed nonempty subvariety. Proof of (a): Let W be a nonempty M -closed subvariety of A. Choose be a generic point x of W over M . Then x is a generic point of W over LM . Also, there exist i and j with x ∈ Wij . Hence, W ⊆ Wij . Each automorphism of L/K extends to an automorphism of LM/M . Since, for aTfixed i, the Wij ’s are conjugate under the action of Gal(L/K), we have W ⊆ j Wij = Ui ⊆ A0 . Proceed by induction to show that W ⊆ A∗ . In particular, each M -rational point of A belongs to A∗ . Therefore, A(M ) = A∗ (M ). Proof of (b): Suppose first A∗ is nonempty. Assume without loss that A = A∗ . Then A = A0 and A is nonempty. For each i we have Ui ⊆ Vi . Conversely, let x be a generic point of Vi over K. Then, there exists an i0 such that x ∈ Ui0 . Hence, x ∈ Vi0 , so Vi = Vi0 . Thus, i = i0 and Vi ⊆ Ui . It follows that Ui = Vi . This implies that Vi is absolutely irreducible. Otherwise there would be atTleast two Wij ’s and hence we would have the contradiction dim(Ui ) = dim( j Wij ) < dim(Vi ) (Lemma 10.1.2). Thus, Vi is a variety which is K-closed, hence also M -closed. The converse follows from (a). 

21.2 Ci -Fields and Weakly Ci -Fields Recall that a form f (X0 , . . . , Xn ) with coefficients in a field K defines a projective Zariski K-closed set in Pn (Section 10.7). If x = (x0 , . . . , xn ) 6= (0, . . . , 0) is a nontrivial zero of f , then the (n + 1)-tuple (ax0 , . . . , axn ) defines the same point of Pn for each a ∈ K × . Since projective hypersurfaces are the simplest projective sets, they occupy a special place in diophantine investigations. We follow the lead of a problem due to Artin: Does each form f ∈ Fq [X0 , . . . , Xd ] of degree d have a nontrivial zero? Chevalley’s affirmative solution [Chevalley1] motivated Lang [Lang1] to explore the concept of a Ci -field: Definition 21.2.1: The field K is called Ci,d if each form f ∈ K[X0 , . . . , Xn ] of degree d with di ≤ n has a nontrivial zero in K n+1 . Call it Ci if it is Ci,d for each d ∈ N. In this case, K is Cj for each j ≥ i.  Example 21.2.2: Every algebraically closed field K is C0 . Indeed, every form f ∈ K[X0 , . . . , Xn ] of positive degree with n ≥ 1 has a nontrivial zero in  K n+1 . Lemma 21.2.3 (Chevalley-Warning): Let f1 , . . . , fm ∈ Fq [X1 , . . . , Xn ] be polynomials with deg(f1 · · · fm ) < n. Put, A = V (f1 , . . . , fm ). Then |A(Fq )| ≡ 0 mod p.
Qm Proof: Consider g(X) = j=1 1−fj (X)q−1 . If x ∈ A(Fq ), then fj (x) = 0, j = 1, . . . , m. Hence, g(x) = 1. If x ∈ Fnq r A(Fq ), then fj (x) 6= 0 for some

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j. Therefore, fj (x)q−1 = 1 and g(x) = 0. This gives X (1) |A(Fq )| + pZ = g(x). x∈Fn q

P Rewrite g(X) as ci X1i1 · · · Xnin where I is a finite set and ci ∈ F× q for each i ∈ I. Then   X  X X  X (2) g(x) = ci xi11 · · · xinn . x∈Fn q

i∈I

x1 ∈Fq

xn ∈Fq

P Let i ∈ I. If i = (0, . . . , 0), then x1 ∈Fq xi11 = 0+pZ. Otherwise, i1 +· · ·+in ≤ Pm deg(g) ≤ j=1 deg(fj )(q − 1) < n(q − 1), so there is a j with 1 ≤ ij < q − 1. Put k = ij . Since the polynomial X k − 1 has at most k roots in F× q , there is × k an a ∈ Fq with a 6= 1. Hence, X x∈Fq

xk =

X

(ax)k = ak

x∈Fq

X

xk ,

x∈Fq

P so x∈Fq xk = 0. Thus, the right hand side of (2) is 0. By (1), |A(Fq )| ≡ 0 mod p.  If f1 , . . . , fm are forms, (0, . . . , 0) ∈ A(Fq ). Hence, by Lemma 21.2.3, V (Fq ) has at least p points. An application of this argument to the case when m = 1 gives Chevalley’s result: Proposition 21.2.4: Every finite field is C1 . Ax applied ultraproducts to deduce from Chevalley’s result that each perfect PAC field with Abelian absolute Galois group is C1 (Theorem 22.9.6). He left unsolved this question: Problem 21.2.5 (Ax): Is every perfect PAC field C1 ? If Ax’s problem has an affirmative solution, then each form of degree d over Q in d+1 variables has a nonempty Q-closed subvariety (Lemma 21.3.1). This motivates the introduction of the weak Ci condition. We use it to prove some results about PAC fields which are Ci . Definition 21.2.6: A field K is called weakly Ci,d , if for each form f ∈ K[X0 , . . . , Xn ] of degree d, with di ≤ n, the Zariski K-closed set V (f ) of Pn contains a subvariety W which is Zariski K-closed. If K is weakly Ci,d for  each d ∈ N, we say that K is weakly Ci . Remark 21.2.7: By definition, every Ci,d field is also weakly Ci,d . In addition, every perfect PAC field K which is weakly Ci,d is also Ci,d . Perfectness here guarantees that a Zariski K-closed variety is defined over K (Lemma 10.2.3). 

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Example 21.2.8: Separably closed fields. Let K be a field of positive characteristic p. Suppose K has an infinite sequence a1 , a2 , a3 , . . . of p-independent elements over K p . Then, a1 , a2 , a3 , . . . are p-independent over Ksp (Lemma Pqi 2.7.3). Hence, for each i and for all x0 , . . . , xqi ∈ Ks the relation j=0 aj xpj = 0 implies x0 , . . . , xqi = 0. Therefore, Ks is not Ci although Ks is PAC (Section 11.1). On the other hand, each form f ∈ K[X0 , X1 ] has a nontrivial zero ˜ 2 . Hence, Ks is weakly C0 . (x0 , x1 ) ∈ K For example, take K = Fp (t1 , t2 , t3 , . . .) with t1 , t2 , t3 , . . . algebraically independent over Fp . Then t1 , t2 , t3 , . . . are p-independent over K p (proof of  Lemma 2.7.2). Thus, Ks is weakly Ci for no positive integer i. We explore the behavior of the weakly Ci property under field extensions. Recall: An extension F of K is primary if Ks ∩ F = K (Lemma 2.6.13). Lemma 21.2.9: A field K is weakly Ci,d if and only if every form f ∈ K[X0 , . . . , Xn ] of degree d with di ≤ n has a nontrivial zero x such that K(x) is a primary extension of K. Proof: Suppose first K is weakly Ci,d . Let f be as stated in the lemma. Then V (f ) contains a K-closed variety W . It is defined over a purely inseparable extension K 0 of K. Let p be a generic point of W over K 0 . Then p is represented by a zero (x0 , . . . , xn ) of f where x0 , say, is transcendental over K 0 (p) = K 0 ( xx10 , . . . , xxn0 ) and K 0 (p)/K 0 is regular. Therefore, Ks ∩ K(x0 , . . . , xn ) = K. Conversely, suppose that f (x) = 0 and that K(x)/K is a primary extension. Denote the K-closed subset of Pn generated by x by W . Since K(x) is linearly disjoint from Ks over K, the set W is Ks -irreducible (lemma 10.2.1). It follows that W is a variety (Lemma 10.2.4).  The form that appears in the next lemma might aptly be called a weakly normic form. Lemma 21.2.10: Assume the field K is not separably closed. Let e0 be an integer. Then there exist an integer e > e0 and a form h ∈ K[Y1 , . . . , Ye ] of degree e satisfying this: Suppose K(y1 , . . . , ye ) is a primary extension of K. Then (3)

h(y1 , . . . , ye ) = 0 =⇒ y1 = · · · = ye = 0.

Proof: By assumption, there exists a nontrivial Galois extension L/K. Denote its degree by l and its Galois group by G. Let w1 , . . . , wl be a basis for L/K and consider the form Y (w1σ X1 + · · · + wlσ Xl ), g(X) = σ∈G

of degree l and with coefficients in K. Suppose K(y)/K is a primary extension and g(y) = 0. Then there exists σ ∈ G with w1σ y1 + · · · + wlσ yl = 0. Moreover, K(y) is linearly disjoint from L over K. Hence, y1 = · · · = yl = 0.

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Next consider the form g2 = g(g(X1 , . . . , Xl ), g(Xl+1 , . . . , X2l ), . . . , g(X(l−1)l+1 , . . . , Xl2 )). It is of degree l2 and it has property (3) for e = l2 . Iterate this procedure to  obtain, finally, a form h of degree exceeding e0 . The next results treat Ci -fields and weakly Ci -fields simultaneously: Lemma 21.2.11: Let K be a Ci -field (resp. weakly Ci -field) and f1 , f2 , . . . , fr forms over K of degree d in n variables with n > rdi . Then f1 , f2 , . . . , fr have a common nontrivial zero in K (resp. some primary extension of K). Proof: First suppose K is separably closed. Then K is Ci for no i but is weakly C0 (Example 21.2.8). By the projective dimension theorem (Section 10.7), V (f1 , . . . , fr ) is a nonempty Zariski closed subset of Pn−1 . Thus, ˜ which is a primary extension of K. f1 , . . . , fr have a nontrivial zero in K, Assume now K is not separably closed. Then Lemma 21.2.10 gives a weakly normic form h ∈ K[Y1 , . . . , Ye ] of degree e > r satisfying n h e i (4) n −1 ≥ n + 1. r r For each positive integer j, let Xj = (Xj1 , . . . , Xjn ) be a vector of variables. Let h1 (X1 , . . . , Xq ) = h(f1 (X1 ), . . . , fr (X1 ), . . . , f1 (Xq ), . . . , fr (Xq ), 0, . . . , 0)     with q = re . Then h1 is a form of degree d1 = de in n1 = n re variables. If (x1 , . . . , xq ) is a nontrivial zero with coordinates in K (resp. a primary extension of K), then there exists j between 1 and q with xj 6= 0 and f1 (xj ) = · · · = fr (xj ) = 0. Replace h by h  1 in  the above definition and define a form h2 of degree d2 = d2 e in n2 = n nr1 variables. Continue by induction to define, for every   variables. positive integer k, a form hk of degree dk = dk e in nk = n nk−1 r Every nontrivial zero of hk in K (resp. in a primary extension E of K) defines a common nontrivial zero of f1 , . . . , fr in K (resp. in E). By assumption, a zero of that type exists if k satisfies nk > dik . Thus, it suffices to prove that nk > dik for all large k. The lemma is done, then, if we choose k such that nk > dik . Indeed, hn i n k > nk − n. (5) nk+1 = n r r Use (5) for nk rather than for nk+1 and substitute in the right hand side of (5) to obtain nk+1 > ( nr )2 nk−1 − n( nr + 1). Continue in this manner, inductively, to obtain the inequality  n k  n k−1  n k−2  n1 − n + + ··· + 1 nk+1 > r r r (6)  h i    e n k 1 nb −n +n , = b r r

21.2 Ci -Fields and Weakly Ci -Fields

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with b = nr − 1. Use (4) to see that the right hand side of (6) bounds nk+1 1 n k > di1ei b (( rdni )k + dnik ). Since n > rdi , the b (( r ) + n). Therefore, di k+1

right side tends to infinity with k. Consequently, nk > dik for all large k, as claimed.  Proposition 21.2.12: Let K be a Ci -field (resp. weakly Ci -field) and E is an extension of K of transcendence degree j. Then E is Ci+j (resp. weakly Ci+j ). Proof: It suffices to prove the proposition for simple transcendental extensions and then for finite separable extensions. Part A: E = K(t), t is transcendental over K. Let f ∈ K(t)[X0 , . . . , Xn ] be a form of degree d with n ≥ di+1 . Assume without loss the coefficients of f lie in K[t]. Denote the maximum of the degrees of the coefficients of f by r. Let s be a positive integer with (n − di+1 + 1)s > di (r + 1) − (n + 1). Consider the (n + 1)(s + 1) variables Yjk , j = 0, . . . , n and k = 0, . . . , s. There exist forms f0 (Y), . . . , fds+r (Y) over K of degree d with f(

s X

Y0k tk , . . . ,

k=0

s X

Ynk tk ) =

ds+r X

k=0

fl (Y)tl .

l=0

Since (n + 1)(s + 1) > di (ds + r + 1), Lemma 21.2.11 gives a nonzero y in K (n+1)(s+1) (resp. K(y)/K Ps is a primary extension) with f0 (y) = · · · = fds+r (y) = 0. Put xj = k=0 yjk tk , j = 0, . . . , n. These elements are not all zero and f (x) = 0. When K is Ci , x is K(t)-rational. Therefore, K(t) is Ci+1 . When K is weakly Ci , we may choose y with K(y) algebraically independent from K(t) over K. By Lemma 2.6.15(a), K(y, t) is a primary extension of K(t). Hence, K(x, t)/K(t) is primary. It follows from Lemma 21.2.9 that K(t) is weakly Ci+1 . Part B: E/K is finite. Let f ∈ E[X0 , . . . , Xn ] be a form of degree d, with n ≥ di . With w1 , . . . , we a basis for E/K, introduce new variables Zjk , j = 0, . . . , n and k = 1, . . . , e. Then there exist forms f1 (Z), . . . , fe (Z) over K of degree d with f(

e X k=1

Z0k wk , . . . ,

e X k=1

Znk wk ) =

e X

fk (Z)wk .

k=1

Since (n + 1)e > edi , Lemma 21.2.11 again gives a nonzero z such that f1 (z) = · · · = fP e (z) = 0 and K(z) = K (resp. K(z)/K) is primary). The e elements xj = k=1 zjk wk , j = 0, . . . , n, satisfy f (x) = 0 and not all of them are zero. If K is Ci , then E(x) = E. Therefore, E is Ci . If K is weakly Ci , then E(z)/E is primary (Lemma 2.6.15(a)). Hence,  E(x)/E is primary. Consequently, E is weakly Ci .

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Chapter 21. Problems of Arithmetical Geometry

Weakly Ci -fields have a large class of subfields that are also weakly Ci . Note, however, that this result does not hold for Ci fields (Exercise 5). Proposition 21.2.13: If L is a primary extension of a field K and L is weakly Ci , then K is also weakly Ci . Proof: Since a primary extension L(x) of L is, under the hypotheses, a primary extension of K, this is immediate from Lemma 21.2.9. 

21.3 Perfect PAC Fields which are Ci First we note a relation between weakly Ci -fields, Ci -fields and the fields ˜ K(σ): Lemma 21.3.1: Let K be a countable Hilbertian field. Then the following are equivalent: (a) K is weakly Ci . ˜ (b) For each e ∈ N and for almost all σ ∈ Gal(K)e , K(σ) is Ci . Proof: If (a) holds, then (Proposition 21.2.12) every algebraic extension of K is weakly Ci . ˜ By Theorem 18.6.1, K(σ) is a perfect PAC field for almost all σ ∈ e Gal(K) . Since a weakly Ci PAC field is Ci (Remark 21.2.7), this gives (b). Assume now (b) holds. Let f ∈ K[X0 , . . . , Xn ] be a form of degree d with di ≤ n. Apply the decomposition-intersection procedure to the subset A = V (f ) of Pn (Section 21.1). Let L be the Galois splitting field of A over K. Choose generators σ10 , . . . , σe0 for Gal(L/K). Then there exist σ1 , . . . , σe ∈ ˜ is Ci . In Gal(K) that respectively extend σ10 , . . . , σe0 , and for which K(σ) ˜ ˜ particular, L ∩ K(σ) = K and A has a K(σ)-rational point. By Lemma 21.1.1, A∗ is nonempty. With K replacing M in Lemma 21.1.1 conclude that A contains a nonempty subvariety which is K-closed. Hence, K is weakly Ci .  We do not expect a general field extension of a Ci -field to be Ci (Exercise 6). But, under simple conditions this is true for weakly Ci fields: Proposition 21.3.2: The following conditions on a countable Hilbertian field K are equivalent: (a) K is weakly Ci . (b) Every field extension F of K is weakly Ci . (c) Every perfect PAC field extension F of K is Ci . Proof of “(a) =⇒ (b)”: Assume without loss F is also countable. If F/K is algebraic, then F is weakly Ci (Proposition 21.2.12). Suppose F/K is transcendental. Let K 0 be a purely transcendental extension of K with F algebraic over K 0 . By Theorem 13.2.1, K 0 is Hilbertian. By Lemma 21.3.1, ˜ the field K(σ) is Ci for each e ∈ N and almost all σ ∈ Gal(K)e . Hence, by ˜ 0 (σ) is Ci for each e ∈ N and almost all σ ∈ Gal(K 0 )e . It Theorem 20.7.1, K

21.3 Perfect PAC Fields which are Ci

461

follows from Lemma 21.3.1 that K 0 is weakly Ci . Consequently (Proposition 21.2.12), F is weakly Ci . Proof of “(b) =⇒ (c)”: Use Remark 21.2.7. ˜ Proof of “(c) =⇒ (a)”: By Theorem 18.6.1 and assumption (c), K(σ) is Ci e for each e ∈ N and for almost all σ ∈ Gal(K) . Hence, by Lemma 21.3.1, K  is weakly Ci . Proposition 21.2.4 says that every finite field is C1 . Hence, by Proposition 21.2.12, every algebraic extension of a finite field is C1 . Every other field contains either Q or Fp (t) for some p and transcendental element t. Each of the latter fields is countable and Hilbertian (Theorem 13.4.2). An application of Proposition 21.3.2 to i = 1 therefore gives this reformulation of Ax’s problem 21.2.5: Corollary 21.3.3: The following conditions are equivalent: (a) Every field is weakly C1 . (b) Each of the fields Q and Fp (t) is weakly C1 . (c) Every perfect PAC field is C1 . Lemma 21.3.4: Let K be a weakly Ci -field and F an extension of K. Then F is weakly Ci+1 . If, in addition, F is perfect and PAC, then F is Ci+1 . Proof: The second statement follows from the first by remark 21.2.7. To prove the first statement, suppose first F/K is algebraic. By Proposition 21.2.12, F is weakly Ci . Hence, F is weakly Ci+1 . Now suppose F/K is transcendental. Use Proposition 21.2.13 and replace K by a smaller field to assume K is countable. Choose t ∈ F transcendental over K. Then K(t) is Hilbertian (Theorem 13.4.2) and weakly Ci+1 (Proposition 21.2.12). It follows from Proposition 21.3.2 that F is weakly  Ci+1 . Lemma 21.3.5: Let F be a field. (a) Suppose F contains an algebraically closed field. Then F is weakly C1 . (b) Suppose F has a positive characteristic. Then F is weakly C2 . (c) Suppose Gal(F ) is procyclic. Then F is weakly C1 . ˜ (d) Suppose F contains Q(σ) with σ ∈ Gal(Q). Then F is weakly C2 . Proof of (a): By Example 21.2.2, each algebraically closed field is C0 . Hence, by Lemma 21.3.4, F is weakly C1 . Proof of (b): By assumption, F contains a finite field. The latter is C1 (Proposition 21.2.4). Hence, by Lemma 21.3.4, F is weakly C2 . Proof of (c): Let K0 be the prime field of F . Consider a form f ∈ F [X0 , . . . , Xd ] of degree d. The coefficients of f are algebraic over a finitely ˜ ∩ F ) is procyclic. Suppose V (f ) generated extension K of K0 . Also, Gal(K ˜ has a subvariety W which is K ∩ F -closed. Then W is also F -closed. So,

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Chapter 21. Problems of Arithmetical Geometry

assume without loss, F is algebraic over K. Finally, replace F by Fins , if necessary, to assume F is perfect (Proposition 21.2.13). Now K is either a finite field, or algebraic over Q, or transcendental over K0 . In all cases (K, 1) is a Hilbertian pair in the sense of Section 20.5 (Theorem 13.4.2). By Lemma 20.5.3, there is a regular ultraproduct E of the ˜ ˜ ∩E ∼ fields K(σ) with K =K F . By Lemma 20.5.2(b), E is pseudo finite. For each positive integer d the C1,d property of a field is elementary. Since every finite field has this property (Proposition 21.2.4), so does every pseudo finite field (Proposition 20.10.2). Therefore, E is C1 . Consequently, by Proposition 21.2.13, F is weakly C1 . Proof of (d): Use (c) and Lemma 21.3.4.



Suppose F in Lemma 2.3.5 is PAC and perfect. Then we may replace “weakly Ci ” by “Ci ” in each of the statement of that lemma. This gives the main result of this section: Theorem 21.3.6: Let F be a perfect PAC field. ˜ Then F is C1 . (a) Suppose F contains an algebraically closed field K. (b) Suppose F has a positive characteristic. Then F is C2 . (c) Suppose Gal(F ) is procyclic. Then F is C1 . ˜ (d) Suppose F contains a field Q(σ), where σ ∈ Gal(Q). Then F is C2 . Remark 21.3.7: Theorem 22.9.6 strengthens part (c) of Theorem 21.3.6 relaxing the condition “Gal(F ) is procyclic” to “Gal(F ) is Abelian.” J´anos Koll´ar proves in [Koll´ ar3, Thm. 1] that every PAC field of char acterisitc 0 is C1 . This settles Problem 21.2.5 in characterisitic 0.

21.4 The Existential Theory of PAC Fields In Section 28.10 it is shown that the theory of PAC fields is undecidable. It is therefore of interest to observe that the decomposition-intersection procedure gives a decision procedure for the existential theory of PAC fields: Definition 21.4.1: (1)

Call a sentence of L(ring) existential if it has the form h_^

(∃X1 ) · · · (∃Xn )

i

i [fij (X) = 0 ∧ gi (X) 6= 0]

j

with fij , gi ∈ Z[X], and i, j range over finite sets.



Replace each inequality gi (X) 6= 0 in (1) by the equivalent formula (∃Yi )[Yi gi (X) − 1 = 0] to assume the sentence has the form (2)

h_^

(∃X1 ) · · · (∃Xn )

i

j

i fij (X) = 0 .

21.5 Kronecker Classes of Number Fields

463

The bracketed expression in (2) defines a Zariski Q-closed subset A of An . Rewrite (2) as (3)

(∃X1 ) · · · (∃Xn )[X ∈ A].

To test if (3) is true in every PAC field of characteristic 0, apply the decomposition-intersection procedure (Section 21.1, effective by Proposition 19.5.6). Let L be the Galois splitting field of A over Q and τ1 , . . . , τe generators for Gal(L/Q). The set of σ ∈ Gal(Q)e whose restriction to L is τ has positive measure ˜ (= 1/[L : Q]e ). Hence, by Theorem 18.6.1, there is a σ such that M = Q(σ) is PAC and L ∩ M = Q. If A∗ is empty, then A has no M -rational point (Lemma 21.1.1). Thus, there is a PAC field of characteristic 0 for which (3) is false. If A∗ is nonempty, then (Lemma 21.1.1) each Q-component W of A∗ is a subvariety of A defined over Q (effectively computable). That variety has an M -rational point in each PAC field M of characteristic 0. In this case, (3) is true in every PAC field of characteristic 0. Now we check the PAC fields of positive characteristic. Proposition 10.4.2 gives us a finite set of primes S, such that for each p 6∈ S the variety W (as above) is well defined and remains a variety when considered as a Zariski Fp -closed set. For each p ∈ S, repeat the decomposition-intersection procedure for A over Fp . If A∗ is nonempty in each of these cases, then (3) is true in every PAC field. If, however, there exists p ∈ S for which A∗ is empty over Fp , then, as above, there exists a PAC field M such that A(M ) is empty. In this case (3) fails in some PAC field. We summarize: Theorem 21.4.2: Both the existential theory of PAC fields of a given characteristic and the existential theory of all PAC fields are primitive recursively decidable.

21.5 Kronecker Classes of Number Fields Our next example discusses the “Kronecker conjugacy” of polynomials. We place it in the framework of classical algebraic number theory and mention some results and open problems, with partial reformulation in terms of group theory. It is customary to credit Kronecker (in 1880) for the impetus to investigate the decomposition of primes in number field extensions [Jehne]. Throughout this section fix a global field K. Definition 21.5.1: Let L be a finite separable extension of K. Denote the set of all p ∈ P (K) that have a prime divisor P ∈ P (L) of relative degree ¯ P = Kp) ¯ by V (L/K). Call a finite separable extension M of K 1 (i.e. L Kronecker conjugate to L (over K) if V (L/K) and V (M/K) differ by

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Chapter 21. Problems of Arithmetical Geometry

only finitely many elements. The set of all finite separable extensions M of K which are Kronecker conjugate to L is called the Kronecker class of L/K. We denote it by K(L/K).  Lemma 21.5.2 (Dedekind, Kummer): Let R be a Dedekind ring with quotient field K. Consider a finite separable extension L = K(x) of K. Suppose x is integral over R and let f = irr(x, K). If Rp [x] is the integral closure of Rp in L (by lemma 6.1.2, this holds for almost all p) and f (X) ≡ f1 (X)e1 · · · fr (X)er mod p is the factorization of f (X) as a product of powers of distinct monic irreducible polynomials modulo p, then pS = Pe11 · · · Perr where P1 , . . . , Pr are distinct prime ideals of S with f (Pi /p) = deg(fi ), i = 1, . . . , r. Proof: See [Lang5, p. 27] or [Janusz, p. 32].



The arithmetic condition for Kronecker conjugacy is equivalent to a Galois theoretic condition: Lemma 21.5.3: Let L1 and L2 be finite separable extensions of K and L a finite Galois extension of K which contains both L1 and L2 . Then L1 and L2 are Kronecker conjugate over K if and only if, with G = Gal(L/K), (1)

[ σ∈G

Gal(L/L1 )σ =

[

Gal(L/L2 )σ .

σ∈G

Proof: For i = 1, 2 let xi ∈ OLi be a primitive element of the extension Li /K, and let fi = irr(xi , K). By Lemma 21.5.2, V (Li /K) differs from the ¯ p by only finitely many set of primes p for which (∃X)[fi (X) = 0] is true in K primes. Let θ be the sentence (∃X)[f1 (X) = 0] ↔ (∃X)[f2 (X) = 0]. Then, Kronecker conjugacy of L1 and L2 over K is equivalent to the truth of ¯ p . By the transfer theorem (Theorem 20.9.3) applied to test θ in almost all K sentences, this is equivalent to the truth of θ in L(σ), for all σ ∈ Gal(L/K). But this is just a restatement of (1).  If Li /K is Galois, i = 1, 2, and L1 and L2 are Kronecker conjugate over K, Lemma 21.5.3 implies L1 = L2 . This is Bauer’s theorem (see also Exercise 5 of Chapter 6). But, in general, Kronecker conjugacy does not imply conjugacy of the fields L1 and L2 over K. This is demonstrated in the following example:

21.5 Kronecker Classes of Number Fields

465

Example 21.5.4: Kronecker conjugate extensions of Q of different degrees q  [Schinzel, Lemma 4]. Let L = Q(cos 2π 2 cos 2π . Then 7 ) and M = Q 7 L/Q is a cyclic extension of degree 3, M/L is a quadratic extension, M is not Galois over Q, but M is Kronecker conjugate to L over Q. We give proofs of these statements. 3 2 2πi/7 and η = Part A: irr(2 cos 2π 7 , Q) = X + X − 2X − 1. Let ζ = e 2π −1 2 cos 7 = ζ + ζ . Then Q(ζ)/Q is a cyclic extension of degree 6 and η is fixed by complex conjugation. Hence, Q(η)/Q is a cyclic extension of degree 3. The conjugates of η over Q are ηi = ζ i + ζ −i , i = 1, 2, 3 and they satisfy

η1 η2 η3 = 1,

(2)

η1 η2 + η1 η3 + η2 η3 = −2,

η1 + η2 + η3 = −1

Hence, irr(η, Q) = X 3 + X 2 − 2X − 1. √ Part B: η 6∈ L. Assume η = θ2 with θ in L. Then irr(θ, Q) is a polynomial of degree 3 with integral coefficients. The element θ is a zero of f (X) = X 6 + X 4 − 2X 2 − 1. Therefore, irr(θ, Q) divides f (X): X 6 + X 4 − 2X 2 − 1 = (X 3 + aX 2 + bX + 1)(X 3 + cX 2 + dX − 1) with a, b, c, d ∈ Z. Thus, a + c = 0, d − b = 0, ad + bc = 0, ac + b + d = 1 and c − a + bd = −2. Eliminate b and c to come down to the equations 2d = a2 + 1 and 2a = d2 + 2. Then d ≥ a from the first and a > d from the second, a contradiction. ˆ /Q). By Part B, Part C: The structure of the Galois group G = Gal(M √ √ M = Q( η1 ) is a quadratic extension of L. Similarly, Q( η2 ) is a quadratic √ extension of L which we claim to be different from Q( η1 ). Otherwise, there exists x ∈ L such that η2 = x2 η1 . By (2), η3−1 = (xη1 )2 , a contradiction to q √ ˆ = L(√η1 , √η2 ) is an extension Part B. Also η3 = η1−1 η2−1 implies that M ˆ /L) ∼ of L of degree 4 with Gal(M = Z/2Z × Z/2Z. The generator of the ˆ /L) by permuting the three subgroups cyclic group Gal(L/Q) acts on Gal(M of order 2 in a cyclic way. So, G is the semidirect product of Z/3Z with Z/2Z × Z/2Z. Part D: The Kronecker conjugacy of L and M . [ σ∈G

ˆ /M )σ = Gal(M

3 [ i=1

By part C we have

[ ˆ /L) = ˆ /L)σ . ˆ /Q(√ηi )) = Gal(M Gal(M Gal(M σ∈G

Hence, by Lemma 21.5.3, M and L are Kronecker conjugate.



Lemma 21.5.3 immediately gives L1 ⊆ L2 if L1 /Q is Galois and L1 and L2 are Kronecker conjugate — just as we saw in Schinzel’s example. Call a field M of a Kronecker class K(L/K) minimal if it contains no proper subfields of the same class.

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Chapter 21. Problems of Arithmetical Geometry

Lemma 21.5.5: Let L1 and L2 be minimal fields of a Kronecker class over ˆ 2 over K are equal. ˆ 1 and L K. Then their Galois closures L ˆ 1 and L ˆ 2 . By Proof: Let N be a finite Galois extension of K that contains L Lemma 21.5.3 [ [ Gal(N/L1 )σ = Gal(N/L2 )σ , σ∈G

σ∈G

ˆ 2 to conclude that with G = Gal(N/K). Restrict both sides to L [ σ∈G2

ˆ 2 /L1 ∩ L ˆ 2 )σ = Gal(L

[

ˆ 2 /L2 )σ , Gal(L

σ∈G2

ˆ 2 /K). By Lemma 21.5.3, L1 ∩ L ˆ 2 is Kronecker conjugate to with G2 = Gal(L ˆ2 ⊆ L ˆ2. L2 and therefore to L1 . Minimality of L1 implies that L1 = L1 ∩ L ˆ ˆ ˆ ˆ  Therefore, L1 ⊆ L2 ; and, by symmetry, L1 = L2 . Corollary 21.5.6: Each Kronecker class contains only finitely many minimal fields. Remark 21.5.7: On the size of Kronecker classes. Fix a number field K. (a) Call the number of nonconjugate minimal fields of K(L/K) the width and denote it by ω(L/K). [Jehne, §4] gives L with ω(L/K) arbitrarily large. The proof incorporates Exercise 7 and the theorem of ScholzShafarevich that each p-group is a Galois group over K. (b) On the other hand, [Jehne, Theorem 3] gives examples in the number field case where K(L/K) is infinite. Indeed, this is the case if Aut(L/K) contains either a nontrivial automorphism of odd order, or a cyclic group of order 8, or a quaternion group of order 8. (c) For a quadratic extension L/K [Jehne, §6] conjectures that K(L/K) consists only of L. This conjecture has been proved by Saxl. To prove the conjecture, assume M is another field in K(L/K). By Lemma 21.5.3, L ⊂ M . Replace M by a smaller field to assume M is a minˆ be the Galois closure of M/K, G = Gal(M ˆ /K), imal extension of L. Let M ˆ ˆ H = Gal(M /L), and U = S Gal(M /M ). Then (G : H) = 2, U is a maximal subgroup of H, and H = g∈G U g . By [Jehne, Thm. 5], H has a unique minimal normal subgroup N which is non-Abelian and simple. Moreover, U N = H. So, V = U ∩ N is a proper subgroup of N . The minimality of N r implies N / G. Choose σ ∈ G S H. ViewSσ as an automorphism of N acting as conjugation. Then, N = h∈H V h ∩ h∈H V σh . But this contradicts the main result of [Saxl]: Let N be a finite simple group and V a proper subgroup. Then there exists a ∈ N lying in no Aut(N )-orbit of V . Saxl’s proof uses the classification of finite simple groups, proving the result case by case. The case where N = PSL(2, pr ) with p 6= 2 appears already in [Jehne, Thm. 5]. The case N = An with n ≥ 5 is due to [Klingen3].

21.6 Davenport’s Problem

467

(d) Saxl’s solution of Jehne’s conjecture has an interesting arithmetic consequence: Let d be a square√free integer and f ∈ Z[X] an irreducible polynomial without a root
in Q( d). Then there are infinitely many prime numbers p such that dp = 1 but f (X) ≡ 0 mod p has no solution. √ ˜ M = Q( d, x), M ˆ the Galois Otherwise, let x be a root of f (X) in Q,
ˆ /Q). Then, for almost p with d = 1 the closure of M/Q, and G = Gal(M p equation f (X) ≡ 0 mod p has a solution. By the transfer principle, [ √ ˆ /M )σ . ˆ /Q( d)) = Gal(M Gal(M σ∈G

√ Thus, M ∈ K(Q( d)/Q), in contrast to Jehne-Saxl.



Another problem of Jehne is still open (see the notes at the end of the chapter for a profinite version): Problem 21.5.8 ([Jehne, §7]): Are there fields K ⊂ L ⊂ M , with K global, L/K finite separable, M/K infinite separable, such that M0 ∈ K(L/K) for every intermediate field L ⊂ M0 ⊂ M of finite degree over K?

21.6 Davenport’s Problem Let K be a global field of characteristic p and let f, g ∈ K[Y ] be nonconstant polynomials which are not p-powers. For each prime p ∈ P (K) for which f is defined modulo p, consider the value set of f : ¯ p }. Vp (f ) = {f (y) | y ∈ K If g(Y ) = f (aY + b) with a, b ∈ K and a 6= 0, then g and f are said to be strictly linearly related. In this case, (1) Vp (f ) = Vp (g) for almost all p ∈ P (K). Call the pair (f, g) exceptional if f and g satisfy (1) but they are not strictly linearly related. Remark 21.6.1: Vp (X 8 ) = Vp (16X 8 ) for every prime number p. To prove this statement, verify the identity

(2) X 8 − 16 = (X 2 − 2)(X 2 + 2) (X − 1)2 + 1 (X + 1)2 + 1 .

−1

= p4 = 1 for By the multiplicativity of the Legendre symbol, p2 −2 p p each odd prime number p. Hence, at least one of the first three factors on the right hand side of (2) is divisible by p for some value of X. It follows that 16 is an 8th power modulo every p. Thus, for each x ∈ Fp there is a y ∈ Fp with x8 = 16y 8 , as claimed. Consider now a polynomial h(X) = cn X n + cn−1 X n−1 + · · · + c0 in Q[X] with n ≥ 1 and cn 6= 0. Then Vp (h(X 8 )) = Vp (h(16X 8 )) for each p which

468

Chapter 21. Problems of Arithmetical Geometry

divides no denominator of c0 , . . . , cn−1 , cn . Assume there were a, b ∈ Q with a 6= 0 and h(16X 8 ) = h((aX + b)8 ). Comparing the coefficients of X 8n on both sides this would give 16 = a8 , which is a contradiction. It follows,  (h(X 8 ), h(16X 8 )) is an exceptional pair. Problem 21.6.2: (a) Are all exceptional pairs in Q[X] of the form (h(X 8 ), h(16X 8 )) with h ∈ Q[X] (M¨ uller)? (b) For a given global field K find all exceptional pairs (f, g) with f, g ∈ K[X] (Davenport). A complete solution of this problem is still unavailable. Transfer principles, however, allow us to reduce (1) to a condition on a finite Galois extension of K(t), in complete analogy with the Galois theoretic version of Kronecker conjugacy of global field extensions. From this point we derive certain relations between f and g (e.g. deg(f ) = deg(g) if the degrees are relatively prime to char(K)). In the special case where K = Q and deg(f ) is a prime, we prove there is no g ∈ Q[X] such that (f, g) is exceptional. But we mention the existence of exceptional pairs (f, g) with deg(f ) prime over other number fields. Clearly (1) is equivalent to the truth of the sentence   (3) (∀T ) (∃X)[f (X) = T ] ↔ (∃Y )[g(Y ) = T ] ¯ p . By Theorem 20.9.3, this is equivalent to the truth of in almost all fields K ˜ (3) in K(σ), for almost all σ ∈ Gal(K). First we eliminate (∀T ) to reduce the ] problem to a test sentence over K 0 = K(t). To this end we choose x, y ∈ K(t) with f (x) = t and g(y) = t and put E = K(t, x) and F = K(t, y). Assume that E/K(t) and F/K(t) are separable extensions. When p = char(K) > 0 ˜ this means, f (X) and g(X) are not a pth powers in K[X]. Lemma 21.6.3: Let N a finite Galois extension of K(t) containing E and F . Put G = Gal(N/K(t)). Then (1) is equivalent to [

(4)

ν∈G

Gal(N/E)ν =

[

Gal(N/F )ν .

ν∈G

˜ Proof: Put K = K(t). If (3) holds in K(σ) for almost all σ ∈ Gal(K), then ˜ 0 (σ) for almost all σ ∈ Gal(K 0 ) (Theorem 20.7.2). In particular, (3) holds in K formula in the outer brackets of (3) hold for t. That is, the sentence 0

(5)

(∃X)[f (X) = t] ↔ (∃Y )[g(Y ) = t]

˜ 0 (σ) for almost all σ ∈ Gal(K 0 ). Therefore, (5) holds in N (σ) for is true in K all σ ∈ G. Consequently, (4) is true. Conversely, suppose that the elements τ ∈ G that fix a root of f (X) − t are exactly the elements that fix a root of g(Y ) − t.

21.6 Davenport’s Problem

469

˜ We prove (3) holds in K(σ) for all σ ∈ Gal(K). Indeed, let σ ∈ Gal(K) ˜ and a ∈ K(σ). Extend the K-specialization t → a to a homomorphism ˜ The residue field N ¯ = ϕ of the integral closure R of K[t] in N into K. ϕ(R) is a normal extension of K(a). Denote the decomposition group {τ ∈ G | τ (Ker(ϕ)) = Ker(ϕ)} of ϕ by H. Then ϕ induces an epimorphism ϕ∗ of ¯ /K(a)) through the formula (Section 6.1) H onto Aut(N (6)

ϕ∗ (τ )(ϕ(z)) = ϕ(τ (z)) for each τ ∈ H and z ∈ R.

Observe that the roots of f (X) = t and g(X) − t are in R. Hence, ϕ maps the roots of f (X) − t (resp. g(Y ) − t) onto the roots of f (X) − a ˜ (resp. g(Y ) − a). Therefore, if b ∈ K(σ) satisfies f (b) = a, then there exists a zero x of f (X) = t with ϕ(x) = b. Note that H ∩ Gal(N/K(x)) is the decomposition group of ϕ over K(x). Hence, by Lemma 6.1.1(a), the ¯ /K(b)). restriction of ϕ∗ to H ∩Gal(N/K(x)) maps that subgroup onto Aut(N In particular, there exists τ ∈ H ∩ Gal(N/K(x)) with (7)

ϕ∗ (τ ) = resN¯ (σ).

Since τ (x) = x, the assumptions give a zero y of g(Y ) = t with τ (y) = y. Thus, c = ϕ(y) is a zero of g(Y ) = a for which (6) and (7) give σ(c) = c.  Remark 21.6.4: The switch from the surjectivity of ϕ∗ on H to its surjectivity on H ∩ Gal(N/K(x)) is a subtlety in the proof of Lemma 21.6.3. From the surjectivity of ϕ∗ on H alone we could choose τ ∈ H such that (7) holds. It follows that there exist zeros x and x0 of f (X) = t with τ (x) = x0 and ϕ(x) = ϕ(x0 ) = b. If f (X) − a is separable, then this implies that x = x0 , and conclude the proof easily. Nothing, however, in the assumptions guaranties separability of f (X)−a. This same subtlety occurs in the stratification procedure of Chapter 30.  Following Section 21.5 we say that fields E and F (and the polynomials f and g) satisfying (4) are Kronecker conjugate over K(t). This is independent of N . If E and F are minimal extensions of K(t) with this property, then as in Lemma 21.5.5, they have the same Galois closure over K(t). The next results, however, allow us the same conclusion under a less restrictive assumption than minimality. Throughout let deg(f ) = m, deg(g) = n. Corollary 21.6.5: Suppose E and F are Kronecker conjugate over K(t) ˆ (resp. Fˆ ) and [E : K(t)] and [F : K(t)] are not divisible by char(K). Let E be the Galois closure of E/K(t) (resp. F/K(t)). Then [E : K(t)] = [F : K(t)] ˆ = Fˆ . and E Proof: By assumption, p∞ is tamely ramified in all fields conjugate to E or to F . Hence, p∞ is tamely ramified in the composite N of all these fields [Cassels-Fr¨ohlich, p. 31]. Let P∞ be a prime divisor of N that lies over p∞ .

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Then the decomposition group of P∞ is cyclic [Cassels-Fr¨ohlich, p. 30]. Let τ be a generator of it. Then p∞ is unramified in N (τ ). Since p∞ is totally ramified in E (Example 2.3.11), the two fields are linearly disjoint over K(t) (Lemma 2.5.8). In particular, the polynomial f (X) − t (a root of which generates E over K(t)) remains irreducible over N (τ ). Thus, the group hτ i operates transitively on the roots x1 , . . . , xm of f (X) − t. Therefore, τ is an m-cycle on x1 , . . . , xm . By symmetry, τ is an n-cycle on y1 , . . . , yn , the roots of g(X) − t. If m < n, then τ m would fix each of x1 , . . . , xm but would move each of y1 , . . . , yn . This contradicts (4). Thus, m ≥ n; and by symmetry, m = n. Next restrict relation (4) to Fˆ to conclude that E ∩ Fˆ and F are Kronecker conjugate over K(t). By the first part of the corollary, [E ∩ Fˆ : K(t)] = ˆ ⊆ Fˆ . By [F : K(t)] = [E : K(t)]. Therefore, E = E ∩ Fˆ ⊆ Fˆ . Hence, E ˆ = Fˆ . symmetry E  Lemma 21.6.6: Assume the hypotheses of Corollary 21.6.5. Then the polynomial f (X) − g(Y ) is reducible in K[X, Y ]; and there are constants a, b ∈ K and distinct roots x1 , . . . , xk of f (X) − t with 1 ≤ k < deg(f ) and ay + b = x1 + · · · + xk .

(8)

Proof: Assume f (X) − g(Y ) is irreducible in K[X, Y ]. Then, since y is transcendental over K, f (X)−t = f (X)−g(y) is irreducible in F [X]. Lemma 13.3.2 gives σ ∈ Gal(N/F ) that moves each root of f (X) − t. This is a contradiction, since σ(y) = y. Thus, f (X) − g(Y ) is reducible. Now consider a nontrivial factorization f (X) − g(Y ) = p(X, Y )q(X, Y ) in K[X, Y ]. By Corollary 21.6.5, degX (p) + degX (q) = degX (f (X) − g(Y )) = deg(f (X) − g(Y )) = deg(p) + deg(q). Since degX (p) ≤ deg(p) and degX (q) ≤ deg(q), it follows that degX (p) = deg(p) = k, with 1 ≤ k < deg(f ). Write p(X, Y ) as (9)

c00 X k + (c10 Y + c11 )X k − 1 + p2 (Y )X k−2 + · · · + pk (Y ),

with c00 , c10 , c11 ∈ K, c00 6= 0, and p2 , . . . , pk ∈ K[Y ]. Since, f (X) − t = p(X, y)q(X, y), the k roots of p(X, y), say x1 , . . . , xk , are roots of f (X) − t. Since f (X) − t is irreducible and separable over K(t), x1 , . . . , xk are distinct. −1 By (9), ay+b = x1 +· · ·+xk , where a = −c−1 00 c10 and b = −c00 c11 , as claimed.  We summarize: Proposition 21.6.7: Let K be a global field and f, g ∈ K[X] nonconstant polynomials of degree not divisible by char(K). Suppose Vp (f ) = Vp (g) for almost all p ∈ P (K). Then:

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471

(a) deg(f ) = deg(g). (b) f (X) − t and g(X) − t are Kronecker conjugate over the rational function field K(t). (c) The splitting fields of f (X) − t and g(X) − t over K(t) coincide. (d) f (X) − g(Y ) is reducible in K[X, Y ]. (e) There exist constants a, b ∈ K and distinct roots x1 , . . . , xk of f (X) − t with 1 ≤ k < deg(f ) and ay + b = x1 + · · · + xk . Note that conditions (a)-(e) are all necessary for f and g to be strictly linearly related. Here is a case when they are also sufficient. Let K be a field. Call a polynomial f ∈ K[X] decomposable over K if f (X) = f1 (f2 (X)) with f1 , f2 ∈ K[X], deg(f1 ) > 1, and deg(f2 ) > 1. (Then, deg(f ) = deg(f1 ) deg(f2 ).) Otherwise, f is indecomposable over K (e.g. if f is of prime degree). Theorem 21.6.8 ([Fried4]): Let f, g ∈ Q[X]. Suppose Vp (f ) = Vp (g) for almost all prime number p and f is indecomposable. Then g and f are strictly linearly related. Proof: We give an elementary proof in the case that deg(f ) = l is a prime. The proof in the general case depends on deeper group theoretic assertions, specifically the theory of doubly transitive permutation groups [Fried4]. Suppose without loss f is monic and let c be the leading coefficient of g. By Proposition 21.6.7(d) f (X) − g(Y ) = r(X, Y )s(X, Y )

(10)

Pm with r, s P ∈ Q[X, Y ] and m = deg(r) ≥ 1, n = deg(s) ≥ 1. Write r = i=0 ri n and s = i=0 si with ri , si ∈ Q[X, Y ] homogeneous of degree i. By (10), X l − cY l = rm (X, Y )sn (X, Y ).

(11) l

−1 is irreducible in Q[X] [Lang7, p. 184], one of the factors rm or Since ZZ−1 sm must be linear. Thus, either m = 1 or n = 1. Suppose for example m = 1. Then r(X, Y ) = a1 X + a2 Y + a3 with a1 , a2 , a3 ∈ Q and, say, a1 6= 0. Let a = − aa21 and b = − aa31 . Substituting X in (10) by aY + b gives f (aY + b) = g(Y ), as claimed. 

Let n be one of the integers 7, 11, 13, 15, 21, or 31. [Fried8] and [Fried11] give indecomposable polynomials f and g of degree n and a number field K such that (f, g) is an exceptional pair over K. For indecomposable polynomials, these numbers are the only exceptional degrees: a consequence of the classification of finite simple groups.

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21.7 On Permutation Groups We gather in this section various definitions and results about permutation groups which enter the proof of the Schur Conjecture in the next section. Remark 21.7.1: The affine linear group. (a) Let G be a group. Suppose G acts on a set X from the left. For each x ∈ X put Gx = {σ ∈ G | σx = x}. Call G transitive if for all x, y ∈ X there is a σ ∈ G with σx = y. Call G regular if Gx = 1 for each x ∈ X. Note that ifTG is both transitive and regular, then |G| = |X|. Finally, call G faithful if x∈X Gx = 1. (b) Let G be a permutation group of a set X and H a subgroup of G. Suppose H is regular and transitive. Then H ∩Gx = 1 and G = HGx = Gx H for each x ∈ X. Indeed, if η ∈ H ∩ Gx , then ηx = η, so η = 1. For each σ ∈ G there is an η ∈ H with ηx = σx. Hence, σ = η · η −1 σ ∈ HGx . Thus, G = HGx . Applying the same argument to σ −1 we find that G = Gx H. If in addition H is normal in G, then G = Gx n H. (c) Denote the cyclic group of order n by Cn and the dihedral group of order 2n by Dn . The latter group is generated by elements σ, τ with defining relations τ 2 = 1, σ n = 1, and σ τ = σ −1 . (d) Let l be a prime number. Denote the group of all affine maps of Fl by AGL(1, Fl ). It consists of all maps σa,b : Fl → Fl with a ∈ F× l and b ∈ Fl defined by σa,b (x) = ax + b. Thus, |AGL(1, Fl )| = (l − 1)l. The subgroup L = {σ1,b | b ∈ Fl } is the unique l-Sylow subgroup of AGL(1, Fl ). Thus, L∼ = Cl and L is normal. Its action on Fl is transitive and regular. Hence, by (b), AGL(1, Fl ) = AGL(1, Fl )x n L. Moreover, AGL(1, Fl )x ∼ = F× l . Thus, AGL(1, Fl ) is a metacyclic group. It particular, AGL(1, Fl ) is solvable. Claim: Each σ ∈ AGL(1, Fl ) r L fixes exactly one x ∈ Fl . Indeed, b .  σ = σa,b with a 6= 1. Hence, σx = x if and only if x = 1−a The following lemma establishes a converse to Remark 21.7.1(d): Lemma 21.7.2: Let l be an odd prime number and G a nontrivial finite solvable group acting faithfully and transitively on a set X of l elements. Let L be a minimal normal subgroup of G. Then: (a) L is transitive and regular. (b) L ∼ = Cl and G = Gx n L for each x ∈ X. (c) L is its own centralizer in G. (d) G/L is a cyclic group of order dividing l − 1 and Gx ∼ = G/L for each x ∈ X. (e) L is the unique l-Sylow subgroup of G. (f) G is isomorphic as a permutation group to a subgroup of AGL(1, Fl ). (g) Each ρ ∈ G r L belongs to exactly one Gx . (h) Let I be a subgroup of G which contains L and P a p-Sylow subgroup of I with p 6= l. Suppose P / I. Then P = 1.

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473

(i) Suppose (G : L) = 2 and let x ∈ X. Then G is the dihedral group Dl generated by the involution τ of Gx and a generator λ of L with the relation λτ = λ−1 . Proof of (a) and (b): Since G is solvable, L is Abelian. S Choose a system P Y of representatives of the L-orbits of X. Then X = · y∈Y Ly and l = y∈Y |Ly|. Given y, y 0 ∈ Y , choose σ ∈ G with σy = y 0 . Then σ(Ly) = σLσ −1 σy = Ly 0 . Thus, all L-orbits have the same length n. Therefore, l = |Y |n. By definition, L 6= 1. Choose λ ∈ L, λ 6= 1. Then there is an x ∈ X with λx 6= x, because G is faithful. Hence, Lx < L. Thus, n = |Lx| = (L : Lx ) > 1. Since l is prime, n = l and |Y | = 1. Therefore, L is transitive. Next consider λ ∈ L and x ∈ X with λx = x. For each x0 ∈ X choose 0 λ ∈ L with λ0 x = x0 . Then λx0 = λλ0 x = λ0 λx = λ0 x = x0 , so λ = 1. Thus, Lx = 1 and L is regular. Hence, by Remark 21.7.1(b), G = Gx n L. In addition, |L| = (L : Lx ) = l. Therefore, L ∼ = Cl . Proof of (c): Assume σ ∈ CG (L). Fix x ∈ X and write σ = λρ with λ ∈ L and ρ ∈ Gx . Since L is Abelian, ρ ∈ CG (L). For each x0 ∈ X choose λ0 ∈ L with λ0 x = x0 . Then ρx0 = ρλ0 x = λ0 ρx = λ0 x = x0 . Therefore, ρ = 1 and σ ∈ L, as asserted. Proof of (d): Conjugation of L by elements of G embeds G/L into Aut(L) (by (c)). The latter is isomorphic to F× l . Hence, G/L is a cyclic group of order dividing l − 1. By (b), Gx ∼ = G/L. Proof of (e): By (d), l2 - |G|, so L is an l-Sylow subgroup of G. Since L is normal, it is the unique l-Sylow subgroup of G. Proof of (f) and (g): Let λ be a generator of L. Choose x0 ∈ X. By (a) and (b), X = {x0 , λx0 , . . . , λl−1 x0 }. The bijection X → Fl mapping λb x0 onto b for b = 0, . . . , l −1 identifies L as a permutation group of Fl such that λb 0 = b for all b ∈ Fl . For an arbitrary x ∈ Fl we have λb x = λb λx 0 = λb+x 0 = x + b. Let σ ∈ G and a ∈ Fl with σλσ −1 = λa . Suppose first σ0 = 0. Then σx = σλx 0 = σλx σ −1 0 = λax 0 = ax. In the general case there is a b ∈ Fl with σ0 = λb 0. Then (λ−b σ)λ(λ−b σ)−1 = λa , so by the preceding case, λ−b σx = ax. Therefore, σx = ax + b. This gives an embedding of G into AGL(1, Fl ) as permutation groups. In particular, each σ ∈ G r L belongs to exactly one Gx (Remark 21.7.1(d)). Proof of (h): Under the assumptions of (h), P ∩ L = 1 and P, L / I. Hence, P ≤ CG (L) = L. Therefore, P = 1. Proof of (i): Suppose (G : L) = 2. Choose a generator λ of L and a generator τ of Gx . Then τ 2 = 1, λl = 1, and λτ = λt for some t ∈ Fl satisfying t2 = 1.  By (c), τ ∈ / L = CG (L). Hence, t = −1. Consequently, G ∼ = Dl .

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Definition 21.7.3: Primitive permutation groups. Let G be a transitive permutation group of a nonempty set X. A block of G is a subset Y of X with the following property: For each σ ∈ G either σY = Y or Y ∩ σY = ∅. Let GY = {σ ∈ G | σY = Y }. Choose S a system of representatives R for the left cosets of GY in G. Then X = · ρ∈R ρY and G acts transitively on the set {ρY | ρ ∈ R} = {σY | σ ∈ G}. The set X itself, the empty set, and each {x} with x ∈ X are trivial blocks. Call G imprimitive if it has a nontrivial block, otherwise G is primitive.  Lemma 21.7.4: Let G be a transitive permutation group of a set X and let x ∈ X. Then G is primitive if and only if Gx is maximal in G. Proof: Suppose first G is imprimitive. Then G has a nontrivial block Y . Since G is transitive, we may assume x ∈ Y . Then Gx ≤ GY . Since Y ⊂ X and G is transitive, GY < G. Now choose y ∈ Y , y 6= x, and σ ∈ G with σx = y. Then σY = Y but σx 6= x, so σ ∈ GY r Gx . Therefore, Gx is not maximal. Conversely, suppose G has a subgroup H with Gx < H < G. Put Y = Hx. Then H = GY . Moreover, Y is a G-block. Indeed, let σ ∈ G and y, y 0 ∈ Y with σy = y 0 . Then σηx = η 0 x for some 0 η, η ∈ H. Hence, (η 0 )−1 ση ∈ Gx < H, so σ ∈ H. Therefore, σY = Y . Further, σY 6= Y for each σ ∈ G r H. Hence, Y ⊂ X. By definition, x ∈ Y . In addition, σx 6= x for each σ ∈ H r Gx . Hence, Y contains more than one element. Thus, Y is a nontrivial G-block. Consequently, G is imprimitive.  Lemma-Definition 21.7.5: Let G be a permutation group of a set A. We say G is doubly transitive if one of the following equivalent conditions hold: (a) G is transitive on A and there exists a ∈ A such that Ga is transitive on A r{a}. (b) G is transitive on A and for each a ∈ A, Ga is transitive on A r{a}. (c) Let a1 , a2 , b1 , b2 ∈ A with a1 6= a2 and b1 6= b2 . Then there is a τ ∈ G with τ a1 = b1 and τ a2 = b2 . Proof: Statements (a) and (b) are equivalent because G is transitive. Thus, it suffices to prove that (b) and (c) are equivalent: Suppose (b) holds. Let a1 , a2 , b1 , b2 be as in (b). Choose a ∈ A and α, β ∈ G with αa1 = a and βb1 = a. Then αa2 6= a and βb2 6= a. Hence, there is a σ ∈ Ga with σαa2 = βb2 . Put τ = β −1 σα. Then τ a1 = b1 and τ a2 = b2 . Conversely, suppose (c) holds. Let a ∈ A and b, b0 ∈ A r{a}. Then there  is a σ ∈ G with σa = a and σb = b0 . Remark 21.7.6: Doubly transitive subgroups of AGL(1, Fp ). If x1 , x2 , y1 , y2 are elements in Fp with x1 6= x2 and y1 6= y2 , then there are unique a ∈ F× p and b ∈ Fp such that ax1 + b = y1 and ax2 + b = y2 . Thus, AGL(1, Fp ) is doubly transitive.

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475

Conversely, let G be a doubly transitive subgroup of AGL(1, Fp ). Then for each (a, b) ∈ (F× p , Fp ) there is a σ ∈ G with σ(1) = a + b and σ(0) = b.  Then σ(x) = ax + b for all x ∈ Fp . It follows that G = AGL(1, Fp ). One of the main tools in the proof of Schur’s Proposition below is the group ring Z[H] of a finite Abelian group P H (Section 16.6). Each ρ ∈ Z[H] has a unique presentation as a sum η∈H cη η with coefficients cη ∈ Z. Define the support of ρ to be the set Supp(ρ) = {η ∈ H | cη 6= 0}. Then hSupp(ρ)i = hηP ∈ H | cη 6= 0i. 0 0 Let ρ0 = η∈H cη η be another element of Z[H]. Suppose cη , cη ≥ 0
P P 0 0 for all η. Then ρρ = τ ∈H ηη 0 =τ cη cη 0 τ and there are no cancellations among the coefficients of ρρ0 . Hence, Supp(ρρ0 ) = Supp(ρ)Supp(ρ0 ). 0 Consider a prime number p. We write ρ ≡ ρ0 mod Pp if cη ≡p cη mod p p for all η ∈ H. Since Z[H] is a commutative ring, ρ ≡ η∈H cη η mod p. Proposition 21.7.7 (Schur): Let G be a primitive permutation group of A = {1, 2, . . . , n}. Suppose G contains an n-cycle ν. Then G is doubly transitive or n is a prime number. Proof: We assume G is not doubly transitive, n is a composite number, and draw a contradiction. Denote the identity map of A by ε. Let H = hνi. Claim A: G = HG1 = G1 H and H ∩ G1 = {ε}. Indeed, H is transitive. Hence, for each σ ∈ G there is an η ∈ H with η(1) = σ(1). Therefore, σ = η · η −1 σ ∈ HG1 . Consequently, G = HG1 . Likewise, G = G1 H. It follows that each σ ∈ G can be uniquely written as σ = ησ1 with η ∈ H and σ1 ∈ G1 . Likewise, there are unique η 0 ∈ H and σ10 ∈ G1 with σ = σ10 η 0 P Claim B: Consider Z[H] as a subring of Z[G]. Let γ = σ∈G1 σ and R = Sr {ρ ∈ Z[H] | γρ = ργ}. Then there is a partition H = · j=1 Hj with H1 = {ε} Pr P and r ≥ 3 such that R = j=1 Zαj where αj = η∈Hj η. Let A1 , . . . , Ar with A1 = {1} be the G1 -orbits of A. Since G is not doubly transitive, G1 is not transitive on {2, . . . , n} (Lemma-Definition 21.7.5). Hence, r ≥ 3. Next define a map f : H → A by f (η) = η(1). Check that f is bijective and G1 ηG1 ∩ H = f −1 (G1 · η(1)) for each η ∈ H. For each j choose ηj ∈ H with S ηj (1) ∈ Aj . Let Hj = G1 ηj G1 ∩ H = f −1 (G1 · ηj (1)) = f −1 (Aj ). Then r H = · j=1 Hj and H1 = {ε}. P Now consider an element ρ = η∈H cη η in Z[H] with cη ∈ Z for each η ∈ H. Then ρ ∈ R if and only if X X X X (1) cη ση = cη ησ. σ∈G1 η∈H

σ∈G1 η∈H

By Claim A, the ση (resp. ησ) on the left (resp. right) hand side of (1) are distinct. Hence, (1) holds if and only if for all σ, σ 0 ∈ G1 and η, η 0 ∈ H the

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equality ση = η 0 σ 0 implies cη = cη0 . In other words, cη is constant as η ranges cηPis constantPon each Hj . Thus, there are on G1 ηG1 ∩ H. In particular, Pr r c1 , . . . , cr ∈ Z with ρ = j=1 cj η∈Hj η = j=1 cj αj . This concludes the proof of Claim B. Claim C: Let P ρ ∈ R. Put K = hSupp(ρ)i. Then K = {ε} or K = H. WritePρ = η∈H cη η. By Claim B, cη are constant on Hj , j = 1, . . . , r. Put κ = η∈Supp(ρ) η. Then Supp(κ) = Supp(ρ) and the coefficients of κ (which are 0 or 1) are also constant on each Hj . Hence, by Claim B, κ ∈ R. This means κγ = γκ. Since the coefficients of both κ and γ are nonnegative, Supp(κ)Supp(γ) = Supp(γ)Supp(κ). Therefore, KG1 = G1 K. This implies that KG1 is a subgroup of G which contains G1 . Since G is primitive, G1 is a maximal subgroup of G (Lemma 21.7.4). Hence, KG1 = G1 or KG1 = G. Thus, |K| · |G1 | = |G1 | or |K| · |G1 | = |G| = |H| · |G1 | (by Claim A). Therefore, K = {ε} or K = H, as claimed. Now choose a prime divisor p of n. Denote the unique subgroup of H of order p by P . Since n is composite, P is a proper subgroup of H. Claim D: For each j, Hj r P is a union of cosets of P . Indeed, P = {η ∈ H | η p = 1}. Hence, for each η0 ∈ H the set {η ∈ H | η p = η0 } is either empty or a coset of P . In the latter case, it consists of p elements. If we prove for each η0 ∈ H with η0 6= ε that (2) p|#{η ∈ Hj | η p = η0 }, then {η ∈ Hj | η p = η0 } is either empty or coincides with {η ∈ H | η p = η0 }. In the latter case it is a coset of P . This will P prove Claim D. To prove (2), consider an element ρ = η∈H cη η in Z[H]. For each η ∈ H integer between 0 and p − 1 satisfying c0η ≡ cη mod p. let c0η be the P unique 0 0 0 Put ρ = η∈H cη η. If cη is constant on each Hj , then so is cη . So, by 0 0 Claim B, ρ ∈ R implies ρ ∈ R. By definition, Supp(ρ ) ⊆ Supp(ρ). Also, 0 0 ρ1 ≡ ρ2 mod p, with ρ1 , ρP 2 ∈ Z[H], is equivalent to ρ1 = ρ2 . p 0 Specifically, let ρ = η∈Hj η . Then Supp(ρ ) ⊆ Supp(ρ) ⊆ {τ p | τ ∈ H}. The right hand side is a proper subgroup of H (because p|n). Hence, hSupp(ρ0 )i < H.

(3) In addition, ρ ≡

P

η∈Hj

η

p


0 ≡ αjp mod p. Hence, ρ0 = αjp . By definition,

R is a subring of Z[H]. As αj is in R (Claim B), so is αjp . Therefore, ρ0 ∈ R. It follows from (3) and Claim C that Supp(ρ0 ) ⊆ hSupp(ρ0 )i ≤ {ε}. In particular, the coefficient of η0 in ρ0 is 0. In other words, the coefficient of η0 in ρ is divisible by p. But the latter coefficient is exactly the right hand side of (2). Consequently, (2) is true. Claim E: Suppose 1 < j ≤ r and |Hj ∩ P | ≤ p−1 2 . Then αj η = αj for all η ∈ P. P P Put α = αj , π = η∈P η, and µ = η∈Hj ∩P η. By Claim D, Hj r P = Sm Pm · i=1 λi P with λ1 , . . . , λm ∈ H. Put λ = i=1 λi . Then

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477

P P P (4) λ ∈ Z[H] and α = η∈Hj η = η∈Hj r P η + η∈Hj ∩P η = λπ + µ For each η ∈ P , multiplication from the right by η yields a bijection of P onto itself. Hence, πη = π. Thus, in order to prove that αη = η, it suffices (by (4)) to prove that µ = 0. P P To this end observe that π 2 = η∈P πη = η∈P π = pπ. Similarly, πµ = kπ with k = |Hj ∩ P |. Since Z[H] is a commutative ring, this implies α2 − 2kα = pλ2 π + µ2 − 2kµ. Hence, (µ2 − 2kµ)0 = (α2 − 2kα)0 . By Claim B, α is in R. Therefore, by the proof of Claim D, (α2 − 2kα)0 ∈ R. Thus, (µ2 − 2kµ)0 ∈ R. In addition, Supp(µ2 − 2kµ)0 ⊆ Supp(µ2 − 2kµ) ⊆ P . Since p 6= n,
P is a proper subgroup of H. Therefore, by Claim C, Supp (µ2 − 2kµ)0 ⊆ {ε}. It follows that (5)

µ2 − 2kµ = pρ + lε

Pk with ρ ∈ Z[H], ε ∈ / Supp(ρ), and l ∈ Z. By definition, µ = i=1 ηi , where η1 , . . . , ηk are distinct elements of H. Each row of the matrix B = (ηi ηj )1≤i,j≤k consists of distinct elements of H. Hence, each element of H appears at most k times as an entry of B and therefore as a summand in P k 2 2 i,j=1 ηi ηj = µ . Thus, the coefficients of µ − 2kµ lie between −2k and k. Since 2k < p, this implies that ρ in (5) is 0. It follows that (5) may be rewritten as µ2 = 2kµ + lε. Thus, ηi ηj ∈ {η1 , . . . , ηk , ε} for all i, j. This means that Supp(µ) ∪ {ε} = {η1 , . . . , ηk , ε} is a subgroup of P . But k + 1 < p. Hence, Supp(µ) ∪ {ε} = {ε}. As j > 1 ε∈ / Supp(µ) (Claim B). Thus, Supp(µ) = ∅. Consequently, µ = 0, as claimed. Pr Sr End of proof: By Claim B, · j=2 Hj ∩ P = P r{ε}. Hence, j=2 |Hj ∩ P | = p − 1. Since r ≥ 3, there is a j ≥ 2 with |Hj ∩ P | ≤ p−1 2 . Hence, by Claim E, P is contained in the subgroup L = {σ ∈ G | γαj σ = γαj } of G. By Claim B, γαj = αj γ. By definition of γ, γσ = γ for each σ ∈ G1 . Therefore, G1 ≤ L and P G1 ≤ L. But P ∩ G1 = 1. Hence, G1 < L. By the primitivity of G, L = G. In particular, (γαj )−1 ∈ L. Therefore, γαj = ε. It follows that εσ = ε for each σ ∈ G, that is G is trivial. This contradiction concludes the proof of the proposition.  The following result completes Schur’s Proposition: Proposition 21.7.8 (Burnside): Let G be a transitive permutation group of a set A of p elements for some prime number p. Then G is doubly transitive or isomorphic (as a permutation group) to a subgroup of AGL(1, Fp ). Proof: For each a ∈ A we have (G : Ga ) = |A| = p. Hence, p divides |G|. By Cauchy’s theorem, G has an element π of order p. Decompose π as a product of disjoint cycles. The order of π, namely p, is the least common multiple of the lengths of these cycles. Thus, π must be a cycle of length p. We may therefore identify A with Fp such that π(a) = a + 1 for each a ∈ Fp . Denote the vector space of all functions from Fp to Fp by V . The operations in V are defined as follows: (f + g)(x) = f (x) + g(x) and (af )(x) =

478

Chapter 21. Problems of Arithmetical Geometry

Pp−1 af (x). For each f ∈ V there is a unique polynomial h(X) = i=0 ai X i in Pp−1 Fp [X] with i=0 ai xi = f (x) for all x ∈ Fp . Indeed, the latter condition gives a Vandermonde system of equations for a0 , . . . , ap−1 which determines them uniquely. Define the degree of f to be deg(h) . Then let Vk be the subspace of all f ∈ V with deg(f ) ≤ k. The action of G on Fp naturally gives an action of G on V : (σf )(x) = f (σ −1 x). The rest of the proof is divided into four parts: Part A: A basis of Vk . Let k be an integer between 0 and p − 1. Then 1, X, . . . , X k give a basis of Vk . Thus, dim(Vk ) = k + 1. Conversely, suppose f0 , . . . , fk are elements of V with deg(fj ) = j, then they are linearly independent, so they form a basis of Vk . Suppose now 1 ≤ k ≤ p − 1. Consider f ∈ V of degree k given by f (x) = ak xk + ak−1 xk−1 + · · · + a0 with ai ∈ Fp and ak 6= 0. Then (πf )(x) = ak (π −1 x)k + ak−1 (π −1 x)k−1 + · · · + a0 = ak (x − 1)k + ak−1 (x − 1)k−1 + · · · + a0 = ak xk + (−kak + ak−1 )xk−1 + lower terms. Hence, f (x) − (πf )(x) = kak xk−1 + lower terms. Thus, deg(πf ) = k and deg(f − πf ) = k − 1. Inductively define fk = f and fj−1 = fj − πfj for j = k, k − 1, . . . , 1. Then deg(fj ) = j, so f0 , . . . , fk form a basis of V . Since f0 , . . . , fk are linear combinations of f, πf, . . . , π k f , the latter k functions also form a basis of V all of its elements have degree k. Part B: The fixed space of G0 . Assume from now on G is not doubly transitive. Then Fp has at least three G0 -orbits, A0 , A1 , A2 . Let gi : Fp → Fp be the characteristic function of Ai , i = 0, 1, 2. Then σgi = gi for each σ ∈ G0 and g0 , g1 , g2 are linearly independent. Thus, at most one of the functions g0 , g1 , g2 is of degree 0 and each of them belongs to the subspace U = {f ∈ V | σf = f for each σ ∈ G0 } of V . We prove that U contains a function f with 1 ≤ deg(f ) ≤ p − 2. If 1 ≤ deg(gi ) ≤ p − 2 for some i between 0 and 2, take f = gi . Otherwise, either deg(g0 ) = deg(g1 ) = deg(g2 ) = p−1 or, say, deg(g0 ) = 0 and deg(g1 ) = deg(g2 ) = p−1. In the former case there are b1 , b2 , c1 , c2 ∈ F× p with deg(b1 g0 + b2 g1 ) ≤ p − 2 and deg(c1 g0 + c2 g2 ) ≤ p − 2. Since the functions b1 g0 + b2 g1 and c1 g0 + c2 g2 are linearly independent, at least one of them is not of degree 0 (because dim(V0 ) = 1), so has degree between 1 and p − 2. In the latter case there are b1 , b2 ∈ F× p with 1 ≤ deg(b1 g1 + b2 g2 ) ≤ p − 2. Let k = deg(f ). Part C: The subspaces Vk and Vk−1 are G-invariant. By part A, f, πf, . . . , π k f form a basis of Vk . Hence, hπi leaves Vk invariant. In addition, hπi is transitive on Fp . Hence, G = hπiG0 = G0 hπi. Moreover, σf = f for each σ ∈ G0 . Therefore, G0 Vk = Vk . Consequently, GVk = Vk .

21.8 Schur’s Conjecture

479

Next consider the G-invariant subspace W =

nX σ∈G

cσ · σf |

X

o

cσ = 0

σ∈G

of V . We claim that W ⊆ Vk−1 . Indeed, write each σ ∈ G as σ = σ0 π i with σ0 ∈ G0 . Then, in the notation of Part A, (σfP )(x) = (π i f )(x) = −i k f (π Px) = f (x − i) = ak x + lower terms. Hence, if σ∈G cσ = 0, then deg( σ∈G cσ ·σf ) ≤ k−1, as claimed. In particular, dim(W ) ≤ dim(Vk−1 ) = k. Finally, note that f − πf, πf − π 2 f, . . . , π k−1 f − π k f are in W and they are linearly independent. Hence, dim(W ) = k. Therefore, W = Vk−1 . It follows that Vk−1 is G-invariant. Part D: End of proof. Define the product gh of two functions g, h ∈ V by (gh)(x) = g(x)h(x). Then, σ(gh) = σ(g)σ(h) for each σ ∈ G and V1 = {g ∈ V | gh ∈ Vk for all h ∈ Vk−1 }. Since both Vk−1 and Vk are G-invariant, so is V1 . In particular, applying σ −1 to the identity map of Fp , we find a ∈ F× p and b ∈ Fp with σ(x) = ax + b for all x ∈ Fp . This gives the desired embedding  of G into AGL(1, Fp ). The combination of Schur’s result and Burnside’s result gives the following characterization of transitive proper subgroups of AGL(1, Fp ): Corollary 21.7.9: Let G be a transitive permutation group of a set A of n elements. Suppose G contains a cycle of length n but G is not doubly transitive. Then n is a prime number p and G is isomorphic (as a permutation group) to a subgroup of AGL(1, Fp ). In particular, G is solvable. Proof: By Schur (Proposition 21.7.7), n = p is a prime number. By Burnside (Proposition 21.7.8), G is isomorphic to a subgroup of AGL(1, Fp ). It follows from Remark 21.7.1(d) that G is solvable. 

21.8 Schur’s Conjecture Let K be a field and f ∈ K[X]. We say that f permutes K, f is a permutation polynomial on K, or f is bijective on K if the map x 7→ f (x) is a bijection of K onto itself. Let R be a ring, f ∈ R[X], and P a maximal ideal of R. We say f is a permutation polynomial modulo P , if the reduction of f modulo P is a permutation polynomial on R/P . In this section we consider a global field K and a polynomial f ∈ OK [X] with char(K) - deg(f ) which is a permutation polynomial modulo infinitely many prime ideals of OK . We prove a conjecture of Schur: f is composed of linear polynomials and Dickson polynomials Dn (a, X). The latter are defined in Z[a, X] by induction on n: D0 (a, X) = 2, D1 (a, X) = X, and (1) Dn (a, X) = XDn−1 (a, X) − aDn−2 (a, X) for n ≥ 2.

480

Chapter 21. Problems of Arithmetical Geometry

Thus, D2 (a, X) = X 2 − 2a, D3 (a, X) = X 3 − 3aX, and D4 (a, X) = X 4 − 4aX 2 + 2a2 . Call Dn (a, X) the Dickson polynomial of degree n with parameter a. Lemma 21.8.1: Let K be a field, a ∈ K, and n a nonnegative integer. Use X and Z for variables. Then: (a) Dn (a, Z + aZ −1 ) = Z n + an Z −n . (b) Suppose f ∈ K(Z) satisfies f (Z + aZ −1 ) = Z n + an Z −n . Then f (X) = Dn (a, X). (c) bn Dn (a, X) = Dn (b2 a, bX). (d) For n ≥ 3 we have Dn (a, X) = X n − naX n−2 + g(X) with g ∈ Z[a, X] of degree at most n − 3. Proof of (a): Check (a) for n = 0 and n = 1. Next assume (a) by induction for n − 2 and n − 1. Then Dn (a, Z + aZ −1 ) = (Z + aZ −1 )Dn−1 (a, Z + aZ −1 ) − aDn−2 (a, Z + aZ −1 ) = (Z + aZ −1 )(Z n−1 + an−1 Z 1−n ) − a(Z n−2 + an−2 Z 2−n ) = Z n + an Z −n Proof of (b): Put X 0 = Z + aZ −1 . Then f (X 0 ) = Dn (a, X 0 ) (by (a)) and [K(Z) : K(X 0 )] ≤ 2. Hence, X 0 is a variable. Therefore, f (X) = Dn (a, X). Proof of (c) and (d): Carry out induction on n.



Definition 21.8.2: Linear relation of polynomials. Let K be a field and f, g ∈ K[X] polynomials. We say f and g are linearly related over K if there exist a, b ∈ K × and c, d ∈ K with f (X) = ag(bX + c) + d.  Lemma 21.8.3: Let K be a field, f ∈ K[X], K 0 an extension of K, and n a positive integer. Suppose char(K) - n and f (X) is linearly related over K 0 to Dn (a0 , X) for some a0 ∈ K 0 . Then f (X) is linearly related over K to Dn (a, X) for some a ∈ K. Proof: We prove the lemma only for n ≥ 3. The case n ≤ 2 may be checked directly. By assumption, there are α0 , β 0 ∈ (K 0 )× and a0 , γ 0 , δ 0 ∈ K 0 with f (X) = α0 Dn (a0 , β 0 X + γ 0 ) + δ 0 . Put a = (β 0 )−2 a0 , α = α0 (β 0 )n , γ = (β 0 )−1 γ 0 , and δ = δ 0 . By Lemma 21.8.1(c), f (X) = αDn (a, X + γ) + δ. By Lemma 21.8.1(d), f (X) = α(X + γ)n − nαa(X + γ)n−2 + lower terms

= αX n + nαγX n−1 + α n2 γ 2 − na X n−2 + lower terms

Hence, α ∈ K, nαγ ∈ K, and α n2 γ 2 − na ∈ K. By assumption, n 6= 0 in K. Therefore, γ ∈ K and a ∈ K. Finally, δ = f (0) − αDn (a, γ) ∈ K.  Consequently, f (X) is linearly related to Dn (a, X) over K.

21.8 Schur’s Conjecture

481

Special cases of Dickson polynomials give rise to well known families of polynomials. For a = 0 and n ≥ 1 induction or Lemma 21.8.1(a) give Dn (0, X) = X n . The special case a = 1 gives the Chebyshev polynomials: Tn (X) = Dn (1, X). We extract properties of Chebyshev polynomials from Lemma 21.8.1: Lemma 21.8.4: Let K be a field and n a nonnegative integer. Then: (a) T0 (X) = 2, T1 (X) = X, and Tn (X) = XTn−1 (X) − Tn−2 (X) for n ≥ 2. (b) Tn (Z + Z −1 ) = Z n + Z −n . (c) Suppose f ∈ K(Z) satisfies f (Z + Z −1 ) = Z n + Z −n . Then f = Tn . (d) For each n ≥ 3 there is a g ∈ Z[X] of degree at most n − 3 satisfying Tn (X) = X n − nX n−2 + g(X). ˜ there are distinct (e) Suppose char(K) - n and n > 2. Then, for each a ∈ K ˜ x1 , x2 ∈ K with Tn (x1 ) = Tn (x2 ) = a. ˜ satisfying y 2 − ay + 1 = 0. Then y + y −1 = a. Proof of (e): Choose y ∈ K ˜ with z n = y, i = 1, . . . , n. Since char(K) - n, there are distinct z1 , . . . , zn ∈ K i Since n ≥ 3, there are i, j such that xi = zi + zi−1 and xj = zj + zj−1 are  distinct. They satisfy Tn (xi ) = Tn (xj ) = a. Remark 21.8.5: Automorphisms of K(z). Let K be a field and z an indeterminate. Recall that PGL(2, K) is the quotient of GL(2,
K) by the group of scalar matrices. Denote the image of a matrix ac db ∈ GL(2, K)      0 0 in PGL(2, K) by ac db . Thus, ac db = ac0 db 0 means that there is an
0 0
eb e ∈ K × with ac0 db 0 = ea 11.7.4 identifies PGL(2, K) with ec ed . Remark a b Aut(K(z)/K), where the action of c d on K(z) is defined by the formula: a b az+b  c d z = cz+d . Lemma 21.8.6: Let K be an algebraically closedfield and n ≥ 3 an   integer  with char(K) - n. Consider the elements σ = 10 ζ0n and τ = 01 10 of PGL(2, K). Put ∆ = hσ, τ i. Then: (a) σ n = τ 2 = 1, σ τ = σ −1 , and ∆ ∼ = Dn . (b) Let ∆1 = hσ1 , τ1 i be a subgroup of PGL(2, K) with σ1n = τ12 = 1 and σ1τ1 = σ1−1 (so, ∆1 ∼ = Dn ). Then there is a λ ∈ PGL(2, K) with hσ1 iλ = λ hσi, τ1 = τ , and ∆λ1 = ∆. (c) Let z be an indeterminate. Then the fixed field of τ (resp. ∆) in K(z) is K(z + z −1 ) (resp. K(z n + z −n )). Proof of (a): Put ζ = ζn . All we need to do is to verify the relation σ τ = σ −1 : 

0 1

1 0



1 0

0 ζ



0 1

  1 ζ = 0 0

   0 1 0 = . 1 0 ζ −1

Proof of (b): Denote the image of σ ∈ GL(2, K) in PGL(2, K) by [σ]. σ1 ] = σ 1 , Since K is algebraically closed, there are σ ˜1 , τ˜1 ∈ GL(2, K) with [˜ ˜1n = τ˜12 = 1. The minimal polynomial of σ ˜1 divides X n − 1. [˜ τ1 ] = τ1 , and σ ˜1 is conjugate Since n 6= char(K), X n − 1 has n distinct roots in K. Hence, σ

482

Chapter 21. Problems of Arithmetical Geometry


to a diagonal matrix a0 d0 with an = dn = 1. Therefore, [˜ σ1 ] is conjugate 1 0  −1 k to 0 a−1 d with a d = ζ and gcd(k, n) = 1. Replace ζ by ζ k , if necessary, to assume [˜ σ1 ] = σ.
Let τ1 = wy xz with wy xz ∈ GL(2, K). Since τ1 σ = σ −1 τ1 , there is an a ∈ K × with       w x w x 1 0 a 0 . = y z y z 0 ζ 0 aζ −1 Therefore, w = aw, xζ = ax, y = aζ −1 y, and zζ = aζ −1 z. Assume w 6= 0. Then a = 1, x = 0, y = 0, and z = 0 (use n 6= 2). This contradiction proves that w = 0. Hence,  x 6= 0 and y 6= 0. Therefore, a = ζ and z = 0. Consequently, τ1 = y0 x0 .  √ Put ρ = 0x √0y . Then, ρ−1 τ1 ρ = τ and ρ−1 σρ = σ. So, there exists λ ∈ PGL(2, K) as claimed. Proof of (c): Let x = z + z −1 and t = z n + z −n . Denote the fixed field of τ (resp. ∆) in K(z) by E1 (resp. E2 ). Then [K(z) : E1 ] = 2 and [K(z) : E2 ] = 2n. Deduce from τ z = z −1 and τ z −1 = z that τ x = x. Hence, K(x) is contained in E1 . On the other hand, z is a root of X 2 − xX + 1. Therefore, [K(z) : K(x)] ≤ 2. Combined with the preceding paragraph, this implies K(x) = E1 . Similarly, K(t) ⊆ E2 and z is a root of the equation X 2n − tX n + 1 = 0.  Hence, K(t) = E2 . Proposition 21.8.7 (M¨ uller): Let K be an algebraically closed field of characteristic p, f ∈ K[X] a polynomial of prime degree l 6= p, and t an indeterminate. Suppose G = Gal(f (X) − t, K(t)) is solvable. Then either G∼ = Dl and f is linearly related to = Cl and f is linearly related to X l or G ∼ the Chebyshev polynomial Tl . Proof: By assumption, f (X) − t is a separable polynomial of degree l which is irreducible. List its zeros in K(t)s as x1 , . . . , xl . Then F = K(x1 , . . . , xl ) is the Galois closure of K(xj )/K(t), G = Gal(F/K(t)) acts faithfully and transitively on {x1 , . . . , xl }. When l = 2, G ∼ = C2 and f (X) is linearly related to X 2 (use that 2 6= p). So, assume from now on l > 2. Then, we may apply Lemma 21.7.2 with {x1 , . . . , xl } replacing X and Hj = Gal(F/K(xj )) replacing Gxj , j = 1, . . . , l. In particular, let L be a minimal normal subgroup of G. Then, by Lemma 27.7.2(b,e), L ∼ = Cl and L is the unique l-Sylow subgroup of G. Let m = [F : K(xj )] and g = genus(F/K). In the remaining parts of the proof we apply Riemann-Hurwitz to prove that g = 0 and m ≤ 2. We then show: If m = 1, then G ∼ = Cl and f is linearly related to X l . If m = 2, ∼ then G = Dl and f is linearly related to Tl .

21.8 Schur’s Conjecture

483

Part A: The Riemann-Hurwitz formula. Denote the set of all prime divisors of F/K which ramify over K(t) by R. It includes all P prime divisors ). Hence, Diff(F/K(t)) = of F/K which ramify over K(x j q∈R dq q and P Diff(F/K(xj )) = q∈R dq,j q (Section 3.6). Here dq (resp. dq,j ) are the different exponents of q over K(t) (resp. K(xj )). Since K is algebraically closed, deg(q) = 1 for each q ∈ R. Therefore, Riemann-Hurwitz formula for F/K(t) and F/K(xj ) becomes: X (2a) dq 2g − 2 = −2ml + q∈R

(2b)

2g − 2 = −2m +

X

dq,j ,

j = 1, . . . , l.

q∈R

Subtract the sum of the l equations (2b) from (2a): (3)

2(g − 1)(1 − l) =

X q∈R

dq −

l X

dq,j ).

j=1

Now consider q ∈ R. Denote the inertia group of q over K(t) by Iq . Then the inertia group of q over K(xj ) is Hj ∩ Iq . To compute the qth term in (3) we distinguish between two cases. Case A1: Iq acts intransitively on {x1 , . . . , xl }. By Lemma 21.7.2(a), L 6≤ Iq . Since L ∼ = Cl , we have L∩Iq = 1. Thus, Iq ∼ = LIq /L ≤ G/L. Since G/L is / L. Hence, by Lemma cyclic, Iq is cyclic. Choose a generator τ of Iq . Then τ ∈ 21.7.2(g), there is a k between 1 and l with τ ∈ Hk and therefore Iq ≤ Hk . This implies that q|K(xk ) is unramified over K(t). Therefore, dq = dq,k (Lemma 3.5.8). For j 6= k we have Iq ∩ Hj = 1 (Lemma 21.7.2(g)). Hence, q is unramified over K(xj ) and dq,j = 0 (Remark 3.5.7(b)). Consequently, the qth term in the right hand side of (3) is 0. Case A2: Iq acts transitively on {x1 , . . . , xl }. Then for each j we have (Iq : Iq ∩ Hj ) = l. Hence, l divides |Iq |. Therefore, by Lemma 21.7.2(e), L ≤ Iq . Let P be the trivial group if p = 0 and a p-Sylow subgroup of Iq if p 6= 0. Then P is a normal subgroup of Iq [Cassels-Fr¨ohlich, p. 29, Thm. 1(ii)]. Hence, by Lemma 21.7.2(h), P is trivial. Thus, Iq is cyclic [Cassels-Fr¨ohlich, p. 31, Cor. 1]. In particular, each Sylow subgroup of Iq other than L is normal, hence trivial (Lemma 21.7.2(h)). Therefore, Iq = L. Denote the fixed field of L, hence of Iq , in F by E. By Lemma 21.7.2(b), E/K(t) is Galois of degree m, K(xj )E = F , and K(xj ) ∩ E = K(t). By the preceding paragraph, q is totally ramified over E and q|E is unramified over K(t). Hence, q is unramified over K(xj ) and q|K(xj ) is totally and tamely ramified over K(t). Therefore, dq = l − 1 and dq,j = 0 (Remark 3.5.7). There are exactly m prime divisors of F/K lying over q|K(t) . Hence, they contribute m(l − 1) to the right hand side of (3).

484

Chapter 21. Problems of Arithmetical Geometry

Part B: Computation of g. Consider a prime divisor p of K(t)/K. Let q be a prime divisor of F/K over p. Suppose Iq acts transitively on {x1 , . . . , xl }. By Case A2, p totally ramifies in K(x1 ). Hence, there are only finitely many such p. Denote their number by r. Conversely, suppose Iq acts intransitively on {x1 , . . . , xl }. Case A1 gives k with q|K(xk ) unramified over K(t). Choose ρ ∈ G with ρxk = x1 . Then ρq|K(x1 ) is unramified over K(t). Hence, p is not totally ramified in K(x1 ). Thus, by Case A1, only the r prime divisors p with Iq transitive contribute to the right hand side of (3). The contribution of each of them is, by Case A2, m(l − 1). Thus, (3) simplifies to 2(l − 1)(1 − g) = rm(l − 1) and furthermore to (4)

2(1 − g) = rm.

By Example 2.3.11, the infinite prime divisor p∞ of K(t)/K totally ramifies in K(x1 ). Hence, r ≥ 1. Therefore, by (4), g = 0 and rm = 2. It follows that, either m = 1 and r = 2 or m = 2 and r = 1. Part C: Suppose m = 1 and r = 2. Then K(x1 ) = F is a Galois extension of degree l of K(t). Hence, G ∼ = Cl . Since r = 2, K(t) has a finite place pc , mapping t to c ∈ K, which totally ramifies in F . In particular, pc has only one prime divisor in F . Hence, f (X) − c has only one root in K. It follows that there are a ∈ K × and b ∈ K with f (X) − c = (aX + b)l . Thus, f (X) is linearly related to X l . Part D: Suppose m = 2 and r = 1. By Lemma 21.7.2(i), G is the dihedral group Dl generated by the involution τ of H1 and a generator σ of L with the relation σ τ = σ −1 . By part B, g = 0. Since K is algebraically closed, each prime divisor of F/K has degree 1. Hence, F = K(z) for some z (Example 3.2.4). Lemma with σ 0 =  1 0  21.8.6(b) gives an0 automorphism  0 λ1 of Aut(F/K) −1 −1 0 λσλ = 0 ζ (and ζ = ζl ) and τ = λτ λ = 1 0 . Put z = λz, x0 = λx1 ,   and t0 = λt. Then, K(x0 ) is the fixed field in K(z 0 ) of 01 10 and K(t0 ) is the D E    fixed field in K(z 0 ) of 10 ζ0 , 01 10 . By Lemma 21.8.6(c), the former field is K(z 0 + (z 0 )−1 ) and the latter field is K((z 0 )l + (z 0 )−l ). Therefore, there are κ1 ∈ Aut(K(x0 )/K) and κ2 ∈ Aut(K(t0 )/K) with x0 = κ1 (z 0 + (z 0 )−1 ) and t0 = κ2 ((z 0 )l + (z 0 )−l ). obius transformations having coefficients in K. Identify λ, κ1 , κ2 with M¨ Put µ = λ−1 κ1 and ν = λ−1 κ2 . Then x1 = µ(z 0 + (z 0 )−1 ) and t = ν((z 0 )l + (z 0 )−l ). Hence, ν −1 ◦ f ◦ µ(z 0 + (z 0 )−1 ) = (z 0 )l + (z 0 )−l . Since z 0 is transcendental over K, Lemma 21.8.4(c) implies ν −1 ◦ f ◦ µ = Tl . Part E: Conclusion of the proof. It remains to prove that µ and ν are linear polynomials. Assume first ν is not a polynomial. Then ν −1 is not a polynomial. Thus, −1 −1 (∞) = ac . ν (X) = aX+b cX+d with a, b, c, d ∈ K and c 6= 0. It follows that ν Use Lemma 21.8.4(e) to find distinct e1 , e2 ∈ K with Tl (ei ) = ac . Then

21.8 Schur’s Conjecture

485

f (µ(ei )) = ν(Tl (ei )) = ν( ac ) = ∞. Since f is polynomial, µ(ei ) = ∞, i = 1, 2. This contradiction to the injectivity of µ on K ∪ {∞} proves that ν is a linear polynomial. Finally, assume that µ is not a polynomial. Then e = µ(∞) ∈ K. Hence, by the preceding paragraph, Tl (∞) = ν −1 (f (µ(∞))) = ν −1 (f (e)) ∈ K, which is a contradiction. Consequently, µ is a linear polynomial.  Lemma 21.8.8: Let K be an algebraically closed field, f ∈ K[X] a polynomial of degree n, x a transcendental element over K, and t = f (x). Suppose char(K) - n. Then Gal(f (X) − t, K(t)) contains an n-cycle. Proof: By assumption, f (X) − t has n distinct roots x1 , . . . , xn in K(t)s . Put x = x1 and F = K(x1 , . . . , xn ). Let v∞ be the valuation of K(t)/K with v∞ (t) = −1. By Example 2.3.11, v∞ is totally ramified in K(x). Thus, v∞ has a unique extension w to K(x). ˆ (By Example 3.5.1, Denote the completion of E = K(t) at v∞ by E −1 ∼ ˆ and E(x)/ ˆ ˆ ˆ E E = K((t )).) By Lemma 3.5.3, f (X) − t is irreducible over E ˆ ˆ is a totally ramified extension of degree n. Since char(K) - n, E(x)/E is ˆ n ) is a compositum ˆ Then Fˆ = E(x ˆ 1 ) · · · E(x tamely ramified. Put Fˆ = F E. ˆ ˆ ˆ of tamely ramified extensions of E. Hence, F /E is tamely ramified [CasselsFr¨ohlich, p. 31, Cor. 2]. Since K is algebraically closed, the residue degree ˆ is 1. Therefore, Fˆ /E ˆ is totally and tamely ramified. Therefore, of Fˆ /E ˆ ˆ ˆ ˆ Gal(F /E) is cyclic [Cassels-Fr¨ ohlich, p. 31, Cor. 1]. It follows that E(x)/ E ∼ ˆ ˆ ˆ ˆ ˆ is a Galois extension. Hence, F = E(x) and Gal(f (X) − t, E) = Gal(F /E) ˆ is transitive. is cyclic of order n. It follows that Gal(f (X) − t, E) ˆ Choose a generator σ of Gal(f (X) − t, E). It decomposes into disjoint cycles whose lengths sum up to n. Since hσi is transitive, there is only one cycle. Thus, σ is an n-cycle. Finally, note that σ ∈ Gal(f (X) − t, K(t)).  Let K be a field and f ∈ K[X] a polynomial with deg(f ) > 1. Then f = f1 ◦ f2 ◦ · · · ◦ fr with fi ∈ K[X] indecomposable and deg(fi ) > 1. Call each fi a decomposition factor of f over K. Lemma 21.8.9: Let K be a field, t an indeterminate, and f ∈ K[X] an indecomposable polynomial over K of degree n with char(K) - n. Then Gal(f (X) − t, K(t)) is a primitive group. Proof: Let x be a root of f (X) − t in K(t)s and F the Galois closure of K(x)/K(t). By Lemma 21.7.4, it suffices to prove that Gal(F/K(x)) is a maximal subgroup of Gal(F/K(t)). In other words, we have to prove that K(x)/K(t) is a minimal extension. Assume there is a field E which lies strictly between K(t) and K(x). Let d = [K(x) : E] and m = [E : K(t)]. Then d > 1 and m > 1. Let v∞ be the valuation of K(t)/K with v∞ (t) = −1. Then v∞ is totally ramified in K(x). Thus, v∞ has a unique extension w0 to K(x), and w0 satisfies w0 (x) = −1 and has the same residue field as v∞ , namely K (Example 2.3.11). Let w be uroth the restriction of w0 to E. Then the residue field of E at w is K. By L¨

486

Chapter 21. Problems of Arithmetical Geometry

(Remark 3.6.2(a)), E is a rational function field over K. Therefore, there is a y ∈ E with w(y) = −1 and v(y) ≥ 0 for all other valuations v of E/K. Therefore, w0 (y) = −d and v(y) ≥ 0 for all other valuations v of K(x)/K. It follows that y = h(x) for some h ∈ K[X]. The degree of the pole divisor of y as an element of K(x) is d. Therefore, [K(x) : K(y)] = d. Since K(y) ⊆ E, this implies E = K(y). Therefore, deg(h) = d. Repeat the above arguments for t, y replacing y, x to find a polynomial g ∈ K[X] of degree m with t = g(y). This gives f (x) = t = g(y) = g(h(x)). Hence, f (X) = g(h(X)). This contradiction to the indecomposability of f proves that K(x)/K(t) is minimal.  Let K be a field and f ∈ K[X] a polynomial of positive degree. Put f ∗ (X, Y ) =

f (X) − f (Y ) . X −Y

Proposition 21.8.10: Let K be an algebraically closed field and f ∈ K[X] a polynomial of degree l. Suppose char(K) - l, f is indecomposable over K, and f ∗ (X, Y ) is reducible. Then l is prime and f (X) is linearly related to X l or to Tl (X). Proof: Let t be an indeterminate. Since char(K) - l, f (X) − t has l distinct roots x1 , . . . , xl in K(t)s . Hence, F = K(x1 , . . . , xl ) is a Galois extension of K(t). Put G = Gal(F/K(t) and Hj = Gal(F/K(xj )), j = 1, . . . , l. View G as a permutation group of {x1 , . . . , xl }. Since f (X) − t is irreducible in K(t)[X], G is transitive. Since f is indecomposable, G is primitive (Lemma 21.8.9). Further, by Lemma 21.8.8, G contains an l-cycle. Since f ∗ (X, Y ) is reducible, l−1 ≥ 2. By Gauss, f ∗ (x1 , Y ) is reducible in K(x1 )[Y ] of degree l − 1. Therefore, H1 is intransitive on the roots x2 , . . . , xl of f ∗ (x1 , Y ). This means that G is not doubly transitive (Lemma-Definition 21.7.5). It follows from Corollary 21.7.9 that G is solvable. Consequently, by  Proposition 21.8.7, f is linearly related to X l or to Tl (X). Lemma 21.8.11 ([Fried-MacCrae, Thm. 3.5]): Let K be a field, f ∈ K[X] a polynomial, and L a field extension of K. Suppose char(K) - deg(f ) and f is decomposable over L. Then f is decomposable over K. Pn Pm Proof: There are g(X) = i=0 ai X i and h(X) = j=0 bj X j in L[X] with f (X) = g(h(X)) and deg(g), deg(h) < deg(f ). Assume without loss L is algebraically closed. Divide g and f by am , if necessary, to assume am = 1. −1/n Then replace X by bn X, if necessary, to assume bn = 1. It follows that f , g, and h are monic. Finally, choose β ∈ L with h(β) = 0. Then replace X by X + β to assume b0 = 0. We prove that g, h ∈ K[X]. Indeed, f (X) = (X n + bn−1 X n−1 + · · · + b1 X)m (5)

+

m−1 X i=0

ai (X n + bn−1 X n−1 + · · · + b1 X)i .

21.8 Schur’s Conjecture

487

The highest power of X involved in the sum on the right hand side of (5) is X (m−1)n . Hence, for each k between 1 and n − 1, only the first term on the right hand side of (5) contributes to the coefficient of X mn−k . Therefore, this coefficient is mbn−k + pk (bn−k+1 , . . . , bn−1 , 1) with pk ∈ Z[Xn−k+1 , . . . , Xn ]. Observe that deg(f ) = mn, so char(K) - m. Induction on k proves that bn−k ∈ K. Thus, h ∈ K[X]. Finally, suppose am−1 , . . . , am−r+1 have been proved to be in K. Then am−r h(X)m−r + am−r−1 h(X)m−r−1 + · · · + a0 is a polynomial in K[X]. Its leading coefficient is am−r . Hence, am−r ∈ K. Consequently, g ∈ K[X].  Let K be a field and f ∈ K[X]. We say f is injective (resp. surjective) on K if the map x 7→ f (x) of K into itself is injective (resp. surjective). Proposition 21.8.12: Let K be a perfect field and f ∈ K[X] a polynomial of degree l with char(K) - l. (a) Suppose K is PAC and f is injective on K and indecomposable over K. Then l is a prime number and f is linearly related over K to a Dickson polynomial of degree l. (b) Suppose K is pseudo finite and f is injective or surjective on K. Then each composition factor of f is linearly related over K to a Dickson polynomial of a prime degree. Proof of (a): By assumption, there are no x, y ∈ K with f (x) = f (y) and x 6= y. Hence, there are no x, y ∈ K with f ∗ (x, y) = 0 and x 6= y. By ˜ Therefore, by Proposition Proposition 11.1.1, f ∗ (X, Y ) is reducible over K. ˜ to X l or to Tl (X). Both 21.8.10, l is prime and f is linearly related over K l X and Tl (X) are Dickson polynomials. It follows from Lemma 21.8.3 that f is linearly related over K to a Dickson polynomial of degree l. Proof of (b): Every injective (resp. surjective) polynomial on a finite field F of degree n is bijective. Since this is an elementary statement, it holds for every pseudo finite field (Proposition 20.10.4). In particular, f is bijective on K. Suppose f (X) = g(h(X)) with g, h ∈ K[X]. Then g is surjective on K and h is injective. By the preceding paragraph both g and h are bijective on K. It follows that each composition factor q of f over K is bijective on K. By (a), each composition factor of f is linearly related to a Dickson polynomial of a prime degree.  Theorem 21.8.13 (Schur’s Conjecture): Let K be a global field and f ∈ K[X] a polynomial with char(K) - deg(f ). Suppose f is a permutation polynomial modulo p for infinitely many prime divisors p of K. Then each composition factor of f is linearly related over K to a Dickson polynomial of a prime degree. Proof: The statement “f is a permutation polynomial” is elementary. By Example 20.10.7(a) there is a pseudo finite field F containing K such that f is

488

Chapter 21. Problems of Arithmetical Geometry

a permutation polynomial on F . By Proposition 21.8.12(b), each composition factor of f is linearly related over F , hence over K (Lemma 21.8.3) to a Dickson polynomial of a prime degree.  Theorem 21.8.14: Let K be a finite field and f ∈ K[X] a polynomial with char(K) - deg(f ). Suppose f permutes infinitely many finite extensions L of K. Then each composition factor of f is linearly related over K to a Dickson polynomial of a prime degree. Proof: Take a nonprincipal ultraproduct F of those finite extensions of K on which f is a permutation polynomial. Then f is a permutation polynomial on F . By Example 20.10.7(a), F is pseudo finite and K ⊆ F . By Proposition 21.8.12(b), each composition factor of f is linearly related over F , hence over K, to a Dickson polynomial of a prime degree.  We end this section with examples which give converses to Proposition 21.8.12 and Theorem 21.8.14: Example 21.8.15: Cyclic polynomials. Let n be a positive integer and F a finite field of order q. Then F× q is a cyclic group of order q − 1. Hence, the map x 7→ xn is bijective on F if and only if gcd(n, q − 1) = 1. Suppose n ≥ 3 is odd. Let q1 , . . . , qr be the prime divisors of n. Then both 1 and 2 are relatively prime to qi , i = 1, . . . , r. Choose an integer a with a ≡ 1 mod qi , i = 1, . . . , r. Then both a and a + 1 are relatively prime to n. Dirichlet’s theorem (Corollary 6.3.2) gives infinitely many prime numbers p ≡ a+1 mod n. By the preceding paragraph, X n is a permutation polynomial modulo p for each of these p.  The next lemma will be used to give more examples of Dickson polynomials which are permutation polynomials modulo infinitely many primes: Lemma 21.8.16: Let n be a positive integer, q a power of a prime number, and a ∈ Fq . Suppose gcd(n, q 2 − 1) = 1. Then Dn (a, X) permutes Fq . Proof: Put K = Fq and L = Fq2 . Then L× is a cyclic group of order q 2 − 1. Hence, the map z 7→ z n is bijective on L. Now consider x1 , x2 ∈ K with Dn (a, x1 ) = Dn (a, x2 ). Choose z1 , z2 ∈ L with zi + azi−1 = xi , i = 1, 2. Then Dn (a, z1 + az1−1 ) = Dn (a, z2 + az2−1 ). By Lemma 21.8.1(a), z1n + an z1−n = z2n + an z2−n . Denote the common value of the two sides of the latter equality by b. The product of the two solutions of the equation X + an X −1 = b is an . Both z1n and z2n are solutions. Hence, z1n = z2n or z1n = an z2−n . By the preceding paragraph, z1 = z2 or z1 = az2−1 . In both cases, x1 = x2 . Consequently, Dn (a, X) is a permutation polynomial on K.  Example 21.8.17: Permutation polynomials modulo infinitely many primes. (a) Let n be a positive integer, K a number field, and a ∈ OK . Suppose gcd(6, n) = 1 and K ∩ Q(ζn ) = Q. Then Dn (a, X) is a permutation polynomial modulo infinitely many prime ideals of OK .

21.9 Generalized Carlitz’s Conjecture

489

Indeed, list the prime divisors of n as q1 , . . . , qr . Choose an integer b with b ≡ 2 mod qi , i = 1, . . . , r. Then gcd(b, n) = 1 and gcd(b2 − 1, n) = 1. Denote the Galois closure of K(ζn )/Q by L. Then choose σ ∈ Gal(L/K) with σζn = ζnb . Chebotarev density theorem (Theorem 6.3.1) gives infinitely   many prime ideals P of OL with L/Q = σ. Let P be one of them. Then P the decomposition group of P over Q is in Gal(L/K). Put p = P|K . Then p = N p is a prime number. By the proof of Corollary 6.3.2, p ≡ b mod n. ¯ p |2 − 1, n) = 1. Consequently, by Lemma Hence, by the choice of b, gcd(|K 21.8.16, Dn (a, X) is a permutation polynomial modulo p. (b) Let n be a positive integer, p a prime number, q a power of p, K a finite separable extension of Fq (t) which is regular over Fq , and a ∈ OK . Suppose gcd(p(q 2 − 1), n) = 1. Then Dn (a, X) is a permutation polynomial modulo infinitely many prime divisors of K/Fq . Indeed, let L be the Galois closure of K/Fq (t), Fqm the algebraic closure of Fq in L, and ϕ(n) the Euler totient function. Then q ϕ(n) ≡ 1 mod n. Consider a large positive integer k. Choose τ in Gal(L/K) whose restriction to Fqm coincides with the restriction of Frobkϕ(n)+1 . Denote the conjugacy q class of τ in Gal(L/Fq (t)) by Con(τ ). Proposition 6.4.8 gives a prime divisor L/Fq (t)
p of Fq (t) with deg(p) = kϕ(n) + 1 and = Con(τ ). Hence, there is a p L/Fq (t) prime divisor Q of L with = τ . Denote the restriction of Q to K by Q ¯ q |2 − 1 = q 2(kϕ(n)+1) − 1 ≡ q. Then deg(q) = deg(p) = kϕ(n) + 1. Hence, |K 2 q − 1 mod n. Consequently, by Lemma 21.8.16, Dn (a, X) is a permutation polynomial modulo q. 

21.9 Generalized Carlitz’s Conjecture Let q be a power of a prime number p and f ∈ Fq [X] a polynomial of degree n. Suppose f is a permutation polynomial of Fqk for infinitely many k but f is not a p-power in Fq [X]. Theorem 21.8.14 describes the composition factors of f when p - n. In the general case the generalized Carlitz’s Conjecture states that gcd(n, q − 1) = 1. In particular, for p 6= 2, the conjecture predicts that n is odd (Carlitz’s Conjecture). The paper [Fried-Guralnick-Saxl] gives more precise information about permutation polynomials than Carlitz’s Conjecture does. Not only that it proves that conjecture for p > 3, but it gives information about Gal(f (X) − ˜ q (t)). Suppose without loss that f is indecomposable over Fq . Then three t, F are three cases: (a) p - n and G is cyclic or isomorphic to the dihedral group Dn . This is essentially contained in Section 21.8. (b) n = pm and G ∼ = H n Fm p , where H is a subgroup of GL(m, Fp ) acting m linearly on Fp . a

a

(c) p ∈ {2, 3}, n = p (p2 −1) with a ≥ 3 odd, and G normalizes the simple group PSL(2, Fpa ).

490

Chapter 21. Problems of Arithmetical Geometry

The proof that no other cases arise uses the classification of finite simple groups and is beyond the scope of this book. One can find an elementary proof of Carlitz’s Conjecture in [Lenstra]. That proof argues with decomposition and inertia groups over Fq (t). Here we follow [Cohen-Fried] which, following a suggestion of [Lenstra, Remark 1], takes place over Fq ((t)). The first step in the proof translates the assumption “f permutes infinitely many finite extensions of Fq ” into a statement of a general nature: Lemma 21.9.1: Let f ∈ Fq [X] be a polynomial which permutes infinitely (Y ) many finite extensions of Fq . Then f ∗ (X, Y ) = f (X)−f has no absolutely X−Y irreducible factor which belongs to Fq [X, Y ]. Proof: Assume h ∈ Fq [X, Y ] is an absolutely irreducible factor of f ∗ (X, Y ). By Corollary 5.4.2 there is an n0 such that for every integer n > n0 there are distinct x, y ∈ Fqn with h(x, y) = 0. Then f (x) = f (y). Hence, f does not permutes Fqn , in contradiction to our assumption.  The converse of Lemma 21.9.1 is also true (see Exercise 14). However, it is not used in proof of the generalized Carlitz Conjecture. Lemma 21.9.2: Let N/M be a finite cyclic extension, σ an element of Gal(M ) whose restriction to N generates Gal(N/M ), and h ∈ M [X] a separable irreducible polynomial which becomes reducible over N . Then σx 6= x for every root x of h. Proof: Assume h has a root x with σx = x. By assumption, N ∩ Ms (σ) = M . Hence, N ∩ M (x) = M , so N and M (x) are linearly disjoint over M . Therefore, [N (x) : N ] = [M (x) : M ]. It follows that h = irr(x, M ) is irreducible over N , a contradiction.  Lemma 21.9.3: Let L/K be a finite cyclic extension, f ∈ K[X] a polynomial of degree at least 2 with a nonzero derivative f 0 , and z an indeterminate. (Y ) Suppose each irreducible factor of f ∗ (X, Y ) = f (X)−f in K[X, Y ] is reX−Y ducible over L. Then f (X) − z is separable and each ρ ∈ Gal(K(z)) whose restriction to L generates Gal(L/K) fixes exactly one root of f (X) − z. Proof: By assumption, (f (X) − z)0 = f 0 (X) 6= 0. All roots of f (X) − z are transcendental over K while all roots of f 0 (X) are algebraic over K. Hence, f (X) − z is relatively prime in K(z)[X] to its derivative. Therefore, f (X) − z is separable. Let Fˆ be the finite Galois extension of E = K(z) generated by L(z) and all roots of f (X) − z. Choose a generator ρ0 of L/K and let G∗ = {ρ ∈ Gal(Fˆ /E) | ρ|L = ρ0 }. ∗ ∗ For each root S x ∗of f (X) − z let Gx = {ρ ∈ G | ρx = x}. We have to prove ∗ that G = · x Gx , where x ranges over all roots of f (X) − z. We divide the rest of the proof into two parts.

21.9 Generalized Carlitz’s Conjecture

491

Part A: Disjointness. Let x and y be distinct roots of f (X)−z in Fˆ . Then f (x) = z and f (y) = z, so x, y are transcendental over K. By the above, (y) = f (X)−z f ∗ (X, y) = f (X)−f X−y X−y ∈ K(y)[X] is a separable polynomial and x is ∗ a root of f (X, y). Let h ∈ K[y, X] be an irreducible factor of f ∗ (X, y) with h(x) = 0. In particular, h is primitive. By assumption, h(X) is reducible in L[y, X]. Hence, h(X) is reducible in L(y)[X]. Now consider ρ ∈ G∗y . Then ρ ∈ Gal(Fˆ /K(y)) and ρ|L(y) generates the cyclic group Gal(L(y)/K(y)). By Lemma 21.9.2, ρ fixes no root of h. In particular, ρx 6= x. Thus, G∗x ∩ G∗y = ∅. Part B: Every ρ ∈ G∗ fixes a root of f (X) − z. Indeed, extend ρ0 in the unique possible way to a generator ρ∗0 of Gal(L(z)/K(z)). Then G∗ = {ρ ∈ Gal(Fˆ /E) | ρ|L(z) = ρ∗0 }. Hence, |G∗ | = [Fˆ : L(z)]. Next let x be a root of f (X)−z in Fˆ and extend ρ∗0 in the unique possible way to a generator ρ∗1 of Gal(L(x)/K(x)). Then G∗x = {ρ ∈ Gal(Fˆ /K(x)) | ρ|L(x) = ρ∗1 }. Hence, |G∗x | = [Fˆ : L(x)]. K(x) n

L(x)

Fˆ

n

K(z)

L(z)

K

L

Finally, let x1 , . . . , xn be the distinct roots of f (X) − z.S Then n = n deg(f ) = [K(xi ) : K(z)] = [L(xi ) : L(z)]. By Part A, | i=1 G∗xi | = Pn ∗ n[Fˆ : L(xi )] = [L(xi ) : L(z)][Fˆ : L(xi )] = [Fˆ : L(z)] = |G∗ |. i=1 |Gxi | =S n ∗  Hence, G = i=1 G∗xi . Thus, each ρ ∈ G∗ fixes some xi . Lemma 21.9.4: Let K be a field, f ∈ K[X] a monic polynomial of degree n > 1, r a divisor of n with char(K) - r, and z an indeterminate. Put ˜ Then E(x) contains an E = K((z√−1 )) and let√x be a root of f (X) − z in E. rth root r z and [E( r z) : E] = r. Proof: Let v be the unique discrete complete valuation of E/K such that v(z −1 ) = 1 (Example 3.5.1). Extend v to E(x) in the unique possible way. By Lemma 3.5.2, both E and E(x) satisfy Hensel’s lemma. Now write f (X) = X n + an−1 X n−1 + · · · + a0 . Put g(Y ) = 1 + an−1 Y + · · · + a1 Y n−1 + a0 Y n and y = x−1 . Then xn g(y) = f (x). Consider the polynomial h(T ) = T r −g(y) ∈ E(x)[T ]. Then h(1) = −an−1 y−· · ·−a1 y n−1 − a0 y n and h0 (1) = r 6= 0. By assumption, xn +an−1 xn−1 +· · ·+a0 = z. Hence,

492

Chapter 21. Problems of Arithmetical Geometry

v(x) < 0, so v(y) > 0. It follows that, v(h(1)) > 0 and v(h0 (1)) = 0. Hensel’s lemma gives t ∈ E(x) with tr = g(y). Finally, let w = xn/r t. Then w ∈ E(x) and wr = xn tr = xn g(y) = f (x). Moreover, E(w)/E is a totally ramified extension of degree r (Example 2.3.8).  Theorem 21.9.5 (Lenstra): Let p be a prime number, q a power of p, and f ∈ Fq [X] a polynomial of degree n > 1. Suppose f 6= f1p for all f1 ∈ Fq [X] and f permutes infinitely many finite extensions of Fq . Then gcd(n, q − 1) = 1. Proof: Put K = Fq . Since f is not a pth power of a polynomial in K[X] (Y ) and K is perfect, the derivative of f is nonzero. Put f ∗ (X, Y ) = f (X)−f . X−Y ∗ By Lemma 21.9.1, no absolutely irreducible factor of f (X, Y ) belongs to K[X, Y ]. Choose a finite, necessarily cyclic, extension L such that all absolutely irreducible factors of f ∗ (X, Y ) belong to L[X, Y ]. Then every irreducible factor of f ∗ (X, Y ) over K is reducible over L. Let z be an indeterminate. Put E = K((z −1 )) and F = L((z −1 )). Then E/K is a regular extension and EL = F (Example 3.5.1). Hence, res: Gal(F/E) → Gal(L/K) is an isomorphism. By Lemma 21.9.3, f (X) − z is separable. Choose a finite Galois extension Fˆ of E which contains F and all roots of f (X) − z in Es . Now assume gcd(n, q − 1) > 1. Choose a common prime divisor r of n and q √ − 1. Then p 6= √ r and K contains all roots of 1 of order r. Put E 0 = E( r z) and F 0 = F ( r z). By Lemma 21.9.4, E 0 /E and F 0 /F are cyclic extensions of degree r. Therefore, both maps res: Gal(F 0 /F ) → Gal(E 0 /E) and res: Gal(F 0 /E 0 ) → Gal(F/E) are isomorphisms. In addition, Lemma 21.9.4 implies that E 0 , F 0 ⊆ Fˆ . We may therefore choose σ, τ ∈ Gal(Fˆ /E) such that hresF 0 σi = Gal(F 0 /E 0 ) and hresF 0 τ i = Gal(F 0 /F ). Put ρ = στ . Then resL ρ generates Gal(L/K). E(x)

F (x)

E0

F0

E

F

K

L

Fˆ

By Lemma 21.9.3, ρ fixes a root x of f (X) − z in Fˆ . By Lemma 21.9.4, E(x) contains a root of z of order r, so E 0 ⊆ E(x). Hence, resE 0 ρ = id. On the other hand, resE 0 ρ = resE 0 τ 6= id. This contradiction proves that gcd(n, q − 1) = 1. 

Exercises

493

Exercises 1. Let K be a perfect field and V a Zariski K-closed set with V (K) infinite. Use the decomposition-intersection procedure and Corollary 10.5.3 to prove that V contains a curve defined over K. 2. For each positive integer i, give an example of a PAC field K which is Ci+1 but not Ci . Hint: Take K to be imperfect, follow Example 21.2.8 and use Propositions 21.2.4 and 21.2.12. 3. Let f ∈ Q[X0 , . . . , Xl ] be an irreducible form of prime degree l. Prove that V (f ) contains a variety defined over Q. Hint: The intersection of l hyperplanes in Pl is nonempty. ˜ 4. Prove that almost all σ ∈ Gal(Q) satisfy this: Let f ∈ Q(σ)[X 0 , . . . , Xn ] ˜ be a polynomial of degree d ≤ n with a Q(σ)-zero. Then V (f ) contains a ˜ curve defined over Q(σ). Hint: Use Chevalley-Warning, the transfer theorem and Exercise 1. 5. Give an example of fields K0 and K such that K0 is algebraically closed ˜ ) where τ is the in K, K is C1 but K0 is not. Suggestion: Take K0 to be Q(τ conjugation on C. Then use Lemmas 20.5.3 and 21.3.5. 6. Show that Fp (t), with t a transcendental, is not C1 . Suggestion: Choose a nonsquare a ∈ Fp and prove that X 2 + aY 2 + tZ 2 has no nontrivial zero in Fp (t). 7. Let M and L be finite separable extensions of a global field K. Suppose that for almost all p ∈ P (K) the number of primes P ∈ P (L) lying over p and having a relative degree 1 is equal to the number of primes P ∈ P (M ) lying over p and having relative degree 1. Prove that [L : K] = [M : K] ˆ of L/K is equal to the Galois closure M ˆ of M/K and the Galois closure L [Kronecker]. Hint: Follow these steps. (a) Choose primitive elements x and y for L/K and M/K, respectively, integral over OK . Put f = irr(x, K) and g = irr(y, K). Use Lemma 21.5.2 ¯p to prove that for almost all p ∈ P (K) the numbers of roots of f and g in K are equal. (b) Let N be a finite Galois extension of K that contains both L and M . Use the transfer theorem (or alternatively the Chebotarev density theorem) to prove that every σ ∈ Gal(N/K) fixes the same numbers of roots of f and of g. ˆ and the elements τ ∈ (c) Apply (b) to the elements σ ∈ Gal(N/L) ˆ Gal(N/M ) and prove that deg(f ) = deg(g). ˆ all roots of g belong to N (σ), so M ˆ ⊆ L. ˆ (d) Observe: For σ ∈ Gal(N/L), 8. S LetxG be S a finitex group and H, I, N subgroups. Suppose G = H n N and I = = I n N . Deduce that H is not a proper x∈G x∈G H . Prove that GS subgroup of I. Hint: Observe that x∈G (IN )x = G.

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S 9. (a) Let A / B ≤ G be a tower of finite groups. Prove that | x∈G Ax | ≤ (G : B)|A|. (b) Let K ⊆ L be a finite separable extension of a global field. Observe: The Dirichlet S density δ of the set V (L/K) (Section 20.5) is equal to the Haar measure of σ∈Gal(K) Gal(L)σ . (c) Use (a), (b) and Lemma 21.5.3 to prove that if M is a finite Galois extension of L which is Kronecker conjugate to L over K, then [M : L] ≤ δ −1 [Klingen2]. 10. Exercise 1(c) of Chapter 13 gives a pair of polynomials f (X) = X 4 + 2X 2 , g(X) = −4X 4 − 4X 2 − 1 for which f (X) − g(Y ) is reducible. Show however, that f (X) − t and g(Y ) − t are not Kronecker conjugate over Q(t). Hint: Use Corollary 21.6.5. 11. Prove for nonzero integers a, b1 , . . . , bn that the following two statements are equivalent. (a) There exist ε1 , . . . , εn ∈ {0, 1} and c ∈ Z satisfying a = bε11 . . . bεnn c2 . (b) For almost all primes p, if b1 , . . . , bn are quadratic residues modulo p, then so is a. Hint: Combine the transfer theorem (or directly the Chebotarev density theorem) with Kummer theory for quadratic extensions over Q. 12. Let n be a positive integer, q a power of a prime number p, and a ∈ Fq . Suppose gcd(n, q − 1) > 1. Find distinct x, y ∈ Fq with Dn (a, x) = Dn (a, y). Combine this with Theorem 21.8.13 to supply a proof of the generalized Carlitz’s Conjecture (Theorem 21.9.5) in the case where p - deg(f ). Hint: Choose a common prime divisor l of n and q − 1. Then choose x = z + az −1 and y = ζz + a−1 ζz with ζ, z ∈ F1 , ζ l = 1, and ζ 6= 1. 13. Let p, l be distinct prime numbers, q a power of p, and a ∈ F× q . Suppose l|p2 − 1. Then Dl (a, X) permutes only finitely many fields Fqk . Hint: Use the hint of Exercise 12 to reduce to the case l|q + 1 and l 6= 2. Then (1)

(l−1)/2 Y Dl (a, X) − Dl (a, Y ) = (X 2 − αi XY + Y 2 + βi2 a), X −Y i=1

where αi = ζli + ζl−i and βi = ζli − ζl−i . [Schinzel2, p. 52]. Observe that ζlq = ζl−1 , so αi , βi2 ∈ Fq . Next note that each of the factors on the right hand side of (1) is absolutely irreducible. Therefore, it has Fqk -zeros off the diagonal if k is large. 14. Prove the following generalization of a theorem of MacCluer (See also [Fried6, Thm. 1]): Let K be a pseudo finite field, f ∈ K[X] a separable polynomial of degree n > 0, x an indeterminate. Put t = f (x). Suppose no (Y ) K-irreducible factor of f ∗ (X, Y ) = f (X)−f is absolutely irreducible. Then X−Y f permutes K.

Notes

495

Hint: Let x1 , . . . , xn be the n distinct roots of f (X) = t. Put F = ˆ be the algebraic closure of K in F , T = {τ ∈ K(x1 , . . . , xn ). Let K Gal(F/K(t) | P resKˆ τ = resKˆ σ}, and Ti = T ∩ Gal(F/K(xi )), i = 1, S n. . . , n. n Prove: |T | = i=1 |Ti | and Ti ∩ Tj = ∅ for i 6= j. Conclude that T = · i=1 Ti . ˆ ˜ ∪ {∞} with ψ(t) = a. Then Now consider a ∈ K. Find a K-place ψ: F → K choose σ in the decomposition group of ψ which belongs to T . There is an i with σxi = xi . Hence, b = ψ(xi ) is in K and satisfies f (b) = a. Consequently, f is surjective on K. Since K is pseudo finite, f is also injective.

Notes The decomposition-intersection procedure has been used by several authors. For example, [Greenleaf] uses it to prove that for each d > 0 almost all fields Qp are C2,d . Also, it appears in the first version [Fried-Sacerdote], of the stratification procedure. [Klingen4] gives a comprehensive survey on Kronecker classes of number fields. Problem 21.5.8 has a negative solution if the following statement holds for each infinite profinite group S G of rank ≤ ℵ0 : For each closed subgroup H of G of infinite index the set g∈G H g does not contain an open neighborhood of 1. There is an attempt in [Klingen2, Thm. 6] to give a negative answer to Problem 21.5.8. Unfortunately there is an error in the proof: A closed subgroup H of a profinite group G of countable rank may have uncountably many, rather that countably many (as claimed in [Klingen2]), conjugates in G. For example, there are uncountably many involutions in G(Q), all conjugate. Thus, Problem 21.5.8 is still open. Davenport’s original problem [Fried-Jarden3, Problem 19.26] asked if Vp (f ) = Vp (g) for nonconstants f, g ∈ Q[X] and almost all p implies f and g are linearly related. Remark 21.6.1 supplies counter examples to that problem. Davenport’s present problem 21.6.2 is a modification of the older one. Lemma 21.7.2 overlaps with a result attributed to Galois [Huppert, p. 163, Satz 3.6]. More on Dickson’s polynomials can be found in [Schinzel2, §1.4-1.5] and [Lidl-Mullen-Turnwald]. The original proof of Proposition 21.7.7 appears in [Schur2]. Our proof is an elaboration of [Lidl-Mullen-Turnwald, p. 126]. Likewise, the original proof of Proposition 21.7.8 appears in [Burnside1]. We have elaborated [LidlMullen-Turnwald p. 127]. [Schur1] proves that every polynomial f ∈ Z[X] of prime degree which is a permutation polynomial modulo infinitely many prime numbers p is a ˜ to Cyclic polycomposition of polynomials which are linearly related over Q nomials or Chebyshev polynomials. Schur conjectured that his result holds for an arbitrary number field K. [Fried1] uses the theory of Riemann surfaces to prove Schur’s conjecture. [Turnwald, Thm. 2] uses algebraic methods

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to generalize Fried’s result to arbitrary global field (See also [Lidl-MullenTurnwald]). Our proof of Schur’s conjecture (Theorem 21.8.13) is based on a result of M¨ uller (Proposition 21.8.7). Our version of the proof of Cohen-Fried proof of the generalized Carlitz’s Conjecture (Theorem 21.9.5) uses improvements of Bensimhoun and Haran. Partially building on ideas of [Lenstra], [Guralnick-M¨ uller, §8] further generalizes the generalized Conjecture to nonconstant separable morphisms between smooth projective curves over Fq .

Chapter 22. Projective Groups and Frattini Covers Every profinite group is a Galois group of some Galois extension (Leptin, Proposition 2.6.12). However, not every profinite group is an absolute Galois group. For example, the only finite groups that appear as absolute Galois groups are the trivial group and Z/2Z (Artin [Lang7, p. 299, Cor. 9.3]). This raises the question of characterizing the absolute Galois groups among all profinite groups. This question is still wide open. A more restrictive question finds a complete solution in projective groups: By Theorem 11.6.2, the absolute Galois group of a PAC field is projective. Conversely, for each projective group G there exists a PAC field K such that Gal(K) ∼ = G (Theorem 23.1.2). Thus, a profinite group G is projective if and only if it is isomorphic to the absolute Galois group of a PAC field. In this chapter we define a projective group as a profinite group G for which every embedding problem is weakly solvable. By Gruenberg’s theorem (Lemma 22.3.2), it suffices to weakly solve only finite embedding problems. This leads to the second characterization of projective groups as those profinite groups which are isomorphic to closed subgroups of free profinite groups (Corollary 22.4.6). Projective groups also appear as the universal Frattini covers of profinite groups (Proposition 22.6.1). Thus, as a preparation for decidability and undecidability results about families of PAC fields, we introduce in this chapter the basis properties of Frattini covers.

22.1 The Frattini Group of a Profinite Group Consider a profinite group G. Denote the intersection of all maximal open subgroups of G by Φ(G). Here we call a subgroup M maximal if there is no subgroup M 0 of G such that M < M 0 < G. The characteristic (and therefore normal) closed subgroup Φ(G) is called the Frattini group of G. We characterize the elements of Φ(G) as “dispensable generators” of G: Lemma 22.1.1: Let G be a profinite group. An element g ∈ G belongs to Φ(G) if and only if there is no proper closed subgroup H of G for which hH, gi = G. Also, if H is a closed subgroup for which H · Φ(G) = G, then H = G. Proof: Let g ∈ Φ(G) and let H be a closed subgroup of G for which hH, gi = G. If H 6= G, then H is contained in a maximal open subgroup M of G. Therefore, G = hH, gi ≤ M < G, a contradiction. Conversely, suppose an element g ∈ G satisfies the above condition. Let M be a maximal open subgroup of G. If g 6∈ M , then G = hM, gi.

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Hence, M = G, a contradiction. Therefore, g ∈ M , and since M is arbitrary, g ∈ Φ(G). Finally, let H be a closed subgroup of G such that H · Φ(G) = G. If H < G, then H is contained in a maximal open subgroup M of G and M = M · Φ(G) = G, a contradiction. Thus, H = G.  Lemma 22.1.2: The Frattini group of a profinite group G is pronilpotent. Proof: Assume first G is finite. Let P be a p-Sylow group of Φ(G). For g ∈ G the group P g is also a p-Sylow subgroup of Φ(G). Hence, there exists a ∈ Φ(G) such that P g = P a . Therefore, ga−1 is in the normalizer, NG (P ), of P in G. Thus, G = NG (P )Φ(G). By Lemma 22.1.1, G = NG (P ). Hence, P is normal in G, and therefore in Φ(G). It follows that Φ(G) is the direct product of its p-Sylow groups. That is, G is nilpotent [Huppert, p. 260]. In the general case observe that for any open normal subgroup N of G, Φ(G)/N ∩ Φ(G) ∼ = N Φ(G)/N ≤ Φ(G/N ). Since Φ(G/N ) is a nilpotent finite group, so is Φ(G)/N ∩ Φ(G). Therefore, Φ(G) = lim Φ(G)/N ∩ Φ(G) is a ←− pronilpotent group.  Lemma 22.1.3: Let A be a minimal normal subgroup of a finite group G. If A ≤ Φ(G), then A is p-elementary Abelian for some prime number p. Proof: By Lemma 22.1.2, A is a nilpotent group. Hence, the center Z(A) of A is nontrivial [Huppert, p. 260] and normal in G. Thus, Z(A) = A and A is Abelian. As such, A is a direct product of its Sylow subgroups and each of them is normal in G. Therefore, A is an Abelian p-group for some prime number p. The subgroup of A consisting of all elements x with xp = 1 is  normal in G. Consequently, A = (Z/pZ)m for some m ≥ 0. As an operation, taking the Frattini subgroup of a group has functorial properties: Lemma 22.1.4: (a) Let θ: G → H be an epimorphism of profinite groups. Then θ(Φ(G)) ≤ Φ(H). If Ker(θ) ≤ Φ(G), then θ(Φ(G)) = Φ(H). (b) Suppose U is a closed subgroup of G and N is a closed subgroup of Φ(U ), normal in G. Then N ≤ Φ(G). (c) Let N beQ a closed normal subgroup of G. Then Φ(N ) ≤ Φ(G). (d) Q Let G = i∈I Gi be a direct product of profinite groups. Then Φ(G) = i∈I Φ(Gi ). Proof of (a): The inverse image of each maximal open subgroup of H is a maximal open subgroup of G. Hence, θ(Φ(G)) ≤ Φ(H). Proof of (b): Assume N 6≤ Φ(G). Then G has a maximal open subgroup M with N 6≤ M . Thus, M N = G. Hence, U = U ∩ M N = (U ∩ M )N . Since N ≤ Φ(U ), Lemma 22.1.1 implies U = U ∩ M . Therefore, N ≤ M , contrary to the assumption.

22.2 Cartesian Squares

499

Proof of (c): Apply (b) to N and Φ(N ) instead of to U and N . Proof of (d): Each Q Gi is a closed normal subgroup of G. Therefore, (c) gives Φ(Gi ) ≤ Φ(G) and i∈I Φ(Gi ) ≤ Φ(G). Q Conversely, let M be a maximal open subgroup ofQ Gj . Then M × i6=j Gi is a maximal open subgroup of G. Thus Q Φ(G) ≤ M × i6=j Gi . Running over ) × all M , this gives Φ(G) ≤ Φ(G j i6=j Gi . Since this is true for each j ∈ I, Q we conclude that Φ(G) ≤ i∈I Φ(Gi ).  Remark 22.1.5: The map θ: Φ(G) → Φ(H) in Lemma 22.1.4(a) need not be surjective. Let G = hbi n hai be the semidirect product of a cyclic group hai of order 5 with a cyclic group H = hbi of order 4 with the relation ab = a2 . Let θ: G → H be the quotient map. Both hai and hbi are maximal subgroups of G, so Φ(G) = 1. On the other hand, H has order 4, so Φ(H) is of order 2. Thus, θ(Φ(G)) < Φ(H). 

22.2 Cartesian Squares The usual direct product of group theory has a useful generalization: Let α: B → A and γ: C → A be homomorphisms of profinite groups. Let B ×A C = {(b, c) ∈ B × C | α(b) = γ(c)}. Since A is Hausdorff, B ×A C is a closed subgroup of B × C. It is therefore a profinite group called the fiber product of B and C over A. There are natural projection maps prB : B ×A C → B and prC : B ×A C → C defined by prB (b, c) = b and prC (b, c) = c, respectively. It is possible to change A, B, and C such that the projection maps will be surjective: Put A0 = {a ∈ A | ∃(b, c) ∈ B × C: α(b) = γ(c) = a}, B0 = α−1 (A0 ), and C0 = γ −1 (A0 ). Then A0 (resp. B0 , C0 ) is a closed subgroup of A (resp. B, C), B0 ×A0 C0 = B ×A C, and the projections maps of B0 ×A0 C0 are surjective. Proposition 22.2.1: Consider a commutative diagram of profinite groups (1)

D

δ

γ

β

 B

/C

α

 /A

where β, α, γ, and δ are homomorphisms. The following statements are equivalent: (a) There exists an isomorphism θ: D → B ×A C with β ◦ θ−1 = prB and δ ◦ θ−1 = prC . (b) Let G be a profinite group and ϕ: G → B and ψ: G → C homomorphisms. Suppose α◦ϕ = γ◦ψ. Then there exists a unique homomorphism

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Chapter 22. Projective Groups and Frattini Covers

π: G → D which makes the following diagram commutative: (2)

G0 @PPP 00@@ PPP ψ 00 @@@πPPPP PPP 00 @ δ P'/ 0 C ϕ 0 D 00 00 β γ 0   B α /A

Proof of “(a)=⇒(b)”: Without loss assume D = B ×A C, β = prB , and δ = prC . Define a map π: G → D by π(g) = (ϕ(g), ψ(g)). Then π is the unique homomorphism which makes (2) commutative. Proof of “(b)=⇒(a)”: Let G = B ×A C, ϕ = prB , and ψ = prC . Then (b) gives a unique π that makes (2) commutative. Define θ: D → B ×A C by θ(d) = (β(d), δ(d)) for d ∈ D. Apply the uniqueness property to maps from D to D and from G to G. It implies the map π ◦ θ (resp. θ ◦ π) is the identity, and θ satisfies (a).  Definition 22.2.2: A commutative diagram (1) satisfying the conditions of Proposition 22.2.1 is said to be a cartesian square.  Lemma 22.2.3: Let (1) be a cartesian square of homomorphisms of profinite groups. (a) If b ∈ B and c ∈ C satisfy α(b) = γ(c), then there exists a unique d ∈ D with β(d) = b and δ(d) = c. (b) If α (resp. γ) is surjective, then δ (resp. β) is surjective. Proof of (a): Assume without loss that D = B ×A C. If b ∈ B and c ∈ C satisfy α(b) = γ(c), then d = (b, c) is the unique element of D satisfying β(d) = b and δ(d) = c. Proof of (b): Consider c ∈ C. By assumption, there is a b ∈ B with α(b) = γ(c). By (a), there is a d ∈ D with δ(d) = c. Thus, δ is surjective.  Lemma 22.2.4: Let (1) be a commutative square of epimorphisms of profinite groups. Then (1) is cartesian if and only if Ker(α ◦β) = Ker(β)×Ker(δ). Proof: First suppose (1) is a cartesian square. Clearly Ker(β) · Ker(δ) ≤ Ker(α ◦ β). Assume without loss, D = B ×A C, β = prB , and δ = prC . Let (b, c) ∈ Ker(α ◦ β). Then α(b) = 1 = γ(c). Hence, (1, c) and (b, 1) belong to D. Moreover, (b, c) = (1, c)·(b, 1) ∈ Ker(β)·Ker(δ). Finally, Ker(β)∩Ker(δ) = 1. Consequently, Ker(α ◦ β) = Ker(β) × Ker(δ). Now suppose Ker(α ◦ β) = Ker(β) × Ker(δ). Define a homomorphism θ: D → B ×A C by θ(d) = (β(d), δ(d)). If θ(d) = (1, 1), then d ∈ Ker(β) ∩ Ker(δ) = 1. Thus, θ is injective.

22.2 Cartesian Squares

501

To prove that θ is surjective, consider (b, c) ∈ B ×A C. Then α(b) = γ(c). There exist d1 , d2 ∈ D with β(d1 ) = b and δ(d2 ) = c. Then α(β(d1 d−1 2 )) = −1 = d α(β(d1 )) · γ(δ(d2 ))−1 = α(b)γ(c)−1 = 1. By assumption, d1 d−1 2 3 d4 for some d3 ∈ Ker(β) and d4 ∈ Ker(δ). Let d = d3 d1 = d4 d2 . Then β(d) = b and δ(d) = c. Therefore, θ(d) = (b, c).  Lemma 22.2.5: Let (1) be a cartesian square of epimorphisms of profinite groups. Then β maps Ker(δ) isomorphically onto Ker(α). Proof: Apply Lemma 22.2.4: Ker(α) = β(Ker(α ◦ β)) = β(Ker(β) × Ker(δ)) = β(Ker(δ)). In addition, Ker(δ) ∩ Ker(β) = 1. Hence, β: Ker(δ) → Ker(α) is an isomorphism.  Lemma 22.2.6: Let (1) and (2) be commutative diagrams of homomorphisms of profinite groups. Suppose (1) is cartesian and all maps except possibly π are surjective. Then: (a) If π is surjective, then Ker(α ◦ ϕ) = Ker(ϕ)Ker(ψ). (b) If Ker(α ◦ ϕ) ≤ Ker(ϕ)Ker(ψ), then π is surjective. Proof of (a): The condition α◦ϕ = γ ◦ψ implies Ker(ϕ)Ker(ψ) ≤ Ker(α◦ϕ). Now suppose π is surjective. Let g ∈ Ker(α ◦ ϕ). Put b = ϕ(g). Then α(b) = 1. Hence, there is a d ∈ D with β(d) = b and δ(d) = 1. Choose an h ∈ G with π(h) = d. Then ψ(h) = δ(π(h)) = δ(d) = 1 and ϕ(gh−1 ) = bβ(π(h))−1 = bβ(d)−1 = 1. Therefore, g = gh−1 · h ∈ Ker(ϕ)Ker(ψ). Proof of (b): Suppose Ker(α ◦ ϕ) ≤ Ker(ϕ)Ker(ψ). Consider d ∈ D. Put b = β(d) and c = δ(d). Choose g, h ∈ G with ϕ(g) = b and ψ(h) = c. Then α(ϕ(gh−1 )) = α(ϕ(g))γ(ψ(h))−1 = α(b)γ(c)−1 = α(β(d))γ(δ(d))−1 = 1. Thus, there are g1 ∈ Ker(ϕ) and h1 ∈ Ker(ψ) with gh−1 = g1 h1 . The element g1−1 g = h1 h of G satisfies ϕ(g1−1 g) = b and ψ(h1 h) = c. Therefore, by Lemma 22.2.3(a), π(g1−1 g) = d.  Example 22.2.7: Fiber products. Here are four examples where fiber products naturally appear. The verification that the occurring commutative squares are cartesian follows from Lemma 22.2.4: (a) Let M and M 0 be Galois extensions of a field K. Put L = M ∩ M 0 and N = M M 0 . Then Gal(N/K) is the fiber product of Gal(M/K) and Gal(M 0 /K) over Gal(L/K) with respect to the restriction maps. (b) Similarly, let G be a profinite group and K, L, M, N closed normal subgroups. Suppose K ∩L = N and KL = M . Then G/N = G/K ×G/M G/L with respect to the quotient maps. (c) Let ϕ: B → A be an epimorphism of profinite groups, C = Ker(ϕ), ¯ = B/B0 , and B0 a closed normal subgroup of B. Suppose B0 ∩C = 1. Put B ¯ ¯ ¯ A0 = ϕ(B0 ), A = A/A0 , α: A → A and β: B → B the canonical maps, and ¯ → A¯ the map induced from ϕ. Then B ∼ ¯ ×A¯ A. ϕ: ¯ B =B

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Indeed, if b ∈ B satisfies α(ϕ(b)) = 1, then ϕ(b) ∈ A0 . Hence, there exists b0 ∈ B0 with ϕ(b0 ) = ϕ(b). Therefore, b = b0 · b−1 0 b ∈ Ker(β)Ker(ϕ). ¯ A. It follows from Lemma 22.2.4 that B ∼ B × = ¯ A Suppose now B0 ∩ C is not necessarily trivial. Then by the preceding ¯ ×A¯ A. Hence, for each a ∈ A and b ∈ B satisfying paragraph, B/B0 ∩ C ∼ =B ϕ(β(b)) ¯ = α(a) there is a b0 ∈ B (not necessarily unique) with β(b0 ) = β(b) and ϕ(b0 ) = a. (d) Condition (1b) of Section 17.3 on a family C of finite groups is equivalent to “C is closed under fiber products”. Thus, a formation is a family of finite groups which is closed under taking quotients and fiber products.  The homomorphism π of diagram (2) need not be surjective even if all other maps are surjective. Here is a condition for this to happen: Lemma 22.2.8: Let ϕ: G → B and ψ: G → C be epimorphisms of profinite groups. Then there is a commutative diagram (2), unique up to an isomorphism such that (1) is a cartesian square with β, α, γ, δ, π epimorphisms. Proof: Let M = Ker(ϕ)Ker(ψ), N = Ker(ϕ) ∩ Ker(ψ), A = G/M , D = G/N , and π: G → G/N the quotient map. Now find β, α, γ, δ that make (1) cartesian and (2) commutative.  The following lemma gives a useful criterion for δ in the cartesian diagram (1) to have a group theoretic section: Lemma 22.2.9: Let (1) be a cartesian diagram of homomorphisms of profinite groups. Suppose there exists a homomorphism ϕ: C → B with α◦ϕ = γ. Then there exists a monomorphism δ 0 : C → D such that δ ◦ δ 0 = idC . Proof: Let ψ = idC . Then α ◦ ϕ = γ ◦ ψ. Hence, by Proposition 22.2.1(b), there exists a homomorphism δ 0 : C → D such that δ◦δ 0 = idC (and β◦δ 0 = ϕ), as claimed. Note that the existence of ϕ implies that δ is surjective. 

22.3 On C-Projective Groups Embedding problems and projective groups have already appeared, for example, in Sections 16.5 and 17.7. Now we consider the subject in detail: Definition 22.3.1: Embedding problems. An embedding problem for a profinite group G is a pair (1)

(ϕ: G → A, α: B → A)

in which ϕ and α are epimorphisms of profinite groups. We call Ker(α) the kernel of the problem. We call the problem finite if B is finite. We say (1) splits if α has a group theoretic section. That is, there is a homomorphism α0 : A → B with α ◦ α0 = idA .

22.3 On C -Projective Groups

503

Embedding problem (1) is said to be solvable (resp. weakly solvable) if there exists an epimorphism (resp. homomorphism) γ: G → B with α ◦ γ = ϕ. The map γ is a solution (resp. weak solution) to (1). Suppose C is a Melnikov formation of finite groups (Section 17.3). Let G be a pro-C group. Then call (1) a C-embedding problem (resp. pro-C embedding problem), if B is a C-group (resp. pro-C). Call G C-projective if every pro-C embedding problem (1) for G is weakly solvable.  Lemma 22.3.2 (Gruenberg): Let C be a Melnikov formation of finite groups and G be a pro-C group. Suppose every finite C-embedding problem for G is weakly solvable. Then G is C-projective. Proof: Consider embedding problem (1) for G with B a pro-C group. Let C = Ker(α). The transition from the finite case to the general case in Part C of the proof forces us to consider pairs (1) in which ϕ is a homomorphism which is not necessarily surjective. Part A: Suppose ϕ: G → A is a homomorphism, B ∈ C, and α: B → A is an epimorphism. Then there is a homomorphism β: G → B with α ◦ β = ϕ. Indeed, B and A0 = ϕ(G) are in C. Hence, so is C and therefore also B0 = α−1 (A0 ). By assumption, there is a homomorphism β: G → B0 with α ◦ β = ϕ. Part B: C is finite. Choose an open normal subgroup U of B with C ∩U = 1. Since B/U ∈ C, Part A gives a homomorphism β: G → B/U for which

1

/C

1

/C

G   ϕ     α /B /A     β    α¯  / B/U / A/α(U )

/1 /1

is a commutative diagram. The right square is cartesian (Example 22.2.7(c)). Therefore (Proposition 22.2.1(b)), there is a homomorphism γ: G → B with α ◦ γ = ϕ. Part C: The general case. Apply Zorn’s lemma to Part B. Let Λ be the set of pairs (L, λ) where L is a closed normal subgroup of B contained in C and λ: G → B/L is a homomorphism which makes the following diagram commutative: (2)

1

/ C/L

G { { { λ { ϕ {{ }{{  /A / B/L αL

/ 1.

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Chapter 22. Projective Groups and Frattini Covers

−1 Here αL is the epimorphism induced by α. The pair (C, αC ◦ ϕ) belongs to 0 0 Λ, so Λ is nonempty. Partially order Λ by (L , λ ) ≤ (L, λ) if L0 ≤ L and the following triangle is commutative:

G vv

λ0 vvv

B/L0

vv {vv

λ

 / B/L.

Suppose Λ0 = {(Li , λi ) | i ∈ I} is a descending chain in Λ. Then lim B/Li = ←− T B/L with L = i∈I Li . The λi ’s define a homomorphism λ: G → B/L with (2) commutative. Therefore, (L, λ) is a lower bound for Λ0 . By Zorn’s Lemma, Λ has a minimal element (L, λ). It suffices to prove that L = 1. Assume L 6= 1. Then B has an open normal subgroup N with L 6≤ N . Therefore, L0 = N ∩ L is a proper open subgroup of L which is normal in B. Part B gives a homomorphism λ0 : G → B/L0 which makes the following diagram commutative: (3)

G ww w λ ww λ ww {ww  / B/L0 / B/L 0

1

/ L/L0

/1

Therefore, (L0 , λ0 ) is an element of Λ less than (L, λ), a contradiction.



Remark 22.3.3: A variant of Lemma 22.3.2. Consider an arbitrary embedding problem (1) for a profinite group G. Put C = Ker(α). Suppose for all L, L0 ≤ C such that L, L0 / B and L0 open in L and for each homomorphism λ: G → B/L there is a homomorphism λ0 : G → B/L0 making (3) commutative. The proof of Lemma 22.3.2 proves that (1) is weakly solvable.  Lemma 22.3.4: Let C be a Melnikov formation of finite groups and G a pro-C group. Suppose every C-embedding problem (1) in which Ker(α) is a minimal normal subgroup of B is weakly solvable. Then G is C-projective. Proof: By Lemma 22.3.2, it suffices to weakly solve each C-embedding problem (1). Put C = Ker(α). If C is minimal normal in B, then (1) is solvable by assumption. Suppose C is not minimal normal in B and proceed by induction on |C|. By assumption, there is a nontrivial normal subgroup L of B which is

22.3 On C -Projective Groups

505

properly contained in C. Consider the diagram L

G

 /B

α

 /A

/1

α ¯

/A

/1

ϕ

1

/C

1

 / C/L

π

 / B/L

in which α ¯ is induced by α. Since |C/L| < |C|, the induction hypothesis gives a homomorphism β: G → B/L with α ¯ ◦ β = ϕ. Also |L| < |C|. Using the induction hypothesis again, there is a homomorphism γ: G → B with π ◦ γ = β. Therefore, α ◦ γ = ϕ.  Proposition 22.3.5: Let C be a full formation of finite groups and G a pro-C group. Suppose each C-embedding problem (1) for G, where Ker(α) is a minimal Abelian p-elementary normal subgroup of B, is weakly solvable. Then G is C-projective. Proof: Let C = Ker(α). By Lemma 22.3.4, it suffices to prove that every embedding problem (1) for G, where B ∈ C and C is a minimal normal subgroup of B, is weakly solvable. We distinguish between two cases: Case A: Suppose C 6≤ Φ(B). Then B has a maximal subgroup B1 with C 6≤ B1 . Let α1 be the restriction of α to B1 . Then B1 C = B and α1 (B1 ) = α(CB1 ) = α(B) = A. Consider the commutative diagram G ϕ

1

/C O

/B O

α

 /A

1

/ C1

/ B1

α1

/A

/1 /1

with C1 = C ∩ B1 . Since C is full, B1 ∈ C. An induction hypothesis on |B| gives a homomorphism γ: G → B1 with α1 ◦ γ = ϕ, hence α ◦ γ = ϕ. Case B: C ≤ Φ(B). By Lemma 22.1.3, C is a minimal Abelian p-elementary normal subgroup of B. Hence, by assumption, (1) is solvable.  The following lemma gives the basic example for projective C-groups: Lemma 22.3.6: Let C be a Melnikov formation of finite groups. Then each free pro-C group F is C-projective. Proof: Choose a basis X for F . Consider a C-embedding problem (ϕ: F → A, α: B → A). Choose x1 , . . . , xn ∈ X such that ϕ(x1 ), . . . , ϕ(xn ) generate

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Chapter 22. Projective Groups and Frattini Covers

A. Then choose b1 , . . . , bn ∈ B with α(bi ) = ϕ(xi ), i = 1, . . . , n. Map X into B by xi 7→ bi , i = 1, . . . , n and x 7→ 1 for x ∈ X r{x1 , . . . , xn }. This map extends to a homomorphism γ: F → B with α ◦ γ = ϕ. Consequently, F is C-projective. 

22.4 Projective Groups A profinite group G is projective if every embedding problem for G is weakly solvable (i.e. G is projective with respect to the formation of all finite groups). Lemma 22.4.1: If C is a full formation of finite groups and G is C-projective, then G is projective. Proof: By Proposition 22.3.5, it suffices to prove that every finite embedding problem (ϕ: G → A, α: B → A)

(1)

with Ker(α) = (Z/pZ)m and p a prime number is weakly solvable. If p - |A|, then Schur-Zassenhaus [Huppert, p. 126] gives a homomorphism α0 : A → B with α ◦ α0 = idA . Thus, α0 ◦ ϕ is a weak solution of (1). Now suppose p divides |A|. Then A contains an isomorphic copy of Z/pZ (Cauchy). Since A is in C, so is Z/pZ. Hence, (Z/pZ)m ∈ C. Therefore, B ∈ C. By assumption, (1) has a weak solution.  Remark 22.4.2: Suppose G is projective and π: F → G is an epimorphism of profinite groups. Then there exists an embedding π 0 : G → F such that π ◦ π 0 is the identity map (Take A = G, B = F , α = π, and ϕ = idG in (1).) Therefore, π 0 is injective, so G is isomorphic to a closed subgroup of F . Corollary 22.4.3 below uses this observation to give a criterion for projectivity.  Corollary 22.4.3: A profinite group G is projective if and only if every α short exact sequence 1 → C → B −→ G → 1 in which C is an Abelian p-elementary group splits. Proof: By Remark 22.4.2, we have only to prove sufficiency. By Lemma 22.3.5 it suffices to give a weak solution to each finite embedding problem (ϕ: G → A, α: B → A) with Ker(α) = (Z/pZ)m for some prime number p and a positive integer m. Complete the embedding problem to a commutative diagram 1

/ (Z/pZ)m

/ B ×A G

/G

/ (Z/pZ)m

 /B

 /A

/1

ϕ

1

α

/1

.

22.4 Projective Groups

507

By assumption, the upper row splits. Hence, there is a homomorphism γ: G → B with α ◦ γ = ϕ.  Remark 22.4.4: Cohomological interpretation of projectivity. In cohomological terms Corollary 22.4.3 signifies that a profinite group G is projective if and only if its cohomological dimension is bounded by 1 [Ribes, p. 211].  Corollary 22.4.5: Let C be a full formation of finite groups and F a free pro-C-group. Then F is projective. Proof: By Lemma 22.3.6, F is C-projective. Hence, by Lemma 2¸2.4.1, F is projective.  Corollary 22.4.6: A profinite group G is projective if and only if it is isomorphic to a closed subgroup of a free profinite group. Proof: By Corollary 17.4.8, G is a quotient of a free profinite group F . So, by Remark 22.4.2, G is isomorphic to a closed subgroup of F . Conversely, let H be a closed subgroup of a free profinite group F . Let ϕ: H → A and α: B → A be epimorphisms, with B finite. Lemma 1.2.5(c) gives an open subgroup H 0 of F containing H and an epimorphism ϕ0 : H 0 → A extending ϕ. By Proposition 17.6.2, H 0 is free. Hence, by Corollary 22.4.5, H 0 is projective. Therefore, there exists a homomorphism γ 0 : H 0 → B with α ◦ γ 0 = ϕ0 . Denote the restriction of γ 0 to H by γ. Then α ◦ γ = ϕ. It follows that H is projective.  Proposition 22.4.7: Let G be a projective group and H a closed subgroup. Then H is projective. Moreover, H is either trivial or infinite. In particular, G is torsion free. Proof: By Corollary 22.4.6, G is isomorphic to a closed subgroup of a free profinite group F . Hence, H is also isomorphic to a closed subgroup of F . A second application of Corollary 22.4.6 now proves that H is projective. Assume G has a nontrivial finite subgroup H. By Cauchy’s theorem, H contains an isomorphic copy of Z/pZ for some p. By the preceding paragraph, Z/pZ is projective. It follows from Corollary 22.4.3 that the natural map Z/p2 Z → Z/pZ has a group theoretic section. Hence, Z/p2 Z ∼ = Z/pZ × Z/pZ. This is a contradiction.  Proposition 22.4.8: Let G be a projective group and C a full formation of finite groups. Then the maximal pro-C quotient of G is projective. In particular, for each p the maximal pro-p quotient of G is projective. ¯ be the quotient map of G onto its maximal proProof: Let π: G → G ¯ → C quotient (Definition 17.3.2). Consider a C-embedding problem (ϕ: G ¯ A, α: B → A) for G. Since G is projective, there is a homomorphism γ: G → B such that α ◦ γ = ϕ ◦ π. Since B ∈ C and C is full, γ(G) ∈ C. Hence, γ ¯ → B; that is γ¯ ◦ π = γ. It follows factors through a homomorphism γ¯ : G ¯ that α ◦ γ¯ = ϕ. By Lemma 22.4.1, G is projective. 

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Chapter 22. Projective Groups and Frattini Covers

Lemma 22.4.9: For profinite groups G and H there exists a unique profinite group G ∗ H with the following properties: (a) G and H are closed subgroups of G ∗ H, G ∩ H = 1, hG, Hi = G ∗ H. (b) Each pair α: G → C and β: H → C of homomorphisms of profinite groups uniquely extends to a homomorphism γ: G ∗ H → C. We call G ∗ H the free product of G and H. Proof: Let F be the free product of G and H in the category of abstract groups. It is the unique abstract group that contains G and H such that G ∩ H = 1, hG, Hi = F , and every pair α: G → C and β: H → C of homomorphisms of abstract groups uniquely extends to a homomorphism γ: F → C [Kurosh, §33]. Denote the collection of normal subgroups N of F of finite index such that N ∩ G and N ∩ H are, respectively, open in G and H, by N . Then N is directed (Section 17.2). Apply Lemma 17.2.1 to the profinite completion Fˆ of F with respect to N . For an open normal subgroup N0 of G the maps G → G/N0 and H → 1 extend to a homomorphism F → G/N0 whose kernel N satisfies N ∩ G = N0 and N ∩ H = H, so N ∈ N . It follows that the canonical map g 7→ (gN )N ∈N embeds G into Fˆ . Similarly H naturally embeds into Fˆ . Since G and H generate F , they also generate Fˆ . Let α: G → C and β: H → C be homomorphisms of profinite groups. Extend them to a homomorphism γ: F → C of abstract groups. Then γ −1 (C0 ) ∈ N for each open normal subgroup C0 of C. By Lemma 17.2.2, γ uniquely defines a homomorphism γˆ : Fˆ → C of profinite groups which extends both α and β. In particular, the maps idG : G → G and H → 1 extend to a homomorphism Fˆ → G. Hence, G ∩ H = 1.  Proposition 22.4.10: The free product G ∗ H of projective groups is a projective group. Proof: Let (ϕ: G ∗ H → A, α: B → A) be an embedding problem for G ∗ H. Then there are homomorphisms γG : G → B and γH : H → B with α ◦ γG = ϕ|G and α ◦ γH = ϕ|H . The extension γ: G ∗ H → B of γG and γH is a weak solution of the embedding problem. 

22.5 Frattini Covers Frattini covers allow us to form profinite groups that inherit properties from one of their finite quotients: Definition 22.5.1: Frattini covers. A homomorphism ϕ: H → G of profinite groups is called a Frattini cover if it satisfies one, hence all, of the following equivalent conditions (Lemma 22.1.1): (1a) ϕ is surjective and Ker(ϕ) ≤ Φ(H). (1b) A closed subgroup H0 of H is equal to H if and only if ϕ(H0 ) = G. (1c) A subset S of H generates H if and only if ϕ(S) generates G.

22.5 Frattini Covers

509

In particular, ϕ maps the set of all open maximal open subgroups of H onto the set of all maximal open subgroups of G, so ϕ(Φ(H)) = Φ(G).  Lemma 22.5.2: Let G be a profinite group and X a subset of G which contains 1 and converges to 1. Then the following holds: (a) X is closed. (b) Every closed subset X0 of X which does not contain 1 is finite. (c) Let ϕ: H → G be an epimorphism of profinite groups. Then H has a closed subset X 0 which contains 1 and converges to 1 such that ϕ maps X 0 homeomorphically onto X. Proof of (a): Let g ∈ G r X. Then g 6= 1. Hence, G has an open normal subgroup N0 with g ∈ / N0 , so xN0 6= gN0 for each x ∈ X ∩ N0 . In addition, / X r N0 . Therefore, there exists an open normal X r N0 is finite and g ∈ subgroup N1 of G with N1 ≤ N0 and xN1 6= gN1 for each x ∈ X r N1 . By the above, xN1 6= gN1 for all x ∈ X ∩ N1 . Thus, the set gN1 is an open neighborhood of g which is disjoint from X. Consequently, X is closed. Proof of (b): By (a), G r X0 is an open neighborhood of 1. Hence, it contains an open normal subgroup N . By assumption, X r N is finite. Therefore, X0 is finite. Proof of (c): By Lemma 1.2.7, there exists a continuous map ϕ0 : G → H such that ϕ ◦ ϕ0 = idG . Put G0 = ϕ0 (G) and X 0 = ϕ0 (X). Then G0 and X 0 are closed subsets of H and ϕ maps G0 (resp. X 0 ) homeomorphically onto G (resp. X). Now consider an open normal subgroup N of H. Then X 0 r N is a closed subset of X 0 which does not contains 1. Hence, ϕ(X 0 r N ) is a closed subset of X which does not contain 1. By (b), ϕ(X 0 r N ) is finite. Hence, X 0 r N is finite. Consequently, X 0 converges to 1. Alternatively, we could use Exercise 1 of Chapter 17.  Corollary 22.5.3: If ϕ: H → G is a Frattini cover, then rank(H) = rank(G). Proof: Choose a system of generators X for G in the following way: If rank(G) < ∞, then |X| = rank(G). If rank(G) = ∞, then X converges to 1 (Proposition 17.1.1) and 1 ∈ X. In the former case choose a subset X 0 of H which ϕ maps bijectively onto X. In the latter case Lemma 22.5.2 gives a subset X 0 of H which converges to 1 and which ϕ maps homeomorphically onto X. By Definition 22.5.1(1c), X 0 generates H. Therefore, in both cases,  rank(H) = |X 0 | = |X| = rank(G). The following rules follow directly from Definition 22.5.1: ψ

ϕ

Lemma 22.5.4: Let H −→ G −→ A be homomorphisms of profinite groups. (a) If ϕ and ψ are Frattini covers, then ϕ ◦ ψ is a Frattini cover. (b) If ϕ is a Frattini cover and ϕ ◦ ψ is surjective, then ψ is surjective.

510

Chapter 22. Projective Groups and Frattini Covers

(c) If ϕ ◦ ψ is a Frattini cover and ψ is surjective, then both ϕ and ψ are Frattini covers. (d) If both ϕ and ϕ ◦ ψ are Frattini covers, then so is ψ. Lemma 22.5.5: Consider a cartesian square of epimorphisms (Definition 22.2.2): D

δ

γ

β

 B

/C

α

 /A

If δ is a Frattini cover, then so is α. Proof: By Lemma 22.2.5, β(Ker(δ)) = Ker(α). Since δ is a Frattini cover, Ker(δ) ≤ Φ(D). By Lemma 22.1.4(a), β(Φ(D)) ≤ Φ(B). Hence, Ker(α) ≤ Φ(B) and α is a Frattini cover.  For each profinite group epimorphism, restriction of the domain gives a Frattini cover: Lemma 22.5.6: Let ϕ: H → G be an epimorphism of profinite groups. Then H has a closed subgroup H 0 such that ϕ|H 0 : H 0 → G is a Frattini cover. Proof: Let {Hi | i ∈ I} be a decreasing chain of closed T subgroups of H with ϕ(Hi ) = G for each i ∈ I. By Lemma 1.2.2(c), ϕ( i∈I Hi ) = G. Applying Zorn’s Lemma, we conclude that H has a minimal closed subgroup H0 with  ϕ(H0 ) = G. The restriction of ϕ to H0 is a Frattini cover of G. We combine the fiber product construction with Lemma 22.5.6: Lemma 22.5.7: Let α: B → A and γ: C → A be Frattini covers. Then there is a commutative diagram /C D@ @@ ϕ @@ β @@ γ   B α /A δ

in which β, δ, and ϕ are Frattini covers. Proof: Let D1 = B ×A C (Section 22.2), α1 = prB , δ1 = prC , and ϕ1 = γ ◦ δ1 = α ◦ β1 . This gives a cartesian square, but the maps in it may not be Frattini covers. Apply Lemma 22.5.6 to find a closed subgroup D of D1 such that ϕ = ϕ1 |D is a Frattini cover. Lemma 22.5.4(d) implies that β = β1 |D and δ = δ1 |D are Frattini covers.  Let (2)

(ϕ: G → A, α: B → A)

22.5 Frattini Covers

511

be an embedding problem for a profinite group. Call (2) a Frattini embedding problem if α is a Frattini cover. The following proposition shows that in order to solve an arbitrary embedding problem for G, it suffices to solve Frattini embedding problem followed by a split embedding problem: Proposition 22.5.8: Let (2) be an embedding problem for a profinite group G. Suppose the following two conditions are satisfied: (a) Every Frattini embedding problem (ϕ: G → A, α0 : B0 → A) where B0 is a subgroup of B is solvable. (b) Every split embedding problem (γ0 : G → B0 , α0 : B 0 → B0 ) for G with Ker(α0 ) ∼ = Ker(α) is solvable. Then (2) is solvable. Proof: Choose a closed subgroup B0 of B such that α0 = α|B0 : B0 → A is a Frattini cover (Lemma 22.5.6). Thus, (3)

(ϕ: G → A, α0 : B0 → A)

is a Frattini embedding problem. By (a), there is an epimorphism γ0 : G → B0 with α0 ◦ γ0 = ϕ. Put C = Ker(α) and let B 0 = B0 nC with B0 acting on C by conjugation. Then π: B 0 → B given by π(b0 , c) = b0 c with b0 ∈ B0 and c ∈ C is an epimorphism [Nobusawa]. Let α0 : B 0 → B0 be the projection on B0 . Then α ◦ π = α0 ◦ α0 and (4)

(γ0 : G → B0 , α0 : B 0 → B0 )

is a split embedding problem for G with Ker(α0 ) = C. Condition (b) gives an epimorphism γ 0 : G → B 0 with α0 ◦ γ 0 = γ0 : 0

γ G B 0 Ao AA γ ooooo 0 o A π  AA ooo  ϕ α0 Ao  γ  woooo A  α  /B /3 A B0 α0

Then γ = π ◦ γ 0 solves embedding problem (2).



Proposition 22.5.8 is useful when G is projective: Proposition 22.5.9: Let G be a projective group. Then: (a) Every embedding problem (2) in which α is a Frattini cover is solvable. (b) If every (finite) split embedding problem for G is solvable, then every (finite) embedding problem for G is solvable. (c) If ψ: G → G is a Frattini cover, then ψ is an isomorphism. Proof of (a): Since G is projective, there exists a homomorphism γ: G → B with α ◦ γ = ϕ. Then α(γ(G)) = ϕ(G) = A. By (1b), γ(G) = B. Thus, γ is a solution of (2).

512

Chapter 22. Projective Groups and Frattini Covers

Proof of (b): Use (a) and Proposition 22.5.8. Proof of (c): Take B = A = G, ϕ = idG , and α = ψ to see that there exists an epimorphism γ: G → G with ψ ◦ γ = idG . Thus, ψ is an isomorphism.  Many properties of projective groups are determined by their quotients modulo their Frattini groups: Corollary 22.5.10: Let G and H be profinite groups with H projective. (a) Each epimorphism θ0 : H/Φ(H) → G/Φ(G) has a lift to an epimorphism θ: H → G. (b) If in addition G is projective, then each isomorphism θ0 : H/Φ(H) → G/Φ(G) has a lift to an isomorphism θ: H → G. Proof of (a): Let πH : H → H/Φ(H) and πG : G → G/Φ(G) be the quotient maps. Then apply Proposition 22.5.9 to find an epimorphism θ with πG ◦ θ = θ 0 ◦ πH . Proof of (b): Suppose θ0 is an isomorphism. Note that both πG and πH are Frattini covers. Hence, by Lemma 22.5.4, θ is a Frattini cover. Since G is projective, a symmetrical argument gives a Frattini cover θ0 : G → H. Since θ0 ◦ θ: H → H is also a Frattini cover, Proposition 22.5.9(c) implies that θ0 ◦ θ is an isomorphism. Consequently, θ is an isomorphism.  Let C be an arbitrary formation of finite groups. The proof of Proposition 22.5.9(b) does not extend to C-projective groups, because the proof of Proposition 22.5.8 involves a subgroups B0 of a C-group B which need not be a C-group. Nevertheless, the result itself is true: Proposition 22.5.11: Let C be a formation of finite groups and G a Cprojective group. Suppose every finite split C-embedding problem (2) for G such that Ker(α) is a minimal normal subgroup of B is solvable. Then every finite C-embedding problem for G is solvable. Proof: Let (2) be a finite C-embedding problem for G. First suppose C = Ker(α) is a minimal normal subgroup of B but (2) does not necessarily split. Since G is C-projective, there is a homomorphism γ: G → B with α ◦ γ = ϕ. ¯ = G/Ker(γ). Let π: G → G ¯ be the quotient map and γ¯ : G ¯ → B Put G ¯ and ϕ: ¯ G → A the homomorphisms induced by γ and ϕ, respectively. Then α ◦ γ¯ = ϕ. ¯ This leads to a commutative diagram G π

 /G ¯ ¯ B ×A G w w γ ¯ ww w ϕ ¯ β w w  {www α  /A B ψ

22.6 The Universal Frattini Cover

513

in which ψ and β are the projections on the coordinates. By Lemma 22.2.5, β maps Ker(ψ) isomorphically onto C. Hence, Ker(ψ) is a minimal normal ¯ Lemma 22.2.9 gives a group theoretic section to ψ. subgroup of B ×A G. ¯ The Therefore, by assumption, there is an epimorphism ζ: G → B ×A G. epimorphism β ◦ ζ: G → B solves (2). Finally suppose C is not a minimal normal subgroup of B. Then C ¯ = B/C0 , has a proper nontrivial subgroup C0 which is normal in B. Put B ¯ B → B ¯ the quotient map, and α ¯ → A the epimorphism induced β: ¯: B ¯ → A) whose by α. This gives a C-embedding problem (ϕ: G → A, α ¯: B kernel C/C0 has a smaller order than |C|. An induction hypothesis gives an ¯ with α epimorphism ϕ: ¯ G→B ¯ ◦ ϕ¯ = ϕ. This gives an embedding problem ¯ B → B) ¯ β: ¯ whose kernel C0 also has a smaller order than (ϕ: ¯ G → B, |C|. Again, an induction hypothesis gives an epimorphism γ: G → B with β¯ ◦ γ = ϕ. ¯ The epimorphism ϕ solves (2). 

22.6 The Universal Frattini Cover The study of projective groups and the study of Frattini covers have a common subject: “universal Frattini covers”: Starting from a profinite group G, we partially order the epimorphisms of profinite groups onto G (also called covers of G). Let θi : Hi → G, i = 1, 2, be covers. We write θ2 ≥ θ1 and say that θ2 is larger than θ1 if there is an epimorphism θ: H2 → H1 with θ1 ◦ θ = θ2 . If θ is an isomorphism, then θ1 is said to be isomorphic to θ2 . ˜ → G is called a projective cover if G ˜ is a An epimorphism ϕ: G ˜ projective group. In this case, G is called a projective cover of G. ˜ → G, unique Proposition 22.6.1: Each profinite group G has a cover ϕ: ˜ G up to an isomorphism, called the universal Frattini cover and satisfying the following equivalent conditions: (a) ϕ˜ is a projective Frattini cover of G. (b) ϕ˜ is the largest Frattini cover of G. (c) ϕ˜ is the smallest projective cover of G. In particular, each projective group is its own universal Frattini cover. Proof: By Corollary 17.4.8, there exists an epimorphism ϕ: F → G with F a free profinite group. Every closed subgroup of F is projective (Corollary ˜ of F such that 22.4.6). Apply Lemma 22.5.6 to produce a closed subgroup G ˜→G ˜ G ϕ˜ = resG˜ (ϕ) is a Frattini (and projective) cover of G. The cover ϕ: appears throughout this proof. Proof of (a) =⇒ (b): Let θ: G1 → G be a Frattini cover. Proposition 22.5.9 ˜ → G1 with θ ◦ γ = ϕ. ˜ Thus, θ ≤ ϕ. ˜ gives an epimorphism γ: G Proof of (b) =⇒ (a): Let ϕ0 : G0 → G be a Frattini cover that is larger than ˜ G0 → G ˜ such that any Frattini cover. Thus, there exists an epimorphism θ: 0 ˜ ˜ ϕ˜ ◦ θ = ϕ . By Lemma 22.5.4(c), θ is a Frattini cover. By “(a) =⇒ (b)”,

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Chapter 22. Projective Groups and Frattini Covers

ϕ˜ is also a maximal Frattini cover of G. Thus, there exists a Frattini cover ˜ → G0 with ϕ0 ◦ θ0 = ϕ. ˜ →G ˜ is a projective ˜ It follows that θ˜ ◦ θ0 : G θ0 : G Frattini cover. By Proposition 22.5.9(c), θ˜ ◦ θ0 is an isomorphism. Hence, ˜ → G0 is an isomorphism. Consequently, G0 is projective and (a) holds. θ0 : G Proof of (a) =⇒ (c): Let ϕ: P → G be a projective cover. Proposition 22.5.9 ˜ with ϕ˜ ◦ γ = ϕ. Thus, ϕ˜ ≤ ϕ. gives an epimorphism γ: P → G Proof of (c) =⇒ (a): Let ϕ0 : G0 → G be a projective cover which is smaller ˜ is projective, there is an than any other projective cover of G. Since G ˜ G ˜ → G0 with ϕ0 ◦ θ˜ = ϕ. ˜ By Lemma 22.5.4(c), ϕ0 is a epimorphism θ: Frattini cover.  Lemma 22.6.2: Let C be a full formation of finite groups and G a pro˜ of G is a pro-C-group and C-group. Then the smallest projective cover G ˜ rank(G) = rank(G). Proof: Let m = rank(G). By Corollary 22.4.5, Fˆm (C) is a projective cover ˜ is a quotient of Fˆm (C). The equality of the ranks is a special of G. Hence, G case of Corollary 22.5.3.  The next lemma characterizes the quotients of the universal Frattini cover of a profinite group: ˜ → G be the universal Frattini cover of a profinite Lemma 22.6.3: Let ϕ: ˜ G ˜ if and only if H is a group G. Then a profinite group H is a quotient of G Frattini cover of a quotient of G. ˜ → H is an epimorphism. Lemma 22.2.8 gives a Proof: Suppose ψ: G commutative diagram of epimorphisms ˜ ?OO G // ? OOO //??? OOOϕ˜O // ???π OOOO O //  δ O'/ G ψ / D // // β γ /   H α /A where the square is cartesian. By Lemma 22.5.4(c), δ is a Frattini cover. Thus (Lemma 22.5.5), so is α. Conversely, suppose α: H → A is a Frattini cover and γ: G → A is an ˜ and Proposition 22.5.9 to find an epimorphism. Apply the projectivity of G ˜ epimorphism ψ: G → H. 

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515

22.7 Projective Pro-p-Groups Fix a prime number p for the whole section. By Example 17.3.3, the family of finite p-groups is full. Thus (Corollary 22.4.5), each free pro-p-group is projective. We prove here a converse to this statement: Lemma 22.7.1: Let m be a cardinal number and V = Fm p be the direct product of m copies of the additive group of Fp . Then rank(V ) = m. Proof: Suppose first m is finite. Consider V as a vector space over Fp . Then rank(V ) = dim(V ) = m. Now suppose m is infinite. Choose a set I of cardinality m Q and for each i ∈ Q I an isomorphic copy Fi of Fp . Each open subgroup H of i∈I Fi contains i∈J Fi for some cofinite subset J of I. The cardinality of the set of cofinite is m. For each such J only finitely many open subgroups Q subsets of I Q F contain of i∈J Fi . Hence, the cardinality of all open subsets of Q i∈I i F is m, as claimed.  i∈I i Lemma 22.7.2: Let I be a set, G = (Fp )I , and N a closed subgroup of G. Then G = N × N 0 for some closed subgroup N 0 of G. Proof: Let ϕ: G → G/N be the quotient map. Lemma 22.5.6 gives a closed subgroup N 0 of G such that ϕ0 = ϕ|N 0 : N 0 → G/N is a Frattini cover. Thus, Ker(ϕ0 ) ≤ Φ(N 0 ) ≤ Φ(G) = Φ(Fp )I = 1 (Lemma 22.1.4(c,d)). Consequently, G = N × N 0.  Lemma 22.7.3: Let m be a cardinal number, G = Fm p , and H a closed subgroup of G. Then H ∼ = Fkp for some cardinal number k ≤ m. Proof: Choose a set J of cardinality |G|. Let U be the set of all triples (U, I, ϕ) where U ≤ H, I ⊆ J, and ϕ: H/U → FIp is an isomorphism. Define a partial ordering on U by the following rule: (U 0 , I 0 , ϕ0 ) ≤ (U, I, ϕ) if U 0 ≤ U , I ⊆ I 0 and the following diagram is commutative: (1)

H/U 0 ϕ0



πU 0 ,U

/ H/U ϕ

0

FIp

ρI 0 ,I

 / FIp

.

Here πU 0 ,U is the quotient map and ρI 0 ,I is the projection. The triple (H, ∅, id) belongs T to U. Suppose S {(Uα , Iα , ϕα ) | α ∈ A} is a descending chain in U. Let U0 = α∈A Uα , I0 = α∈A Iα , and ϕ0 = lim ϕα ←− (we assume here 0 ∈ / A). Then (U0 , I0 , ϕ0 ) is an element of U which is smaller or equal to each (Uα , Iα , ϕα ). Zorn’s lemma gives a minimal element (U, I, ϕ) in U. Assume U is not trivial. Lemma 22.7.2 gives a closed subgroup V of G with V × U = G. Then V is a proper subgroup of G. Hence, V is contained in an open subgroup

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G0 of G of index p. Put H 0 = G0 ∩ H and U 0 = G0 ∩ U . Then H/H 0 ∼ = G/G0 ∼ = Fp and H/U 0 ∼ = H/U × H/H 0 . Since |G| ≥ |H/U | = 2|I| > |I|, there is a j ∈ J r I. Put I 0 = I ∪ {j}. Then it is possible to lift ϕ to an 0 isomorphism ϕ0 : H/U 0 → FIp such that (1) is commutative. This contradicts the minimality of (U, I, ϕ) and proves that U is trivial. It follows that H ∼ = (Fp )I . By Lemma 22.7.1 and Corollary 17.1.5, |I| = rank(H) ≤ rank(G) = m, as claimed.  For a profinite group G let Gp = hg p | g ∈ Gi

 and [G, G] = [g1 , g2 ] | g1 , g2 ∈ G .

Both closed subgroups of G are characteristic. Lemma 22.7.4: Let G be a pro-p-group of rank m. Then Φ(G) = Gp [G, G] and G/Φ(G) is isomorphic to the vector space Fm p . Proof: Maximal subgroups of finite p-groups are normal subgroups of index p. Therefore, maximal open subgroups of G are open normal Q subgroups of index p. This gives a canonical embedding G/Φ(G) → G/N , where N ranges over all open normal subgroups of G of index p. By Lemma 22.7.3, G/Φ(G) ∼ = Fm p with m = rank(G/Φ(G)) = rank(G) (Corollary 22.5.3). Now let G0 = Gp [G, G]. Since G/Φ(G) ∼ = Fm p , the canonical map G → p G/Φ(G) maps each g [a, b] to 1. Hence, G0 ≤ Φ(G). On the other hand let U be an open normal subgroup of G that contains G0 . Then G/U is an Abelian elementary p-group. In particular, U is the intersection of all open normal subgroups of index p that contain U . Hence, Φ(G) ≤ U . Since G0 / G, the intersection of all such U is G0 . Thus, Φ(G) ≤ G0 . Consequently,  Φ(G) = G0 . Remark 22.7.5: Subgroups of finite index. Let m be an infinite cardinal number. By Lemma 22.7.1, the cardinality of the set of open subgroups of m m Fm p of index p is m. On the other hand, |Fp | = 2 . Hence, any basis B of m m Fp , where Fp is considered as a vector space over Fp , has cardinality 2m . For each v ∈ B the subgroup generated by B r{v} has index p. All these subgroups are distinct. Therefore, Fm p has subgroups of index p which are not open. In contrast, Serre proved that every subgroup of finite index of a finitely generated pro-p group G is open [Serre11, p. 32, Exercises 5 and 6] (see also [Ribes-Zalesski, Thm. 4.2.8]) and wrote he did not know if that statement holds for arbitrary finitely generated profinite groups. Recently [Nikolov-Segal] gave an affirmative answer to Serre’s question: (2) Every subgroup N of a finite index of a finitely generated profinite group G is open. The proof of (2) depends on properties of certain “verbal subgroups”: Consider a group theoretic word w(X1 , . . . , Xm ) in the variables X1 , . . . , Xm . For each group G let w(G) be the subgroup generated (in the sense of abstract

22.7 Projective Pro-p-Groups

517

groups) by w(x) with x ∈ Gm . Given a positive integer d, we say w is dlocally finite if every group H which is generated (in the abstract sense) by d elements and satisfies w(H) = 1 is finite. The existence of d-locally finite words is proved in the introduction of [Nikolov-Segal]: (3) For each finite group A which is generated by d elements there exists a d-locally finite word w with w(A) = 1. T To prove (3) consider the free group F on x1 , . . . , xd . Let N = Ker(θ), where θ ranges over all homomorphisms from F to A. Then N is a normal subgroup of a finite index. By Corollary 17.5.8, N is finitely generated. Let y1 , . . . , yk be generators of N . For each i write yi = wi (x), where wi is a word in x1 , . . . , xd . Then consider the word w(X) = w1 (X1 ) · · · wk (Xk ), where Xi = (Xi1 , . . . , Xid ) and X = (X1 , . . . , Xk ). Each of the generators yi = wi (x) of N belongs to w(F ), so N ≤ w(F ). Conversely, let x01 , . . . , x0d ∈ F . Then the map xi 7→ x0i , i = 1, . . . , d, extends to a homomorphism κ: F → F . Let θ: F → A be an arbitrary homomorphism. Then θ ◦ κ is also a homomorphism from F to A. Hence, θ(wi (x0 )) = θ(wi (κ(x))) = θ ◦ κ(wi (x)) = θ ◦ κ(yi ) = 1 for i = 1, . . . , k. It follows that w(F ) ≤ N . Consequently, w(F ) = N . Now let H be a group with d generators and w(H) = 1. Let π: F → H be an epimorphism. Then π(w(F )) = w(H) = 1. Hence, |H| ≤ (F : w(F )) < ∞. Consequently, w is d-locally finite, as desired. The key result in the proof of (2) bounds the number of factors in the elements of w(H): (4) Let d be a positive integer and w(X1 , . . . , Xm ) a d-locally finite word. Then there exists a positive integer r such that for each finite group A generated by d elements and for each a ∈ w(A) there are b1 , . . . , br ∈ Am and β1 , . . . , βr ∈ {±1} such that a = w(b1 )β1 · · · w(br )βr [Nikolov-Segal, Thm. 2.1]. The proof of (4) uses the classification of finite simple groups. Compactness arguments generalize (4) to profinite groups: (5) Let d be a positive integer and w(X1 , . . . , Xm ) a d-locally finite word. Then there exists a positive integer r such that for each profinite group G generated (in the profinite sense) by d elements and for each g ∈ w(G) there are b1 , . . . , br ∈ Gm and β1 , . . . , βr ∈ {±1} such that g = w(b1 )β1 · · · w(br )βr . In particular, w(G) is closed. Now consider a profinite group G generated (in the profinite sense) by elements x1 , . . . , xd and let N be a subgroup of finite index. Without loss assume that N is normal in G. Then G/N is finite. By (3), there exists a d-locally finite word w(X1 , . . . , Xm ) such that w(G/N ) = 1. It follows that w(G) ≤ N . Let G0 be the abstract subgroup of G generated by x1 , . . . , xd . Then w(G0 ) / G0 , w(G0 /w(G0 )) = 1, and G0 /w(G0 ) is generated (in the sense of abstract groups) by d elements. Thus, G0 /w(G0 ) is finite and (G0 : w(G0 )) < ∞. Therefore, (G0 w(G) : w(G)) = (G0 : G0 ∩ w(G)) ≤ (G0 : w(G0 )) < ∞.

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By (5), w(G) is closed, so G0 w(G) is closed. Since G0 is dense in G, we have G0 w(G) = G. It follows that w(G) has a finite index in G. Consequently, w(G) is open, so also N is open, as claimed.  Proposition 22.7.6 (Tate): A pro-p-group G is projective if and only if it is pro-p free. Proof: By Corollary 22.4.5, it suffices to show that G projective implies G is pro-p free. Let m = rank(G). By Lemma 22.7.4, G/Φ(G) ∼ = Fm ∼ = Fˆm (p)/Φ(Fˆm (p)). p

Therefore, by Corollary 22.5.10(b), G ∼ = Fˆm (p).



Corollary 22.4.6 immediately gives an analog of Schreier’s theorem (Proposition 17.5.6) for discrete free groups: Corollary 22.7.7 (Tate): A closed subgroup of a free pro-p-group is pro-p free. Proof: Let G be a closed subgroup of a free pro-p group F . By Proposition 22.7.6, F is projective. By Proposition 22.4.7, G is projective. Therefore, by Proposition 22.7.6, G is free pro-p.  Corollary 22.7.8: Let G be a pro-p group of rank m. Then the universal Frattini cover of G is Fˆm (p). ˜ be the universal Frattini cover of G. By Lemma 22.6.2, G ˜ is Proof: Let G ˜ is projective. Hence, by Proposition a pro-p group of rank m. In addition, G 22.7.6, G ∼  = Fˆm (p). Here is a generalization of Lemma 22.1.4(c) for pro-p groups: Lemma 22.7.9: Let G be a pro-p group and H a closed subgroup. Then Φ(H) ≤ Φ(G). Proof: By Lemma 22.7.4, Φ(H) = H p [H, H] ≤ Gp [G, G] = Φ(G).



Proposition 22.7.10: Every pro-p group G which is small is finitely generated. Proof: By Lemma 22.7.4, Φ(G) is the intersection of all open subgroups of G of index p. By assumption, there are only finitely many of them. Hence, G/Φ(G) is a finite group. In particular, G/Φ(G) is finitely generated. It follows from Lemma 22.1.1 that G is also finitely generated.  Corollary 22.7.11: Let A be an Abelian projective pro-p group. Then either A is trivial or A ∼ = Zp . Proof: By Proposition 22.7.6, A is pro-p free. Since A is Abelian, rank(A) ≤ 1. Otherwise, A would have a non-Abelian finite quotients of rank 2. So, either A = 1 or A ∼  = Zp . The following result surveys the structure of an arbitrary Abelian pro-p group:

22.7 Projective Pro-p-Groups

519

Proposition 22.7.12: Let A be an additive Abelian pro-p group. Then: (a) A is a Zp -module. (b) Suppose A is finitely generated. Then A ∼ = Z/pk1 Z ⊕ · · · ⊕ Z/pkm Z ⊕ Zrp , with k1 ≥ · · · ≥ km ≥ 1 and r is a nonnegative integer. The sequence k1 , . . . , km , r are uniquely determined by A. (c) Suppose A is finitely generated and torsion free. Let B be a subgroup. Then A and B are free Zp modules, A has a Zp -basis a1 , . . . , an and there are positive integer α1 , . . . , αm with 0 ≤ m ≤ n satisfying this: α1 a1 , . . . , αm am form a Zp -basis of B and αi |αi+1 , i = 1, . . . , m − 1. 0 0 (d) In general, A ∼ = Zm p × A where m is a cardinal number and A is the intersection of all kernels of the epimorphisms of A onto Zp . (e) If A is torsionfree, then A ∼ = Zm p for some cardinal number m. Proof of (a): Let B be an open subgroup of A.TThen, (A : B) = pn with n ∞ a nonnegative integer. Hence, pn A ≤ B. Thus, n=1 pn A = 0. n n n The rule (z + p Z)(a + p A) = za + p A defines a continuous action of Z/pn Z on A/pn A. Going to the limit, the first paragraph gives an action of Zp on A. Proof of (b): Statement (b) is a special case of the main theorem on finitely generated modules over principal ideal domains [Lang7, §III, Thms. 7.3 and 7.5]. Proof of (c): See [Lang7, §III, Thm. 7.8]. Proof of (d): See [Geyer-Jensen1, p. 337, (4)]. Proof of (e): See [Ribes-Zalesskii, Thm. 4.3.3].



Example 22.7.13: The rank of local Galois groups. Let K be a local field, that is, K is a finite extension of Qp or of Fp ((t)). Denote the maximal tamely ramified extension of K by Ktr . Let G = Gal(K) and P = Gal(Ktr ). Then P is a closed normal pro-p group of G. By Lemma 22.1.4(c), Φ(P ) ≤ Φ(G). By Lemma 22.7.4, Φ(P ) = P p [P, P ]. Hence, rank(G) = rank(G/[P, P ]) (Corollary 22.5.3). This observation, is implicitly used in [Jannsen, Section 3] to determine the number of generators of G. Indeed, if K is a finite extension of Qp , then by results of Iwasawa and local class field theory, rank(G/[P, P ]) = [K : Qp ] + 3. Hence, rank(G) = [K : Qp ] + 3. In particular, rank(Gal(Qp )) = 4. If char(K) = p, then the rank of G/[P, P ] and therefore also of Gal(K) is infinite. ˜ by F . It is proved in [Jarden-Ritter] Denote the fixed field of Φ(G) in K that F is the compositum of all Galois extensions N of K containing Ktr such that Gal(N/Ktr ) ∼  = Z/pZ.

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Chapter 22. Projective Groups and Frattini Covers

22.8 Supernatural Numbers The next definition generalizes the index and the order of finite groups to profinite groups: Definition 22.8.1: Supernatural numbers. A supernatural number is a Q formal product n = pnp , where np is either a nonnegative integer or ∞, and p ranges over all primes. Let m and n be supernatural numbers. We say that m divides n if mp ≤ np for every prime p. Define the product of supernatural numbers ni , for i ∈ I, by the formula Y P YY pni,p = p i∈I ni,p . i∈I p

p

P

Here i∈I ni,p = ∞ if ni,p > 0 for infinitely many i ∈ I or ni,p = ∞ for at least one i ∈ I. Similarly, there is a greatest common divisor and a least common multiple of supernatural numbers: Y Y pmin(mp ,np ) and lcm(m, n) = pmax(mp ,np ) . gcd(m, n) = p

p

Define the index of a closed subgroup H in a profinite group G to be
(G : H) = lcm (G : U ) | U is an open subgroup of G that contains H . Then define the order of G to be
#G = (G : 1) = lcm (G : U ) | U is an open subgroup of G . Note that the index generalizes the usual index, (G : H), if H is an open subgroup of G.  Example 22.8.2: Order of three classical Abelian groups. Y Y Y ˆ= p∞ ; and # Z/pZ = p. #Zp = p∞ ; #Z p

p



p

The usual rules for indices of finite groups hold for profinite groups: Lemma 22.8.3: (a) Let K ≤ H ≤ G be profinite groups. Then (G : K) = (G : H)(H : K). (b) Let {Hi | i ∈ I} beTa directed family of closed subgroups of a profinite group G. Put H = i∈I Hi . Then (G : H) = lcmi∈I (G : Hi ). (c) Let hGi , πji ij,i∈I be an inverse limit of profinite groups such that πji is an epimorphism for j ≥ i. Then the inverse limit G = lim Gi satisfies ←− #G = lcmi∈I {#Gi }.

22.8 Supernatural Numbers

(d) #

Q

Gi =

i∈I

Q

i∈I

521

#Gi .

Proof of (a): Let N be the set
of all open normal subgroups of G. Then (G : H) = lcm (G : HN ) | N ∈ N and (H : K) = lcm (H : K(H ∩ N )) | N ∈
N . The identities K(H ∩ N ) = H ∩ KN and (HN : KN ) = (H : H ∩ KN ) imply (H : K) = lcm{(HN : KN ) | N ∈ N }.

(1) In addition, (2)

(G : KN ) = (G : HN )(HN : KN ).

Let p be a prime number. If p∞ |(G : K), then (1) and (2) imply p∞ |(G : H) or p∞ |(H : K). Suppose therefore that p∞ - (G : K). Let n1 , n2 , and n3 be the maximal exponents of p-powers that divide (G : K), (G : H), and (H : K), respectively. Then n1 , n2 , n3 < ∞. Also, there exists a group N ∈ N such that pn1 (resp. pn2 , pn3 ) is the greatest power of p dividing (G : KN ) (resp. (G : HN ), (HN : KN )). By (2), n1 = n2 + n3 , as desired. Proof of (b): By (a), (G : Hi )|(G : H) for each i ∈ I. Hence, lcmi∈I (G : Hi )|(G : H). Conversely, let U be an open subgroup of G that contains H. Then there is an i ∈ I with Hi ≤ U (Lemma 1.2.2(a)). Therefore, (G : U )|(G : Hi ), so (G : U )| lcmi∈I (G : Hi ). Thus, lcmU (G : U )| lcmi∈I (G : Hi ) with U ranging over all open subgroups containing H. This gives (b). Proof of (c): Let πi : G → Gi beTthe canonical epimorphism and Ni = Ker(πi ). Then (G : Ni ) = #Gi and Ni = 1. Apply (b): #G = lcm #Gi . Proof of (d): Denote Q the family ofQfinite subsets of I by F. Partially order F by inclusion. Then Gi = lim i∈F Gi , where F ranges over F. Hence, ←− by (a) and (c), Y Y Y Y Gi = lcm # Gi = lcm #Gi = #Gi .  # i∈I

F ∈F

i∈F

F ∈F

i∈F

i∈I

Lemma 22.8.4: Let G be a profinite group and H a closed subgroup. Suppose p - (G : H) for all but finitely many p and p∞ - (G : H) for all p. Then H is open. Proof: For each p let pnp be Q the maximal power of p which divides (G : H).  By assumption, (G : H) = pnp < ∞, so H is open. Lemma 22.8.5: Let A be a finitely generated Abelian group and B a closed subgroup of infinite index. Then B is contained in an open subgroup of index p for infinitely many p or B is contained in a closed subgroup C with A/C ∼ = Zp for some p. Proof: Divide out by B, if necessary, to assume B = 0. We have to prove: Z/pZ is a factor of A for infinitely many p or Zp is a factor of A for some p.

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Proposition 22.7.12(b) presents A as M Z/pkp,1 Z ⊕ · · · ⊕ Z/pkp,m(p) Z ⊕ Zkpp , (3) A= p

with nonnegative integers kp , m(p) and positive integers kp,i . Each of the direct summands in (3) appears also as a factor group of A. In particular, if m(p) > 0, then Z/pkp,1 Z is a factor of A. Hence, so is Z/pZ. By Lemma 22.8.4, m(p) > 0 for infinitely many p or there is a p with kp > 0. This together with the preceding arguments concludes the proof of the lemma.  Remark 22.8.6: Degree of algebraic extensions. Let L/K be an algebraic extension. Define [L : K] to be the least common multiple of all degrees [E : K] with E ranging over all finite subextensions of L/K. Call [L : K] the degree of L/K. When L/K is separable, Galois correspondence implies that [L : K] = (Gal(K) : Gal(L)). Hence, we may rephrase Lemma 22.8.3 in field theoretic terms: (4a) Let K ⊆ L ⊆ M be fields with M/K separable algebraic. Then [M : K] = [M : L][L : K]. (4b) Let {L Qi | i ∈ I} be a family of separable algebraic extensions of K. Put L = i∈I Li . Then [L : K] = lcmi∈I [Li : K]. (4c) Q Suppose in (4b) the Li are linearly disjoint over K. Then [L : K] = i∈I [Li : K]. Now suppose L/K is an arbitrary algebraic extension. Let E be a finite subextension. Denote the maximal separable extension of K in E (resp. L) by E0 (resp. L0 ). Then E is linearly disjoint from L0 over E0 , so [E : E0 ] = [L0 E : L0 ]. Thus, [L : K] = [L : L0 ][L0 : K]. This and similar considerations show the separability assumption in (4a) is redundant. 

22.9 The Sylow Theorems The concept of Sylow groups carries over to profinite groups: Definition 22.9.1: Let p be a prime number. A closed subgroup P of a profinite group G is said to be a p-Sylow group of G if P is a pro-p-group and p - (G : P ).  Proposition 22.9.2: Let G be a profinite group and p a prime number. Then: (a) G has a p-Sylow group and every pro-p subgroup of G is contained in a p-Sylow group of G. (b) The p-Sylow groups of G are conjugate. (c) If θ: G → H is an epimorphism of profinite groups and P is a p-Sylow group of G, then θ(P ) is a p-Sylow group of H. (d) If N is a closed normal subgroup of G and P a p-Sylow group of G, then N ∩ P is a Sylow group of N .

22.9 The Sylow Theorems

523

Q (e) #G = p #Gp where Gp is a p-Sylow group of G and p ranges over all prime numbers. Proof of (a): Let N be the set of all open normal subgroups of G and H a pro-p-subgroup of G (H may be trivial). For each N ∈ N denote the set of all p-Sylow groups of G/N which contain HN/N by P(N ). By Sylow’s theorem for finite groups, P(N ) is finite and nonempty. If M ∈ N and M ≤ N , then the quotient map G/M → G/N maps p-Sylow groups of G/M onto p-Sylow groups of G/N . It therefore defines a canonical map of P(M ) into P(N ). By Lemma 1.1.3, the inverse limit of the sets P(N ) is nonempty. Thus, there is a (PN )N ∈N such that PN ∈ P(N ) and PM is mapped onto PN for all M, N ∈ N with M ≤ N . The inverse limit P = lim PN is a pro-p-subgroup ←− of G = lim G/N that contains H. If N ∈ N , then P N/N = PN . Hence, ←− p - (G : P N ). Therefore, p - (G : P ). It follows that P is a p-Sylow group of G that contains H. Proof of (b): Let P and P 0 be p-Sylow groups of G. If N ∈ N , then P N/N and P 0 N/N are p-Sylow groups of the finite group G/N . Hence, they are conjugate. It follows from Lemma 1.2.2(e) that P and P 0 are conjugate. Proof of (c): The group Q = θ(P ) is a pro-p-group and (H : Q) = (G : θ−1 (Q)) divides (G : P ), which is relatively prime to p. Hence, Q is a pSylow group of H. Proof of (d): Use (a) to choose a p-Sylow group Q of N and a p-Sylow group P 0 of G which contains Q. Then (N ∩ P 0 : Q) divides (P 0 : Q) (Lemma 22.8.3(a)), hence #P 0 , so (N ∩ P 0 : Q) is a power of p. On the other hand, (N ∩ P 0 : Q) divides (N : Q), so (N ∩ P 0 : Q) is relatively prime to p. It follows that (N ∩ P 0 : Q) = 1 and N ∩ P 0 = Q. By (b), there exists g ∈ G with P g = P 0 . Since N /G, we have (N ∩P )g = −1 N ∩ P 0 = Q, so N ∩ P = Qg is a p-Sylow group of N . Q Proof of (e): The formula #G = l #Gl follows from the formula #G =  (G : Gl ) · #Gl for each prime l (Lemma 22.8.3(a)). Finite nilpotent groups are direct products of their unique p-Sylow groups as p runs over all prime numbers. The next proposition is an immediate consequence of this and Lemma 22.1.2: Proposition 22.9.3: Each pronilpotent group N is the direct product of its p-Sylow groups. In particular, if G is a profinite group, Φ(G) is a direct product of its p-Sylow groups and each of them is normal in G. Proof: For each prime number p the group N has a unique p-Sylow group, Np , and it is characteristic in N , since this is the case for the finite quotients of N . The groups Np generate N and the intersection of Np with the Q group generated by all other Sylow groups of N is trivial. It follows that N = Np . Since each p-Sylow group of Φ(G) is characteristic, Lemma 22.1.2 gives the second part of the proposition. 

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Chapter 22. Projective Groups and Frattini Covers

Proposition 22.9.4: Let A be an Q Abelian projective group. Then there is a set S of prime numbers with A ∼ = p∈S Zp . Q Proof: By Proposition 22.9.3, A ∼ = p Ap , where Ap is the p-Sylow group of A. By Proposition 22.4.7, Ap is projective. Hence, by Corollary 22.7.11, Ap is either trivial or is isomorphic to Zp .  We conclude this section with some applications to fields: Corollary 22.9.5: Let K be an algebraic extensionQ of a finite field. Then there exists a set S of prime numbers with Gal(K) ∼ = p∈S Zp . ˆ Hence, Gal(K) is proProof: By Section 1.5, Gal(K) is a subgroup of Z. jective (Corollary 22.4.6). It follows from Proposition 22.9.4 that Gal(K) ∼ = Q Z for some set S of prime numbers.  p p∈S Theorem 22.9.6 ([Ax2, Thm. D]): Let K be a perfect PAC field with Gal(K) Abelian. Then K is C1 . Proof: By Theorem 11.6.2, Gal(K) is projective. Hence, by Proposition 22.9.4, Gal(K) is procyclic. It follows from Theorem 21.3.6(c) that K is C1 .  And here is an application to Hilbertian fields: Theorem 22.9.7 ([Uchida, Thm. 3(ii)]): Let K ⊆ L ⊆ N be fields. Suppose K is Hilbertian, N is a pronilpotent extension of K, and [L : K] is divisible by two distinct prime numbers. Then L is Hilbertian. Proof: By Proposition 22.9.3, Gal(N/K) is the direct product of its Sylow of [L : K], let Nq be the fixed field in subgroups Gp . Choose a prime divisor q Q N of Gq , and Nq0 the fixed field in N of p6=q Gp . Then Nq Nq0 = N , L 6⊆ Nq , and L 6⊆ Nq0 . By the diamond theorem (Theorem 13.8.3), L is Hilbertian. 

22.10 On Complements of Normal Subgroups Let N be a closed normal subgroup of a profinite group G. A closed subgroup H of G is called a complement to N in G if N ∩ H = 1 and N H = G. The Schur-Zassenhaus theorem, whose finite version appears in the proof of Lemma 22.4.1, gives the existence of a complement, unique up to conjugation in G, under a simple index condition. Now suppose N, M / G, H ≤ G, and M ≤ N ∩ H. We say, H is a complement to N in G/M if H/M is a complement to N/M in G/M . In other words, N ∩ H = M and N H = G. Lemma 22.10.1 (Schur-Zassenhaus): Let N be a closed normal subgroup of a profinite group G. Suppose gcd(#N, (G : N )) = 1. Then N has a complement in G. Furthermore, all complements to N in G are conjugate. Proof: First we prove the uniqueness part of the lemma: Let H and H 0 be two complements to N in G. Let M be an open normal subgroup of G. Then

22.10 On Complements of Normal Subgroups

525

(HM : M ) = (G : N M )|(G : N ) and (N M : M )|#N . Hence, (HM : M ) and (N M : M ) are relatively prime. Therefore, N M ∩ HM = M . It follows that HM/M is a complement to N M/M in G/M . Similarly, H 0 M/M is a complement to N M/M in G/M . By the uniqueness part of the finite group version of Schur-Zassenhaus [Huppert, p. 128], HM/M and H 0 M/M are conjugate. Consequently, by Lemma 1.2.2(e), H and H 0 are conjugate. The proof of the existence of a complement is divided into two parts: Part A: N is finite. Then G has an open normal subgroup M with N ∩ M = 1. Apply Schur-Zassenhaus [Huppert, p. 126] to the finite group G/M to find a complement H to N M in G/M . Then H is a complement to N in G. Part B: N is infinite. With U0 = G let {Uα | α < λ} be a wellTordering of the open normal subgroups of G. For each β ≤ λ let Mβ = N ∩ α<β Uα . We apply a transfinite induction to find a decreasing sequence {Hβ | β ≤ λ} of closed subgroups of G such that Hβ is a complement to N in G/Mβ . Indeed, let H0 = G. Assume the statement is true for β. Use Part A and that Mβ+1 is open in Mβ to find a complement Hβ+1 to Mβ in Hβ /Mβ+1 . Then Hβ+1 is a complement to N in G/Mβ+1 . Now assume γ ≤ Tλ is a limit ordinal and Hβ has been defined for each β
AK = H

and A ∩ K = 1.

The rest of the proof splits into two parts as G is finite or not: Part A: G is finite [Brandis, §2]. For x, y ∈ G with Hx = Hy there is a unique k(x, y) ∈ K with xy −1 A = k(x, y)A. The function k(x, y) satisfies the following rules (use normality of A in G): (2a) k(x, x) = 1, k(y, x) = k(x, y)−1 , and k(x, y)k(y, z) = k(x, z). (2b) x−1 k(x, y)y ∈ A and k(xg, yg) = k(x, y) for each g ∈ G. (2c) k(xa, y) = k(x, y) for each a ∈ A. Let S be a set of representative systems of right cosets of G modulo H. Thus, for all Φ ∈ S and x ∈ G we have |Φ ∩ Hx| = 1. Define a convolution product between Q Φ1 , Φ2 ∈ S: (3) Φ−1 r1−1 k(r1 , r2 )r2 , where the product ranges over all pairs 1 ∗ Φ2 = (r1 , r2 ) with r1 ∈ Φ1 , r2 ∈ Φ2 , and r1 ∈ Hr2 .

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By (2b), each of the factors of the product belongs to the Abelian group A, so that the product is well defined. Use (2) to check the following rules for Φ1 , Φ2 , Φ3 ∈ S: (4a) Φ−1 1 ∗ Φ1 = 1. −1 −1 (4b) (Φ−1 1 ∗ Φ2 ) · (Φ2 ∗ Φ3 ) = Φ1 ∗ Φ3 . −1 −1 (4c) (Φ1 ∗ Φ2 )−1 = Φ2 ∗ Φ1 . g (4d) (Φ1 g)−1 ∗ (Φ2 g) = (Φ−1 1 ∗ Φ2 ) for each g ∈ G. −1 −(G:H) (4e) (Φ1 a) ∗ Φ1 = a for each a ∈ A. Each Φ ∈ S defines a map ϕ : G → A by ϕ(g) = (Φg)−1 ∗Φ. By (4b) and (4d), ϕ(g1 g2 ) = ϕ(g1 )g2 ϕ(g2 ). That is, ϕ is a crossed homomorphism. In particular, Ker(ϕ) = {g ∈ G | ϕ(g) = 1} is a subgroup of G (which is not necessarily normal). Since ϕ(a) = a−(G:H) for each a ∈ A, the restriction of ϕ to A is an automorphism. In particular, Ker(ϕ)∩A = 1. Also, since A is an Abelian group, ϕ(ga) = ϕ(g)a ϕ(a) = ϕ(g)ϕ(a), for a ∈ A. Moreover, for each g ∈ G there is an a ∈ A with ϕ(a) = ϕ(g)−1 . Thus, g = (ga)·a−1 ∈ Ker(ϕ)A. Consequently, Ker(ϕ) is a complement to A in G. Part B: G is infinite. Since A is finite, K is open in H. Therefore, there exists an open normal subgroup N of G with H ∩ N ≤ K. Consider the finite quotient group G/N . Observe that AN ∩ KN = N : if x = an1 = kn2 , with a ∈ A, k ∈ K, and n1 , n2 ∈ N , then k −1 a = n2 n−1 1 ∈ H ∩ N . Hence, a ∈ A ∩ K = 1 and x = n1 ∈ N . By Part A, there exists an open subgroup L of G with AN L = G and AN ∩ L = N . Also, A ∩ L = (A ∩ AN ) ∩ L = A ∩ N ≤ A ∩ K = 1. Thus, L is a complement to A in G.  Remark 22.10.3: Uniqueness. Assume in Lemma 22.10.2 that all complements to A in H are conjugate. If G is finite, then Gasch¨ utz’ theorem [Huppert, p 121] asserts that all complements to A in G are conjugate. As in the proof of Lemma 22.10.1, one can deduce for arbitrary G that A has a unique (up to conjugacy) complement in G.  We apply Lemma 22.10.2 to prove another criterion for projectivity: Proposition 22.10.4: A profinite group G is projective if and only if for each p all p-Sylow groups of G are free pro-p groups. Proof: If G is projective and P is a p-Sylow subgroup of G, then P is projective (Proposition 22.4.7). Thus, P is a free pro-p group (Proposition 22.7.6). Conversely, suppose all p-Sylow groups of G are free pro-p groups. By Corollary 22.4.3, it suffices to prove that each exact sequence 1 → A → E → G → 1, in which A is a finite Abelian p-group, splits. From Proposition 22.9.2, E has a p-Sylow group P that contains A. Then P/A, which is isomorphic to a p-Sylow group of G, is by assumption a free pro-p group. In particular, P/A is projective (Corollary 22.4.5). Hence, A has a complement in P (Remark 22.4.2). By definition, the index (E : P ) is relatively prime to p. Consequently, by Lemma 22.10.2, A has a complement in E. 

22.10 On Complements of Normal Subgroups

527

We apply these results to Frattini covers: Proposition 22.10.5: Let H be a Frattini cover of a profinite group G. Then #G and #H have the same prime divisors. Proof: G is a homomorphic image of H. Hence, each prime divisor p of #G divides #H. Conversely, assume p divides #H but not #G. By assumption, there is a Frattini cover ϕ: H → G. Thus, Ker(ϕ) ≤ Φ(H). Hence, p - (H : Φ(H)), so p|#Φ(H). Therefore, Φ(H) has a nontrivial p-Sylow subgroup P . By Proposition 22.9.3, P is normal in H. Thus, Schur-Zassenhaus (Lemma 22.10.1) gives a closed subgroup B of H with P ∩ B = 1 and P B = H. By Lemma 22.1.1, B = H. Consequently, P = 1, a contradiction.  of profinite groups Corollary 22.10.6: Let G1 , G2 , G3 , . . . be a sequence Q∞ of mutually relatively prime orders, and let G = i=1 Gi . (a) If each Gi is projective, then so is G. Q∞ ˜ ˜∼ (b) The universal Frattini covers satisfy G = i=1 G i. Proof: A p-Sylow group of Gi is a p-Sylow group of G. Thus, (a) is a consequence of Proposition 22.10.4. ˜ By Proposition 22.10.5, Q∞ ˜the groups Gi have mutually relatively prime ˜i → is a projective group. In Q addition, if ϕ˜i : G orders. Hence, by (a), i=1 G i ∞ Gi is the universal Frattini cover, i = 1, 2, . . ., and ϕ˜ = i=1 ϕ˜i , then Ker(ϕ) ˜ =

∞ Y i=1

Ker(ϕ˜i ) ≤

∞ Y i=1

˜ i ) = Φ( Φ(G

∞ Y

˜i) G

i=1

Q∞ ˜ (Lemma 22.1.4(d)). Thus, i=1 G i is a Frattini cover of G. It follows from Q∞ ˜  Proposition 22.6.1 that i=1 Gi is the universal Frattini cover of G. For finite groups the next result improves Proposition 22.10.5: ˜ be the universal Frattini cover of a finite group Proposition 22.10.7: Let G ˜ G. Then #G and #Φ(G) have the same prime divisors. ˜ then p|#G. ˜ Therefore, p divides #G (ProposiProof: If p divides #Φ(G), tion 22.10.5). ˜ has a nontrivial p-Sylow group P . Conversely, if p divides #G, then G ˜ Since G is projective, so is P . Therefore, P is infinite (Proposition 22.4.7). ˜ ˜ a quotient of G, is finite. Therefore, Φ(G) ˜ ∩P On the other hand, G/Φ( G), ˜ is nontrivial. Consequently, p divides #Φ(G).  Remark 22.10.8: The finiteness assumption in Proposition 22.10.7 is essential. The Frattini group of Fˆe , for e ≥ 2, is trivial (e.g. Corollary 24.10.4(d)). Since each p divides #Fˆe and Fˆe is its own universal Frattini cover (Proposition 22.6.1), Proposition 22.10.7 fails for arbitrary profinite groups.  A Corollary of Proposition 22.10.7 is that the condition on C in Lemma 22.4.1 to be full is necessary. More precisely, suppose a Melnikov formation C

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contains no simple Abelian groups. Then each free pro-C group is C-projective (Lemma 22.3.6) but not projective. This follows from the following result: Proposition 22.10.9: Let C be a Melnikov formation of finite groups and G a pro-C group. Suppose G is projective. Then Cp ∈ C for each p dividing the order of G. Proof: Let p be a prime divisor of #G. Then there exists an epimorphism ϕ: G → A onto a finite group A having an order divisible by p. Let α: A˜ → A be the universal Frattini cover of A. Since G is projective, there is an epimorphism γ: G → A˜ with α ◦ γ = ϕ (Proposition 22.5.9(a)). In particular, ˜ is also a pro-C group. In addition, A˜ is a pro-C group. It follows that Φ(A) ˜ is pronilpotent (Lemma 22.1.2) and its p-Sylow subgroup Φ(A) ˜ p is Φ(A) ˜ nontrivial (Proposition 22.10.7). Thus, Cp is a quotient of A, hence belongs to C. 

22.11 The Universal Frattini p-Cover Fix a prime number p for the whole section. Call a profinite group G p-projective if each p-Sylow group of G is projective, hence a free pro-p group (Proposition 22.7.6). In cohomological terms this means the pth cohomological dimension of G is at most 1. The following result, a p analog of Remark 22.4.2, justifies the name p-projective: π

Proposition 22.11.1: Let 1 → P → H −→ G → 1 be a short exact sequence of profinite groups. Suppose G is p-projective and P is a pro-p group. Then the short exact sequence splits. Proof: Choose a p-Sylow group Hp of H which contains P (Proposition 22.9.2(a)). Then Gp = π(Hp ) is a p-Sylow group of G (Proposition 22.9.2(c)). By assumption, Gp is projective. Hence, the short exact sequence 1 → P → π Hp −→ Gp → 1 splits. This means that Hp has a subgroup G0p which π maps isomorphically onto Gp . In other words, P ∩ G0p = 1 and P G0p = Hp . The rest of the proof breaks up into two parts. Part A: P is finite. First suppose P is Abelian. By definition, utz (Lemma 22.10.2), P has a gcd(#P, (H : Hp )) = 1. Hence, by Gasch¨ complement in H. Now suppose P is non-Abelian. Then, its center Z is a nontrivial Abelian ¯ = H/Z, P¯ = P/Z, and π ¯ → G be the normal subgroup of H. Let H ¯: H ¯ reduction of π modulo Z. Then #P < #P . An induction hypothesis gives a splitting of π ¯ . Thus, H has a subgroup H0 with P ∩ H0 = Z and P H0 = H. π Hence, 1 → Z → H0 −→ G → 1 is a short exact sequence. The preceding paragraph gives a subgroup H1 of H0 with H1 ∩ Z = 1 and H1 Z = H0 . This subgroup is a complement to P in H. Part B: P is infinite. We repeat the arguments of Part B of the proof of Lemma 22.10.1. We find a decreasing transfinite sequence {Pβ | β ≤ m}

22.11 The Universal Frattini p-Cover

529

T of closed normal subgroups of H with P0 = P , Pβ = α<β Pα for each limit ordinal β ≤ m, and Pm = 1. By transfinite induction we construct a decreasing sequence {Hβ | β ≤ m} of closed subgroups of H with P ∩Hβ = Pβ and P Hβ = H. Let β < m. Suppose P has a complement Hβ in H/Pβ . Since Pβ /Pβ+1 is finite and Hβ /Pβ ∼ = H/P ∼ = G, Part A gives a complement Hβ+1 to Pβ in Hβ /Pβ+1 . Then Hβ+1 is a complement to P in Pβ+1 . Now consider a limit ordinal T γ ≤ m. Suppose Hβ has been constructed for each β < γ. Put Hγ = β<γ Hβ . Then Hγ is a complement to P in H/Pγ (use Lemma 1.2.2(b)). Transfinite induction gives a complement to P in H/Pm , that is, in H.  An embedding problem with a pro-p kernel is said to be a p-embedding problem. Corollary 22.11.2: Let G be a p-projective profinite group. Then every p-embedding problem for G is weakly solvable. Proof: Let (ϕ: G → A, α: B → A) be an embedding problem for G. Let H = B ×A G and let π: H → G be the projection on the second coordinate. Then, Ker(π) ∼ = Ker(α) (Lemma 22.2.5) is a pro-p group. Proposition 22.11.1 gives a section π 0 : G → H of π. Let β: H → B be the projection onto the first coordinate. Then β ◦ π 0 is a week solution of the given embedding problem.  Lemma 22.11.3: Every closed subgroup of a p-projective group is p-projective. Proof: Let G be a p-projective group and H a closed subgroup. Choose a p-Sylow subgroup Hp of H and a p-Sylow subgroup Gp of G which contains Hp (Proposition 22.9.2(a)). Since Gp is projective, so is Hp (Proposition 22.4.7). Hence, H is p-projective.  Certain quotients of p-projective groups are p-projective: Lemma 22.11.4: Let G be a p-projective group and N a closed normal subgroup. Suppose p - #N . Then G/N is p-projective. Proof: Let Gp be a p-Sylow subgroup of G. By assumption, Gp ∩ N = 1, hence Gp N/N ∼ = Gp , so Gp N/N is projective. In addition, Gp N/N is a p-Sylow subgroup of G/N (Proposition 22.9.2(c)). Consequently, G/N is p-projective.  A p-cover of a profinite group G is an epimorphism ϕ: H → G with a pro-p kernel. Call ϕ a Frattini p-cover if in addition ϕ is Frattini. Construction 22.11.5: p-projective Frattini p-cover. Let G be a profinite ˜ → G its universal Frattini cover. Put K = Ker(ϕ). group and ϕ: ˜ G ˜ Then, ˜ K ≤ Φ(G). Hence, K is pronilpotent (Lemma 22.1.2). Thus, by Proposition

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Q 22.9.3, K = Kl , where l ranges over all prime numbers and Kl is the ˜ Since G ˜ is unique l-Sylow subgroup of K. In particular, Kl is normal in G. is projective, hence pro-l free. projective (Proposition 22.6.1), each K l Q ˜ Put Put Kp0 = l6=p Kl . Then Kp0 is a closed normal subgroup of G. ¯ = K/K 0 , and ϕ: ˆ → G the epimorphism induced by ϕ. ˆ = G/K ˜ 0, K ˆ G ˜ Then G p

p

∼ Kp is pro-p free. In addition, K ¯ ≤ Φ(G). ˆ Therefore, ¯ = Ker(ϕ) ¯ = K ˆ and K ˆ ˆ ϕ: ˆ G → G is a Frattini p-cover. Finally, by Lemma 22.11.4, G is p-projective.  We apply the existence of a p-projective Frattini p-cover of a profinite group to prove a converse to Corollary 22.11.2: Lemma 22.11.6: Let G be a profinite group. Suppose each finite p-embedding problem for G is weakly solvable. Then G is p-projective. Proof: Let (1)

(ϕ: G → A, α: B → A)

be an embedding problem for G with C = Ker(α) pro-p. First suppose C is finite. Choose an open normal subgroup N of B with ¯ = B/N , and C¯ = N C/N . Let α ¯ → A, ¯ N ∩ C = 1. Let A¯ = B/N C, B ¯: B ∼ ¯ ¯ ¯ πB : B → B, and πA : A → A be the quotient maps. Then Ker(α) ¯ =C=C and the square α / A B πB

 ¯b

πA

α ¯

 / A¯

is cartesian (Example 22.2.7(b)). By assumption, there exists a homomor¯ with α phism γ¯ : G → B ¯ ◦ γ¯ = πA ◦ ϕ. Hence, there exists a homomorphism γ: G → B such that α ◦ γ = ϕ. Now consider the general case. Suppose L and L0 are closed normal subgroups of B, L0 ≤ L ≤ C, and L0 is open in L. Then L/L0 is a finite p-group. By the first case, each embedding problem for G with kernel L/L0 is weakly solvable, so by Remark 22.3.3, (1) is weakly solvable. It follows that each cover of G with a pro-p kernel has a section. In parˆ → G (Construction ticular, this holds for the universal Frattini p-cover ϕ: ˆ G ˆ Since G ˆ is p-projective, 22.11.5). Thus, G is isomorphic to a subgroup of G. so is G (Lemma 22.11.3).  Lemma 22.11.7: Let G be a p-projective group. Then every p-embedding problem (2)

(ϕ: G → H, α: B → A)

22.12 Examples of Universal Frattini p-covers

531

in which α is a Frattini p-cover is solvable. Furthermore, each Frattini p-cover ψ: H → G is an isomorphism. Proof: By Corollary 22.11.2, there is a homomorphism γ: G → B with α ◦ γ = ϕ. Then α(γ(G)) = ϕ(G) = B, so γ(G) = B. Thus, γ solves embedding problem (2). In particular, there is an epimorphism γ: G → H with ψ ◦ γ = id. Therefore, ψ is an isomorphism.  Proposition 22.11.8: Let ϕ0 : G0 → G be a p-cover of a profinite groups. The following conditions are equivalent. (a) ϕ0 is a p-projective Frattini p-cover. (b) ϕ0 is a maximal Frattini p-cover. (c) ϕ0 is a minimal p-projective p-cover. Proof of “(a) =⇒ (b)”: Suppose η: H → G is a Frattini p-cover. Lemma 22.11.7 gives an epimorphism γ: G0 → H with η ◦ γ = ϕ0 . Hence, ϕ0 is a maximal Frattini p-cover. Proof of “(a) =⇒ (c)”: Suppose η: H → G is a p-projective p-cover. Lemma 22.11.7 gives an epimorphism γ 0 : H → G0 with ϕ0 ◦ γ 0 = η. Hence, ϕ0 is a minimal p-projective p-cover. ˆ → G be the p-projective Frattini p-cover Proof of “(b) =⇒ (a)”: Let ϕ: ˆ G that Construction 22.11.5 gives. The maximality of ϕ0 gives an epimorphism ˆ with ϕˆ ◦ γ = ϕ0 . Since both ϕ0 and ϕˆ are Frattini, so is γ (Lemma γ: G0 → G 22.5.4(c)). ˆ → G0 with ϕ0 ◦ γ 0 = ϕ. ˆ Lemma 22.11.7 gives an epimorphism γ 0 : G Again, by Lemma 22.5.4(c), γ 0 is Frattini. By Lemma 22.5.4(a), γ 0 ◦ γ is Frattini. By construction, Ker(γ 0 ◦ γ) ≤ Ker(ϕ0 ). Hence, Ker(γ 0 ◦ γ) is a pro-p group. Therefore, by Lemma 22.11.7, γ 0 ◦ γ is an isomorphism, so γ and γ 0 are isomorphisms. In particular, G0 is p-projective. Proof of “(c) =⇒ (a)”: Consider again the p-projective Frattini p-cover ˆ → G that Construction 22.11.5 gives. The minimality of ϕ0 gives an ϕ: ˆ G ˆ → G0 with ϕ0 ◦γ 0 = ϕ. ˆ By Lemma 22.5.4(c), ϕ0 is Frattini. epimorphism γ 0 : G  Remark 22.11.9: Universal Frattini p-cover. We call each cover ϕ0 : G0 → G satisfying the equivalent conditions of Proposition 22.11.8 a universal Frattini p-cover of G. Construction 22.11.5 establishes the existence of a universal Frattini p-cover of G. The proof of Proposition 22.11.8 shows it is unique. That is, given another universal Frattini p-cover ϕ00 : G00 → G there is an isomorphism γ: G0 → G00 with ϕ00 ◦ γ = ϕ0 . ˆ of Construction 22.11.5 as the Occasionally we refer also to the group G universal Frattini p-cover of G. 

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Chapter 22. Projective Groups and Frattini Covers

22.12 Examples of Universal Frattini p-covers The universal Frattini cover of a pro-p group P of rank e is Fˆe (p) (Corollary 22.7.8). It coincides with the universal Frattini p-cover of P . We generalize this observation and prove that if a finitely generated profinite group A with p - #A acts on P , then this action lifts to an action on Fˆe (p) and A n Fˆe (p) is the universal Frattini p-cover of A n P . Although Aut(P ) is a profinite group (Remark 16.10.5), it is not necessarily pro-p. For example, Aut(Z/pZ) = Z/(p − 1)Z. However, Aut(P ) contains an open normal pro-p subgroup. This follows from the following result: Lemma 22.12.1: Let P be a finitely generated pro-p group. Then K = {κ ∈ Aut(P ) | κ(g)Φ(P ) = gΦ(P ) ∀g ∈ P } is a pro-p group. Proof: Since Φ(P ) is an open characteristic subgroup of P , each κ ∈ Aut(P ) defines an automorphism κ ¯ ∈ Aut(P/Φ(P )) by κ ¯ (gΦ(P )) = κ(g)Φ(P ). The group K is the kernel of the homomorphism κ 7→ κ ¯ of Aut(P ) into Aut(P/Φ(P )). The rest of the proof naturally breaks up into two parts: Part A: P is finite. Choose generators x1 , . . . , xe of P . Let T = {(x1 f1 , . . . , xe fe ) | f1 , . . . , fe ∈ Φ(P )}. For each κ ∈ K and (x1 f1 , . . . , xe fe ) ∈ T there are f10 , . . . , fe0 ∈ Φ(P ) with κ(xi fi ) = xi fi0 , i = 1, . . . , e. By Lemma 22.1.1, x1 f1 , . . . , xe fe generate P , so if κ(x1 f1 , . . . , xe fe ) = (x1 f1 , . . . , xe fe ), then κ = 1. This means, K acts regularly on T . Thus, each K-orbit of T has the same number of elements as K. Denote the number of these orbits by k. Then |T | = k|K|. On the other hand, |T | = |Φ(P )|e is a p-power. Hence, K is a p-group, as claimed. Part B: The general case. Consider an open characteristic subgroup N of P . Denote reduction modulo N by a bar. Let g )Φ(P¯ ) = g¯Φ(P¯ ) ∀¯ g ∈ P¯ }. KN = {κ ∈ Aut(P¯ ) | κ(¯ By Part A, KN is a finite p-group. Each κ ∈ Aut(P ) defines κN ∈ Aut(P¯ ) by κN (¯ g ) = κ(g). Suppose g )Φ(P ) = g¯Φ(P ). By Lemma 22.1.4(c), Φ(P ) ≤ Φ(P¯ ). κ ∈ K. Then κN (¯ g )Φ(P¯ ) = g¯Φ(P¯ ). Therefore, κN ∈ KN . Hence, κN (¯ By Remark 16.10.5, the intersection of all N as above is trivial. Thus, the maps κ 7→ κN define an embedding of K into lim KN . The latter group ←− is pro-p. Consequently, K is a pro-p group. 

22.12 Examples of Universal Frattini p-covers

533

Proposition 22.12.2: Let P be a pro-p group of finite rank e and A a finitely generated profinite group. Suppose A acts on P and p - #A. Put G = A n P . Then the action of A on P lifts to an action of A on Fˆe (p) and ˆ = A n Fˆe (p) is the universal Frattini p-cover of G. G Proof: The action of A on P is defined by a homomorphism ψ: A → Aut(P ) with y ψ(a) = y a for each y ∈ P . Let B = ψ(A). Choose generators y1 , . . . , ye for P and generators x1 , . . . , xe for Fˆe (p). Extend the map xi 7→ yi , i = 1, . . . , e, to an epimorphism ϕ1 : Fˆe (p) → P . Consider b ∈ B. For each i choose x0i ∈ Fˆe (p) with ϕ1 (x0i ) = yib . Then lift b to a homomorphism ˆb: Fˆe (p) → Fˆe (p) satisfying xˆb = x0 . By definition, ϕ1 (Fˆe (p)ˆb ) = P . Since i i ˆ ϕ1 is a Frattini cover, Fˆe (p)b = Fˆe (p). By Proposition 16.10.6(a), ˆb is an ˆ automorphism of Fˆe (p). It satisfies, ϕ1 (xb ) = ϕ1 (x)b for each x ∈ Fˆe (p). In particular, ˆb leaves Ker(ϕ1 ) invariant. ˆ be the closed By Remark 16.10.5, Aut(Fˆe (p)) is a profinite group. Let B ˆ ˆ ˆ subgroup of Aut(Fe (p)) generated by all b. Then each β in B leaves Ker(ϕ1 ) ˆ there exists a unique β¯ ∈ Aut(P ) with invariant. Therefore, for each β ∈ B ¯ ¯ ϕ1 ◦ β = β ◦ ϕ1 . Let ϕ2 (β) = β. If β = ˆb for some b ∈ B, then β¯ = b. Thus, ˆ → B defined by ϕ2 (β) = β¯ is an epimorphism satisfying ϕ2 (ˆb) = b for ϕ2 : B each b ∈ B. Consider β ∈ Ker(ϕ2 ) and x ∈ Fˆe (p). Then, ϕ1 (xβ ) = ϕ1 (x)ϕ2 (β) = ϕ1 (x), so xβ Ker(ϕ1 ) = xKer(ϕ1 ). We have already noticed that ϕ1 : Fˆe (p) → P is a Frattini cover. Thus, Ker(ϕ1 ) ≤ Φ(Fˆe (p)), so xβ Φ(Fˆe (p)) = xΦ(Fˆe (p)). It follows that Ker(ϕ2 ) is contained in the pro-p group K of Lemma 22.12.1, so Ker(ϕ2 ) is a pro-p group. Since p - #A and therefore p - #B, SchurZassenhaus (Lemma 22.10.1) gives a section ϕ02 to ϕ2 . The homomorphism ϕ02 ◦ ψ: A → Aut(Fˆe (p)) defines an action of A on ˆ Fe (p) compatible with its action on P through ϕ1 . Construct the semidirect ˆ → G with ˆ = A n Fˆe (p) and extend ϕ1 to an epimorphism ϕ: G product G ˆ ϕ(Fe (p)) = P and ϕ(a) = a for each a ∈ A. ˆ Thus, G ˆ is Since p - #A, the group Fˆe (p) is the p-Sylow group of G. ˆ ˆ p-projective. In addition, Ker(ϕ) = Ker(ϕ1 ) ≤ Φ(Fe (p)) ≤ Φ(G) (Lemma ˆ → G is a universal 22.1.4(c)). Hence, ϕ is Frattini. Consequently, ϕ: G Frattini p-cover.  Examples 22.12.3: (a) Let p be an odd prime number. The dihedral group Dpn of order 2pn n is generated by two elements σ, τ with the defining relations σ p = 1, τ 2 = 1, and σ τ = σ −1 (Remark 21.7.1). Thus, Dpn = ±1 n Cpn , where Cpn is a multiplicative copy of Z/pn Z. By Proposition 22.12.2, the universal Frattini p-cover of Dpn is Dp∞ = ±1 n C, where C is a multiplicative copy of Zp . We denote the generator of ±1 by t and the image of 1 ∈ Zp in C by s. Then st = s−1 . (b) Proposition 22.12.2 has a global analog. Let G = A n B be a semidirect product of profinite groups with gcd(#A, #B) = 1. Then, the universal

534

Chapter 22. Projective Groups and Frattini Covers

˜ = A˜ n B, ˜ where A˜ and B ˜ are the universal Frattini Frattini cover of G is G covers of A and B, respectively [Ribes2, p. 392, Thm. 3.2]. 

22.13 The Special Linear Group SL(2, Zp ) Algebraic groups over Zp are a source of interesting examples of Frattini pcovers. We describe one example in detail and refer the reader to the general result: Remark 22.13.1: The group SL(2, Zp ). Consider the group SL(2, Zp ) of all 2 × 2 matrices with entries in Zp and determinant 1. For each positive integer n, reduction modulo pn defines an epimorphism πn : SL(2, Zp ) → n n SL(2, T∞ Z/p Z). Let Γn = Ker(πn ) = {s ∈ SL(2, Zpn) | s ≡ 1 mod p }. Since lim SL(2, Z/p Z). Thus, SL(2, Zp ) is a n=1 Γn = 1, we have SL(2, Zp ) = ←− profinite group.
Each element s of Γn has the form 1 + pn u with u = ac db in M (2, Zp ) (the additive group of 2 × 2 matrices with entries in Zp ). Then 1 = det(s) = 1 + pn (a + d) + p2n (ad − bc), so trace(u) = a + d ≡ 0 mod p. Hence, the map λ: 1 + pn u 7→ u mod p is a homomorphism of Γn onto the hypersurface sl(2, Fp ) = {¯ u ∈ M (2, Fp ) | trace(¯ u) = 0} of M (2, Fp ) whose kernel is Γn+1 . Note that dim(sl(2, Fp )) = 3. Thus, to prove that λ is surjective, it suffices to show that its image contains three linearly the following
three elements: u1 =
independent
elements. Consider 0 1 0 0 1 0 = = , u , and u 2 3 0 0 1 0 0 −1+bn+1 p+bn+2 p2 +··· where bn+1 , bn+2 , . . . are integers between 0 and p − 1 such that (1 + pn )−1 = 1 − pn + bn+1 pn+1 + bn+2 pn+2 + · · ·. Then       0 1 0 0 1 0 , λ(1 + pn u2 ) = , λ(1 + pn u3 ) = λ(1 + pn u1 ) = 0 0 1 0 0 −1 are linearly independent, as desired. It follows that Γn /Γn+1 ∼ = (Z/pZ)3 . 3 3(n−1) . Therefore, Γ1 = lim Γ1 /Γn Thus, (Γn : Γn+1 ) = p and (Γ1 : Γn ) = p ←− is a pro-p group. Consequently, π1 : SL(2, Zp ) → SL(2, Fp ) is a p-cover.  Proposition 22.13.2: Suppose p ≥ 5. Then the map π1 : SL(2, Zp ) → SL(2, Fp ) is a Frattini p-cover. Proof: Consider a closed subgroup H of SL(2, Zp ) which π1 maps onto SL(2, Fp ). By Remark 22.13.1, it suffices to prove that H = SL(2, Zp ). For this it suffices to prove that πn (H) = SL(2, Z/pn Z) for all n. Inductively assuming this holds for n, it suffices to consider s ≡ 1 mod pn in SL(2, Zp ) and to find h ∈ H with s ≡ h mod pn+1 . Write s = 1 + pn u with u ∈ M (2, Zp ). By Remark 22.13.1, trace(u) ≡ 0 mod p.

22.13 The Special Linear Group SL(2, Zp )

535

Claim: Each v ∈ sl(2, Fp ) is the sum of matrices with zero squares. Indeed, the square of each of the matrices  v1 =

0 0

1 0



 v2 =

0 1

0 0



 v3 =

1 1

−1 −1



is zero. Moreover, v1 , v2 , v3 are linearly independent, so they form a basis of sl(2, Fp ). Thus, each v ∈ sl(2, Fp ) is a sum of the vi ’s, each taken several times. Pk Using the claim, we may write u = i=1 ui with u2i ≡ 0 mod p, i = 1, . . . , k. Suppose we haveQfound hi ∈ H with 1 + pn uiQ≡ hi mod pn+1 , k k i = 1, . . . , k. Then h = i=1 hi ∈ H and 1 + pn u ≡ i=1 (1 + pn ui ) ≡ h mod pn+1 . Thus, we may assume u2 ≡ 0 mod p. The induction hypothesis, applied to 1 + pn−1 u, gives h0 ∈ H and v ∈ M (2, Zp ) with h0 = 1 + pn−1 u + pn v. Let h = hp0 . Then   p (pn−1 u + pn v)2 + · · · + (pn−1 u + pn v)p 2   p (u + pv)2 + · · · + pp(n−1) (u + pv)p . = 1 + pn (u + pv) + p2(n−1) 2

h = 1 + p(pn−1 u + pn v) +

First suppose n = 1. Then, using u2 ≡ 0 mod p, we find that (u + pv)2 ≡ p(uv + vu) mod p2 . Hence, (u + pv)p ≡ p2 (uv + vu)2 (u + pv)p−4 ≡ 0 mod p2 , because p ≥ 5. Thus, h ≡ 1 + pu mod p2 . If n ≥ 2, then 1 + 2(n − 1) ≥ n + 1. Therefore, h ≡ 1 + pn u mod pn+1 .  Corollary 22.13.3: Suppose p ≥ 5. Put  u=

1 0

1 1

 and

u0 =



1 1

0 1

 .

(a) Considered as elements of SL(2, Fp ), the matrices u and u0 have order p. Moreover, SL(2, Fp ) = hu, u0 i. (b) Considered as elements of SL(2, Zp ), the matrices u and u0 generate SL(2, Zp ).
Proof: By Proposition 22.13.2, it suffices to prove (a). Indeed, uk = 10 k1
and (u0 )k = k1 01 for every positive integer k. Thus, both u and u0 are of order p.
Multiplying a matrix s = ac db in SL(2, Fp ) on the left with uk (resp. (u0 )k ) amounts to adding the second line (resp. first line) multiplied by k to the first line (resp. second line). Assuming c 6= 0, three successive row
0
0
operations of this type map ac db to 1c bd , then to 10 b1 , and finally to
1 0 0 0 1 . Thus, s ∈ hu, u i.

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Chapter 22. Projective Groups and Frattini Covers

If c = 0, then a 6= 0. Adding the first line to the second one, we make c 6= 0. Then we proceed as before.  Dividing out SL(2, Fp ) by its center (which consists of the matrices ±1), produces the projective special linear group PSL(2, Fp ). When p ≥ 5, this group is simple [Lang7, p. 539, Thm. 8.4]. Likewise, PSL(2, Zp ) = SL(2, Zp )/ ± 1. These groups fit into the following commutative diagram: (1)

SL(2, Zp )

π1

γ

γ ˜

 PSL(2, Zp )

/ SL(2, Fp )

π ¯1

 / PSL(2, Fp ),

where both γ and γ˜ are the quotient maps with respect to ±1 and π ¯1 is induced by π1 . Corollary 22.13.4: Suppose p ≥ 5. Then each of the maps in diagram (1) is a Frattini cover. Proof: First note that γ is Frattini. Indeed, suppose H is a subgroup of SL(2, Fp ) which is mapped onto PSL(2, Fp ). Then (SL(2, Fp ) : H) ≤ 2. Each of the generators u and u0 of SL(2, Fp ) given by Corollary 22.13.3 has order p. Hence, u, u0 ∈ H. Thus, H = SL(2, Fp ), which proves our claim. By Proposition 22.13.2, π1 is a Frattini cover. Hence, γ ◦ π1 is Frattini (Lemma 22.5.4(a)). Therefore, by Lemma 22.5.4(c), both π ¯1 and γ˜ are Frattini.  Remark 22.13.5: Group schemes. Let G be a simple simply-connected affine group scheme. Then reduction modulo p gives a Frattini cover G(Zp ) → G(Fp ) for all p ≥ 3 except for SL(2, Z3 ). When p = 2, the map is Frattini in all but at most ten cases [Weigel4, Thm. B].  Example 22.13.6: A C-projective nonprojective group. Let C be a formation of finite groups. If C is full, then each C-projective group is projective (Lemma 22.4.1). If C is not full, this is not necessarily the case, as the following example shows: Let C be the Melnikov formation consisting of all finite groups whose composition groups are only A5 . By [Huppert, p. 183, Satz 6.14(3)], PSL(2, F5 ) ∼ = A5 . By Lemma 22.3.6, the free pro-C group F = Fˆ2 (C) on two generators is C-projective. By Lemma 22.13.3, PSL(2, F5 ) is generated by two elements. Let ϕ: F → PSL(2, F5 ) be a epimorphism. By Corollary 22.13.4, the quotient map γ: SL(2, F5 ) → PSL(2, F5 ) is a Frattini cover. Assume F is projective. Then there exists an epimorphism ψ: F → SL(2, F5 ) with γ ◦ ψ = ϕ (Proposition 22.5.9(a)). Hence, SL(2, F5 ) is a Cgroup, so Z/2Z ∼ = Ker(γ) is a C-group. This means that Z/2Z ∼ = A5 . We conclude from this contradiction that F is not projective. 

22.14 The General Linear Group GL(2, Zp )

537

22.14 The General Linear Group GL(2, Zp ) Although the p-cover π1 : SL(2, Zp ) → SL(2, Fp ) is Frattini if p ≥ 5 (Proposition 22.13.2), it is far from being a universal Frattini p-cover. If it were, Γ1 = Ker(π1 ) would be a free pro-p group of finite rank (indeed 3). By Nielsen-Schreier (Proposition 17.5.7), the rank of the open subgroups of Γ1 would tend to infinity with the index. This would contradict the result below saying the rank is bounded: Notation 22.14.1: Let n be a positive integer and G = GL(n, Zp ). For each positive integer k let Gk = {g ∈ G | g ≡T 1 mod pk }. Then Gk is an open ∞  normal subgroup of G, Gk+1 ≤ Gk , and k=1 Gk = 1. Lemma 22.14.2: Let p be an prime number and k a positive integer. Suppose k ≥ 2 if p = 2. Then: (a) Every element of Gk+1 is a pth power of an element of Gk . (b) The map λk : Gk /Gk+1 → Gk+1 /Gk+2 given by gGk+1 7→ g p Gk+2 for g ∈ Gk is an isomorphism. ¯, where a ¯ is the residue of a modulo p, is an (c) The map (1 + pk a)Gk+1 → a p isomorphism µk : Gk /Gk+1 ∼ = M (n, Fp ). Thus, Gk+1 = Gk and Gk /Gk+1 may be viewed as an Fp -vector space of dimension n2 . (d) Gk is a pro-p group, Φ(Gk ) = Gk+1 , and rank(Gk ) = n2 . Proof of (a): Consider a ∈ M (n, Zp ). We find x ∈ M (n, Zp ) with (1 + pk x)p = 1 + pk+1 a. To this end put x1 = a. Then (1 + pk x1 )p ≡ 1 + pk+1 a mod pk+2 . (Note the use of k ≥ 2 when p = 2.) Let r ≥ 1. By induction assume we have found xr ∈ M (n, Zp ) which commutes with a such that (1 + pk xr )p ≡ 1 + pk+1 a mod pk+r+1 . Thus, there is a b ∈ M (n, Zp ) with (1 + pk xr )p = 1 + pk+1 a + pk+r+1 b. Then b commutes with a. Put xr+1 = xr − pr (1 + pk xr )−p+1 b. Then xr+1 commutes with a and
p

(1 + pk xr+1 )p = 1 + pk xr − pk+r (1 + pk xr )−p+1 b

≡ (1 + pk xr )p − pk+r+1 (1 + pk xr )p−1 (1 + pk xr )−p+1 b ≡ (1 + pk xr )p − pk+r+1 b ≡ 1 + pk+1 a mod pk+r+2 . This completes the induction. The sequence xr converges to an element x ∈ M (n, Zp ) which commutes with a and satisfies (1 + pk x)p = 1 + pk+1 a, as needed. Proof of (b): Consider an element g = 1 + pk a of Gk with a ∈ M (n, Zp ). Then g p ≡ 1 + pk+1 a mod pk+2 (again, we use k ≥ 2 if p = 2), so there is a b ∈ M (n, Zp ) with g p = 1 + pk+1 a + pk+2 b = (1 + pk+1 a)(1 + pk+2 (1 + pk+1 a)−1 b) ∈ Gk+1 . If in addition, a ≡ 0 mod p, then g p ∈ Gk+2 , so λ is well defined.

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Chapter 22. Projective Groups and Frattini Covers

If λk (gGk+1 ) = 1, then a ≡ 0 mod p, so g ∈ Gk+1 . This means that λk is injective. Finally, by (a), λk is surjective. Proof of (c): The only nontrivial point is proving that µk is an isomorphism, is the multiplicativity of µk . This follows as in the proof of (b): (1 + pk a)(1 + pk b) = 1 + pk (a + b) + p2k ab ∈ (1 + pk (a + b))Gk+1 . p It follows from the isomorphism Gk /Gk+1 ∼ = M (n, Fp ) that Gk+1 ≤ Gk . p Combining this with (a), we get Gk+1 = Gk . 2 Proof of (d): For each l ≥ k, (c) implies Gl /Gl+1 ∼ = (Z/pZ)n , so Gk /Gl is a finite p-group. Thus, Gk = lim Gk /Gl is a pro-p group. Also, [Gk , Gk ] ≤ ←− Gk+1 . Hence, Φ(Gk ) = Gpk [Gk , Gk ] = Gk+1 [Gk , Gk ] = Gk+1 (use (c) and Lemma 22.7.4). Consequently,

rank(Gk ) = rank(Gk /Gk+1 ) = dimFp (Gk /Gk+1 ) = n2 (Corollary 22.5.3).



Lemma 22.14.3: Let P be a pro-p group and e a positive integer. Inductively define an ascending sequence of closed normal subgroups Pi of P by P1 = P and Pi+1 = Pip . For each i ≥ 1 suppose: (1a) Pi /Pi+1 ∼ = Fep . (1b) The map xPi+1 7→ xp Pi+2 is an isomorphism Pi /Pi+1 ∼ = Pi+1 /Pi+2 . Then rank(Q) ≤ e for every closed subgroup Q of P . Proof: Let N be an open normal subgroup of P . Then (P : N ) = pl for some T∞ T∞ l positive integer l. Hence, i=1 Pi ≤ P p ≤ N . It follows that i=1 Pi = 1. Let Q be a closed subgroup of P . Using compactness arguments, it suffices to prove rank(QPn /Pn ) ≤ e for each n. Dividing out P by P n , we may assume that P is finite, (1a) holds for i = 1, . . . , n − 1, and that (1b) holds for i = 1, . . . , n − 2. Put K = Q ∩ P2 and k = rank(K). Then Q/K ∼ = QP2 /P2 is a vector space over Fp (by (1a)) of dimension, say, d. Choose x1 , . . . , xd ∈ Q with Q = hx1 , . . . , xd iK. Then QP2 = hx1 , . . . , xd iP2 . Since P2 satisfies the same assumptions as P and has a smaller order, an induction hypothesis gives k ≤ e. Since P3 = P2p and P2 /P3 is Abelian, P3 = Φ(P2 ) (use Lemma 22.7.4). Hence, by Lemma 22.1.4, Φ(K) ≤ P3 . Let π: P/P2 → P2 /P3 be the isomorphism given by π(xP2 ) = xp P3 (use (1b)). Note that xpi ∈ Q ∩ P2 = K. In addition, the elements x1 , . . . , xd are linearly independent modulo K, so they are linearly independent modulo P2 . Hence, xp1 , . . . , xpd are linearly independent modulo P3 . Thus, xp1 , . . . , xpd are linearly independent modulo Φ(K). Therefore,
k = rank(K) = dimFp (K/Φ(K)) ≥ dimFp hxp1 , . . . , xpd iΦ(K)/Φ(K)
≥ dimFp hxp1 , . . . , xpd iP3 /P3
= dimFp π(hx1 , . . . , xd iP2 /P2 ) = dimFp (QP2 /P2 ) = dimFp (Q/K) = d.

Exercises

539

Also, there are y1 , . . . , yk−d ∈ K with K = hxp1 , . . . , xpd , y1 , . . . , yk−d iΦ(K). Then K = hxp1 , . . . , xpd , y1 , . . . , yk−d i, so Q = hx1 , . . . , xd iK = hx1 , . . . , xd , y1 , . . . , yk−d i. Consequently, rank(Q) ≤ k ≤ e, as claimed.



We are now in a position to prove that the rank of closed subgroups of GL(n, Zp ) is bounded: Proposition 22.14.4: Let m be maximum rank(A) with A ranging on all subgroups of GL(n, Fp ). Then rank(H) ≤ n2 + m for every closed subgroup H of GL(n, Zp ). Proof: Use Notation 22.14.1. By Lemmas 22.14.2 and 22.14.3, rank(H ∩ G1 ) ≤ n2 . Using GL(n, Zp )/G1 ∼ = GL(n, Fp ), we find that rank(H) ≤ rank(H ∩G1 )+rank(H/H ∩G1 ) ≤ n2 +rank(HG1 /G1 ) ≤ n2 +m, as claimed.  Remark 22.14.5: Characterization of p-adically linear profinite groups. The converse of Proposition 22.14.4 is also true: Each pro-p group P with a bound on the rank of its closed subgroups is isomorphic to a subgroup of GL(n, Zp ) for some n [Dixon-du.Sautoy-Mann-Segal, p. 155, Thm. 7.19]. Moreover, suppose a profinite group G has an open pro-p subgroup P whose closed subgroups have a bounded rank. Then the induced representation from P to G will embed G into GL(n0 , Zp ) with n0 possibly larger than n. See also [Lubotzky2, Thm. 1].  The group SL(2, Zp ) is a Frattini cover of PSL(2, Fp ) (Corollary 22.13.4), but as noted at the beginning of this section, it is not the universal Frattini cover of PSL(2, Fp ). Problem 22.14.6: Describe the universal Frattini p-cover of PSL(2, Zp ) (hence of PSL(2, Fp )).

Exercises 1. Use fiber products and Proposition 22.3.5 to prove the following: Let C be a full formation of finite groups and G a pro-C-group. Suppose every short exact pro-C sequence 1 → K → H → G → 1 in which K is a minimal Abelian p-elementary subgroup of H splits. Then G is C-projective. 2. Let θ: H → G be a Frattini cover of profinite groups. Prove that θ−1 (Φ(G)) = Φ(H). Deduce that if Φ(G) = 1, then G ∼ = H/Φ(H). 3. Let G and H be closed subgroups of a free profinite group. Prove that G/Φ(G) ∼ = H/Φ(H) implies G ∼ = H.

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Chapter 22. Projective Groups and Frattini Covers

4. Let F be a profinite group and G, H closed subgroups. Suppose for each pair α: G → C, β: H → C of homomorphisms of profinite groups there is a unique homomorphism γ: F → C which extends both α and β. Prove that G ∩ H = 1, hG, Hi = F , and there is an isomorphism F → G ∗ H whose restriction to both G and H is the identity map. Hint: To prove that F = hG, Hi, extend the inclusion maps of G and H into hG, Hi to a homomorphism γ: F → hG, Hi. Then compose each of the three maps with the inclusion hG, Hi → F . 5. Let n be a positive integer relatively prime to the order of a profinite group G. Prove that the map ϕn that takes g to g n for g ∈ G is surjective (although it may not be a homomorphism). Hint: ϕn induces a surjective map G/N → G/N for each open normal subgroup N of G. 6. Let C be a full family of finite groups. Suppose a prime p divides the order of a pro-C-group. Use the Sylow theorems to prove that every p-group belongs to C. 7. A celebrated theorem of Artin says: If the absolute Galois group Gal(K) of a field K is finite, then |Gal(K)| = 1 or 2 [Lang7, p. 299, Cor. 9.3]. Use the Sylow theorems to generalize this Q as follows: Let K be a field and #Gal(K) = pnp . Prove the following statements: (a) n2 = 0, 1, or ∞. (b) n2 = 1 only if char(K) = 0. (c) If p 6= 2, then np = 0 or ∞. 8. [Ershov2] suggests the following alternative proof to Douady’s theorem (Proposition 17.1.1): (a) Let P be a pro-p-group and m a cardinal number with P/Φ(P ) ∼ = Fm p (Lemma 22.7.1). Apply Corollary 22.5.10 to prove that there is an epimorphism π: Fˆm (p) → P . The image of a basis of Fˆm (p) is a set of generators of P that converges to 1. (b) Let G be a profinite group. For each prime number p choose a p-Sylow group Gp of G. Prove that the collection of Gp generates G. (c) For eachSp use (a) to choose a set of generators Sp converging to 1. Prove that S = Sp is a set of generators of G which converges to 1. 9. (Haran) Let ϕ: G → B and ψ: G → C be epimorphisms of profinite groups. Suppose that Φ(B) = 1 and that ψ is a Frattini cover. Apply Lemma 22.5.5 to show there exists a unique epimorphism π: C → B such that π ◦ ψ = ϕ. 10. [Lim] With A and B subsets of a profinite group G, denote the closed subgroup generated by all commutators [a, b] = a−1 b−1 ab with a ∈ A and b ∈ B by [A, B]. The lower central series of G is the descending sequence of closed characteristic groups G = G1 ≥ G2 ≥ G3 ≥ . . . defined inductively by Gi+1 = [Gi , G], i = 1, 2, . . . . Prove the equivalence of the following three conditions (Proposition 22.9.3):

Exercises

541

(a) G is pronilpotent. (b) G is the direct product of its p-Sylow groups. T∞ (c) i=1 Gi = 1. Hint: Show that if ϕ: G → N is an epimorphism, then ϕ([A, B]) = [ϕ(A), ϕ(B)] for all subgroups A and B of G. Then use the equivalence of the conditions for G finite. 11. [Ershov2, p. 361]) Call a Melnikov formation C of finite groups admissible if C is closed under taking finite Frattini covers. Prove: (a) Every full family D of finite groups is admissible. (b) For each e ∈ N, the family of all D ∈ D of rank at most e is admissible. Hint: Use Fˆe (D) to obtain D with rank(D) ≤ e. ˆ with 12. Let g be an element of a profinite group G and α an element of Z p-component α(p) = 0 for each prime p that divides #g = #hgi. Prove that g α = 1 (Exercise 12 of Chapter 1). Q∞ 13. Let G = i=1 Gi be a direct product of profinite groups of relatively prime orders. Let N be a closed Q∞ normal subgroup of G. Prove that N = Q ∞ ∼ (N ∩ G ) and G/N = i i=1 i=1 Gi /(N ∩ Gi ) are again direct products of profinite groups of relatively prime orders. Q∞ Hint: If N is open, there exists a positive integer k with i=k+1 Gi ≤ N . Therefore, each n ∈ N can be presented Q∞ as a product n = g1 . . . gk g with gi ∈ N ∩ Gi , i = 1, . . . , k, and g ∈ i=k+1 Gi . For each i with 1 ≤ i ≤ k ˆ with component αi (p) = 1 if p|#Gi and αi (p) = 0 consider the element αi ∈ Z αi otherwise. Then gi = n (Exercise 12). 14. Let S be a non-Abelian simple group and let S˜ be its universal Frattini cover. (a) Observe that a nontrivial profinite group H is a quotient of S˜ if and only if there exists a Frattini cover π: H → S such that Ker(π) = Φ(H) (Lemma 22.6.3). (b) Deduce from (a) that each closed normal subgroup N of H, with N 6= H, is contained in Ker(π). In particular, H = [H, H] is its own commutator. 15. Let G be a profinite group and H a closed subgroup. Prove that the universal Frattini cover of H is isomorphic to a closed subgroup of the universal Frattini cover of G. The same applies to the universal Frattini p-covers of G ˜ → G be the universal Frattini cover of G. Use Lemma and H. Hint: Let ϕ: ˜ G ˜ → H is a ˜ of ϕ˜−1 (H) such that res ˜ : H 22.5.6 to find a closed subgroup H H ˜ is isomorphic to the universal Frattini cover of H. Frattini cover. Then H 16. (Efrat-V¨olklein) Let G be an infinite profinite group. Suppose each nontrivial closed subgroup of G is open. Follow the following steps to prove that G ∼ = Zp for some prime number p: (a) Use the Sylow theorems to prove that G is a pro-p-group for some p.

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(b) Reduce to the case where G has a normal open subgroup N isomorphic to Zp . (c) Choose g ∈ G r N and prove that g acts trivially on N . (d) Then prove G = hgi ∼ = Zp .

Notes Our earliest source material for projective groups is [Gruenberg]. Investigation of PAC fields from 1980 on brought refinement to the study of projective groups. A family C of finite groups is almost full if C is closed under taking quotients, direct products, and subgroups but not necessarily under forming extensions. For example, the families of finite Abelian groups and finite nilpotent groups are almost full, but neither full nor Melnikov. Nevertheless, the theory of almost full families, as appears in [Fried-Jarden3], is not very exciting. In contrast, the theory of Melnikov formation is rather rich. We return to the latter theory in Chapter 25. There is a vast literature on free products of profinite groups. We mention here only a few results: Let C be a full formation of profinite groups. QC Denote the free product of finitely many pro-C groups Gi , i ∈ I, by ∗ i∈I Gi . This is the unique pro-C group G containing isomorphic copies of Gi satisfying: Given homomorphisms αi of Gi into a pro-C group C, there is a unique homomorphism α: G → C extending each αi . Then Ggi ∩ Gj 6= 1 for g ∈ G implies i = j and g ∈ Gi [Herfort-Ribes, Thm. B’]. More on free products of finitely many pro-C groups can be found in [Ribes-Zalesskii, §9.1]. Q [Binz-Neukirch-Wenzel] defines a free pro-C product G = ∗ i∈I Gi of groups converging to 1 for full families C and proves that each open subgroup of G is again a free pro-C product of groups converging to 1. Finally, [Haran3] and [Melnikov4] define free products of groups indexed by profinite spaces, that is inverse limits of discrete finite spaces. Proposition 22.5.9(b) is called “Jarden’s lemma” in [Matzat2, p. 231]. The proof that appears in [Matzat2], is a simplified version of the proof of Proposition 22.5.11. We deduce Proposition 22.5.9(b) from Proposition 22.5.8. The latter appears in a less explicit form in [Malle-Matzat, p. 270, Thm. 1.8], where it is attributed to Ikeda and Nobusawa. Several authors have independently developed the theory of Frattini covers. We mention here [Cossey-Kegel-Kovacs], [Ershov-Fried], [Ershov2], [Haran-Lubotzky], and [Cherlin-v.d.Dries-Macintyre]. A cohomological proof that a closed subgroup of a projective group is projective (Corollary 22.4.7) can be found in [Ribes, p. 204]. [Cossey-KegelKovacs] use wreath products to prove this. PAC fields provide still another proof (Exercise 1 of Chapter 23). Likewise, [Ribes, p. 235, Thm. 6.5 and p. 211, Prop. 3.1] gives cohomological proofs to Propositions 22.10.4 and 22.11.1, respectively. Our proof

Notes

543

is based on Brandis’ non-cohomological proof of Gasch¨ utz theorem (Lemma 22.10.2). It is traditional to call Lemma 22.10.1 after Schur and Zassenhaus although they proved it only in the finite case where either H or G/H is solvable. For an arbitrary finite group the proof requires the Feit-Thompson theorem that every finite group of odd order is solvable. Proposition 22.11.1 is applied in [Kuhlmann-Pank-Roquette] to split short exact sequences of Galois groups over Henselian fields: Let (K, v) be a Henselian field. Let Gu = {σ ∈ Gal(K) | v(σx − x) ≥ 0 for each  x ∈

Ks with v(x) ≥ 0} be the inertia group and Gr = {σ ∈ Gal(K) | v σx x −  1 > 0 for all x ∈ Ks× } the ramification group of Gal(K). Denote the

fixed field of Gu (resp. Gr ) in Ks by Ku (resp. Kr ). Then the following two exact sequences split: 1 −→ Gal(Kr /Ku ) −→ Gal(Kr /K) −→ Gal(Ku /K) −→ 1 1 −→ Gal(Kr ) −→ Gal(K) −→ Gal(Kr /K) −→ 1 Frattini p-covers are discussed in [Fried13, Part II, p. 126]. Proposition 22.12.2 with outline of the proof is mentioned on [Fried13, p. 134]. Corollary 22.13.4 has special significance when p = 5. Then PSL(2, F5 ) ∼ = A5 [Huppert, p. 183, Satz 6.14]. Thus, PSL(2, Z5 ) is a Frattini 5-cover of A5 . The universal Frattini 5-cover of A5 is discussed in [Fried13, pp. 129–136].

Chapter 23. PAC Fields and Projective Absolute Galois Groups The absolute Galois group of a PAC field is projective (Theorem 11.6.2). But not every field with projective absolute Galois group is PAC (Example 23.1.4). Nevertheless, for each projective absolute Galois group G there is a PAC field K with Gal(K) ∼ = G (Corollary 23.1.2). Moreover, if rank(G) ≤ ℵ0 , we may choose K to be algebraic over Q (Theorem 23.2.3). It follows that the elementary theory of PAC fields of characteristic 0 is determined by the PAC fields which are algebraic over Q (Corollary 23.2.6). Sections 23.3 -23.7 treat the theory of perfect PAC fields of corank bounded by a positive integer e that contain a fixed countable Hilbertian ˜ 1 , . . . , σe ) which field K. Although the models of this theory of the form K(σ e are not e-free are indexed by a set σ ∈ G(K) of measure zero, they nevertheless determine the theory (Theorem 23.6.3). In analogy to the theory of perfect e-free PAC fields containing K (Section 20.5), each sentence is equivalent to a basic sentence modulo the theory of perfect PAC fields of corank ≤ e (as in Proposition 20.6.6). If K has elimination theory, this leads to a recursive decision procedure for the theory. Frattini covers play a decisive role in the proofs.

23.1 Projective Groups as Absolute Galois Groups We characterize absolute Galois groups of PAC fields among all profinite groups as projective groups: Theorem 23.1.1: Let L/K be a Galois extension, G a projective group, and α: G → Gal(L/K) an epimorphism. Then K has an extension E which is perfect, PAC, and linearly disjoint from L, and there exists an isomorphism γ: Gal(E) → G such that α ◦ γ = resL . Proof: Proposition 1.3.3 gives a Galois extension F1 /E1 with the following properties: E1 is an extension of K, linearly disjoint from L, F1 is an extension of L, and there is an isomorphism γ1 : Gal(F1 /E1 ) → G with α ◦ γ1 = resF1 /L . Let E2 be a regular extension of E1 which is PAC (Proposition 13.4.6). Then resF1 E2 /F1 : Gal(F1 E2 /E2 ) → Gal(F1 /E1 ) is an isomorphism. In particular, Gal(F1 E2 /E2 ) is a projective group. Remark 22.4.2 gives an embedding θ: Gal(F1 E2 /E2 ) → Gal(E2 ) such that resE2,s /F1 E2 ◦ θ is the identity on Gal(F1 E2 /E2 ). Let E be the maximal purely inseparable extension of the fixed field in E2,s of θ(Gal(F1 E2 /E2 )). Put γ = γ1 ◦ resE/F ˜ 1. Then γ: Gal(E) → G is an isomorphism and α ◦ γ = resE/L . In particular, ˜ resE/L : Gal(E) → Gal(L/K) is an epimorphism, so E and L are linearly ˜ disjoint over K. Since E is an algebraic extension of E2 , it is PAC (Corollary 11.2.5). 

23.1 Projective Groups as Absolute Galois Groups

545

The special case L = K gives a converse to Ax’ theorem 11.6.2: Corollary 23.1.2 ([Lubotzky-v.d.Dries, p. 44]): Given a projective group G and a field K, there is an extension F of K which is perfect and PAC with Gal(F ) ∼ = G. The combination of Theorem 11.6.2 and Corollary 23.1.2 leads to the already announced characterization of absolute Galois groups of PAC fields among all profinite group: Corollary 23.1.3: A profinite G is projective if and only if G is isomorphic to the absolute Galois group of a PAC field K. Example 23.1.4: Projectivity of the Galois group does not imply PAC. For ˆ is a projective group (Corollary 22.4.5), each prime number p, Gal(Fp ) ∼ =Z  but Fp is not a PAC field (Proposition 11.1.1). Remark 23.1.5: On the characterization of absolute Galois groups. Let G be a profinite group. By Leptin (Corollary 1.3.4), G is always isomorphic to a Galois group Gal(L/K) for some Galois extension L/K. However, G is not always an absolute Galois group. For example, by Artin-Schreier, if G is finite, then G is an absolute Galois group if and only if G is trivial, or G∼ = Z/2Z [Lang7, p. 299, Cor. 9.3]. Corollary 23.1.3 is another example of characterizing a class of projective groups as a class of absolute Galois groups of fields with a given arithmetic property (namely PAC). More examples of this nature are given in [Haran-Jarden2] (PRC versus real projectivity) and in [Haran-Jarden4] (PpC versus p-projectivity) and various generalizations thereof. However, the question of characterizing absolute Galois groups by topological and group theoretic means is still wide open.  Here is a model theoretic application of Theorem 23.1.1: Theorem 23.1.6: Let θ be a sentence of L(ring) and e a positive integer. Then: (a) θ is true in every perfect PAC field of infinite corank if and only if θ is true in every perfect PAC field. (b) θ is true in every perfect PAC field of finite corank at least e if and only if θ is true in every perfect PAC field of finite corank. Proof of (a): Suppose that θ is true in every perfect PAC field of infinite corank and let K be a perfect PAC field of finite corank. We prove that θ is true in K. By Theorem 11.6.2, Gal(K) is projective. By Corollary 22.4.5, Fˆω (p) is projective for each p. Therefore, by Proposition 22.4.10, the free product Gal(K) ∗ Fˆω (p) is projective. Since Gal(K) is a quotient of Gal(K) ∗ Fˆω (p), Theorem 23.1.1 produces a perfect PAC field Lp , regular over K, with Gal(Lp ) ∼ ∗ Fˆω (p). = Gal(K) Q Let L = Lp /D be a nonprincipal ultraproduct of the fields Lp . Then L is a regular perfect PAC extension of K. Moreover, every finite quotient H of

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Chapter 23. PAC Fields and Projective Absolute Galois Groups

Gal(K) is a Galois group over each of the fields Lp , so H is a Galois group over L (Remark 20.4.5(d)). Conversely, if H is a finite quotient of Gal(L), then H is a finite quotient of Gal(Lp )) for infinitely many p (Corollary 7.7.2). Choose p larger than |H| to find an epimorphism Gal(K)∗ Fˆω (p) → H with the factor Fˆω (p) in the kernel. Thus, H is a quotient of Gal(K). Therefore, Gal(K) and Gal(L) have the same finite quotients. Since Gal(K) is finitely generated, Gal(K) ∼ = Gal(L) (Proposition 16.10.7(b)). Since L is a regular extension of K, the restriction map of Gal(L) is an epimorphism onto Gal(K). Hence, by Proposition 16.10.6(a), it is an isomorphism. It follows from Corollary 20.3.4 that L is an elementary extension of K. By assumption, θ is true in each of the fields Lp . Hence, θ is true in L. Therefore, θ is true in K. Proof of (b): Same as the proof of (a) except, choose Lp with Gal(Lp ) ∼ = Gal(K) ∗ Fˆe (p).  Remark 23.1.7: In contrast to Theorem 23.1.6, the theory of PAC fields does not coincide with the theory of PAC fields of finite corank. Remark 28.10.3 gives an example of a sentence θ which holds in every perfect PAC field of finite corank, but does not hold in every perfect PAC field. 

23.2 Countably Generated Projective Groups For our model theoretic applications it suffices to consider only projective groups of rank ≤ ℵ0 . This section realizes each such group as the absolute Galois group of a perfect PAC field which are algebraic over a given countable Hilbertian field K. We conclude that the perfect PAC fields algebraic over K determine the elementary theory of all perfect PAC fields over K. Every finite embedding problem over K has a solution over a finite extension of K. This is the content of the following result: Lemma 23.2.1 ([Kuyk1, Thm. 3]): Let K be a Hilbertian field, L a finite Galois extension, G a finite group, and α: G → Gal(L/K) an epimorphism. Then there is a tower K 0 ⊆ L0 ⊆ M 0 of Galois extensions with the following properties: K 0 /K is a finite separable extension, L ∩ K 0 = K, LK 0 = L0 , and there is an isomorphism β: Gal(M 0 /K 0 ) → G with α ◦ β = resM 0 /L . Proof: Embed G in Sn for some positive integer n. Lemma 16.2.7 gives a Galois extension M1 /K with Gal(M1 /K) = Sn and M1 ∩ L = K. Put M 0 = LM1 . Then, Gal(M 0 /K) ∼ = Gal(L/K) × Sn . Let G∗ be the subgroup 0 of Gal(M /K) consisting of the pairs (σ, α(σ)) for σ ∈ G. The projection (σ, α(σ)) 7→ σ, for σ ∈ G, gives an isomorphism β: G∗ → G. Let K 0 be the fixed field of G∗ in M 0 and L0 = LK 0 . Then G∗ = Gal(M 0 /K 0 ) and restriction of elements of G∗ to L is given by projection of (σ, α(σ)) ∈ G∗ onto the second factor. Consequently, α ◦ β = resM 0 /L . 

23.2 Countably Generated Projective Groups

547

Proposition 23.2.2: Let K be a countable Hilbertian field, L a finite Galois extension of K, G a profinite group of rank at most ℵ0 , and α: G → Gal(L/K) a homomorphism. Then there are a PAC field E, separably algebraic over K, a Galois extension F of E, and an isomorphism β: Gal(F/E) → G with a commutative diagram Gal(F/E) u β uuu res u u uu  u zu / Gal(L/K). G α

Proof: Example 17.1.7(a) gives a sequence α

α

α

2 1 0 · · · −→ G2 −→ G1 −→ G0

of finite groups and epimorphisms with G0 = α(G), G = lim Gi and α = ←− lim αi . Order the nonconstant absolutely irreducible polynomials of K[T, X] ←− which are separable in X in a sequence f1 , f2 , f3 , . . . . Apply Lemma 23.2.1 to inductively construct a sequence of finite Galois extensions, Ln /Kn , with the following properties: (1a) K0 is the fixed field in L of α(G) and L0 = L . (1b) Kn is a finite separable extension of K with a zero of fn . (1c) Kn ⊆ Kn+1 , Ln ⊆ Ln+1 , Ln ∩ Kn+1 = Kn . (1d) There exists a commutative diagram Gal(Ln+1 /Kn+1 ) βn+1

res

/ Gal(Ln /Kn ) βn



Gn+1

αn

 / Gn

in which βn and βn+1 are isomorphisms. S∞ The union E = n=0 Kn is a separableSalgebraic extension of K. By ∞ Theorem 11.2.5 it is a PAC field. Also, F = n=0 Ln E is a Galois extension of E and Ln ∩ E = Kn , n = 0, 1, 2, . . . . Thus, the diagram of (1d) induces a commutative diagram Gal(Ln+1 E/E) β˜n+1

res

β˜n



Gn+1

/ Gal(Ln E/E)

αn

 / Gn

in which β˜n and β˜n+1 are isomorphisms. Take β to be the inverse limit of  {β˜n }.

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Chapter 23. PAC Fields and Projective Absolute Galois Groups

If G is projective, we may apply Remark 22.4.2 to the epimorphism res: Gal(E) → Gal(F/E) and then replace E by the maximal purely inseparable extension of the fixed field of the image of Gal(F/E) in Es to get the following result: Theorem 23.2.3: Let K be a countable Hilbertian field, L a finite Galois extension of K, G a profinite group of rank at most ℵ0 , and α: G → Gal(L/K) a homomorphism. Then there exists a perfect PAC field E, algebraic over . K, and an isomorphism β: Gal(E) → G with α ◦ β = resE/L ˜ Now, for the model theoretic applications: Proposition 23.2.4: Let K be a countable Hilbertian field and F a countable perfect PACQextension of K. Then F is K-elementarily equivalent to ∞ an ultraproduct n=1 En /D of perfect PAC fields with En algebraic over K and Gal(En ) ∼ = Gal(F ) for each n ∈ N. Proof: Let L1 ⊆ L2 ⊆ L3 ⊆ · · · be an ascending sequence of finite Galois extensions of K whose union is Ks . For each n, Theorem 23.2.3 gives a perfect PAC field En , algebraic over K, and an isomorphism βn which makes the following diagram commutative, n = 1, 2, . . .: Gal(En ) q q q βn q res qqq q q xq  Gal(F ) res / Gal(Ln /K) Q Let D be a nonprincipal ultraproduct of N. Put E ∗ = En /D and F ∗ = F N /D. Lemma 20.3.1 gives an isomorphism β which makes the following diagram commutative: Gal(E ∗ ) r β rrr r res r r yrrr  Gal(F ∗ ) α / Gal(K) Since E ∗ and F ∗ are perfect PAC fields, the elementary equivalence theorem  (Theorem 20.3.3) implies E ∗ ≡K F ∗ . Consequently, E ∗ ≡K F . Theorem 23.2.5: Let K be a countable Hilbertian field and P a class of projective groups with this property: If E and F are elementary equivalent PAC fields and Gal(F ) ∈ P, then Gal(E) ∈ P. Then a sentence θ of L(ring, K) is true in all perfect PAC fields F with K ⊆ F and Gal(F ) ∈ P if and only if θ is true in all perfect PAC fields E, algebraic over K, with Gal(E) ∈ P. Proof: Let F be a perfect PAC field containing K with Gal(F ) ∈ P. By Skolem-L¨owenheim (Proposition 7.4.2), F has a countable Q elementary subEi /D with Ei a field F0 that contains K. By Proposition 23.2.4, F0 ≡K

23.3 Perfect PAC Fields of Bounded Corank

549

perfect PAC field, algebraic over K, and Gal(Ei ) ∼ = Gal(F0 ), for each i. By assumption, Gal(Ei ) ∈ P. Hence, θ is true in Ei for each i. Therefore, θ is true in F .  Corollary 23.2.6: Let K be a countable Hilbertian field. Then a sentence θ of L(ring, K) is true in each perfect PAC field which contains K if and only if θ is true in each perfect PAC field which is algebraic over K.

23.3 Perfect PAC Fields of Bounded Corank Throughout the rest of this chapter, K is a fixed countable Hilbertian field and e is a fixed positive integer. We introduce four classes of fields: PAC(K, ≤ e), the class of perfect PAC fields of corank ≤ e that contain K; PAC(K, e), the class of perfect PAC fields of corank e that contain K; ˆ e }, and NF(K, e) = {E ∈ PAC(K, e) | Gal(E) 6= F FMP(K, e) = {E ∈ PAC(K, e) | only finitely many primes divide #Gal(E)}. Since each projective group is the absolute Galois group of a perfect PAC field that contains K (Corollary 23.1.2), (1)

FMP(K, e) ⊂ NF(K, e) ⊂ PAC(K, e) ⊂ PAC(K, ≤ e).

The first three classes of fields in (1) have no characterization by axioms of L(ring, K) (i.e. they are nonelementary). Indeed, for each prime number p consider the projective group Fˆe (p) (Corollary 22.4.5) and choose a perfect PAC field QEp that contains K with Gal(Ep ) ∼ = Fˆe (p) (Corollary 23.1.2). / Let E = Ep /D. Then Ep ∈ FMP(K, e) and Gal(E) is trivial, so E ∈ PAC(K, e). By Corollary 7.7.2, a sentence of L(ring, K) which holds in a sequence of fields holds in each ultraproduct of those fields. It follows that none of the classes FMP(K, e), or NF(K, e), or PAC(K, e) is elementary. Nevertheless, PAC(K, ≤ e) is elementary. Thus, there is a set of axioms ∆(K, ≤ e) of L(ring, K) such that for each structure E of L(ring, K), E |= ∆(K, ≤ e) if and if E ∈ PAC(K, ≤ e). Indeed, by Proposition 20.4.4(a), the PAC axioms are elementary. Next, for each positive integer n list the finite groups of order at most n! of rank at most e as Gn,1 , . . . , Gn,r(n) . Then include in ∆(K, ≤ e) all statements saying that for each separable polynomial f ∈ K[X] of degree n the Galois group Gal(f, K) is isomorphic to Gn,i for some i between 1 and r(n). By Remark 20.4.5(d) each of these statements is elementary. If K is an explicitly given field with elimination theory (Section 19.2), then ∆(K, ≤ e) can be explicitly given. As previously, if M is a class of structures for a given language L, denote the theory of M in L by Th(M). This is the set of sentences of L which are true in every structure in M. Applying this functor to the tower (1) we get (2) Th(PAC(K, ≤ e)) ⊆ Th(PAC(K, e)) ⊆ Th(NF(K, e)) ⊆ Th(FMP(K, e)). Theorem 23.6.3 shows all four sets in (2) coincide. Moreover, if K is an explicitly given field with elimination theory, Th(PAC(K, ≤ e)), and therefore each of these theories, is decidable.

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Chapter 23. PAC Fields and Projective Absolute Galois Groups

23.4 Basic Elementary Statements Let K be a countable Hilbertian field, e a positive integer, and E, F perfect ˜ = F ∩K. ˜ A sufficient PAC field extensions of K of corank at most e with E∩K condition for E and F to be K-equivalent is the existence an isomorphism ϕ: Gal(F ) → Gal(E) such that resF˜ /K˜ = resE/ ˜ K ˜ ◦ ϕ (Theorem 20.3.3). To establish the existence of ϕ, it suffices to prove that a finite group A of rank at most e is a Galois group over E if and only if A is a Galois group over F . The proof of the second condition on ϕ is much more difficult, one has to prove that the same embedding problems over finite extensions of K for Gal(E) and Gal(F ) are solvable. We show that the solvability of embedding problems can be expressed by elementary statements and establish a procedure to check the solvability of embedding problems over PAC fields. Let L be a finite separable extension of K and h ∈ L[X] a monic irreducible separable Galois polynomial of degree m. Let A be a subgroup of Gal(h, L) (again, considered as a permutation group) and α: B → A an α epimorphism of finite groups. Let θL (h, B −→ A) be the following statement about a field E that contains K: (1a) There is a K-embedding of L in E (whose image we identify with L); (1b) Gal(h, E) = A; (1c) there is a monic irreducible normal separable polynomial g ∈ E[X] of degree |B| whose splitting field over E contains the splitting field of h; and (1d) there is an isomorphism β: Gal(g, E) → B such that the following diagram is commutative (2)

/B

β

Gal(g, E) HH HH H res HHH H$

A



α

α

Remark 23.4.1: Interpretation of θL (h, B −→ A) as an elementary statement. Condition (1a) which implies that Gal(h, E) is a subgroup of Gal(h, L), gives sense to (1b). Let L1 be the splitting field of h over L, and let L0 be the fixed field of A (regarded as a subgroup of Gal(L1 /L)) in L1 . Then (1b) implies L1 ∩ E = L0 . We produce a sentence θ in the language L(ring, K) such that α (3) E |= θ if and only if θL (h, B −→ A) is true in E, for each field extension E of K. Suppose L = K(x0 ) and p(X) = irr(x0 , K). Then (1a) is equivalent to “p(X) has a root x in E.” Identify x with x0 . Working in L[X] modulo p(X), we may interpret L(ring, L) in L(ring, K). Thus, it suffices to produce θ in L(ring, L).

23.4 Basic Elementary Statements

551

For (1b), view A as a subgroup of Sm and use the proof of Proposition 20.4.4 to rewrite “Gal(h, E) ∼ = A” as an elementary statement ϕ(x) about E, where ϕ(X) is a formula in L(ring, L) with the free variable X. The coefficients of h are expressed as polynomials in x of degree less than deg(p). In order to interpret (1c) in L(ring, L), consider B as a subgroup of Sn with n = |B|. Then the conjunction of (1c) and (1d) is equivalent to the following statement: “there exists a monic irreducible separable Galois polynomial g(X) of degree n in E[X] with Galois group B and roots z1 , . . . , zn in Ks satisfying this: if y1 , . . . , ym are the roots of h, then there exists a polynomial q ∈ E[X] of degree at most n − 1 with y1 = q(z1 ) and yα(τ )(1) = q(zτ (1) ) for each τ ∈ B.” Finally, use congruences modulo g and h, respectively, to eliminate reference to the elements z1 , . . . , zn and y1 , . . . , ym in the resulting elementary statement.  α

Call θL (h, B −→ A) a basic statement. If rank(B) ≤ e, call it an ebasic statement. Denote the Boolean algebra in L(ring, K) generated by all e-basic statements by Be . Remark 23.4.2: Interpretation of embedding problems as elementary statements. Let h ∈ K[X] be a monic irreducible separable normal polynomial, B a finite subgroup and α: B → Gal(h, K)

(4)

α

an epimorphism. Remark 23.4.1 interprets θK (h, B −→ Gal(h, K)) as an elementary statement in L(ring) with parameters in K, namely the coefficients α of h. By (1c) and (1d), θK (h, B −→ Gal(h, K)) is true in K if and only if the embedding problem (4) for Gal(K) is solvable. Consequently, solvability of embedding problems over K is elementary.  The next lemma interprets the truth of special statements belonging to Be in some perfect PAC field in terms of Frattini covers. Section 23.5 shows that any Boolean combination of basic statements is equivalent modulo Th(PAC(K, ≤ e)) to one of these special statements (given by (5)). Lemma 23.4.3: Let α

θL (h, B −→ A)

α0

i and θL (h, Bi −→ A),

i = 1, . . . , r,

be e-basic statements and θ the statement (5)

α

θL (h, B −→ A) ∧

r ^ i=1

Consider also the following assertion:

α0

i ¬θL (h, Bi −→ A).

552

Chapter 23. PAC Fields and Projective Absolute Galois Groups

(C) For no i, 1 ≤ i ≤ r, does there exist a commutative diagram ϕ0

ϕi /B ¯i o i Bi0 B@ @@ } @@ }} ψ }} @ 0 α @@ }   ~}} αi A

(6)

in which ϕ0i is a Frattini cover and ϕi and ψ are epimorphisms. Then: (a) If there exists a perfect PAC field E, with K ⊆ E, in which θ is true, then (C) holds. (b) If (C) holds, then there exists E ∈ FMP(K, e) in which θ is true. Proof of (a): Suppose there is a perfect PAC field E which contains K in which θ is true. Let M be the splitting field of h over E. Then A = Gal(h, E) = Gal(M/E). By (1), E has a finite Galois extension F that contains M and there exists a commutative diagram /B

β

(7)

Gal(F/E) II II II res III I$

A



α

Assume for some i, there exists a commutative diagram (6) in which ϕ0i is a Frattini cover and ϕi and ψ are epimorphisms. Combine (6) with (7) to obtain the following commutative diagram (8)

Gal(E) res

 ϕ0 β / B ϕi / B ¯i o i Bi0 Gal(F/E) @@ RRR } @ RRR RRR @@α@ ψ }}}} RRR @@ 0 res } RRR  ~}} αi ) A Since E is a PAC field, Gal(E) is projective (Theorem 11.6.2). Also, ϕi ◦β◦res is an epimorphism. Since ϕ0i is a Frattini cover, there is an epimorphism ρ: Gal(E) → Bi0 that completes (8) to a commutative diagram (Proposition 22.5.9(a)). The fixed field, Fi0 , of Ker(ρ) is therefore a finite Galois extension of E that contains M , and the isomorphism βi0 : Gal(Fi0 /E) → Bi0 induced by ρ makes the following diagram commutative Gal(E)

res

0

βi / Gal(F 0 /E) i J JJ JJ J res JJJ J$

A

/ B0 i ~ ~ ~ ~~ 0 ~~ αi

23.4 Basic Elementary Statements

553

α

That is, θL (h, Bi0 −→ A) is true in E. This contradicts the assumption that θ is true in E. Proof of (b): Conversely, suppose (C) holds. Let p be a prime number which divides none of the orders of B, B10 , . . . , Br0 . Consider the group H = B × (Z/pZ)e . Let b1 , . . . , be be generators of B. For i = 1, . . . , e, let vi be the element of (Z/pZ)e with 1 in the ith coordinate and 0 in all other coordinates. Since the orders of bi and vi are relatively prime, (b1 , v1 ), . . . , (be , ve ) generate ˜ → H be the universal H. Since (Z/pZ)e is of rank e, so is H. Let γ˜ : H ˜ Frattini cover of H. Then H is a projective group of rank e (Proposition ˜ is divisible exactly by the prime numbers 22.6.1). Moreover, the order of H dividing the order of the finite group H (Proposition 22.10.5). Let M0 be the splitting field of h over L and let L0 be the fixed field of A in M0 . Theorem 23.2.3 gives a perfect PAC field E, algebraic over L, and a commutative diagram (9)

Gal(E) UU UUUU UUUUres UUUU ϕ UUUU UUUU  */ / / ˜ H B A H pr α γ ˜

in which ϕ is an isomorphism. In particular, E ∈ FMP(K, e). We prove that θ is true in E. ˜ and F the Let M be the fixed field of Ker(α ◦ pr ◦ γ˜ ◦ ϕ) = Ker(res) in E fixed field of Ker(pr ◦ γ˜ ◦ ϕ) in Ks . Then M is the splitting field of h over E and (9) induces the commutative diagram (7) in which β is an isomorphism. α Thus, θL (h, B −→ A) is true in E. α Assume there is an i with θL (h, Bi −→ A) true in E. In particular, E has 0 a Galois extension F containing M that gives rise to a commutative diagram Bi0 o @@ @@ @ α0i @@ 

β0

A

Gal(F 0 /E) uu uu uures u u zu u

¯i = Gal(F¯ /E), in which β 0 is an isomorphism. Then, with F¯ = F ∩ F 0 , B −1 0 0 −1 ψ = resF¯ /M , ϕi = resF/F¯ ◦β , ϕi = resF 0 /F¯ ◦β , we obtain a commutative diagram (6) in which ϕi , ϕ0i , and ψ are epimorphisms. If ϕ0i is shown to be a Frattini cover, then (C) is false, and this contradiction will conclude the lemma. ˜ ∼ ˜ × Fˆe (p), so By Corollary 22.7.8 and Corollary 22.10.6, Gal(E) ∼ =H =B ˆ Fe (p) is the p-Sylow group of Gal(E) (Proposition 22.10.5). Denote the fixed ˜ by N . Then Gal(N/E) ∼ ˜ Since the orders of B = field of Fˆe (p) in E = B. 0 Gal(F/E) and Bi = Gal(F /E) are relatively prime to p, F F 0 ⊆ N . Hence,

554

Chapter 23. PAC Fields and Projective Absolute Galois Groups

res: Gal(N/E) → Gal(F/E) is a Frattini cover. Hence, res: Gal(F F 0 /E) → Gal(F/E) is also Frattini (Lemma 22.5.4(c)). Since F ∩ F 0 = F¯ , Gal(F F 0 /E)

res

res

 Gal(F/E)

/ Gal(F 0 /E) res

res

 / Gal(F¯ /E)

is a cartesian square (Example 22.2.7(a)). By Lemma 22.5.5, res: Gal(F 0 /E) → Gal(F¯ /E) (and hence ϕi ) is Frattini. Consequently, θ is true in E.  Since B, B1 , . . . , Br are finite groups, there is an effective check for condition (C). If, in addition, K is a presented field with elimination theory and L is a presented finite separable extension of K, then L is also a presented field with elimination theory (Definition 19.2.8). This allows an explicit check of the statement that A is a subgroup of the permutation group Gal(h, L) (Lemma 19.3.2). Corollary 23.4.4: Suppose K is a presented field with elimination theory and θ is given by (5). Then there is a primitive recursive procedure for deciding the existence of a field E ∈ PAC(K, ≤ e) in which θ is true. Proof: By Lemma 23.4.3, the truth of (C) is equivalent to the existence of a field E ∈ PAC(K, ≤ e) in which θ is true.  Corollary 23.4.5: If there exists a perfect PAC field E that contains K in which θ is true, then there exists a field E 0 ∈ F M P (K, e) in which θ is true. Corollary 23.4.6: Suppose A is a subgroup of Gal(h, L) of rank at most e and r = 0 in (5). Then there exists a field E ∈ F M P (K, e) in which θ is true. Proof: In this case Condition (C) is empty.



23.5 Reduction Steps This section shows that any Boolean combination of e-basic statements is equivalent modulo Th(PAC(K, ≤ e)) to an elementary statement given by (3) of Section 23.4. As in that section, K is a countable Hilbertian field. α

Lemma 23.5.1: Let L be a finite separable extension of K, θL (h, B −→ A) an e-basic statement, and q ∈ L[X] a separable polynomial. Suppose h divides q. Then there are epimorphisms αi : Bi → Ai , of finite groups with α is equivalent modulo rank(Bi ) ≤ e, i = 1, . . . , r, such that Wr θL (h, B −→ αA) i Ai ). Th(PAC(K, ≤ e)) to the disjunction i=1 θL (q, Bi −→

23.5 Reduction Steps

555

Proof: Let M and N be the splitting fields of h and q over L, respectively. Denote the fixed field of A in M by M0 . Then M ⊆ N . Let N1 , . . . , Nr be the fields satisfying (1)

Ni ⊆ N

and M0 = M ∩ Ni ,

i = 1, . . . , r.

For each i, let Ai = Gal(N/Ni ) (regarded as a subgroup of Gal(q, L)). Find all subgroups Bij of B ×Ai of rank ≤ e with these properties: The restriction, αij , of the projection onto Ai and the restriction of prB to Bij are surjective and the diagram (2)

Bij

αij

/ Ai

prB

Gal(N/Ni )

res

 B

res

 Gal(M/M0 )

 /A

α

α

is commutative. We prove θL (h, B −→ A) is equivalent modulo Th(PAC(K, ≤ W αij e)) to i,j θL (q, Bij −→ Ai ). α Indeed, let E ∈ PAC(K, ≤ e) be a field in which θL (h, B −→ A) is true. Then M0 = M ∩ E and there exists a commutative diagram (3)

Gal(F/E)

/B

β

res

α

 Gal(M E/E)

 A

in which F is a Galois extension of E and β is an isomorphism. Thus, Ni = N ∩ E is one of the fields for which (1) holds. Also, Fi = N F is a finite Galois extension of E. N

NE

Fi

M

M Ni

ME

F

M0

Ni

E

L In particular, rank(Gal(Fi /E)) ≤ rank(Gal(E)) ≤ e. Hence, Gal(Fi /E) is isomorphic to one of the groups Bij satisfying (2). Moreover, there is a commutative diagram (4)

Gal(Fi /E) res

βij

 Gal(N E/E)

/ Bij αij

 Ai

556

Chapter 23. PAC Fields and Projective Absolute Galois Groups

in which βij is an isomorphism and αij is an epimorphism. This means that αij E |= θL (q, Bij −→ Ai ). αij Conversely, suppose E ∈ PAC(K, ≤ e) and E |= θL (q, Bij −→ Aj ) for one of the groups Bij satisfying (2). Then there exists a Galois extension Fi of E and an isomorphism βij : Gal(Fi /E) → Bij with (4) commutative. Denote the fixed field in Fi of the kernel of the projection map Bij → B by F . This gives a commutative diagram Gal(F/E)

β

res

/B α

 Gal(M/M0 )

 A α

in which β is an isomorphism. Consequently, E |= θL (h, B −→ A).



αi

Lemma 23.5.2: Let θL (h, Bi −→ A), be an e-basic statement, i = 1, . . . , r. ρj Then Vr there areαie-basic statements θL (h, Cj −→ A), j = 1, . . . , s, such that θL (h, Bi −→ A) is equivalent modulo Th(PAC(K, ≤ e) to the disjunction Wi=1 ρj s j=1 θL (h, Cj −→ A). Proof: Consider the subgroups C of B1 × · · · × Br of rank at most e, with these properties: The projection pri of B1 × · · · × Br onto Bi maps C surjectively onto Bi , i = 1, . . . , r; and αi ◦ pri = αi0 ◦ pri0 on C, for all i and i0 between 1 and r. List these groups as C1 , . . . , Cs and define ρj to be αi ◦ pri restricted to Cj , j = 1, . . . , s (this is independent of i). Then (Cj , ρj ), j = 1, . . . , s, satisfy the conclusion of the lemma. Note: If no such group C exists, then the conjunction has no model in PAC(K, ≤ e). In this case, regard the empty disjunction as a false sentence modulo Th(PAC(K, ≤ e)).  Proposition 23.5.3: Let θi be an e-basic statement, i = 1, . . . , r, and P a Boolean polynomial in r variables. Put θ = P (θ1 , . . . , θr ). The following holds: (a) If θ has a model E ∈ PAC(K, ≤ e), then it has a model E 0 ∈ FMP(K, e); and (b) If K is a presented field with elimination theory, then it is possible to decide (primitive recursively) if θ has a model E ∈ PAC(K, ≤ e). Proof: Write θ in normal form as _^ ^  αij αik θLij (hij , Bij −→ Aij ) ∧ ¬θLik (hik , Bik −→ Aik ) i

j

k

where i, j and k range over finite sets. The theorem follows from Corollaries 23.4.4, 23.4.5 and 23.4.6 if we can reduce to the case that θ is of the form given by (5) of Section 23.4.

23.5 Reduction Steps

557

First observe that if (a) and (b) are true for each of the disjuncts, then they are true for θ. We may therefore assume that θ is ^

α

i θLi (hi , Bi −→ Ai ) ∧

i∈I

^

αj

¬θLj (hj , Bj −→ Aj ),

j∈J

where I and J are disjoint finite sets. The reduction of θ to the form (5) of Section 22.2 is divided into five parts: Part A: Reduction to the case that αi = id. for some i ∈ I. Let N be a finite Galois extension of K that contains Li and the splitting field of hi over Li , for each i ∈ I; and Lj and the splitting field of hj over Lj , for each j ∈ J. Choose an irreducible polynomial h ∈ K[X] with a root that generates N over K. Let {Nr | r ∈ R} be the set of all fields between K and N with Ar = Gal(N/Nr ) generated by e elements. If E ∈ PAC(K, ≤ e), then N ∩ E = Nr for some r ∈ R. The latter equality is equivalent to id E |= θNr (h, Ar −→ Ar ). Therefore, θ is equivalent modulo Th(PAC(K, ≤ e)) to the nonempty disjunction id

_

θNr (h, Ar −→ Ar ) ∧

r∈R

^

α

i θLi (hi , Bi −→ Ai ) ∧

i∈I

^

αj



¬θLj (hj , Bj −→ Aj )

j∈J

Again, it suffices to consider each of the disjuncts separately. Hence, assume θ is id

θNr (h, Ar −→ Ar ) ∧

^

α

i θLi (hi , Bi −→ Ai ) ∧

i∈I

^

αj

¬θLj (hj , Bj −→ Aj )

j∈J

Part B: Reduction to the case with Nr = Li = Lj for all i ∈ I and j ∈ J. Suppose θ has a model E in PAC(K, ≤ e) and let i ∈ I. Denote the splitting ˆ i . Let L0 be the fixed field of Ai in L ˆ i . Then L0 = field of hi over Li by L i i ˆ ˆ ˆ Li ∩ E = Li ∩ N ∩ E = Li ∩ Nr ⊆ Nr for each i ∈ I where N is the field of ˆ j and Part A. We may thus replace Li by Nr for each i ∈ I. Similarly define L 0 0 Lj for each j ∈ J. Assume Lj 6⊆ Nr for some j. Then θ and the conjunction αj without ¬θLj (hj , Bj −→ Aj ) have the same models in PAC(K, ≤ e). Hence, αj we may drop ¬θLj (hj , Bj −→ Aj ) from the conjunction. Therefore, we may replace Lj by Nr for each j ∈ J. Thus, without loss, assume θ is ^

α

i θL (hi , Bi −→ Ai ) ∧

i∈I

where I is nonempty.

^ j∈J

αj

¬θL (hj , Bj −→ Aj ),

558

Chapter 23. PAC Fields and Projective Absolute Galois Groups

Part C: Reduction to the case where all hi and hj are the same. Let αi Ai ) is h = lcm(hi , hj | i ∈ I, j ∈ J). By Lemma 23.5.1, θL (hi , Bi −→ equivalent modulo the theory PAC(K, ≤ e) to a disjunction of the form W αk θ (h, D −→ C ). If this disjunction is empty, then θ has no model. k k k L Otherwise, consider each of the disjuncts separately to assume that θ is ^

α

^

i θL (h, Bi −→ Ai ) ∧

i∈I

αj

¬θL (hj , Bj −→ Aj ).

j∈J αj

Similarly θL (hj , Bj −→ Aj ) is equivalent to a disjunction as above. Thus, we may assume θ is ^

α

i θL (h, Bi −→ Ai ) ∧

i∈I

^

αj

¬θL (h, Bj −→ Aj ),

j∈J

where I is nonempty. Part D: Reduction to the case that all Ai ’s and Aj ’s are the same. Suppose there are i1 , i2 ∈ I with Ai1 6= Ai2 . Since it is impossible for a field E ∈ PAC(L, ≤ e) to exist with Gal(h, E) = Ai1 and Gal(h, E) = Ai2 , θ has no model. We may therefore assume that Ai = A for all i ∈ I. In particular, if θ holds in E ∈ PAC(L, ≤ e), then Gal(h, E) = A, because I 6= ∅. If there αj is a j ∈ J with Aj 6= A, then E |= ¬θL (h, Bj −→ Aj ) and the corresponding conjunct can be dropped from the conjunction. Thus, we may assume Aj = A for each j ∈ J and θ is ^

α

i θL (h, Bi −→ A) ∧

i∈I

^

αj

¬θL (h, Bj −→ A).

j∈J

V αi Part E: Conclusion. The conjunction i∈I θL (h, BiW −→ A) is equivalent ρs modulo Th(PAC(K, ≤ e)) to a disjunction of the form s∈S θL (h, Cs −→ A) (Lemma 23.5.2). If S is empty, then θ has no model. Otherwise, consider each of the disjuncts separately to assume θ is ρ

θL (h, C −→ A) ∧

^

αj

¬θL (h, Bj −→ A).

j∈J

This concludes the proof of the proposition.



23.6 Application of Ultraproducts This section applies ultraproducts to show that each sentence of L(ring, K) is equivalent modulo Th(PAC(K, ≤ e)) to a boolean combination of basic elementary statements. With this the results of Section 23.4 give the proof of the main theorem:

23.6 Application of Ultraproducts

559

Lemma 23.6.1: Let E and F be models of Th(PAC(K, ≤ e)). Suppose π

π

E |= θL (h, B −→ A) ⇐⇒ F |= θL (h, B −→ A) π

for each e-basic statement θL (h, B −→ A). Then E and F are K-elementarily equivalent. Proof: Our assumption implies, in particular, that Ks ∩ E ∼ =K Ks ∩ F . ˜ ∩E ∼ ˜ ∩ F . Identify K ˜ ∩ E with Since E and F are perfect fields, K =K K ˜ ∩ F and denote it by M . For each positive integer n, let the composite of K all extensions of E (resp. M and F ) of degrees at most n be En (resp. Mn and Fn ). Since Gal(E) and Gal(F ) are finitely generated, En , Mn and Fn are finite Galois extensions of E, M and F , respectively. Moreover, Mn ⊆ En and Mn ⊆ Fn . Use the primitive element theorem to prove that there exists a finite separable extension L of K, with L contained in M , and a separable polynomial h ∈ L[X] with splitting field Ln such that Ln ∩ M = L and Ln M = Mn . Then Gal(h, L) = Gal(h, M ) = Gal(h, E) = Gal(h, F ). The res e-basic statement θL (h, Gal(En /E) −→ Gal(h, L)) is true in E. Hence, it is also true in F . Thus, F has a finite Galois extension Fn0 with a commutative diagram

(1)

/ Gal(En /E) Gal(Fn0 /F ) QQQ m m QQQ mmm res QQ( vmmm res Gal(Mn /M ) β

in which β is an isomorphism. In particular, from the definition of En , Fn is the composite of extensions of F of degree at most n. Thus, Fn0 ⊆ Fn , and [En : E] = [Fn0 : F ] ≤ [Fn : F ]. Interchange the roles of E and F to get the inequality [Fn : F ] ≤ [En : E]. Therefore, [Fn : F ] = [En : E], Fn0 = Fn and (1) turns out to be the diagram

(2)

/ Gal(En /E) Gal(Fn /F ) QQQ m m QQQ m m mres m m res QQ( vm Gal(Mn /M ) β

This proves that the finite set Bn of all isomorphisms β for which the diagram (2) is commutative, is nonempty. The groups Gal(Fn+1 /Fn ) and Gal(En+1 /En ) are the intersections of all subgroups of Gal(Fn+1 /F ) and Gal(En+1 /E), respectively, of index at most n. Hence, if γ ∈ Bn+1 , then γ(Gal(Fn+1 /Fn )) = Gal(En+1 /En ). Therefore, γ induces an element of Bn . Thus, the collection {Bn }∞ n=1 forms an inverse system of finite nonempty sets.

560

Chapter 23. PAC Fields and Projective Absolute Galois Groups

By Lemma 1.1.3, lim Bn is nonempty. Each element in the inverse limit de←− fines an isomorphism ϕ such that the following diagram is commutative β / Gal(E) Gal(F ) MMM q q MM qq res M& xqq res Gal(M )

By Theorem 20.3.3, E ≡K F .



Since the language L(ring, K) is countable, each model of Th(PAC(K, ≤ e)) has a countable submodel (Skolem-L¨ owenheim). Thus, the equivalence classes of Th(PAC(K, ≤ e)) have a complete system of representatives {Ei | i ∈ I} with |I| ≤ 2ℵ0 and Ei is countable for each i ∈ I. For a sentence θ of L(ring, K), let S(θ) = {i ∈ I | Ei |= θ}. Let S be the Boolean algebra of I which is generated by all sets of the form π π S(θL (h, B −→ A)) with θL (h, B −→ A) an e-basic statement. Proposition 23.6.2: For each θ ∈ L(ring, K), S(θ) ∈ S. Proof: Assume S(θ) 6∈ S. Then, there are ultrafilters E and F ofQ I with E ∩ S = FQ∩ S, S(θ) ∈ E, but S(θ) 6∈ F (Lemma 7.6.2). Let E = Ei /E and F = Ei /F. Then the same e-basic statements hold in E and in F . By Lemma 23.6.1, E ≡K F . Since θ is true in E, but false in F , this is a contradiction.  For the main result, use the notation of Section 23.1: Theorem 23.6.3: Let K be a countable Hilbertian field. Then: (a) Th(FMP(K, e)) = Th(PAC(K, ≤ e)). (b) If K is a presented field with elimination theory, then Th(PAC(K, ≤ e)) is (recursively) decidable [Cherlin-v.d.Dries-Macintyre]. Proof of (a): By (2) of Section 23.3, Th(PAC(K, ≤ e)) ⊆ Th(FMP(K, e)). Conversely, let η ∈ Th(FMP(K, e)). Assume η ∈ / Th(PAC(K, ≤ e)). Then there exists a model E ∈ PAC(K, ≤ e)) with E |= ¬η. By Proposition 23.6.2, η is equivalent modulo Th(PAC(K, ≤ e)) to a Boolean combination θ of ebasic statements. In particular, E |= ¬θ. By Proposition 23.5.3, ¬θ has a model E 0 ∈ FMP(K, e). It follows that E 0 |= ¬θ. This contradicts the assumption that η ∈ Th(FMP(K, e)) and proves that η ∈ Th(PAC(K, ≤ e)). Proof of (b): Let K be a presented field with elimination theory. Consider a sentence η ∈ L(ring, K). By Proposition 23.6.2, Th(PAC(K, ≤ e)) |= η ↔ θ for some Boolean combination θ of e-basic statements. By G¨odel’s completeness theorem (Corollary 8.2.6), Th(PAC(K, ≤ e))i ` η ↔ θ. Use the system of axioms ∆(K, ≤ e) for the class PAC(K, ≤ e) (Section 23.3) and to find θ in finitely many steps (Lemma 8.7.1). Then apply Corollary 23.4.4 to decide if θ belongs to Th(PAC(K, ≤ e)). Whatever the conclusion is, it applies to η. This gives (b). 

Notes

561

Exercises 1. Combine Corollary 23.1.2, Theorem 11.6.2, Artin’s theorem about fields with finite absolute Galois groups, and Corollary 11.2.5 to give an alternative proof to the second part of Corollary 22.4.6: Every closed subgroup of a projective group G is projective and G is torsion free. 2. [Wheeler, Thm 2.3] Let K be a perfect field. Denote the class of all regular extensions of K by Reg(K). Apply Proposition 11.3.5 and Exercise 13 of Chapter 2 to prove that the following conditions on E ∈ Reg(K) are equivalent: (a) E is existentially complete in Reg(K) (i.e. if E ⊆ F and F ∈ Reg(K), then E is existentially closed in F ); and (b) E is perfect and PAC and the map res: Gal(E) → Gal(K) is the universal Frattini cover of Gal(K). 3. In the notation of Exercise 2, apply Theorem 20.3.3 to prove that fields E and E 0 that are existentially complete fields in Reg(K) are K-elementarily equivalent. 4.

Let f and g be polynomials in Q[X]. α (a) Find a Boolean combination of basic statements, θQ (h, B −→ A), which is equivalent modulo Th(PAC(Q, ≤ e)) to the sentence [(∃X)f (X) = 0] ↔ [(∃X)g(X) = 0]. (b) How can you decide if the above sentence is true for all field E ∈ PAC(Q, ≤ e)? 5. Let e = 1, 2, or 3. Does there exists an irreducible polynomial f ∈ Q[X] √ of degree 4 such that for all E ∈ PAC(Q, ≤ e), 2 ∈ E if and only if f has a root in E? Hint: Consider first the case where e = 1. Note that Gal(f, Q) is a transitive subgroup of S4 and distinguish between the different cases. Compare also with Section 21.5. Deduce the cases where e = 2 or 3 from the first case. 6.

Use Example 21.5.4 to show that the sentence (∃X)[X 3 + X 2 − 2X − 1 = 0] ←→ (∃X)[X 6 + X 4 − 2X 2 − 1 = 0]

holds for all E ∈ PAC(Q, ≤ 1) but that it does not belong to Th(PAC(Q, ≤ 2)).

Notes [Kuyk1] uses wreath products to prove Lemma 23.2.1, whereas [Jarden2] applies the free generators theorem (Theorem 18.5.6) to achieve the same result. The proof we give here is based on the field crossing argument.

Chapter 24. Frobenius Fields The embedding property (Proposition 17.7.3) for free profinite groups is essential to the primitive recursive procedure for perfect PAC fields with free absolute Galois groups (Theorem 30.6.2). Since this is only a special case of a general result, we focus attention here on PAC fields whose absolute Galois groups have the embedding property: the Frobenius fields. The “field crossing argument” (e.g., the proofs of Lemma 6.4.8, Part G in Section 6.5, Part C of Proposition 16.8.6, Lemma 20.2.2, and Lemma 23.2.1) applies to give an analog, for Frobenius fields, of the Chebotarev density theorem (Proposition 24.1.4). The remainder of the Chapter concentrates on the embedding property. For example, if a profinite group has the embedding property, so does its universal Frattini cover (Proposition 24.3.5). We further show that every profinite group G has a smallest cover E(G) with the embedding property (Proposition 24.4.5). In particular, for G finite, E(G) is finite and unique (Proposition 24.4.6). The construction of E(G) leads to a decision procedure for projective groups with the embedding property (Corollary 24.5.3).

24.1 The Field Crossing Argument The decidability algorithms of Chapters 20 and 23 are based on the elementary equivalence theorem for PAC fields (Theorem 20.3.3). In addition to the PAC assumption, this theorem also requires certain infinite embedding problems for the absolute Galois groups of the fields involved to be solvable. The resulting decision procedure are recursive. Establishing a primitive recursive decision procedures, requires a finite version of the embedding properties of the absolute Galois groups. We achieve this in this section by using the field crossing argument: Let K be a field, S/R be a ring cover (Definition 6.1.3) with K ⊆ R, and F/E the corresponding field cover. We say S/R is regular (resp. finitely generated, Galois) over K if E/K is regular (resp. R is finitely generated over K, F/E is Galois). Let ϕ: S → Ks be a K-homomorphism. The decomposition group of ϕ is (Remark 6.1.6) Dϕ = {σ ∈ Gal(F/E) | ϕ(x) = 0 ⇐⇒ ϕ(σx) = 0, for all x ∈ S}. Lemma 24.1.1: Let K be a PAC field, S/R a regular finitely generated Galois ring cover over K, and F/E the corresponding Galois field cover. Denote the algebraic closure of K in F by L. Let E 0 be a field between E and F with res: Gal(F/E 0 ) → Gal(L/K) surjective. Suppose there is an epimorphism γ˜ : Gal(K) → Gal(F/E 0 ) with resF/L ◦ γ˜ = resKs /L . Denote the fixed field

24.1 The Field Crossing Argument

563

of Ker(˜ γ ) in Ks by M . Then there exists a K-epimorphism ϕ: S → M satisfying: (1a) ϕ(R) = K and Dϕ = Gal(F/E 0 ). (1b) If char(K) = p and y1 , . . . , ym ∈ R are p-independent over E p with m bounded by [K : K p ], then ϕ can be chosen with ϕ(y1 ), . . . , ϕ(ym ) being p-independent over K p . Proof: The field M is a Galois extension of K which contains L and γ˜ induces an isomorphism γ: Gal(M/K) → Gal(F/E 0 ) with resF/L ◦ γ = resM/L . By assumption, F is a regular extension of L, so E 0 M ∩ F = E 0 L and res: Gal(E 0 M/E 0 ) → Gal(M/K) is an isomorphism. Put F 0 = F M and consider Gal(F 0 /E 0 ) as a subgroup of Gal(F/E 0 ) × Gal(M/K). Then Gal(F 0 /E 0 ) = {(σ1 , σ2 ) ∈ Gal(F/E 0 ) × Gal(M/K) | resL (σ1 ) = resL (σ2 )}. Let ∆ = {(γ(σ), σ) | σ ∈ Gal(M/K)}. Denote the fixed field of ∆ in F 0 by D. Then ∆ ⊆ Gal(F 0 /E 0 ) and ∆ ∩ Gal(M/K) = ∆ ∩ Gal(F/E 0 ) = 1. In addition, for each (σ1 , σ2 ) ∈ Gal(F 0 /E 0 ), (σ1 , σ2 ) = (γ(σ2 ), σ2 ) · (γ(σ2 )−1 σ1 , 1) ∈ ∆ · Gal(F/E 0 ), and with σ1 = γ(τ1 ), τ1 ∈ Gal(M/K) (σ1 , σ2 ) = (γ(τ1 ), τ1 ))(1, τ1−1 σ2 ) ∈ ∆ · Gal(M/K). Thus, ∆ · Gal(M/K) = ∆ · Gal(F/E 0 ) = Gal(F 0 /E 0 ). It follows that DM = DF = F 0 , D ∩ F = E 0 , and D ∩ M = K. In particular, D is a regular extension of K. DL D F0 w w ww wF w w E0 E0L E0M

E

EL

EM

K

L

M

The integral closure U of R in D is finitely generated over R [Lang4, p. 120], hence over K. Since K is PAC, there exists a K-epimorphism ψ: U → K. By Lemma 2.5.10, the integral closure V of U in F 0 is U M = U ⊗K M . In particular, S ⊆ V . Thus, ψ extends to an M -epimorphism ψ 0 : V → M . Since [F 0 : D] = [M : K], the decomposition group Dψ0 is ∆ (Lemma 6.1.4). Let ϕ be the restriction of ψ 0 to S. Then Gal(F/E 0 ) = resF Dψ0 ≤ Dϕ ≤ Gal(F/E 0 ). Hence, Dϕ = Gal(F/E 0 ). This proves (1a).

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Chapter 24. Frobenius Fields

Moreover, since ϕ(R) = K, ϕ(S) = M0 is a Galois extension of K in M with Gal(M0 /K) ∼ = Dϕ = Gal(F/E 0 ) ∼ = Gal(M/K) (Remark 6.1.4). Therefore, ϕ(S) = M . If char(K) = p and y1 , . . . , ym are as in (1b), then, since D/E is separable, they are p-independent over Dp . By Proposition 10.11, we may choose ψ so that ψ(y1 ), . . . , ψ(ym ) are p-independent over K p .  Definition 24.1.2: A profinite group G has the embedding property if each embedding problem (2)

(ζ: G → A, α: B → A)

where ζ and α are epimorphisms and B ∈ Im(G) (i.e. B is a finite quotient of G) is solvable. That is, there exists an epimorphism γ: G → B with α◦γ = ζ. Note that each quotient of a group B in Im(G) belongs to Im(G). Hence, by induction on the order of Ker(α), it suffices to consider only embedding problems (2) where B ∈ Im(G) and Ker(α) is a minimal normal subgroup of B.  In addition to the groups Fˆe (C) with C a formation of finite groups, Section 24.6 contains many examples of groups with the embedding property. Definition 24.1.3: (a) A field K is called a Frobenius field if K is PAC and Gal(K) has the embedding property. (b) Let S/R be a regular finitely generated Galois ring cover over K and F/E the corresponding field cover. Denote the algebraic closure of K in F by L. We say that S/R satisfies the decomposition group condition if for each subgroup H of Gal(F/E) with (3)

H ∈ Im(Gal(K))

and

resL (H) = Gal(L/K)

there exists a K-homomorphism ϕ: S → Ks satisfying (1).



Proposition 24.1.4 ([Fried-Haran-Jarden]): A field K is Frobenius if and only if every regular finitely generated Galois ring cover over K satisfies the decomposition group condition. Proof: Suppose first K is Frobenius. Let S/R, F/E, and L be as in Definition 24.3(b). Let H be a subgroup of Gal(F/E) satisfying (3). Let E 0 be the fixed field of H in F . The embedding property of Gal(K) gives an epimorphism γ˜ : Gal(K) → Gal(F/E 0 ) with resF/L ◦ γ˜ = resKs /L . Lemma 24.1.1 then gives a homomorphism ϕ: S → Ks satisfying (1). Thus S/R satisfies the decomposition condition. Conversely, suppose every regular finitely generated Galois ring cover over K satisfies the decomposition group condition. Consider a variety V over K. Let F be its function field. Choose an integrally closed domain R, finitely generated over K, containing the coordinate ring of V and having F as its quotient field. Take S = R, E = F , and H = 1 in the decomposition

24.2 The Beckmann-Black Problem

565

group condition to assure the existence of a K-homomorphism ϕ: R → K. Thus V (K) is nonempty. This proves K is PAC. Next we prove Gal(K) has the embedding property. Let L be a finite Galois extension of K and α: G → Gal(L/K) an epimorphism, with G ∈ Im(Gal(K)). Consider a set X = {xσ | σ ∈ G} of |G| algebraically independent elements over L. Let G act on X from the right in the obvious manner, and on L by aσ = aα(σ) . This defines a faithful action of G on the rational function field F = L(X). Denote the fixed field of G in F by E. Then E ∩ L = K. Since F/L is regular, so is EL/L. Therefore, E/K is a finitely generated regular extension (Lemma 10.5.1). Let S/R be a finitely generated regular ring cover over K with F/E the corresponding field cover (Definition 6.1.3). The decomposition group condition gives a K-epimorphism ϕ: S → M , where M is a Galois extension of K containing L, ϕ(R) = K, and Dϕ = Gal(F/E). Assume without loss resL (ϕ) is the identity map of L. Let ϕ∗ : Gal(M/K) → Gal(F/E) be the isomorphism induced by ϕ (Lemma 6.1.4). Identify Gal(F/E) with G and resF/L with α to find ϕ∗ ◦ resKs /M : Gal(K) → G is an epimorphism and α ◦ ϕ∗ ◦ resKs /M = resL .



For general profinite groups neither the projective property nor the embedding property implies the other (Examples 24.6.1 and 24.6.7). We call a profinite group G superprojective if it is both projective and has the embedding property. Proposition 24.1.5: Let F be a Frobenius field. Then Gal(F ) is superprojective. Proof: By definition, F is PAC. Hence, by Theorem 11.6.2, Gal(F ) is projective. By definition, Gal(F ) has the embedding property. Consequently, Gal(F ) is superprojective.  A special case of Theorem 23.1.1 yields a converse to Proposition 24.1.5: Proposition 24.1.6: Let L/K be a Galois extension, L0 a field between K and L and G is a superprojective group. Suppose Gal(L/L0 ) is a quotient of G. Then there exists a field extension F of K which is perfect and Frobenius such that L ∩ F = L0 and Gal(F ) ∼ = G.

24.2 The Beckmann-Black Problem Let L/K be a finite Galois extension of fields with Galois group G and t an indeterminate. The Beckmann-Black Problem for K, L, and G asks whether K(t) has a Galois extension F with the following properties: (1a) Gal(F/K(t)) ∼ = G. (1b) F/K is a regular extension.

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Chapter 24. Frobenius Fields

(1c) There is a prime divisor p of F/K with decomposition field K(t) and residue field L. The following observation shows that the Beckmann-Black problem is apparently stronger than the regular inverse Galois problem (Section 16.2): Lemma 24.2.1 ([D`ebes, Prop. 1.2]): Let K0 be a field and G a finite group. Suppose each field extension K of K0 has the following property: If G is realizable over K, then G is regular over K. Then G is regular over K0 (although G need not be realizable over K0 ). Proof: Let L = K0 (xσ | σ ∈ G) with xσ algebraically independent over K0 . Define an action of G on L by (xσ )τ = xστ and aσ = a for a ∈ K0 . Let K be the fixed field of G in L. Then Gal(L/K) = G. By assumption, K(t) has a Galois extension F with Galois group G such that F/K is regular. Then L(t) = K0 (t, xσ | σ ∈ G) is a purely transcendental extension of K0 , Gal(LF/L(t)) ∼ = G, and LF/L is a regular extension. Thus, LF/K0 is a regular extension.  The Beckmann-Black problem has an affirmative solution over PAC fields, even in a stronger form: Theorem 24.2.2 ([D`ebes, Thm. 3.2]): Let K be a PAC field, t an indeterminate, and G a finite group. For i = 1, . . . , n let Gi be a subgroup of G and Li a Galois extension of K with Galois group Gi . Then K(t) has a Galois extension F satisfying this: (2a) Gal(F/K(t)) ∼ = G. (2b) F/K is a regular extension. (2c) For each i there exists a prime divisor pi of F/K with decomposition group over K(t) equal to Gi and with residue field Li . Moreover, p1 , . . . , pn are distinct. Proof: Proposition 16.12.2 gives F with properties (2a) and (2b). Put R = K[t] and E = K(t). Choose a finitely generated ring cover S/R over K with corresponding field cover F/E. For each i let Ei be the fixed field of Gi in F . Then apply Lemma 24.1.1 with E 0 , L, M replaced by Ei , K, Li . This gives a K-epimorphism ϕi : S → Li with ϕi (R) = K and Dϕi = Gi . By Proposition 3.3.4, S is a Dedekind domain. In particular, the local ring of S at Ker(ϕi ) is a valuation ring. Hence, ϕi defines a prime divisor pi of F/K with the same decomposition group. Finally, to ensure that pi is different from p1 , . . . , pi−1 , choose yj ∈ −1 ]. None E r Opj , j = 1, . . . , i − 1, and replace R by Ri = R[y1−1 , . . . , yi−1 of the places ϕ1 , . . . , ϕi−1 will be finite on Ri . Therefore, pi 6= pj for all j < i.  Problem 24.2.3: Prove or disprove: Every field K with an affirmative solution for the regular inverse Galois problem has an affirmative solution for the Beckmann-Black problem.

24.3 The Embedding Property and Maximal Frattini Covers

567

24.3 The Embedding Property and Maximal Frattini Covers ˜ of a profinite group G with the We prove that the universal Frattini cover G ˜ is superprojective. embedding property has the embedding property. Thus, G Lemma 24.3.1: Let G be a group with the embedding property and N a family of open normal subgroups of G with this property: If N 0 is an open normal subgroup of G and G/N 0 ∼ = G/N for some N ∈ N then N 0 ∈ N . Denote the intersection of all N ∈ N by N0 . Then G/N0 has the embedding property. Proof: Let ϕ: G/N0 → A and α: B → A be epimorphisms with B ∈ Im(G/N0 ). That is, G has an open normal subgroup M that contains N0 and there is an isomorphism β: G/M → B. Choose N1 , . . . , Nn ∈ N with J = N1 ∩ · · · ∩ Nn ≤ M . Let γ: G/J → G/M and π: G → G/N0 be the quotient maps. The embedding property for G gives an epimorphism θ1 : G → G/J with α◦β◦γ◦θ1 = ϕ◦π. Put Ni0 = θ1−1 (Ni /J), i = 1, . . . , n. Then, Ker(θ1 ) = N10 ∩ · · · ∩ Nn0 and G/Ni ∼ = G/Ni0 , i = 1, . . . , n. By assumption, N10 , . . . , Nn0 ∈ N , so N0 ≤ Ker(θ1 ). Thus, θ1 induces an epimorphism θ0 : G/N0 → G/J with θ0 ◦ π = θ1 . The map θ = β ◦ γ ◦ θ0 is an epimorphism of G/N0 onto B with α ◦ θ = ϕ. G qqq q q qq qqq q  q θ1 qq q G/N0 q q h h qq hhh { qqq hθ0hhhhhhh θ {{{ q q h q {{ hh qhqhqhhhhh {{ xq  } { sh /B /A G/J γ / G/M α β It follows that G/N0 has the embedding property.



Lemma 24.3.2: (a) If a profinite group G has the embedding property, then so does G/Φ(G). ˜ → G be the universal Frattini (b) Let G be a profinite group and let ϕ: G ˜ and G/Φ( ˜ ˜ ∼ cover. Suppose Φ(G) = 1. Then Ker(ϕ) = Φ(G) G) = G. ˜ (c) Suppose Φ(G) = 1 and G has the embedding property. Then G has the embedding property. Proof of (a): For M a maximal open subgroup of G, the intersection N (M ) of all conjugates of M in G is an open normal subgroup of G. The collection N = {N (M ) | M ranges over maximal open subgroups of G} consists of all maximal open normal subgroups T of G. It satisfies the hypotheses of Lemma 24.3.1. Furthermore Φ(G) = N (M ). Thus, (a) is a special case of Lemma 24.3.1.

568

Chapter 24. Frobenius Fields

˜ Thus, ϕ(Φ(G)) ˜ ≤ Proof of (b): Since ϕ is a Frattini cover, Ker(ϕ) ≤ Φ(G). ˜ ≤ Ker(ϕ). ConseΦ(G) = 1 (Lemma 22.1.4(a)), and this implies Φ(G) ˜ quently, Ker(ϕ) = Φ(G). Proof of (c): Combine (a) and (b).



Here is an important collection of groups with the embedding property: Lemma 24.3.3: Let C be a formation of finite groups and F is a free pro-Cgroup. Then F has the embedding property. Proof: Proposition 17.7.3 gives the result when F has finite rank. Assume F has infinite rank. Let ϕ: F → A and α: B → A be epimorphisms with B ∈ Im(F ). In particular, B belongs to C. Let X be a basis of F . Then X2 = X r Ker(ϕ) is a finite set. For each x ∈ X2 choose γ2 (x) ∈ B with α(γ2 (x)) = ϕ(x). Next choose a finite subset X1 of X ∩ Ker(ϕ) with |X1 | = |Ker(α)| and a bijective map γ1 : X1 → Ker(α). Define γ: X → B as follows: γ(x) = γ1 (x) if x ∈ X1 , γ(x) = γ2 (x) if x ∈ X2 , and γ(x) = 1 otherwise. Then γ extends to an epimorphism γ: F → B with α ◦ γ = ϕ.  Corollary 24.3.4: Each profinite group G has a cover H with the embedding property such that rank(G) = rank(H). If G is finite and rank(G) = e, e then H can be chosen with |H| ≤ |G||G| . Proof: By Corollary 17.4.8, G is the quotient of a free profinite group F with rank(F ) = rank(G). By Lemma 24.3.3, F has the embedding property. Now suppose G is finite and rank(G) = e. Then there are at most |G|e epimorphisms of F onto G. List the kernels of these epimorphisms as N1 , . . . , Nk , with k ≤ |G|e . Put H = F/(N1 ∩ · · · ∩ Nk ). Then H covers G, and Lemma 24.3.1 shows H has the embedding property. Finally, |H| ≤ Qk k  i=1 (F : Ni ) = |G| . Finally, we give a stronger version of the converse of Lemma 24.3.2(c): Proposition 24.3.5 ([Haran-Lubotzky]): If a profinite group G has the em˜ That is, G ˜ is bedding property, then so does its universal Frattini cover G. superprojective. ˜ → A and α: B → A be epimorphisms with B ∈ Im(G), ˜ and Proof: Let ϕ: G ˜ let ϕ: ˜ G → G be the universal Frattini cover of G. We prove the existence of ˜ → B with α ◦ θ = ϕ in cases: an epimorphism θ: G Case A: B ∈ Im(G).

By Lemma 22.2.8, ϕ and ϕ˜ are part of a commutative

24.4 The Smallest Embedding Cover of a Profinite Group

569

diagram of epimorphisms ˜ O?O G // ??OOO // ?π? OOOϕ˜O // ??? OOOO /  η2 OO' /G ϕ // H // // η1 γ2 /   A γ1 / C in which the square from H to C is Cartesian. Since G has the embedding property and A ∈ Im(G), there is an epimorphism ζ: G → A with γ1 ◦ ζ = γ2 . As the square is Cartesian, there is a homomorphism η: G → H with η2 ◦ η = IdG and η1 ◦ η = ζ. Applying Lemma 22.5.4(c) to ϕ˜ = η2 ◦ π we conclude that η2 is a Frattini cover. Since η2 (η(G)) = G, this implies η(G) = H. Therefore, η ◦ η2 = IdH . The embedding property of G gives an epimorphism β: G → B with ˜ onto B with α ◦ β = η1 ◦ η. Thus, θ = β ◦ ϕ˜ is an epimorphism of G α ◦ θ = α ◦ β ◦ ϕ˜ = η1 ◦ η ◦ η2 ◦ π = η1 ◦ π = ϕ, as needed. ˜ Part B: B ∈ Im(G). Lemma 22.6.5 gives a Frattini cover β: B → B 0 0 with B ∈ Im(G). The maps α and β define a commutative diagram of epimorphisms B0 APPP 00AAA PPPP α 00 πAA PPPP PPP 00 AA P( 00 D δ /A β 0 00 λ 00 δ0   / A0 B0 0 λ

where the square from D to A0 is Cartesian (Lemma 22.2.8). Part A gives ˜ → B 0 with λ0 ◦ θ0 = λ ◦ ϕ. Hence, there exists a an epimorphism θ0 : G ˜ homomorphism ζ: G → D with δ 0 ◦ ζ = θ0 and δ ◦ ζ = ϕ. Applying Lemma 22.5.4(c) to the Frattini cover β = π ◦ δ 0 we may conclude that both π and ˜ = θ0 (G) ˜ = B 0 , ζ is surjective. Since δ 0 are Frattini covers. Since δ 0 (ζ(G)) ˜ G is projective (Proposition 22.6.1) and π is a Frattini cover, there is an ˜ → B with π ◦ θ = ζ (Proposition 22.5.9(a)). Therefore, epimorphism θ: G ˜ has the embedding property. α ◦ θ = δ ◦ π ◦ θ = δ ◦ ζ = ϕ. It follows that G 

24.4 The Smallest Embedding Cover of a Profinite Group ˜ with the embedding property Every profinite group G has a smallest cover G ˜ (Proposition 24.4.5). By Chatzidakis, G is unique. We prove the uniqueness only under the assumption that rank(G) ≤ ℵ0 (Lemma 24.4.8).

570

Chapter 24. Frobenius Fields

˜ uses properties of double covers: Let G1 and The construction of G G2 be profinite groups. Consider the collection P = P(G1 , G2 ) of triples (π1 , π2 , A) with πi : Gi → A an epimorphism, i = 1, 2. Partially order P by (π1 , π2 , A) ≤ (π10 , π20 , A0 ) if there exists an epimorphism π: A0 → A which makes the diagram (1)

G11 B G2 11BBB π10 π20 |||

11 BB || 11 BB ~|||

1

π1 11 A0

π2 11

11 π

1  

A

commutative. Write (π1 , π2 , A) ∼ (π10 , π20 , A0 ) if π is an isomorphism. Then ≤ induces a partial ordering on the quotient of P by ∼. Let pri : G1 × G2 → Gi be projection onto the ith coordinate, i = 1, 2. Define H = H(G1 , G2 ), dual to P, to be the collection of closed subgroups H of G1 × G2 with pri (H) = Gi , i = 1, 2. Partially order H by inclusion. The collections P and H are related by a map T : P → H defined by T (π1 , π2 , A) = G1 ×A G2 (Section 22.2). Lemma 24.4.1: The map T : P → H induces an order reversing bijection between P/ ∼ and H. Proof: Let (π1 , π2 , A), (π10 , π20 , A0 ) ∈ P. Suppose there exists an epimorphism π such that Diagram (1) is commutative, then G1 ×A0 G2 ≤ G1 ×A G2 . Conversely, assume that G1 ×A0 G2 ≤ G1 ×A G2 . Each a0 ∈ A0 is of the form πi0 (gi ) for some gi ∈ Gi , i = 1, 2. Define π(a0 ) to be π1 (g1 ). This gives a well defined map π: A0 → A making Diagram (1) commutative. Indeed, suppose h1 ∈ G1 and π10 (h1 ) = a0 . Then (g1 , g2 ), (h1 , g2 ) ∈ G1 ×A0 G2 ≤ G1 ×A G2 , so π1 (g1 ) = π2 (g2 ) = π1 (h1 ). Thus, T induces an order reversing injection from P/ ∼ into H. To see that T is surjective let H ∈ H and ηi = pri |H , i = 1, 2. Lemma 22.2.8 produces a commutative diagram of epimorphisms H7  777  η 7  77   77η2 η1   G1 ×A G 2 7 J  J t JJ 777  η¯ ttt  η¯2 J 77 1  t JJ 7 J$  zttt G1 J G JJJ t 2 t t JJJ t ttπt π1 JJJ J% ytttt 2 A

24.4 The Smallest Embedding Cover of a Profinite Group

571

with η¯i the restriction of pri to G1 ×A G2 , i = 1, 2. For (g1 , g2 ) ∈ H we have η¯i (η(g1 , g2 )) = ηi (g1 , g2 ) = gi , i = 1, 2. Therefore, η(g1 , g2 ) = (g1 , g2 ) and  H = G1 ×A G2 = T (π1 , π2 , A). Lemma 24.4.2: (a) Each H ∈ H contains a minimal element of H. (b) For each (π1 , π2 , A) ∈ P there exists a maximal element (π10 , π20 , A0 ) ∈ P with (π1 , π2 , A) ≤ (π10 , π20 , A0 ). Proof: Let T {Hα } be a descending chain T in H. By Lemma 1.2.2(c), T pri ( Hα ) = pri (Hα ) = Gi , i = 1, 2, so Hα ∈ H. Thus, statement (a) follows from Zorn’s lemma. Statement (b) follows from (a) by Lemma 24.4.1.  The universal Frattini cover of a profinite group G is the smallest projective cover of G (Proposition 22.6.1). Thus, every Frattini cover of G is a quotient of each projective cover of G. The next definition introduces two types of covers, the “embedding covers” and “I-covers” which relate to each other as projective covers relate to Frattini covers: Definition 24.4.3: Call an epimorphism ϕ: H → G of profinite groups an embedding cover of G if H has the embedding property. In this case we also say that H is an embedding cover of G. Call an epimorphism ε: E → G of profinite groups an I-cover if for each embedding cover ϕ: H → G there exists an epimorphism γ: H → E with ε ◦ γ = ϕ. Note that If ε: E → G and ζ: F → E are I-covers, then so is ε ◦ ζ. Conversely, if ε and ζ are epimorphisms and ε ◦ ζ is an I-cover, then so is ε. Finally, an epimorphism ε: E → G which is both an embedding cover and an I-cover is a smallest embedding cover.  Lemma 24.4.4: If a profinite group G does not have the embedding property, then G has an I-cover ε: E → G with a nontrivial finite kernel. Proof: Suppose each I-cover of G with a finite kernel is an isomorphism. We show that G has the embedding property. Indeed, let α: B → A and α0 : G → A be epimorphisms with B ∈ Im(G). Lemma 24.4.2 gives a commutative diagram of epimorphisms /G 00 00 0 0 β ε 0   00 0 β B PPP / D @ 00α PPP @@@ 00 PPP π@ 0 α PPP@@@00 PP'  A E

ε

with (β, β 0 , D) a maximal element of P(B, G), and E = B ×D G. We prove ε is an I-cover.

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Chapter 24. Frobenius Fields

Let ϕ: H → G be an embedding cover. Then B ∈ Im(H). Therefore, there exists an epimorphism ϕ0 : H → B with β ◦ ϕ0 = β 0 ◦ ϕ. Since the square from D to E is Cartesian, there exists a homomorphism η: H → E with ε0 ◦ η = ϕ0 and ε ◦ η = ϕ. In particular, η(H) ∈ H(B, G). It follows from the minimality of E in H(B, G) (Lemma 24.4.1) that η(H) = E. Therefore, ε: E → G is an I-cover. By hypothesis ε is an isomorphism. Hence, β is an isomorphism and β −1 ◦ β 0 is an epimorphism from G to B satisfying α ◦ β −1 ◦ β 0 = α0 . Thus, G has the embedding property.  Proposition 24.4.5: Every profinite group G has a smallest embedding cover. Proof: Choose a free profinite group F and a cover ϕ: F → G (Proposition 17.4.8). Let K = Ker(ϕ) and m be a cardinal number greater than |F |. By transfinite induction construct a descending sequence {Lκ }κ≤m of closed normal subgroups of F in the following way. First choose L0 = K. Next suppose Lκ has been chosen such that the quotient map F/Lκ → F/K is an I-cover. If Lκ has a proper open subgroup M which is normal in F and the quotient map F/M → F/Lκ is an I-cover, choose Lκ+1 = M , otherwise let Lκ+1 = Lκ . In both cases the quotientTmap F/Lκ+1 → F/K is an I-cover. If λ ≤ m is a limit ordinal, let Lλ = κ<λ Lκ . Again, the canonical map F/Lλ → F/K is an I-cover. Indeed, suppose ϕ: H → F/K is an embedding cover. By transfinite induction construct epimorphisms ϕκ : H → F/Lκ , for κ < λ, which commute with the canonical maps. Their limit ϕλ = lim ϕκ ←− is an epimorphism ϕλ which gives ϕ when composed with the quotient map F/Lλ → F/K. Each open subgroup of Lκ is the intersection of an open subgroup of F with Lκ . Thus, in the definition of Lκ+1 , the case that Lκ+1 6= Lκ cannot occur more than |F | times (Proposition 17.1.2). Since |F | < m, there is a κ0 < m with Lκ = Lκ+1 for each κ ≥ κ0 . In particular, L = Lm is a normal closed subgroup of F contained in K, the quotient map π: F/L → F/K is an I-cover and there is no open normal subgroup M of F properly contained in L such that the quotient map F/M → F/L is an I-cover. To prove that π is a smallest embedding cover, it suffices to verify that F/L has the embedding property. If not, then Lemma 24.4.4, gives an Icover ε: E → F/L with a nontrivial finite kernel. Since F has the embedding property (Lemma 24.3.3), F → F/L is an embedding cover. Hence, there is an epimorphism θ: F → E such that ε ◦ θ is the quotient map. Let M = Ker(θ). Then M / F , M < L, and the quotient map F/M → F/K is an I-cover. This contradicts the preceding paragraph.  Lemma 24.4.6: Let ε: E → G be a smallest embedding cover of a profinite group G and C a formation of finite groups. (a) If G is a pro-C-group, then so is E. (b) rank(E) = rank(G).

24.4 The Smallest Embedding Cover of a Profinite Group

573

(c) If ε0 : E 0 → G is another smallest embedding cover of G, then there exist epimorphisms θ: E → E 0 and θ0 : E 0 → E with ε0 ◦ θ = ε and ε ◦ θ0 = ε0 . In particular, Im(E) = Im(E 0 ). (d) If G is small, then θ and θ0 of (c) are isomorphisms. Thus, in this case ε: E → G is, up to isomorphism, uniquely determined. We denote E by E(G). (e) If G is a finite group, then there is a primitive recursive procedure to construct E(G) and ε from G. Proof: To prove (a) and (b), choose a cover ϕ: Fˆm (C) → G with m = rank(G) (Proposition 17.4.8). Since Fˆm (C) has the embedding property (Lemma 24.3.3), there is an epimorphism θ: Fˆm (C) → E. Hence, E is a pro-C-group and rank(E) = rank(G). Assertion (c) follows from the definition of a smallest e-embedding cover. Assertion (d) is a consequence of (c) and Proposition 16.10.6(b). Finally, let G be a finite group of rank e. Then G has an embedding e cover H of order at most |G||G| (Corollary 24.3.4). Note that E(G) is the finite cover of G of a smallest order with the embedding property. Thus, we may find E(G) by checking the finitely many groups of order bounded by e  |G||G| that cover G. Small profinite groups are characterized by their finite quotients (Proposition 16.10.7(b)). Thus, the smallest embedding cover of such groups is unique (Lemma 24.4.6(d)). A back and forth argument partially generalizes this to profinite groups of countable rank: Lemma 24.4.7: Let G and G0 be profinite groups of at most countable rank, A a finite group, and ϕ: G → A, ϕ0 : G0 → A epimorphisms. Suppose Im(G) = Im(G0 ) and both G and G0 have the embedding property. Then there exists an isomorphism θ: G → G0 with ϕ0 ◦ θ = ϕ. Proof: Choose descending sequences Ker(ϕ) ≥ K1 ≥ K2 ≥ K3 ≥ · · · and Ker(ϕ0 ) ≥ K10 ≥ K20 T≥ K30 ≥ · · · of T open normal subgroups of G ∞ ∞ and G0 , respectively, with i=1 Ki = 1 and i=1 Ki0 = 1. Inductively define descending sequences Ker(ϕ) ≥ L1 ≥ M1 ≥ L2 ≥ M2 ≥ · · ·, and Ker(ϕ0 ) ≥ L01 ≥ M10 ≥ L02 ≥ M20 ≥ · · · of open normal subgroups of G and G0 , respectively, and isomorphisms θi0 : G0 /L0i → G/Li , θi : G/Mi → G0 /Mi0 , i = 1, 2, . . . satisfying this: (2a) L1 = K1 , Li = Ki ∩ Mi−1 , i = 2, 3, . . . . (2b) Mi0 = Ki0 ∩ L0i , i = 1, 2, . . . . ¯ G/L1 → A is induced by ϕ and ϕ0 : G0 /L01 → A is (2c) ϕ¯ ◦ θ0 = ϕ0 , where ϕ: 0 induced by ϕ . (2d) For each n ≥ 2 the following diagram, where the vertical arrows are

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defined as the quotient maps, is commutative. (3)

G0

G

0 πn

 G0 /L0n

πn

0 θn

λ0n

 / G/Ln

λn

  0 o θn−1 G/Mn−1 G0 /Mn−1 µ0n

 G0 /L0n−1

µn 0 θn−1

 / G/Ln−1

Indeed, G/L1 ∈ Im(G) = Im(G0 ). Hence, there exists an epimorphism G0 → G/L1 with ϕ◦ζ ¯ 10 = ϕ0 . Put L01 = Ker(ζ10 ) and let θ10 : G0 /L01 → G/L1 be the isomorphism induced by ζ10 . Then ϕ¯ ◦ θ10 = ϕ¯0 . Let now n ≥ 2. Inductively suppose the bottom line of (3) has al0 0 = Kn−1 ∩ L0n−1 . Then ready been constructed. Following (2b), put Mn−1 0 0 0 G /Mn−1 ∈ Im(G ) = Im(G). Hence, there is an epimorphism ζn−1 : G → 0 ◦ µ0n ◦ ζn−1 : G → G/Ln−1 is the quotient map. Put G0 /Mn−1 such that θn−1 0 be the isomorphism Mn−1 = Ker(ζn−1 ) and let θn−1 : G/Mn−1 → G0 /Mn−1 which ζn−1 induces. Then the lower square of (3) is commutative. Similarly one constructs an isomorphism θn0 : G0 /L0n → G/Ln making the upper square of (3) commutative. This concludes the induction The isomorphisms θi , i = 1, 2, . . ., define an isomorphism θ: G → G0 with ϕ0 ◦ θ = ϕ.  ζ10 :

Corollary 24.4.8: Let G be a profinite group of rank at most ℵ0 . Suppose ε: E → G and ε0 : E 0 → G0 are smallest embedding covers of G. Then E ∼ = E0. Proof: By Lemma 24.4.6(b), rank(E) = rank(G) = rank(E 0 ). Hence, both rank(E) and rank(E 0 ) are at most ℵ0 . Also, Im(E) = Im(E 0 ) (Lemma 24.4.6(c)). Since both E and E 0 have the embedding property, Lemma 24.4.7 applied to A = 1 implies E ∼  = E0. Actually, the uniqueness of the smallest embedding cover of a small group (Lemma 24.4.6(d)) and Corollary 24.4.8 are special cases of a result of Chatzidakis asserting the uniqueness of the smallest embedding cover of each profinite group (see Notes to this chapter).

24.5 A Decision Procedure Let A and B be finite collections of finite groups. This section establishes a decision procedure for the existence of a superprojective group that has the elements of A among its quotients but none of the elements of B. This is a key ingredient in a decision procedure for Frobenius fields (Chapter 30).

24.5 A Decision Procedure

575

Lemma 24.5.1: (a) Let A and A0 be finite quotients of a group Γ with the embedding property. Then each minimal element H of H(A, A0 ) is a quotient of Γ. (b) Let A1 , . . . , Am be finite quotients of a group Γ with the embedding property and e a positive integer. Suppose rank(Ai ) ≤ e, i = 1, . . . , m. Then A1 × · · · × Am contains a finite quotient H of Γ of rank bounded by e with pri (H) = Ai , i = 1, . . . , m. Proof of (a): Let H be a minimal element of H(A, A0 ). Then the projections η: H → A and η 0 : H → A0 are surjective. By Lemma 24.4.1, there exist a group B and epimorphisms α: A → B and α0 : A0 → B such that H = A×B A0 . Let ϕ: Γ → A be an epimorphism. Since Γ has the embedding property and since A0 ∈ Im(Γ) there is an epimorphism ϕ0 : Γ → A0 with α ◦ ϕ = α0 ◦ ϕ0 . So, there is a homomorphism γ: Γ → H with η ◦ γ = ϕ and η 0 ◦ γ = ϕ0 . Since the image γ(Γ) is contained in H(A, A0 ), the minimality of H implies γ is surjective. Proof of (b): First consider the case m = 2. Choose a minimal element H of H(A1 , A2 ). By (a), H ∈ Im(Γ). Also, Ai ∈ Im(Fˆe ), i = 1, 2, and Fˆe has the embedding property (Lemma 24.3.3). Therefore, by (a), H ∈ Im(Fˆe ). Thus, rank(H) ≤ e. An induction on m proves (b).  Proposition 24.5.2 ([Haran-Lubotzky, p. 200]): Let C be a full formation of finite groups. Consider groups A1 , . . . , Am , B1 , . . . , Bn in C of rank at most e with m ≥ 1 and n ≥ 0. Let A be the collection of all subgroups A of A1 × · · · × Am with rank(A) ≤ e and pri (A) = Ai , i = 1, . . . , m. Then the following conditions are equivalent: (a) There exists a superprojective pro-C-group Γ with A1 , . . . , Am ∈ Im(Γ) and B1 , . . . , Bn 6∈ Im(Γ). (b) There exists a superprojective pro-C-group Γ of rank at most e with A1 , . . . , Am ∈ Im(Γ) B1 , . . . , Bn 6∈ Im(Γ). (c) There exists A ∈ A such that none of the groups B1 , . . . , Bn is a Frattini cover of a quotient group of E(A). Proof of “(a) =⇒> (c)”: Lemma 24.5.1 gives A ∈ A ∩ Im(Γ). Since Γ is an embedding cover of A, the smallest embedding cover E = E(A) of A is a ˜ of E is the smallest projective quotient of Γ. The universal Frattini cover E cover of E (Proposition 22.6.1). Therefore, as Γ is a projective cover of E, ˜ is a quotient of Γ. Thus, none of the groups B1 , . . . , Bn is a quotient of E ˜ It follows from Lemma 22.6.5 that no B1 , . . . , Bn is a Frattini cover of a E. quotient of E. Proof of “(c) =⇒ (b)”: By Proposition 24.4.6, E = E(A) is a pro-C-group ˜ is a pro-C-group and rank(E) ˜ ≤ e. and rank(E) ≤ e. Lemma 22.6.4 implies E ˜ ˜ Also, E is superprojective (Proposition 24.3.5). Clearly A1 , . . . , Am ∈ Im(E) ˜ and (by Lemma 22.6.5), none of the groups B1 , . . . , Bm is a quotient of E. 

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Corollary 24.5.3: Let C be a primitive recursive full formation of finite groups and e a positive integer. There is a primitive recursive procedure to decide whether given groups A1 , . . . , Am , B1 , . . . , Bn ∈ C of rank at most e satisfy the conditions of Proposition 24.5.2. Proof: There is a primitive recursive construction of the collection A of Lemma 24.5.2 corresponding to A1 , . . . , Am . Also, there is a primitive recursive construction of E(A) for each A ∈ A (Proposition 24.4.6(e)). Thus, there is a primitive recursive check for the validity of (c) of Proposition 24.5.2. 

24.6 Examples This section presents examples that, in particular, show the independence of the concepts introduced in Sections 24.1 - 24.5. Example 24.6.1: Finite simple groups. Each finite simple groups G has the ˜ embedding property (but is not projective). The universal Frattini cover, G, of G is superprojective (Proposition 24.3.5).  Lemma 24.6.2: Let G be a profinite group of at most countable rank. Then G has the embedding property if and only if for each pair of epimorphisms ϕ: G → A and ψ: G → A onto a finite group A, there exists an epimorphism θ: G → G with ψ ◦ θ = ϕ. Proof: Necessity is given by Lemma 24.4.7. Conversely, suppose G has the property stated in the lemma relative to each finite group A. Let ϕ: G → A and α: B → A be epimorphisms. Assume that B ∈ Im(G). Then there exists an epimorphism π: G → B. By assumption, there exists an epimorphism θ of G onto G with α ◦ π ◦ θ = ϕ. The epimorphism γ = π ◦ θ of G onto B satisfies α ◦ γ = ϕ. Thus, G has the embedding property.  Example 24.6.3: [Ershov3, p. 511]. Lemma 24.4.7 requires A to be a finite group. Suppose G = Fˆω = A and ϕ = Id. Let {x1 , x2 , . . .} be a basis of Fˆω and define the epimorphism ψ: Fˆω → Fˆω by ψ(x1 ) = 1 and ψ(xi+1 ) = xi , i = 1, 2, . . . . Then there exists no epimorphism θ: Fˆω → Fˆω with ψ ◦ θ = id because any x ∈ Fˆω with θ(x) = x1 would give x = ψ(x1 ) = 1, so x1 = 1, a contradiction.  Example 24.6.4: Z/pm Z modules with and without the embedding property. Consider the group G = (Z/pm1 Z) × · · · × (Z/pmr Z) with 1 ≤ m1 ≤ · · · ≤ mr = m. Thus, G is a finitely generated Z/pm Z module. We show G has the embedding property if and only if it is a free Z/pm Z module (i.e. m1 = m). First suppose m1 < m. Let ϕ: G → Z/pm1 Z be the projection onto the first factor. Let ψ: G → Z/pm1 Z be the projection onto the rth factor of G multiplied by pm−m1 (We identify Z/pm1 Z with the group of all elements of Z/pm Z divisible by pm−m1 .) Assume G has the embedding property. Then there exists an epimorphism θ: G → G with ψ ◦ θ = ϕ. Let θ(1, 0, . . . , 0) =

24.6 Examples

577

(a1 , a2 , . . . , ar ). Then ψ ◦ θ(1, 0, . . . , 0) = pm−m1 ar . Since (1, 0, . . . , 0) is of order pm1 , we have pm1 ar = 0. As m1 < m, this implies pm1 −1 ψ ◦ θ(1, 0, . . . , 0) = pm−1 ar = 0. On the other hand ϕ(1, 0, . . . , 0) = 1 + pm1 Z has order pm1 , so pm1 −1 (pm−m1 ar ) 6= 0. This contradiction proves that G does not have the embedding property. Now, suppose m1 = · · · = mr = m. Choose a basis x1 , . . . , xr for G as a Z/pm Z module. Let ϕ: G → A and ψ: G → A be epimorphisms. Then there are generators y1 , . . . , yr of G with ψ(yi ) = ϕ(xi ), i = 1, . . . , r (e.g. by Gasch¨ utz’s lemma, Lemma 17.7.2). Extend the map (x1 , . . . , xr ) →  (y1 , . . . , yr ) to an automorphism θ of G with ψ ◦ θ = ϕ. Example 24.6.5: Necessity of the condition Φ(G) = 1 in Lemma 24.3.2(c). Consider H = Z/p2 Z×Z/p2 Z and G = Z/pZ×Z/p2 Z. Then the epimorphism ε: H → G which is multiplication by p on the first coordinate and the identity on the second coordinate has kernel isomorphic to Z/pZ. By Example 24.6.4, ˜ = H ˜ = H is the smallest embedding cover of G. By Corollary 22.7.8, G ˜ ˆ F2 (p) and G is a superprojective group (Lemma 24.3.3). Note that Φ(G) = Φ(Z/p2 Z) = Z/pZ 6= 1 (Lemma 22.1.4(d)). Thus, the condition Φ(G) = 1 is necessary for the conclusion of Corollary 24.3.2(c) to hold.  Lemma 24.6.6: (a) Let G be a small profinite group with the embedding property. Let N and N 0 be open normal subgroups with G/N 0 ∼ = G/N . Then N ∼ = N 0. (b) Let A and B be finitely generated profinite groups. Suppose B acts nontrivially on A. Choose an isomorphic copy B 0 of B. Then B 0 ×(BnA) does not have the embedding property. Proof of (a): Let π: G → G/N and π 0 : G → G/N 0 be the quotient maps and α: G/N 0 → G/N an isomorphism. Lemma 24.6.2 gives an epimorphism θ: G → G with α ◦ π 0 ◦ θ = π. Since, G is finitely generated, θ is an automorphism (Corollary 16.10.8). Hence, θ(N ) = N 0 and N ∼ = N 0. Proof of (b): Set N = B n A, N 0 = B 0 × A and G = B 0 × (B n A) in (a).  Example 24.6.7: A PAC field which is not a Frobenius field. The dihedral group Dn is the semidirect product of Z/nZ and {±1}, where −1 acts on Z/nZ by the rule x 7→ −x. For n > 2, this action is nontrivial. Hence, by Lemma 24.6.6(b), G = Z/2Z × Dn does not have the embedding property. Suppose in addition, n = p is a prime. Then both ±1 and Z/pZ are maximal subgroups of Dp . So, Φ(G) = Φ(Dp ) = 1. Apply Lemma 24.3.2(c) to con˜ does not have the embedding property. Thus, G ˜ is a projective clude that G group which is not superprojective. Corollary 23.1.2, gives a PAC field F ˜ Thus, F is not a Frobenius field. with G(F ) ∼  = G. Each algebraic extension of a PAC field is a PAC field (Corollary 11.2.5). The remainder of this section concludes with examples showing the analogous result for Frobenius fields is false.

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Chapter 24. Frobenius Fields

Example 24.6.8: Products of symmetric groups without the embedding property and without nontrivial nilpotent normal subgroups. Let m and n be distinct positive integers exceeding 4, and let G = Sm ×Sn . The subgroups N1 = Am × Sn and N2 = Sm × An are normal, and G/N1 ∼ = Z/2Z ∼ = G/N2 . If we prove N1 6∼ = N2 , we can deduce from Lemma 24.6.6 that G does not have the embedding property (Lemma 24.25(a)). Assume, there is an isomorphism ϕ: N1 → N2 . Then pr2 (ϕ(Am × 1)) / An , where pr2 is the projection on the second factor of Sm × An . Since Am and An are simple nonisomorphic groups, pr2 (ϕ(Am × 1)) = 1. Thus, ϕ(Am × 1) = Am × 1. Hence, Am × 1 is of index 2 in ϕ−1 (Sm × 1). Therefore, pr2 (ϕ−1 (Sm × 1)) is a normal subgroup of Sn of order at most 2. But Sn has no normal subgroup of order 2. It follows that pr2 (ϕ−1 (Sm × 1)) = 1 and ϕ−1 (Sm × 1) ≤ Am × 1. This is a clear contradiction. Note in addition that if N is a normal nilpotent subgroup of G, then so  are its projections onto Sm and Sn . This implies N = 1. Lemma 24.6.9: Q ∞ (a) Suppose G = i=1 Gi is a direct product of profinite groups with the embedding property and G1 , G2 , . . . are of pairwise relatively prime orders. Then G has the embedding property. (b) Each projective pronilpotent group N has the embedding property. Proof of (a): LetQϕ: G → A and Q∞α: B → A be epimorphisms with B ∈ ∞ Im(G). Then A = i=1 Ai , B = i=1 Bi and ϕ and α are the direct product of epimorphisms ϕi : Gi → Ai and αi : Gi → Ai with Bi ∈ Im(Gi ), i = 1, 2, . . . (Exercise 13 of Chapter 22). Since Gi has the embedding property, the result follows. Proof of (b): By Proposition 22.9.3, N is the direct product of its p-Sylow groups. Each p-Sylow group is p-free (Proposition 22.10.4). Therefore, each p-Sylow group has the embedding property (Lemma 24.3.3). Thus, N satisfies the hypotheses of (a), and it too has the embedding property.  Lemma 24.6.10: Let G be a small profinite group with the embedding property and N a closed characteristic subgroup of G. Then G/N has the embedding property. Proof: Let ϕ and ψ be epimorphisms of G/N onto a finite group A. Let π: G → G/N be the quotient map. Lemma 24.6.2 gives an epimorphism θ: G → G with ψ◦π◦θ = ϕ◦π. By Proposition 16.10.6, θ is an automorphism. Since N is characteristic, θ(N ) = N . Hence, θ induces an automorphism θ¯ of G/N with ψ ◦ θ¯ = ϕ. If follows from Lemma 24.6.2 that G/N has the embedding property.  Example 24.6.11: [Haran-Lubotzky, p. 199] Neither projective subgroups nor overgroups of finite index of superprojective groups must be superprojective. We give a sequence of projective groups, G1 < G2 < G3 , with G1 open and normal in G3 , G1 , and G3 superprojective, and G2 not superprojective.

24.7 Non-projective Smallest Embedding Cover

579

Consider a finite group G without the embedding property, having no nontrivial nilpotent normal subgroups. For example G = Sm × Sn with m, n distinct integers larger than 4 (Example 24.6.8). Embed G into a simple group S. For example first embed G in Sk , with k ≥ 3. Then embed Sk into Ak+2 by f (π) = π for π ∈ Sk even and f (π) = (k+1 k+2)π for π odd. ˜ G2 = Let ϕ: S˜ → S be the universal Frattini cover of S. Put G3 = S, ˜ Since G3 is projective (Proposition 22.6.1), so are ϕ−1 (G) and G1 = Φ(S). G1 and G2 (Proposition 22.4.7). Furthermore, G1 is pronilpotent (Lemma 22.1.2). It therefore has the embedding property (Lemma 24.6.9(b)). In addition, S˜ has the embedding property (Example 24.6.1 and Proposition 24.3.5). We show G2 does hot have the embedding property. First note that since Φ(S) = 1, we have Ker(ϕ) = G1 . Thus, G1 is open in G3 . If N is an open normal nilpotent subgroup of G2 , then ϕ(N ) is a normal nilpotent subgroup of G. Hence, it is trivial, and N is contained in G1 . Thus, G1 is a maximal open normal nilpotent subgroup of G2 . In particular, G1 is an open characteristic subgroup of G2 . By Corollary 22.5.3, S˜ is finitely generated. Hence, by Corollary 17.6.3, G2 is also finitely generated. Since, however, G does not have the embedding property, Lemma 24.6.10 implies  that G2 also does not have it. Corollary 24.6.12: There exists a tower of PAC fields, K ⊂ L ⊂ M , satisfying this: M is a finite Galois extension of K, both K and M are Frobenius fields, but L is not. Proof: Let G1 < G2 < G3 be as in Example 24.6.11. Corollary 23.1.2 ˜ 2 ) and gives a perfect PAC field K with Gal(K) ∼ = G3 . Choose L = K(G ˜ M = K(G1 ). 

24.7 Non-projective Smallest Embedding Cover Although “projective cover” and “embedding cover” play analogous roles in the theory of profinite groups, their roles are not completely symmetric. Whereas the smallest projective cover of a profinite group with the embedding property has the embedding property (Proposition 24.3.5), there are projective groups whose smallest embedding covers are not projective (Lemma 24.7.3): Lemma 24.7.1: Let N be a non trivial closed normal subgroup of a profinite group G. For each prime l choose an l-Sylow group Gl of G. Suppose N 6= 1. Then there is an l with N ∩ Gl 6= 1. Proof: Choose an l which divides the order of N . Then its l-Sylow group  N ∩ Gl (Proposition 22.9(d)) is nontrivial. Lemma 24.7.2: Let G be a profinite group and ε: E → G the smallest embedding cover of G. Then the orders of G and E are divisible by the same prime numbers.

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Proof: Denote the set of all prime numbers that divide #G by S. Let C be the formation of all finite groups whose orders are divisible only by prime numbers belonging to S. Then C is full. Lemma 17.4.8 gives an epimorphism ϕ: F → G, where F is a free pro-C group. By Lemma 24.3.3, F has the embedding property. Hence, there is an epimorphism γ: F → E such that ε ◦ γ = ϕ. It follows that #G and #E are divisible by the same prime numbers.  Proposition 24.7.3 ([Chatzidakis3, Prop. 1.2]): Let p be an odd prime number and let H = H2 n Hp be the profinite group defined by the following rules: H2 = ha, bi is the free pro-2-group on a, b, Hp = hci ∼ = Zp , and ca = c−1 , cb = c. Then H is projective but its smallest embedding cover is not projective. Proof: For each prime number l, every l-Sylow group of H is l-free. Hence, H is projective (Proposition 22.10.4). We prove in four parts that the smallest embedding cover π: E → H is not projective. Part A: H is generated by two elements, namely a and bc. Indeed, choose a generator u for Zp and a generator v for Z2 . Since bc = cb, the map (u, v) 7→ (b, c) extends to an epimorphism Z2 ×Zp → hb, ci. Since uv generates Z2 × Zp , bc generates hb, ci. Hence, H = ha, b, ci = ha, bci, as claimed. Part B: H does not have the embedding property. Indeed, consider the Klein group A2 = ha0 , b0 i of order 4 defined by the relations a20 = b20 = 1 and a0 b0 = b0 a0 . The group A2 acts on the cyclic group Ap = hc0 i of order and cb00 = c0 . The semidirect product A = A2 n A0 is a p by ca0 0 = c−1 0 quotient of H via the map (a, b, c) → (a0 , b0 , c0 ). Consider the epimorphism α: A → A2 defined by α(a0 ) = b0 , α(b0 ) = a0 , and α(c0 ) = 1. Its kernel is hc0 i. Consider the epimorphism η: H → A2 defined by η(a) = a0 , η(b) = b0 , and η(c) = 1. Assume there is an epimorphism θ: H → A with α ◦ θ = η. Then η maps the p-Sylow group hci of H onto the p-Sylow group hc0 i of A, so θ(c) = ci0 , where i is relatively prime to p. Also, θ(a) = cj0 b0 . Apply θ to the relation a−1 ca = c−1 to get ci0 = c−i 0 . Hence, p|2i, a contradiction. Therefore, θ does not exist and H does not have the embedding property, as claimed. Part C: Ep is Abelian. Indeed, let F be the free profinite group on two generators x, y. Use Part A to define an epimorphism ϕ: F → H by ϕ(x) = a and ϕ(y) = bc. Let U = ϕ−1 (hci). Then F/U ∼ = H/hci ∼ = H2 is the free pro-2 ∼ group of rank 2, N = Ker(ϕ) ≤ U , and U/N = hci ∼ = Zp . Suppose U0 is a closed normal subgroup of F , U0 ≤ U , and F/U0 is a pro-2 group. Then rank(F/U0 ) = 2, so the quotient map F/U0 → F/U is an isomorphism (Lemma 17.4.11). Hence, U0 = U . Thus, U is the smallest closed normal subgroup of F such that F/U is a pro-2 group. As such, U is a characteristic subgroup of F .

24.8 A Theorem of Iwasawa

581

Let V be the smallest closed normal subgroup of U such that U/V is an Abelian pro-p group. Then V is characteristic in U , hence in F . Since U/N ∼ = hci ∼ = Zp , the group N contains V . Let ϕ0 : F/V → H be the epimorphism which ϕ induces. Since F has the embedding property, so does F/V (Lemma 24.6.10). Hence, there exists an epimorphism γ: F/V → E with π ◦ γ = ϕ0 . Note that U/V is a p-Sylow group of F/V . Thus, γ maps U/V onto a p-Sylow group Ep of E. Since U/V is Abelian, so is Ep . Part D: Conclusion of the proof. Since N/V is contained in U/V and the latter group is pro-p, the intersection of N/V with (F/V )2 is trivial. Thus, ϕ0 is injective on (F/V )2 . Hence, π is injective on E2 = γ((F/V )2 ). The only prime numbers which divide the order of F/V are 2 and p. By Part B, π is not injective (otherwise E ∼ = H and H would have the embedding property, in contradiction to Part A). Thus, Ker(π) 6= 1. By Lemma 24.7.2, 2 and p are the only prime numbers dividing #E. Hence, by Lemma 24.7.1, Ker(π) ∩ Ep 6= 1, so π is not injective on Ep . Since π(Ep ) = Hp ∼ = Zp and Ep is Abelian, this implies that Ep is not pro-p free. Consequently, E itself is not projective.  Remark 24.7.4: Solution of two open problems of [Fried-Jarden3]. The example, E → H, which Proposition 24.7.3 gives solves [Fried-Jarden3, Prob˜ is its own smallest lem 23.16(a)] negatively. Since H is projective, H = H projective cover. On the other hand, as E = E(H) is not projective, so the smallest projective cover of E is different from E. Thus, Proposition 24.7.3 also gives a negative answer to [Fried-Jarden3, Problem 23.16(b)]. 

24.8 A Theorem of Iwasawa The back and fourth argument of Lemma 24.4.7 led Iwasawa to characterize Fˆω as a profinite group of at most countable rank for which every finite embedding problem is solvable [Iwasawa, p. 567]. Here are some generalizations of that characterization: Theorem 24.8.1: Let C be a formation of finite groups and F a pro-C group of at most countable rank. Then F ∼ = Fˆω (C) if and only if Im(F ) = C and F has the embedding property. Proof: By Lemma 24.3.3, Fˆω (C) has the embedding property. In addition, Im(Fˆω (C)) = C. Thus, our theorem is a special case of Lemma 24.4.7.  Corollary 24.8.2: Let C be a formation of finite groups and F a pro-C group of at most countable rank. Each of the following condition suffices (and also necessary) for F to be isomorphic to Fˆω (C): (a) Every C-embedding problem for F is solvable. (b) F is C-projective and every split C-embedding problem for F is solvable. Proof: For each B ∈ C the embedding problem (F → 1, B → 1) is finite and splits, hence solvable. Thus, B ∈ Im(F ).

582

Chapter 24. Frobenius Fields

If either every C-embedding problem for F is solvable or F is C-projective and each split-embedding problem for F is solvable, then F has the embedding property (Proposition 22.5.9). Therefore, by Theorem 24.8.1, F ∼ =  Fˆω (C). The special case where C is the formation of all finite groups gives Iwasawa’s theorem: Corollary 24.8.3 ([Iwasawa, p. 567]): Let F be a profinite group of at most countable rank. Suppose every finite embedding problem for F is solvable. Then F ∼ = Fˆω . Denote the maximal prosolvable extension of a field K by Ksolv . Corollary 24.8.4: Let K be a Hilbertian field. Suppose Gal(K) is projective and of rank at most ℵ0 (e.g. K is countable). Then Gal(Ksolv /K) ∼ = Fˆω (solv). Proof: Put G = Gal(Ksolv /K). Consider a finite embedding problem (1)

(ϕ: G → A, α: B → A)

where B is solvable and C = Ker(α) is a minimal normal subgroup of B. Then C is Abelian. If (1) splits, it has a solution, by Proposition 16.4.5. Hence, by Proposition 22.5.9, (1) has a solution even if (1) does not split. It follows from Corollary 24.8.2 that G ∼  = Fˆω (solv). Example 24.8.5: Countable Hilbertian fields with projective absolute Galois groups. (a) Consider the maximal Abelian extension Qab of Q. By Theorem 16.11.3, Qab is Hilbertian. Class field theory implies Gal(Qab ) is projective [Ribes, p. 302]. Hence, by Corollary 24.8.4, Gal(Qsolv /Qab ) ∼ = Fˆω (solv). A ∼ long standing conjecture of Iwasawa predicts that Gal(Qsolv ) = Fˆω . Suppose we could prove that every non-Abelian finite simple group is GAR over Qab (Definition 16.8.1). Then, by Matzat (Proposition 16.9.1), every finite embedding problem for Gal(Qab ) would be solvable. Therefore, by Corollary 24.8.3, Gal(Qab ) ∼ = Fˆω . (b) Let K be a countable PAC Hilbertian field. Then Gal(K) is projective (Theorem 11.6.2). Thus, by Corollary 24.8.3, Gal(Ksolv /K) ∼ = Fˆω (solv). This is also a consequence of the much stronger result (2)

Gal(K) ∼ = Fˆω

which was stated as Problem 24.41 of [Fried-Jarden3]. This was proved for the first time when char(K) = 0 in [Fried-V¨ olklein2]. In the general case one considers a finite split embedding problem (3)

(ϕ: Gal(K) → A, α: B → A).

24.9 Free Profinite Groups of at most Countable Rank

583

and the associated embedding problem (4)

(ϕ ◦ res: Gal(K(t)) → A, α: B → A).

When K is PAC (or even ample) (3) is solvable ([Pop1, Main Theorem A] or [Haran-Jarden6, Thm. B]). When K is Hilbertian, one uses Lemma 13.1.1 to specialize the solution of (4) to a solution of (3). It follows from Corollary 24.8.2 that Gal(K) ∼ = Fˆω .  Remark 24.8.5(b) gives evidence to the following generalization of Iwasawa’s conjecture: Conjecture 24.8.6 ([Fried-V¨ olklein2, p. 470]): Let K be a Hilbertian field with Gal(K) projective. Then each finite embedding problem for Gal(K) is solvable.

24.9 Free Profinite Groups of at most Countable Rank We prove for a Melnikov formation C that if an open subgroup M of a closed normal subgroup of a free pro-C group is pro-C, then M has the embedding property (Corollary 24.9.4). If in addition, rank(M ) ≤ ℵ0 and Im(M ) = C, then M ∼ = Fˆω (C). This is an analog of Weissauer’s theorem (Theorem 13.9.1) about extensions of Hilbertian fields. Let G be a profinite group and S a finite simple group. Denote the intersection of all open normal subgroups H of G with G/H ∼ = S by MG (S). First suppose, S is the cyclic group Cp of a prime order p. Let Np be the intersection of all closed normal subgroup H of G with G/H a p-group. Then G/Np is the maximal pro-p quotient of G (Definition 17.3.2), MG (S)/Np = Φ(G/Np ), and G/MG (S) ∼ = Cpm for some cardinal number m (Lemma 22.7.1). Now suppose S is non-Abelian and G has m open normal subgroups N with G/N ∼ = S m . In both cases denote m by = S. By Lemma 18.3.11, G/MG (S) ∼ rG (S). For a fixed G, rG is a function from the set of all finite simple groups to the set of all cardinal numbers at most max(ℵ0 , rank(G)). We call rG the S-rank function of G. We write rG ≤ rH for profinite groups G, H if rG (S) ≤ rH (S) for each finite simple group S. Recall subgroups N1 , . . . , Nr of G are µ-independent Trthat open Qnormal r when G/ i=1 Ni ∼ = i=1 G/Ni (Lemma 18.3.7). For r = 2, this is equivalent to N1 N2 = G. Lemma 24.9.1: Let G be a profinite group, S a finite simple group, and N1 , . . . , Nr µ-independent open normal subgroups with G/Ni ∼ = S, i = 1, . . . , r. 0 be µ-independent open normal subgroups with G/Nj0 ∼ (a) Let N10 , . . . , Nr+s = S, j = 1, . . . , r + s. Then there are distinct j1 , . . . , js in {1, . . . , r + s} such that N1 , . . . , Nr , Nj01 , . . . , Nj0s are µ-independent.

584

Chapter 24. Frobenius Fields

∼ (b) Let L1 , . . . , Ls be open Trof G with G/Lj = S, j = Ts normal subgroups 1, . . . , s. Put L = j=1 Lj and N = i=1 Ni . Let M be an open subgroup of N with M / G and r = rG/M (S). Suppose L and N are µ-independent. Then L and M are µ-independent. Tr+s Tr Proof of (a): Put N = i=1 Ni and N 0 = j=1 Nj0 . Suppose first S ∼ = Cp . Then G/N ∩ N 0 ∼ = Ftp for some t ≥ r + s. Now use Steinitz’s Replacement 0 be two sets of linearly independent Theorem: Let v1 , . . . , vr and v10 , . . . , vr+s vectors in a vector space V . Then there are j1 , . . . , js in {1, . . . , r + s} such that v1 , . . . , vr , vj0 1 , . . . , vj0 s are linearly independent. Now suppose S is non-Abelian. Choose j1 , . . . , js with N1 , . . . , Nr , By Example 18.3.11, these subgroups are µNj01 , . . . , Nj0s distinct. independent. Proof of (b): By assumption, N L = G and N/M = MG/M (S). Also, G/M L, being a quotient of G/L, is a direct product of isomorphic copies of S. Since rG/M (S) = rG/N (S), we have N ≤ M L. Thus, M L = G, so L and M are µ-independent.  The following result gives two simple properties of rG (S): Lemma 24.9.2: ¯ is a homomorphic image of G, then rG¯ ≤ rG . (a) If G (b) Let G1 , G2 be finite groups. Then rG1 ×G2 = rG1 + rG2 . ¯ be an epimorphism. Then Proof of (a): Let ϕ: G → G MG (S) ≤ ϕ−1 (MG¯ (S)). Hence, rG¯ (S) ≤ rG (S) for each finite simple group S. Proof of (b): We have to prove that rG1 ×G2 (S) = rG1 (S) + rG2 (S) for each finite simple group S. For S = Cp use that the dimension of a direct sum of vector spaces is the sum of their dimensions. For S non-Abelian use Lemma 18.3.10.  Example 24.6.11 shows, by illustration with finite groups, it is rare that each open subgroup of a profinite group with the embedding property has the embedding property. However: Lemma 24.9.3: Suppose each open normal subgroup of a profinite group G has the embedding property. Then each closed normal subgroup of G has the embedding property. Proof: Let N be a closed normal subgroup of G and (1)

(ϕ: N → A, α: B → A)

a finite embedding problem for G with B ∈ Im(N ) and C = Ker(α) minimal normal in G (Definition 24.1.2). Let N1 = Ker(ϕ) and let M be an open

24.9 Free Profinite Groups of at most Countable Rank

585

normal subgroup of N with N/M ∼ = B. By Lemma 1.2.5, G has an open normal subgroup G0 such that N ∩ G0 ≤ N1 ∩ M . Then K = N G0 is an open normal subgroup of G, ϕ extends to a homomorphism λ: K → A by λ(ng0 ) = ϕ(n) for g0 ∈ G0 and n ∈ N , and K/M G0 ∼ = N/M ∼ = B. By assumption, K has the embedding property. Thus, there exists an epimorphism θ: K → B with α ◦ θ = λ. Put K1 = Ker(λ) and K2 = Ker(θ). Then N ∩ K1 = N1 , N K1 = K, K/K2 ∼ = B, and K1 /K2 ∼ = C. In particular, N1 and K2 are normal in K. Hence, N1 K2 /K2 is a normal subgroup of K/K2 which is contained in K1 /K2 . The latter group is minimal in K/K2 , so either N1 K2 = K1 or N1 ≤ K2 . Case 1: N1 K2 = K1 . Then N K2 = K. Hence, θ(N ) = θ(K) = B and θ|N : N → B solves embedding problem (2). Case 2: N1 ≤ K2 . Then L = N K2 is normal in K and L/K2 ∼ = N/N1 ∼ = A. Hence, L ∩ K1 = K2 . Thus, (2)

B∼ = L/K2 × K1 /K2 ∼ = A × C. = K/K2 ∼

Since C Q is a minimal normal subgroup of B, it is isomorphic to a direct s product i=1 Si of isomorphic copies of a single finite simple group S (Remark 16.8.4). Let r = rA (S) and E1 , . . . , Er µ-independent open normal subgroups of N which contain N1 and with N/Ei ∼ = S, i = 1, . . . , r. Since B ∈ Im(N ), Lemma 24.9.2 implies, rN (S) ≥ rB (S) = r + s. Thus, N has µ-independent open normal subgroups F1 , . . . , Fr+s with N/Fj ∼ = S, j = 1, . . . , r + s. Lemma 24.9.1(a) gives distinct j1 , . . . , js in {1, . . . , r + s} such that E1 , . . . , Er , Fj1 , . . . , Fjs are µ-independent. Put F = Fj1 ∩ · · · ∩ Fjs . Then N/F ∼ = Ss ∼ = C, and N1 and F are µ-independent (Lemma 24.9.1(b)). Thus, N/N1 ∩ F ∼ = N/N1 × N/F ∼ = A × C = B. We conclude that embedding problem (1) is solvable.  Corollary 24.9.4: Let C be a Melnikov formation of finite groups, F a free pro-C-group, N a closed normal subgroup of G, and M an open subgroup of N . Suppose M is a pro-C group. Then (a) M has the embedding property. (b) If, in addition, rank(F ) ≤ ℵ0 and Im(M ) = C, then M ∼ = Fˆω (C). Proof of (a): Choose an open normal subgroup D of F with D ∩ N ≤ M (Lemma 1.2.5). Then E is an open subgroup of F and M / E. Moreover, D, M , and N are pro-C groups, so E is also a pro-C group. By Proposition 17.6.2, E is pro-C free. Hence, each open normal subgroup of E is also pro-C free, so has the embedding property (Lemma 24.3.3). It follows from Lemma 24.9.3 that M has the embedding property. Proof of (b): Use (a) and Theorem 24.8.1.



Remark 24.9.5: We prove in Proposition 24.10.3 that if M in Corollary 24.9.4(b) is properly contained in N , then the condition Im(M ) = C is satis fied and M ∼ = Fˆω (C).

586

Chapter 24. Frobenius Fields

Example 24.9.6: Free profinite groups have closed subgroups without the embedding property. Indeed, if G is a projective, but not superprojective group (Example 24.6.7), then G is isomorphic to a closed subgroup of a free profinite group (Corollary 22.4.6). But G does not have the embedding property. 

24.10 Application of the Nielsen-Schreier Formula Let C be a Melnikov formation of finite groups and F a free pro-C group with 2 ≤ rank(F ) ≤ ℵ0 . Then every closed normal subgroup N of F is a pro-C group with the embedding property (Corollary 24.9.4). Yet, N need not be pro-C free, because Im(N ) may be a proper subset of C. For example, if C is the formation of all finite groups and N is the minimal closed subgroup such that F/N is a pro-p group, then Cp is not a quotient of N , so N is not free. We establish sufficient conditions for Im(N ) = C to hold, and therefore for N to be pro-C free. They are based on the Nielsen-Schreier’s rank formula (Proposition 17.6.2) for an open subgroup E of F which is pro-C: rank(E) = 1 + (F : E)(rank(F ) − 1). Lemma 24.10.1 ([Lubotzky-v.d.Dries, p. 35]): Let F be a free pro-C-group with 2 ≤ rank(F ) ≤ ℵ0 and N a closed normal subgroup F of infinite index. Suppose N is contained in an open subgroup H of F with (1)

rank(H/N ) < 1 + (F : H)(e − 1).

Then N ∼ = Fˆω (C). Proof: By Corollary 24.9.4(b) it suffices to prove that each group G ∈ C is a quotient of N . Indeed, there exists an open subgroup E of H with N ≤ E / F and (H : E) ≥ rank(G). Applying Corollary 17.6.3 to the group H/N and its open subgroup E/N we conclude that rank(E/N ) ≤ 1 + (H : E)(rank(H/N ) − 1). By (1), rank(H/N ) ≤ (F : H)(e − 1). Hence, rank(E/N ) ≤ 1 + (F : E)(e − 1) − (H : E). By Proposition 17.6.2, E is C-free. So, Proposition 17.7.4 gives a commutative diagram of epimorphisms E rrr r r π rrr xrrr α  / E/N (E/N ) × G γ

where π is the quotient map and α(π(x), g) = π(x) for each x ∈ E. In particular, for each g ∈ G there is an x ∈ E with γ(x) = (1, g). Then π(x) = α(γ(x)) = 1, so x ∈ N . Thus, γ(N ) = 1 × G and G is a quotient of N.  The following result is an analog of Kuyk’s result (Theorem 16.11.3) on Abelian extensions of Hilbertian fields:

24.10 Application of the Nielsen-Schreier Formula

587

Proposition 24.10.2 ([Jarden-Lubotzky1, Lemma 1.4]): Let F be a free pro-C group of rank m and N a closed normal subgroup of F . Suppose 2 ≤ m ≤ ℵ0 and F/N is Abelian and infinite. Then N ∼ = Fˆω (C). Proof: First suppose m < ℵ0 . Put A = F/N and let π: F → A be the quotient map. Choose a prime number p dividing the order of A. The map a 7→ ap maps A onto its subgroup Ap . Put H = π −1 (Ap ). Then (A : Ap ) > 1 and rank(H/N ) = rank(Ap ) ≤ rank(A) ≤ m < 1 + (A : Ap )(m − 1) = 1 + (F : H)(m − 1). Hence, by Lemma 24.10.1, N ∼ = Fˆω (C). Now consider the case where m = ℵ0 . By Corollary 24.9.4(b), it suffices to prove that each C-group G is a quotient of N . Indeed, choose an epimor¯ = ϕ(N ). phism ϕ: F → Fˆe (C) with e ≥ rank(G). Put F¯ = Fˆe (C) and N ¯ ¯ ¯ ¯ ¯ Then F /N is Abelian. If (F : N ) < ∞, then N is free of rank at least e ¯ ) = ∞, then N ¯ ∼ (Proposition 17.6.2). If (F¯ : N = Fˆ (C), by the preceding ¯ , hence also of N . paragraph. In both cases G is a quotient of N  The condition on M in the next proposition is the group theoretic version of the condition on the field M in Weissauer’s extension theorem of Hilbertian fields (Theorem 13.9.1(b)). Proposition 24.10.3 (Lubotzky-Melnikov-v.d.Dries): Let F be a free proC-group with 2 ≤ rank(F ) ≤ ℵ0 , N a closed normal subgroup of infinite index, and M a proper open subgroup of N . Suppose M is pro-C. Then M∼ = Fˆω (C). Proof: The proof splits into two parts according to the rank of F . Part A: rank(F ) = e is finite. Choose an open normal subgroup H of F with N ∩ H ≤ M (Lemma 1.2.5(a)) and put E = M H. Then H and M are pro-C, hence so is also E. Moreover, E is open in F and M / E. Then E is C-free and M is a closed normal subgroup of E. By Proposition 17.6.2, (F : N E)(e − 1) = rank(N E) − 1. Hence, rank(E/M ) = rank(N E/N ) ≤ 1 + (F : N E)(e − 1) < 1 + (N : M )(F : N E)(e − 1) ≤ 1 + (N E : E)(rank(N E) − 1). It follows from Lemma 24.10.1, with N E replacing F and E replacing H, that M ∼ = Fˆω (C). Part B: rank(F ) = ℵ0 . By Corollary 24.9.4(b) it suffices to prove that each group G ∈ C is a quotient of M . Indeed, there is an open subgroup E of F with N ∩ E = M . Let E0 be a normal subgroup of finite index in F which is contained in E0 , and let X be a basis of F . Choose a finite subset Y of X with |Y | > rank(G) and X r Y ⊆ E0 . Then, the smallest closed normal subgroup L of F which contains X r Y is contained in E. By Lemma

588

Chapter 24. Frobenius Fields

17.4.9(a), F/L is isomorphic to the free pro-C-group with basis Y . Also, M L is an open proper subgroup of the closed normal subgroup N L of F . N

NL

NE

M

ML

E

M ∩L

L

E0

F

There are two possibilities: if (F : N L) < ∞, then M L/L is a free proC-group of rank exceeding rank(G) (Proposition 17.6.2); if (F : N L) = ∞, then Part A, with F/L replacing F , gives M L/L ∼ = Fˆω (C). In both cases G ∈ Im(M L/L) = Im(M/M ∩ L) ⊆ Im(M ).  Theorem 25.4.7(a) generalizes Proposition 24.10.3 to the case where m = rank(F ) ≥ ℵ0 . Proposition 24.10.3 has consequences analogous to Theorem 16.11.6: Proposition 24.10.4: Let F be a free pro-C-group with 2 ≤ rank(F ) ≤ ℵ0 and N a nontrivial closed normal subgroup of F of infinite index. Then (a) N has infinite rank. (b) N is non-Abelian (in particular the center of F is trivial). (c) If C contains a nonsolvable group, then N is not prosolvable. (d) If C does not consist of p-groups for some prime p, then N is not pronilpotent. In particular, the Frattini subgroup of F is trivial. (e) If C is the formation of all finite p-groups, then N ∼ = Fˆω (p). Proof: Since N is nontrivial, it has an open normal proper subgroup M . In particular, M is a pro-C group. By Proposition 24.10.3, M is isomorphic to Fˆω (C). So, rank(M ) = ℵ0 . Therefore, rank(N ) = ℵ0 . This proves (a). Proof of (b): Each Melnikov formation is nonempty and closed under extensions. In particular, C contains non-Abelian groups. Thus, M is non-Abelian. Therefore, also N is non-Abelian. Proof of (c): If C contains nonsolvable group, M is not prosolvable, so also N is not prosolvable. Proof of (d): Suppose there are distinct prime numbers p and q dividing the orders of groups in C. By (c), we may assume C contains only solvable groups. Hence, Cp and Cq occur as composition factors of groups in C, so Cp , Cq ∈ C. Let Cp operate on (Cq )p by permuting the factors of the as the cycle (12 · · · p). Denote the corresponding semidirect product by E. Then, E ∈ C and has Cp as a nonnormal p-Sylow subgroup. Hence, E is nonnilpotent. Since E is a quotient of M , M is not pronilpotent. Therefore, also N is not pronilpotent. By Lemma 22.1.2, Φ(F ) is pronilpotent, so ϕ(F ) must be trivial.

24.10 Application of the Nielsen-Schreier Formula

589

Proof of (e): By Corollary 22.7.7, N is a free pro-p group. Since rank(N ) is  infinite, N ∼ = Fˆω (p). Theorem 25.4.7 generalizes Proposition 24.10.4 to the case where rank(F ) is an arbitrary infinite cardinal number. Proposition 24.10.5 ([Jarden-Lubotzky1, Thm. 1.9]): Let F be a free proC group with 2 ≤ rank(F ) ≤ ℵ0 . Let N1 and N2 be normal subgroups of F neither of which contains the other. Suppose N = N1 ∩ N2 has infinite index. Then N ∼ = Fˆω (C). Proof: The proof has three parts. Part A: Reduction steps. By assumption, N is a proper normal subgroup of both N1 and N2 . Hence, if (N1 : N ) < ∞ or (N2 : N ) < ∞, then N ∼ = Fˆω (C) (Proposition 24.10.3). We may therefore assume (N1 : N ) = ∞ and (N2 : N ) = ∞. For i = 1, 2 choose proper open normal subgroup Ni0 of Ni which contain N . By Proposition 24.10.3, Ni0 ∼ = Fˆω (C). Also, N10 6≤ N2 (otherwise N = N10 has a finite index in N1 ). Similarly, N20 6≤ N10 . Next observe: N1 N2 /N10 N20 ∼ = N1 /N10 × N2 /N20 . Hence, by Proposition 24.10.3, N10 N20 is pro-C free. Further, both N10 and N20 are normal in N10 N20 and N10 ∩ N20 = N . Replacing F, N1 , N2 respectively by N10 N20 , N10 , N20 we may assume N1 ∼ = N2 ∼ = Fˆω (C) and N1 N2 = F. Part B: Suppose rank(F ) < ∞. For i = 1, 2 choose an open normal subgroup Mi of Ni of large index which contains N such that

(2)


2 + (N1 : M1 ) + (N2 : M2 ) (rank(F/N ) − 1) < 1 + (N1 : M1 )(N2 : M2 )(rank(F/N ) − 1).

By Corollary 17.6.3, rank(Mi /N ) ≤ 1 + (Ni : Mi )(rank(Ni ) − 1), i = 1, 2. The assumptions made in Part A imply F/N ∼ = N1 /N ×N2 /N . Hence, Ni /N is a quotient of F/N and (F : M1 M2 ) = (N1 : M1 )(N2 : M2 ). N1

N 1 M2

F

M1

M1 M2

M1 N 2

N

M2

N2

590

Chapter 24. Frobenius Fields

Therefore, rank(M1 M2 /N ) ≤ rank(M1 /N ) + rank(M2 /N ) ≤ 2 + (N1 : M1 )(rank(N1 ) − 1) + (N2 : M2 )(rank(N2 ) − 1)
≤ 2 + (N1 : M1 ) + (N2 : M2 ) (rank(F/N ) − 1) < 1 + (N1 : M1 )(N2 : M2 )(rank(F/N ) − 1) = 1 + (F : M1 M2 )(rank(F/N ) − 1). By Lemma 24.10.1, N ∼ = Fˆω (C). Part C: Suppose rank(F ) = ℵ0 . By Corollary 24.9.4, it suffices to prove each G ∈ C is a quotient of N . So, choose an epimorphism ϕ: F → Fˆe (C) ¯1 = ϕ(N1 ) and with e ≥ max(2, rank(G)) and such that none of the groups N ¯ N2 = ϕ(N2 ) contains the other. Denote the image of elements and subgroups ¯2 is pro-C free with rank ¯1 ∩ N of F under ϕ with a bar. By Parts A and B, N at least e. Each element of N1 commutes modulo N with all elements of N2 . Hence, ¯1 ∩ N ¯2 commutes modulo N ¯ with each element of N ¯1 and each element of N ¯ ¯ ¯ ¯ ¯ ¯ ¯2 /N ¯ ¯ N2 . Thus, N1 ∩ N2 /N lies in the center of F /N . In particular, N1 ∩ N ¯ is Abelian. Hence, by Propositions 17.6.2 and 24.10.2, N is pro-C free with ¯ , hence also of N . rank at least e. Therefore, G is a quotient of N  Corollary 25.4.5 generalizes Proposition 24.10.5 to the case where m is an arbitrary cardinal number at least 2. Corollary 24.10.6 ([Jarden-Lubotzky1, Cor. 1.10]): Let F be a free proC group with 2 ≤ rank(F ) ≤ ℵ0 and let N be a closed normal subgroup of F . Suppose G/N is pronilpotent of infinite index and two distinct prime numbers divide (G : N ). Then N ∼ = Fˆω (C). Proof: By Proposition 22.9.3, G/N is the direct product of its p-Sylow groups and each of them is normal. Hence, G has closed normal subgroups N1 and N2 with N1 ∩ N2 = N . It follows from Proposition 24.10.5 that  N∼ = Fˆω (C). Corollary 25.4.6 generalizes Corollary 24.10.6 to the case where m is an arbitrary cardinal number greater or equal to 2. Example 24.10.7 (v. d. Dries): The commutator subgroup of Fˆe . Let F = Fˆe and denote the commutator subgroup of F by [F, F ]. It is the smallest closed subgroup of F which contains all commutators a−1 b−1 ab. It is normal and of infinite index. The quotient group F/[F, F ] is the free proabelian group of ˆ e (Lemma 17.4.10). If B is a closed subgroup rank e. That is, F/[F, F ] ∼ =Z Q Q ˆe = Zep , then B = (B ∩ Zep ), where the p ranges over all prime of Z numbers. The group B ∩ Zep is a Zp -submodule of the free Zp -module Zep . Since Zp is a principal ideal domain, B ∩ Zep is also a free Zp -module of rank

Exercises

591

at most e [Lang7, p. 146, Thm. 7.1]. Therefore, rank(B) ≤ e. Assuming e ≥ 2, choose B as H/[F, F ] with H an open subgroup of F of index at least 2. Then rank(B) < 1 + (F : H)(e − 1). It follows from Lemma 24.10.1 that  [F, F ] ∼ = Fˆω if e ≥ 2. Example 24.10.8: Necessity of hypothesis (1) in Lemma 24.10.1. Let B ⊂ C be nonempty Melnikov formations of finite groups (e.g. B is the family of p-groups and C is the family of all finite groups). Consider the free pro-Cgroup F = Fˆe (C). Suppose there is a B ∈ B with rank(B) ≤ e. Let N be the intersection of all open normal subgroups E of F with F/E ∈ B. Then N is a closed normal subgroup of F and F/N ∼ = Fˆe (B) (Lemma 17.4.10). By Proposition 17.6.2, rank(E) = rank(E/N ) for each open subgroup E of F that contains N By Corollary 17.6.5, Fˆe (B) is infinite, so (F : N ) = ∞. On the other hand, none of the groups G ∈ C r B is a quotient of N . Thus, N ∼ 6 Fˆω (C). Hence, neither hypothesis (1) nor the conclusion of Lemma = 24.10.1 hold.  Example 24.10.9 ([Lubotzky-v.d.Dries, p. 34]): A nonfree prosolvable profinite group which satisfies the Nielsen-Schreier index formula for open subgroups. Let p0 < p1 < p2 < · · · be a sequence of prime numbers. With e > 2, inductively construct a descending sequence Fˆe = N0 > N1 > N2 > . . . of closed characteristic subgroups of Fˆe . Given Nn ∼ = Fˆen with ˆ en = 1 + (Fe : Nn )(e − 1), let Nn+1 be theTunique open normal subgroup of ∞ Nn with Nn /Nn+1 ∼ = (Z/pn Z)en . Let N = i=0 Nn . Then rank(Nn /N ) = en for each n. Hence, rank(H/N ) = 1 + (F : H)(e − 1) for each open subgroup H of Fˆe that contains N Q (Exercise 13). Clearly G = Fˆe /N is a prosolvable ∞ group. But, since #G = n=1 penn is a supernatural product of finite prime powers, G is certainly nonfree. 

Exercises 1. Let G be a profinite group. Suppose every finite embedding problem (ϕ: G → A, α: B → A) where Ker(α) is a minimal normal subgroup of B is solvable. Prove that G has the embedding property. 2. Prove that each inverse limit of profinite groups with the embedding property has the embedding property. 3. Let α: B → A, β: C → B be epimorphisms of finite groups. Prove that If α ◦ β is an I-cover, then so is β. 4. Let β 0 : B → D be an I-cover of finite groups and β: G → D an epimorphism of profinite groups. Combine the construction appearing in the proof of Lemma 24.4.4 with Exercise 3 to construct an I-cover ε: E → G and an epimorphism ε0 : E → B such that β 0 ◦ ε0 = β ◦ ε. 5. Let α, α0 : C → A be epimorphisms of finite groups. Construct I-covers ε, ε0 : E → C with E finite and α ◦ ε = α0 ◦ ε0 .

592

Chapter 24. Frobenius Fields

6. Let ϕ: H → A be an embedding cover of a finite group A and let α: B → A be an epimorphism. Suppose B is a quotient of H and rank(B) ≤ ℵ0 . Construct an epimorphism γ: H → B with α ◦ γ = ϕ. 7. Let G be a profinite group and F a smallest embedding cover of G (Propo˜ is a quotient of F˜ . sition 24.4.5). Prove that E(G) 8. Let G be a group of even order 2k which is generated by elements of order 2. Suppose that 4 divides k. Show that the right regular representation of G maps G into a subgroup of Ak . Deduce that for m, n > 2, the group Sm × Sn can be embedded into the simple group Am!n! . Hint: Factor the image in Sk of each involution of G into a product of disjoint 2-cycles. 9. For a profinite group G, define the descending Frattini sequence inductively: Φ0 (G) = G and Φn+1 (G) = Φ(Φn (G)). Prove that if G is pro-p. then T ∞ n=1 Φn (G) = 1. Hint: If N is an open subgroup of G, then there is a normal sequence, N = Nn / · · · / N1 / N0 = G, of groups such that (Ni : Ni + 1) = p, i = 0, . . . , n − 1. Show that Ni ≤ Φi (G). 10. Let N be a closed normal subgroup of a finitely generated pro-p-group G such that N 6≤ Φ(G). Apply Lemma 22.7.4 to conclude that rank(G/N ) < rank(G). 11. Let G be a pro-p-group of rank e such that rank(H) = 1+(G : H)(e−1) for each open subgroup H. Prove that G ∼ = Fˆe (p). [Lubotzky1, Thm. 4.2] Hint: Present G as a quotient Fˆe (p)/N , where N is a closed normal subgroup of Fˆe (p). If N 6= 1, apply Exercise 2 to find an integer n such that N ≤ Φn (Fˆe (p)) but N n ≤ Φn+1 (Fˆe (p)). Now apply Exercise 10. 12. Give an example of a profinite group G of rank at most α0 such that Im(G) consists of all finite groups but G is not free. 13. Let G be a profinite group of rank e. Suppose that G has a descending sequence G = G0 > G1 > G1 > . . . of open subgroups, with trivial intersection, such that rank(Gn ) − 1 = (G : Gn )(e − 1) for each n. Use Corollary 17.5.2 to prove that rank(H) − 1 = (G : H)(e − 1) for each open subgroup H of G.

Notes Lemma 24.1.1 gives a homomorphism ϕ whose decomposition group D(ϕ) is a given subgroup Gal(F/E 0 ) of Gal(F/E). Even if Gal(F/E 0 ) is a cyclic group with a generator σ, there is no distinguished element in D(ϕ) which is made equal to σ. Thus, Lemma 24.1.1 is an analog to Frobenius density theorem, which preceded the Chebotarev density theorem. This explains our choice of the name Frobenius field. An appropriate analog to the Chebotarev density theorem can be found in [Jarden9].

Notes

593

The proof of Lemma 24.2.2 is a simplified version of the proof of [D`ebes, Prop. 1.2]. A proof of the Beckman-Black problem in characteristic 0 appears in [Colliot-Th´el`ene] and in general in [Moret-Bailly] and [Haran-Jarden7, Thm. 2.2]. See [D`ebes] for a history of the problem and results over nonample fields. [Haran-Lubotzky] proves the existence and uniqueness of the smallest embedding covers of each finitely generated profinite group. We generalize their methods and prove the same result for arbitrary profinite groups (Proposition 24.4.5). Chatzidakis translated the conditions entering in the definition of the smallest embedding cover of a profinite group G into a countably many sorted first order language (see [Chatzadakis1] and its reproduction [Chatzidakis4].) This, together with the translation of the data about the finite quotients of E(G) gives an “ω-stable” theory T . An application of the existence (Morley) and uniqueness (Shelah) of the prime model of T allows Chatzidakis to prove the existence and the uniqueness of the smallest embedding cover of G. Proposition 24.4.5 gives a pure profinite group proof of existence. It is desirable to prove the uniqueness along the same lines. The decision problem solved by Proposition 24.4.6 is raised in [FriedHaran-Jarden, Section 1]. The problem of finding a PAC field which is not Frobenius is raised in [Fried-Haran-Jarden, Problem 1.9] and is first answered in [Ershov-Fried] by proving the universal Frattini cover of S2 × S3 is projective but not superprojective. This is a special case of our Lemma 24.6.6(b). Corollary 24.6.12 solves Problem 1.8 of [Fried-Haran-Jarden]. ^ = E(H) ˜ for [Haran-Lubotzky, Cor. 2.12] states without proof that E(H) every finitely generated profinite group H. [Fried-Jarden3, Problem 23.16] states this and a related question as an open problem. Following [Chatzidakis3], Section 24.7 solves the open problem and refutes [Haran-Lubotzky, Cor. 2.12].

Chapter 25. Free Profinite Groups of Infinite Rank This chapter studies free pro-C groups of infinite rank and their closed subgroups, where C is a Melnikov formation of finite groups. The data needed to characterize a pro-C group F of rank m as Fˆm (C) becomes more complicated with increasing m. We distinguish between three cases. If m < ∞, it suffices to know that each C-group of rank at most m is an image of F (Lemma 17.7.1). If m = ℵ0 , each finite embedding problem for F has to be solvable (Corollary 24.8.2). When m > ℵ0 , each C-embedding problem for F must have m distinct solutions (Lemma 25.1.2 and Theorem 25.1.7). As a first application of this result we prove that a closed subgroup H of Fˆm (C) which is pro-C and does not lie too deep is isomorphic to Fˆm (C) (Proposition 25.2.2). By “not lying too deep” we mean that there are less than m open subgroups of Fˆm (C) containing H. When m = ℵ0 , this means that H is open. In the general case, H may well be of infinite index. This basic result then leads to a group theoretic analog of Haran’s diamond theorem for Hilbertian field (Theorem 13.8.3): Let N1 , N2 , M be closed subgroups of Fˆm (C). Suppose m is infinite, N1 , N2 are normal, M is pro-C, N1 ∩ N2 ≤ M , but neither N1 nor N2 are contained in M . Then M ∼ = Fˆm (C) (Theorem 25.4.3). As a result, several types of closed subgroups of Fˆm (C), with m infinite, are proved to be free pro-C of rank m. For example, let M be a proper open subgroup of a normal closed subgroup N of Fˆm (C). Suppose M is pro-C. Then M ∼ = Fˆm (C) (Theorem 25.4.7). Each closed normal subgroup N of Fˆe (C) of infinite index, with 2 ≤ e < ∞, is contained in a closed normal subgroup E which is isomorphic to Fˆω (C) (Proposition 25.8.3). So, if M is a pro-C proper open subgroup of N , then M∼ = Fˆω (C) (Proposition 25.8.4). Not every closed normal subgroup of Fˆm (C) is free pro-C. For example, for each non-Abelian group S in C there is a closed normal subgroup N of Fˆm (C) such that S is the only simple quotient of N . More generally, let f be a function from the set of finite simple groups to the set of cardinal numbers at most m. Suppose f (S) = 0 for each S ∈ / C and f (Cp ) is either 0 or m. Then there exists a closed normal subgroup N of Fˆm (C) with rN = f (Proposition 25.7.7). Here rN is the S-rank function of N introduced in Section 24.9. Conversely, rN satisfies the same conditions as f for each closed normal subgroup N of Fˆm (C) (Proposition 25.7.4). Moreover, the function rN characterizes the normal closed subgroup N up to an isomorphism (Theorem 25.7.3). For example, N ∼ = Fˆm (C) if and only if rN (S) = m for all simple S in C. A class of subgroups of Fˆm (C) that lie more deeply than closed normal

25.1 Characterization of Free Profinite Groups by Embedding Problems

595

subgroups is that of accessible subgroups. They are intersections of normal transfinite sequences of closed subgroups of Fˆm (C). Like closed normal subgroups, accessible subgroups of pro-C groups are pro-C (Lemma 25.9.1). By Lemma 25.9.3, each accessible subgroup N of a pro-C group is the intersection of a normal sequence of closed subgroups of at most countable rank. Accessible subgroups of Fˆm (C) are characterized as homogeneous pro-C groups (Theorem 25.9.11). Here we call a pro-C group G of infinite rank homogeneous if every pro-C embedding problem (ϕ: G → A, α: B → A) satisfying (1a) Ker(α) is contained in every open normal subgroup of B, and (1b) rank(G) = ℵ0 and B finite, or rank(A) < rank(G) and rank(B) ≤ rank(G) is solvable . Like closed normal subgroups, accessible subgroups are uniquely determined up to an isomorphism by their S-rank functions (Proposition 25.7.2). However, in contrast to closed normal subgroups, the only restriction on these functions is that their values at S ∈ / C are 0 (Proposition 25.9.10).

25.1 Characterization of Free Profinite Groups by Embedding Problems In order to characterize profinite groups of rank exceeding ℵ0 by their finite quotients, in addition to the embedding property, we must add an hypothesis concerning the cardinality of the set of solutions of finite embedding problems. Lemma 25.1.1: Let F be a profinite group and (1)

(ϕ: F → A, α: B → A)

a finite embedding problem for F . If rank(F ) < ∞, then the number of solutions to (1) is at most |B|rank(F ) . If rank(F ) = ∞, then this number is at most rank(F ). Proof: The finite case is clear. Suppose rank(F ) = ∞. To each solution, γ: F → B, of (1), associate the open normal subgroup Ker(γ) of F . If x1 , . . . , xn generate F modulo Ker(γ), then γ is determined by its values on x1 , . . . , xn . There are at most |B|n possibilities for these values, so the map γ 7→ Ker(γ) has finite fibers. By Proposition 17.1.2, F has rank(F ) open normal subgroups. Thus, there are at most rank(F ) possible epimorphisms γ that solve embedding problem (1).  If α in (1) is an isomorphism, then (1) has a unique solution, α−1 ◦ ϕ. The next result gives the number of solutions of (1) when Ker(α) 6= 1 and F is free of infinite rank.

596

Chapter 25. Free Profinite Groups of Infinite Rank

Lemma 25.1.2: Let C be a formation of finite groups and F a free pro-C group with basis X. Suppose B ∈ C and Ker(α) 6= 1 in embedding problem (1). If |B| < |X| < ∞, then (1) has at least 2|X|−|B| solutions. If X is infinite, then (1) has exactly |X| solutions. Proof: Choose a set theoretic section α0 : A → B to α: B → A with α0 (1) = 1. Thus, α ◦ α0 = idA . The proof splits up into two parts. Part A: |B| < |X| < ∞. List the elements of ϕ(X) r{1} as a1 , . . . , ae and the elements of Ker(α) as be+1 , . . . , bn . Put bi = α0 (ai ), i = 1, . . . , e. Then n ≤ |A|−1+|Ker(α)| ≤ |A|·|Ker(α)| = |B| and B = hb1 , . . . , bn i. Now choose distinct elements x1 , . . . , xn of X with ϕ(x1 ) = a1 , . . . , ϕ(xe ) = ae . Define a map γ0 : X → B by γ0 (xi ) = α0 (ϕ(xi )) for i = 1, . . . , e, γ0 (xj ) = bj α0 (ϕ(xj )) for j = e + 1, . . . , n, and γ0 (x) = bα0 (ϕ(x)) for x ∈ X r{x1 , . . . , xn } with an arbitrary b ∈ Ker(α). The number of possible γ0 is |Ker(α)||X|−n , which is at least 2|X|−|B| . Each γ0 extends to a homomorphism γ: F → B. By definition, α(γ(x)) = ϕ(x) for each x ∈ X, so α◦γ = ϕ and α(γ(F )) = ϕ(F ) = A. Moreover, for each j between e+1 and n either ϕ(xj ) = 1 or there is an i between 1 and e with ϕ(xj ) = ϕ(xi ). In the former case bj = γ(xj ) ∈ hγ(F )i. In the latter case bj = γ0 (xj )α0 (ϕ(xj ))−1 = γ(xj )α0 (ϕ(xi ))−1 = γ(xj )γ(xi )−1 ∈ hγ(F )i. Thus, Ker(α) ≤ hγ(F )i. Consequently, hγ(F )i = B, as desired. Part B: X is infinite. Let X0 = X ∩ Ker(ϕ) and X1 = X r X0 . Since Ker(α) 6= 1 and |X| = |X0 |, the number of surjective maps γ0 : X0 → Ker(α) that map all but finitely many elements of X0 to 1 is equal to |X|. Extend each γ0 to a map γ1 : X → B by γ1 (x) = γ0 (x) if x ∈ X0 and γ1 (x) = α0 (ϕ(x)) if x ∈ X1 . Then γ1 maps almost every element of X to 1 and satisfies α(γ1 (x)) = ϕ(x) for each x ∈ X. Hence, γ1 uniquely extends to an epimorphism γ: F → B with α ◦ γ = ϕ. Distinct γ0 ’s induce distinct γ ’s. It follows that embedding problem (1) has at least |X| solutions. Combining this with Lemma 25.1.1, we conclude that the number of solutions is exactly |X|.  If γ: F → B is a solution of embedding problem (1), we call Ker(γ) a solution group of (1). Lemma 25.1.3: Let C be a formation of finite groups and F a free pro-C group of rank m. Suppose B ∈ C and Ker(α) 6= 1. Then the number of solution groups of (1) lies between is exactly m if m is infinite.

2m−|B| |Aut(B)|

and

|B|m |Aut(B)|

if |B| < m < ∞ and

Proof: Denote the set of all solutions of (1) by Γ and the set of all solution groups of (1) by N . Then Ker(γ) ∈ N for each γ ∈ Γ. Two solutions γ, γ 0 ∈ Γ have the same kernel if and only if there is a θ ∈ Aut(B) with γ 0 = θ ◦ γ. So, the fibers of the map γ 7→ Ker(γ) of Γ onto N have cardinality |Aut(B)|. Now apply Lemmas 25.1.1 and 25.1.2 to verify the assertion of the lemma. 

25.1 Characterization of Free Profinite Groups by Embedding Problems

597

Lemma 25.1.4: Let C be a formation of finite groups, F a pro-C group, and m a cardinal number. Suppose every C-embedding problem (1) with B 6= 1, where Ker(α) is a minimal normal subgroup of B, has at least m solutions. Then every C-embedding problem (1) with Ker(α) and nontrivial has at least m solutions. Proof: Put C = Ker(α). By assumption (1) has m solutions if C is a finite minimal normal subgroup of B. Otherwise, choose a non trivial proper subgroup C0 of C which is normal in B. Then (ϕ: F → A, α ¯ : B/C0 → A), with α ¯ induced by α, is a pro-C-embedding problem for F with B/C0 nontrivial. Since Ker(¯ α) = C/C0 is nontrivial and has a smaller order than ¯ ◦ βi = ϕ. C, induction gives m epimorphisms βi : F → B/C0 with α For each i consider the pro-C-embedding problem (βi : F → B/C0 , π: B → B/C0 ), where π is the quotient map. Its kernel C0 is nontrivial and has a smaller order than C. Induction gives an epimorphism γi : F → B with π ◦ γi = βi .  Each of the γi solves (1) and they are all distinct. Lemma 25.1.5: Let C be a formation of finite groups and F a pro-C group of infinite rank. Suppose every finite C-embedding problem for F with a nontrivial kernel has rank(F ) solutions. Then every pro-C-embedding problem (2)

(ϕ: F → G, α: H → G)

in which Ker(α) is finite and rank(G) < rank(F ) is solvable. Proof: Assume without loss that Ker(α) 6= 1. As in the proof of Lemma 25.1.4, an induction on |Ker(α)| reduces the lemma to the case where Ker(α) is a minimal nontrivial closed normal subgroup of H. Choose an open normal subgroup H1 of H with H1 ∩ Ker(α) = 1. This gives a commutative diagram (3)

F PPP PPP ϕ PPP PPP P α P'/ H G η

 B

ξ

α0

 /A

in which B = H/H1 and A = G/α(H1 ) are finite groups and the square from H to A is cartesian. In particular, Ker(α0 ) ∼ = Ker(α) is nontrivial. Let B be the set of all epimorphisms β: F → B with α0 ◦ β = ξ ◦ ϕ. Put B0 = {β ∈ B | Ker(ϕ) ≤ Ker(β)}. Denote the set of all epimorphisms ζ: G → B with α0 ◦ ζ = ξ by G. For each β ∈ B0 there is a unique ζ ∈ G with ζ ◦ϕ = β. Moreover, the map B0 → G given by β 7→ ζ is bijective. By Lemma

598

Chapter 25. Free Profinite Groups of Infinite Rank

25.1.1, either G is finite or G is infinite and |G| ≤ rank(G). By assumption, rank(G) < rank(F ) = |B|, so |B0 | < |B|. This gives an epimorphism β: F → B with (4)

α0 ◦ β = ξ ◦ ϕ and Ker(ϕ) 6≤ Ker(β).

Since the square from H to A in (3) is cartesian, there is a homomorphism γ: F → H which makes the following diagram commutative. F 0 @PPP 00@@ PPP ϕ 00 @@@γPPPP PPP 00 @ α P'/ 0 H G β 0 00 00 η ξ 0   B α0 / A Thus, α(γ(F )) = ϕ(F ) = G and η(γ(F )) = β(F ) = B. Claim: γ is surjective. Indeed, since γ(F ) ≤ H, Lemma 24.4.1 gives a commutative diagram of epimorphisms /G // // η¯ ξ0 /  /// ξ  α00 / A0 / B QQQ @ // QQQ QQQ @@@π // @ Q α0 QQQ @@// QQQ  ( A

γ(F )

α ¯

where α ¯ and η¯ are, respectively, the restrictions of α and η to γ(F ), and the square from γ(F ) to A0 is cartesian. Since the square in (3) is cartesian, η maps Ker(α) isomorphically onto Ker(α0 ). By assumption, Ker(α) is a minimal closed normal subgroup of H. Hence, Ker(α0 ) is a minimal normal subgroup of B. Therefore, either π is an isomorphism or α00 is an isomorphism. In the latter case α ¯ is an isomorphism. But then η¯ ◦ (¯ α)−1 ◦ ϕ = β and therefore Ker(ϕ) ≤ Ker(β), contrary to (4). Thus, π is an isomorphism. Consequently, by Lemma 24.4.1, γ(F ) = H. The map γ is the desired solution to embedding problem (2).  Proposition 25.1.6 ([Melnikov3], [Chatzidakis4, §3.2, Cor.]): Let C be a formation of finite groups and F and F 0 pro-C groups. Suppose rank(F ) = rank(F 0 ) are infinite and every finite C-embedding problem for F and for F 0 with a nontrivial kernel has rank(F ) solutions. Then F ∼ = F 0. Proof: Let rank(F ) = m. Order all open normal subgroups of F and F 0 in transfinite sequences {Lα | α < m} and {L0α | α < m}, respectively,

25.1 Characterization of Free Profinite Groups by Embedding Problems

599

with L0 = F and L00 = F 0 . Apply transfinite induction to construct two descending sequences {Nβ | β < m} and {Nβ0 | β < m} of closed normal subgroups of F and F 0 , respectively, satisfying: (5a) For all β < m there are isomorphisms θβ : F/Nβ → F 0 /Nβ0 with commutative diagrams θγ

F/Nγ  F/Nβ

θβ

/ F 0 /Nγ0  / F 0 /Nβ0

for β < γ < m, where the vertical arrows are the quotient maps. (5b) rank(F/Nβ ) < m for each β < m. (5c) Nβ ≤ Lα and Nβ0 ≤ L0α for all α < β < m. Consider δ < m. Suppose Nγ , Nγ0 , and θγ have been defined for each γ < δ such that (5) holds for each β < γ. There are two cases to consider: Case A: δ = γ+1. Let Mδ = Lγ ∩Nγ . Lemma 25.1.5 gives an epimorphism θ0 making the following diagram commutative:

1

F0        θ 0  F 0 /N 0 γ     −1 θγ     / F/Mδ / F/Nγ

/ Nγ /Mδ

/ 1.

For Mδ0 = Ker(θ0 ), this induces a commutative diagram F/Mδ o  F/Nγ

θ¯0

θγ

F 0 /Mδ0  / F 0 /Nγ0

in which θ¯0 is an isomorphism. Define Nδ0 to be L0γ ∩ Mδ0 . Now apply Lemma 25.1.5 to F to obtain a closed normal subgroup Nδ of F which is contained in Mδ and for which F/Nδ  F/Mδ

θδ

(θ¯0 )−1

/ F 0 /N 0 δ  / F 0 /M 0

δ

600

Chapter 25. Free Profinite Groups of Infinite Rank

is commutative with θδ an isomorphism. Then rank(F/Nδ ) = rank(F 0 /Nδ0 ) < m. T Case B: δ is a limit ordinal. In this case let Nδ = γ<δ Nγ and Nδ0 = T 0 γ<δ Nγ . The collection of isomorphisms θγ defines an isomorphism θδ which satisfies (5). T T Finally, note by (5c) that β<m Nβ = 1 and β<m Nβ0 = 1. Hence, the isomorphisms θβ define the desired isomorphism θ: F → F 0 .  Theorem 25.1.7: Let C be a formation of finite groups, m an infinite cardinal, and F a pro-C group. Suppose rank(F ) ≤ m and each finite pro-C embedding problem (6)

(ϕ: F → A, α: B → A)

where B 6= 1 and Ker(α) is a minimal normal subgroup of B has m solutions. Then F ∼ = Fˆm (C). Proof: Choose a simple group S in C. Then (F → 1, S → 1) is a Cembedding problem with a minimal normal kernel. By assumption, it has m solutions. Only finitely many solutions have the same kernel. Hence, F has at least m open normal subgroups N with F/N ∼ = S, so rank(F ) ≥ m. Consequently, rank(F ) = m. By Lemma 25.1.4, each C-embedding problem (6) for F with Ker(α) 6= 1 has m solutions. By Lemma 25.1.2, each C-embedding problem for Fˆm (C) with a nontrivial kernel has m solutions. Therefore, by Proposition 25.1.6,  F ∼ = Fˆm (C). Here is a criterion for freeness that applies to projective groups: Lemma 25.1.8: Let C be a formation of finite groups, m an infinite cardinal, and M a C-projective group of rank at most m (equivalently, M is a closed subgroup of Fˆm (C)). Suppose every split C-embedding problem for M with a nontrivial kernel has m solutions. Then M ∼ = Fˆm (C). Proof: Consider a C-embedding problem (7)

(ϕ: M → A, α: B → A)

with Ker(α) 6= 1. By Theorem 25.1.7, it suffices to construct m solutions to (7). The construction follows that of Proposition 22.5.8. Since M is C-projective, there is a homomorphism γ: M → B with ¯ = M/Ker(γ). Let π: M → M ¯ be the quotient map α ◦ γ = ϕ. Put M ¯ ¯ and γ¯ : M → B and ϕ: ¯ M → A the homomorphisms induced by γ and ¯ with the projections ϕ, respectively. Construct the fiber product B ×A M ¯ ¯ ¯ ψ: B ×A M → M and β: B ×A M → B. Lemma 22.2.9 gives a section to ψ. Thus, ¯ , ψ: B ×A M ¯ →M ¯) (8) (π: M → M is a finite split C-embedding problem for M with Ker(ψ) ∼ = Ker(α) 6= 1. By assumption, (8) has distinct solutions γi with i ranging on a set I of cardinality m.

25.2 Applications of Theorem 25.1.7

601

For each i ∈ I, β ◦ γi is a solution of (7). If i 6= j, there is an x ∈ M with γi (x) 6= γj (x). Also, ψ(γi (x)) = π(x) = ψ(γj (x)). Hence, β(γi (x)) 6=  β(γj (x)). Consequently, β ◦ γi 6= β ◦ γj .

25.2 Applications of Theorem 25.1.7 The characterization of Fˆm with m ≥ ℵ0 by the m-fold solvability of each finite embedding problem with a nontrivial kernel has many applications. For example, the absolute Galois group of an ultraproduct of fields with free absolute Galois groups is free (Theorem 25.2.3). Secondly, the completion of the free abstract group of rank m with respect to the family of all normal subgroups of finite index is Fˆ2m (Theorem 25.3.4). But most important is the generalization of Proposition 17.6.2 from open subgroups of Fˆm to closed subgroups which do not lie too deep (Proposition 25.2.2). To be more precise, recall that the rank of a profinite group G is the smallest cardinality of a set of generators of G which converges to 1. When rank(G) is infinite, it is also the cardinality of the set of all open normal subgroups of G (Proposition 17.1.2). In particular, if N is a closed normal subgroup of G and G/N is not finitely generated, then rank(G/N ) is the cardinality of the set of all open normal subgroups of G which contain N . We generalize the latter observation to closed subgroups of G which are not necessarily normal. Let H be a closed subgroup of G. Denote the set of all open subgroups of G which contains H by OpenSubgr(G/H). Define the weight of G/H to be 1 if H is open and |OpenSubgr(G/H)| if (G : H) = ∞. Here are some properties of the weight function which we use in the sequel: Lemma 25.2.1: Let G be a profinite group, H a closed subgroup, and m an infinite cardinal number. (a) Suppose H / G T and rank(G/H) = ∞. Then weight(G/H) = rank(G/H). (b) Suppose H = i∈I Hi , |I| < m, and Hi is an open subgroup of G for each i ∈ I. Then weight(G/H) < m. (c) Let N be the largest closed normal subgroup of G which is contained in H. Suppose weight(G/H) < m. Then rank(G/N ) < m. (d) Suppose H2 ≤ H1 ≤ G are closed subgroups with weight(G/H1 ) < m and weight(H1 /H2 ) < m. Then weight(G/H2 ) < m. Proof of (a): The intersection of all conjugates of an open subgroup of G which contains H is open, normal, and contains H. Conversely, every open normal subgroup of G is contained in only finitely many open subgroups of G. Hence, weight(G/H) = rank(G/H). Proof of (b): If I is finite, then H is open in G and weight(G/H) = 1 < m. Suppose I is infinite. Then ℵ0 ≤ |I| < m. Since the cardinality of the set of allTfinite subsets of I is |I|, we may replace {Hi | i ∈ I} by the collection { i∈I0 Hi | I0 is a finite subset of I}, if necessary, to assume that the set {Hi | i ∈ I} is closed under finite intersections. By

602

Chapter 25. Free Profinite Groups of Infinite Rank

S Lemma 1.2.2(a), OpenSubgr(G/H) = i∈I OpenSubgr(G/Hi ). It follows S that, weight(G/H) = | i∈I OpenSubgr(G/Hi )| ≤ ℵ0 · |I| < m. Proof of (c): Let Hi , i ∈ I, be the open subgroups of G which contain H. For each i ∈ I let Ni be the G which is contained T largest open T normal T ofT T T subgroup in Hi . Then, N = g∈G H g = g∈G i∈I Hig = i∈I g∈G Hig = i∈I Ni . Therefore, by (a) and (b), rank(G/N ) < m. Proof of (d): Denote the set of open subgroups of G which contain H1 by OpenSubgr(G/H1 ). By assumption, |OpenSubgr(G/H1 )| < m, so each E ∈ OpenSubgr(H1 /H2 ) is the intersection of less than m open subgroups of G. By (b), weight(G/E) < m. The intersection of all E ∈ OpenSubgr(H1 /H2 ) is H2 . There are less than m such E’s. Therefore, by (b), weight(G/H2 ) < m.  Proposition 25.2.2: Let C be a Melnikov formation of finite groups, F a free pro-C group of infinite rank m, and N a closed pro-C subgroup of F with weight(F/N ) < m. Then N ∼ = Fˆm (C). Proof: Consider a C-embedding problem for N (1)

(ϕ: N → A, π: B → A)

where B 6= 1 and C = Ker(π) is a minimal normal subgroup of B. By Theorem 25.1.7, it suffices to construct m solutions to (1). Indeed, N 0 = Ker(ϕ) is an open normal subgroup of N . Choose an open normal subgroup E0 of F with N0 = N ∩ E0 ≤ N 0 . Put E 0 = N 0 E0 and E = N E0 . Then E is an open subgroup of F and E/E0 ∼ = N/N0 is in C. Since E0 is normal in F , it is a pro-C group. Hence, E is a pro-C group (because C is Melnikov). By Proposition 17.6.2, E ∼ = Fˆm (C). Also, N ∩ E 0 = N 0 and 0 ˜ E → A by ϕ(ne ˜ 0 ) = ϕ(n) for N E = E. Extend ϕ to an epimorphism ϕ: 0 0 0 n ∈ N and e ∈ E . Then Ker(ϕ) ˜ =E. Now let β be an ordinal number less than m. Inductively suppose (1) has distinct solutions ψα , α < β. Then Ker(ψα ) is an open normal subgroup T of N and M = α<β Ker(ψα ) is a closed normal subgroup of N with M ≤ N 0 and rank(N/M ) < m (Lemma 25.2.1(b)). Combined with weight(E/N ) < m and Lemma 25.2.1(d), this gives weight(E/M ) < m. Let M0 be the maximal closed normal subgroup of E which is contained in M . By Lemma 25.2.1(c), rank(E/M0 ) < m. Lemma 25.1.2 supplies m solutions to the embedding problem (ϕ: ˜ E→ ˜ E → B such that π ◦ ψ˜ = ϕ˜ A, π: B → A). One of them is an epimorphism ψ: ˜ does not contain M0 . Thus, ψ(E ˜ 0 ) = C. It follows that and Eβ = Ker(ψ) ˜ 0 ) is a nontrivial normal subgroup of B which is contained in C. Since C ψ(M ˜ ). Combined with π(ψ(N ˜ )) = ϕ(N ) = ˜ 0 ) = C, so C ≤ ψ(N is minimal, ψ(M ˜ ˜ A, this gives ψ(N ) = B. Put ψβ = ψ|N . Then ψβ solves (1). In addition, Ker(ψβ ) = N ∩ Eβ does not contain M . Therefore, Ker(ψβ ) 6= Ker(ψα ),  hence ψβ 6= ψα for all α < β. This concludes the induction step.

25.2 Applications of Theorem 25.1.7

603

Theorem 25.2.3 ([Chatzidakis4, Thm. 3.4]): Let {Ki | i ∈ I} be a set of fields with free absolute Galois groups and D an ultrafilter on I. Put K = Q Ki /D and ei = rank(Gal(Ki )). Then: (a) Gal(K) is a free profinite group. (b) If for each e ∈ NQthe set {i ∈ I | e < ei < ∞} belongs to D, then rank(Gal(K)) = | 2ei /D|; Q (c) otherwise rank(Gal(K)) = | ei /D|. Proof: First suppose there is a positive integer e with {i ∈ I | ei = e} ∈ D ∼ for a set of i’s in D. Then Gal(K) = Fˆe (Lemma 17.7.1 and Remark 20.4.5(d)) Q and rank(Gal(K)) = e = | ei /D|. This concludes the case that Gal(K) has finite rank. Now suppose there does not exist e as above. Then {i ∈ I | e < ei } ∈ D set of all Galois for each e, so rank(Gal(K)) ≥ ℵ0 . Let Un (resp. Ui,n ) be the Q extensions of K (resp. Ki ) of degree at most n. Then Un = Ui,n /D. Consider i ∈ I with ei < ∞ and a finite group A. The number of finite Galois extensions of Ki with Galois group isomorphic to A is at most |A|ei . Denote the number of finite groups of order at most n by g(n). Then |Ui,n | ≤ g(n)nei ≤ 2ei k(n) , where k(n) is the smallest integer greater than log n+log g(n) . On the other hand, if ei ≥ 2, then Ki has exactly 2ei quadratic log 2 extensions. Proof of (b): Suppose Q for each e ∈ N the set Ie = {i ∈ I | e < ei < ∞} belongs to D. Then 2ei /D is infinite, so for n ≥ 2, |

Y

2ei /D| ≤ |

Hence, |Un | = |

Q

Y

Ui,n /D| ≤ |

Y

2ei /D|k(n) = |

Y

2ei /D|.

2ei /D|. By Proposition 17.1.2, rank(Gal(K)) = |

Q

2ei /D|.

Proof of (c): Suppose I∞ = {i ∈ I | ei is infinite } belongs to D. For each i ∈ I∞ let Xi be a basis of Gal(Ki ). Then, for each n ≥ 2, |Ui,n | is equal to the number of finite subsets of Xi of cardinality atQmost n. That is, Q |Ui,n | = ei . Thus, |Un | = | ei /D|, so rank(Gal(K)) = | ei /D|. Proof of (a): Consider a finite Galois extension L/K, a finite group B, and an epimorphism α: B → Gal(L/K) with a nontrivial kernel. Let M be the set of all finite Galois extensions M of K for which M contains L and there exists a commutative diagram (2)

Gal(M/K) u uu res uu u uu  u zu / Gal(L/K) B α β

in which β is an isomorphism.

604

Chapter 25. Free Profinite Groups of Infinite Rank

Q Write L in the form L = Li /D where Li /Ki is a Galois extension and D = {i ∈ I | Gal(Li /Ki ) ∼ = Gal(L/K)} is in D. For each i ∈ D define a set Mi with respect to the extension Li /Ki as M is defined with respect to L/K. Since the existence of the commutative diagram (2) is an elementary Q statement on fields (Section 23.4), M = Mi /D. As mentioned in the proof of Lemma 25.1.3, |Aut(B)| · |M| (resp. |Aut(B)| · |Mi |) is the number of solutions of the embedding problem (res: Gal(K) → Gal(L/K), α: B → Gal(L/K)) (resp. (res: Gal(Ki ) → Gal(L i /Ki ), αi : B → Gal(Li /Ki )), where αi are epimorphisms such that Q αi /D = α). By Theorem 25.1.7, it suffices to prove that |Aut(B)| · |M| = rank(Gal(K)). Distinguish between two cases. First suppose Ie ∈ D for each e. Let i ∈ Ie . Then ei < ∞. By Lemma 25.1.3, 2ei −|B| ≤ |Aut(B)| · |Mi | ≤ |B|ei .  |B| Hence, |B|ei ≤ 2(ei −|B|)l for all large ei , where l = log log 2 + 1. In addition, Q ei −|B| /D is an infinite cardinal number, so 2 |

Y

2ei −|B| /D| ≤ |

Y

|Aut(B)||Mi |/D|≤|

Y

2ei −|B| /D|l = |

Y

2ei −|B| /D|.

Q Q Q Q As |M| = | Mi /D| and | 2ei −|B| /D| = |2|B| 2ei −|B| /D| = | 2ei /D|, we deduce from (b) that rank(Gal(K)) = |Aut(B)| · |M|, as desired. Now suppose I∞ ∈ D. Then, {i ∈ I |Q |Aut(B)| · |Mi | = ei } ∈ D Q(Lemma 25.1.2). Hence, by (c), |Aut(B)| · |M| = | |Aut(B)| · |Mi |/D| = | ei /D| = rank(Gal(K)). Our proof is complete in both cases. 

25.3 The Pro-C Completion of a Free Discrete Group Let C be a formation of finite groups and F a free abstract group with basis X 0 . Suppose there is a C ∈ C with rank(C) ≤ |X 0 |. Consider the completion Fˆ = Fˆ (C) = lim F/N with N ranging over all normal subgroups of F ←− satisfying F/N ∈ C. Denote the canonical map from F into Fˆ (C) by θ. Put X = θ(X 0 ). Then X generates Fˆ (C). Lemma 25.3.1: Let G be a pro-C group. Then: (a) For each map ϕ00 : X 0 → G with hϕ00 (X 0 )i = G there is a unique epimorphism ϕ: Fˆ → G with ϕ ◦ θ|X 0 = ϕ00 . (b) θ maps X 0 bijectively onto X. (c) Each map ϕ0 of X into a pro-C group G with hϕ0 (X)i = G uniquely extends to an epimorphism ϕ: Fˆ (C) → G. (d) If X is infinite, then rank(Fˆ (C)) = 2|X| . Proof: We repeat the arguments of the proof of Proposition 17.4.2. Proof of (a): Consider an open normal subgroup M of G. Let π: G → G/M be the quotient map. Then π ◦ ϕ00 : X 0 → G/M extends to an epimorphism

25.3 The Pro-C Completion of a Free Discrete Group

605

ϕ0M : F → G/M . Put M 0 = Ker(ϕ0M ) and let ϕ¯M : F/M 0 → G/M be the induced isomorphism. Thus, F/M 0 ∼ = G/M ∈ C. So, θ induces an isomorphism ˆ , where M ˆ = hθ(M 0 )i. Let π ˆ be the quotient ˆ : Fˆ → Fˆ /M θM : F/M 0 → Fˆ /M −1 ◦π ˆ : Fˆ → G/M . map and ϕM = ϕ¯M ◦ θM The collection of all ϕM defines an epimorphism ϕ: Fˆ → G with ϕ ◦ θ|X 0 = ϕ00 . π ˆ / ˆ Fˆ /M FˆO , O ,, ,, θM θ ,, , FO 3 ,, / F/M 0 33 , 33 ,,ϕM 33 ,, 3 , X 0 ϕ0M 33 ,,, ϕ¯M 33 , 33,, ϕ00 3,    π / G/M G

Since hθ(X 0 )i = F , the map ϕ is unique. Proof of (b): Consider distinct x0 , y 0 ∈ X 0 . Choose nontrivial C ∈ C with r = rank(C) ≤ |X 0 |. Let c1 , . . . , cr be generators of C. Define a map ϕ00 : X 0 → C as follows. First suppose r = 1. Put ϕ00 (x0 ) = c1 and ϕ00 (z 0 ) = 1 for all z 0 ∈ X 0 r{x0 }. Now suppose r ≥ 2. Choose distinct elements x01 , . . . , x0r ∈ X 0 with x01 = x0 and x0r = y 0 . Then put ϕ00 (x0i ) = ci , i = 1, . . . , r. In both cases ϕ00 (x0 ) 6= ϕ00 (y 0 ). Part (a) gives an epimorphism ϕ: Fˆ → C with ϕ ◦ θ|X 0 = ϕ00 . In particular, ϕ(θ(x0 )) 6= ϕ(θ(y 0 )). Thus, θ(x0 ) 6= θ(y 0 ). Proof of (c): Use (a). Proof of (d): The rank of Fˆ is the number of all open normal subgroups of Fˆ (Proposition 17.1.2). The latter is the number of normal subgroups N of F with F/N ∈ C. By (c), this number is 2|X| .  Following Lemma 25.3.1, we identify X 0 with X via θ and consider X as a subset of both F and Fˆ (C). In its latter role we call X a free set of generators for Fˆ (C). Note, however, that unlike a basis of a free profinite group (Definition 17.4.1), X need not converge to 1. This is reflected by the exponentially large rank, 2|X| , of Fˆ (C). Lemma 25.3.2: Let x0 ∈ X. Put Y = {x0 } ∪ {x−1 0 x | x ∈ X and x 6= x0 }. Then Y is a basis of F and a free set of generators of Fˆ (C). Proof: The map x 7→ x−1 0 x for x 6= x0 and x0 7→ x0 of X onto Y is bijective and Y generates both F and Fˆ (C). So, it extends to an epimorphism ϕ: Fˆ (C) → Fˆ (C) (Lemma 25.3.1).

606

Chapter 25. Free Profinite Groups of Infinite Rank

Note that w(ϕ(x1 ), . . . , ϕ(xn )) is a nonempty reduced word in Y whenever w(x1 , . . . , xn ) is a nonempty reduced word in X, so Y is a basis of F . The map x−1 0 x 7→ x for x 6= x0 and x0 7→ x0 therefore extends to an epimorphism ϕ0 : Fˆ (C) → Fˆ (C) which is the inverse of ϕ (use the uniqueness in Lemma 25.3.1). Consequently, ϕ is an isomorphism and Y is a free set of generators of Fˆ (C).  Lemma 25.3.3: Suppose X is infinite. Then the number of solutions of each C-embedding problem for Fˆ (C) having a nontrivial kernel is 2|X| . Proof: Consider a C-embedding problem (1)

(ϕ: Fˆ (C) → A, α: B → A)

S with Ker(α) 6= 1. Then Fˆ (C) = · a∈A ϕ−1 (a). Since A is finite and X infinite, there is an a ∈ A with |ϕ−1 (a) ∩ X| = |X|. Choose x0 ∈ X with −1 (ϕ(x0 )) ∩ X| = |X|. Applying ϕ(x0 ) = a. Then |Ker(ϕ) ∩ x−1 0 X| = |ϕ Lemma 25.3.2, we may assume |Ker(ϕ) ∩ X| = |X|. Put X0 = Ker(ϕ) ∩ X and X1 = X r X0 . Now choose a set theoretic section α0 : A → B to α. Since Ker(α) 6= 1, the number of surjective maps γ0 : X0 → Ker(α) is 2|X| . Each γ0 extends to a map γ1 : X → B by setting γ1 (x) = α0 (ϕ(x)) for x ∈ X1 . Then γ1 further extends to a solution γ: Fˆ (C) → B of (1). This gives at least 2|X| distinct solutions. On the other hand, Lemma 25.1.1 bounds the number of these solutions by the number of open normal subgroups of Fˆ (C), which is at most 2|X| because rank(Fˆ (C)) = 2|X| (Lemma 25.3.1(d)). Thus, (1) has exactly 2|X| solutions.  Theorem 25.3.4 ([Melnikov1]): Let F be the free abstract group of an infinite rank m. Then Fˆ (C) ∼ = Fˆ2m (C) Proof: By Lemma 25.3.1, rank(Fˆ (C)) = 2m . By Lemma 25.3.1(d), each finite C-embedding problem has 2m solutions. Therefore, by Theorem 25.1.7,  Fˆ (C) ∼ = Fˆ2m (C).

25.4 The Group Theoretic Diamond Theorem The diamond theorem for Hilbertian fields (Theorem 13.8.3) has an analog for free profinite groups of infinite rank. We start with a technical result which gives sufficient conditions for a closed subgroup M of Fˆm to be isomorphic to Fˆm in terms of twisted wreath products: Proposition 25.4.1: Let m be an infinite cardinal, F = Fˆm , and M a closed subgroup of F . Suppose for each open subgroup E of F containing M and each open normal subgroup D of E there exist (1a) an open subgroup F0 of E with M ≤ F0 ,

25.4 The Group Theoretic Diamond Theorem

607

(1b) an open normal subgroup L of F with L ≤ F0 ∩ D, and (1c) a closed normal subgroup N of F with N ≤ M ∩ L such that no finite embedding problem (2)

(ϕ: ¯ F/N → F/L, α ¯ : A¯ wrF0 /L F/L → F/L)

where A¯ 6= 1, and α ¯ , ϕ¯ are the quotient maps is solvable. Then M ∼ = Fˆm . Proof: The proof naturally breaks up into several parts: Part A: Preliminaries. Consider a finite nontrivial group A and a finite group G1 acting on A. Let α1 : G1 n A → G1 be the projection map and µ: M → G1 an epimorphism. In order to prove that M ∼ = Fˆm , it suffices to construct m distinct solutions to the embedding problem (µ: M → G1 , α1 : G1 n A → G1 )

(3)

(Lemma 25.1.8). To this end choose an open normal subgroup D of F with M ∩ D ≤ Ker(µ) (Lemma 1.2.5). Let E = M D. Then let F0 , L, and N be as in (1). Then M ·(F0 ∩D) = F0 . Also, let ϕ: F → F/L = G and ϕ0 : F0 → F0 /L = G0 be the quotient maps: F0

E

M ∩D

F0 ∩ D

D

M ∩L

L

M

F

Ker(µ)

N

Extend µ to an epimorphism ϕ1 : F0 → G1 by setting ϕ1 (ad) = µ(a) for a ∈ M and d ∈ F0 ∩ D. Since L ≤ F0 ∩ D ≤ Ker(ϕ1 ), the epimorphism ϕ1 decomposes as ϕ1 = ϕ¯1 ◦ ϕ0 in the following commutative diagram: (4)

F0 ϕ0

G0 n A

α0

ρ

 G1 n A

 / G0

ϕ1

ϕ ¯1 α1

 / G1

Here G0 acts on A via ϕ¯1 , α0 is the quotient map, ρ|G0 = ϕ¯1 , and ρ|A = idA .

608

Chapter 25. Free Profinite Groups of Infinite Rank

Part B: Epimorphisms onto a wreath product. Let α: A wrG0 G → G be the wreath product. Fix a set I of cardinality m. For each i ∈ I we construct an epimorphism ψi : F → A wrG0 G with α ◦ ψi = ϕ. To this end choose a basis X of F of cardinality m. Put X0 = X r L and X1 = X ∩ L. Then X0 is finite and |X1 | = m. Since |IndG G0 (A) × I| = m, G there is a bijection IndG0 (A) × I → X1 which we write as (f, i) 7→ xf,i . Thus, X = X0 ∪· {xf,i | f ∈ IndG G0 (A), i ∈ I}. Define ψi : X → A wrG0 G by ψi (xf,k ) = f if k = i, ψi (xf,k ) = 1 if k 6= i, and ψi (x0 ) = ϕ(x0 ) for x0 ∈ X0 (here we identify G with a subgroup of AwrG0 G via α). Then ψi converges to 1 and α◦ψi (x) = ϕ(x) for each x ∈ X. In particular, hα(ψi (X))i = G. In addition, ψi (X1 ) = IndG G0 (A) = Ker(α). Therefore, ψi extends to an epimorphism ψi : F → A wrG0 G with α ◦ ψi = ϕ.

(5)

Let π: IndG G0 (A) → A be the epimorphism given by f 7→ f (1). By σ0 definition, π(f ) = f (1)σ0 = f (σ0 ) = f σ0 (1) for f ∈ IndG G0 (A) and σ0 ∈ G0 . Hence, π extends to an epimorphism π: G0 n IndG G0 (A) → G0 n A which is the identity map on G0 . Part C: If ψ: F → A wrG0 G is an epimorphism with α ◦ ψ = ϕ, then π(ψ(N )) = A. Indeed, ϕ(N ) = 1 and N / F , so ψ(N ) ≤ IndG G0 (A) and ψ(N ) / A wrG0 G. Hence, ψ(N ) is a normal G-invariant subgroup of IndG G0 (A). Thus, A1 = π(ψ(N )) is a normal G0 -invariant subgroup of A. Therefore, G0 acts on A/A1 . This gives three commutative diagrams: (6)

F0 F vN qq rr v q r q v r ψ| ψ| F0 r q N v ϕ ϕ0 vv ϕ|N qqq rrr vv v xqqq α { xrrr    α α / /G /1 G G0 IndG G0 n IndG G0 (A) G0 (A) ψ

A wrG0 λ



(A/A1 ) wrG0 G

π

π α ¯

/G

 G0 n A

α0

/ G0

 A

/1

in which λ is the epimorphism induced by the quotient map A → A/A1 . Now, ψ(N ) ≤ π −1 (A1 ) = {f ∈ IndG G0 (A) | f (1) ∈ A1 } and ψ(N ) is a (A). Hence, G-invariant subgroup of IndG G0 \ σ {f ∈ IndG ψ(N ) ≤ G0 (A) | f (1) ∈ A1 } σ∈G

=

\ σ∈G

{f ∈ IndG G0 (A) | f (σ) ∈ A1 } = Ker(λ).

25.4 The Group Theoretic Diamond Theorem

609

¯ F/N → (A/A1 ) wrG G It follows that λ ◦ ψ induces an epimorphism ψ: 0 that solves (2) with A¯ = A/A1 . By assumption, this cannot happen, unless A1 = A, as claimed. Part D: Solutions of (3). Fix i ∈ I. By (5) and the middle diagram of (6) with ψi replacing ψ, ρ ◦ π ◦ ψi maps F0 into G1 n A and α0 ◦ π ◦ ψi |F0 = ϕ0 . By (4), α1 ◦ ρ ◦ π ◦ ψi |F0 = ϕ¯1 ◦ α0 ◦ π ◦ ψi |F0 = ϕ¯1 ◦ ϕ0 = ϕ1 , so ρ ◦ π ◦ ψi |M is a homomorphism of M into G1 n A satisfying α1 ◦ (ρ ◦ π ◦ ψi |M ) = ϕ1 |M . By Part C, π ◦ ψi (N ) = A. Therefore, ρ ◦ π ◦ ψi (M ) ≥ ρ ◦ π ◦ ψi (N ) = ρ(A) = A = Ker(α1 ). It follows that ρ ◦ π ◦ ψi |M : M → G1 n A is surjective. Hence, it solves (3). Part E: The solutions are distinct. Consider distinct i, j in I. We have to prove ρ ◦ π ◦ ψi |M 6= ρ ◦ π ◦ ψj |M . To this end put Aˆ = A × A. Let π1 : Aˆ → A and π2 : Aˆ → A be the projections on the coordinates. Then G0 acts on Aˆ coordinatewise. This gives a wreath product α ˆ : Aˆ wrG0 G → G and a commutative diagram (7)

π∗

2 / A wrG0 G Aˆ wrG0 G MMM MMMαˆ α π1∗ MMM MMM   &/ α A wrG0 G G

ˆ in which πr∗ is the identity on G and πr∗ (fˆ) = πr ◦ fˆ for each fˆ ∈ IndG G0 (A), r = 1, 2. ˆ X → Aˆ wrG G by Use the basis X of F from Part B to define a map ψ: 0   (f, 1) k = i ˆ f,k ) = (1, f ) k = j ψ(x  1 k 6= i, j ˆ 0 ) = ϕ(x0 ) for each x0 ∈ X0 (again, we identify G with a subgroup and ψ(x ˆ F → Aˆ wrG G of Aˆ wrG0 G via α ˆ ). Then ψˆ extends to an epimorphism ψ: 0 ∗ ∗ ˆ ˆ ˆ with α ˆ ◦ ψ = ϕ. Furthermore, π1 ◦ ψ = ψi and π2 ◦ ψ = ψj . ˆ ) = A, ˆ where π ˆ ˆ By Part C (with Aˆ replacing A), π ˆ ◦ψ(N ˆ : IndG G0 (A) → A is ˆ the map given by fˆ 7→ fˆ(1). Thus, there is an x ∈ N with π ˆ ◦ ψ(x) = (a, 1), ˆ = π ◦ π1∗ and π2 ◦ π ˆ = π ◦ π2∗ . Hence, with 1 6= a ∈ A. Clearly, π1 ◦ π ˆ ˆ = π1 ◦ π ˆ ◦ ψ(x) = π1 (a, 1) = a 6= 1 while π ◦ψj (x) = 1. π ◦ψi (x) = π ◦π1∗ ◦ ψ(x) Since ρ is the identity on A, ρ ◦ π ◦ ψi (x) 6= ρ ◦ π ◦ ψj (x). Consequently,  ρ ◦ π ◦ ψi 6= ρ ◦ π ◦ ψj . Proposition 25.4.1 gives a new proof of Proposition 17.6.2 in the case where C is the formation of all finite groups and rank(Fˆ ) ≥ ℵ0 . This proof is devoid of combinatorial constructions for discrete free groups.

610

Chapter 25. Free Profinite Groups of Infinite Rank

Proposition 25.4.2: Let m be an infinite cardinal and M an open subgroup of Fˆm . Then M ∼ = Fˆm . Proof: Put F = Fˆm . Let E be an open subgroup of F containing M and D an open normal subgroup of E. Choose an open normal subgroup L of F in M ∩ D. Put F0 = M and N = L. Then consider an embedding problem (ϕ: ¯ F/N → F/L, α ¯ : A¯ wrF0 /L F/L → F/L) with A¯ a nontrivial ¯ the projection onto F/L. Since ϕ¯ is finite group on which F0 /L acts and α an isomorphism while α ¯ is not, the embedding problem has no solution. We  conclude from Proposition 25.4.1 that M ∼ = Fˆm . Theorem 25.4.3 (Group theoretic diamond dheorem [Haran5, Thm. 3.2]): Let m be an infinite cardinal, M1 , M2 closed normal subgroups of Fˆm , and M a closed subgroup of Fˆm . Suppose M1 ∩ M2 ≤ M , M1 6≤ M and M2 6≤ M . Then M ∼ = Fˆm . Proof: We put F = Fˆm and use Proposition 25.4.2 to assume (F : M ) = ∞. Then we first prove the theorem under an additional assumption: (8) Either M1 M2 = F or (M1 M : M ) > 2. The proof of the theorem in this case utilizes Proposition 25.4.1. It has two parts. Part A: Construction of L, F0 , and N . Let D / E be open subgroups of F with M ≤ E. Choose an open normal subgroup L of F in D and put F0 = M L. Let G = F/L and ϕ: F → G the quotient map. Put G0 = ϕ(M ) = F0 /L, G1 = ϕ(M1 ), and G2 = ϕ(M2 ). Then (9a) G1 , G2 / G. Moreover, choosing L sufficiently small, the following holds: (9b) G1 , G2 6≤ G0 (use M1 , M2 6≤ M ). (9c) (G : G0 ) > 2 (use (F : M ) = ∞). (9d) G1 G2 = G or (G1 G0 : G0 ) > 2 (use (8)). This implies: (10) G2 6≤ G1 G0 or (G1 G0 : G0 ) > 2. Indeed, suppose both G2 ≤ G1 G0 and G1 G2 = G. Then G = G1 G0 , so by (9c), (G1 G0 : G0 ) > 2. Now let N = L ∩ M1 ∩ M2 . Then N ≤ M . Part B: An embedding problem. Suppose G0 acts on a nontrivial finite ¯ Put H = A¯ wrG G. Consider the embedding problem group A. 0 (11)

(ϕ: F → G, π: H → G)

where π is the quotient map. We have to prove that (11) has no solution which factors through F/N . Assume ψ: F → H is an epimorphism with π ◦ ψ = ϕ and ψ(N ) = 1. For i = 1, 2 put Hi = ψ(Mi ). Then Hi / H and π(Hi ) = ϕ(Mi ) = Gi .

25.4 The Group Theoretic Diamond Theorem

611

We use (10) to find h1 ∈ H1 and h2 ∈ H2 with π(h1 ) = 1 and [h1 , h2 ] 6= 1. / G1 G0 . Suppose first G2 6≤ G1 G0 . Then there is an h2 ∈ H2 with π(h2 ) ∈ By (9b), G1 6≤ G0 , so Lemma 13.7.4(b) provides the required h1 ∈ H1 . Now / G0 . suppose (G1 G0 : G0 ) > 2. We use (9b) to find h2 ∈ H2 with π(h2 ) ∈ Lemma 13.7.4(a) gives the required h1 ∈ H1 . Having chosen hi choose γi ∈ Mi with ψ(γi ) = hi . Then ϕ(γ1 ) = π(h1 ) = 1. So, γ1 ∈ L. Then [γ1 , γ2 ] ∈ [L, M2 ] ∩ [M1 , M2 ] ≤ L ∩ (M1 ∩ M2 ) = N . So, [h1 , h2 ] = [ψ(γ1 ), ψ(γ2 )] ∈ ψ(N ) = 1. This contradiction proves that ψ as above does not exist. Conclusion of the proof: In the general case we use M1 6≤ M to conclude that (M1 M : M ) ≥ 2. The case (M1 M : M ) > 2 is covered by the special case proved above. Suppose (M1 M : M ) = 2. Choose an open subgroup K2 of F containing M but not M1 M . Thus, K2 ∩ M1 M = M . Put K = K2 M1 M . Then (K : K2 ) = (M1 M : M ) = 2, hence K2 / K. Observe: M1 K2 = K and K2 ∩ M1 ≤ K2 ∩ M1 M = M ≤ K. Furthermore, K2 6≤ M , because (K2 : M ) = ∞. By Proposition 25.4.2, K ∼ = Fˆm , so the first alternative of (8) applies  with K replacing F and K2 replacing M2 . Consequently, M ∼ = Fˆm . Theorem 25.4.3 gives a diamond theorem for the category of pro-C groups, when C is a Melnikov formation: Theorem 25.4.4: Let C be a Melnikov formation of finite groups, m an infinite cardinal number, and M1 , M2 , M closed subgroups of F = Fˆm (C). Suppose M is pro-C, M1 , M2 / F , M1 ∩ M2 ≤ M , but M1 6≤ M and M2 6≤ M . Then M ∼ = Fˆm (C). ˆ = M ˆ (C) the intersection of all open normal Proof: Let Fˆ = Fˆm and N F subgroups K of Fˆ with Fˆ /K ∈ C. Lemma 17.4.10 gives an epimorphism ˆ 2 = ϕ−1 (M2 ), and ˆ . Put M ˆ 1 = ϕ−1 (M1 ), M ϕ: Fˆ → F with Ker(ϕ) = N −1 ˆ 1, M ˆ 2 / Fˆ , N ˆ ≤M ˆ1 ∩ M ˆ2 ≤ M ˆ but M ˆ 1 6≤ M ˆ and ˆ = ϕ (M ). Then M M ˆ . By Theorem 25.4.3, M ˆ ∼ ˆ 2 6≤ M M = Fˆm . ˆ with M ˆ /L ∈ C. Then Suppose L is an open normal subgroup of M ∼ ˆ ˆ ˆ ˆ ˆ ˆ ˆ N /L ∩ N = LN /L and LN /L / M /L, so N /L ∩ N ∈ C. By Lemma 17.4.10, ˆ = N ˆ , hence N ˆ ≤ L. Thus, N ˆ ≤ M ˆ (C). On the other hand, L∩N M ˆ . It follows that N ˆ = M ˆ (C). ˆ /N ˆ ∼ M = M is pro-C. Hence, MMˆ (C) ≤ N M ˆ ∼ By the first paragraph, M = = Fˆm . Therefore, by Lemma 17.4.10, M ∼ ˆ /M ˆ (C) ∼  M = Fˆm (C), as claimed. M A special case of the diamond theorem generalizes Proposition 24.10.5: Corollary 25.4.5 ([Jarden-Lubotzky1, Thm. 1.9]): Let C be a Melnikov formation of finite groups, m ≥ 2 a cardinal number, and N1 , N2 closed normal subgroups of Fˆm (C) neither of which contains the other. Suppose N = N1 ∩ N2 has infinite index in Fˆm (C). Then N ∼ = Fˆm (C). Proof: The case where m ≤ ℵ0 is covered by Proposition 24.10.5. The case  where m ≥ ℵ0 is a special case of Theorem 25.4.4.

612

Chapter 25. Free Profinite Groups of Infinite Rank

Corollary 25.4.6 ([Jarden-Lubotzky1, Cor. 1.10]): Let C be a Melnikov formation of finite groups, m ≥ 2 a cardinal number, and N a closed normal subgroup of Fˆm (C). Suppose Fˆm (C)/N is pronilpotent of infinite index and two distinct prime numbers divide (Fˆm (C) : N ). Then N ∼ = Fˆm (C). Proof: By Proposition 22.9.3, Fˆm (C)/N is the direct product of its p-Sylow groups and each of them is normal. Hence, Fˆm (C) has closed normal subgroups N1 and N2 neither of which contains the other such that N1 ∩N2 = N .  By Proposition 25.4.5, N ∼ = Fˆm (C). We use the diamond theorem to generalize Proposition 24.10.4 to closed subgroups of free pro-C groups of arbitrary infinite rank: Theorem 25.4.7: Let C be a Melnikov formation of finite groups, m an infinite cardinal number, and N a closed normal nontrivial subgroup of F = Fˆm (C). (a) Let M be a proper open subgroup of N . Suppose M is pro-C. Then M∼ = Fˆm (C). (b) rank(N ) = m. (c) N is non-Abelian. In particular, the center of F is trivial. (d) If C contains a nonsolvable group, then N is not prosolvable. (e) If C does not consist of p-groups for some p, then N is not pronilpotent. In particular, the Frattini subgroup of F is trivial. (f) Suppose C is the formation of all finite p-groups. Then N ∼ = Fˆm (p). Proof of (a): The case where (F : M ) < ∞ is covered by Proposition 17.6.2. Suppose (F : M ) = ∞. Then choose an open normal subgroup L of F with N ∩ L ≤ M . Our assumptions imply N 6≤ M and L 6≤ M . By Theorem 25.4.4, M ∼ = Fˆm (C). Proof of (b): Since N is nontrivial, it has a proper open normal subgroup M . By (a), M ∼ = Fˆm (C), so rank(M ) = m. It follows that rank(N ) = m. Proof of (c), (d), (e), (f): Repeat the proof of Proposition 24.10.4(b),(c), (d),(e), respectively.  Our first application of Theorem 25.4.7 generalizes Proposition 24.10.2 to free profinite groups of arbitrary rank: Corollary 25.4.8: Let C be a Melnikov formation of finite groups, m ≥ 2 a cardinal number, and N a closed normal subgroup of F = Fˆm (C). Suppose F/N is infinite and Abelian. Then N ∼ = Fˆm (C). Proof: The case where m is finite is covered by Proposition 24.10.2. Assume m ≥ ℵ0 . Choose x ∈ F r N . Put M = hN, xi. Then M is a normal subgroup of F and M/N is a nontrivial procyclic group. Choose a proper open subgroup M0 of M which contains N . By Theorem 25.4.7(a), M0 ∼ = Fˆm (C). Moreover, ∼ ˆ  M0 /N is procyclic. By Proposition 25.2.2, N = Fm (C). ˆ We do not know if the diamond theorem holds for Fe with 2 ≤ e < ∞:

25.5 The Melnikov Group of a Profinite Group

613

Problem 25.4.9: Let C be a Melnikov formation of finite groups, e ≥ 2 an integer, and M, M1 , M2 closed subgroups of Fˆe (C). Suppose M is pro-C of infinite index, M1 , M2 / F , M1 ∩ M2 ≤ M , but M1 , M2 6≤ M . Is M ∼ = Fˆω (C)? Remark 25.4.10: Lior Bary-Soroker adjusted the method of proof of Theorem 25.4.3 and gave an affirmative answer to Problem 25.4.9 [Bary-Soroker1]. 

25.5 The Melnikov Group of a Profinite Group The Frattini group Φ(G) of a profinite group G has been defined in Section 22.1 as the intersection of all maximal open subgroups of G. It is a characteristic closed subgroup of G characterized by the following property: If H is a closed subgroup of G and H ·Φ(G) = G, then H = G. Numerous applications of the Frattini group in this book prove this concept to be indispensable for the study of closed subgroups of profinite groups. Likewise the Melnikov group of G is vital for the study of closed normal subgroups of G. Define M (G) to be the intersection of all maximal open normal subgroups of G. Again, M (G) is a closed characteristic subgroup of G. If N is a maximal open normal subgroup of G, then G/N is a finite T simple group. Thus, M (G) = S MG (S), where S ranges over all finite simple groups and where is the intersection of all N with G/N ∼ = S. Q MGrG(S) (S) ∼ (Section 24.9). In particular, M (G) = 1 if Therefore, G/M (G) = S S and only if G is a direct product of finite simple groups. Each maximal open normal subgroup N of G is contained in a maximal subgroup M of G. Indeed, N is the intersection of all conjugates of M , so Φ(G) ≤ N . Therefore, Φ(G) ≤ M (G). If G is a pro-p group, then every maximal subgroup is normal of index p, so Φ(G) = M (G). Here are some basic properties of M (G): Lemma 25.5.1: Let G and H be profinite groups. (a) Let ϕ: G → H be an epimorphism. Then ϕ(M (G)) ≤ M (H). (b) If H / G and H · M (G) = G, then H = G. (c) If H / G, then M (H) ≤ M (G). then ϕ−1 (H1 ) Proof of (a): If H1 is a maximal open normal subgroup of H, T is a maximal open normal subgroup of G. Thus, M (G) ≤ ϕ−1 (H1 ) = T −1 ϕ ( H1 ) = ϕ−1 (M (H)). Consequently, ϕ(M (G)) ≤ M (H). Proof of (b): Assume H 6= G. Then H is contained in a maximal open normal subgroup N of G. Hence, H · M (G) ≤ N < G, a contradiction. Proof of (c): Assume M (H) 6≤ M (G). Then G has a maximal open normal subgroup N which does not contain M (H), so N · M (H) = G. Hence, (N ∩ H)M (H) = H. By (b), N ∩ H = H. Thus, M (H) ≤ H ≤ N , a contradiction. 

614

Chapter 25. Free Profinite Groups of Infinite Rank

Example 25.5.2: The Frattini group can be a proper subgroup of the Melnikov group. For example, A3 is the unique nontrivial proper normal subgroup of S3 . Hence, M (S3 ) = A3 . On the other hand, both A3 and {1, (1 2)}  are maximal subgroups of S3 , so Φ(S3 ) = 1. Q Lemma 25.5.3: Let G = i∈I Si be a direct product of finite simple groups and N a closed normal subgroup. Then: is Abelian}, In = {i ∈ I | Si is non-Abelian}, Ga = (a) Q Put Ia = {i ∈ I | Si Q S , and G = n i∈Ia i i∈In Si . Then G = Ga × Gn and N = (Ga ∩ N ) × (Gn ∩ N ). (b) N is a direct product of finite simple groups and (c) G = K × N for some closed normal subgroup K. (d) Let ϕ: G → H be an epimorphism. Then G has a normal subgroup K which ϕ maps isomorphically onto H. Thus, H is a direct product of finite simple groups. Proof of (a): The statement G = Ga ×Gn is clear. Consider x ∈ N and write it as x = yz with y ∈ Ga and z ∈ Gn . For each i ∈ I let πi : G → Si be the ith projection. Let j ∈ In with Sj 6≤ N . By Lemma 18.3.9(a), πj (N ) = 1, so πj (z) = πj (x) = 1. Therefore, z ∈ N , hence y ∈ N . Consequently, N = (Ga ∩ N ) × (Gn ∩ N ). Proof of (b) and (c): Put Nn = Gn ∩ N and Na = Ga ∩ N . For each prime number p let Q Gp (resp. Np ) be the unique p-Sylow group of G (resp. N ). Then Na = p Np , where p ranges over all prime numbers. By (a), G = Ga × Gn and N = Na × Nn . By Lemma 18.3.9, Nn is a direct product of finite simple groups and Gn has a closed normal subgroup Kn with Gn = Kn × Nn . Lemma 22.7.2 gives Q for each p a closed subgroup Kp of Gp with Gp = Kp × Np . Put Ka = p Kp . By Lemma 22.7.3, Np is a direct product of copies of Cp . Put K = Ka × Kn . Then N is a direct product of finite simple groups, K is a closed normal subgroup of G, and G = K × N . Proof of (d): Let N = Ker(ϕ) and K the normal complement that (c) gives. Then ϕ maps K isomorphically onto H. By (b), K is a direct product of finite simple groups. Hence, so is H.  As a corollary, we get a property of the Melnikov group which the Frattini group does not have (Remark 22.1.5): Lemma 25.5.4: Let ϕ: G → H be an epimorphism of profinite groups. Then ϕ(M (G)) = M (H). Proof: Since H/ϕ(M (G)) is a quotient of G/M (G), it is a direct product of finite simple groups (Lemma 25.5.3(d)). Therefore, ϕ(M (G)) is the intersection of maximal open normal subgroups, so M (H) ≤ ϕ(M (G)). On the other hand, ϕ(M (G)) ≤ M (H) (Lemma 25.5.1(a)). Consequently, ϕ(M (G)) = M (H). 

25.6 Homogeneous Pro-C Groups

615

A Frattini cover is an epimorphism ϕ: G → H of profinite groups such that Ker(ϕ) ≤ Φ(G). Likewise, we call ϕ a Melnikov cover if Ker(ϕ) ≤ M (G). Alternatively, ϕ has the following property: G1 / G and ϕ(G1 ) = H implies G1 = G. The latter definition immediately gives: Lemma 25.5.5: Let α: B → A and β: C → B be epimorphisms of profinite groups. Then α ◦ β is a Melnikov cover if and only if both α and β are. Lemma 25.5.6: Let ϕ: G → H be a Melnikov cover of profinite groups. Then the quotient map ϕ: ¯ G/M (G) → H/M (H) is an isomorphism. Proof: By Lemma 25.5.4, ϕ(M (G)) = M (H), so ϕ¯ is well defined. By definition, Ker(ϕ) ≤ M (G). Therefore, ϕ¯ is an isomorphism. 

25.6 Homogeneous Pro-C Groups One of the central concepts of Chapter 24 is the embedding property. Among others each closed normal subgroup of a free profinite group F has the embedding property (Lemma 24.9.4). Closed normal subgroups of infinite index of F have an additional property called homogeneity. Its definition uses Melnikov covers: Fix a Melnikov formation C of finite groups for the whole section. Let G be a pro-C group. Call an embedding problem (1)

(ϕ: G → A, α: B → A)

for G Melnikov if α is a Melnikov cover. Now suppose m = rank(G) ≥ ℵ0 . Call G C-homogeneous if each Melnikov pro-C embedding problem (1) which satisfies one of the following conditions is solvable: (2a) m = ℵ0 and B is finite. (2b) m > ℵ0 , rank(A) < m, and rank(B) ≤ m. Lemma 25.6.4 replaces “rank(A) < m” in (2b) by the weaker condition “rank(M (A)) < m”. This leads to the characterization of C-homogeneous group G by its rank and the quotient G/M (G) (Proposition 25.7.2). By lemma 25.6.3, each nontrivial closed normal subgroup N of infinite index of Fˆm (C) with m ≥ 2 is C-homogeneous of rank max(ℵ0 , m). Thus, N is uniquely determined among all closed normal subgroups of G, up to isomorphism, by the quotient N/M (N ). Lemma 25.6.1: Let G be a profinite group of an uncountable rank m, H a closed subgroup, and N a closed normal subgroup of H. Suppose rank(H/N ) < m. Then G has a closed normal subgroup M such that rank(G/M ) < m and M ∩ H ≤ N . Proof: Let Ni with i ranging on a set I be all open normal subgroups of H which contain N . By Lemma 17.1.2, |I| ≤ max(ℵ0 , rank(H/N )) < m. For each i ∈ I choose an open normal subgroup T Mi of G with Mi ∩ H ≤ Ni (Lemma 1.2.5(a)). By Lemma 25.2.1(b), M = i∈I Mi satisfies the requirements. 

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Chapter 25. Free Profinite Groups of Infinite Rank

Proposition 25.6.2: Let F be a free pro-C group of infinite rank m and (3)

(ϕ: F → A, α: B → A)

a pro-C embedding problem with rank(A) < m and rank(B) ≤ m. Then (3) is solvable. In particular, F is C-homogeneous. Proof: The case where B is finite is covered by Lemma 25.1.2. Assume S B is infinite. Let X be a basis of F . Then, ϕ(X) r{1} = N (ϕ(X) r N ) where N ranges over the set N of all open normal subgroups of A. For N ∈ N , the group ϕ−1 (N ) is open and normal in N , hence X r ϕ−1 (N ) is finite, so ϕ(X) r N ⊆ ϕ(X − ϕ−1 (N )) is also finite. By Proposition 17.1.2, |N | = rank(A) < m. It follows that, |ϕ(X)| < m. Choose a subset X0 of X with ϕ(X0 ) = ϕ(X) and |X0 | < m. Put X1 = X r X0 . Then |X1 | = m. Lemma 1.2.7 gives a continuous set theoretic section to α, that is, a continuous map α0 : A → B with α ◦ α0 = idA . Then, the set Z0 = α0 (ϕ(X0 )) converges to 1. If rank(Ker(α)) < ∞, choose a finite set of generators Z1 for Ker(α). If rank(Ker(α)) = ∞, choose a set of generators Z1 of Ker(α) converging to 1. In the latter case, |Z1 | = rank(rank(α)) ≤ rank(B) ≤ m (Lemma 17.1.5). Thus, in each case, Z0 ∪ Z1 is a set of generators of B which converges to 1 of cardinality at most m. Finally, choose a surjective map γ1 : X1 → Z1 . Extend γ1 to a map γ: X → Z0 ∪ Z1 by setting γ(x) = α0 (ϕ(x)) for x ∈ X0 . By construction, γ converges to 1. Since X is a basis of F , γ extends to an epimorphism γ: F → B. In addition, α(γ(x)) = ϕ(x) for each x in X0 or in X1 . Hence, α ◦ γ = ϕ and γ is a solution of (3)  Lemma 25.6.3: Let F be a free pro-C group of rank m ≥ 2, and N a nontrivial closed normal subgroup of infinite index. Then N is C-homogeneous with rank(N ) = max(ℵ0 , m). Proof: If m ≤ ℵ0 , then rank(N ) = ℵ0 (Proposition 24.10.4(a)). If m ≥ ℵ0 , then rank(N ) = m (Theorem 25.4.7(b)). Now consider a pro-C Melnikov embedding problem (4)

(ϕ: N → A, α: B → A)

satisfying this: (5a) either rank(N ) = ℵ0 and B is finite, (5b) or rank(N ) > ℵ0 , rank(A) < m, and rank(B) ≤ m. Put N1 = Ker(ϕ) and distinguish between three cases. In each case construct a closed normal subgroup L1 of F with N ∩ L1 ≤ N1 and put L = N L1 . Then L / F , so L is a pro-C group. These subgroups should satisfy additional properties: Case A: rank(N ) = ℵ0 , m < ℵ0 and B is finite. Then (N : N1 ) = |A| < ∞. Use Proposition 17.6.2 to choose L1 open with rank(L) ≥ rank(B). Then L is pro-C free of finite rank.

25.6 Homogeneous Pro-C Groups

617

Case B: rank(N ) = ℵ0 , m = ℵ0 and B is finite. Again, (N : N1 ) < ∞. Choose L1 open. Then L ∼ = Fˆω (C) (Proposition 17.6.2). Case C: rank(N ) = m > ℵ0 , rank(A) < m, and rank(B) ≤ m. Then rank(N/N1 ) = rank(A) < m. Use Lemma 25.6.1 to choose L1 with rank(F/L1 ) < m. Then rank(F/L) < m. By Proposition 25.2.2, L ∼ = Fˆm (C). In each case extend ϕ to an epimorphism ϕ: ˜ L → A by ϕ(nl ˜ 1 ) = ϕ(n) for n ∈ N and l1 ∈ L1 . In Case A, Proposition 17.7.3 gives an epimorphism γ: L → B with α ◦ γ = ϕ. ˜ In Cases B and C, Proposition 25.6.2 supplies γ. Thus, γ(N ) is a normal subgroup of B with α(γ(N )) = A. Since α: B → A  is Melnikov, γ(N ) = B. Consequently, γ|N is a solution of (4). The next lemma relaxes the assumption “A is finite” (resp. rank(A) < m) in (5) to “M (A) is finite” (resp. rank(M (A)) < m): Lemma 25.6.4: Let G be a C-homogeneous pro-C group of infinite rank m and (6)

(ϕ: G → A, α: B → A)

a Melnikov pro-C embedding problem. Suppose (7a) either m = ℵ0 , M (A) is finite, and rank(B) ≤ ℵ0 , (7b) or m > ℵ0 , rank(M (A)) < m, and rank(B) ≤ m. Then (6) is solvable. Proof: Put C = Ker(α) and break up the proof into two parts: Part A: C is finite. By assumption, C ≤ M (B). Hence, M (B)/C = M (A) (Lemma 25.5.4). Thus, (8a) either m = ℵ0 , M (B) is finite, and rank(B) ≤ ℵ0 , (8b) or m > ℵ0 , rank(M (B)) < m, and rank(B) ≤ m. If (8a) holds, choose an open normal subgroup L of B with M (B)∩L = 1. If (8b) holds, choose a closed normal subgroup L of B with M (B) ∩ L = 1 and rank(B/L) < m (Lemma 25.6.1). Put K = α(L). Let πA : A → A/K and πB : B → B/L be the quotient maps and let ϕ¯ = πA ◦ϕ. Let α ¯ : B/L → A/K be the homomorphism induced by α. Then Ker(¯ α) = CL/L ≤ M (B)L/L = M (B/L) (Lemma 25.5.4), so α ¯ is a Melnikov cover. If (8a) holds, B/L is finite. If (8b) holds, then rank(A/K) ≤ rank(B/L) < m. The homogeneity of G gives an epimorphism ξ which makes the following diagram commutative: (9)

C  CL/L

G ϕ ξ  α /B /A πB πA   α¯  / B/L / A/K

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Chapter 25. Free Profinite Groups of Infinite Rank

Since Ker(α) ∩ Ker(πB ) = C ∩ L = 1, the right square in (9) is cartesian. Hence, there is a homomorphism γ: G → B with πB ◦ γ = ξ and α ◦ γ = ϕ. We have to prove that γ is surjective. Indeed, α induces an isomorphism α0 : B/M (B) → A/M (A) (Lemma 25.5.6). Let µA : A → A/M (A) be the quotient map. Put ψ = (α0 )−1 ◦ µA ◦ ϕ. Then consider the diagram (10)

ϕ /AN G1 TCTTT NNN 11CC TTTTT ψ NNNµA TTTT NNN 11 CCCγ TTTT NN& 11 CC! ) µB α0 11 / / A/M (A) B/M (B) 11ξ B 11 π ¯B 11 πB    µ ¯B / B/M (B)L B/L

where µB , π ¯B , and µ ¯B are the quotient maps and ξ = πB ◦ γ. It is commutative and all maps except possibly γ are surjective. Since M (B) ∩ L = 1, the square in (10) is cartesian (Example 22.2.7(b)). Since ψ is surjective, ψ(M (G)) = M (B/M (B)) = 1 (Lemma 25.5.4), soM (G) ≤ Ker(ψ). Similarly, ξ(M (G)) = M (B/L)), so ξ −1 (M (B/L)) = M (G)Ker(ξ). Thus, µB )) = ξ −1 (M (B)L/L) Ker(¯ µB ◦ ξ) = ξ −1 (Ker(¯ = ξ −1 (M (B/L)) = M (G)Ker(ξ) ≤ Ker(ψ)Ker(ξ). We conclude from Lemma 22.2.6(b) that γ is surjective. Part B: C is arbitrary. Use Zorn’s lemma to find a minimal closed normal subgroup N of B contained in C with an epimorphism γN : G → B/N such that αN ◦ γN = ϕ. Here αN : B/N → A is the epimorphism induced by α. We have to prove that N = 1. Assume N is nontrivial. Then N has an open subgroup N 0 which is normal in B. Let πN 0 ,N : B/N 0 → B/N be the quotient map. Then Ker(πN 0 /N ) = N/N 0 is finite. By Lemma 25.5.5, πN 0 ,N and αN are Melnikov, so M (B/N ) ∼ = M (A) (Lemma 25.5.6). Thus, either m = ℵ0 and M (B/N ) is finite or m > ℵ0 and rank(M (B/N )) < m. Part A gives an epimorphism γN 0 : G → B/N 0 with πN 0 /N ◦ γN 0 = γN . It satisfies αN 0 ◦ γN 0 = ϕ. This contradiction to the minimality of N proves that N = 1.  The proof of the next result uses transfinite induction on ordinal numbers and distinguishes between “even” and “odd” ordinals. By definition, an ordinal number α is even (resp. odd) if α = α0 + k, where α0 is a limit ordinal and k is an even (resp. odd) finite ordinal. In particular, each limit ordinal is even.

25.6 Homogeneous Pro-C Groups

619

Proposition 25.6.5: Let πi : Gi → A be a Melnikov cover of pro-C groups, i = 1, 2. Suppose G1 and G2 are C-homogeneous of the same infinite rank m. In addition suppose either m = ℵ0 and M (A) is finite or m > ℵ0 and rank(M (A)) < m. Then there is an isomorphism ϕ: G1 → G2 with π2 ◦ ϕ = π1 . Proof: Put Ni,0 = Ker(πi ), i = 1, 2. Since πi is Melnikov, Ni,0 ≤ M (Gi ). Also, M (Gi )/Ni,0 ∼ = M (A) (Lemma 25.5.4). Thus, either m = ℵ0 and M (Gi )/Ni,0 is finite or m > ℵ0 and rank(M (Gi )/Ni,0 ) < m. Let ϕ0 : G1 /N1,0 → G2 /N2,0 be the isomorphism induced by π1 and π2 . We have to lift ϕ0 to an isomorphism ϕ: G1 → G2 . To do so list the open normal subgroups of G1 as K1,α with α ranging over all even ordinals at most m. List the open normal subgroups of G2 as K2,α with α ranging over all odd ordinal numbers less than m. By transfinite induction construct for each α ≤ m a closed normal subgroup Ni,α of Gi which is contained in M (Gi ). For each even ordinal number α ≤ m construct an isomorphism ϕα : G1 /N1,α → G2 /N2,α . Finally, for each odd ordinal α < m construct an isomorphism ψα : G2 /N2,α → G1 /N1,α . This data should satisfy the following conditions: T (11a) Ni,α ≥ Ni,β for α < β ≤ m and Ni,0 ∩ α<β Ki,α ≤ Ni,β for β ≤ m. (11b) For each α < m either m = ℵ0 and M (Gi )/Ni,α is finite or m > ℵ0 and rank(M (Gi )/Ni,α ) < m. (11c) ϕβ lifts ϕα for α < β ≤ m. −1 (11d) Let α be an even ordinal. Then ψα+1 lifts ϕ−1 α and ϕα+2 lifts ψα+1 . To carry out the induction step consider β < m and assume all objects have been defined for α < β. Suppose first β = α + 1 is a successor ordinal. Assume α is even. Put N1,β = K1,α ∩ N1,α . Let π2,α : G2 → G2 /N2,α and π1,β,α : G1 /N1,β → G1 /N1,α be the quotient maps. Then (12)

(π2,α : G2 → G2 /N2,α , ϕα ◦ π1,β,α : G1 /N1,β → G2 /N2,α )

is a pro-C embedding problem for G2 . Its kernel is N1,α /N1,β ≤ M (G1 /N1,β ), so (12) is Melnikov. Taking (11b) into account, Lemma 25.6.4 gives an epimorphism ψ˜β : G2 → G1 /N1,β which solves (12). Put N2,β = Ker(ψ˜β ). Let ψβ : G2 /N2,β → G1 /N1,β be the isomorphism induced by ψ˜β . Then ψβ lifts ϕ−1 α . If m = ℵ0 , then M (G1 )/N1,β is finite whereas if m > ℵ0 , then rank(M (G1 )/N1,β ) < m (use (11b)). Thus, (11b) holds for i = 2 and β replacing α. The case with α odd is treated in the same way as the case with α even. Consider the case where β is a limit ordinal. Let ϕβ : G1 /N1,β → G2 /N2,β be is an isomorphism the limit of the isomorphisms ϕα , with α < β. Then ϕβ T which lifts all ϕα and ψα−1 with α < β. Put Ni,β = α<β Ni,α . Then

620

Chapter 25. Free Profinite Groups of Infinite Rank

T Ni,0 ∩ α<β Ki,α ≤ Ni,β . If β < m, then m > ℵ0 and the set of all ordinals α < β has cardinality less than m. Hence, by (11b) T and Lemma 25.2.1(a,b), rank(M (Gi )/Ni,β ) < m. If β = m, then Ni,m ≤ α<m Ki,α = 1, so Ni,m = 1.  The isomorphism ϕ = ϕm is the desired lifting of ϕ0 .

25.7 The S-rank of Closed Normal Subgroups Let C be a Melnikov formation of finite groups and F a free pro-C group of infinite rank m. We classify the closed normal subgroups N of F up to isomorphism by their S-rank functions rN (Theorem 25.7.3). Moreover, let f be a function from the set of all simple finite groups to the set of cardinal numbers at most m. Then there is a closed normal subgroup N of F with / C, 0 ≤ f (S) ≤ m if S is rN = f if and only if this holds: f (S) = 0 if S ∈ non-Abelian, and f (Cp ) is either 0 or m (Propositions 25.7.4 and 25.7.7). Lemma 25.7.1: Let F be a free pro-C group of infinite rank m and S a finite simple group. Then rF (S) = 0 if S ∈ / C and rF (S) = m if S ∈ C. Proof: Suppose S ∈ C. An application of Lemma 25.1.2 to the embedding problem (F → 1, S → 1) gives rS (F ) ≥ m, so rS (F ) = m.  Proposition 25.7.2 ([Melnikov2, Thm. 3.2]): Let G1 and G2 be C-homogeneous pro-C groups of infinite rank m. Suppose rG1 = rG2 . Then: (a) There is an isomorphism ϕ0 : G1 /M (G1 ) → G2 /M (G2 ). (b) Every isomorphism as in (a) can be lifted to an isomorphism ϕ: G1 → G2 . (c) Let G be a C-homogeneous pro-C group of infinite rank m. Suppose the following holds for each finite simple group S: rG (S) = m if S ∈ C and / C. Then G ∼ rG (S) = 0 if S ∈ = Fˆm (C). Proof: Statement (a) follows from the definitions: G1 /M (G1 ) ∼ =

Y S

S rG1 (S) =

Y

S rG2 (S) ∼ = G2 /M (G2 )

S

where S ranges over all finite simple groups (Section 25.5). Now consider an isomorphism ϕ0 as in (a). Put A = G2 /M (G2 ), π2 : G2 → A the quotient map, and π1 : G1 → A the compositum of the quotient map G1 → G1 /M (G1 ) with ϕ0 . Then both π1 and π2 are Melnikov covers and M (A) = M (G2 /M (G2 )) = 1. Proposition 25.6.5 gives ϕ as in (b). Statement (c) follows from Lemma 25.7.1 by (b).  Theorem 25.7.3 ([Melnikov1, Thm. 3.1]): Let C be a Melnikov formation of finite groups, m ≥ 2 a cardinal number, and N, N 0 closed normal subgroups of Fˆm (C). Assume N, N 0 are of infinite index if m < ℵ0 . (a) Suppose rN = rN 0 . Then N ∼ = N 0. (b) Suppose rN (S) = m for each simple group S ∈ C. Then N∼ = Fˆmax(m,ℵ0 ) (C).

25.7 The S -rank of Closed Normal Subgroups

621

Proof of (a): If N is trivial, then rN 0 = rN = 0, so N 0 is also trivial. Assume therefore both N and N 0 are nontrivial. Put m∗ = max(m, ℵ0 ). By Proposition 25.6.3, both N and N 0 are Chomogeneous of rank m∗ . Hence, by Proposition 25.7.2, N ∼ = N 0. Proof of (b): Apply Proposition 25.7.2(c) to N .



Proposition 25.7.4: Let C be a Melnikov formation of finite groups and F a free pro-C group of rank m ≥ 2. Let N be a closed normal subgroup of infinite index of F and S a finite simple group. Then, / C, (a) rN (S) = 0 if S ∈ (b) 0 ≤ rN (S) ≤ max(m, ℵ0 ) if S ∈ C is non-Abelian, and (c) either rN (S) = 0 or rN (S) = max(m, ℵ0 ) if S = Cp ∈ C. Proof: Statements (a) and (b) are clear. To prove (c), assume S = Cp ∈ C and N has an open normal subgroup M with N/M ∼ = Cp . We have to prove rN (S) = max(m, ℵ0 ). Indeed, F has an open normal subgroup D with N ∩ D ≤ M . Put E = N D. Then E is open in F , so E ∼ = Fˆm0 (C) with m ≤ m0 < ∞ if m is finite 0 and m = m if m is infinite (Proposition 17.6.2). Also, E/M D ∼ = N/M ∼ = Cp , hence ME (Cp ) ≤ M D (Section 25). Lemma 17.4.10 gives an epimorphism ϕ: E → Fˆm0 (p) with Ker(ϕ) ≤ M D. Then N ∩ Ker(ϕ) ≤ N ∩ M D = M , so ϕ(N )/ϕ(M ) ∼ = N/M ∼ = Cp . In particular ϕ(N ) is a closed normal nontrivial subgroup of Fˆm0 (p). By Corollary 22.7.7, ϕ(N ) is a free pro-p group. Distinguish between two cases. First suppose m is finite. Fix a positive integer e. Since (F : N ) = ∞, we may choose D with (F : E) large enough such that m0 ≥ e (Proposition 17.6.2). If (Fˆm0 (p) : ϕ(N )) = ∞, then, by Theorem 24.10.4, rank(ϕ(N )) = ℵ0 . If (Fˆm0 (p) : ϕ(N )) < ∞, then rank(ϕ(N )) ≥ m0 ≥ e (Proposition 17.6.2). It follows that in both cases, rank(ϕ(N )) = ℵ0 and rN (Cp ) ≥ rank(ϕ(N )) = ℵ0 . Therefore, rN (Cp ) = ℵ0 . Now assume m is infinite. Then rank(ϕ(N )) = m (Theorem 25.4.7(b)). Hence, m ≥ rN (Cp ) ≥ rϕ(N ) (Cp ) = m (Lemma 25.7.1). Consequently,  rN (Cp ) = m. Theorem 25.7.5: Let C be a Melnikov formation of finite groups, m ≥ 2 a cardinal number, and N a closed normal subgroup of Fˆm (C) which is of infinite index if m < ℵ0 . Suppose: (1a) Cp is a quotient of N for every prime number p. (1b) S max(m,ℵ0 ) is a quotient of N for every non-Abelian simple group S ∈ C. Then N ∼ = Fˆmax(m,ℵ ) (C). 0

Proof: Let F = Fˆmax(m,ℵ0 ) (C). By Lemma 25.7.1, by (1), and by Proposition 25.7.4, rN = rF . Hence, by Theorem 25.7.2(c) N ∼ = F.

622

Chapter 25. Free Profinite Groups of Infinite Rank

Corollary 25.7.6: Let C be a Melnikov formation of finite groups, 2 ≤ m ≤ ℵ0 a cardinal number, and N a closed normal subgroup of Fˆm (C) which is of infinite index if m < ℵ0 . Suppose: (2a) Cp is a quotient of N for every prime number p. (2b) S q is a quotient of N for every non-Abelian simple group S ∈ C and every positive integer q. Then N ∼ = Fˆω (C). Proof: By (2b), rN (S) = ℵ0 for every non-Abelian simple group S ∈ C, so we may apply Theorem 25.7.5.  Proposition 25.7.7: [Melnikov1, Thm. 3.3] Let C be a Melnikov formation of finite groups and F a free pro-C group of infinite rank m. Consider a function f from the set of all finite simple groups to the set of all cardinal numbers at most m. Suppose each finite simple group S satisfies the following condition: (3a) f (S) = 0 if S ∈ / C, and (3b) either f (S) = 0 or f (S) = m if S = Cp ∈ C. Then F has a closed normal subgroup N with rN = f . Proof: Denote the set of all simple groups in C by S. Consider S ∈ S. If S = Cp put L(S) = F . If S is non-Abelian, use F/MF (S) ∼ = S m (Lemma 25.7.1) to choose a closed normal subgroup L(S) of F containing MF (S) with L(S)/MF (S) ∼ = S f (S) . The rest of the proof naturally divides into three parts. T Part A: Let K = S∈S L(S). Then K · MF (S) = L(S) for all S ∈ S. Consider S ∈ S. List the elements Tn of S in a sequence S0 , S1 , S2 , . . . with = S. Write K = L(S), K = S 0 0 n i=0 L(Si ) for n = 1, 2, 3, . . ., and K = T T∞ 0 K = L(S ). For each n ≥ 1, Kn−1 /Kn ∼ = Kn−1 ·L(Sn )/L(Sn ) ≤ 0 i S ∈S i=0 F/L(Sn ), so the only composition factor of Kn−1 /Kn is Sn . It follows that S is not a composition factor of L(S)/K, so S is not a composition factor of L(S)/K · MF (S). On the other hand, the only possible composition factor of L(S)/K · MF (S) is S. Therefore, K · MF (S) = L(S). Part B: Construction of N . Let L be the set of all closed normal subgroups L of F with L · MF (S) = L(S) for each S ∈ S. By Part A, L T is nonempty. chain in L. Put L = Suppose {Li | i ∈ I} is a descending i∈I Li . By T Lemma 1.2.2(b), L · MF (S) = i∈I Li · MF (S) = L(S), so L ∈ L. Zorn’s Lemma gives a minimal element N in L. Part C: rN = f . Let S be a finite simple group. First suppose, S ∈ / S. Since N is a pro-C group, S is not a factor of N . Hence, rN (S) = 0. Next suppose S = Cp ∈ S. Then N · MF (S) = F and F/MF (S) is a quotient of N . Hence, rN (S) ≥ m (Lemma 25.7.1). On the other hand rN (S) ≤ rank(N ) ≤ m. Therefore, rN (S) = m. Finally, suppose S is in S but non-Abelian. Then N/N ∩ MF (S) ∼ = L(S)/MF (S) ∼ = S f (S) , so MN (S) ≤ N ∩ MF (S). Lemma 18.3.9 applied

25.8 Closed Normal Subgroups with a Basis Element

623

to the closed normal subgroup N ∩ MF (S)/MN (S) of N/MN (S) gives a closed normal subgroup N0 of F with N0 ∩ (N ∩ MF (S)) = MN (S) and N0 · (M ∩ MF (S)) = N . Thus, N0 · MF (S) = L(S). The minimality of N implies N0 = N , so MN (S) = N ∩ MF (S). Thus, N/MN (S) ∼ = L(S)/MF (S).  Consequently, rN (S) = f (S). As a corollary, we classify the closed normal subgroups of a free pro-C groups as C-homogeneous groups with special S-rank functions: Theorem 25.7.8: Let C be a Melnikov formation of finite groups and G a pro-C group of infinite rank m. Then G is isomorphic to a closed normal subgroup of Fˆm (C) if and only if G is C-homogeneous and for each prime number p either rG (Cp ) = 0 or rG (Cp ) = m. Proof: First suppose G is isomorphic to a closed normal subgroup N of Fˆm (C). By Lemma 25.6.3, N is C-homogeneous. Moreover, for each p, either rG (Cp ) = 0 or rG (Cp ) = m (Proposition 25.7.4). Conversely, suppose G is C-homogeneous and for each p either rG (Cp ) = 0 or rG (Cp ) = m. In addition, for each finite simple group S, rG (S) ≤ rank(G) ≤ m and rG (S) = 0 if S ∈ / C. By Proposition 25.7.7, Fˆm (C) has a closed normal subgroup N with rN = rG . It follows from Proposition 25.7.2 that G ∼  = N.

25.8 Closed Normal Subgroups with a Basis Element Let C be a Melnikov formation of finite groups and F0 a free abstract group. Then the canonical map of F0 into its pro-C completion F is injective (Proposition 17.5.11). It is rare that a closed normal subgroup N of F of infinite index intersects F0 nontrivially. But, when it happens, N is a free pro-C group (Proposition 25.8.1). This implies that every closed normal subgroup N of Fˆe (C) of infinite index with 2 ≤ e < ℵ0 is contained in a closed normal subgroup N0 which is isomorphic to Fˆω (C) (Proposition 25.8.3). As a consequence we prove an analog of Theorem 25.4.7(a) for free profinite groups of finite rank. Our first result is a substantial generalization of the fact (Fˆe )0 ∼ = Fˆω proved for e ≥ 2 in Example 24.10.7: Proposition 25.8.1 ([Melnikov1, p. 10]): Let C be a Melnikov formation of finite groups, F a free pro-C-group of rank m ≥ 2, X a basis of F , F0 the abstract subgroup of F generated by X, and N a closed normal subgroup of F of infinite index. Suppose F0 ∩ N 6= 1. Then N ∼ = Fˆmax(m,ℵ0 ) (C). Proof: By Remark 17.6.1, F0 is free on X. Put N0 = N ∩ F0 . The rest of the proof splits up into two parts according to the rank of F : Part A: m is finite. Lemma 17.6.4(b) gives an open normal subgroup E of F containing N such that rank(E/N ) < 1 + (F : E)(m − 1). By Lemma 24.10.1, N ∼ = Fˆω (C).

624

Chapter 25. Free Profinite Groups of Infinite Rank

Part B: m is infinite. By Theorem 25.7.3(b), it suffices to prove rN (S) = m for each finite simple group S ∈ C. Indeed, Lemma 17.6.4(a) gives an open normal subgroup E of F which contains N and a basis Y of E with an element y ∈ Y ∩ N . Distinguish now between two cases: Case B1: S is non-Abelian. Choose a set I of cardinality m, put G = S I and choose g ∈ G with gi 6= 1 for each i ∈ I. Since S is simple, the smallest normal subgroup of S which contains gi is S itself. Hence, by Lemma 18.3.9, the smallest closed normal subgroup of G which contains g is G itself. By Corollary 17.6.1(b), rank(G) = m. Choose a set G0 of generators of G of cardinality m which converges to 1 (Proposition 17.1.2). Construct a surjective map ϕ0 : Y → G0 ∪ {g} with ϕ0 (y) = g. Then ϕ0 extends to an epimorphism ϕ: E → G (Definition 17.4.1). In particular, ϕ(N ) is a closed normal subgroup of G which contains g. By the choice of g, ϕ(N ) = G, so rS (N ) ≥ rS (G) = m. Therefore, rS (N ) = m. Case B2: S = Cp . Choose a map ϕ0 : Y → Cp which maps y onto a generator of Cp and all other elements of Y onto 1. Extend ϕ0 to an epimorphism ϕ: E → Cp . Then, ϕ(N ) = Cp . Hence, by Proposition 25.7.4(c),  rN (Cp ) = m. Lemma 25.8.2: The intersection of finitely many nontrivial normal subgroups of a free abstract group is nontrivial. Proof: It suffices to consider two nontrivial normal subgroups A, B of a free abstract group F and to prove that A ∩ B 6= 1. Assume A ∩ B = 1. Choose a ∈ A and b ∈ B with a, b 6= 1. Then hai ∼ = Z, hbi ∼ = Z, and ab = ba. Hence, ∼ C = ha, bi = Z × Z. On the other hand, C, as a subgroup of F , must be free (Proposition 17.5.6). It follows from this contradiction that A ∩ B 6= 1.  Proposition 25.8.3 ([Melnikov1, Thm. 4.2]): Let C be a Melnikov formation of finite groups, e ≥ 2 an integer, and N a closed normal subgroup of Fˆe (C) of infinite index. Then N is contained in a closed normal subgroup M of Fˆe (C) which is isomorphic to Fˆω (C). Proof: Put F = Fˆe (C). If rN (S) = ℵ0 for each simple group S ∈ C, then N∼ = Fˆω (C) (Theorem 25.7.3(b)). Suppose there is a simple group S ∈ C with r = rN (S) finite. Choose a basis x1 , . . . , xe of F . Let F0 be the abstract subgroup of F generated by x1 , . . . , xe . By Remark 17.6.1, F0 is free and x1 , . . . , xe form a basis for F0 . Let E be an open normal subgroup of F which contains N . Then E is a free pro-C group with f = rank(E) = 1 + (F : E)(e − 1) (Proposition 17.6.2). Choose (F : E) large enough to imply f − 1 ≥ rank(S r+1 ). Then S r+1 is a quotient of Fˆf −1 (C), so rFˆf −1 (C) (S) ≥ r + 1. Put E0 = F0 ∩ E. By Lemma 17.2.1, (F0 : E0 ) = (F : E) and E is the closure of E0 in F . By Proposition 17.5.7, E0 is a free abstract group with a

25.9 Accessible Subgroups

625

finite basis y1 , . . . , yf . Hence, E is the free pro-C group with basis y1 , . . . , yf (Lemma 17.4.6). Let K be the closed normal subgroup of E generated by y1 . Assume KN is open in E. Let ϕ: E → Fˆf −1 (S) be an epimorphism with ϕ(y1 ) = 1. Then ϕ(N ) = ϕ(KN ) is an open subgroup of Fˆf −1 (S). Hence, ϕ(N ) is a free pro-S group of rank at least f − 1. Therefore, rN (S) ≥ rϕ(N ) (S) ≥ rFˆf −1 (C) (S) > rN (S). It follows from this contradiction that (E : KN ) = ∞. Let z1 , . . . , zd be representatives for the left cosets of E0 in F0 . Then zT1 , . . . , zd are representatives for the left cosets of E in F . Therefore, L = d zj is a closed normal subgroup of F . Hence, so is LN . j=1 K the normal abstract subgroup of E0 generated by y1 . Put Let K Td 0 be z L0 = j=1 K0 j . F0

F

E0

E

K0

K

KN

L0

L

LN N

By Lemma 25.8.2, L0 6= 1. It follows from Proposition 25.8.1 that LN ∼ = Fˆω (C).  Proposition 25.8.3 may be used to reduce questions about closed normal subgroups of Fˆe (C) to closed normal subgroups of Fˆω (C). For example, we prove a result which would also follow from an affirmative answer to Problem 25.4.9: Proposition 25.8.4: Let C be a Melnikov formation of finite groups, e ≥ 2 an integer, F = Fˆe (C), N a closed normal subgroup of F , and M a proper open subgroup of N . Suppose (F : N ) = ∞ and M is pro-C. Then M ∼ = Fˆω (C). Proof: Choose a closed normal subgroup E of F which contains N and is isomorphic to Fˆω (C) (Proposition 25.8.3). By Theorem 25.4.7(a), with E  replacing F , we have M ∼ = Fˆω (C).

25.9 Accessible Subgroups Projective groups have been defined as profinite groups for which every finite embedding problem is solvable. They are characterized as closed subgroups of free profinite groups. Likewise, homogeneous groups have been defined as

626

Chapter 25. Free Profinite Groups of Infinite Rank

profinite groups of infinite rank for which every Melnikov embedding problem of a certain type is solvable. Here we characterize homogeneous groups as “accessible subgroups” of free profinite groups. We call a closed subgroup M of a profinite group G accessible if M is the intersection of a descending transfinite normal sequence of closed subgroups T α ranges over all of G. Thus, M = α<µ Nα , where µ is an ordinal number, T ordinals less than µ, N0 = G, Nα+1 / Nα , and Nβ = α<β Nα for each limit ordinal β < µ. When µ is finite, M is just a subnormal subgroup of G. Here are some basic properties of accessible subgroups: Lemma 25.9.1: Let M be an accessible subgroup of a profinite group G. Then: (a) Each accessible subgroup of M is an accessible subgroup of G. (b) Let H be a closed subgroup of G. Then H ∩ M is an accessible subgroup of H. (c) Let ϕ: G → H be an epimorphism. Then ϕ(M ) is an accessible subgroup of H. (d) Suppose G is finite. Then M is a subnormal subgroup of G. (e) Suppose G is a pro-C group for some Melnikov formation C of finite groups. Then M is pro-C. Proof: Statements (a) and (b) follow directly from the definition. For (c) use Lemma 1.2.2(c) to prove that ϕ maps each descending normal transfinite sequence of closed subgroups of G onto a descending normal transfinite sequence of closed normal subgroups of H. In particular, ϕ maps the intersection of the former sequence onto the intersection of the latter sequence. Now suppose G is finite. Let m be the maximal length of a normal sequence of subgroups of G. Then each normal transfinite sequence of subgroups of G collapses into a normal sequence of at most m subgroups. In particular, M is the smallest subgroup of that sequence. Thus, M is subnormal in G. This proves (d). Finally, let G and C be as in (e). If G is finite, induction on the length of a normal sequence from G to M proves that M is a C-group. In the general case consider an open normal subgroup N of M . Choose ¯ = G/H and let an open normal subgroup H of G with H ∩ M ≤ N . Put G ∼ ¯ ϕ: G → G be the quotient map. Then ϕ(M )/ϕ(N ) = M/N . By (c) and (d), ϕ(M ) is subnormal, so ϕ(M ) ∈ C. Hence, M/N ∈ C. Consequently, M is pro-C.  It is somewhat surprising that each accessible subgroup of a profinite group G is the intersection of a countable normal sequence of subgroups of G. In order to see this we introduce notation which will also be useful in the next chapter: Let S be a subset of a profinite group G. Denote the smallest closed normal subgroup of G containing S by [S]G . Thus, [S]G is the intersection

25.9 Accessible Subgroups

627

of all closed normal subgroups of G which contain S. Alternatively, [S]G = hsg | s ∈ S, g ∈ Gi. When S = {s1 , . . . , sn }, write [s1 , . . . , sn ]G instead of [S]G . For example, let K be a field and σ1 , . . . , σe ∈ Gal(K). Then, in the notation of Section 18.10, Gal(Ks [σ1 , . . . , σe ]) = [σ1 , . . . , σe ]Gal(K) . Here are some obvious rules for this notation: Lemma 25.9.2: Let G be a profinite group and S a subset of G. (a) Suppose H / G and S ⊆ H. Then [S]G ≤ H. (b) Suppose H ≤ G and S ⊆ H. Then [S]H ≤ [S]G . (c) Suppose S ⊆ S 0 ⊆ G. Then [S]G ≤ [S 0 ]G . (d) Suppose ϕ: G → H is an epimorphism. Then ϕ([S]G ) = [ϕ(S)]H . Again, let S be a subset of a profinite group G. Inductively define (1) (i+1) = G, [S]G = [S]G , and [S]G = [S][S](i) . This gives a normal

(0) [S]G

G

(2)

(3)

descending sequence of closed subgroups of G: G.[S]G .[S]G .[S]G .· · · ⊇ S. All rules of Lemma 25.9.2 inductively extend to the generalized operation. (i) (i) In particular, if ϕ: G → H is an epimorphism, then ϕ([S]G ) = [ϕ(S)]H . Lemma 25.9.3: The following holds for a closed subgroup M of a profinite group G. T∞ (i) (a) If M is accessible, then i=1 [M ]G = M . (b) Suppose for each epimorphism ϕ: G → H with H finite, ϕ(M ) is subnormal in H. Then M is accessible. (c) Let ϕ: G → H be a Melnikov cover. Suppose M is accessible and ϕ(M ) = H. Then M = G. T∞ (i) Proof: Put Mi = [M ]G , i = 1, 2, 3, . . ., and Mω = i=1 Mi . Then M ≤ Mω . Proof of (a): First suppose G is finite. By Lemma 25.9.1(d), M is subnormal. Thus, there is a finite normal sequence G . H1 . H2 . · · · . Hr = M . It follows that M1 ≤ H1 . Hence, M2 = [M ]M1 ≤ M1 ∩ H2 ≤ H2 . Repeating this argument r times gives Mr ≤ Hr = M . Therefore, Mω = M . ¯ be an epimorphism with G ¯ finite. In the general case let ϕ: G → G Denote images under ϕ by a bar. Using induction on i, it follows from ¯ i , i = 1, 2, . . ., hence Mω = M ¯ ω (Lemma Lemma 25.9.2(d), that Mi = M ¯ ¯ 1.2.2(c)). By Lemma 25.9.1(c), M is an accessible subgroup of G. Hence, by ¯ =M ¯ ω = Mω . Since ϕ is arbitrary and both M and Mω the finite case, M are closed, M = Mω . ¯ be an epimorphism with G ¯ finite. Using the Proof of (b): Let ϕ: G → G ¯ is subnormal. bar notation as in the proof of (a), our assumption is that M ¯ ¯ ¯ω = M. Hence, by (a) and its proof, Mω = Mω = M . It follows that, M Therefore, M is accessible. Proof of (c): By induction assume Mi = G. Then Mi+1 / G and ϕ(Mi+1 ) = H, so Mi+1 = G. This completes the induction. T ∞  Now we may use (a) to conclude that M = i=1 Mi = G.

628

Chapter 25. Free Profinite Groups of Infinite Rank

Lemma 25.9.4: Intersection of accessible subgroups of a profinite group is accessible. Proof: Let G be a profinite group T and Mi , i ∈ I, a set of accessible subgroups. We have to prove N = i∈I Mi is accessible. (j)

(j)

Put Nj = [N ]G and Mij = [Mi ]G . By induction suppose Nj−1 ≤ Mi,j−1 . Then Nj = [N ]Nj−1 ≤ [Mi ]Mi,j−1 = MijT(Lemma 25.9.2), so this ∞ inclusion holds for all j. By Lemma 25.9.3(a), j=1 Mi,j = Mi . Hence, T∞ T T∞ T T∞ i∈I i∈I Mi = N . Therefore j=1 Nj ≤ j=1 Mi,j = j=1 Nj = N and N is accessible.  Given an epimorphism ϕ: G → A of profinite groups, there is a closed subgroup B of G such that ϕ|B : G → A is a Frattini cover (Lemma 22.5.6). Accessible subgroups give rise to an analog of that result for Melnikov covers: Proposition 25.9.5: Let ϕ: G → A be an epimorphism of profinite groups. Then G has an accessible subgroup B such that ϕ|B : B → A is a Melnikov cover. chain of accessible subgroups of G Proof: Let Ni , i ∈ I, be a descending T with ϕ(Ni ) = A. Then N = i∈I Ni is an accessible subgroup (Lemma 25.9.4) and ϕ(N ) = A (Lemma 1.2.2(c)). Thus, by Zorn’s Lemma, G has a minimal accessible subgroup B with ϕ(B) = A. Each proper closed normal subgroup B0 of B is accessible. By minimality, ϕ(B0 ) < A. Consequently,  ϕ|B is a Melnikov cover of A. Lemma 25.9.6 ([Ribes-Zalesskii, Thm. 8.4.2]): Let C be a Melnikov formation of finite groups, F a free pro-C group of infinite rank m, and N an accessible subgroup with weight(F/N ) < m. Then N ∼ = Fˆm (C). Proof: By Lemma 25.9.1(e), N is a pro-C group. Hence, by Proposition  25.2.2, N ∼ = Fˆm (C). The computation of the rank of an accessible subgroup of a free profinite group in Proposition 25.9.9 and Theorem 25.9.12 uses special profinite groups which we construct in the following two lemmas: Lemma 25.9.7: Let T be a finite simple non-Abelian group, p a prime number, and W a finite dimensional faithful Fp [T ]-module. Then W has a faithful composition factor. Proof: We prove the lemma by induction on n = dimFp (W ). If W is irreducible, W is the desired composition factor. Otherwise, W has a submodule V of dimension m with 0 < m < n. Choose an Fp -basis of V and extend it to an Fp -basis of W . The action of T on W defines an embedding ρ of
T into the subgroup H of GLn (Fp ) consisting of all matrices A0 B with C A ∈ GLm (Fp ), B ∈ Mm×(n−m) (Fp ), and C ∈ GLn−m (Fp ). Let N be the
B with Im and normal subgroup of H consisting of all matrices I0m In−m

25.9 Accessible Subgroups

629

In−m being the unit matrices. Then N is nilpotent, so ρ(T ) ∩ N = 1. Thus, ρ induces an embedding
ρ¯: T → H/N . But H/N is isomorphic to the group of all matrices A0 B0 with A ∈ GLm (Fp ) and B ∈ GLn−m (Fp ), that is, to GLm (Fp )×GLn−m (Fp ). Take t ∈ T with (A, B) = ρ¯(t) 6= 1. Assume without loss A 6= 1. Let α: GLm (Fp ) × GLn−m (Fp ) → GLm (Fp ) be the projection on the first coordinate. Then α ◦ ρ¯(t) 6= 1. This means Ker(α ◦ ρ¯) < T . Since T is simple, α ◦ ρ¯ is injective. It follows that V is a faithful Fp [T ]-module. Induction gives a faithful composition factor of V , hence of W .  Lemma 25.9.8: Let I be a set and S, T finite simple groups. Suppose S 6= T if both S and T are Abelian. Then there exists a pro-{S, T } group B with open normal subgroups L, M satisfying this: (a) B/M ∼ = T and M is the unique maximal normal subgroup of B (thus, M = M (B)). Q (b) L ≤ M and L ∼ = i∈I Li where each Li is isomorphic to a direct product of finitely many isomorphic copies of S. (c) If I is infinite, rank(B) = |I|. (d) If S or T are non-Abelian, then L = M . Proof: We construct a finite group Tˆ, an epimorphism π: Tˆ → T , a finite group V , and an action of Tˆ on V satisfying this: (2a) M (Tˆ) = Ker(π). (2b) The trivial group and V itself are the only subgroups of V which are Tˆ-invariants. (2c) V is the direct product of at least two and at most finitely many isomorphic copies of S. (2d) If S or T are non-Abelian, then Tˆ = T and π is the identity map. Having constructed TˆQ, π, and V , we choose for each i ∈ I an isomorphic copy Li of V and let L = i∈I Li . The action of Tˆ on each Li gives an action of Tˆ on L. Let B = Tˆ n L. Then, each Li is a minimal normal subgroup of B. π Now let M be the kernel of the combined epimorphism B → Tˆ −→ T . ∼ Thus, L ≤ M and B/M = T . If I is infinite, then rank(B) = rank(L) = |I| (Corollary 17.1.6(b)). To prove M (B) = M , we have to show that M is the unique maximal open normal subgroup of B. So, let K be a maximal open normal subgroup of B. Assume M 6≤ K. Since M/L = M (B/L) (by (2a)), this assumption implies L 6≤ K, so there is a j ∈ I with Lj 6≤ K. Thus, K ∩ Lj is a proper subgroup of Lj which is normal in B. The minimality of Lj implies K ∩ Lj = 1. Since K is maximal normal, KLj = B. Therefore, B/K ∼ = Lj . This is a contradiction, because B/K is simple while Lj is not (by (2c)). Finally, (2d) implies (d). The construction of Tˆ, π, and V splits up into three cases. Case A: S is non-Abelian. Put Tˆ = T and let π: Tˆ → T be the identity map. Let V = S T be the group of all functions f : T → S with multi-

630

Chapter 25. Free Profinite Groups of Infinite Rank

Q plication rule (f g)(t) = f (t)g(t). Thus, V = t∈T St where St = {f ∈ V | f (T r{t}) = 1} is isomorphic to S (Remark 13.7.5). Define the action 0 of T on V by f t (t0 ) = f (tt0 ). Then Stt = S(t0 )−1 t for all t, t0 ∈ T . Thus, T acts transitively on the set Q {St | t ∈ T }. Since S is non-Abelian, the only normal subgroups of V are t∈T0 St with T0 ⊂ T (Lemma 18.3.9). Hence, the only normal subgroups of V which are invariant under the action of T are the trivial group and V itself. Case B: S = Cp and T is non-Abelian. Again, put Tˆ = T and let π: Tˆ → T be the identity map. Then T has a faithful Fp [T ]-module W of finite dimension n over Fp with n > 1. For example, Fp [T ] itself is such a module. Lemma 25.9.7 gives a faithful composition factor V of W . Thus, V is an irreducible Fp [T ]-module. Its dimension is greater than 1, otherwise T is a subgroup of F× p , so T is Abelian, which is a contradiction. The module V will be a finite group with a T -action satisfying (2b). Case C: S = Cp and T = Cq with p 6= q. Choose a positive integer k with p 6≡ 1 mod q k , so that ζqk ∈ / Fp . Let Tˆ = Cqk and π: Tˆ → T an epimorphism. ˆ Then Ker(π) = M (T ). Now put V = Fp [ζqk ]. Multiplication by ζqk makes V an Fp [Cqk ]-module with dimFp (V ) > 1. Moreover, V is an irreducible module. Indeed, f = irr(ζqk , Fp ) is the characteristic polynomial of ζqk as an automorphism of / Fp , hence deg(f ) > 1. Any proper nontrivial V . By assumption, ζqk ∈ submodule of V leads to a factorization of f over Fp . Hence, there is no such submodule. This concludes the construction of Tˆ and V and the proof of the lemma.  Every subgroup of a finite p-group is subnormal [Huppert, p. 308], so every closed subgroup N of a pro-p group F is accessible (Lemma 25.9.3(b)). If F is pro-p free, then so is N . But there are no constraints on the rank of N . This explains the exclusion of the formation of p-groups from the next result, which partially generalizes Lemma 25.6.3. Proposition 25.9.9 ([Melnikov2, Thm. 3.4] and [Ribes-Zalesskii, Prop. 8.5.10]): Let C be a Melnikov formation of finite groups, F a free pro-C group of rank m ≥ 2, and N a nontrivial accessible subgroup of infinite index. Suppose C is not the formation of p-groups for a single prime number p. Then N is C-homogeneous with rank(N ) = max(ℵ0 , m). If in addition, rN (S) = mmax(ℵ0 ,m) for each finite simple group S ∈ C, then N∼ = Fˆmax(ℵ0 ,m) (C). Proof: In contrast to the situation of Lemma 25.6.3 we do not know rank(N ) in advance, so we first solve enough embedding problems of F and then compute rank(N ):

25.9 Accessible Subgroups

631

Part A: An embedding problem. Consider a pro-C Melnikov embedding problem (3)

(ϕ: N → A, α: B → A)

for N satisfying one of the following conditions: (4a) either m < ℵ0 and B is finite, (4b) or m = ℵ0 , rank(B) ≤ ℵ0 , and A is finite, (4c) or m > ℵ0 , rank(B) ≤ m, and rank(A) < m. Put N1 = Ker(ϕ). In each of the above three cases we construct a closed normal subgroup L1 of F with N ∩ L1 ≤ N1 and put L = N L1 . Then L is accessible in F , so L is a pro-C group (Lemma 25.9.1(e)). These subgroups should satisfy additional properties: If (4a) holds, use Proposition 17.6.2 to choose L1 open with rank(L) ≥ rank(B). Then L is pro-C free. If (4b) holds, choose L1 open. Then L ∼ = Fˆω (C) (Proposition 17.6.2). If (4c) holds, use Lemma 25.6.1 to choose L1 with rank(F/L1 ) < m. Then rank(F/L) < m. By Proposition 25.2.2, L ∼ = Fˆm (C). In each case extend ϕ to an epimorphism ϕ: ˜ L → A by ϕ(nl ˜ 1 ) = ϕ(n) for n ∈ N and l1 ∈ L1 . In Case (4a), Proposition 17.7.3 gives an epimorphism γ: L → B with α ◦ γ = ϕ. In Cases (4b) and (4c) Lemma 25.6.2 supplies such γ. Thus, γ(N ) is a subnormal subgroup of B with α(γ(N )) = A. Since α: B → A is Melnikov, γ(N ) = B (Lemma 25.9.3(c)). Therefore, γ|N is a solution of (3). Part B: Computation of rank(N ). By assumption, N is nontrivial. Hence, there is a finite simple group T and an epimorphism ϕ: N → T . We choose a finite simple group S in C in the following way. If T is non-Abelian, choose S = T . If T is Abelian, then T ∼ = Cq for some prime number q. By assumption, C is not the formation of finite q-groups, so we may choose S 6= T in C. Let I be a set of cardinality max(ℵ0 , m), B be the profinite group which Lemma 25.9.8 supplies, M = M (B), and α: B → B/M ∼ = T the quotient map. Then B is a pro-C group of rank max(ℵ0 , m) and α is a Melnikov cover. Part A gives an epimorphism γ: N → B with α ◦ γ = ϕ. Thus, rank(N ) ≥ rank(B). Consequently, rank(N ) = max(ℵ0 , m). Part C: The final statement. Suppose now rN (S) = max(ℵ0 , m) for each finite simple group S ∈ C. Then N ∼ = Fˆmax(ℵ0 ,m) (C) (Proposition 25.7.2(c)).  The next result says that, in contrast to closed normal subgroups of Fˆm (C), there are no constraints on the S-rank of an accessible subgroup except the obvious ones: Proposition 25.9.10 ([Ribes-Zalesskii Thm. 8.5.3]): Let C be a Melnikov formation of finite groups, m ≥ 2 a cardinal number, and f a function from the set of finite simple groups to the set of cardinal numbers at most

632

Chapter 25. Free Profinite Groups of Infinite Rank

max(ℵ0 , m). Suppose f (S) = 0 if S ∈ / C. Then Fˆm (C) has an accessible subgroup N with rN = f . Proof: Put F = Fˆm (C). If m < ℵ0 replace F by an accessible subgroup H which is isomorphic to Fˆω (C). For example, choose H to be a proper open normal subgroup of a closed normal subgroup of infinite index of F (Proposition 24.10.3). Thus, assume without loss, m ≥ ℵ0 . Denote the set of all simple groups in C by S. Consider the direct product Q G = S∈S S f (S) . By Corollary 17.1.6(b), rank(G) ≤ m. Thus, there is an epimorphism ϕ: F → G (Proposition 17.4.8). Proposition 25.9.5 gives an accessible subgroup N of F such that ϕ|N : N → G is a Melnikov cover. By Lemma 25.5.6, N/M (N ) ∼ = G/M (G) ∼ = G. It follow that rN = rG = f .  The results accumulated in the last sections culminate with the classification of C-homogeneous groups: Theorem 25.9.11 ([Melnikov2, Thm. 3.2]): Let C be a Melnikov formation of finite groups and G a pro-C group of infinite rank m. Suppose C is not the formation of all finite p-groups for a single prime number p. Then, G is C-homogeneous if and only if G is isomorphic to a nontrivial accessible subgroup of Fˆm (C). Proof: By Proposition 25.9.9, each accessible subgroup of Fˆm (C) is Chomogeneous. Conversely, suppose G is C-homogeneous. Then rG (S) ≤ m for each finite simple group S, so Proposition 25.9.10 gives an accessible subgroup N of Fˆm (C) with rN = rG . In other words, N/M (N ) ∼ = G/M (G). If N is open in Fˆm (C), then N is C-free (Proposition 17.6.2), hence C-homogeneous (Proportion 25.6.2). If the index of N in Fˆm (C) is infinite, then N is Chomogeneous of rank m (Proposition 25.9.9). In both cases we conclude from Proposition 25.7.2 that G ∼  = N. Let C be a formation of finite groups and G a pro-C group. We call G virtually free pro-C if G has an open subgroup which is free pro-C. Theorem 25.9.12 ([Ribes-Zalesskii, Cor. 8.5.15]): Let C be a Melnikov formation of finite groups, m ≥ 2 a cardinal number, N a nontrivial accessible subgroup of Fˆm (C) of infinite index. Suppose C contains a non-Abelian simple group. Then N is virtually free pro-C. Proof: Put m∗ = max(ℵ0 , m). Claim: Suppose M is an open normal subgroup of N with T = N/M a simple group. Let S be a finite simple group belonging to C. Suppose S or T are non-Abelian. Then rM (S) = m∗ . If T is non-Abelian, then M ∼ = Fˆm∗ (C). Indeed, M is accessible, so M is C-homogeneous and rank(M ) = m∗ (Proposition 25.9.9). Lemma 25.9.8 gives a Melnikov cover α: B → N/M where Ker(α) is a direct product of m∗ isomorphic copies of S. Let ϕ: N → N/M be the

Notes

633

quotient map. Since N is C-homogeneous (Theorem 25.9.11), there is an epimorphism γ: N → B with α ◦ γ = ϕ. It maps M onto Ker(α), so rM (S) = m∗ . If T is non-Abelian, then the latter conclusion holds for every finite simple group S ∈ C. Hence, by Proposition 25.7.2(c), M ∼ = Fˆm∗ (C). This concludes the proof of the claim. Now choose a maximal open normal subgroup M of N . Thus, N/M is simple. If N/M is non-Abelian, M ∼ = Fˆm∗ (C), by the Claim. Otherwise, choose a non-Abelian simple group S in C. By the Claim, rM (S) = m∗ . In particular, M has an open normal subgroup L with M/L ∼ = S. An application  of the Claim with M, L replacing N, M gives L ∼ = Fˆm∗ (C). Remark 25.9.13: C-Homogeneous Melnikov covers. Let C be a Melnikov formation of finite groups. We call an epimorphism ϕ: H → G of pro-C groups a C-homogeneous cover if H is C-homogeneous and refer to rank(H) as the rank of the cover. Suppose C contains a non-Abelian simple group. Let G be a nontrivial C-group and m ≥ max(ℵ0 , rank(G)) a cardinal number. Then G has a pro-C homogeneous Melnikov cover of rank m. Indeed, choose an epimorphism ϕ: ˜ Fˆm (C) → G. Proposition 25.9.5 gives ˆ an accessible subgroup H of Fm (C) such that ϕ = ϕ| ˜ H : H → G is a Melnikov cover. By Proposition 25.9.9, H is C-homogeneous of rank m. Let ϕ0 : H 0 → G be another C-homogeneous Melnikov cover of rank m. If m = ℵ0 and G is finite or rank(M (G)) < m, there is an isomorphism θ: H → H 0 satisfying ϕ0 ◦ θ = ϕ (Proposition 25.6.5). In other words, in this case, the C-homogeneous Melnikov cover ϕ: H → G of G is unique. In the general case, H/M (H) ∼ = G/M (G) ∼ = H 0 /M (H 0 ) (Lemma 25.5.6). Hence, rH1 = rH2 . So, by Proposition 25.7.2, H ∼ = H 0 . However, it is not clear whether there exists an isomorphism θ: H → H 0 with ϕ0 ◦ θ = ϕ. For example, let ϕ: Fˆω → Fˆω and ϕ0 : Fˆω → Fˆω be Melnikov covers.  Does there exist θ ∈ Aut(Fˆω ) with ϕ0 ◦ θ = ϕ?

Notes The basic source for this Chapter is [Melnikov1]. Other sources are [Melnikov2], [Chatzidakis1], [Jarden-Lubotzky1], and [Haran5]. [Ribes-Zalesskii, Chap8] gives an excellent presentation to a great part of the Chapter. [Melnikov1, Prop. 2.1] proves Proposition 25.2.2 under the assumption N / F . Likewise [Ribes-Zalesskii, Thm. 8.7.1] proves Theorem 25.4.7 under the weaker assumption that M is a proper open normal subgroup of N . The notation M (G) for the Melnikov group, albeit not the name, appears in [Melnikov2]. See also [Ribes-Zalesskii, Sec. 8.5]. Likewise, the notion of Homogeneous pro-C groups is due to [Melnikov2]. The definition for a C-homogeneous group G that appears in [RibesZalesskii, p. 322] replaces Condition (2a) of Section 25.6 by the seemingly stronger condition:

634

Chapter 25. Free Profinite Groups of Infinite Rank

(1) m = ℵ0 , rank(B) ≤ ℵ0 , and A is finite. Nevertheless, one can break up the epimorphism α: B → A into a sequence of epimorphisms B/L T∞ n+1 → B/Ln with Ln open normal in B, L0 = Ker(α), Ln+1 ≤ Ln , and n=0 Ln = 1. An application of (2a) of Section 25.6 and induction gives a compatible sequence of epimorphisms γn : G → B/Ln . This gives a solution of embedding problem (1) of Section 25.6. Thus, our definition is equivalent to that of [Ribes-Zalesskii]. The notion of “accessible group” appears in [Melnikov2]. Lemma 25.9.8 is a workout of [Ribes-Zalesskii, Lemma 8.5.8]. Both results consider a set I and finite simple groups S, T which are distinct if both of them are Abelian. Then [Ribes-Zalesskii, Lemma 8.5.8] constructs ∼ a profinite group B with Q an open normal subgroup L such that B/L = T , ∼ M (B) = L, and L = i∈I Li with Li isomorphic to a direct product of finitely many copies of S. When S ∼ = Cp , the proof of [Ribes-Zalesskii, Lemma 8.5.8] chooses each Li as an irreducible Fp [T ]-module of dimension > 1. Case B of the proof of Lemma 25.9.8 supplies the missing construction when T is non-Abelian. However, if T ∼ = Cq and q divides p − 1 all irreducible Fp [T ]-modules are of dimension 1. To overcome this difficulty, Lemma 25.9.8 modifies the construction of B in this case. We then use the modified B in the proof of Proposition 25.9.9 in the same way as [Zalesskii-Ribes] do with the original B. Theorem 25.9.12 assumes C contains a non-Abelian simple group. In contrast, [Ribes-Zalesskii, Cor. 8.5.15] assumes C is not the formation of pgroups with a single p. However, the proof of [Ribes-Zalesskii, Cor. 8.5.15(b)] uses the existence of a non-Abelian simple group in C.

Chapter 26. Random Elements in Profinite Groups Let n be a positive integer and F = Fˆn the free profinite group of rank n. For each e-tuple (x1 , . . . , xe ) in F e we consider the closed (resp. normal closed) subgroup hxi (resp. [x]) generated by x1 , . . . , xn in F . We investigate the probability that hxi (resp. [x]) is an open subgroup of F . Having done so, we strive to prove that with probability 1 each of the groups hxi and [x] are C-free of specific ranks. We mention here only the most striking results: For almost all x ∈ F e the group hxi has an infinite index in F and is isomorphic to Fˆe (Lemma 26.1.7). This settles problem 16.16 of [FriedJarden3]. ˆ n )e the group hxi is of infinite index if e ≤ n and is For almost all x ∈ (Z n ˆ n )e ˆ open in Z if e > n (Theorem 26.3.5). In addition, for almost all x ∈ (Z ˆ e if e ≤ n and to Z ˆ n if e > n (Theorem the group hxi is isomorphic to Z 26.3.6). This generalizes Lemma 18.5.8. Suppose n ≥ 2. Then F e has a subset C of positive measure such that [x] has infinite index for each x ∈ C (Theorem 26.4.5(b)). The proof of this result uses the Golod-Shafarevich inequality. If e > n, then F e has a subset B of positive measure such that [x] = Fˆn (Theorem 26.4.5(a)). Here we must use the classification of finite simple groups. Finally, we prove that for almost all x ∈ C the group [x] is isomorphic to Fˆω (Theorem 26.5.6). The proof uses Melnikov’s criterion for a closed normal subgroup of Fˆn to be free of countable rank (Corollary 25.7.6).

26.1 Random Elements in a Free Profinite Group ˆ to an arbitrary free profinite This section generalizes Lemma 18.5.8 from Z group F of rank at least 2: The closed subgroup of F generated by an e-tuple (x1 , . . . , xe ) chosen at random is of infinite index and isomorphic to Fˆn . Notation 26.1.1:

For a finite group G and a positive integer e let

de (G) = max(m ∈ N | Gm is generated by e elements) = max(m ∈ N | Gm is a homomorphic image of Fˆe )  De (G) = {(x1 , . . . , xe ) ∈ Ge | hx1 , . . . , xe i = G}. The next Lemma implies that de (S) is a positive integer if e is large enough: Lemma 26.1.2 (P. Hall): Let S be a finite simple non-Abelian group, s = rank(S), and e a positive integer. Then: (a) de (S) is the number of open normal subgroups N of Fˆe with Fˆe /N ∼ = S.

636

Chapter 26. Random Elements in Profinite Groups

(b) de (S) = rFˆe (S) = rFˆe (S) (S) (Section 24.9). (c) de (S) =

|De (S)| |Aut(S)| . e−1

and |S|e−2rank(S) ≤ de (S) if rank(S) ≤ e. (d) de (S) ≤ |S| (e) de (S) < de+1 (S). Proof of (a) and (b): By the opening discussion of Section 24.9, de (S) = rFˆe (S). By Lemma 16.8.3, de (S) is then the number of normal subgroups N of Fˆe with Fˆe /N ∼ = S. Similar statements hold for the free pro-S group Fˆe (S) or rank e. Proof of (c): Put d =

|De (S)| |Aut(S)| .

Fix a basis z1 , . . . , ze of Fˆe . The map

ψ 7→ (ψ(z1 ), . . . , ψ(ze )) is a bijection between the set of all epimorphisms ψ: Fˆe → S and De (S). Two epimorphisms ψ and ψ 0 have the same kernel if and only if there is an α ∈ Aut(S) with α ◦ ψ = ψ 0 . Hence, Fˆe has exactly d normal subgroups N with Fˆe /N ∼ = S. By (a), d = de (S). Proof of (d): Since S is simple and non-Abelian, mapping each x ∈ S onto the corresponding inner automorphism embeds S into Aut(S). Hence, |S| ≤ |Aut(S)|. By definition, |De (S)| ≤ |S|e . Hence, by (b), de (S) ≤ |S|e−1 . Now suppose s = rank(G) ≤ e. Let x1 , . . . , xs be generators of S. Then x1 , . . . , xs , xs+1 , . . . , xe generate S for all xs+1 , . . . , xe ∈ S. Hence, |Dn (S)| ≥ |S|e−s . Also, each α ∈ Aut(S) is uniquely determined by (α(x1 ), . . . , α(xs )).  Therefore, |Aut(S)| ≤ |S|s . It follows from (c) that de (S) ≥ |S|e−2s . Consider the alternative group Ak on k letters. The probability that a pair (x, y) ∈ Ak generates Ak approaches 1 as k → ∞. In other words, (1)

|D2 (Ak )| −→ 1 (k!)2 /4 k→∞

[Dixon]. The inequality stated in the next lemma is not good enough to prove (1). Nevertheless, it suffices for the application in the proof of Lemma 26.1.4: Lemma 26.1.3: Let k ≥ 7 be an odd integer. Then |D2 (Ak )| ≥ (k − 3)!(k − 7)!. Proof: Let γ = (1 2 3) and let ρ = (b3 b4 · · · bk ) be a cyclic permutation of the set B = {3, 4, . . . , k}. We claim: (2)

Ak = hγ, ρi.

To prove (2), it suffices to prove that H = hγ, ρi contains each 3-cycle. Assume without loss that b3 = 3. Since k is odd, σ = ργ = (1 b4 · · · bk ) ∈ H

and

2

τ = ργ = (2 b4 · · · bk ) ∈ H.

26.1 Random Elements in a Free Profinite Group

637

Hence, for each j ≥ 3 (bj 2 3) = γ σ

j−3

,

(1 bj 3) = γ τ

j−3

,

j−3

(1 2 bj ) = γ ρ

and

belong to H. Finally, let b, c, d be distinct elements of B. Then (2 c b) = (1 2 b)(1 2 c) , (c b 1) = (1 b 3)(1 c 3) , (b 3 c) = (b 2 3)(c 2 3) , and (d b c) = (2 b c)(2 d 3) belong to H. Thus, every 3-cycle of {1, 2, . . . , k} belongs to H. Hence, H = Ak , as asserted. Next check the residues modulo 6 to find a positive integer m with k − 6 ≤ m ≤ k − 3 which is prime to 6. Each cyclic permutation α = (a1 a2 · · · am ) of m integers in A = {4, 5, . . . , k} belongs to Ak . Moreover, (αγ)m = αm γ m = γ m = γ ±1 . Hence, by (2), Ak = hαγ, ρi. There are (k − 3)(k − 4) · · · (k − 2 − m)/m permutations α and (k − 3)! permutations ρ. The former number is ≥ (m − 1)!. Hence, |D2 (Ak )| ≥ (k − 7)!(k − 3)!, as asserted.  (k−3)!(k−7)! k!

Lemma 26.1.4: For each odd integer n ≥ 15, Lk = Ak by 2 elements.

is generated

Proof: By [Huppert, p. 175], Aut(Ak ) ∼ = Sk . So, |Aut(Ak )| = k!. It follows from Lemmas 26.1.2 and 26.1.3 that d2 (Ak ) =

(k − 3)!(k − 7)! |D2 (Ak )| ≥ ≥ 1. k! k!

To prove the latter inequality, verify that 7! ≥ 14 · 13 · 12 and then prove inductively that (n − 7)! ≥ n(n − 1)(n − 2) for each n ≥ 14. Thus, Lk is generated by 2 elements.  Lemma 26.1.5: The probability that an e-tuple of elements of Lk generates Lk approaches 0 as k approaches infinity over the odd positive integers. Proof: In order for an e-tuple of elements of Lk to generate Lk its projection on each of the factors must generate Ak . The probability of the last event is at most 1 − k1e , because an e-tuple of elements which belong to the subgroup Ak−1 of index k, does not generate Ak . Hence, the probability that an e-tuple of elements of Lk generates Lk is at most

(3)

(   (k−3)!(k−7)! ke ) (k−3)!(k−7)! k!ke k! 1 1 1− e = 1− e k k

The expression in the braces approaches the inverse of the basis of the natural logarithms (which is usually denoted by the same letter e which we are using here for another purpose). The exponent of the braces approaches infinity as k approaches infinity. Consequently, the right hand side of (3) approaches 0 as k → ∞. 

638

Chapter 26. Random Elements in Profinite Groups

Lemma 26.1.6: For n ≥ 2 and e ≥ 1, the probability for an e-tuple of elements of Fˆn to generate Fˆn is 0. Proof: Let k ≥ 15 be an odd integer. By Lemma 26.1.4, there is an epimorphism ψ: Fˆn → Lk . If (x1 , . . . , xe ) ∈ (Fˆn )e generates Fˆn , then its image under ψ generates Lk . Hence, the probability for an e-tuple of elements of Fˆn to generate Fˆn is at most the probability for an e-tuple of elements of Lk to generate Lk . By Lemma 26.1.5, the latter probability approaches 0 as k → ∞. Therefore, the former probability is 0.  Proposition 26.1.7: Let F be a free profinite group of rank at least 2 and e a positive integer. Then for almost all (x1 , . . . , xe ) ∈ F e (a) hx1 , . . . , xe i has infinite index [Kantor-Lubotzky], and (b) hx1 , . . . , xe i ∼ = Fˆe [Lubotzky3]. Proof of (a): If rank(F ) is infinite, then so is the rank of each open subgroup. Hence, we may assume F = Fˆn with 2 ≤ n < ∞. By Proposition 17.6.2, each open subgroup of Fˆn is isomorphic to Fˆs for some s. For each s, the group Fˆn has only finitely many open subgroups of index at most s (Lemma 16.10.2). Applying Lemma 26.1.6 to each of these subgroups, we conclude that the probability of an e-tuple to generate an open subgroup of F is 0. This proves our claim. Proof of (b): First note that there is an epimorphism ψ: F → Fˆn with 2 ≤ n < ∞. If x1 , . . . , xe ∈ F and hψ(x1 ), . . . , ψ(xe )i ∼ = Fˆe , then hx1 , . . . , xe i ∼ = ˆ Fe (Lemma 17.14.11). Thus, if (b) holds for the quotient, it holds for F . Therefore, we may assume that F = Fˆn with 2 ≤ n < ∞. There are two cases to consider: Case A: n ≥ e + 3. To prove G = hx1 , . . . , xe i is isomorphic to Fˆe , it suffices to show that each finite group B which is generated by e elements is a quotient of G (Lemma 17.7.1). Since there are only countably many finite groups, it suffices to fix a finite group B with rank(B) ≤ e and to prove that for almost all (x1 , . . . , xe ) ∈ F e the group B is a quotient of hx1 , . . . , xe i. Indeed, Let l = |B|. Then B can be embedded into the symmetric group Sl . Consider the cycle κ = (l + 1 l + 2) of Sl+2 . Define an embedding f of Sl into Al+2 by the following rule: f (π) = π if π ∈ Al and f (π) = πκ if π ∈ / Al . Let k(B) = max(7, l + 2). Then, we can view B as a subgroup of Ak for each k ≥ max(7, l + 2). Let k ≥ 7 be an odd integer. Since Ak is generated by two elements (Lemma 26.1.3), |Dn (Ak )| ≥ |Ak |n−2 (Proof of Part (c) of Lemma 26.1.2). Also, |Aut(Ak )| = |Sk | = 2|Ak | [Suzuki, p. 299, Statement 2.17]. Hence, by Lemma 26.1.2, (4)

dn (Ak ) = d (A )

1 |Dn (Ak )| 1 ≥ |Ak |n−3 ≥ |Ak |e . |Aut(Ak )| 2 2 d (A )

By definition, Akn k is generated by n elements. Hence, Akn k is a quotient of F , with kernel N . Since Ak is simple non-Abelian, F has exactly

26.1 Random Elements in a Free Profinite Group

639

dn (Ak ) open normal subgroups Ni which contain N with F/Ni ∼ = Ak (Lemma 26.1.2(a)). For each i between 1 and dn (Ak ) let ϕi : F → Ak be an epimorphism with kernel Ni and Bk,i = {(x1 , . . . , xe ) ∈ F e | hϕi (x1 ) · · · , , ϕi (xe )i ∼ = B}. Denote the probability that e elements of B generate B by pe (B). Then  |B| e pe (B). (5) µ(Bk,i ) = |Ak | The sets Bk,i with k ≥ k(B) and i = 1, . . . , dn (Ak ) are µ-independent (Example 18.3.11). By (4) and (5) dn (Ak )

X

X

k≥k(B)

i=1

µ(Bk,i ) =

X k≥k(B)

≥

 |B| e |Ak |

dn (Ak )pe (B)

X 1 pe (B)|B|e = ∞, 2

k≥k(B)

because all terms are constant and pe (B) 6= 0, since B is generated by e elements.
S By Lemma 18.3.4, µ k,i Bk,i = 1. Each e-tuple in the union generates a closed subgroup of F which has B as a quotient. Case B: The general case. By (a), almost all (x1 , . . . , xe ) generate a closed subgroup G of F of infinite index, so G is contained in an open subgroup H of F of index at least e + 2. Since k ≥ 2, the group H is free of rank at least e + 3 (Proposition 17.6.2). Hence, by Case A, the probability for G to be contained in H and not to be free is zero. Since F has only countably many open subgroups, the probability for G not to be free is zero. This concludes the proof of (b).  Remark 26.1.8: 

Proposition 26.1.7 solves Problem 16.16 of [Fried-Jarden3].

Remark 26.1.9: Generalization of (1). [Liebeck-Shalev] completes earlier works of [Dixon] and [Kantor-Lubotzky] and prove: The probability that a pair (x, y) of elements of a finite non-Abelian simple group S generates S approaches 1 as |S| approaches infinity. Thus, in Notation 26.1.1, lim

|S|→∞

|D2 (S)| = 1, |S|2

where S ranges over the finite non-Abelian simple groups. The proof of this result uses the classification of finite simple groups.  Corollary 17.6.5 says that the free pro-C group is infinite if C is a Melnikov formation. The following example shows this does not hold for an arbitrary formations:

640

Chapter 26. Random Elements in Profinite Groups

Example 26.1.10: Finite free pro-C groups. Let S be a finite non-Abelian simple group. Put C = {S d | d = 0, 1, 2, . . .}. Claim A: C is a formation of finite groups (Section 17.3). Indeed, for each normal subgroup N of S d there exists another normal subgroup N 0 with S d = N × N 0 . Moreover, N 0 ∼ = S k for some k (Lemma 18.3.9). Thus, C is closed under quotients. That C is closed under fiber products is a consequence of Lemma 18.3.11. Consequently, C is a formation. Claim B: Let n ≥ 2 be an integer and d = dn (S) (Notation 26.1.1). Then S d is the free pro-C group of rank n. Indeed, let F be the free abstract group with basis x1 , . . . , xn . By definition, F has a normal subgroup N with F/N ∼ = S d . Moreover, d is the m maximal number m such that S is a quotient of F . Let π: F → S d be an epimorphism with Ker(π) = N . Put yi = π(xi ), i = 1, . . . , n. We claim that y1 , . . . , yn are free generators of S d in the category of pro-C groups. Indeed, suppose z1 , . . . , zn are generators of S m for some m. Then there is an epimorphism ϕ: F → S m . If N 6≤ Ker(ϕ), then by Claim A, F/(Ker(ϕ) × N ) ∼ = S e for some e > d, in contradiction the to maximality of d. It follows that N ≤ Ker(ϕ), so there exists an epimorphism ϕ: ¯ Sd → Sm with π ◦ ϕ¯ = ϕ. In particular, ϕ(y ¯ i ) = zi , i = 1, . . . , n, as needed.  Here is an application of Proposition 26.1.7 to fields: Example 26.1.11: A non-Hilbertian field K with Ks (σ) PAC and e-free for all e and almost all σ ∈ Gal(K)e . Let K be a PAC field with Gal(K) ∼ = Fˆ2 . For example, starting from a countable Hilbertian field E, almost all fields Es (σ1 , σ2 ) are 2-free PAC (Theorem 20.5.1). By Proposition 26.1.7(b), for each positive integer e and for almost all σ ∈ Gal(K)e the field Ks (σ) is e-free. By Ax-Roquette (Corollary 11.2.5), each of the fields Ks (σ) is PAC. However, by Lemma 16.12.5, K is not Hilbertian because Gal(K) is finitely generated. 

26.2 Random Elements in Free pro-p Groups Theorem 26.1.7 gets a new form if we consider free pro-p-groups instead of free profinite groups: Lemma 26.2.1: Let f ∈ Zp [X1 , . . . , Xn ] be a nonzero polynomial. Then the set A = {(x1 , . . . , xn ) ∈ Znp | f (x1 , . . . , xn ) = 0} has measure 0. Proof: View f as a polynomial in Xn with coefficients in Zp [X1 , . . . , Xn−1 ]. Choose a nonzero coefficient g ∈ Zp [X1 , . . . , Xn−1 ]. An induction hypothesis | g(x1 , . . . , xn−1 ) = 0} has on n implies that B = {(x1 , . . . , xn−1 ) ∈ Zn−1 p r B the set A(x1 ,...,xn−1 ) = {xn ∈ measure 0. For each (x1 , . . . , xn−1 ) ∈ Zn−1 p Zp | (x1 , . . . , xn−1 , xn ) ∈ A} is finite, hence has measure 0. It follows from Fubini [Halmos, p. 147, Thm. A] that µ(A) = 0. 

26.2 Random Elements in Free pro-p Groups

641

Proposition 26.2.2 ([Lubotzky3, Prop. 7]): Let F = Fˆn (p) be the free pro-p-group of rank n and let e be a positive integer. Let Ae = {(x1 , . . . , xe ) ∈ F e | hx1 , . . . , xe i is open in F } Be = {(x1 , . . . , xe ) ∈ F e | hx1 , . . . , xe i ∼ = Fˆe (p)}. Then: (a) If e < n, then µ(Ae ) = 0 and µ(Be ) = 1. (b) 0 < µ(An ) < 1 and µ(Bn ) = 1. (c) If e > n, then 0 < µ(Ae ) < 1 and 0 < µ(Be ) < 1. Proof of (b): By Nielsen-Schreier (Proposition 17.6.2), the rank of each proper open subgroup of F is greater than n. Hence, (1)

An = {(x1 , . . . , xn ) ∈ F n | hx1 , . . . , xn i = F }.

Let V = Fnp ∼ = F/Φ(F ) (Lemma 22.7.4). By definition, x1 , . . . , xn generate F if and only if their reductions v1 , . . . , vn modulo Φ(F ) generate V . The latter happens exactly if v1 , . . . , vn are linearly independent. Hence, µ(An ) is the probability in V n that v1 , . . . , vn are linearly independent. Thus (2)

p n − 1 pn − p pn − pn−1 · · · · pn pn pn      1 1 1 = 1− n . 1 − n−1 · · · 1 − p p p

µ(An ) =

Hence, 0 < µ(An ) < 1. To compute µ(Bn ) let Z = Znp and choose an epimorphism π: F → Z. Consider each element of Z as a column with r entries in Zp . In this notation (z1 · · · zn ) denotes an n × n matrix with entries in Zp . Then ¯n = {(z1 , . . . , zn ) ∈ Z n | hz1 , . . . , zn i ∼ B = Z} = {(z1 , . . . , zn ) ∈ Z n | rankhz1 , . . . , zn i = n} 2

= {(z1 , . . . , zn ) ∈ Znp | rank(z1 · · · zn ) = n} 2

= {(z1 , . . . , zn ) ∈ Znp | det(z1 · · · zn ) 6= 0}. ¯n ) = 1. By Lemma 26.2.1, µ(B ¯n , then rankhx1 , . . . , xn i = If x1 , . . . , xn ∈ F and (π(x1 ), . . . , π(xn )) ∈ B n. Since each closed subgroup of F is a free pro-p-group (Corollary 22.7.7), ¯n ) ⊆ this implies hx1 , . . . , xn i ∼ = F . Hence, (x1 , . . . , xn ) ∈ Bn . Thus, π −1 (B Bn . It follows from the preceding paragraph that µ(Bn ) = 1. Proof of (a): By Nielsen-Schreier, the rank of each open subgroup of F is at least e. Hence, in case (a), Ae = ∅, so µ(Ae ) = 0.

642

Chapter 26. Random Elements in Profinite Groups

To compute µ(Be ) consider the projection τ : F n → F e on the first e coordinates. Suppose (x1 , . . . , xn ) ∈ Bn . Then, rankhx1 , . . . , xe i = e, so hx1 , . . . , xe i ∼ = Fˆe (p). Therefore, (x1 , . . . , xe ) ∈ Be . Thus, Bn ⊆ τ −1 (Be ). By (b) and Proposition 18.2.2, µ(Be ) = 1. Proof of (c): Let ρ: F e → F n be the projection on the first n coordinates. Let (x1 , . . . , xe ) ∈ ρ−1 (An ). Then, (x1 , . . . , xn ) ∈ An . By (1), hx1 , . . . , xn i = F and therefore hx1 , . . . , xe i = F . Thus, ρ−1 (An ) ⊆ Ae . Hence, by (b), 6 Fˆe (p), we have, ρ−1 (An ) ⊆ F e r Be . 0 < µ(An ) ≤ µ(Ae ). Also, since F ∼ = Therefore, µ(Be ) < 1. Next use Nielsen-Schreier formula to choose an open subgroup U of F with l = rank(U ) > e. The rank of each open subgroup of U is also greater than e. Hence, U e ∩ Ae = ∅. Since µ(U e ) > 0, this implies that µ(Ae ) < 1. Finally, let λ: F l → F e be the projection on the first e coordinates. Then Bl ⊆ λ−1 (Be ). Hence, µ(Bl ) ≤ µ(Be ). Applying (b) to U and l instead of to F and e, we conclude that µ(Bl ) > 0, so µ(Be ) > 0. This concludes the proof of (c) and the proposition.  It will be interesting to compute the measure of Ae and Be in the case where F is the free prosolvable group on n generators. The proof of Proposition 26.2.2 does not apply to this case.

ˆn 26.3 Random e-tuples in Z ˆ for almost all a ∈ Z ˆ e . Moreover, hai is open in By Lemma 18.5.8, hai ∼ =Z ˆ if e ≥ 2 and of infinite index if e = 1. Theorems 26.3.5 and 26.3.6 below Z ˆ n. generalize this result to Z We fix positive integers e, n and use l to denote a prime number. Lemma 26.3.1: Suppose e ≥ n. Then the number of e-tuples (a1 , . . . , ae ) ∈ Qn−1 (Fnl )e that generate Fnl is i=0 (le − li ). If e < n, there are none. Proof: Consider the Fl -vector spaces Fnl and Fel . Then dim(Fnl ) = n. If e < n, no e-tuple (a1 , . . . , ae ) ∈ (Fnl )e generates Fnl . Suppose e ≥ n. Put R = {(a1 , . . . , ae ) ∈ (Fnl )e | a1 , . . . , ae generate Fnl } C = {(a(1) , . . . , a(n) ) ∈ (Fel )n | a(1) , . . . , a(n) are linearly independent} Define a map α: (Fnl )e → (Fel )n as follows. With each (a1 , . . . , ae ) ∈ (Fnl )e associate the matrix A = (aij )1≤i≤e, 1≤j≤n . Then let α(a1 , . . . , ae ) = (a(1) , . . . , a(n) ) with a(j) being the jth column of A, j = 1, . . . , n. This map is bijective.

ˆn 26.3 Random e-tuples in Z

643

Suppose (a1 , . . . , ae ) ∈ R. Then a1 , . . . , ae generate Fnl , so the row rank of A is n. Hence, the column rank of A is also n. Since A has exactly n columns, they are linearly independent. Conversely, if the columns of A are linearly independent, then the rows generate Fnl . Consequently, α maps R bijectively onto C. An n-tuple (a(1) , . . . , a(n) ) of (Fel )n belongs to C if and only if the vectors a(1) , . . . , a(i) are linearly independent for i = 1, . . . , n. Equivalently, a(i+1) ∈ Fel rha(1) , . . . , a(i) i for i = 0, . . . , n − 1. If a(1) , . . . , a(i) are linearly independent, then |Fel rha(1) , . . . , a(i) i| = le − li . Therefore, |R| = |C| = Qn−1 e i  i=0 (l − l ). Lemma 26.3.2: Let n be a positive integer, Z one of the groups Zn , Znl , or ˆ n , and Y a subgroup of Z of finite index. Then Y ∼ Z = Z. Proof: Let m = (Z : Y ). Then mZ is isomorphic to Z and mZ ≤ Y . First suppose Z = Zn . Then Z is a free Z-module of rank n. Hence, Y is a free Z-module of rank at most n [Lang7, p. 146, Thm. 7.1]. Similarly, n = rank(Z) = rank(mZ) ≤ rank(Y ). Therefore, rank(Y ) = n and Y ∼ = Zn . ˆ n or Zn . Then mZ is an open subgroup of Z and Y Next suppose Z is Z l is a union of cosets of mZ, so Y is open. ˆ n let N be the set of all subgroups of Zn of finite index. When Z = Z n When Z = Zl add the assumption that the index is a power of l. In both cases Z is the completion of Z with respect to N . Hence, Y is the completion of a subgroup Y0 of Zn of finite index (Lemma 17.2.1). The case Z = Zn ˆ n in the former case and to implies Y0 ∼ = Zn . Therefore, Y is isomorphic to Z n  Zl in the latter case. Denote the number of open subgroups of index n of a profinite group G by an (G). Lemma 16.10.2 says an (G) is finite if G is finitely generated. Lemma 26.3.3: alk (Znl ) ≤ lnk . Proof: Use induction on k and start with k = 1. Each open subgroup of Znl of index l contains lZnl , so al (Znl ) = al (Fnl ). By duality, al (Fnl ) is equal to n −1 which is the number of subspaces of Fnl of dimension 1. The latter is ll−1 n smaller than l . Suppose k ≥ 2. Consider an open subgroup H of Znl of index lk . Then H is contained in an open subgroup M of Znl of index lk−1 . By Lemma 26.3.2, M∼ = Znl . By the first paragraph, M has at most ln open subgroups of index l. By induction, there are at most ln(k−1) possibilities for M . Therefore,  there are at most lnk possibilities for H. Lemma 26.3.4: Suppose e > n. Then almost all e-tuples (a1 , . . . , ae ) in (Znl )e generate an open subgroup of Znl . Proof: For each positive integer k list the open subgroups of Znl of index lk

644

Chapter 26. Random Elements in Profinite Groups

as Hk1 , . . . , Hk,r(k) . By Lemma 26.3.3, r(k) = alk (Znl ) ≤ lnk . Hence, ∞ r(k) X X

µ(Hki ) =

k=1 i=1

∞ X r(k) k=1

lek

≤

∞ X

1 < ∞. (e−n)k l k=1

Denote the set of all a = (a1 , . . . , ae ) ∈ (Znl )e such that ha1 , . . . , ae i ≤ Hki for infinitely many pairs (k, i) by A. By Borel-Cantelli (Lemma 18.3.5(a)), µ(A) = 0. If a ∈ (Znl )e r A, then ha1 , . . . , ae i is contained in only finitely many open subgroups of Znl , so ha1 , . . . , ae i is open.  Theorem 26.3.5 ([Kantor-Lubotzky, Prop. 12]): Given positive integers e ˆ n )e : and n the following statements hold for an e-tuple (a1 , . . . , ae ) ∈ (Z Q e n −1 ˆ is (which is (a) The probability that a1 , . . . , ae generate Z i=e−n+1 ζ(i) positive) if e > n and 0 if e ≤ n. Here ζ(s) is the Riemann zeta function. ˆ n is 1 if (b) The probability that a1 , . . . , ae generate an open subgroup of Z e > n and 0 if e ≤ n. ˆ n. (c) If e < n, then ha1 , . . . , ae i is of infinite index in Z ˆn Proof of (a): Suppose e ≥ n. For each l let πl be the epimorphism of Z n n ˆ onto Fl with kernel Nl = lZ . Put ˆ n )e | πl (a1 ), . . . , πl (ae ) generate Fn }. Al = {(a1 , . . . , ae ) ∈ (Z l Then, by Lemma 26.3.1, µ(Al ) =

n−1 1 Y

lne

(le − li ) =

i=0

e Y i=e−n+1

(1 −

1 ). li

T Put A = l Al . If a = (a1 , . . . , ae ) ∈ A, then for each l we have ˆ n , so hai is contained in no open subgroup of index l. Hence, ha, Nl i = Z ˆ n. ˆ n . Therefore, hai = Z hai is contained in no proper open subgroup of Z Conversely, the latter condition implies a ∈ A. To compute the measure of A suppose first e > n. As l ranges on all ˆ : Nl ) are relatively prime in pairs. prime numbers, the indices len = (Z Hence, the Al are µ-independent (Example 18.3.8). Therefore, (1)

µ(A) =

e Y

n Y Y 1 (1 − i ) = ζ(i)−1 . l i=e−n+1 i=e−n+1 l

In particular, since each i appearing in (1) is at least 2, the ith factor is positive. Qe Q Suppose e = n. Then, in the above notation, µ(A) = i=1 l (1 − l1i ). Q Since l (1 − 1l ) = 0 [LeVeque, p. 99, Thm. 6-11], µ(A) = 0.

ˆn 26.3 Random e-tuples in Z

645

Finally, if e < n, then each Al is empty, so µ(A) = 0. S∞ T Proof of (b): First suppose e > n. Let A0 = m=1 l≥m Al be the set of all ˆn e (a1 , . . . , aT e ) ∈ (Z ) which belong to all but finitely many Al ’s. By the proof of (a), µ( l Al ) 6= 0. Hence, by Borel-Cantelli (Lemma 18.3.5(c)), µ(A0 ) = 1. ˆ n onto Z ˆ n by π Next denote the projection of Z ˆl . Let A00l be the set of all l T n e ˆ (a1 , . . . , ae ) ∈ (Z ) with hˆ πl (a1 ), . . . , π ˆl (ae )i open in Znl . Put A00 = l A00l . By Lemma 26.3.4, µ(A00l ) = 1 for each l, so µ(A00 ) = 1. Consequently, µ(A0 ∩ A00 ) = 1. Consider a = (a1 , . . . , ae ) ∈ A0 ∩ A00 . Put H = ha1 , . . . , ae i. Assume ˆ n : H) = ∞. By Lemma 22.8.4, H is contained in an open subgroup of (Z ˆ Zn of index l for infinitely many l or there is an l such that H is contained ˆ n /N ∼ ˆ n with Z in a closed subgroup N of Z = Zl . The former case contradicts 0 00 ˆ n : H) < ∞. A . It follows that (Z a ∈ A . The latter contradicts a ∈Q Finally, suppose e ≤ n. Then l µ(Al ) = 0 (proof of (a)). Since the Al ’s ˆ n r Al ˆ n )e belong to infinitely many Z are µ-independent, almost all a ∈ (Z ˆ r Al (Lemma 18.3.5(d)). Suppose a is one of them. For each l with a ∈ Z the group hai is contained in an open subgroup of index l. Since there are ˆ n : hai) = ∞. This completes the proof of (b). infinitely many such l’s, (Z Proof of (c): Suppose e < n. Then, for each l, ha1 , . . . , ae i is contained in ˆ n of index l. Hence, ha1 , . . . , ae i has an infinite index all open subgroups of Z ˆ n.  in Z Theorem 26.3.6 ([Jarden-Lubotzky2, Thm. 3.1]): Let e, n be positive inˆ min(e,n) for almost all (a1 , . . . , ae ) ∈ (Z ˆ n )e . tegers. Then ha1 , . . . , ae i ∼ =Z ˆ n and divide the proof into three parts. Proof: We put A = Z Part A: e > n. By Theorem 26.3.5(b), for almost all a ∈ Ae the group hai ˆ n. is open in A. By Lemma 26.3.2, hai ∼ =Z Part B: e = n. Ap = Znp of A.

For each prime number p consider the quotient group

Claim: For almost all a ∈ Anp we have hai ∼ = Znp . Consider each a ∈ Anp as a column of height n whose ith entry is a row ai = (ai1 , ai2 , . . . , ain ) of elements of Zp . In this way identify a with an n × n matrix with entries in Zp . Then hai is a free Zp -module of rank which is equal to the rank of the matrix a. Thus, hai ∼ = Znp if and only if det(a) 6= 0. By Lemma 26.2.1, the latter condition holds for a subset of Mn (Zp ) of measure 1. Therefore, hai ∼ = Znp for almost all a ∈ Anp . By the Claim, for almost all a ∈ An each Abelian group which is generated by n elements is a quotient of hai. Since hai is generated by n elements, ˆ n. Lemma 17.7.1 implies hai ∼ =Z

646

Chapter 26. Random Elements in Profinite Groups

Part C: e < n. Consider now each a = (a1 , . . . , an ) ∈ An as a pair a = (b, c), where b = (a1 , . . . , ae ) ∈ Ae and c = (ae+1 , . . . , an ) ∈ An−e . By Part B, (2)

ˆ n, hai ∼ =Z

holds for almost all a ∈ An . Hence, by Fubini’s theorem [Halmos, p. 147, Thm. A], for almost all b ∈ Ae the set of c ∈ An−e such that (2) holds for a = (b, c) has measure 1. Choose c ∈ An−e for which (2) holds. For each Abelian profinite group B and each (b01 , . . . , b0e ) ∈ B e , we may extend the map (b1 , . . . , be ) 7→ (b01 , . . . , b0e ) to a map of a into B (say ai 7→ 0, i = e + 1, . . . , n), hence to a homomorphism h: hai → B. The restriction of h to hbi is a homomorphism ˆ e.  into B. Consequently, hbi ∼ =Z Remark 26.3.7: Positively generated groups. Let G be a profinite group and k a positive integer. Put Ak = {(x1 , . . . , xk ) ∈ Gk | hx1 , . . . , xk i = G}. Note that a k-tuple (x1 , . . . , xk ) ∈ Gk belongs to Ak if and only if hx1 , . . . , xk iN = G for each open normal subgroup N of G, so Ak is closed. We say that G is positively finitely generated if there is a positive integer k with µ(Ak ) > 0. In this case G is finitely generated. Free profinite groups of rank at least 2 are not positively finitely generated (Proposition 26.1.7(a)). In contrast, Fˆe (p) is positively finitely generated. Indeed, (2) of Section 26.2 gives an explicit positive value for µ(Ae ). ˆ n is positively finitely generated. More generally, evBy Theorem 26.3.5(a), Z ery finitely generated profinite group G which contains an open prosolvable group is positively finitely generated [Mann, Thm. 10]. In particular, every closed subgroup G of GL(n, Zp ) is positively finitely generated. Indeed, G is finitely generated (Proposition 22.14.4) and {g ∈ G | g ≡ 1 mod p} is an open pro-p (hence prosolvable) subgroup of G (Lemma 22.14.2(d)). Finally, for each profinite group and a positive integer n let mn (G) be the number of maximal open subgroups of G of index n. By [Lubotzky-Segal, Thm. 11.1], G is positively finitely generated if and only if there is a c > 0  and a positive integer r with mn (G) ≤ cnr for all n.

26.4 On the Index of Normal Subgroups Generated by Random Elements Let G be a profinite group. For each σ = (σ1 , . . . , σe ) ∈ Ge let [σ] = [σ]G be the closed normal subgroup of G generated by σ1 , . . . , σe (Section 25.9). In other words, [σ] is the intersection of all closed normal subgroups of G which contain hσi. Observe that if H is a closed normal subgroup of G and σ ∈ H e , then [σ]H ≤ [σ]G ≤ H but it may happen that [σ]H < [σ]G . In addition note that if ϕ: G → A is an epimorphism, then ϕ([σ]) = [ϕ(σ)]. Finally, if G is Abelian, then [σ] = hσi.

26.4 On the Index of Normal Subgroups Generated by Random Elements

647

The purpose of this section is to study the index of [σ] in Fˆn when σ is taken at random in Fˆne . For a profinite group G and a closed normal subgroup H we set: Be (G, H) = {σ ∈ Ge | [σ]G = H}, Be (G) = Be (G, G) = {σ ∈ Ge | [σ]G = G} Ce (G) = {σ ∈ Ge | (G : [σ]) = ∞} Lemma 26.4.1: Let G be a profinite group and let H be a closed normal subgroup. (a) Be (G, H) is closed subset of Ge . (b) Suppose rank(G) ≤ ℵ0 . Then Ce (G) is a measurable subset of Ge . Proof of (a): For each open normal subgroup N of G, {σ ∈ Ge | [σ]G N = HN } is an open-closed subset of G. The intersection of all these sets is Be (G, H). Hence, Be (G, H) is closed. S Proof of (b): We have Ce (G) = Ge r H Be (G, H), where H ranges over all open normal subgroup of G. By Proposition 17.1.2, G has at most ℵ0 open  subgroups. By (a), each Be (G, H) is closed. Hence, Ce is measurable. The relation rank of a finite p-group G of rank d is defined as the minimal number r for which there exists a short exact sequence 1 → [y1 , . . . , yr ] → Fˆd (p) → G → 1. The relation rank and the rank of G are related by the Golod-Shafarevich inequality: r > 14 d2 [Huppert, p. 395, Satz 18.1]. This ensures that Ce (G) has positive measure (Theorem 25.4.5): Lemma 26.4.2: Let n ≥ 2 and e ≥ 1 be integers. Put P = Fˆn (p). Then, µ(Ce (P )) > 0. Proof: Assume µ(Ce (P )) = 0. Take an open normal subgroup Q of P of index m such that (1)

em ≤

1 (m(n − 1) + 1)2 . 4

Then Φ(Q) is open in P , so µ(Φ(Q)e ) > 0. Hence, we may choose σ ∈ Φ(Q)e r Ce (P ). Then [σ]P is an open normal subgroup of P , hence also of Q. Therefore, G = Q/[σ]P is a finite p-group. By Nielsen-Schreier (Proposition 17.6.2), rank(Q) = m(n − 1) + 1. Since [σ]P ≤ Φ(Q), the quotient map Q → G is a Frattini cover, so rank(G) = rank(Q) = m(n − 1) + 1 (Corollary 22.5.3). On the other hand, let τ1 , . . . , τm be representatives for the left cosets τ of P modulo Q. Then [σ]P = [σi j | i = 1, . . . , e, j = 1, . . . , m]Q , so relation.rank(G) ≤ em. By Golod-Shafarevich, relation.rank(G) >

1 rank(G)2 , 4

648

Chapter 26. Random Elements in Profinite Groups

so em > 14 (m(n−1)+1)2 . This contradiction to (1) proves that µ(Ce (P )) > 0.  The following result overlaps with Lemma 18.3.11: Lemma 26.4.3: Let G be a profinite group, e a positive integer, and Mi , Nj , i, j = 1, 2, 3, . . . distinct open normal subgroups. Suppose G/Mi is a simple Abelian group, i = 1, 2, 3, . . ., and G/Nj is a simple non-Abelian group, S∞ j = 1, 2, 3, . . . . Then i=1 Mie , N1e , N2e , N2e , . . . are µ-independent. Proof: Let i1 < · · · < ik and n be positive integers. Put M = Mie1 ∩· · ·∩Miek and N = N1e ∩ · · · ∩ Nne . Then no composition factor of Ge /N is Abelian. Hence, (Ge : M ∩ N ) = (Ge : M )(Ge : N ). Next let m be a positive integer. Apply the inclusion-exclusion principle (Lemma 18.3.1(a)) and the preceding paragraph to the sets M1e ∩ e N e , . . . , Mm ∩ N e: µ

m [

m
X Mie ∩ N e = (−1)k−1

i=1

=

X

(−1)k−1

k=1

=µ

µ(Mie1 ∩ · · · ∩ Miek ∩ N e )

1≤i1 <···
k=1 m X

X

m [

µ(Mie1 ∩ · · · ∩ Miek )µ(N e )

1≤i1 <···

Mie µ(N e ).

i=1

S∞ S∞ Taking the limit on m, we conclude that µ( i=1 Mie ∩N ) = µ( i=1 Mie )µ(N ). Finally, by Example 18.3.10, N1e , . . . , Nne are µ-independent, so µ(

∞ [

i=1

Mie ∩ N1e ∩ · · · ∩ Nne ) = µ(

∞ [

Mie )µ(N1e ) ∩ · · · ∩ µ(Nn )e .

i=1

This implies the statement of the Lemma.



ˆ n an epimorphism. Lemma 26.4.4: Let n be a positive integer and ϕ: Fˆn → Z Then every open normal subgroup of Fˆn of prime index contains Ker(ϕ). Proof: Let N be an open normal subgroup of Fˆn of a prime index p. Choose ˆ n → Fn and let ψ = π ◦ ϕ. It suffices to prove that an epimorphism π: Z p Ker(ψ) ⊆ N .
, Assume Ker(ψ) is not contained in N . Then Fˆn / Ker(ψ) ∩ N ∼ = Fn+1 p ˆ which is impossible, because the rank of each quotient of Fn is at most n. Thus, Ker(ψ) ⊆ N .  Theorem 26.4.5 ([Jarden-Lubotzky2, Thm. 1.4]): Let n ≥ 2 be a positive integer. Put F = Fˆn .

26.4 On the Index of Normal Subgroups Generated by Random Elements

649

(a1) If e ≤ n, then µ(Be (F )) = 0. In fact, if e < n, then Be (F ) = ∅ while if e = n, then Be (F ) 6= ∅. (a2) If e > n, then 0 < µ(Be (F )) < 1. (b1) If e ≤ n, then µ(Ce (F )) = 1. In fact, if e < n, then Ce (F ) = F e while if e = n, then Ce (F ) 6= F e . (b2) If e > n, then 0 < µ(Ce (F )) < 1. Proof: The proof has five parts: Part A: In each case µ(Ce (F )) > 0. Choose a prime number p. Let P = Fˆn (p) be the free pro-p group on n generators and π: F → P be an epimorphism. If σ ∈ F e , then π([σ]) = [π(σ)]. Hence, if [π(σ)] has infinite index in P , then [σ] has infinite index in F . Thus, π −1 (Ce (P )) ⊆ Ce (F ). By Lemma 26.4.2, µ(Ce (P )) > 0. Hence, µ(Ce (F )) > 0. Part B: If e < n, then (F : [σ]) = ∞ for each σ ∈ F e . Again, choose a prime number p and let ϕ: F → Znp be an epimorphism. Each open subgroup of Znp is isomorphic to Znp (Lemma 26.3.2), hence can not be generated by less than n elements. Let σ be an arbitrary e-tuple of F . Since ϕ([σ]) = [ϕ(σ)] = hϕ(σ)i, the index of ϕ([σ]) in Znp is infinite. Consequently, (F : [σ]) = ∞. ˆ n. Part C: If e = n, then µ(Ce (F )) = 1. Choose an epimorphism ϕ: F → Z n ˆ By Theorem 26.3.5(b), [σ] = hσi has an infinite index in Z for almost all ˆ n )n . Hence, (F : [σ]) = ∞ for almost all σ ∈ F n . σ ∈ (Z Part D: If e > n, then µ(Be (F )) > 0.

We have to prove that

µ(F e r Be (F )) < 1. Indeed, σ ∈ F e r Be (F ) if and only if σ1 , . . . , σe belong to a normal subgroup N of G with G/N simple. Thus, F e r Be (F ) =

[

[

p F/N ∼ =Z/pZ

Ne ∪

[

[

N e.

S∈SNA F/N ∼ =S

Here SNA is the set of all non-Abelian finite simple groups. Recall that if A and B are independent subsets of a probability space, then 1 − µ(A ∪ B) = (1 − µ(A))(1 − µ(B)). Hence, µ(A ∪ B) < 1 if and only if 1 − µ(A) > 0 and 1 − µ(B) > 0. Also, if B is a union of a sequence Q∞ B1 , B2 , B3 , . . . of independent sets, then 1 − µ(B) = j=1 (1 − µ(Bj )). Thus, P∞ µ(B) < 1 if and only if µ(Bj ) < 1 for each j and j=1 µ(Bj ) < ∞ (Lemma 18.3.2). S S Let in our case A = p F/N ∼ N e and let Bj range over all N e =Z/pZ with F/N ∈ SN A. By Lemma 26.4.3, A, B1 , B2 , B3 , . . . are µ-independent. ˆ n be an epimorphism. By Lemma Moreover, as in Part C, let ϕ: F → Z 26.4.4, each open normal subgroup N of F with F/N ∼ = Z/pZ contains

650

Chapter 26. Random Elements in Profinite Groups

ˆ n )e which Ker(ϕ). Therefore, µ(A) is equal to the measure of all σ ∈ (Z are contained in a maximal subgroup of a prime index. The latter measure is ˆ n . Hence, by ˆ n )e which do not generate Z equal to the measure of all σ ∈Q(Z −1 Theorem 26.3.5(a), 1−µ(A) = e−n 0, where ζ is the Riemann zeta function. To prove the condition on the Bj ’s, let S ∈ SN A. By Example 18.3.11, the sets N e , with N ranging over SNA are µ-independent. For each S ∈ SNA, dn (S) is the number of open normal subgroups N of F with F/N ∼ = S (Lemma 26.1.2(a)). By Lemma 26.1.2(c), dn (S) ≤ |S|n−1 . Hence, (2)

X

X

µ(N e ) =

S∈SNA F/N ∼ =S

X dn (S) |S|e

S∈SNA

≤

X S∈SNA

1 |S|e+1−n

≤

X S∈SNA

1 < ∞. |S|2

The last inequality holds because for each positive integer n there are at most two simple groups of order n ([Kimmerle-Lyons-Sandling-Teaque, Thm. 5.1] proves this result by using the classification of finite simple groups). Consequently, µ(C) > 0. Part E: Conclusion of the proof. If e < n, then F is not generated by e elements. Hence, Be (F ) = ∅. If e = n, then F is generated by e elements and therefore Be (F ) 6= ∅. However, since Be (F ) ⊆ F e r Ce (F ), Part C implies that µ(Be (F )) = 0. This concludes the proof of (a1). Part D of the proof takes care of the left inequality of (a2). In order to prove also the right inequality of (a2) we take a proper open normal subgroup E of F . Then E e ⊆ F e r Be (F ) and µ(E e ) > 0. Therefore, µ(Be (F )) < 1. Parts B and C give (b1). Finally, if e > n, then, by Part D, µ(F e r Ce (F )) ≥ µ(Be (F )) > 0.  Together with Part A we get 0 < µ(Ce (F )) < 1. Remark 26.4.6: Note that the only application of the classification of simple groups in the proof of Theorem 26.4.5 occurs in the proof of Part D, and therefore in the proof of the inequalities µ(Be (F )) > 0 and µ(Ce (F )) < 1. In addition, we apply the classification of finite simple groups in Proposition 26.4.7(b) below.  If e > n, then [σ] is, with a positive probability, of infinite index. Nevertheless, as the next result shows, the quotient F/[σ] is small compared to F. Proposition 26.4.7: Let e > n. Then almost all σ ∈ F e have the following properties: (a) The maximal Abelian quotient of F/[σ] is finite. (b) There are only finitely many open maximal normal subgroups N of F which contain [σ] such that F/N is simple and non-Abelian.

26.5 Freeness of Normal Subgroups Generated by Random Elements

651

ˆ n . Then, for each σ ∈ f e , Proof of (a): Choose an epimorphism ϕ: F → Z n ˆ the maximal Abelian quotient of F/[σ] is Z /hϕ(σ)i. By Lemma 26.3.5(b), ˆ n )e . Hence, for almost all ˆ n for almost all τ ∈ (Z hτ i has a finite index in Z e σ ∈ F , the maximal Abelian quotient of F/[σ] is finite. Proof of (b): By (2) and Borel-Cantelli, almost all σ ∈ F e belong to only finitely many open normal subgroups S with S/N ∈ SNA. This proves (b). 

26.5 Freeness of Normal Subgroups Generated by Random Elements Let e ≥ 1 and n ≥ 2 be integers. Put F = Fˆn . We prove here that [σ] is a free profinite group for almost all σ ∈ F e . The case where the index is finite is well known. Therefore, it suffices to prove that [σ] ∼ = Fˆω for almost all σ ∈ F e which satisfy (F : [σ]) = ∞. A basic tool is the following special case of Corollary 25.7.6: Lemma 26.5.1: Let N be a closed normal subgroup of F satisfying: (1a) Z/pZ is a quotient of N for every p. (1b) S q is a quotient of N for every finite simple group S and every positive integer q. Then N ∼ = Fˆω . Lemma 26.5.2: For almost all σ ∈ F e and for each prime number p the group Z/pZ is a quotient of [σ]. Proof: For each p let ϕp : F → Fˆn (p) be an epimorphism. Then S Ker(ϕp ) has an infinite index in F , hence measure 0. It follows that U = p Ker(ϕp )e is a / U , then for each p, ϕp [σ] is a nontrivial pro-p-group. zero subset of F e . If σ ∈  Therefore, Z/pZ is a quotient of ϕp [σ], hence of [σ]. Lemma 26.5.2 settles Condition (1a) of Lemma 26.5.1. The proof of Condition (1b) uses the notion the “S-rank” of a profinite group (Section 24.9). For a profinite group G and a closed subgroup H let N (G, H) be the family of all closed normal subgroups M of G satisfying HM = G. Denote the intersection of all M ∈ N (G, H) by N(G, H). The following rule is a group theoretic analog of the tower property of linear disjointness of fields (Lemma 2.5.3): Rule 26.5.3: Let G be a profinite group, H an open subgroup of G, and M and N closed normal subgroups of G such that M ≤ N . Then HM = G if and only if HN = G and (H ∩ N )M = N . Lemma 26.5.4: Let G be a profinite group, S a simple non-Abelian group, k a positive integer, and H an open normal subgroup of G with G/H ∼ = Sk. Then, H · N(G, H) = G.

652

Chapter 26. Random Elements in Profinite Groups

Proof: By Lemma 1.2.2(b), it suffices to prove N (G, H) is closed under finite intersections. So, let M and N be open normal subgroups of G with HM = HN = G. By induction on k we prove H(M ∩ N ) = G. Consider first the case k = 1. By assumption, HM/H = G/H ∼ = S is a / H. non-Abelian simple group. Hence, there exist m, m0 ∈ M with [m, m0 ] ∈ Since HN = G there exist h ∈ H and n ∈ N with m0 = hn. Using the identity [m, m0 ] = [m, hn] = [m, n][m, h]n and the relation [m, h]n ∈ H, we conclude that [m, n] ∈ / H. On the other hand, [m, n] ∈ M ∩ N . Hence, M ∩ N 6≤ H. It follows that H(M ∩ N ) = G, because F/H ∼ = S is simple. Now suppose k > 1 and the statement holds for k − 1. Then G has an open normal subgroup E containing H such that G/E ∼ = S and E/H ∼ = k−1 . It satisfies, EM = EN = G. By Rule 26.5.3, H(E ∩ M ) = E and S H(E ∩ N ) = E. By the induction hypothesis, applied to E instead of G, we have H · (E ∩ M ∩ N ) = E. By the case k = 1, E(M ∩ N ) = G. Hence, by Rule 26.5.3, H(M ∩ N ) = G.  Let S be a finite simple group. Recall (Remark 17.4.7) that a proS group is a profinite group whose only composition factor is S. For an arbitrary profinite group G, we denote the intersection of all open normal subgroups M of G with G/M ∼ = S rG (S) = S by MG (S). It satisfies G/MG (S) ∼ (Section 24.9). Let F be a free pro-S group and N a nontrivial closed subgroup of F of infinite index. If S = Z/pZ, then rank(N ) = ∞ (Proposition 24.10.4(a)), so rN (S) = ∞. It is quite surprising that for non-Abelian S there exists N with rN (S) < ∞. This will follow from Lemma 26.5.5(a). Nevertheless, there are only countably many such N (Lemma 26.5.5(d)). Lemma 26.5.5: Let G be a profinite group, S a simple non-Abelian group, and k a nonnegative integer. (a) Let H 0 ≤ G0 be open normal subgroups of G with G0 /H 0 ∼ = S k . Then 0 0 N(G , H ) / G and rN(G0 ,H 0 ) (S) = k. (b) Suppose N is a closed normal subgroup of G with rN (S) = k. Then there exist open normal subgroups H 0 ≤ G0 of G with G0 /H 0 ∼ = S k , H 0 N = G0 , and H 0 ∩ N = MN (S). (c) Suppose G is a pro-S-group. Let N be a closed normal subgroup of G with rN (S) = k. Then there exists open normal subgroups H 0 ≤ G0 of G with N = N(G0 , H 0 ) and G0 /H 0 ∼ = Sk. (d) Suppose G is a pro-S-group. The number of closed normal subgroups of G with rN (S) < ∞ is bounded by max(ℵ0 , rank(G)). In particular, if rank(G) ≤ ℵ0 , then there are at most countably many closed normal subgroups N of G with rN (S) < ∞. Proof of (a): Let N = N(G0 , H 0 ) and M = H 0 ∩ N . The family N (G0 , H 0 ) is closed under conjugation by elements of G. Hence, N / G. By Lemma 26.5.4, H 0 N = G0 . Hence, N/M ∼ = G0 /H 0 ∼ = S k . Therefore, rN (S) ≥ k. If rN (S) > k, then N has an open normal subgroup N0 such

26.5 Freeness of Normal Subgroups Generated by Random Elements

653

that N/N0 ∼ = S and M 6≤ N0 . Since S is simple, this implies M N0 = N . The group N0 need not be normal in G0 . So, consider N1 = N(N, M ). Then N1 ≤ N0 . By the preceding paragraph, N1 / G0 . By Lemma 26.5.4, M N1 = N and therefore H 0 N1 = G0 . By the definition of N , this implies that N ≤ N1 ≤ N0 < N . We conclude from this contradiction that rN (S) = k. Proof of (b): Since N / G, the set of all open N 0 / N such that N/N 0 ∼ = S is closed under conjugation by elements of G. Hence, MN (S) is normal in G and open in N . Choose an open normal subgroup H 0 of G with H 0 ∩N = MN (S). Put G0 = H 0 N . Then G0 /H 0 ∼ = S k , as required. = N/MN (S) ∼ Proof of (c): Put M = MN (S). By assumption, N/M ∼ = S k . Let G0 and 0 0 0 H be as in (b). Then, N ∈ N (G , H ). Hence, N0 = N(G0 , H 0 ) ≤ N . In particular, N0 is a closed normal subgroup of N with H 0 N0 = G0 (Lemma 26.5.4), hence M N0 = N . Since G is a pro-S group, so is N/N0 . If N0 < N , then N contains an open normal subgroup N1 which contains N0 with N/N1 ∼ = S. In particular, M N1 = N . On the other hand, by definition of M , we have M ≤ N1 , hence N1 = N . This contradiction implies N0 = N . Proof of (d): The cardinality of the set of all open normal subgroups of G is at most max(ℵ0 , rank(G)) (Proposition 17.1.2). Now use (c).  Let F be a profinite group. Then Ce (F ) = {σ ∈ F e | (F : [σ]) = ∞}. Theorem 26.5.6 ([Jarden-Lubotzky2, Thm. 2.7]): Let e ≥ 1 and n ≥ 2 be integers. Put F = Fˆn . Then [σ] ∼ = Fˆω for almost all σ ∈ Ce (F ). Proof: By Lemmas 26.5.1 and 26.5.2 it suffices to consider a non-Abelian simple group S and to prove that r[σ] (S) = ∞ for almost all σ ∈ Ce (F ). To this end denote the set of all open normal subgroups of F by E. Each E ∈ E is a free profinite group and rank(E) = 1 + (F : E)(n − 1) (Proposition 17.6.2). Hence, there is an epimorphism hE of E onto the free ¯ If pro-S-group with rank(E) generators. Denote the latter group by E. σ ∈ E e , then [σ] = [σ]F is a normal subgroup of E. Hence, hE ([σ]) is a ¯ (Nevertheless, since [σ] may properly contain [σ]E , normal subgroup of E. we may have [hE (σ)]E¯ < hE ([σ]).) Let ¯ : hE ([σ])) = ∞} Ce (F, E) = {σ ∈ Ce (F ) ∩ E e | (E

C=

[

Ce (F, E).

E∈E

The proof will be concluded, once we prove Claims A and B below: Claim A: r[σ] (S) = ℵ0 for each σ ∈ Ce (F ) r C. Consider σ ∈ Ce (F ) r C. ¯ for each open normal Then (F : [σ]) = ∞ and hE ([σ]) is open normal in E subgroup E of F that contains [σ]. By Proposition 17.6.2, hE ([σ]) is a free pro-S-group. Moreover, ¯ = rank(E) = 1 + (F : E)(n − 1). r¯ = rank(hE ([σ])) ≥ rank(E)

654

Chapter 26. Random Elements in Profinite Groups

Let m = rank(S). By the remarks preceding Lemma 26.5.5 and by Lemma 26.1.2(b), r[σ] (S) ≥ rhE ([σ]) (S) = dr¯(S) ≥ |S|r¯−2m . Since (F : E) is unbounded and n > 1, we find that r[σ] (S) = ℵ0 . ¯ be Claim B: r[σ] (S) = ℵ0 for almost all σ ∈ C. For each E ∈ E let B(E) 0 0 0 0 the set of all pairs (G , H ) such that H ≤ G are open normal subgroups of ¯ be the union of all the sets N(G0 , H 0 )e ¯ and (E ¯ : N(G0 , H 0 )) = ∞. Let B(E) E 0 0 ¯ with (G , H ) ∈ B(E). By the index assumption, each set N(G0 , H 0 )e with ¯ has measure zero. Since B(E) ¯ is countable, B(E) ¯ is a zero (G0 , H 0 ) ∈ B(E) S −1 e ¯ ¯ set in E . It follows that B = E∈E hE (B(E)) is a zero set in F e . If σ ∈ C r B, then there exists E ∈ E such that σ ∈ Ce (F, E) and ¯ Thus, hE ([σ]) is a closed normal subgroup of E ¯ of an infinite / B(E). hE (σ) ∈ 0 0 ¯ index. If rhE ([σ]) (S) < ℵ0 , there exists (G , H ) ∈ B(E) with hE ([σ]) = N(G0 , H 0 ) (Lemma 26.5.5(c)). Since hE (σ) ∈ hE ([σ])e , we have that σ ∈ ¯ This contradiction proves r[σ] (S) ≥ rh ([σ]) (S) = ℵ0 . B(E). E This concludes the proof of Claim B.  Corollary 26.5.7: Let F = Fˆn with n ≥ 2. (a) If e ≤ n then [σ] ∼ = Fˆω for almost all σ ∈ F e . (b) If e > n, then [σ] ∼ = Fˆω for a set of σ ∈ F e of a positive measure (but less than 1). Proof: By Theorem 26.4.5(b), µ(Ce (F )) = 1 if e ≤ n and 0 < µ(Ce (F )) < 1 if e > n. Now apply Theorem 26.5.6 .  Corollary 26.5.8 ([Jarden-Lubotzky2, Cor. 2.9]): For each e ≥ 1 and for almost all σ ∈ F e , [σ] is a free profinite group. Proof: If σ ∈ F e r Ce (F ), then [σ] is open in F and is therefore a free profinite group (Proposition 17.6.2). By Theorem 26.5.6, [σ] is free for almost  all σ ∈ Ce (F ). Hence, [σ] is free for almost all σ ∈ F e . Example 26.5.9: Exceptional σ. Let F = Fˆn with n ≥ 2. Denote the intersection of all open normal subgroups N of F such that F/N is a solvable group by F solv . The index of F solv in F is infinite. By Lemma 17.4.10, each simple quotient of F solv is non-Abelian. Hence, F solv 6= Fˆω . Let N be the set of all open normal subgroups N of F solv with F solv /N simple. Denote the intersection of all N ∈ N by M . Then F solv /M is the /N direct product of simple non-Abelian groups. Choose σ ∈ F solv with σ ∈ for all N ∈ N . Then, [σ] = F solv . Thus, the conclusion of Corollary 26.5.8  does not hold for all σ ∈ F e .

Notes Proposition 26.1.7(a) appears in [Kantor-Lubotzky, Prop. 11]. Our proof follows [Lubotzky3, Thm. 1(a)].

Chapter 27. Omega-free PAC Fields Let K be a countable Hilbertian field. Then each finite extension of K is Hilbertian (Proposition 12.3.3). Thus, we could be tempted to think of Hilbertian algebraic extensions of K as “small”. On the other hand, Ks (σ) is PAC for almost all σ ∈ Gal(K)e (Theorem 18.6.1), so PAC algebraic extensions of K could be thought of as “large”. In this chapter we construct an abundance of algebraic extensions N of K which are both small and large in that sense, thus they are both Hilbertian and PAC (Theorem 27.4.8). Indeed, we construct N as an ω-free PAC field and conclude that N is Hilbertian (Corollary 27.3.3). Moreover, we prove that ω-free and PAC is equivalent to the following stronger form of Hilbertianity: For every variety V over N with a generic point x and for every polynomial f ∈ N [x, Y ] which is irreducible over N (x) there exists a ∈ V (N ) such that f (a, Y ) is irreducible over N (a version of Lemma 27.2.1 which holds when char(K) = 0). These algebraic results have model theoretic interpretations: We extend L(ring) by adding a sequence of predicate to the language and denote the extended language by LR (ring). Every field can be naturally considered as a structure for L(ring). A field extension L/K is an extension of structures for LR (ring) if and only if K is algebraically closed in L. We prove that if a field K of characteristic 0 is existentially closed within the language LR (ring) in every field L in which K is algebraically closed, then K is PAC, and ω-free (Proposition 27.2.2). It follows that ω-free PAC fields of characteristic 0 are the models of a model companion of a theory TR of fields in the language LR (ring) (Theorem 27.2.3).

27.1 Model Companions The absolute Galois group of a countable field is isomorphic to Fˆω if and only if each finite embedding problem for Gal(K) is solvable (a special case of Corollary 24.8.2). An arbitrary field which has this property is said to be ω-free. This section prepares an augmentation of the theory of fields in Section 27.2 to a theory in which ω-free PAC fields play an existential completeness role analogous to the role played by algebraically closed fields for the ordinary theory of fields. Definition 27.1.1: Let T be a theory in a language L. A theory T˜ in L is called the model companion of T if the following holds: (1a) Each model of T˜ is a model of T . (1b) Each model of T can be embedded in a model of T˜. (1c) T˜ is model complete; that is, A ≺ B whenever A and B are models of T˜ with A ⊆ B (Section 9.1).

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A theory T is said to have the amalgamation property, if whenever two models B and C of T contain a common model A there are embeddings f : B → D and g: C → D into a common model D, such that f and g coincide on A. Call the theory T˜ a model completion of T if in addition to (1) (2) T has the amalgamation property. Example 27.1.2: If T is the theory of fields, then the theory T˜ of algebraically closed fields is the model completion of T (Corollary 9.3.2). Examples of model companions of theories which are not model completions appear in the next section.  First we characterize the models of T˜ among the models of T : Definition 27.1.3: Let A ⊆ B be structures for a language L with domains A ⊆ B. Denote the language L augmented with new constant symbols for the elements of A by L(A). Then A is existentially closed in B if each existential sentence θ of L(A) (Definition 21.4.1) which is true in B is also true in A. Note that this definition is equivalent to the one we gave in Section 7.3.  Lemma 27.1.4: Let A ⊆ B be structures for a language L. Then A is existentially closed in B if and only if B can be embedded in a structure A∗ for L which is an elementary extension of A (Section 7.3). Proof: The sufficiency of the condition is obvious. Now assume that A is existentially closed in B. Consider the set T of all sentences ϕ(b1 , . . . , bn ) of the language L(B) where ϕ(X1 , . . . , Xn ) is a quantifier free formula of L and b1 , . . . , bn are elements of B such that B |= ϕ(b1 , . . . , bn ). We show that the union of T with the theory of A in the language L(A) has a model A∗ . Then each b ∈ B corresponds to a distinguished element b∗ of A∗ . The correspondence b 7→ b∗ is an embedding of B into A∗ and A ≺ A∗ as models of L. The construction of A∗ applies the compactness theorem (Proposition 7.7.6). By that theorem, it suffices to construct a model for T0 ∪ Th(A) for each finite subset T0 of T . The conjunction of all sentences in T0 has the form ϕ(b1 , . . . , bn ) as above. Since A is existentially closed in B, there exists a1 , . . . , an ∈ A such that A |= ϕ(a1 , . . . , an ). Interpret b1 , . . . , bn in A, respectively, as a1 , . . . , an , and interpret all other elements of B in A arbitrarily. This structure of L(B) gives the desired model for T0 ∪ Th(A).  Definition 27.1.5: If A ⊆ B are structures for a language L, we call B an extension of A (Section 7.3). If in addition B is a model of a theory T , we call B a T -extension of A. A theory T is said to be inductive if the union of each ascending transfinite sequence of models of T is also a model of T . For example, the theories

27.1 Model Companions

657

of groups, rings and fields are inductive. But the theory of finite fields is not inductive.  Lemma 27.1.6: Every model complete theory is inductive. Proof: Let T be a model complete theory and let {Aα | α < γ} be an ascending transfinite sequence of models of T . Then Aα ≺ Aβ for each α ≤ β < γ. Hence, the union A of this transfinite sequence is an elementary extension of each of its members (Lemma 7.4.1(a)). In particular, A is a model of T .  By definition, each model companion T˜ of a theory T is model complete. Hence, by Lemma 27.1.6, T˜ is an inductive theory. A sentence θ of a language L is said to be of type ∀∃ if θ is of the form (∀X1 ) · · · (∀Xm )(∃Y1 ) · · · (∃Yn )ϕ(X, Y), (abbreviate to (∀X)(∃Y)ϕ(X, Y)) where ϕ(X, Y) is a quantifier free formula of L. The use of the compactness theorem in the next lemma parallels its use in Lemma 27.1.4. For a theory T of a language L, the collection T∀∃ of sentences of L of type ∀∃ valid in Mod(T ) is called the set of logical consequences of T of type ∀∃. In particular, each model of T is a model of T∀∃ . Lemma 27.1.7: Let T be a theory of a language L and let A be a structure for L. Then A ∈ Mod(T∀∃ ) if and only if A is existentially closed in some T -extension B. Proof: Suppose first that A is existentially closed in some T -extension B and let (∀X)(∃Y)ϕ(X, Y) be a sentence in T∀∃ where X = (X1 , . . . , Xm ) and Y = (Y1 , . . . , Yn ). Then B |= (∀X)(∃Y)ϕ(X, Y). Denote the domain of A by A. For each x ∈ Am , B |= (∃Y)ϕ(x, Y). Since A is existentially closed in B, we have A |= (∃Y)ϕ(x, Y). It follows that A ∈ Mod(T∀∃ ). Conversely, suppose that A ∈ Mod(T∀∃ ). Let S be the set of all sentences (∀Y)ϕ(a, Y) of the augmented language L(A), where ϕ(X, Y) is a quantifier free formula of L and a ∈ Am such that A |= (∀Y)ϕ(a, Y). We use the compactness theorem (Proposition 7.7.6) to show that S ∪ T has a model: Since S is closed under finite conjunctions (up to logical equivalence), it suffices to prove that T ∪ {(∀Y)ϕ(a, Y)} has a model, where (∀Y)ϕ(a, Y) is as in the last paragraph. If not, then Mod(T ) |= ¬(∃X)(∀Y)ϕ(X, Y). Hence, Mod(T ) |= (∀X)(∃Y)¬ϕ(X, Y). The latter, however, is a ∀∃-sentence of L. It, therefore belongs to T∀∃ . Thus, A |= (∀X)(∃Y)¬ϕ(X, Y), contrary to A |= (∀Y)ϕ(a, Y). Therefore, let B be a model of S ∪ T ; in particular B is a structure for L(A). Let (∃Y)ψ(a, Y) be an existential sentence of L(A) which is true in B, where ψ(X, Y) is a quantifier free formula of L and a ∈ Am . If A 6|= (∃Y)ψ(a, Y), then A |= (∀Y)¬ψ(a, Y). Consequently, (∀Y)¬ψ(a, Y)

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belongs to S and therefore is true in B, a contradiction. Thus, B extends A and A is existentially closed in B.  Lemma 27.1.8: If T is an inductive theory in a language L, then Mod(T ) = Mod(T∀∃ ). Proof: We must prove that each model A of T∀∃ is also a model of T . For this, we inductively construct an ascending sequence A = A0 ⊆ B0 ⊆ A1 ⊆ B1 ⊆ · · · of structures for L such that Bn ∈ Mod(T ), An is existentially closed in Bn and An ≺ An+1 for each n ≥ 0. Suppose the sequence has been constructed up to Bn . Apply Lemma 27.1.4 to find an extension An+1 of Bn with An ≺ An+1 . Thus, A ≺ An+1 , so An+1 ∈ Mod(T∀∃ ). By Lemma 27.1.7, An+1 has a T -extension Bn+1 in which An+1 is existentially closed. Since T is inductive, Bω , the union of the sequence, is in Mod(T ). But it is also an elementary extension of A (Lemma 7.4.1(a)). It follows that A ∈ Mod(T ).  A model A of a theory T is said to be T -existentially closed if A is existentially closed in each of its T -extensions. For example, a field K is algebraically closed if and only if K is T -existentially closed, where T is the theory of fields in L(ring). The next result generalizes this observation to arbitrary theories: Proposition 27.1.9: Let T˜ be a model companion of a theory T of a language L. Then a model A of T is also a model of T˜ if and only if A is T -existentially closed. Proof: First suppose A ∈ Mod(T˜). Let B be a T -extension of A. Then ˜ of T˜. Since T˜ is model complete, B ˜ is an B can be extended to a model B elementary extension of A. Hence, A is existentially closed in B. Conversely, suppose that A is T -existentially closed. By (1b), A has a ˜ Lemma 27.1.7 implies that T˜-extension A˜ and A is existentially closed in A. ˜ ˜ A ∈ Mod(T∀∃ ). Since T is model complete, it is inductive (Lemma 27.1.6). By Lemma 27.1.8, Mod(T˜) = Mod(T˜∀∃ ). Consequently, A ∈ Mod(T˜).  The remainder of this section develops a criterion for the existence of a model companion of an inductive theory: Lemma 27.1.10: If T is an inductive theory of a language L, then each model A of T has a T -existentially closed extension A0 . Proof: Let m be the maximum of the cardinal numbers |A|, |L|, and ℵ0 , where A is the domain of A. Let {ϕα | α < m} be a wellordering of all existential sentences of the augmented language L(A). Define an ascending transfinite sequence {Aβ | β < m} of structures for L: A0 = A; if β = α + 1 is a successor ordinal and ϕα is true in some T -extension B of A Sα , then Aβ = B, otherwise let Aβ = Aα ; and if β is a limit ordinal, Aβ = α<β Aα (∈ Mod(T ), since T is inductive).

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S Now let A(1) = α<m Aα . Then A(1) ∈ Mod(T ), and each existential sentence of L(A) which is true in some T -extension of A(1) is already true in A(1) . Apply the same construction to A(1) to obtain a model A(2) of T with domain A(2) such that each existential sentence of L(A(1) ) which is true in some T -extension is already true in A(2) . Iterate this process countably manyStimes to find an ascending sequence A ⊆ A(1) ⊆ A(2) ⊆ · · · for which ∞  A0 = n=1 A(n) is a T -extension of A which is T -existentially closed. Lemma 27.1.11 (Robinson’s Test for Model Completeness): A theory T in a language L is model complete if and only if each model of T is T -existentially closed. Proof: It suffices to prove that the condition is sufficient. Let A ⊆ B be two models of T . Apply Lemma 27.1.4 to find an extension A2 of B with A ≺ A2 . Continue inductively to find a sequence A1 ⊆ B1 ⊆ A2 ⊆ B2 ⊆ A3 ⊆ B3 ⊆ · · · of models of T , where A1 = A, B1 = B such that An ≺ An+1 and Bn ≺ Bn+1 for each positive integer n. By Lemma 7.4.1(a), the union of these sequences is an elementary extension of both A and B. Hence, A ≺ B. Thus, T is model complete.  Proposition 27.1.12: Let T be an inductive theory in a language L. Denote the class of all T -existentially closed structures by Exis(T ). Then T has a model companion if and only if there exists a theory T˜ of L such that Exis(T ) = Mod(T˜). In this case T˜ is a model companion of T . Proof: First suppose that T˜ is a model companion of T . Then, by Proposition 27.1.9, Exis(T ) = Mod(T˜). Conversely, let T˜ be a theory of L with Exis(T ) = Mod(T˜). We prove that T˜ is a model companion of T . Let B˜ ⊆ C˜ ˜ C˜ ∈ Exis(T ), so B˜ is existentially closed in C. ˜ Thus, be models of T˜. Then B, ˜ ˜ ˜ B is T -existentially closed. By Lemma 27.1.11, T is model complete. Since ˜ T is inductive, each model A of T has a T -existentially closed extension A, ˜ ˜ which is then a model of T . Consequently, T is a model companion of T . 

27.2 The Model Companion in an Augmented Theory of Fields We augment the language L(ring) to a language LR (ring) by adding an n-ary relation symbol Rn for each positive integer n. Let TR be the theory of fields together with the axioms (1)

Rn (X1 , . . . , Xn ) ↔ (∃Z)[Z n + X1 Z n−1 + · · · + Xn = 0],

In a field K, interpret Rn as a relation on K, n = 1, 2, 3, . . ., in the unique way such that (1) holds. Then consider K also as a model of TR . If

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K and L are fields, regarded as models of TR , then K is a substructure of L if and only if K ⊆ L and K is algebraically closed in L. Clearly TR is an inductive theory. We construct a theory T˜R such that Exis(TR ) = Mod(T˜R ). The models of T˜R are 1-perfect ω-free PAC fields. By Proposition 27.1.12, T˜R will be a model companion for TR . Lemma 27.2.1: The following conditions on a field K are equivalent: (a) K is ω-free and PAC. (b) Let V be a variety defined over K with a generic point x, f (x, Y ) ∈ K[x, Y ] an irreducible separable polynomial in Y over K(x), and 0 6= h(x) ∈ K[x]. Then there exists a ∈ V (K) such that h(a) 6= 0 and f (a, Y ) is irreducible in K[Y ]. (c) Let V be a variety defined over K with a generic point x, f1 (x, Y ), . . . , fk (x, Y ) ∈ K[x, Y ] irreducible separable polynomials in Y over K(x), and 0 6= g(x) ∈ K[x]. Then there is an a ∈ V (K) such that f1 (a, Y ), . . . , fk (a, Y ) are irreducible over K and g(a) 6= 0. Furthermore, suppose char(K) = p, the imperfect exponent of K is at least m, and h1 (x), . . . , hm (x) ∈ K[x] are p-independent over K(x)p . Then (a) implies that a can be chosen in (b) and (c) such that h1 (a), . . . , hm (a) are p-independent over K p . Proof of “(a) =⇒ (b)”: By assumption, Im(Gal(K)) consists of all finite groups and Gal(K) has the embedding property (Definition 24.1.2). Since K is also PAC, it is a Frobenius field (Definition 24.1.3). By Lemma 19.7.2 there exists g0 (x) ∈ K[x], g0 (x) 6= 0, such that K[x, g0 (x)−1 ] is integrally closed. Let F be the splitting field of f (x, Y ) over K(x). Choose a root y of f (x, Y ) in F . Let f0 (x) be the coefficient of the highest power of Y in f (x, Y ) and let d0 (x) be the discriminant of y over K(x). Now choose a primitive element z for F/K(x), integral over K[x] and let d1 (x) be its discriminant over K(x). Finally, let g(x) = g0 (x)f0 (x)d0 (x)d1 (x)h(x). Then consider the ring R = K[x, g(x)−1 ]. By Lemma 6.1.2, R[y] and R[z] are the respective integral closures of R in K(x, y) and F . Let L be the algebraic closure of K in F . By Corollary 10.2.2(a), K(x)/K is a regular extension, so resL Gal(F/K(x)) = Gal(L/K). Since K is ω-free, Gal(F/K(x)) ∈ Im(Gal(K)). Thus, R[z]/R satisfies the decomposition group condition (Definition 24.1.3). Since K is Frobenius, Proposition 24.1.4 gives a K-homomorphism ϕ: R[z] → Ks with decomposition group Gal(F/K(x)) such that ϕ(R) = K. In particular, a = ϕ(x) is a point of V (K) and for c = ϕ(z), (2)

[K(c) : K] = [F : K(x)].

If char(K) = p, h1 (x), . . . , hm (x) ∈ K[x] are p-independent over K(x)p and m is at most the imperfect degree of K, then Proposition 24.1.4 allows us to choose a such that h1 (a), . . . , hm (a) are p-independent in K.

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Let b = ϕ(y). Then, [K(b) : K] ≤ [K(x, y) : K(x)] and [K(c) : K(b)] ≤ [F : K(x, y)]. By (2), [K(b) : K] = [K(x, y) : K(x)]. Hence, f (a, Y ) is irreducible in K[Y ]. Proof of “(b) =⇒ (a)”: Condition (b) implies that K is a PAC field. We need only show that every finite embedding problem for Gal(K) is solvable. Let L be a finite Galois extension of K, B a finite groups, and α: B → Gal(L/K) an epimorphism. We have to produce an homomorphism γ: Gal(K) → B satisfying α ◦ γ = resKs /L . Lemma 11.6.1 gives a finitely generated regular extension E = K(x) of K and a finite Galois extension F of E that contains L with B = Gal(F/E) and α = resF/L . Let S/R be a ring cover for F/E with R = K[x, h(x)−1 ] and 0 6= h(x) ∈ K[x] and let z be a primitive element for S/R (Definition 6.1.3). If f ∈ K[x, Z] is an irreducible polynomial in Z over K(x) such that f (x, z) = 0, then (b) gives a K-rational specialization x → a such that f (a, Z) is irreducible in K[Z] and h(a) 6= 0. This extends to an Lhomomorphism ϕ of S onto a field N = K(ϕ(z)) which is Galois over K and contains L. By Lemma 6.1.4, ϕ induces an embedding ϕ∗ of Gal(N/K) into Gal(F/E) such that resF/L ◦ ϕ∗ = resN/L . Since f (a, Z) is irreducible over K, [N : K] = [F : E] and ϕ∗ is bijective. Thus, γ = ϕ∗ ◦ resN : Gal(K) → B is an epimorphism which solves our embedding problem. Proof of “(b) ⇐⇒ (c)”: Use the primitive element theorem.



Recall that a field K is 1-imperfect if either char(K) = 0 or char(K) = p and [K : K p ] = p (Section 2.7). Let TR again be the theory of fields together with the axioms of (1) as at the beginning of this section. Proposition 27.2.2: A field K is an existentially closed model of TR if and only if K is 1-imperfect, ω-free, and PAC. Proof: Let p = char(K). Each of the two parts of the proof proves one direction of the proposition: Part A: Suppose K is an existentially closed model of TR . Part A1: Proof that K is 1-imperfect. We need only treat the case where p > 0. If t is transcendental over K, then K is existentially closed in K(t). Lemma 27.1.4 produces an elementary extension K ∗ of K containing K(t) such that K(t) is algebraically closed in K ∗ . In particular, t1/p 6∈ K ∗ . Hence, K ∗ is imperfect. Therefore, K is imperfect. On the other hand assume the imperfect exponent of K exceeds 1. Then there exist p-independent elements a, b in K. Let x, y be transcendental elements over K such that axp +by p = 1. By Lemma 2.7.4, K is algebraically closed in K(x, y). Hence, K is existentially closed in K(x, y), so there are c, d ∈ K with acp + bdp = 1, contrary to the p-independence of a and b in K. It follows that K is 1-imperfect.

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Part A2: Proof that K is ω-free and PAC. It suffices to prove that K satisfies Condition (b) of Lemma 27.2.1. Let V = V (h1 , . . . , hr ) be a variety defined over K with generic point x, f (x, Y ) ∈ K[x, Y ] an irreducible separable polynomial over K(x), and 0 6= g(x) ∈ K[x]. By Corollary 10.2.2, K(x)/K is regular. Hence, by Lemma 27.1.4, K(x) is algebraically closed in an elementary extension K ∗ of K. Therefore, f (x, Y ) is irreducible over K ∗ . In addition, h1 (x) = · · · = hr (x) = 0. Since irreducibility of a polynomial over a field is an elementary statement, there exists a ∈ V (K) such that f (a, Y ) is irreducible in K[Y ] and g(a) 6= 0. Part B: Suppose K is a 1-imperfect ω-free PAC field. We have to prove that if K is algebraically closed in an extension E, then K is existentially closed in E. Indeed, let ϕ(X1 , . . . , Xn ) be a quantifier free formula of LR (ring) and x1 , . . . , xn elements in E such that E |= ϕ(x). Then K is algebraically closed in K(x) and K(x) |= ϕ(x). Thus, we may assume that E = K(x). By Lemma 27.1.4, it suffices to prove that E is algebraically closed in an elementary extension K ∗ of K. Since K is 1-perfect, K(x) is regular over K (Lemma 2.7.5). Thus, (x) generates a variety V over K (Corollary 10.2.2). Part B1: Data for being algebraically closed. Extend the language L(ring, K) to a language L(ring, K, x) by adding n constant symbols corresponding to x1 , . . . , xn . Call a system, (3)

(f1 (x, Y ), . . . , fk (x, Y ); g1 (x), . . . , gl (x); h1 (x), . . . , hm (x)),

of polynomials with coefficients in K data for being algebraically closed, if the following hold: (4a) f1 (x, Y ), . . . , fk (x, Y ) are separable and irreducible in E[Y ]. (4b) g1 (x), . . . , gl (x) 6= 0. (4c) h1 (x), . . . , hm (x) ∈ K[x] r E p . Suppose that for each choice of the data (3) for being algebraically closed there exists a ∈ V (K) such that (5a) f1 (a, Y ), . . . , fk (a, Y ) are separable and irreducible in K[Y ]; (5b) g1 (a), . . . , gl (a) 6= 0; and (5c) h1 (a), . . . , hm (a) ∈ K r K p . Then (K, a) is a structure for L(ring, K, x) in which the sentences of (4) are true. By the compactness theorem (Proposition 7.7.6) there is a structure (K ∗ , x∗ ) for L(ring, K, x) such that K ∗ is an elementary extension of K, x∗ ∈ V (K ∗ ), f (x∗ , Y ) is separable and irreducible in K ∗ [Y ] for every separable and irreducible polynomial f (x, Y ) in E[Y ], g(x∗ ) 6= 0 for every nonzero g(x) ∈ K[x]; and h(x∗ ) ∈ K ∗ r(K ∗ )p for every h(x) ∈ K[x] r E p . It follows that the K-specialization x → x∗ maps E isomorphically onto a subfield E ∗ of K ∗ such that E ∗ is algebraically closed in K ∗ . If p = 0, we remove all occurrences of hi from Part B1.

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Part B2: Existence of the specialization satisfying (5). The proof of the Proposition is complete if, given data (3) for being algebraically closed, we demonstrate the existence of a ∈ V (K) that satisfies (5). Indeed, let u1 , . . . , ur be a p-basis for E over E p with p = char(K). If r ≥ 2, choose 2(r − 1) elements z2 , w2 , . . . , zr , wr such that z2 , . . . , zr are algebraically independent elements over E and (6)

u1 zip + ui wip = 1,

i = 2, . . . , r.

Consider the field extension F = K(x, z, w) of E, where F = E if r = 1 or if char(K) = 0. By Lemma 2.7.4, E is algebraically closed in F . Therefore, so is K. By Lemma 2.7.5, F is a regular extension of K. 1/p 1/p For each i between 1 and m, E(hi (x)1/p ) ⊆ E(u1 , . . . , ur ). Since E 1/p 6∈ F . Hence, by (6), is algebraically closed in F , hi (x) 1/p

1/p

F ⊂ F (hi (x)1/p ) ⊆ F (u1 , . . . , u1/p r ) = F (u1 ). 1/p

It follows that F (hi (x)1/p ) = F (u1 ). Thus, there exist relations (7)

1/p

si (x, z, w)u1

=

p−1 X

tij (x, z, w)hi (x)j/p ,

i = 1, . . . , m,

j=0

with si , tij ∈ K[x, z, w] and si (x, z, w) 6= 0. Since F is regular over K, (x, u, z, w) generates a variety over K. Hence, by Lemma 27.2.1, there exists a K-rational specialization (x, u, z, w) → (a, b, c, d) such that (5a) and (5b) hold, b1 6∈ K p and si (a, c, d) 6= 0, i = 1, . . . , m. By (7), 1/p si (a, c, d)b1

=

p−1 X

tij (a, c, d)hi (a)j/p ,

i = 1, . . . , m.

j=0

Hence, (5c) is also satisfied. Thus, a belongs to V (K) and satisfies (5).



The results of this section prepare all necessary assumptions for an application of Proposition 27.1.12 to TR : Theorem 27.2.3: The theory TR has a model companion T˜R . A field K is a model of T˜R if and only if K is 1-imperfect, ω-free and PAC. Proof: We show there exists a set of sentences T˜R in L(ring) such that a field K is a model of T˜R if and only if K is 1-imperfect, ω-free and PAC. Indeed, axiomatize 1-imperfect with the sentences p = 0 → (∃X)(∀Y )[Y p 6= X ∧ (∀Z)(∃U0 , . . . , ∃Up−1 )[Z =

p−1 X i=0

Uip X i ]],

664

Chapter 27. Omega-free PAC Fields

one for each prime p. That is, if p = 0 in K there exists an element in K r K p and any two elements of K are p-dependent over K p . Proposition 11.3.2 axiomatizes the PAC property. Finally, by Remark 23.4.2, the statement “each finite embedding problem over K is solvable” is elementary. It follows from Proposition 27.2.2 that Mod(T˜R ) = Exis(T ). Conse quently, by Proposition 27.1.12, T˜R is a model companion of TR . The following example proves that TR does not have the amalgamation property. Thus, although TR has a model companion, it does not have a model completion. Example 27.2.4: Let t, u, v be algebraically independent elements over Fp and put w = u − v. Then t, u, v 1/p are algebraically independent over Fp . Therefore, Fp (t, u) is algebraically closed in Fp (t, u, v 1/p ). Similarly Fp (t, u) is algebraically closed in Fp (t, u, w1/p ). If, however, L is a TR -extension of both Fp (t, u, v 1/p ) and Fp (t, u, w1/p ), then they are both algebraically closed in L. In particular, v 1/p , which is algebraic over Fp (t, u, v) = Fp (t, u, w), belongs to Fp (t, u, w1/p ). This means that v 1/p f (t, u, w1/p ) = g(t, u, w1/p ), for two polynomials f, g with coefficients in Fp , where f 6= 0. Raise this expression to the pth power and substitute t = 0 to obtain (u − w)f (0, up , w) = g(0, up , w). The exponents of the powers of u on the right side are multiples of p, whereas they are congruent to 1 modulo p on the left hand side, a contradiction. 

27.3 New Non-Classical Hilbertian Fields We extend the theory TR of Section 27.2 to a theory TR,Q in an extended language LR,Q (ring). We prove that TR,Q has a model completion T˜R,Q whose models are the ω-imperfect ω-free PAC fields. Augment LR (ring) to a language LR,Q (ring) by adding n-ary relation symbols, Qp,n , one for each prime p and each positive integer n. Let TR,Q be the theory of LR,Q (ring) consisting of TR (Section 27.2) together with the axioms X

(1) Qp,n (X1 , . . . , Xn ) ↔ p = 0 ∧ (∃Ui )[

i

Uip X1i1 · · · Xnin = 0 ∧

_

Ui 6= 0],

i

one for each (p, n) where i ranges over all n-tuples (i1 , . . . , in ) of integers between 0 and p − 1. Given a field K, we may uniquely regard K as a model of TR,Q . Thus, if p = char(K) and x1 , . . . , xn ∈ K, then x1 , . . . , xn are pdependent if and only if Qp,n (x1 , . . . , xn ) holds in K. Therefore, if K is a substructure of L (as models of TR,Q ), then L is linearly disjoint from K 1/p over K. By Lemma 2.6.1, L is separable over K. Since L is a TR -extension of K, K is algebraically closed in L, so L is regular over K (Lemma 2.6.4). If K is a substructure of both L and M , replace M by a K-isomorphic copy, if necessary, to assume that M is algebraically independent from L

27.3 New Non-Classical Hilbertian Fields

665

over K and both are contained in a common field. By Corollary 2.6.8(a), LM is a regular extension of both L and M . It follows that TR,Q has the amalgamation property. Conversely, if L is a regular field extension of K, then L/K is an extension of models of TR,Q . Theorem 27.3.1: The theory TR,Q has a model completion whose models are the ω-imperfect ω-free PAC fields. Proof (Sketch): This is similar to the proof of Theorem 27.2.3 (and its main ingredient Proposition 27.2.2). We elaborate on two points that differ from the arguments of the result for TR : First: If a field K of characteristic p is TR,Q -existentially closed and x1 , . . . , xn are n algebraically independent elements over K, then x1 , . . . , xn are p-independent in E = K(x). Thus, the existential sentence (∃X1 , . . . , ∃Xn )[¬Qp,n (X)] is true in E (take Xi to be xi , i = 1, . . . , n). Hence, it holds in K. Therefore, K is ω-imperfect. Second: If K is an ω-imperfect ω-free and PAC field of characteristic p, and if E = K(x) is a finitely generated regular extension of K, apply Condition (c) of Lemma 27.2.1 to find a K-specialization of x into K such that a given p-independent elements u1 , . . . , um of E are mapped into pindependent elements of K.  Remark 27.3.2: There are two obvious generalizations of Theorems 27.2.3 and 27.3.1. One may augment the theory TR with an axiomatization of a Melnikov formation of finite groups C in the language of fields. This may give a theory with a model companion (respectively, model completion) whose countable models are the 1-imperfect (resp. ω-imperfect) PAC fields K with Gal(K) ∼ = Fˆω (C). Alternatively, one may augment TR so as to bound the imperfect exponent of the fields by m, for some given positive integer m.  We apply Lemma 27.2.1(b) in the case that V runs over the affine spaces An , n = 1, 2, . . ., to rephrase the condition that K is Hilbertian: Corollary 27.3.3 ([Roquette1]): Every ω-free PAC field is Hilbertian. Recall that a field K is Hilbertian if and only if A1 (K) is a nonthin set (Proposition 13.5.3). If K is a number field, then K is Hilbertian but C(K) is a thin set for each curve of genus at least 1 (Remark 13.5.4). We show that ω-free PAC fields are not only Hilbertian but V (K) is nonthin for every variety V over K. Thus, ω-free PAC fields are Hilbertian in a stronger sense. Proposition 27.3.4: A field K is ω-free and PAC if and only if V (K) is nonthin for every variety V over K. Proof: Suppose first K is ω-free and PAC. Let V be a variety defined over K and x a generic point x of V over K. For i = 1, . . . , m consider an element

666

Chapter 27. Omega-free PAC Fields

yi ∈ K(x)s which is integral over K[x] such that K(x, yi )/K is regular and [K(x, yi ) : K(x)] ≥ 2. Let hi ∈ K[X, Y ] be a polynomial which is monic in Y such that hi (x, Y ) = irr(yi , K(x)). Let g ∈ K[X] with g(x) 6= 0. By Lemma 27.2.1(c), there is an a ∈ V (K) such that h1 (a, Y ), . . . , hm (a, Y ) are irreducible over K and g(a) 6= 0. In particular, hi (a, Y ) has no zero in K, i = 1, . . . , m. By Remark 13.5.2(g), V (K) is nonthin. Now suppose V (K) is nonthin for every variety V defined over K. We prove that K satisfies Condition (b) of Lemma 27.2.1. Let V be a variety over K with a generic point x, f (x, Y ) ∈ K[x, Y ] an irreducible separable polynomial in Y over K(x), and h ∈ K[X] with h(x) 6= 0. We use a simplified version of Proof B of Lemma 13.1.2 to find a ∈ V (K) such that h(a) 6= 0 and f (a, Y ) is irreducible over K. Qn Assume without loss, f (x, Y ) is monic. Write f (x, Y ) = i=1 (Y − yi ) with y1 , . . . , yn ∈ K(x)s . Denote the collection Q of all nonempty subsets of {1, . . . , n} by I. For each I ∈ I let fI (Y ) = i∈I (Y − yi ). Since f (x, Y ) is irreducible over K(x), there is a coefficient zI of fI which does not belong to K(x). Thus, there is a polynomial gI ∈ K[X, Z] monic and of degree at least 2 in Z such that gI (x, Z) = irr(zI , K(x)). Since V (K) is nonthin, there is an a ∈ V (K) such that h(a) 6= 0, f (a, Y ) is separable, and gI (a, c) 6= 0 for all c ∈ K (Remark 13.5.2(g)). Assume f (a, Y ) = p(Y )q(Y ) is a nontrivial factorization in K[Y ] with monic p and q. Extend the K-specialization x → a to a K-specialization (x, y) → (a, b), Qn with b = (b1 , . . . , bn ) ∈ Ksn .QThen f (a, Y ) = i=1 (Y − bi ). Thus, there is an I ∈ I such that p(Y ) = i∈I (Y − bi ), the polynomial fI (Y ) maps onto p(Y ), and zI maps onto a coefficient c of p(Y ). Then c lies in K and satisfies gI (a, c) = 0. This contradiction to the choice of a proves that f (a, Y ) is irreducible over K.  Our previous examples of Hilbertian fields include number fields, transcendental finitely generated extensions of fields, fields of power series K0 ((X1 , . . . , Xr )) with r ≥ 2, and composition of pairs of Galois extensions of Hilbertian fields. Corollary 27.3.3 provides new examples of Hilbertian fields. They are large algebraic extensions of given countable Hilbertian fields: Example 27.3.5: Algebraic ω-free PAC fields. Let K be a countable Hilbertian field. Take G = Fˆω in Theorem 23.2.3 to conclude the existence of an ω-free PAC field E algebraic over K. Alternatively take (σ1 , σ2 ) ∈ Gal(K)2 such that Ks (σ) is 2-free PAC field (Theorem 20.8.2). Then use the results of Section 24.10 to provide an algebraic extension E of Ks (σ) such that Gal(E) ∼ = Fˆω (e.g. take E as the maximal Abelian extension of Ks (σ) as in Example 24.10.7). Then E is an ω-free PAC field. By Corollary 27.3.3, E is Hilbertian. Note that E = Ks (σ)ab (with σ = (σ1 , σ2 ) taken at random in Gal(K)2 ) can not be proved to be Hilbertian by the diamond theorem (Theorem 13.8.3) over K. Indeed, assume L and M are Galois extensions of K such that Ks (σab ) 6⊆ L, Ks (σ)ab 6⊆ M , by Ks (σ)ab ⊆ LM . Then Ks (σ) ⊆ L

27.4 An Abundance of ω -Free PAC Fields

667

or Ks (σ) ⊆ M , otherwise Ks (σ) is Hilbertian (by the diamond theorem), so Ks (σ) has a linearly disjoint sequence of quadratic extension (Corollary 16.2.7), in contradiction to Lemma 16.11.2. Suppose for example that Ks (σ) ⊆ L. By Proposition 16.12.6, L = Ks . Hence, Ks (σ)ab ⊆ L, in contradiction to our assumption.  We have mentioned in Example 24.8.5(b) that the converse of Corollary 27.3.3 is also true: Every Hilbertian PAC field is ω-free.

27.4 An Abundance of ω-Free PAC Fields Let K be a countable Hilbertian field. By Theorem 18.10.2, Ks [σ] is PAC for almost all σ ∈ Gal(K)e . Here we prove that Ks [σ] is ω-free for almost all σ ∈ Gal(K)e . This will imply that Ks [σ] is Hilbertian. This result presents a climax of Field Arithmetic. It uses several major results proved in this book and two results whose full proofs are unfortunately beyond the framework of the book. The latter are the regularity of finite groups over complete discrete fields (Proposition 16.12.1) and the stability of fields (Theorem 18.9.3). The former include Borel-Cantelli Lemma, the theorem of Ax-Roquette that every algebraic extension of a PAC field is PAC (Theorem 11.2.5), Weissauer’s theorem about infinite extensions of Hilbert fields (Theorem 13.9.1) Roquette’s theorem that every ω-free PAC field is Hilbertian (Theorem 27.3.3), and Melnikov’s characterization of closed normal subgroups of Fˆω by finite simple quotients (Corollary 25.7.6). The strategy of the proof is for a σ taken at random in Gal(K)e to embed [σ] as a closed normal subgroup of Fˆω , to prove that each of the groups Z/pZ is a quotient [σ], and every finite non-Abelian simple is a quotient of [σ] infinitely often. By Corollary 25.7.6, this will prove that [σ] ∼ = Fˆω . We begin by enhancing some notation introduced in Section 18.10: Let N/K be a Galois extension, L/K a Galois subextension, and σ = (σ1 , . . . , σe ) ∈ Gal(N/K)e . Then L(σ) is the fixed field in L of σ1 , . . . , σe , L[σ] is the maximal Galois extension of K which is contained in L(σ). It follows that L[σ] = L ∩ N [σ]. Lemma 27.4.1: Let K be a Hilbertian field. Then for almost all (σ, τ ) ∈ Gal(K)e+1 , the field Ks [σ, τ ] is properly contained in Ks [σ]. Proof: Corollary 16.2.7 gives a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions of K with Gal(Li /K) ∼ = (Z/2Z)e+1 , i = 1, 2, 3, . . . . For each i let σi1 , . . . , σie , τi be a system of generators for Gal(Li /K). For almost all (σ, τ ) ∈ Gal(K)e+1 there is an i with resLi (σ, τ ) = (σi , τi ) (Lemma 18.5.3(b)). Then K = Li (σ, τ ) = Li [σ, τ ]. Since Gal(Li /K) is not generated by e elements, K ⊂ Li (σi ). Since Li /K is Abelian, Li (σi ) = Li [σi ].  Consequently, Ks [σ, τ ] ⊂ Ks [σ]. Lemma 27.4.2: Let K be a countable Hilbertian field. Then for almost all σ ∈ Gal(K)e , the field Ks [σ] is a Galois extension of an ω-free PAC field.

668

Chapter 27. Omega-free PAC Fields

Proof: Let S be the set of all (σ, τ ) ∈ Gal(K)e+1 with the following properties: (1a) Ks [σ] is PAC. (1b) Ks [σ] properly contains Ks [σ, τ ]. By Theorem 18.10.2 and by Lemma 27.4.1, µ(S) = 1. Hence, by Fubini’s theorem, for almost all σ ∈ Gal(K)e the set of all τ ∈ Gal(K) with (σ, τ ) ∈ S has measure 1 [Halmos, p. 147, Thm. A]. Thus, for almost all σ ∈ Gal(K)e there is a τ ∈ Gal(K) satisfying (1). Let (σ, τ ) ∈ S. Choose a proper finite extension M of Ks [σ, τ ] in Ks [σ]. By Weissauer (Theorem 13.9.1), M is Hilbertian. By Ax-Roquette (Corollary 11.2.5), M is PAC. Hence, by Example 24.8.5(b), M is ω-free.  Proposition 27.4.3: Let K be a field and G a finite group. Then there exists a variety V over K with the following property: If L is a field extension of K and V (L) 6= ∅, then G is regular over L. Proof: Consider the formal power series E = K((t)) in the indeterminate t. By Proposition 16.12.1, G is regular over E. Let f ∈ E[Y, Z] be a Z-stable polynomial which is Galois with respect to Z such that Gal(f (Y, Z), E(Y )) ∼ = G (Proposition 16.2.8). Let x1 , . . . , xn be the elements of E which appear as coefficients of f . Thus, f (Y, Z) = h(x, Y, Z) for some h ∈ K[X, Y, Z]. By Example 3.5.1, E is a regular extension of K. By Lemma 10.2.2(c), x generates a variety W over K in An . Lemma 16.1.4 gives a nonempty Zariski K-open subset W0 of W such that h(a, Y, Z) is absolutely irreducible, Galois with respect to Z, and Gal(h(a, Y, Z), L(Y )) ∼ = G for every field extension L of K and every a ∈ W0 (L). Thus, G is regular over L (Remark 16.2.2). Now choose a polynomial g ∈ K[X] which vanishes on W r W0 but not on W . Let V be the variety generated in An+1 over K by (x, g(x)−1 ). If L is a field extension of K and (a, b) ∈ V (L), then a ∈ W0 (L). Hence, by the preceding paragraph, G is regular over L.  Lemma 27.4.4: Let K be a Hilbertian field and G a finite group. Then there exists a positive integer n and a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions of K such that for each i, Gal(Li /K) ∼ = Sn and Li has a linearly disjoint sequence Li1 , Li2 , Li3 , . . . of Galois extensions with Gal(Lij /Li ) ∼ = G for each j. Proof: Let V be the variety given by Proposition 27.4.3. Choose a generic point x of V over K and put F = K(x). Then F is a regular extension of K (Corollary 10.2.2) of transcendence degree, say, r. Theorem 18.9.3 gives a separating transcendence base t1 , . . . , tr for F/K such that the Galois closure Fˆ of F/K(t) is a regular extension of K and Gal(Fˆ /K) ∼ = Sn for some positive integer n. Since K is Hilbertian, Lemma 16.2.6 gives a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions of K with Gal(Li /K) ∼ = Sn and V (Li ) 6= ∅ for each i.

27.4 An Abundance of ω -Free PAC Fields

669

Let i be a positive integer. By Lemma 27.4.3, G is regular over Li . Hence, by Lemma 16.2.6, Li has a linearly disjoint sequence Li1 , Li2 , Li3 , . . .  with Gal(Lij /Li ) ∼ = G for each i. Lemma 27.4.5: Let K be a Hilbertian field and G a finite group. Suppose G is the normal subgroup of itself generated by e elements. Then, for almost all σ ∈ Gal(K)e , the group G is realizable over Ks [σ]. Proof: Let Li and Lij be the fields which Lemma 27.4.4 gives. For each i and j choose σij ∈ Gal(Lij /Li )e with Lij [σij ] = Li . Then put Aij = σ = σi }. Since the Lij are linearly disjoint over Li {σ ∈ Gal(Li ) | resLijS ∞ with fixed degree, µ( j=1 Aij ) = µ(Gal(Li )) (Lemma 18.5.3(b)). Similarly,
S∞ S µ i=1 Gal(Li ) = 1. Hence, µ( i,j Aij ) = 1. Consider σ ∈ Aij . Then, Lij ∩ Ks [σ] = Li . Hence, Gal(Lij Ks [σ]/Ks [σ]) ∼ = Gal(Lij /Li ) ∼ = G, as claimed.



Lemma 27.4.6: Let S be a finite non-Abelian simple group and n a positive integer. Then, for almost all σ ∈ Gal(K)e , the group S n occurs as a Galois group over Ks [σ]. Qn Proof: Rewrite S n as i=1 Si with Si ∼ = S. Choose s = (s1 , . . . , sn ) ∈ S n with si 6= 1 for all i. Let N be the normal subgroup of G generated by s. Then the projection of N on the ith coordinate is the whole group Si . By Lemma 18.3.9, N = S n . It follows from Lemma 27.4.5 that S n occurs as a Galois group over Ks [σ] for almost all σ ∈ Gal(K)e .  Lemma 27.4.7: Let p be a prime number. Then, for almost all σ ∈ Gal(K)e , the group Z/pZ occurs as a Galois group over Ks [σ]. Proof: Lemma 16.3.6 gives a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions of K with Gal(Li /K) ∼ = Z/pZ for each i. Choose a generator σ ¯i of Gal(Li /K). For almost all σ ∈ Gal(K)e there is an i with σi , . . . , σ ¯i ) (Lemma 18.5.3(b)). Hence, Z/pZ is realizresLi (σ1 , . . . , σe ) = (¯  able over Ks [σ]. We sum up and prove our main result: Theorem 27.4.8 ([Jarden16, Thm. 2.7]): Let K be a countable Hilbertian field. Then, for almost all σ ∈ Gal(K)e , the field Ks [σ] is ω-free and PAC. In particular, Ks [σ] is Hilbertian. Proof: Let S be the set of all σ ∈ Gal(K)e with the following properties: (2a) Ks [σ] is PAC. (2b) K has an algebraic extension M in Ks [σ] which is ω-free. (2c) For each finite non-Abelian group S and for every positive integer n, the group S n occurs over Ks [σ] as a Galois group. (2d) Z/pZ occurs over Ks [σ] as a Galois group for every prime number p.

670

Chapter 27. Omega-free PAC Fields

By Theorem 18.10.2, Lemma 27.4.2, Lemma 27.4.6, and Lemma 27.4.7, µ(S) = 1. Let σ ∈ S and G = Gal(Ks [σ]). Then G is a closed normal subgroup of Gal(M ) and Gal(M ) ∼ = Fˆω . By (2c), (2d), and Corollary 25.7.6, ˆ . G∼ F = ω  It follows from (2a) and Corollary 27.3.3 that Ks [σ] is Hilbertian.

Notes The notions of model completion and model companion are due to A. Robinson. More in this direction can be found in [Hirschfeld-Wheeler]. The predicate symbols Rn of Section 24.6 appear for the first time in [Adler-Kiefe, p. 306] in order to prove that the theory of 1-free perfect PAC fields is the model completion of the theory of perfect fields of corank at most 1. One can find the predicates Qp,n in [K. Schmidt, p. 96]. Schmidt uses an ultraproduct criterion to prove (in the notation of Section 24.7) that the class of TR,Q -existentially closed models is axiomatizable. Hence, TR,Q has a model companion T˜R,Q [K. Schmidt1, p. 107]. He also observes that models of T˜R,Q are PAC and Hilbertian [K. Schmidt1, p. 97]. Theorems 27.2.3 and 27.3.1 are results of a correspondence with K. Schmidt. They also appear in [K. Schmidt2, p. 92]. One can find in [Ershov3, p. 512] results without proof on model completions of various theories of fields in the spirit of Remark 27.3.2. Corollary 27.3.3 is proved in [Jarden9, p. 145] by model theoretic methods.

Chapter 28. Undecidability In contrast to the theories considered so far (e.g. the theory of finite fields, ˜ 1 , . . . , σe ) for fixed e, and the theory of the theory of almost all fields Q(σ perfect PAC fields of bounded corank), the theory of perfect PAC fields is undecidable. This is the main result of this chapter (Corollary 28.10.2). An application of Cantor’s diagonalization process to Turing machines shows that certain families of Turing machines are nonrecursive. An interpretation of these machines in the theory of graphs shows the latter theory to be undecidable. Finally, Frattini covers interpret the theory of graphs in the theory of fields. This applies to demonstrate the undecidability of the theory of perfect PAC fields.

28.1 Turing Machines A Turing machine may be considered as an abstract primitive computer. As with the “machine languages” in use in the 1950’s, description and interpretation of the machine causes no difficulties, but programming the machine requires immense work. Definition: A Turing machine is a system M = hQ, R, L, S, pr, mv, mdi, consisting of these components: a finite set Q = {q1 , . . . , qe } of operational modes; the move commands R (“move right”), L (“move left”) and S (“stay”); and its working functions, pr: Q × {0, 1} → {0, 1} (“print”), mv: Q×{0, 1} → {R, L, S} (“move”) and md: Q×{0, 1} → Q (“next mode”). A tape T = t1 t2 t3 t4 · · · is an infinite strip (sequence) with ti ∈ {0, 1}, i = 1, 2, . . . . The sequence of squares without reference to the ti ’s is called the ribbon of M . Each Turing machine M operates on a tape T , the initial tape, as a mechanical device whose operational modes correspond to the elements of Q. Intuitively the initial tape is “input data” and the functions pr, mv, and md are the “program.” We use the word instant to refer to the time interval required for the sequence of machine operations called a beat to be executed. We denote the value of ty at the instant x by τ (x, y), and let κ(x) be the mode of the machine at instant x. If M is at initial mode q1 , in instant 1 the machine goes through beat 1: it scans square 1; then it prints τ (2, 1) = pr(q1 , t1 ) in square 1; moves one square to the right if mv(q1 , t1 ) = R, and it stays at square 1 if mv(q1 , t1 ) = S; and then it changes the mode to md(q1 , t1 ). At the xth instant with the machine in mode κ(x) and at the yth square it goes through beat x: it reads the value τ (x, y) in square y; then it prints τ (x + 1, y) = pr(κ(x), τ (x, y)) in the yth square; it then moves one square to the right if mv(κ(x), τ (x, y)) = R, it stays at the yth square if mv(κ(x), τ (x, y)) = S, and it moves one square to the left if mv(κ(x), τ (x, y)) = L (unless y = 1, in which case the machine stops); and then it changes the mode to κ(x + 1) = md(κ(x), τ (x, y)).

672

Chapter 28. Undecidability

If the machine does not stop at the end of the xth instant, it proceeds to perform the (x + 1)th beat. At the end of the xth beat the machine has changed the original tape T = T1 to a tape Tx . The action of M is said to be eventually stationary if there exists an instant b when the machine is at square y, mv(κ(b), τ (b, y)) = S, and τ (x, y) = τ (b, y) and κ(x) = κ(b) for each x ≥ b (i.e. each successive beat after b consists of the same operation; and Tb = Tb+1 = · · ·). If M stops at instant c, then we call T 0 = Tb+1 the final tape. The machine M is uniquely determined by its working functions pr, mv, and md. Let p1 < p2 < · · · be the sequence of primes. Encode M as a positive integer as follows: For t ∈ {0, 1} and qi ∈ Q, let  γ(qi , t) =

5 if pr(qi , t) = 0 7 if pr(qi , t) = 1,

  11 if mv(qi , t) = R δ(qi , t) = 13 if mv(qi , t) = L  17 if mv(qi , t) = S, and ε(qi , t) = p10+j

if md(qi , t) = qj .

Now define code(M ) to be 2a(M ) 3b(M ) with a(M ) =

e Y i=1

γ(qi ,0)

p19 10+i

23δ(qi ,0) 29ε(qi ,0)

and b(M ) =

e Y

γ(qi ,1) 23δ(qi ,1) 29ε(qi ,1)

p19 10+i

.

i=1

The map from M to code(M ) is injective, and the set of codes of all Turing machines is primitive recursive (Exercise 2). Moreover, one may reconstruct γ, δ, ε and therefore M from its code.

28.2 Computation of Functions by Turing Machines Denote the set of nonnegative integers by N . For a positive integer n, partition the ribbon R of the Turing machine M into n subribbons, R = R1 ∪· · · · ∪· Rn , where Rk consists of all squares y with y ≡ k mod n. Call each such subribbon an n-track (or just track). In each track we may store an integer z0 ∈ N by printing 1 in the first z0 squares of the track and 0 in the rest of the squares. Let f : N m → N be a function and let k1 , . . . , km+1 ≤ n be distinct positive integers. We say that M computes f (from the n-tracks k1 , . . . , km to the km+1 th n-track) if when M operates on a tape T with x1 , . . . , xm stored in the n-tracks k1 , . . . , km , respectively, and 0 stored in the remaining tracks, then M eventually stops and the final tape T 0 is identical with T except that M has stored f (x1 , . . . , xm ) in the km+1 th n-track. Proposition 28.2.1: For each recursive function f (x1 , . . . , xm ) there exists a Turing machine M which computes f .

28.2 Computation of Functions by Turing Machines

673

Proof: A recursive function arises from a list of basic functions and basic operations (Section 8.5), and the parts of the proof treat each of these separately. The first step in each part is the choice of an integer n and integers k1 , . . . , km+1 with 1 < k1 < · · · < km+1 ≤ n such that M computes f from the n-tracks k1 , . . . , km to the km+1 th n-track without changing the n-tracks k1 , . . . , km (where the arguments for f are stored). In each case we store 1 in the first track and we leave this track unchanged through the whole operation of M . Also, in each case to each mode of operation qi there corresponds a number li , with 1 ≤ li ≤ n, such that whenever M is in mode qi it scans a square in the li th n-track. In particular, the first track throughout the action of M corresponds only to the initial mode q1 and the final mode qe . Thus, if M is in mode qe and it reads 1, then it scans square 1. If e 6= 1, we define mv(q1 , 1) = R and mv(qe , 1) = L. This gives complete control on the action of M at the last instant, where M is at the mode qe and it reads 1. We bother to define pr, mv and md of M only for those arguments which actually occur in their computation. Thus, our definitions appear as commands for a computer (with all commands, that would never be executed, left unstated). Throughout the proof the variable t ranges over {0, 1}. Part A: The zero function f (x) = 0. With n = 3, k1 = 2, k2 = 3 and e = 1, the machine M computes f from the 2nd track to the 3rd. It does this in one beat: pr(q1 , 1) = 1,

mv(q1 , 1) = L,

md(q1 , 1) = q1 .

For example, here is the starting (and final) tape for x = 4 1 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 ··· Part B: The successor function f (x) = x + 1. With n = 3, k1 = 2, k2 = 3 and e = 6, M computes f from the 2nd track to the 3rd. In outline: M moves to the right and scans the squares in the second track printing 1 in the 3rd track square to the right of the second track square. The last time it does this is when it has scanned 0 in the second track. Then, after printing 1 in the next square to the right, it moves only left until it stops: pr(q1 , t) = t, pr(q2 , 1) = 1, pr(q3 , 0) = 1, pr(q2 , 0) = 0, pr(q4 , t) = 1, pr(q5 , t) = t, pr(q6 , t) = t,

mv(q1 , t) = R, mv(q2 , 1) = R, mv(q3 , 0) = R, mv(q2 , 0) = R, mv(q4 , t) = L, mv(q5 , t) = L, mv(q6 , t) = L,

md(q1 , t) = q2 ; md(q2 , 1) = q3 ; md(q3 , t) = q1 ; md(q2 , 0) = q4 ; md(q4 , t) = q5 ; md(q5 , t) = q6 ; md(q6 , t) = q4 .

674

Chapter 28. Undecidability

Example: Computation of f (3). inst\sqr

1

2

3

4

5

6

7

8

9

10 11

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 stop

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

0

0

0

0 1 1 1 1 1 1 1 1 1 1 1 1 1 1

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0

12

mode

0 1 1 1 1 1 1 1 1 1 1 1

1 2 3 1 2 3 1 2 3 1 2 4 5 6 4 5 6 4 5 6 4 5 6

The xth line indicates the tape and the mode of operation at the beginning of the xth instant. The square with the bold figure is that being scanned. Part C: The coordinate projection function f (x1 , . . . , xm ) = xl , 1 ≤ l ≤ m. With n = m+2 and e = 2n+1, M computes f from the n-tracks 2, . . . , m+1 to the (m + 2)th n-track. In outline: as M moves to the right and scans 1 in the (l + 1)th n-track it prints 1 in the next square of the (m + 2)th n-track. When, however, M scans 0 in the (l + 1)th n-track, it starts to move to the left until it stops: pr(qi , t) = t,

mv(qi , t) = R,

md(qi , t) = qi+1 , i = 1, . . . , l, l + 2, . . . , n − 1; pr(ql+1 , 1) = 1, mv(ql+1 , 1) = R, md(ql+1 , 1) = ql+2 ; mv(qn , 0) = R, md(qn , t) = q1 ; pr(qn , 0) = 1, pr(ql+1 , 0) = 0, mv(ql+1 , 0) = L, md(ql+1 , 0) = qn+l ; mv(qi , t) = L, md(qi , t) = qi−1 , i = n + 3, . . . , 2n + 1; pr(qi , t) = t, pr(qn+2 , t) = t, mv(qn+2 , t) = L, md(qn+2 , t) = q2n+1 .

28.2 Computation of Functions by Turing Machines

675

Part D: Iteration. There are two ingredients. First: Given a machine M that computes a function f (x1 , . . . , xm ) from the n-tracks k1 , . . . , km to the km+1 th n-track we may, for any r > n, replace M by a machine M 0 that computes f (x1 , . . . , xm ) from the r-tracks l1 , . . . , lm to the lm+1 th r-track. Second: Given M1 and M2 that, respectively, compute the functions f1 (x1 , . . . , xm ) and f2 (x1 , . . . , xm ), we may construct a machine M that first computes f1 and then computes f2 . Indeed, adjust the computation tracks of f1 and f2 , if necessary, so that the corresponding tracks in M will be disjoint. Let q1 , . . . , qe denote the modes of M1 and qe+1 , . . . , qe+e0 the modes of M2 . Then in the program for M1 replace the values of pr, mv, and md at (qe , 1) by pr(qe , 1) = 1, mv(qe , 1) = S, md(qe , 1) = qe+1 . With the commands for M2 now listed below those for M1 we have M . Each of the remaining parts of the proof considers one of the recursion operations: Part E: Composition, f (x1 , . . . , xm ) = g(h1 (x1 , . . . , xm ), . . . , hl (x1 , . . . , xm )). Suppose that for each i, 1 ≤ i ≤ l, there exists a machine Mi that computes hi (x1 , . . . , xm ) from the ni -tracks ki,1 , . . . , ki,m to the ki,m+1 th ni track and that there exists M0 that computes g(y1 , . . . , yl ) from the n0 -tracks k0,1 , . . . , k0,l to the k0,l+1 th n0 -track. Choose n suitably large and use Part D to construct M that successively computes yi = hi (x1 , . . . , xm ) from the n-tracks 2, . . . , m+1 to the (m+1+i)th n-tracks, i = 1, . . . , l, and then computes g(y1 , . . . , yl ) from the n-tracks m + 2, . . . , m + 1 + l to the (m + 2 + l)th n-track. Altogether M computes f (x1 , . . . , xm ) from the n-tracks 2, . . . , m+1 to the (m + l + 2)nd n-track. Part F: Induction, f (x1 , . . . , xm , 0) = f0 (x1 , . . . , xm ) f (x1 , . . . , xm , y + 1) = g(x1 , . . . , xm , y, f (x1 , . . . , xm , y)). Suppose f0 and g can be computed by Turing machines. Apply Part D to the argument below, with n chosen suitably large. First, M copies y from the (m + 2)nd n-track to the (m + 4)th n-track. Second, M computes f0 (x1 , . . . , xm ) from the n-tracks 2, . . . , m + 1 to the (m + 3)rd track. Third, M successively computes g(x1 , . . . , xm , z, f (x1 , . . . , xm , z)), for z = 0, . . . , y − 1 from the n-tracks 2, . . . , m + 1, m + 5, m + 3 to the (m + 3)rd n-track; it subtracts 1 from the (m + 4)th n-track and adds 1 to the (m + 5)th n-track. The computation ends when 0 has been stored in the

676

Chapter 28. Undecidability

(m + 4)th n-track, and M has computed f (x1 , . . . , xm , y) from the n-tracks 2, . . . , m + 2 to the (m + 3)rd n-track. Note the loop in the computation. The number of times the machine has to repeat the loop is known in advance; it is y + 1. Part G: The minimalization function, f (x1 , . . . , xm ) = min{y ∈ N | g(x1 , . . . , xm , y) = 0}. Assume that g is a computable function for which, given x1 , . . . , xm , there exists y ∈ N such that g(x1 , . . . , xm , y) = 0. A loop appears in M whose number of repetitions is not known in advance. In the yth repetition M computes z = g(x1 , . . . , xm , y) from the tracks 2, . . . , m + 2 to the track m + 3. If z = 0 the machine stops. Otherwise, the value of y in the track m + 2 is increased by 1 and the loop is repeated once more. The computation begins with y = 0. The value of y in the track m + 2  at the end of the computation is f (x1 , . . . , xm ). Remark 28.2.2: Each of the Turing machines constructed in Proposition 28.2.1 uses a different number of tracks for the computation. Thus, if M1 and M2 compute functions of one variable f1 (x) and f2 (x) using, respectively, n1 and n2 tracks with n1 6= n2 , then the argument x has been stored differently in the initial tapes of M1 and M2 . In order to unify the storage of the arguments and the value of a function (dependent only on the number of variables) let 1 < k1 , . . . , km ≤ n + 1 be distinct integers and construct a Turing machine M that first changes a tape with the numbers 1, x1 , . . . , xm stored in the (m + 1)-tracks 1, 2, . . . , m + 1 to a tape with the same numbers now stored in the (n+1)-tracks 1, k1 , . . . , km . Since the actual construction of M is lengthy, we do not carry it out here. This allows us to follow the proof of Proposition 28.2.1 and to construct, for each recursive function f (x1 , . . . , xm ), a machine M that changes a tape on which 1, x1 , . . . , xm are stored in the (m + 1)-tracks 1, 2, . . . , m + 1 to a tape on which 1 and f (x1 , . . . , xm ) are stored in the 1st and 2nd 2-tracks. 

28.3 Recursive Inseparability of Sets of Turing Machines Recursive sets and recursive functions (Section 8.6) are technical tools that interpret decidability of theories. Undecidable theories correspond to nonrecursive sets. In order to prove that a subset A of N (= the set of nonnegative integers) is nonrecursive, one may prove a stronger property, namely that A is “recursively inseparable” from a certain subset B of N . The advantage of the stronger concept lies in Lemma 28.3.2. That lemma gives a convenient tool to deduce the recursive inseparability of sets A0 and B 0 from the recursive inseparability of A and B. Definition 28.3.1: Recursively inseparable sets. We call disjoint subsets A and B of N recursively separable if there exists a recursive set C such that

28.3 Recursive Inseparability of Sets of Turing Machines

677

A ⊆ C and B ∩ C = ∅. Otherwise, A and B are recursively inseparable. In the latter case both A and B are nonrecursive.  Lemma 28.3.2: Let A and B be recursively inseparable subsets of N . Suppose that A0 and B 0 are disjoint subsets of N and that there exists a recursive function f : N → N such that f (A) ⊆ A0 and f (B) ⊆ B 0 . Then A0 and B 0 are recursively inseparable. Proof: Assume that there exists a recursive set C such that A0 ⊆ C and B 0 ∩ C = ∅. Then A ⊆ f −1 (C) and B ∩ f −1 (C) = ∅. Since the characteristic function of f −1 (C) satisfies χf −1 (C) (x) = χC (f (x)), the set f −1 (C) is recursive. Thus, contrary to assumption, A and B are recursively separable.  Definition 28.3.3: Recursively inseparable. Two collections of Turing machines, A and B, are said to be recursively inseparable if code(A) and code(B) are recursively inseparable subsets of N .  We construct recursively inseparable sets of machines by considering the action of each machine on its own code. Definition 28.3.4: The tapes Tn . For n a nonnegative integer, denote the tape on which 1 and n, respectively, are stored in the 1st and 2nd 2-tracks by Tn . Let Hlt(n) (resp. Stat(n)) be the collection of Turing machines which eventually stop (resp. become stationary) when they work on Tn . Finally, denote the collection of Turing machines M which belong to Hlt(code(M )) (resp. Stat(code(M )) by Hlt (resp. Stat).  Lemma 28.3.5: The sets of Turing machines Hlt and Stat are recursively inseparable. Proof: Assume that there exists a recursive set C such that (1)

code(Stat) ⊆ C

and

code(Hlt) ∩ C = ∅.

By definition, χC is a recursive function. By Remark 28.2.2 there exists a machine M0 which eventually stops, when applied to Tx , to produce the final tape TχC (x) . Let qe be the final operation mode and, with no loss, assume that e > 1. Then, from the opening statements of the proof of Proposition 28.2.1, (2)

pr(qe , 1) = 1,

mv(qe , 1) = L,

md(qe , 1) = qd ,

for some d, 1 ≤ d ≤ e. Add modes qe+1 , qe+2 , and qe+3 and change (2) to (3)

pr(qe , 1) = 1,

mv(qe , 1) = R,

md(qe , 1) = qe+1 ,

with the following addition to the operations of M0 (and (3)): pr(qe+1 , 1) = 1, pr(qe+2 , 1) = 1, pr(qe+1 , 0) = 0, pr(qe+3 , 1) = 1,

mv(qe+1 , 1) = L, mv(qe+2 , 1) = L, mv(qe+1 , 0) = L, mv(qe+3 , 1) = S,

md(qe+1 , 1) = qe+2 ; md(qe+2 , 1) = qe+2 ; md(qe+1 , 0) = qe+3 ; md(qe+3 , 1) = qe+3 .

678

Chapter 28. Undecidability

This produces a new machine M1 for which one checks that x ∈ C implies M1 ∈ Hlt(x), otherwise M1 ∈ Stat(x).

(4)

Consider code(M1 ): If code(M1 ) ∈ C, then M1 ∈ Hlt(code(M1 )). Hence, M1 ∈ Hlt and (by (1)) code(M1 ) 6∈ C. If code(M1 ) 6∈ C, then M1 ∈ Stat(code(M1 )). Hence, M1 ∈ Stat and code(M1 ) ∈ C. These contradictions conclude the lemma.  Lemma 28.3.6: The collections Stat(0) and Hlt(0) are recursively inseparable. Proof: that

To each Turing machine M we associate a new machine, M 0 , such M ∈ Stat =⇒ M 0 ∈ Stat(0), M ∈ Hlt =⇒ M 0 ∈ Hlt(0).

(5a) (5b)

In outline, define M 0 so that applied to T0 , it first changes T0 to Tm with m = code(M ), and then it continues as M . Thus, if M ∈ Stat, then M 0 eventually becomes stationary, and if M ∈ Hlt, then M 0 eventually stops. In more detail, if M has modes q1 , . . . , qe , then M 0 has 4m + e modes 0 q10 , . . . , q4m+e . The definitions involving the first 4m modes allow M 0 to change T0 to Tm : pr(q10 , 1) = 1, 0 , 0) = 1, pr(q2i 0 pr(q2i+1 , 0) = 0, 0 , t) = t, pr(q2m+i

mv(q10 , 1) = R, 0 mv(q2i , 0) = R, 0 mv(q2i+1 , 0) = R, 0 mv(q2m+i , t) = L,

md(q10 , 1) = q20 , 0 0 md(q2i , 0) = q2i+1 , i = 1, . . . , m, 0 0 md(q2i+1 , 0) = q2i+2 , i = 1, . . . , m − 1, 0 0 md(q2m+i , t) = q2m+i+1 , i = 1, . . . , 2m.

The definitions involving the last e modes of M 0 essentially give the operations of M : 0 0 0 , t) = pr(qi , t), mv(q4m+i , t) = mv(qi , t), md(q4m+i , t) = md(qi , t), pr(q4m+i 0

i = 1, . . . , e. In the notation of Section 28.1, code(M 0 ) = 2a(M ) 3b(M a(M 0 ) =

4m+e Y i=1

γ(q 0 ,0)

i p19 10+i

0

0

23δ(qi ,0) 29ε(qi ,0)

,

b(M 0 ) =

4m+e Y

γ(q 0 ,1)

i p19 10+i

0

0

)

with 0

23δ(qi ,1) 29ε(qi ,1)

.

i=1

By Exercise 2, the set of all codes of Turing machines is primitive recursive. Therefore, functions code(M ) 7→ a(M 0 ) and code(M ) 7→ b(M 0 ) are primitive recursive. Thus, the function f : N → N defined by f (code(M )) = code(M 0 ) for each Turing machine M , and f (n) = 0 if n is not a code of a machine, is primitive recursive. By (5), f (code(Stat)) ⊆ code(Stat(0)) and f (code(Hlt)) ⊆ code(Hlt(0)). From Lemmas 28.3.2 and 28.3.5, the sets code(Stat(0)) and code(Hlt(0)) are therefore recursively inseparable. 

28.4 The Predicate Calculus

679

28.4 The Predicate Calculus For each positive integer n denote the first order language with countably many m-ary predicate symbols for each m ≤ n which does not contain the equality symbol by Ln (predicate). This section interprets Turing machines in L2 (predicate) and gives two recursively inseparable sets of sentences arising from Stat(0) and Hlt(0) (Proposition 28.4.4): Definition 28.4.1: Stationary relations. For each positive integer m define πm : N → {1, . . . , m} by π(x) = x for x = 1, . . . , m and π(x) = m for x ≥ m. A k-ary relation R of N is said to be m-stationary if (x1 , . . . , xi , . . . , xk ) ∈ R ⇐⇒ (x1 , . . . , πm (xi ), . . . , xk ) ∈ R for all x1 , . . . , xk ∈ N and each i, 1 ≤ i ≤ k. In other words, if R is mstationary, then (x1 , . . . , xi−1 , j, xi+1 , . . . , xk ) ∈ R for some j ≥ m if and only if the k-tuple is in R for all j ≥ m. Thus, if R is m-stationary, then R is also n-stationary for each n ≥ m. Let L0 be a language that extends a language L and let A = hA, . . .i be a structure for L. Call a sentence θ0 of L0 satisfiable in A if there exist interpretations of the relation symbols of L0 r L as relations of A such that θ0 is true in the corresponding structure A0 for L0 . Consider the structure hN, 1,0 i with 0 as the successor function and let L be the corresponding language. A sentence θ in an extended language L0 is said to be stationarily satisfiable in hN, 1,0 i if there exists a positive integer m such that the relations (excluding the successor function) that interpret the additional relation symbols are m-stationary.  Lemma 28.4.2: To each Turing machine M there effectively corresponds a quantifier-free formula λ(U, X, V, Y ) in L2 (predicate) such that: (a) M 6∈ Hlt(0) if and only if (∀X, Y )λ(1, X, X 0 , Y ) is satisfiable in hN, 1,0 i. (b) M ∈ Stat(0) if and only if (∀X, Y )λ(1, X, X 0 , Y ) is stationarily satisfiable in hN, 1,0 i. Proof: The proof divides into parts consisting of the construction of λ and the proofs of (a) and (b) Part A: Construction of λ. Let M be a Turing machine with the operational modes q1 , . . . , qe and the working functions pr, mv, and md. Define functions r, v, and d from {1, 2, . . . , e} × {0, 1} to {0, 1}, {−1, 0, 1}, and {1, 2, . . . , e}, respectively, as follows: r(i, t) = pr(qi , t); v(i, t) = −1, 0, or 1 as mv(qi , t) = L, S, or R, respectively; and md(qi , t) = qd(i,t) . Then λ(1, X, X 0 , Y ) (whose terms will be explained in Parts B and C) consists of a conjunction of tenWformulas: We (1a) i=1 Qi (X) ∧ i6=j ¬[Qi (X) ∧ Qj (X)]. (1b) [T0 (X, Y ) ∨ T1 (X, Y )] ∧ ¬[T0 (X, Y ) ∧ T1 (X, Y )]. V W1 (1c) j=−1 [Vj (X) ∧ i6=j ¬[Vi (X) ∧ Vj (X)]]. (1d) Q1 (1) ∧ T1 (1, 1) ∧ T0 (1, Y 0 ) ∧ Sc(1, 1) ∧ ¬Sc(1, Y 0 ).

680

Chapter 28. Undecidability

Ve

V1

) ∧ Qi (X) ∧ Tt (X, Y ) → Tr(i,t) (X 0 , Y ) ∧ Vv(i,t) (X) ∧ Qd(i,t) (X 0 )]. (1f) ¬Sc(X, Y ) → [T1 (X 0 , Y ) ↔ T1 (X, Y )]. (1g) V1 (X) → [Sc(X, Y ) ↔ Sc(X 0 , Y 0 )] ∧ ¬Sc(X 0 , 1). (1h) V0 (X) → [Sc(X, Y ) ↔ Sc(X 0 , Y )]. (1i) V−1 (X) → [Sc(X, Y 0 ) ↔ Sc(X 0 , Y )]. (1j) ¬[Sc(X, 1) ∧ V−1 (X)]. (1e)

i=1

t=0 [Sc(X, Y

Replace each occurrence of 1 in λ(1, X, X 0 , Y ) by U and each occurrence of X 0 by V to obtain λ(U, X, V, Y ). Part B: Interpretation of λ and proof of (a). The formula λ derives from the effect of M as applied to the tape T0 . We interpret the relation symbols of λ as relations on N as follows: (2a) Qi (x) holds ⇐⇒ the mode of M at instant x is qi . (2b) Vj (x) holds ⇐⇒ the move command at instant x is L, S, R, respectively, if j = −1, 0, 1. (2c) Tt (x, y) holds ⇐⇒ the tape symbol in the yth square at instant x is t. (2d) Sc(x, y) holds ⇐⇒ M scans the yth square at instant x. For example, (1a) interprets that M is in one and only one of the modes q1 , . . . , qe at any instant and (1b) interprets that the tape symbol in the yth square at instant x is either 0 or 1. Continuing in this way, (1d) interprets that M works on T0 and (1j) interprets that if M scans square 1 at instant x, then the next move command is not L; in other words, M never stops. Thus, if M 6∈ Hlt(0), then λ(1, x, x0 , y) is true for each (x, y) ∈ N × N. Conversely, suppose that (∀X, Y )λ(1, X, X 0 , Y ) is satisfiable in hN, 1,0 i. Then there are relations Qi , Vj , Tt , and Sc on N such that λ(1, x, x0 , y) is true in the corresponding extension of hN, 1,0 i for each x, y ∈ N. We must show that these relations then correspond to the operations of M as given by (2). With this conclude from (1j) that M 6∈ Hlt(0). Indeed, if M starts to work at the instant x = 1 on the tape T0 , then from the validity of (1c), (1d), and (1e) for x = 1 it follows that (2) is true for x = 1 and for every y. For example, since the mode of operation of M at the instant x = 1 is q1 and since Q1 (1) is true (by (1d)), (2a) is true for x = 1. Next it follows from (1d) that Q1 (1)∧T1 (1, 1)∧Sc(1, 1) is true. Hence, Vv(1,1) is true (by (1e)). Thus, by the definition of v, a move command L, S, or R at x = 1 implies that V−1 (1), VV 0 (1), or V1 (1), respectively, is true. The converse follows from this and from i6=j ¬(Vi (1) ∧ Vj (1)) (by (1c)). Similarly verify (2c) and (2d) for x = 1. Now assume by induction that (2) is true for x and for each y. It follows from (1a) and (1e) that (2a) is true for x0 ; from (1g)-(1i) that (2d) is true for x0 and for each y; from (1c) and (1e) that (2b) is true for x0 ; and from (1b), (1e), and (1f) that (2c) is true for x0 , and for each y. Part C: Proof of (b). Suppose that M ∈ Stat(0). Define Qi , Vj , Tt , and Sc by (2). Then M 6∈ Hlt(0) and, therefore (Part B), λ(1, x, x0 , y) is true for all

28.4 The Predicate Calculus

681

(x, y) ∈ N × N. By definition, there exist a, b ∈ N such that M scans square a at each instant x ≥ b and the tape and mode remain unchanged. Since M moves at most one square to the right at each instant, none of the squares greater than b has been scanned until instant b. Thus, a ≤ b. Therefore, the relations Qi , Vj , Tt , and Sc are (b + 1)-stationary. Conversely, suppose Qi , Vj , Tt , and Sc are m-stationary relations and λ(1, x, x0 , y) is true for all (x, y) ∈ N × N. Apply M to the tape T0 at instant x = 1. Then, as in Part B, Qi , Vj , Tj , and Sc satisfy (2). Thus, at the latest, M is stationary from the mth instant on some square a < m. Consequently, M ∈ Stat(0).  The next lemma translates stationary satisfiability into satisfiability in finite models: Lemma 28.4.3: Let λ(U, X, V, Y ) be a quantifier-free formula Ln (predicate) without an occurrence of 0 . Denote the sentence

of

(∃U )(∀X)(∃V )(∀Y )λ(U, X, V, Y ) by θ and the sentence (∀X, Y )λ(1, X, X 0 , Y ) by θ0 . (a) If θ has a model, then θ0 is satisfiable in hN, 1,0 i. (b) If θ0 is stationarily satisfiable in hN, 1,0 i, then θ is true in infinitely many finite models. Proof of (a): Let A = hA, R1 , . . . , Rk i be a model of θ. Then there exists a1 ∈ A and for each x ∈ A there exists s(x) ∈ A such that A |= λ(a1 , x, s(x), y) for all y ∈ A. Inductively define a function π: N → A by π(1) = a1 and π(n0 ) = s(π(n)). With an = π(n), n = 1, 2, 3, . . . consider the set A0 = {an | n ∈ N}. Restriction of the relations of A to A0 gives a model, A0 = hA0 , a1 , s, R01 , . . . , R0k i, of the sentence (∀X, Y )λ(a1 , X, s(X), Y ). For each integer r, if Ri is an r-ary relation on A, define the r-ary relation ¯ i on N by R ¯ i = {(x1 , . . . , xr ) ∈ Nr | (π(x1 ), . . . , π(xr )) ∈ R0i }, i = 1, . . . , k. R ¯1, . . . , R ¯ k i, Induction on structure shows that, with N = hN, 1,0 , R N |= ϕ(x1 , . . . , xn ) ⇐⇒ A0 |= ϕ(π(x1 ), . . . , π(xn )) for each quantifier-free formula ϕ(X1 , . . . , Xn ) and for each (x1 , . . . , xn ) ∈ Nn (Note, however, that since π is not necessarily an injection, this would not hold if ϕ would contain the equality symbol.) In particular, N |= λ(1, x, x0 , y) for each (x, y) ∈ N × N. Proof of (b): Suppose that θ0 is satisfiable in hN, 1,0 i such that the interpretation of each relation symbol of λ(U, X, V, Y ) in N is m-stationary. Define a function s from the set A = {1, 2, . . . , m} into itself by s(x) = x0 for x = 1, 2, . . . , m − 1 and s(m) = m. Let π: N → A be the function π(x) = x

682

Chapter 28. Undecidability

for x = 1, . . . , m and π(x) = m for x ≥ m. Restrict the relations of λ to A. This gives a finite model A in which (∀X, Y )λ(1, X, s(X), Y ) is true. Drop 1 and s from A to get a finite model in which θ is true. Since the relations of λ are also n-stationary for each n ≥ m, this gives infinitely many finite models for θ.  Proposition 28.4.4 (B¨ uchi): The set of sentences of L2 (predicate) which are true in infinitely many finite models and the set of sentences of L2 (predicate) which have no models are recursively inseparable. Proof: For each Turing machine M let λ(U, X, V, Y ) be the quantifier-free formula of L2 (predicate) that Lemma 28.4.2 attaches to M . Let θ and θ0 be as in Lemma 28.4.3. If M ∈ Stat(0), then θ0 is stationarily satisfiable in hN, 1,0 i (Lemma 28.4.2(b)). Hence, θ is true in infinitely many finite models (Lemma 28.4.3(b)). If M ∈ Hlt(0), then θ0 is not satisfiable in hN, 1,0 i (Lemma 28.4.2(a)). Hence, θ has no model (Lemma 28.4.3(a)). Moreover, the map f : code(M ) 7→ code(θ) (Section 8.6) is recursive. In addition, f (Stat(0)) (resp. f (Hlt(0))) is contained in the set of codes of sentences true in infinitely many finite models (resp. with no models). By Lemma 28.3.6, Stat(0) and Hlt(0) are recursively inseparable. Therefore, by Lemma 28.3.2, the Proposition holds. 

28.5 Undecidability in the Theory of Graphs We interpret the language L2 (predicate) in the language of graphs and use B¨ uchi’s theorem to construct recursively inseparable theories in this language. A graph in this chapter is a structure Γ = hA, R i with A a set and R a binary symmetric nonreflexive relation on A. Denote the language of the theory of graphs without equality by L(graph). It contains exactly one binary predicate symbol P . Thus, a graph is a structure for L(graph) which satisfies the axiom (∀X, Y )[¬P (X, X) ∧ [P (X, Y ) ↔ P (Y, X)]] Visually we present the relation R of a graph Γ by a diagram which shows the points of A and an edge between each pair of points that belongs to R. For example, the diagram x

.

y

    z .

.

expresses that (x, y) and (y, z) belongs to R but (x, z) does not. A broken line through points x1 , x2 , . . . , xn of A . _ _ _ . _ _ _ . _ _ _ ··· _ _ _ . x2 x3 xn

x1

signifies that each pair (xi , xj ), with i 6= j, belongs to R.

28.5 Undecidability in the Theory of Graphs

683

Proposition 28.5.1 (Lavrov [Ershov-Lavrov-Taimanov-Taitslin, p. 79]): The set of sentences of L(graph) which are true in infinitely many finite graphs and the set of sentences of L(graph) which have no models are recursively inseparable. Proof: The proof divides into parts consisting of the construction and the verification of properties of a recursive map f from the set of sentences of L2 (predicate) into the set of sentences of L(graph) for which the following holds. If θ belongs to the set of sentences S1 of L2 (predicate) which are true in infinitely many finite models, (resp. sentences S2 which have no model), then f (θ) is true in infinitely many finite graphs (resp. f (θ) has no model). Proposition 28.4.4 implies that S1 and S2 are recursively inseparable. Thus, Lemma 28.3.2 with S1 and S2 , respectively, replacing A and B, implies the proposition. Part A: Definition of f . Let θ be a sentence of L2 (predicate) which involves only unary relation symbols U1 , . . . , Um and binary relation symbols B1 , . . . , Bn . Denote the following formula of L(graph) by µ(X): h m+n+3 ^

(∃Y1 ) · · · (∃Ym+n+3 )

P (X, Yi ) ∧

i=1

^

i P (Yi , Yj ) .

i6=j

Let νi (X) be the formula h i+1 ^

(∃T1 ) · · · (Ti+1 )

^

¬µ(Tj ) ∧ P (X, Tj ) ∧

j=1

P (Tk , Tj )∧

k6=j

¬(∃T )[¬µ(T ) ∧ P (X, T ) ∧

i+1 ^

ii

P (T, Tj )

j=1

i = 1, . . . , m. Finally, let βj (X, Y ) be the formula i h^

(∃U )(∃W )(∃Z1 ) · · · (∃Zi )

P (W, Zj ) ∧ P (X, U ) ∧ P (U, W ) ∧ P (Y, W )

j=1

∧¬µ(U ) ∧ ¬µ(W ) ∧

i ^ j=1

^

¬µ(Zj ) ∧

P (Zj , Zk )

j6=k

i ii h ^ P (Z, Zj ) , ∧¬(∃Z) ¬µ(Z) ∧ P (Z, W ) ∧ j=1

i = 1, . . . , n. Now define a recursive map ϕ 7→ ϕ∗ from formulas of L2 (predicate) to formulas of L(graph) by a structure induction: Ui∗ (X) is νi (X), i = 1, . . . , m,

684

Chapter 28. Undecidability

and Bj∗ (X, Y ) is βj (X, Y ), i = 1, . . . , n. Next make the map commute with negation and conjunction. If ϕ(X, Y ) is (∃Y )ψ(X, Y ) and ψ ∗ (X, Y ) has already been defined, then ϕ∗ (X, Y ) is (∃Y )[µ(Y )∧ψ ∗ (X, Y )]. Finally, define f (θ) for a sentence θ of L2 (predicate) as θ∗ ∧ (∃X)µ(X). Part B: Interpretation of finite models for L2 (predicate) as finite graphs. Suppose that θ is true in infinitely many finite models. Let M = hM, U1 , . . . , Um , B1 , . . . , Bn i be one of these. Without loss assume that |M | > m + n + 3. We construct a finite graph Γ = hA, R i, with |A| ≥ |M |, in which f (θ) is true. For each i, 1 ≤ i ≤ m, and for each x ∈ M which belongs to Ui , add i + 1 new elements, t(x, i, 1), . . . , t(x, i, i + 1), to M . For each j, 1 ≤ j ≤ n, and for each pair (x, y) of elements of M which belongs to Bj , add j + 2 new elements, u(x, y, j), w(x, y, j); z(x, y, j, 1), . . . , z(x, y, j, j). Denote the resulting set by A. Define a symmetric nonreflexive relation R on A. Two elements x, y of A relate to each other exactly when they are connected by an edge in one of the following diagrams: y

x

.

.

(1a) for x 6= y and x, y ∈ M ;

x

kkkw . GGG GG kkkwww k k GG kkk www k GG k G ww kkk k w k . _ _ _ _ . _ _ _ _ ... _ _ _ _ .

(1b)

t(x,i,1)

t(x,i,2)

t(x,i,i+1)

for x ∈ M , x ∈ Ui and i = 1, . . . , m; and (1c)

x .

. u(x,y,j)

y .

. _ _ _ _ . _ _ _ _ ... _ _ _ _ .

w(x,y,j)

z(x,y,j,1)

z(x,y,j,j)

for x, y ∈ M , (x, y) ∈ Bj and j = 1, . . . , n. Let Γ = hA, R i be the resulting graph. In particular, no point of M is connected to itself. Thus, if P (x, y) holds for x, y ∈ M , then x 6= y. It follows that the following statements hold: (2a) x ∈ M ⇐⇒ Γ |= µ(x) (indeed an element of A relates to m + n + 3 elements if and only if it is in M ). (2b) If x ∈ M , then x ∈ Ui ⇐⇒ Γ |= νi (x), i = 1, . . . , m. (2c) If x, y ∈ M , then (x, y) ∈ Bj ⇐⇒ Γ |= βj (x, y), j = 1, . . . , n.

28.5 Undecidability in the Theory of Graphs

685

It follows by induction on structure that for each formula ϕ(X1 , . . . , Xr ) of L2 (predicate) and for all x1 , . . . , xr ∈ M , M |= ϕ(x) if and only if Γ |= ϕ∗ (x). In particular, Γ |= f (θ). Consequently, f (θ) is true in infinitely many finite graphs. Part C: Extracting a model for θ from a graph. Conversely, let Γ = hA, R i be a graph in which f (θ) is true. Then the set M = {x ∈ A | Γ |= µ(x)} is nonempty. Define relations Ui and Bj on M : Ui = {x ∈ M | Γ |= νi (x)}, i = 1, . . . , m : and Bj = {(x, y) ∈ M × M | Γ |= βj (x, y)}, j = 1, . . . , n. The structure M = hM, U1 , . . . , Um , B1 , . . . , Bn i is a model of θ.



Lemma 28.5.2: Let θ be a sentence of L(graph). Suppose θ holds in each infinite graph. Then θ is true in almost all finite graphs. Proof: For each positive integer m there are, up to isomorphism, only finitely many graphs hA, R i with |A| ≤ m. Assume ¬θ holds in infinitely many finite graphs. Then, each nonprincipal ultraproduct of those graphs is an infinite graph in which ¬θ is true. We conclude from this contradiction, that θ is true in almost all finite graphs.  By Definition 28.3.1, if A ⊆ B are subsets of N , then A and N r B are recursively inseparable if and only if there exists no recursive subset C of N with A ⊆ C ⊆ B. We use this observation in the proof of the following result: Corollary 28.5.3: Consider the following lattice of sets of sentences of L(graph). S = all satisfiable sentences I = all sentences true in infinitely many finite graphs AA = the theory of sentences true in almost all finite graphs G = the theory of graphs (a) There is no recursive set between I and S. (b) There is no recursive set between G and AA. In particular, none of the four sets is decidable. Proof of (a): The set of satisfiable sentences is the complement of the set of sentences which are false in each graph. Thus, (a) reformulates Proposition 28.5.1. Proof of (b): The map θ 7→ ¬θ is an injective recursive map from the set of sentences of L(graph) into itself. It maps the set of sentences with no

686

Chapter 28. Undecidability

model onto G, and it maps I onto the set of sentences, I 0 , false in infinitely many finite models. By Lemma 28.3.2 and Proposition 28.5.1, I 0 and G are recursively inseparable. Since AA is the complement of I 0 , (b) follows. If one of the sets in the statement of the corollary is recursive, then it is a recursive set between either I and S, or between G and AA. Since this contradicts either (a) or (b), we are done.  Proposition 28.5.4: The four theories of Corollary 28.5.3 are distinct. Proof (Haran): The parts of the proof consist of introducing the equality relation to L(graph) and the proof that G, AA, I, and S are distinct. Part A: Introducing the equality relation. Let ε(X, Y ) be the formula (∀Z)[P (X, Z) ↔ P (Y, Z)]. For each graph Γ = hA, R i define an equivalence relation on A: x ∼ y ⇐⇒ Γ |= ε(x, y). That is, x is equivalent to y if and only if y is related to exactly the same elements of A as is x. Denote the equivalence class of an element x ∈ A by ¯ Then R ¯ = {(¯ x ¯ and denote the set of all equivalence classes by A. x, y¯) ∈ A¯ × A¯ | (x, y) ∈ R} is a well defined symmetric nonreflexive relation on A. ¯ = hA, ¯ Ri ¯ is a graph and Therefore, Γ ¯ |= P (¯ Γ |= P (x, y) ⇐⇒ Γ x, y¯) for each x, y ∈ A. An induction on structure shows that ¯ |= ϕ(¯ Γ |= ϕ(x1 , . . . , xn ) ⇐⇒ Γ x1 , . . . , x ¯n ) for each formula ϕ(X1 , . . . , Xn ) of L(graph). In particular, Γ is elementarily ¯ If Γ is a finite graph, then so is Γ ¯ and |A| ¯ ≤ |A|. Moreover equivalent to Γ. ¯ |= ε(a, b) ⇐⇒ a = b Γ ¯ That is, the equivalence relation in Γ ¯ is the equality for each a, b ∈ A. relation. Part B: G 6= AA.

Consider the sentence θ

(∃X1 )[(∃X2 )P (X1 , X2 ) ∧ (∀Y )[P (X1 , Y ) ↔ ε(X2 , Y )] ∧(∀Z2 )[¬ε(X1 , Z2 ) → (∃Z1 , Z3 )[P (Z1 , Z2 ) ∧ P (Z2 , Z3 ) ∧ ¬ε(Z1 , Z3 ) ∧(∀Y )[P (Z2 , Y ) → ε(Y, Z1 ) ∨ ε(Y, Z3 )]]]]. That is, θ interprets the existence of an element x1 which relates (up to equivalence) to exactly one element, and each element z2 not equivalent to x1

28.6 Assigning Graphs to Profinite Groups

687

relates (up to equivalence) to exactly two nonequivalent elements. Replace Γ ¯ to assume that each equivalence class of Γ contains exactly one element. by Γ Suppose that Γ is a model for θ. We do an induction to show, for each integer n ≥ 0, that A has n + 1 distinct elements x1 , . . . , xn+1 which relate according to the following diagram (3)

.x1

.x2

···

xn−1

.

xn

.

xn+1

.

Let x1 ∈ A be related to exactly one element. Apply the induction assumption to assume that diagram (3) exists without the point xn+1 . But, since xi is related to xi−1 and to xi+1 , for i = 2, . . . , n−1, none of x1 , . . . , xn−2 are related to xn . Hence, xn is related to a new element xn+1 , which gives (3). In particular, A is an infinite set. It follows that ¬θ is true in all finite graphs. Also, θ is true in the graph hN, R i, where R = {(n, n + 1) | n ∈ N}. Consequently, G 6= AA. Part C: AA 6= I. The sentence (∀X, Y )[¬P (X, Y )] interprets the triviality of the graph (i.e. no two points are related). There are infinitely many trivial finite graphs and there are infinitely many nontrivial finite graphs. Hence, AA6=I. Part D: I 6= S. The sentence θ of Part B is satisfiable but false in each finite graph.  Remark 28.5.5: Let IG be the theory of infinite graphs and FG the theory of finite graphs. The arguments of Part A of the proof of Proposition 28.5.4 show that IG = G. Indeed to each point x of a finite graph Γ we can add infinitely many points, each of which is equivalent to x. The new graph so obtained is infinite and elementarily equivalent to Γ. Similarly, each finite graph is elementarily equivalent to an arbitrarily large finite graph. Thus AA = FG. In particular, by Corollary 28.5.3, both IG and FG are undecidable theories. Finally, each finite graph Γ has a recursive theory Th(Γ). The arguments of the preceeding paragraph show that FG = AA ⊂ Th(Γ) ⊂ I. This settles Problem 33 of [Fried-Jarden5] (see also [Fried-Jarden5, Remark 28.5.5]). 

28.6 Assigning Graphs to Profinite Groups In the following sections we interpret the theory of graphs in the theory of perfect PAC fields. With this we prove the undecidability of the latter theory. The interpretation goes through several stages. This section uses a pair of finite groups, (D, W ), as an auxiliary parameter, to assign to each profinite group G a graph ΓG . Sections 28.7 and 28.8 develop conditions that guarantee that a given graph Γ equals ΓG for some profinite group G. Section 28.9 shows that under these conditions a given graph Γ is ΓGal(K) for some perfect PAC field K. And, finally, Section 28.10 concludes the undecidability of the theory of perfect PAC fields as an application of Proposition 28.8.3.

688

Chapter 28. Undecidability

Fix a pair (D, W ) of finite groups with D nontrivial. For each profinite group G define the graph ΓG = (AG , RG ) as follows: AG is the set of open normal subgroups N of G with G/N ∼ = D; and RG is the set of pairs (N1 , N2 ) ∈ AG × AG such that N1 N2 = G and there exists an open normal subgroup M of G with M ≤ N1 ∩ N2 and G/M ∼ = W. Note that if RG is nonempty there must be a sequence G/M → G/N1 ∩ N2 → G/N1 × G/N2 that induces an epimorphism W → D × D. Clearly RG is a symmetric nonreflexive relation on AG . Hence, if AG is nonempty, then ΓG = hAG , RG i is a graph. This notation suppresses the dependence of ΓG on (D, W ). Lemma 28.6.1: Suppose the respective Frattini subgroups, Φ(D) and Φ(W ), of D and W are trivial. Assume also that π: H → G is a Frattini cover of profinite groups. If ΓG is nonempty, then ΓH is nonempty, and ΓG ∼ = ΓH . In ˜ is the universal Frattini cover of G, then Γ ˜ ∼ . particular, if G Γ = G G Proof: From the isomorphisms H/π −1 (N ) ∼ = G/N for open normal subgroups N of G, the map N 7→ π −1 (N ) maps AG injectively into AH . If (N1 , N2 ) ∈ RG , then (π −1 (N1 ), π −1 (N2 )) ∈ RH . Conversely, if M is an open normal subgroup of H and either H/M ∼ =D or H/M ∼ = W , then Ker(π) ≤ Φ(H) ≤ M (Lemma 22.1.4(a)). Hence, M = π −1 (π(M )). Thus, the map AG → AH induces an isomorphism of graphs. 

28.7 The Graph Conditions We develop sufficient conditions on the finite groups D and W that guarantee the surjectivity of the map G 7→ ΓG of profinite groups to graphs. For D a finite group and I any set regard the collection of all functions from I to D as a profinite group, denoted DI (Lemma 1.2.6). Each d ∈ DI represents a vector whose ith coordinate is di , and multiplication is defined componentwise. For each i ∈ I let πi : DI → D be projection on the ith coordinate (i.e. πi (d) = di ). The kernel of πi is Di0 = {d ∈ DI | di = 1}, and with Di = {d ∈ DI | dj = 1 for each j 6= i}, DI = Di × Di0 . In particular, for I = {1, 2}, D × D = DI = D1 × D2 . Recall that every finite group G has a composition series: a sequence of subgroups G = G0 . G1 . · · · . Gr = 1 such that the quotient Gi /Gi+1 is a simple group, i = 0, . . . , r − 1. The Jordan-H¨older theorem states that the length and the set of quotients of a composition series are invariants of G [Huppert, p. 63]. The quotient groups are the composition factors of G. Consider two finite groups D and U such that D 6= 1 and D × D acts on U as a group of automorphism. Let W = (D × D) n U be the semidirect product of U and D × D (Definition 13.7.1) and the associated short split exact sequence: (1)

1

/U

/W o

λ θ

/

D×D

/ 1,

28.7 The Graph Conditions

689

We assume the following properties for D and U : (2a) D and U have no composition factor in common. (2b) For each finite set I and each epimorphism π: DI → D there exists i ∈ I such that Ker(π) = Ker(πi ). (2c) Φ(W ) = Φ(D) = 1. (2d) For each embedding θ0 of D × D into W as a semidirect complement of U (i.e. U · θ0 (D × D) = W and U ∩ θ0 (D × D) = 1) and for each nontrivial normal subgroup N of U none of the factors D1 and D2 acts trivially (via conjugation) on N . Remark 28.7.1: If I is an arbitrary set and π: DI → D is a continuous epimorphism, then there exists a finite subset J of I and an epimorphism ¯ ◦ πJ , where πJ : DI → DJ is the suitable π ¯ : DJ → D such that π = π coordinate projection. By (2b) there exists i ∈ J such that Ker(π) = Ker(πi ).  Definition 28.7.2: A pair of finite groups (D, W ) is said to satisfy the graph conditions if W can be factored as a semidirect product (D × D) n U such that Condition (2) holds.  The following two examples provide the reader with some practice in the mechanics of the graph conditions. Example 28.7.3: [Ershov4]. Let D be a simple non-Abelian group and let H be a finite group whose composition factors are non-Abelian and distinct from D. The action of D × D on U = H D×D is given by permutation of the coordinates: (3)

0

(hδ )δ = hδ0 δ

for h ∈ U ;

δ, δ 0 ∈ D × D.

The semidirect product W = U o (D × D) is the wreath product of D × D with H (Remark 13.7.7). Condition (2a) is immediate. Condition (2b) follows from Lemma 16.8.3(a). Proof of (2c): Since D is simple, Φ(D) = 1. In the notation of (1), Lemma 22.1.4 implies that λ(Φ(W )) ≤ Φ(D×D) = Φ(D)×Φ(D) = 1. Hence, Φ(W )/ U . By Lemma 22.1.2, however, Φ(W ) is nilpotent. If Φ(W ) were nontrivial, U (and therefore H) would have a cyclic composition factor, contrary to the assumptions. Proof of (2d): Suppose that D × D is a subgroup of W in some way. Let N be a nontrivial normal subgroup of U . With no loss assume that N is a minimal normal subgroup of U . We show that if δ ∈ D × D normalizes N , then δ = 1. This will imply (2d). By Lemma 18.3.9, N is equal to one of the groups Hδ0 = {h ∈ U | hδ00 = 1 for all δ 00 6= δ 0 }. with δ 0 ∈ D × D. By assumption, Hδ0 = Hδδ0 = Hδ0 δ−1 . Hence, δ 0 = δ 0 δ −1 and δ = 1, as claimed. 

690

Chapter 28. Undecidability

Example 28.7.4: [Cherlin-v.d.Dries-Macintyre]. Let p and q be distinct odd primes. Consider the dihedral group D = Dp of order 2p generated by β of order 2 and γ of order p with the relation βγβ = γ −1 . Then α = γβ is of order 2. Let τ be the unique epimorphism of D (with Ker(τ ) generated by γ) onto the multiplicative group {±1} so that τ (β) = −1. Let U be the multiplicative cyclic group of order q and let D × D operate on U by the rule (4)

u(x,y) = uτ (x)τ (y) ,

for

x, y ∈ D

and

u ∈ U.

The composition factors of Dp , namely Z/2Z and Z/pZ, are distinct from Z/qZ, the unique composition factor of U . This gives (2a). Proof of (2b): Let I be a finite set and let π: DI → D be an epimorphism. Denote the elements of the ith factor Di of DI that correspond to α, β, and γ by αi , βi , and γi = αi βi . If there exists j ∈ I such that Dj ≤ Ker(π), then consider π as an epimorphism from DI r{j} onto D and apply an induction to prove the existence of i ∈ I r{j} such that Ker(π) = Ker(πi ). Otherwise, for each i ∈ I, Ker(π) ∩ Di , as a proper normal subgroup of Di is either 1 or hγi i. If Ker(π)∩Di = hγi i for all i ∈ I, then D is generated by pairwise commuting copies of Z/2Z. Hence, D is Abelian, a contradiction. If Ker(π) ∩ Di = 1 for some i ∈ I, say for i = 1, then π(D1 ) = D. Hence, for each i ∈ I r{1}, each element of π(Di ) commutes with each element of D. That is, the center of D is nontrivial, a contradiction unless I = {1}. This proves (2b). Proof of (2c): Both hαi and hβi have index p in D. Hence, Φ(D) ≤ hαi ∩ hβi = 1, so Φ(D) = 1. Therefore, Φ(W ) ≤ U . By (1), θ(D × D) has index q in W . Thus, Φ(W ) ≤ U ∩ θ(D × D) = 1. Proof of (2d): Let θ0 be an embedding of D × D into W as a semidirect complement of U . Since the orders of U and D × D are relatively prime, there is a u ∈ U with θ0 (D × D) = θ(D × D)u (Schur-Zassenhaus - Lemma 22.10.1). By (2b), D × D has a unique factorization as a direct product of two copies of D. Thus, θ0 (D × D) = θ(D)u × θ(D)u . Take x ∈ D such that τ (x) = −1 and v ∈ U , v 6= 1. Since q is odd, u

v θ(x) = v u

−1

θ(x)u

= v τ (x)u = (v −1 )u = v −1 6= v.

Hence, none of the direct factors of θ0 (D × D) acts trivially on U .



28.8 Assigning Profinite Groups to Graphs Suppose (D, W ) is a pair of finite groups that satisfy the graph conditions. In particular, there is a split short exact sequence (1)

1

/U

/W o

λ θ

/

D×D

/ 1.

28.8 Assigning Profinite Groups to Graphs

691

Let Γ = hA, Ri. We construct a projective group G such that ΓG ∼ = Γ, with ΓG = hAG , RG i defined in Section 28.6. To that end consider the profinite group DA × W R whose elements are pairs (d, w) with d ∈ DA and w ∈ W R . Consider, also the following coordinate projections πA : DA × W R → DA , πA (d, w) = d; πa (d, w) = da , for a ∈ A; πa : DA × W R → D, πr (d, w) = wr , for r ∈ R. πr : DA × W R → W,

and

The subgroup G = {(d, w) ∈ DA × W R | r = (a, b) ∈ R =⇒ λ(wr ) = (da , db )}, of DA ×W R is closed, and therefore profinite. Moreover, if Γ is a finite graph, then G is a finite group. Lemma 28.8.1: πA (G) = DA and Ker(πA ) ∩ G = U R . Proof: Indeed, if d ∈ DA and r = (a, b) ∈ R, choose wr ∈ W such that λ(wr ) = (da , db ). Then (d, w) defined by this procedure belongs to G and πA (d, w) = d. Thus, πA (G) = DA . Suppose that u ∈ U R . Then λ(ur ) = (1, 1) for each r ∈ R. Hence, (1, u) ∈ Ker(πA ) ∩ G. Conversely, if (d, w) ∈ Ker(πA ) ∩ G, then d = 1 and  λ(wr ) = (1, 1) for each r ∈ R. Thus, Ker(πA ∩ G) = U R . Lemma 28.8.1 shows that the sequence (2)

π0

A DA → 1, 1 → U R → G −→

0 = πA |G , is exact. with πA

Lemma 28.8.2: There exists a continuous homomorphism η: DA → G such 0 ◦ η = Id. (i.e. (2) splits). Also, if a0 ∈ A, r = (a, b) ∈ R and that πA 0 a 6∈ {a, b}, then Da0 , acts trivially on Ur through η. Proof: If d ∈ DA , define η by η(d) = (d, w), where wr = θ(da , db ) for each r = (a, b) ∈ R. Now assume that a0 and r = (a, b) satisfy the hypotheses of the lemma. If d ∈ Da0 , then da = db = 1. Hence, wr = θ(da , db ) = 1. Thus, η(d) = (d, w) commutes with each element of Ur .  Proposition 28.8.3: The graphs hA, Ri and hAG , RG i are isomorphic. Proof: We present the construction and properties of a map from hA, Ri into hAG , RG i in three parts:

692

Chapter 28. Undecidability

Part A: Map of A onto AG . For each a ∈ A put πa0 = πa |G and N (a) = Ker(πa0 ). Then the following diagram commutes: (3)

G@ @@ @@ @@ 0 πa @

0 πA

D

/ DA | | || || πa | } |

By (2), πa0 is surjective. Hence, G/N (a) ∼ = D and N (a) ∈ AG . We regard 0 (Ker(πa0 )) = N as a map from A into AG . By the commutativity of (3), πA Ker(πa ). Since Ker(πa ) = Ker(πb ) implies that a = b, N is injective. Now we show surjectivity of N . Consider an open normal subgroup N of G with G/N ∼ = D and let 0 π : G → D be the corresponding epimorphism. Since the composition factors of π 0 (U R ) are among the compositions factors of U , graph condition (2a) of Section 28.7 implies that U R ≤ Ker(π 0 ) = N . By (2) there exists an 0 epimorphism π: DA → D such that π 0 = π ◦ πA . By Remark 28.7.1, there is an a ∈ A with Ker(π) = Ker(πa ). Therefore, N = Ker(πa0 ) = N (a). Part B: Map of R into RG . (3) and by (2),

Let r = (a, b) ∈ R. Then a 6= b. Hence, by

N (a)N (b) = Ker(πa0 )Ker(πb0 ) 0 −1 0 −1 0 −1 = (πA ) (Ker(πa ))(πA ) (Ker(πb )) = (πA ) (DA ) = G.

In addition, with πr0 = πr |G , the diagram 0

πr /W G FF w FF w w FF ww 0 (πa ,πb0 ) FF ww λ F# w {w D×D

commutes. Thus, Ker(πr0 ) ≤ N (a) ∩ N (b). Therefore, if we show that πr0 is surjective, then G/Ker(πr0 ) ∼ = W and (N (a), N (b)) ∈ RG . It will follow that the map r 7→ (N (a), N (b)) from R into RG is well defined. Indeed, let wr ∈ W and let λ(wr ) = (da , db ). For each a0 ∈ A distinct from both a and b, choose any element da0 ∈ D. Also, for r0 = (a0 , b0 ) ∈ R, r0 6= r, choose any element wr0 ∈ W such that λ(wr0 ) = (da0 , db0 ). This defines (d, w) in G for which πr (d, w) = wr . Thus, πr0 is surjective. If r0 = (a0 , b0 ) is an element of R with (N (a), N (b)) = (N (a0 ), N (b0 )), then N (a) = N (a0 ) and N (b) = N (b0 ). By Part A, a = a0 and b = b0 . Hence, R maps injectively into RG .

28.9 Assigning Fields to Graphs

693

Part C: R maps surjectively onto RG . Let a, b ∈ A with (N (a), N (b)) ∈ RG Then N (a)N (b) = G and there exists an open normal subgroup M of G such that M ≤ N (a) ∩ N (b) with G/M ∼ = W . This gives a commutative diagram of epimorphisms 0

π /W G FF w FF w w FF ww 0 (πa ,πb0 ) FF ww λ0 F# {ww D×D

The remaining parts of the proof show that (a, b) is in R: Part C1: Ker(λ0 ) = U and π 0 (U R ) = U . Indeed, the composition factors of λ0 (U ) are among the composition factors of U . But, since λ0 (U ) is a subgroup of D × D, these are among the composition factors of D. By graph condition (2a) of Section 28.7, λ0 (U ) = 1. Hence, Ker(λ) ≤ Ker(λ0 ). In addition, |Ker(λ)| = |W |/|D × D| = |Ker(λ0 )|. Therefore, Ker(λ0 ) = U . The same argument shows that λ0 (π 0 (U R )) = 1. Hence, π 0 (U R ) ≤ Ker(λ0 ) = U . Therefore, U/π 0 (U R ) is a quotient of (π 0 )−1 (U )/U R , hence of DA (by (2)). Thus, the decomposition factors of U/π 0 (U R ) are among those of both U and D. By the graph condition (2a) of Section 28.7, U = π 0 (U R ). Part C2: An element r ∈ R. Part C1 gives r ∈ R such that the normal subgroup U 0 = π 0 (Ur ) of U is nontrivial. In addition, Part C1 gives a commutative diagram

(4)

1

/ UR π0

1

 /U

0

πA /G / DA GGG G 0 0 (πa ,πG π0 b) GG (πa ,πb ) #   /W / D×D λ0

/1 /1

Note that (πa , πb ) maps Da × Db isomorphically onto D × D. Hence, the 0 (Lemma 28.8.2) induces a section θ0 : D × D → W section η: DA → G of πA 0 0 of λ . In particular, θ embeds D × D into W as a semidirect complement of U. Part C3: (a, b) ∈ R. Write r as (a0 , b0 ). If a 6∈ {a0 , b0 }, then Lemma 28.8.2 shows that Da acts trivially on Ur . By (4), the first factor D1 of D × D acts (via θ0 ) trivially on U 0 . This contradicts graph condition (2d) of Section 28.7. Therefore, a ∈ {a0 , b0 }. Similarly b ∈ {a0 , b0 }, so r = (a, b) or r = (b, a). In either case (a, b) ∈ R. 

694

Chapter 28. Undecidability

28.9 Assigning Fields to Graphs Again, let (D, W ) be a pair of finite groups with D nontrivial and let K be a field. We define AK to be the set of all Galois extensions L/K with Gal(L/K) ∼ = D and RK to be the set of pairs (L1 , L2 ) ∈ AK × AK with L1 ∩ L2 = K for which there exists a Galois extension N/K with L1 L2 ⊆ N and Gal(N/K) ∼ = W . If AK is nonempty, then the structure ΓK = hAK , RK i is a graph, and the map L 7→ Gal(L) is an isomorphism of ΓK onto ΓGal(K) . Lemma 28.9.1: If H is a nontrivial finite group, then rank(H n ) ≥

log n . log |H| n

Proof: The number of epimorphisms of H n onto H is at most |H|rank(H ) . On the other hand, each coordinate projection of H n is an epimorphism. n  Hence, n ≤ |H|rank(H ) . Apply log to both sides for the result. Note that the proof of Lemma 28.8.3 does not use graph condition (2c) of Section 28.7. It is this condition however that assures that a perfect PAC field can be assigned to a graph. Proposition 28.9.2: Let (D, W ) be a pair of finite groups with D 6= 1 that satisfies the graph conditions. Then, for each graph Γ = hA, Ri there exists a perfect PAC field K such that Γ ∼ = ΓK . If Γ is finite, then corank(K) < ∞. Furthermore rank(Gal(K)) ≥ rank(DA ) ≥ log |A|/ log |D|. Proof: Lemma 28.8.3 constructs a profinite group G such that Γ ∼ = ΓG . ˜ the universal Frattini cover of G, Lemma 28.6.1 shows that Γ ˜ ∼ With G G = ΓG . ˜ Apply Corollary 23.1.2 to produce a perfect PAC field K with Gal(K) ∼ = G. ∼ Then, Γ = ΓK . If Γ is a finite graph, then G above is finite. By Corollary 22.5.3, rank(Gal(K)) = rank(G) < ∞. By (2) of Section 28.8, rank(G) ≥ rank(DA ). Now Lemma 28.9.1 gives the result.  Remark 28.9.3: The field K in Proposition 28.9.2 may be chosen to contain any given field K0 . Thus, the undecidability results of Section 28.10 generalize to the appropriate theories. 

28.10 Interpretation of the Theory of Graphs in the Theory of Fields Let (D, W ) be a pair of finite groups with |D| = m and |W | = n. We attach a formula ϕ0 of L(ring) to each formula ϕ of L(graph). For l an integer let fX (T ) = T l + X1 T l−1 + · · · + Xl . If K is a field and a ∈ K l , denote the splitting field of fa (T ) over K by Ka . Let H be a finite group. Use Remark 20.4.5(d) to effectively construct a formula αl,H (X) of L(ring) such that for each field K and each x ∈ K l , αl,H (x) holds in

28.10 Interpretation of the Theory of Graphsin the Theory of Fields

695

K if and only if fx (T ) is separable and Gal(fx (T ), K) ∼ = H. Use α for l = m, n to construct a formula π(X, Y, Z) of L(ring) such that for (x, y, z) ∈ K m × K m × K n , π(x, y, z) is true in K if and only if Kx ∩ Ky = K, Gal(Kx /K) ∼ = Gal(Ky /K) ∼ = D, Gal(Kz /K) ∼ = W , and Kx , Ky ⊆ Kz . 0 Now define ϕ by induction on the structure of ϕ: R(X, Y )0 = (∃Z)π(X, Y, Z), [¬ϕ]0 = ¬[ϕ0 ], 0 0 0 and [(∃X)ϕ] = (∃X)[αm,D (X) ∧ ϕ0 ], [ϕ1 ∨ ϕ2 ] = ϕ1 ∨ ϕ2 0

where X = (X1 , . . . , Xm ), Y = (Y1 , . . . , Ym ), and Z = (Z1 , . . . , Zn ). It follows that for each formula ϕ(X1 , . . . , Xk ) in L(graph), for each field K with AK 6= ∅, and for all a1 , . . . , ak ∈ K m we have: K |= ϕ0 (a1 , . . . , ak ) if and only if Ka1 , . . . , Kak ∈ AK and ΓK |= ϕ(Ka1 , . . . , Kak ). In particular, if θ is a sentence of L(graph): ΓK |= θ ⇐⇒ K |= θ0 .

(1)

Finally, denote the sentence (∃X)αm,D (X) ∧ θ0 by θ00 . Then, for each field K:

(2)

AK 6= ∅

and

ΓK |= θ ⇐⇒ K |= θ00 .

Theorem 28.10.1: Let Q be the set of sentences θ of L(ring) for which θ is true in at least one perfect PAC field of finite corank exceeding e for each positive integer e. Let P be the set of sentences θ of L(ring) such that θ is false in each field. Then P and Q are recursively inseparable. Proof: The map θ 7→ θ00 of sentences of L(graph) to sentences of L(ring) is primitive recursive. Suppose θ is true in infinitely many finite graphs Γ1 , Γ2 , . . . . Then limi→∞ |Γi | = ∞. Let Ki be the perfect PAC of finite corank that corresponds to Γi by Proposition 28.9.2, i = 1, 2, . . . . Apply (2) to see that θ 00 is true in each of the fields K1 , K2 , . . . . Suppose on the other hand, that θ is false in each graph. If there exists a field K in which θ00 is true, then, by (2), θ is true in ΓK , a contradiction. Thus, θ00 is false in each field. The theorem now follows from Lemma 28.3.2 and Lavrov’s theorem (Proposition 28.5.1). 

696

Chapter 28. Undecidability

Corollary 28.10.2: Consider the following diagram of sets of sentences of L(ring): S = all sentences satisfiable in fields SPAC = all sentences satisfiable in perfect PAC fields SFC = all sentences satisfiable in perfect PAC fields of finite corank FC = The theory of perfect PAC fields of finite corank PAC = The theory of perfect PAC fields F = The theory of fields (a) There is no recursive set between SFC and S. (b) There is no recursive set between F and FC. In particular, none of the six sets is decidable. Proof: Let θ be a sentence of L(ring) which is true in a perfect PAC field K of finite corank. Then, for each integer e, θ is true in a perfect PAC field of finite corank exceeding e. Otherwise, there exists e0 such that ¬θ is true in each perfect PAC field of finite corank exceeding e0 . But Theorem 23.1.6(b) shows that ¬θ is true in each perfect PAC field of finite corank, a contradiction. Thus, SFC is equal to the set Q of Theorem 28.10.1. Since S is the complement of the set of sentences of L(ring) which are false in each field, (a) follows from Theorem 28.10.1. Statement (b) follows from (a) by taking negations and complements.  Remark 28.10.3: The six sets of Corollary 28.10.2 are distinct. Indeed, by Proposition 28.5.4, there exists a sentence θ of L(graph) which is true in all finite graphs but is false in an infinite graph Γ0 . If K is a perfect PAC field of finite corank, then Lemma 16.10.2 implies that K has only finitely many Galois extensions L with Gal(L/K) ∼ = D. It follows that AK is a finite set. If AK is empty, then K |= ¬(∃X)αm,D (X). If however AK 6= ∅, then, by (1), K |= θ0 . In both cases K |= [(∃X)αm,D (X)] → θ0 . On the other hand let K 0 be a perfect PAC field with Γ0 ∼ = ΓK 0 . Then K 0 |= (∃X)αm,D (X). But, by (1), K 0 6|= θ0 , so K 0 6|= [(∃X)αm,D (X)] → θ0 . This shows that PAC and FC are distinct. We leave to the reader the remainder of the remark (Exercise 7).  Remark 28.10.4: The theory of algebraically closed fields of characteristic 0 is recursive and it lies between the sets FC and SFC of Corollary 28.10.2. 

Notes

697

Exercises 1. Construct a Turing machine M such that when M is applied to a 2-track tape T with 1 and n, respectively, stored in the 2-tracks 1 and 2, it ends with a 3-track tape T 0 with 1, n, and 0, respectively, stored in tracks 1, 2, and 3. 2. Use Exercise 1 of Chapter 19 to show that the codes of Turing machines (Section 28.1) form a primitive recursive subset of N. 3. (a) Draw an inst-mode diagram, similar to that of Part B of the proof of Proposition 28.2.1 for the coordinate projection function f (x1 , x2 , x3 ) = x2 as applied to (2, 3, 4). (b) Draw an inst-mode diagram for the computation of f (x1 , x2 , x3 ) = x2 + 1 as applied to (2, 3, 4) using Part E of the proof of Proposition 28.2.1. 4. Draw an inst-mode diagram to compute the function f (x, y) = xy at x = y = 3 by induction from f0 (x) = 2x and g(x, y, z) = y + z as in Part F of the proof of Proposition 28.2.1. 5. Let D = Z/pZ, W = Z/pZ × Z/pZ, and G = Z3p , where p is a prime. Describe the graph ΓG of Section 28.6. 6.

In this problem we consider graph condition (2b) of Section 28.7: (a) Suppose that D is generated by nontrivial groups, A1 , . . . , Ae with e > 1, having these properties: Ai is a quotient of D, i = 1, . . . , e; and Ai centralizes Aj (as subgroups of D) for 1 ≤ i 6= j ≤ e. With I = {1, 2, . . . , e}, find an epimorphism π: DI → D such that Ker(π) 6= Ker(πi ), i = 1, . . . , e. (b) Use the argument of Examples 28.7.3 and 28.7.4 to show that D satisfies graph condition (2b) if and only if there exist no groups A1 , . . . , Ae , with e > 1, satisfying the conditions of (a). 7. Finish the proof of Remark 28.10.3 that the 6 sets of Corollary 28.10.2 are distinct.

Notes Sections 28.1-28.5 are based on [Ershov-Lavrov-Taimanov-Taitslin]. The interpretation of the theory of graphs in the theory of perfect PAC fields is due independently to [Cherlin-v.d.Dries-Macintyre] and to [Ershov4]. Both sources use Frattini covers. Corollary 28.5.3 of [Fried-Jarden5] erroneously claims that the theories G, FG, IG, and AA of Section 28.5 are distinct. Indeed, Part C of that corollary incorrectly claims that the sentence (∃X, Y )[P (X, Y )] is true in all infinite graphs, while it is false in each graph with no edges. The correct relations are G = IG ⊂ F G = AA, as indicated in Remark 28.5.5 of the present edition. We are indebted to Eric Rosen for pointing out this error.

Chapter 29. Algebraically Closed Fields with Distinguished Automorphisms Let K be an explicitly given countable Hilbertian field and e a positive integer. Denote the theory of all sentences θ of the language L(ring, K) which ˜ hold in K(σ) for almost all σ ∈ Gal(K)e by Almost(K, e). By Theorem 20.6.7, Almost(K, e) is decidable. Moreover, for each sentence θ of L(ring, K) ˜ ˜ let Truth(θ) = {σ ∈ K(σ) | K(σ) |= θ} and let Prob(θ) be the Haar measure of Truth(θ). Then Prob(θ) is a rational number between 0 and 1 which can be computed if θ is explicitly given. In this chapter we add e unary function symbols Σ1 , . . . , Σe to L(ring, K) and denote the resulting language by L(ring, K, Σ1 , . . . , Σe ). We prove that, in contrast to the decidability result of the preceding paragraph, the theory ˜ σ1 , . . . , σe i for of all sentences θ of L(ring, K, Σ1 , . . . , Σe ) which hold in hK, e almost all σ ∈ Gal(K) is undecidable (Proposition 29.2.3). Moreover, as θ ranges over all sentences of L(ring, K, Σ1 , . . . , Σe ), the probability Prob(θ) ˜ σ1 , . . . , σe i ranges over all definable real numbers between that θ holds in hK, 0 and 1 (Theorems 29.4.5 and 29.4.6). In particular, Prob(θ) takes also transcendental values.

29.1 The Base Field K Throughout this chapter we work over a fixed infinite base field K, finitely generated over its prime field. By Theorem 13.4.2, K is Hilbertian. ˜ We summarize some properties of almost all fields K(σ) for e ≥ 2 that eventually lead to our undecidability results. We denote the group of roots of unity of a field F by U (F ). Proposition 29.1.1: The following statements hold for every integer e ≥ 2 and almost all σ ∈ Gal(K)e : ˜ (a) K(σ) is PAC. ∼ ˜ (b) Gal(K(σ)) = Fˆe . ˜ (c) U (K(σ)) is a finite group. ˜ ˜ ˜ | Ve σi z = z ∧ (∃α ∈ K)[α 6= 0 ∧ σ1 α = zα]}. (d) U (K(σ)) = {z ∈ K i=1 In addition, for every positive integer n the measure of the set of all σ ∈ ˜ Gal(K)e with |U (K(σ))| ≥ n is positive. Proof (a) and (b): Statements (a) and (b) are special cases of Theorem 20.5.1. Statement (c) repeats Theorem 18.11.7. ˜ To prove (d), let L = K(σ). Consider an element z of the right hand side of (d). Let α be as above and denote the degree of the Galois closure of L(α) over L by n. Then σ1 α = zα. Hence, α = σ1n α = z n α. Therefore, z n = 1.

29.1 The Base Field K

699

For the converse, we may assume that (b) and (c) hold. Let z be an element of U (L). Denote the order of z by n. Since Gal(L) is free, L has a cyclic extension N of degree n. By Kummer theory, N is generated over L by a nonzero element α satisfying αn ∈ L. Then σ1 α = ζn α for some primitive root of unity ζn ∈ L and z = ζni for some i. Hence, σ1 αi = zαi . Thus, z belongs to the right hand side of (d). The last part of the proposition follows from the fact that Gal(K(ζn ))e has positive measure.  Lemma 29.1.2 ([Duret, 4.3 and 5.2]): Let a1 , . . . , ak , b1 , . . . , bl be distinct elements of a field L. (a) Let n be a positive integer satisfying char(L) - n and let c be a nonzero element of L. Then, the algebraic subset V of A1+k+l defined by the system of equations X + ai = Yin ,

i = 1, . . . , k;

X + bj = cZjn ,

j = 1, . . . , l

is an absolutely irreducible curve which is defined over L. (b) Suppose char(L) = p > 0 and a1 , . . . , ak , b1 , . . . , bl are linearly independent over Fp . Let c ∈ L. Then the algebraic subset V of A1+k+l defined by the system of equations ai X = Yip − Yi , i = 1, . . . , k bj X + c = Yjp − Yi , j = 1, . . . , l is an absolutely irreducible curve which is defined over L. Proof of (a): Let x be an indeterminate. Choose algebraic elements yi , zj over L(x) with x+ai = yin and x+bj = czjn for all i and j. Since x+a1 , . . . , x+ ˜ they are multiplicaak , x + b1 , . . . , x + bl are distinct prime elements of L[x], × n ˜ tively linearly independent modulo (L(x) ) . Thus, by Kummer theory, ∼ ˜ y, z)/L(x)) ˜ Gal(L(x, = (Z/nZ)k+l [Lang7, p. 295, Thm. 8.2]. Hence, Yin − ˜ y1 , . . . , yi−1 ) for i = 1, . . . , k (x + ai ) is an irreducible polynomial over L(x, n ˜ y, z1 , . . . , zj−1 ) for and cZj − (x + bj ) is an irreducible polynomial over L(x, ˜ then the L-specialization ˜ j = 1, . . . , l. Therefore, if (ξ, η, ζ) ∈ V (L), x→ξ ˜ can be successively extended to an L-specialization (x, y, z) → (ξ, η, ζ). It follows that V is an absolutely irreducible curve with generic point (x, y, z). ˜ y, z) : L(x)] ˜ ≤ [L(x, y, z) : L(x)] ≤ nk+l . Finally, note that nk+1 = [L(x, ˜ Hence, L(x, y, z) and L(x) are linearly disjoint over L(x), so L(x, y, z) is a regular extension of L. Consequently, V is defined over L (Corollary 10.2.2). Proof of (b): Replace Kummer theory by Artin-Schreier theory [Lang7, p. 296, Thm. 8.3]. Note that the assumption about the linear independence ˜ = over Fp of a1 , . . . , ak , b1 , . . . , bl implies that the additive group ℘(L(x)) p k+l ˜ ˜ in the additive subgroup of L(x) generated {u −u | u ∈ L(x)} has index p ˜  by a1 x, . . . , ak x, b1 x + c, . . . , bl x + c and ℘(L(x)).

700

Chapter 29. Algebraically Closed Fields with Distinguished Automorphisms

29.2 Coding in PAC Fields with Monadic Quantifiers Every first order language L naturally extends to a language Ln , the language of n-adic quantifiers. It is the simplest extension of L which allows for each m ≤ n quantification over certain m-ary relations on the underlying sets of structures of L. To obtain Ln from L adjoin for each m ≤ n a sequence of m-ary variable symbols Xm1 , Xm2 , Xm3 , . . . . The variable symbols of L are taken here as x1 , x2 , x3 , . . . . An atomic formula of Ln is either an atomic formula of L or a formula (xi1 , . . . , xim ) ∈ Xmj , where m ≤ n and i1 , . . . , im , j are positive integers. As usual we close the set of formulas of Ln under negation, disjunction, conjunction, and quantification on variables. A structure for Ln (or an n-adic structure for L) is a system hA, Q1 , . . . , Qn i, where A is a structure for L and, for each m ≤ n, Qm is a nonempty collection of m-ary relations on the underlying set of A (which we also denote by A). The structure is weak if for each m, all relations in Qm are finite. We interpret the variables xi as elements of A and the variables Xmj as elements of Qj . Thus, “(x1 , . . . , xm ) ∈ Xmj ” means “(x1 , . . . , xm ) belongs to Xmj ”, “∃xi ” means “there exists an element xi in A”, and “∃Xmj ” means “there exists an element Xmj in Qm ”. Theories of Ln , also called n-adic theories, are often undecidable. Thus, whenever we “interpret” such a theory in another theory (e.g. a theory of PAC fields), the latter also turns out to be undecidable. To be more precise let T ad T ∗ be theories of languages L and L∗ , respectively. An interpretation of T in T ∗ is a recursive map θ 7→ θ∗ of sentences of L onto sentences of L∗ such that T |= θ if and only if T ∗ |= θ∗ . Obviously, if T is undecidable, then so is T ∗ . We are mainly interested in the case where L = L(ring, K) is the language of rings enriched by constant symbols for each element of K. For integers q ≥ 2 and p, and for a field F we say that hypothesis H(p, q) holds in F if at least one of the following conditions holds: (1a) char(F ) = p, p - q, ζq ∈ F , and (F × )q 6= F × . (1b) char(F ) = p, p|q, and ℘(F ) = {up − u | u ∈ F } 6= F . Similarly we say that a class, F, of n-adic structures over fields satisfies hypothesis H(p, q) if for each structure hF, Q1 , . . . , Qn i in F, F is a field that satisfies hypothesis H(p, q). For the next lemma consider a class F of weak monadic structures (i.e. weak 1-adic structures) over PAC fields that contain K and satisfy condition H(p, q) for some p and q. To each hF, Qi in F we associate another monadic structure hF, Q0 i and denote the class of all hF, Q0 i’s by F 0 . The definition of hF, Q0 i is divided into two cases: Case A: p - q.

Q0 is the collection of all sets

D(A, x) = {a ∈ A | (∃y ∈ F )[y 6= 0 ∧ a + x = y q ]} with A ∈ Q and x ∈ F .

29.2 Coding in PAC Fields with Monadic Quantifiers

Case B: p|q.

701

Q0 is the collection of all sets

n h x io = yp − y E(A, u, x) = a ∈ A | (∃y ∈ F ) u+a with A ∈ L and u, x ∈ F . In both cases each A0 ∈ Q0 is contained in some A ∈ Q. Lemma 29.2.1: (a) For each structure hF, Qi in F the collection Q0 consists of all subsets of the sets A ∈ Q. (b) The monadic theory Th(F 0 ) is interpretable in Th(F). Proof: We treat each of the above cases separately. Case A: Let hF, Qi be a structure in F. Choose an element c ∈ F r F q . Let A ∈ Q and let A0 be a subset of A. By assumption, A is finite and F is PAC. Hence, by Proposition 29.1.2(a), there exist x ∈ F and ya ∈ F × for each a ∈ A such that a + x = yaq for all a ∈ A0 and a + x = cyaq for all a ∈ A r A0 . Then A0 = D(A, x), because (F × )q ∩ c(F × )q = ∅. This proves (a). Now define a map ϕ 7→ ϕ∗ from formulas of L1 onto formulas of L1 by induction on the structure of ϕ. If ϕ is an atomic formula of L, let ϕ∗ = ϕ. If ϕ is the formula a ∈ X, define ϕ∗ to be the formula q a ∈ AX ∧ (∃yX )[yX 6= 0 ∧ a + xX = yX ]

where xX , yX are variable symbols on elements and AX is a variable symbol on sets attached to the variable X. Next let the star operation commute with negation, disjunction, conjunction, and quantification on elements. Finally, if ψ ∗ has been defined for a formula ψ and ϕ is the formula (∃X)ψ, then define ϕ∗ to be (∃AX )(∃xX )ψ ∗ . One verifies by induction on the structure of a formula ϕ(z, X1 , . . . , Xn ) that for each monadic structure hF, Qi in F, for A1 , . . . , An ∈ Q, and x1 , . . . , xn ∈ F we have (2)

hF, Qi |= ϕ∗ (z, A1 , x1 , . . . , An , xn ) ⇐⇒ hF, Q0 i |= ϕ(z, D(A1 , x1 ), . . . , D(An , xn )).

In particular, if θ is a sentence of L1 , then θ is true in hF, L0 i if and only if θ∗ is true in hF, Li. Case B: Again, let hF, Qi be a structure in F. Choose an element c ∈ F r ℘(F ), let A ∈ Q, and let A0 be a subset of A. Since F is an infinite field, P there exists u ∈ F such that a∈A α(a) u+a 6= 0 for every function α: A → Fp 1 which is not identically zero. In other words, the elements u+a with a ranging on A are linearly independent over Fp . Now apply Proposition 29.1.2(b) to

702

Chapter 29. Algebraically Closed Fields with Distinguished Automorphisms

x find x ∈ F and for each a ∈ A an element ya ∈ F such that u+a = yap − ya for x 0 p 0 each a ∈ A and u+a = ya −ya +c for each a ∈ A r A . Thus, E(A, u, x) = A0 . This proves (a). The proof of (b) is carried out as in Case A. 

Our next construction allows us to replace monadic structures by certain n-adic structures. As before we start from a class F of weak monadic structures over PAC fields that satisfies hypotheses H(p, q). For each structure hF, Qi ∈ F and every m ≤ n let Qm be the collection of all subsets of A1 × · · · × Am , where A1 , . . . , Am ∈ Q. Denote the class of n-adic structures hF, Q1 , . . . , Qn i obtained in this way by Fn . Lemma 29.2.2: Th(Fn ) is interpretable in Th(F). Proof: The interpretation of Th(Fn ) in Th(F) goes through two auxiliary theories. Specifically, in addition to F 0 and Fn , we associate with F one more auxiliary class F˜n , and make the following interpretations: Th(F˜n ) Th(F 0 ) Th(F). Th(Fn ) The interpretation of Th(F 0 ) in Th(F) is done in Lemma 29.2.1(b). It remains to perform the two first interpretations: Part A: Interpretation of Th(Fn ) in Th(F˜n ). Our first step is to blow up m-ary relations, with m ≤ n, to n-ary relations. We omit each m-ary variable symbol Xmj with m < n from Ln and rename Xnj as Yj . Denote the language obtained in this way by L˜n . In addition to the atomic formulas of L the only atomic formulas of L˜n are (xi1 , . . . , xin ) ∈ Yj where i1 , . . . , in , j are positive integers. We define a map ϕ 7→ ϕ˜ from formulas of Ln onto formulas L˜n by induction on the structure of ϕ. If ϕ is atomic formula of L, let ϕ˜ = ϕ. If ϕ is the formula (xi1 , . . . , xim ) ∈ Xmj , with m ≤ n, define ϕ˜ to be the formula (xi1 , . . . , xim , xim , . . . , xim ) ∈ Ym+nj . Next let the tilde operation commute with negation, disjunction, and quantification on elements. Finally, if ψ˜ has been defined for a formula ψ and ϕ is the formula (∃Xmj )ψ, we define ϕ˜ to ˜ be the formula (∃Ym+nj )ψ. ˜ Let Fn be the class of all structures hF, Qn i of L˜n with hF, Qi ∈ F. For each hF, Qi ∈ F and each substitution f of the variables of Ln we define a substitution f˜ of the variables of L˜n as follows: If f (Xmj ) = R, then f˜(Ym+nj ) = {(a1 , . . . , am , am , . . . , am ) | (a1 , . . . , am ) ∈ R}. The value of the symbol variables on elements remains unchanged. It follows by induction on the structure of a formula ϕ(x, Xm1 j1 , . . . , Xmr jr ) of Ln that hF, Q1 , . . . , Qn i |= ϕ(f (x), f (Xm1 j1 ), . . . , f (Xmr jr )) if and only if ˜ f˜(x), f˜(Ym1 +nj1 ), . . . , f˜(Ymr +njr )). hF, Qn i |= ϕ( In particular, this holds for each sentence θ of Ln . Thus, Th(Fn ) is interpretable in Th(F˜n ).

29.2 Coding in PAC Fields with Monadic Quantifiers

703

Part B: Interpretation of Th(F˜n ) in Th(F 0 ). For each hF,PQi in F conn sider the bilinear map π: F n × F n → F defined by π(c, x) = i=1 ci xi . For n each c ∈ F , A1 , . . . , An ∈ Q, and B ⊆ π(c, A1 × · · · × An ), the set S(c, A1 , . . . , An , B) = {(x1 , . . . , xn ) ∈ A1 × · · · × An | π(c, x) ∈ B} belongs to Qn . Conversely, let A1 , . . . , An ∈ Q. Pn Since F is infinite and A1 , . . . , An are finite, there exists c ∈ F n with i=1 (xi − x0i )ci 6= 0 for all distinct x, x0 ∈ A1 × · · · × An . Then the map x 7→ π(c, x) from A1 × · · · × An into F is injective. Hence, if we start from a subset R of A1 × · · · × An and define B = {π(c, x) | x ∈ R}, then R = S(c, A1 , . . . , An , B). This representation of Qn allows us to interpret Th(F˜n ) in Th(F 0 ): To each formula ϕ of L˜n we associate a formula ϕ0 of L1 . The definition proceeds by induction on the structure of ϕ. If ϕ is the formula (x1 , . . . , xn ) ∈ Y where Y = Yj for some j, ϕ0 is the formula n ^

xi ∈ AY,i ∧ π(cY , ∗) is injective on AY,1 × · · · × AY,n ∧ π(cY , x) ∈ BY ,

i=1

where cY = (cY,1 , . . . , cY,n ) are variables symbols on elements, and AY,1 , . . . , AY,n , BY are 1-ary variable symbols on sets attached to Y . As before, make the prime operation commute with negation, disjunction, and quantification on elements. Finally, if ϕ is the formula (∃Y )ψ, where Y is as above and ψ is a formula for which ψ 0 has been defined, then ϕ0 is the formula (3)

(∃AY,1 ) · · · (∃AY,n )(∃BY )(∃cY,1 ) · · · (∃cY,n )   π(cY , ∗) is injective on AY,1 × · · · × AY,n ∧ π(cY , x) ∈ BY ∧ ψ ∗ .

For all formulas ϕ(x, Y1 , . . . , Ys ) of Ln where x = (xi1 , . . . , xir ), for all hF, Qi ∈ F, for all Ai1 , . . . , Ain ∈ Q, for all ci1 , . . . , cin ∈ F such that the map π(ci , ∗): Ai1 × · · · × Ain → F is injective, and for all subsets Bi of π(ci , Ai1 × · · · × Ain ), i = 1, . . . , s, put Si = S(ci , Ai1 , . . . , Ain , Bi ). Let b1 , . . . , br ∈ F . An induction on the structure of ϕ proves that hF, Qn i |= ϕ(b, S1 , . . . , Ss ) if and only if hF, Q0 i |= ϕ0 (b, c1 , A1 , B1 . . . , cs , As , Bs ), where ci = (ci1 , . . . , cin ) and Ai = (Ai1 , . . . , Ain ). In particular, if θ is a sentence of Ln , then θ ∈ Th(F˜n ) if and only if θ0 ∈ Th(F 0 ), as desired. 

704

Chapter 29. Algebraically Closed Fields with Distinguished Automorphisms

Proposition 29.2.3: Let F be a class of weak monadic structures over PAC fields that satisfies hypotheses H(p, q) and the following assumption: (4) For each positive integer n there exists hF, Qi ∈ F and there exists A ∈ Q of cardinality at least n. Then Th(F) is undecidable. Proof: Applying Lemma 29.2.2 to n = 2, it suffices to prove that Th(F2 ) is undecidable. By Remark 28.5.5, the theory of finite graphs is undecidable, so it suffices to interpret that theory in Th(F2 ). To each sentence θ of the language L(graph) of graphs recursively associate the following sentence θ∗ of L2 : (∀A ∈ Q1 )((∀R ∈ Q2 )[R ⊆ A × A, and R is symmetric and nonreflexive → (A, R) |= θ]. By (3), θ is true in each finite symmetric graph if and only if θ∗ is true in each hF, Qi ∈ F. Thus, the map θ 7→ θ∗ is an interpretation of the theory of  finite graphs into Th(F2 ), as desired.

˜ σ1 , . . . , σe i’s 29.3 The Theory of Almost all hK, We combine the methods developed in Section 29.2 with the algebraic background of Section 29.1 to obtain undecidability results for theories over PAC fields. Recall that we are working over a fixed infinite base field K, finitely generated over its prime field. For each e ≥ 1 we extend the languages ˜ to languages L(ring, K) and L(ring, K) L = L(ring, K, Σ1 , . . . , Σe )

and

˜ Σ 1 , . . . , Σe ) L˜ = L(ring, K,

by adding e unary function symbols Σ1 , . . . , Σe . Every e-tuple (σ1 , . . . , σe ) of automorphisms of Ks over K uniquely extends to an e-tuple of automor˜ σi is a structure for L. ˜ ˜ also denoted by σ1 , . . . , σe . Thus, hK, phisms of K, ˜ Σ1 , . . . , Σe ) be the set of all sentences θ of L˜ which are true Let Almost(K, ˜ σi for almost all σ ∈ Gal(K)e . in hK, ˜ Each elementary statement on K(σ) in L has a natural interpretation as ˜ σi in L. ˜ More precisely, we associate with each formula a statement on hK, ϕ(x1 , . . . , xn ) of L a formula ϕ∗ (x1 , . . . , xn ) of L˜ such that for all σ ∈ Gal(K)e e ˜ and a ∈ K(σ) ˜ ˜ σi |= ϕ∗ (a). K(σ) |= ϕ(a) ⇐⇒ hK, In particular, if θ is a sentence of L, then θ ∈ Almost(K, e) if and only if θ∗ ∈ ˜ Σ1 , . . . , Σe ). The star operation leaves each atomic formula of L Almost(K, unchanged and commutes with disjunction and negation. If ϕ is Vea formula (∃x)ψ and ψ ∗ has been defined, then ϕ∗ is the formula (∃x)[ i=1 σi x = x ∧ ψ ∗ ].

˜ σ1 , . . . , σe i’s 29.3 The Theory of Almost all hK,

705

The truth set of a sentence θ of L˜ is defined to be ˜ σi |= θ}. Truth(θ) = {σ ∈ Gal(K)e | hK, It is a measurable set. Indeed, if ϕ(x1 , . . . , xn ) is a quantifier free formula and ˜ then the truth of ϕ(a1 , . . . , an ) in hK, ˜ σi depends only on the a1 , . . . , an ∈ K, restriction of σ to the normal closure N of K(a1 , . . . , an )/K, hence on the restriction of σ to the maximal Galois extension N0 of K in N . Therefore, Truth(ϕ(a1 , . . . , an )) is an open-closed set. For an arbitrary formula ϕ(x, y) we have [

Truth ϕ(x, y) . Truth (∃y)ϕ(x, y) = ˜ y∈K

We conclude by induction on the structure of ϕ that Truth(θ) is even a Borel subset of Gal(K)e . The measure of Truth(θ) may be considered as the probability of θ to ˜ σi’s. We write Prob(θ) = µ(Truth(θ)). be true among the hK, ˜ Σ1 , . . . , Σe ) is an undecidable theTheorem 29.3.1: For e ≥ 2, Almost(K, ory. Proof: Let S=

\

˜ Truth(θ) ∩ {σ ∈ Gal(K)e | U (K(σ)) is finite}.

˜ 1 ,...,Σe ) θ∈Almost(K,Σ

˜ Σ1 , . . . , Σe ). By By definition, µ(Truth(θ)) = 1 for each θ ∈ Almost(K, ˜ ˜ Proposition 29.1.1, U (K(σ)) is finite for almost all σ ∈ K(σ). Since ˜ Σ1 , . . . , Σe ) is countable, µ(S) = 1. By Proposition 29.1.1, each Almost(K, σ ∈ S has these properties: ˜ (1a) K(σ) is PAC. ˜ (1b) Gal(K(σ)) = Fˆe . ˜ (1c) U (K(σ)) is a finite group. ˜ ˜ ˜ | Ve σi z = z ∧ (∃α ∈ K)[α 6= 0 ∧ σ1 α = zα]}. (1d) U (K(σ)) = {z ∈ K i=1

˜ By (1b), F extension F 0 of degree 2. If char(K) 6= 2, √ = K(σ) has an 0 ×r (F × )2 . If char(K) = 2, then F 0 = F (x) then F = F ( u) with u ∈ F 2 r where x − x = a and a ∈ F ℘(F ). In each case F satisfies hypothesis H(char(K), 2). ˜ ˜ and F = {hK(σ), Qσ i | σ ∈ S}. Then F is a set Let Qσ = {U (K(σ))} ˜ of weak monadic structures with |U (K(σ))| unbounded (Proposition 29.1.1). By Proposition 29.2.3, F is undecidable. Condition (1d) suggests an interpretation of Th(F) in ˜ Σ1 , . . . , Σe ): replace z ∈ X by Almost(K, e ^ i=1

σi z = z ∧ (∃a)[a 6= 0 ∧ σ1 z = za].

706

Chapter 29. Algebraically Closed Fields with Distinguished Automorphisms

If ϕ∗ is an interpretation of a formula ϕ of L1 , then ϕ∗ is also the interpretation of (∃X)ϕ. An induction on the structure of formulas in L1 proves the following statement: Let ϕ(z1 , . . . , zm , X1 , . . . , Xn ) be a formula of L1 ˜ Then ϕ∗ (z1 , . . . , zn ) has its free variables among the and a1 , . . . , am ∈ K. z1 , . . . , zm ’s and for all σ ∈ S ˜ ˜ ˜ σi |= ϕ∗ (a). ˜ . . . , U (K(σ))) ⇐⇒ hK, hK(σ), Qσ i |= ϕ(a, U (K(σ)), In particular, a sentence θ of L1 belongs to Th(F) if and only if θ∗ is in ˜ Σ1 , . . . , Σe ). Almost(K, ˜ Σ1 , . . . , Σe ) is undecidable.  It follows that Almost(K, Problem 29.3.2: Is T (K, 1) undecidable?

29.4 The Probability of Truth Sentences As before K is an infinite finitely generated field extension of its prime field. We survey here a stronger undecidability result for Almost(Σ1 , . . . , Σe ) than that proved in Section 29.3. Moreover, we represent the set of all Prob(θ), with θ ranging over all sentences of L(ring, K, e), as the set of all “definable numbers” between 0 and 1. The stronger decidability result involves Arithmetic. This is the complete theory of the structure N = hN, +, ·, 1i. Lemma 29.4.1 ([Cherlin-Jarden, Prop. 2.4]): Let F be a class of weak monadic structures over PAC fields that satisfies Hypotheses H(p, q). Suppose for each hF, Qi ∈ F the cardinality of the sets A ∈ Q is unbounded. Then Th(N ) is interpretable in Th(F). Moreover, there is a recursive map ϕ(x) 7→ ϕ∗ (X) from formulas of arithmetic to formulas of L1 satisfying this: for all hF, Qi ∈ F and A ∈ Q we have N |= ϕ(|A|) if and only if hF, Qi |= ϕ∗ (A). It is well known Arithmetic is undecidable [Ershov-Lavrov-TaimanovTaitslin, Thm. 3.2.4]. Thus, each F satisfying the assumptions of Lemma 29.4.1 is undecidable. The encoding of Arithmetic in Almost(Σ1 , . . . , Σe ) uses some facts about ˜ torsion of elliptic curves over the fields K(σ). The first of them is already cited in Section 18.11: Proposition 29.4.2 ([Geyer-Jarden1, Thm. 1.1]): For e ≥ 2 and for almost ˜ is finite. all σ ∈ Gal(K)e the set Etor (K(σ)) The second one is an easier result: Proposition 29.4.3 ([Cherlin-Jarden, Prop. 1.4]): For each e ≥ 1, for almost all σ ∈ Gal(K)e , and for every n ∈ N there exists an elliptic curve E ˜ over K which has a K(σ)-rational point of order n. Finally, we need an analog of Proposition 29.1.1 for elliptic curves:

29.4 The Probability of Truth Sentences

707

Proposition 29.4.4 ([Cherlin-Jarden, Cor. 1.6]): For each e ≥ 1, for almost ˜ all σ ∈ Gal(K)e , and for every elliptic curve defined over K(σ), e ^  ˜ ˜ | ˜ σ1 a = a + z . Etor (K(σ)) = z ∈ E(K) σi z = z ∧ (∃a ∈ E(K)) i=1

A real number r is said to be arithmetically definable if there exists k a formula ϕ(x, y) of N such that for all k, m ∈ N, r > m if and only if N |= ϕ(k, m). For example, every rational number is arithmetically definable. Theorem 29.4.5 ([Cherlin-Jarden, Thm. 5.3]): Suppose K is an explicitly given finitely generated extension of its prime field and e is a positive in˜ Σ1 , . . . , Σe ). Then Prob(θ) is an teger. Let θ be a sentence of L(ring, K, arithmetically definable real number. The proof of Theorem 29.4.5 uses manipulation of recursive functions and convergent series of real numbers and explicit computation in Galois groups over K (Lemma 19.3.2) but none of the results 29.4.2–cs29.4.4. Those tools are needed for the following converse of Theorem 29.4.5: Theorem 29.4.6 ([Cherlin-Jarden, Thms. 6.5 and 7.2]): Let K be an infinite finitely generated field over its prime field and let e ≥ 2 be an integer. Then for every definable real number r between 0 and 1 there exists a sentence θ of L(ring, K, Σ1 , . . . , Σe ) such that Prob(θ) = r. Remark 29.4.7: Sentences with transcendental probabilities. We have already mentioned in the introduction to this chapter that the probability of ˜ a sentence in L(ring, K) to be true in K(σ) is a rational number. In con˜ Σ1 , . . . , Σe ) to be true in trast, the probability of a sentence of L(ring, K, ˜ hK, σ1 , . . . , σe i may take a transcendental value. By Theorem 29.4.6 it suffices to give an example of a definable transcendental number between 0 and 1. The example we give is π4 which can P∞ i+1 1 be expressed as i=0 (−1) 2i+1 . This expression follows from the relaP∞ π i+1 xi tion tan 4 = 1 and the formula arctan x = i=0 (−1) 2i+1 . For each Pn 1 . Then positive integer n consider the partial sum sn = i=1 (−1)i+1 2i+1 s1 , s3 , s5 , . . . is a monotonically descending sequence which converges to π4 . Write s2n+1 = abnn with an and bn relatively prime positive integers. Then each of the sequences a1 , a2 , a3 , . . . and b1 , b2 , b3 , . . . is recursive. By a theorem of G¨odel, each recursive subset of N is representable [Shoenfield, p. 128]. Thus, there are formulas α(x) and β(x) of Arithmetic such that N |= α(a) if and only if a is one of the ai ’s and N |= β(b) if and only if b is one of the bi ’s. We may therefore write a formula ϕ(x, y) of Arithmetic such that N |= ϕ(a, b) if and only if ab > s1 or there are positive integers m, n satisfying s2m+1 ≤ ab < s2n+1 . Thus, N |= ϕ(a, b) if and only if ab > π4 . Consequently, π  4 is a definable number.

Chapter 30. Galois Stratification Chapter 30 extends the constructive field theory and algebraic geometry of Chapter 19, in contrast to Chapter 20, to give effective decision procedures through elimination of quantifiers. Such an elimination of quantifiers requires formulas outside of the theory L(ring). We call these more general formulas “Galois formulas”. These formulas include data for a stratification of the affine space An into K-normal basic sets A. Each coordinate ring K[A] is equipped with a Galois ring cover C and a collection of conjugacy classes Con of subgroups of the Galois group Gal(K(C)/K(A)). Each successive elimination of a quantifier contributes to the subgroups that appear in Con. This leads to the primitive recursiveness of the theory of all Frobenius fields which contain a given field K with elimination theory (Theorem 30.6.1).

30.1 The Artin Symbol For K, a fixed field, denote the class of all perfect Frobenius fields M (Section 24.1) that contain K by Frob(K). We refer to the situation where K is a presented field with elimination theory (Definition 19.2.8) as the explicit case. The results of this chapter hold without restriction on K; but, in the explicit case they become effective in the sense of Chapter 19. Thus, the existence theorems that are established in the general case become effective in the explicit case. Let A be a normal K-basic set and C an integral domain extending K[A] (Section 19.6). We call C/A a (Galois) ring/set cover over K if C/K[A] is a (Galois) ring cover (Definition 6.1.3). In the explicit case suppose A is a presented K-basic set and F is a presented finite separable extension of K(A). Then Lemma 19.7.2 effectively produces a normal K-basic open subset A0 of A and an integral domain C such that K(C) = F and C/A0 is a ring/set cover over K. Let C/A be a Galois ring/set cover over K with K[A] = K[x1 , . . . , xn , g(x)−1 ] and let z be a primitive element for the ring cover C/K[A]. Denote the Galois group Gal(K(C)/K(A)) by Gal(C/A) and consider a field M that contains K. If (a1 , . . . , an ) ∈ A(M ), then the K-specialization x → a uniquely extends to a homomorphism ϕ0 of K[A] into M . By Lemma 6.1.4, ϕ0 extends to a homomorphism ϕ from C into a Galois extension N = M (ϕ(z)) = M · ϕ(C) ¯ of M . Denote the quotient fields of ϕ(K[A]) and ϕ(C), respectively, by E ¯ ¯ ¯ and F . Then F /E is a Galois extension. Let D(ϕ) = {σ ∈ Gal(C/A) | (∀u ∈ C)[ϕ(u) = 0 =⇒ ϕ(σu) = 0]}

30.1 The Artin Symbol

709

be the decomposition group of ϕ. Each σ ∈ D(ϕ) induces an element σ ¯ of ¯ by the formula σ Gal(F¯ /E) ¯ (ϕ(u)) = ϕ(σu) for each u ∈ C. By Lemma ¯ that maps σ to σ ¯ is an isomorphism. 6.1.4, the map ϕ0 : D(ϕ) → Gal(F¯ /E) Furthermore, ϕ0 maps the subgroup DM (ϕ) = {σ ∈ Gal(C/A) | (∀u ∈ C)[ϕ(u) ∈ M =⇒ ϕ(σu) = ϕ(u)]} of D(ϕ) onto Gal(F¯ /F¯ ∩M ). If M = K, then DM (ϕ) = D(ϕ). In the general case, the composition of the isomorphism resF¯ : Gal(N/M ) → Gal(F¯ /F¯ ∩ M ) with (ϕ0 )−1 is an isomorphism ϕ∗ : Gal(N/M ) → DM (ϕ), where ϕ(ϕ∗ (σ)(u)) = σ(ϕ(u)) for all σ ∈ Gal(N/M ) and u ∈ C. As ϕ ranges over all possible extensions of ϕ0 to C, the group DM (ϕ) ranges over a conjugacy class of subgroups of Gal(C/A). We refer to this class as the Artin symbol of a in Gal(C/A) and denote it by Ar(C/A, M, a). If D/A is another Galois ring/set cover such that C ⊆ D and a ∈ A(M ), then Ar(C/A, M, a) = resK(C) Ar(D/A, M, a). Indeed, in the notation above let z1 be a primitive element for D/K[A]. Extend ϕ to a homomorphism ϕ1 of D into N1 = M (ϕ1 (z1 ). Then N1 is a Galois extension of M which contains N and resK(C) DM (ϕ1 ) ≤ DM (ϕ). Since in the commutative diagram Gal(N1 /M )

ϕ∗ 1

resN

 Gal(N/M )

ϕ

∗

/ DM (ϕ1 ) 

resK(C)

/ DM (ϕ)

both ϕ∗1 and ϕ∗ are isomorphisms and resN is surjective, so is resK(C) . Thus, resK(C) DM (ϕ1 ) = DM (ϕ), which proves our claim. Whenever there is no confusion, we omit reference to the cover from the Artin symbol and write it as Ar(A, M, a). By definition, Ar(A, M, a) is a conjugation domain of subgroups of Gal(C/A). Hence, if H ∈ Ar(A, M, a), then Ar(A, M, a) = {H σ | σ ∈ Gal(C/A)}. If n = 0, then A0 consists of one point, O, the origin. If A = A0 , then K(A) = K and C = L is a finite Galois extension of K. In this case ϕ0 is the identity map, ϕ is an automorphism of L over K and Ar(A, M, O) = {Gal(L/L ∩ M )σ | σ ∈ Gal(L/K)}. In the general definition replacement of A by an open subset A0 does not affect the Artin symbol. Indeed, let h ∈ K[X1 , . . . , Xn ] be a polynomial that does not vanish on A. Put A0 = A r V (h) and C 0 = C[h(x)−1 ], where x is a generic point of A. Then C 0 /A0 is also a Galois ring/set cover. If a ∈ A0 (M ), then Ar(A0 , M, a) = Ar(A, M, a). More generally, if A0 is a K-normal basic set contained in A with a generic point x0 , then the specialization x → x0 uniquely extends to a Khomomorphism τ0 of K[A] into K[A0 ] (Remark 19.6.4). With z a primitive

710

Chapter 30. Galois Stratification

element for C/K[A] let p(Z) be the image of irr(z, K(A)) in K[A0 ][Z] under τ0 . Then any root z 0 of p(Z) is a primitive element for a Galois ring/set cover C 0 /A0 and τ0 extends to a homomorphism τ : C → C 0 with τ (z) = z 0 . The cover C 0 /A0 is said to be induced by C/A, and τ induces an isomorphism τ ∗ : Gal(C 0 /A0 ) → D(τ ). Claim: If a ∈ A0 (M ), then τ ∗ (Ar(A0 , M, a)) ⊆ Ar(A, M, a). Indeed, let ˜ that extends the specialization x0 → ψ be a homomorphism of C 0 into M a, and let N = M (ψ(z 0 )). By definition, τ (τ ∗ (σ)u) = στ u for all σ ∈ Gal(C 0 /A0 ) and u ∈ C. Hence, τ ∗ (DM (ψ)) ≤ DM (ψ ◦ τ ). Since both groups are isomorphic to Gal(N/M ), we have τ ∗ (DM (ψ)) = DM (ψ ◦ τ ). A conjugacy domain of subgroups of Gal(C/A) is a collection of subgroups of Gal(C/A) which is closed under conjugation by elements of Gal(C/A). Consider a family H of finite groups and let Con(A, H) be a conjugacy domain of subgroups of Gal(C/A) belonging to H. Note that if Ar(A, M, a)∩ Con(A, H) 6= ∅, then Ar(A, M, a) ⊆ Con(A, H). Assume, as above, that C 0 /A0 is induced by τ : C → C 0 from C/A. Define Con(A0 , H) = {H ≤ Gal(C 0 /A0 ) | H ∈ H

and τ ∗ (H) ∈ Con(A, H)}.

We say that the conjugacy domain Con(A0 , H) of Gal(C 0 /A0 ) is induced by Con(A, H). If a ∈ A0 (M ), then Ar(A0 , M, a) ⊆ Con(A0 , H) if and only if Ar(A, M, a) ⊆ Con(A, H).

30.2 Conjugacy Domains under Projections Continue the conventions of Section 30.1. Let n ≥ 0 and let π: An+1 → An be projection onto the first n coordinates. If n = 0, then π maps each point of A1 onto the only point, O, of A0 . Fix for the whole section a family H of finite groups. Let A ⊆ An+1 and B ⊆ An be normal K-basic sets such that π(A) = B. Suppose A is equipped with a Galois ring cover and a conjugacy domain Con(A, H). Suppose B is equipped with a Galois ring cover. We construct a conjugacy domain Con(B, H) for B for which, with some additional conditions on A and B, the following holds: (1) For each perfect Frobenius extension M of K and each b ∈ B(M ), Ar(B, M, b) ⊆ Con(B, H) if and only if there exists a ∈ A(M ) such that π(a) = b and Ar(A, M, a) ⊆ Con(A, H). There are two cases: either dim(A) = dim(B) + 1 or dim(A) = dim(B). Lemmas 30.2.1 and 30.2.3 treat the first case, Lemma 30.2.5 the second. We start with the first case. Let C/A and D/B be Galois ring/set covers over K with these properties: A ⊆ An+1 ; B = π(A) (so, K[B] ⊆ K[A]); K(A) = K(B)(y) with y transcendental over K(B); and K(D) contains the algebraic closure L of K(B) in K(C). In addition, let z be a primitive element

30.2 Conjugacy Domains under Projections

711

for the ring cover C/K[A] and let x be a generic point for B. Suppose (x, y) is a generic point for A. Lemma 30.2.1: In the above notation, there exists a polynomial h ∈ K[X1 , . . . , Xn ], not vanishing on B, such that for C 0 = C[h(x)−1 ], A0 = A r V (h), D0 = D[h(x)−1 ], and B 0 = B r V (h), the pair (C 0 /A0 , D0 /B 0 ) of Galois ring/set covers satisfies these conditions: (2a) D0 ∩ L/K[B 0 ] is a ring cover. (2b) π(A0 ) = B 0 . (2c) For each field extension M of K, for each transcendental element y 0 over ^ (y 0 ) with ϕ(x) ∈ B 0 (M ) M and for each K-homomorphism ϕ: C 0 → M 0 0 and ϕ(y) = y let N = M [ϕ(D ∩ L)] and F = M (y 0 , ϕ(z)). Then [K(C): L(y)] = [F : N (y 0 )] and F/N is a regular extension. Moreover, in the explicit case when A, B, C and D are presented, h can be computed effectively. (3)

K(D)

K(C)

L

L(y)

N

N (y 0 )

K(B)

K(A)

M

M (y 0 )

F

Proof: Let K[B] = K[x, g1 (x)−1 ], K[A] = K[x, y, g2 (x, y)−1 ], and S = L ∩ D. Find a polynomial f ∈ S[Y, Z], irreducible over L, with f (y, z) = 0. Since L(y, z) = K(C) is a regular extension of L, f (Y, Z) is absolutely irreducible. Bertini-Noether (Proposition 10.4.2) produces a nonzero element u ∈ S with this property: if ϕ is a homomorphism of C into a field and ϕ(u) 6= 0, then the polynomial f ϕ = f ϕ (Y, Z) is absolutely irreducible and has the same degree in Z as f (Y, Z). Choose h ∈ K[X1 , . . . , Xn ] with h(x) = g1 (x)k NL/K(B) (u) for some integer k ≥ 0. Further, a multiplication of h by an appropriate polynomial assures that with D0 and B 0 given in the statement of the lemma, (D0 ∩ L)/K[B 0 ] is a ring cover (Remark 6.1.5). In order to prove that the pair C 0 /A0 and D0 /B 0 of Galois ring/set covers satisfies (2) we have only to check (2c). Indeed, h(ϕ(x)) 6= 0. Hence, ϕ(u) 6= 0, and f ϕ (y 0 , Z) is irreducible over N (y 0 ). Therefore [F : N (y 0 )] = degZ f ϕ (y 0 , Z) = degZ ϕ(y, Z) = [K(C): L(y)]. Finally, since f ϕ is absolutely  irreducible and F = N (y 0 , ϕ(z)), the extension F/N is regular. We refer to the properties listed in (2) by stating that the pair (C 0 /A0 , D0 /B 0 ) of Galois ring /set covers is specialization compatible. Notation 30.2.2: For a finite group G, denote the family of all subgroups of G by Subgr(G). Consider a Galois ring/set cover C/A, a subfield E of K(A),

712

Chapter 30. Galois Stratification

and a Galois extension L of E contained in K(C). If Con(A, H) is a conjugacy domain of subgroups of Gal(C/A) belonging to H, then resL (Con(A, H)) denotes the collection of groups obtained by restricting elements of Con(A, H) to L.  Lemma 30.2.3: Suppose the pair (C/A, D/B) of Galois ring/set covers of Lemma 30.2.1 is specialization compatible. Let Con(A, H) be a conjugacy domain of subgroups of Gal(C/A) belonging to a family H of finite groups. Define Con(B, H) = {G ≤ Gal(D/B) | G ∈ H

and resL G ∈ resL (Con(A, H))}.

where L is the algebraic closure of K(B) in K(C). Let I = Subgr(Gal(D/B)) ∪ Subgr(Gal(C/A)). Then for each Frobenius field M that contains K with (4)

Im(Gal(M )) ∩ I = H ∩ I,

and for each b ∈ B(M ), (5) Ar(B, M, b) ⊆ Con(B, H) if and only if there exists a ∈ A(M ) such that π(a) = b and Ar(A, M, a) ⊆ Con(A, H). Proof: Suppose first that there exists a ∈ A(M ) with π(a) = b and ˜ such Ar(A, M, a) ⊆ Con(A, H). Let ϕ be a K-homomorphism of C into M that ϕ(x, y) = a. By Section 30.1, resL(y) (DM (ϕ)) = DM (resL(y) ϕ). Also for S = D ∩ L, S · K[A]/K[A] is a Galois ring cover and L(y) is the quotient field of S · K[A]. Put ψ = resL(y) ϕ. By definition, resL DM (ψ) ⊆ DM (resS ψ). Since M (ϕ(S · K[A])) = M (ϕ(S)), both groups have the same order, so they coincide. Thus, resL (DM (resL(y) ϕ)) = DM (resL ϕ). Therefore, resL (DM (ϕ)) = DM (resL ϕ). Since ϕ(x) = b, this gives resL (Ar(A, M, a)) ⊆ Ar(S/B, M, b). Since the left hand side of the inclusion is a conjugacy domain and the right hand side is a conjugacy class of subgroups of Gal(L/K(B)) they are equal: resL (Ar(A, M, a)) = resL (Ar(B, M, b)). If G ∈ Ar(B, M, b), then G ∈ Im(Gal(M )) ∩ Subgr(Gal(C/A)) and resL G ∈ resL (Ar(A, M, a)) ⊆ Con(A, H). By (4), G ∈ H. Hence, G ∈ Con(B, H). It follows that Ar(B, M, b) ⊆ Con(B, H). Suppose now that M is a Frobenius field which contains K and satisfies (4) and b is a point of B(M ) with Ar(B, M, b) ⊆ Con(B, H). The existence of a ∈ A(M ) with π(a) = b and Ar(A, M, a) ⊆ Con(A, H) falls into two parts. Part A: Specialization of (x, y) to a point of transcendence degree 1 over M . Without loss assume K(D) = L. Take a transcendental element y 0 over M and extend the K-specialization x → b to a homomorphism ϕ of C into the algebraic closure of M (y 0 ) such that ϕ(y) = y 0 . Recall that K[A] =

30.2 Conjugacy Domains under Projections

713

K[x, y, g2 (x, y)−1 ] and z is a primitive element for the ring cover C/K[A]. Since π(A) = B, we have g2 (b, y 0 ) 6= 0. Let z 0 = ϕ(z), N = M · ϕ(D), R = M [y 0 , g2 (b, y 0 )−1 ] = M [ϕ(K[A])], E = M (y 0 ), and F = E(z 0 ) (see Diagram (3)). Then R[z 0 ]/R is a Galois ring cover over M , with F/E the corresponding field cover. By the specialization compatibility assumption on (C/A, D/B), [F : N (y 0 )] = [K(C): L(y)]. Therefore, in the following commutative diagram, 1

/ Gal(K(C)/L(y)) O

/ Gal(C/A) O

ϕ∗

1

/ Gal(F/N (y 0 ))

res

ϕ∗

/ Gal(F/E)

/ Gal(D/B) O

/1

ϕ∗

res

/ Gal(N/M )

/1

the left vertical arrow is an isomorphism. Part B: Application of the Frobenius property. By definition, ϕ∗ (Gal(N/M )) ∈ Ar(B, M, b) ∈ Con(B, H), so there is an H ∈ Con(A, H) with resK(D) H = ϕ∗ (Gal(N/M )). In particular, H ∈ Im(Gal(M )). The commutativity of the diagram in Part A shows that resK(D) (ϕ∗ (Gal(F/E))) = ϕ∗ (Gal(N/M )). Since the left vertical arrow is surjective, H ≤ ϕ∗ (Gal(F/E)). Hence, there is a subgroup H 0 of Gal(F/E) such that ϕ∗ (H 0 ) = H. Since all maps denoted by ϕ∗ are injective, H 0 ∈ Im(Gal(M )) and resN H 0 = Gal(N/M ). By Lemma 30.2.1, N is the algebraic closure of M in F . Since M is a Frobenius field, Proposition 24.1.4 gives an M -epimorphism ψ of R[z 0 ] onto a Galois extension F 0 of M that contains N such that, in the notation of Section 30.1, ψ(y 0 ) = c ∈ M and DM (ψ) = ψ ∗ (Gal(F 0 /M )) = H 0 . By definition, ϕ∗ (DM (ψ)) ≤ DM (ψ ◦ ϕ). But, since both DM (ψ ◦ ϕ) and DM (ψ) are isomorphic to Gal(F 0 /M ), H = ϕ∗ (H 0 ) = ϕ∗ (DM (ψ)) = DM (ψ◦ ϕ). The point a = (b, c) = ψ ◦ ϕ(x, y) belongs to A(M ) and π(a) = b. Hence, H ∈ Ar(A, M, a). In addition, H ∈ Con(A, H). Consequently, Ar(A, M, a) ⊆ Con(A, H). This concludes the proof of the lemma.  Remark 30.2.4: Note for later application that the only use made of the assumption that M is a Frobenius field in the Galois stratification procedure appears in Part B of the proof of Lemma 30.2.3.  Let C/A and D/B be Galois ring/set covers over K with K[A] integral over K[B] such that π(A) = B. Let E (resp. F ) be the maximal separable extension of K(B) in K(A) (resp. K(C)). Both extensions K(A)/E and K(C)/F are purely inseparable. Hence, K(A) and F are linearly disjoint over E and F · K(A) = K(C). If char(K) = p 6= 0, let q be a power of p such that K(A)q ⊆ E and K(C)q ⊆ F . Then K(C)q /K(A)q is a Galois

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Chapter 30. Galois Stratification

extension and K(C)q · E = F . Therefore, F/E is also a Galois extension and res: Gal(C/A) → Gal(F/E) is an isomorphism. Assume F ⊆ K(D). K(D)

F

K(C)

E

K(A)

K(B) Let Con(A, H) be a conjugacy domain of subgroups of Gal(C/A) belonging to H. Define (6)

Con(B, H) = {H σ | H ≤ Gal(K(D)/E), H ∈ H, resF H ∈ resF (Con(A, H)),

σ ∈ Gal(D/B)}.

Lemma 30.2.5: In the above notation, for each perfect field M that contains K and satisfies Im(Gal(M ))∩Subgr(Gal(D/B)) ⊆ H and for each b ∈ B(M ) (7) Ar(B, M, b) ⊆ Con(B, H) if and only if there exists a ∈ A(M ) with π(a) = b and Ar(A, M, a) ⊆ Con(A, H). Proof: Denote the integral closures of K[B] in E and F , respectively, by R and S. Then R ⊆ K[A] and S ⊆ D ∩ C. Also, let x and y be generic points of A and B such that π(x) = y. Suppose that Ar(B, M, b) ⊆ Con(B, H). Then there exists an H ≤ Gal(K(D)/E) that belongs to Ar(B, M, b) and there is G ∈ Con(A, H) with resF H = resF G. Since H ∈ Ar(b, M, b), there is a K-homomorphism ϕ of D into a Galois extension N of M such that ϕ(y) = b and ϕ∗ (Gal(N/M )) = H. In particular, ϕ(R) ⊆ M . Since M is perfect and K(C)/F is purely inseparable, there is a unique K-homomorphism ψ of C into N with ψ|S = ϕ|S . Since K(A)/E is purely inseparable, a = ψ(x) belongs to A(M ) and π(a) = b. Also, if we denote M [ψ(C)] by N0 , then resF (ψ ∗ (Gal(N0 /M ))) = resF H. Since resF : Gal(C/A) → Gal(F/E) is an isomorphism, ψ ∗ (Gal(N0 /M )) = G. Therefore, G ∈ Ar(A, M, a) and Ar(A, M, a) ⊆ Con(A, H). For the converse use the condition Im(Gal(M )) ∩ Subgr(Gal(D/B)) ⊆ H to reverse the above arguments.  The condition of specialization compatibility from Lemma 30.2.3 and the special assumptions of Lemma 30.2.5 require us to go through a stratification procedure in order to apply these lemmas to a given set of Galois ring/set covers:

30.3 Normal Stratification

715

Lemma 30.2.6: Let n ≥ 0 and let {Ct /At | t ∈ T } be a finite collection of Galois ring/set covers over K with At ⊆ An+1 , t ∈ T . Let B ⊆ An be a normal K-basic set with B ⊆ π(At ), t ∈ T . Then there exist a nonempty Zariski K-open subset B 0 of B and a Galois ring/set cover D/B 0 over K and 0 for each t ∈ T there exist a subset I(t) of I and Galois ring/set covers Cti /A0ti over K, i ∈ I(t), with S these properties: (a) π −1 (B 0 ) ∩ At = · i∈I(t) A0ti and π(A0ti ) = B 0 . 0 /A0ti is induced (Section 30.1) by Ct /At . (b) Cti (c) If dim(A0ti ) = dim(B 0 ), then K[A0ti ] is integral over K[B 0 ] and K(D) 0 ). contains the maximal separable extension of K(B 0 ) in K(Cti 0 0 (d) If dim(Ati ) = dim(B ) + 1, then K(D) contains the algebraic closure of 0 0 ) and the pair (Cti /A0ti , D/B 0 ) of Galois ring/set covers K(B 0 ) in K(Cti is specialization compatible. Moreover, in the explicit case, if Ct /At and B are presented, then I(t), 0 Cti /A0ti , i ∈ I(t), and D/B 0 can be effectively computed. Proof: Let t ∈ T . Apply Proposition 19.7.3 to find a stratification of At ∩ S π −1 (B) into a disjoint union · i∈J(t) Ati of normal K-basic sets. In particular, π(Ati ) ⊆ B, for i ∈ J(t). Let I(t) = {j ∈ J(t) | dim(π(Atj )) = dim(B)} and I 0 (t) = J(t) r I(t). Then [ [ [  B0 = (B r π(Atj )) ∪ π(Atj ) t∈T

j∈I(t)

j∈I 0 (t)

is of dimension smaller than B. Find a polynomial f ∈ K[X1 , . . . , Xn ] that vanishes on B0 but not on B. Note that for each i ∈ I(t) either dim(Ati ) = dim(B) or dim(Ati ) = dim(B) + 1. Let Cti be the ring cover of Ati induced by Ct . If dim(Ati ) = dim(B), let Fti be the separable closure of K(B) in K(Cti ). If dim(Ati ) = dim(B) + 1, let Fti be the algebraic closure of K(B) in K(Cti ). Choose a finite Galois extension P of K(B) containing all Fti , i ∈ I(t), t ∈ T . Choose a generic point x for B. Lemmas 19.7.2 and 30.2.1 produce a multiple g ∈ K[X1 , . . . , Xn ] of f that vanishes on B0 but not on B such that with 0 = Cti [g(x)−1 ] the following conditions hold: A0ti = Ati r V (g) and Cti 0 (8a) B = B r V (g) has a Galois ring cover D with K(D) = P . (8b) If dim(Ati ) = dim(B), then K[A0ti ] is integral over K[B 0 ]. 0 /A0ti , D/B 0 ) of Galois (8c) If dim(Ati ) = dim(B) + 1, then the pair (Cti set/ring covers is specialization compatible. These conditions imply the conditions of the lemma. 

30.3 Normal Stratification The condition of specialization compatibility from Lemma 30.2.3 and the special assumptions of Lemma 30.2.5 require us to go through a stratification procedure in order to apply these lemmas to a given set of Galois ring/set covers:

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Chapter 30. Galois Stratification

Lemma 30.3.1: Let n ≥ 0 and let {Ct /At | t ∈ T } be a finite collection of Galois ring/set covers over K with At ⊆ An+1 , t ∈ T . Let B ⊆ An be a normal K-basic set with B ⊆ π(At ), t ∈ T . Then there exist a nonempty Zariski K-open subset B 0 of B and a Galois ring/set cover D/B 0 over K and 0 for each t ∈ T there exist a subset I(t) of I and Galois ring/set covers Cti /A0ti over K, i ∈ I(t), with S these properties: (a) π −1 (B 0 ) ∩ At = · i∈I(t) A0ti and π(A0ti ) = B 0 . 0 /A0ti is induced (Section 30.1) by Ct /At . (b) Cti (c) If dim(A0ti ) = dim(B 0 ), then K[A0ti ] is integral over K[B 0 ] and K(D) 0 ). contains the maximal separable extension of K(B 0 ) in K(Cti 0 0 (d) If dim(Ati ) = dim(B ) + 1, then K(D) contains the algebraic closure of 0 0 ) and the pair (Cti /A0ti , D/B 0 ) of Galois ring/set covers K(B 0 ) in K(Cti is specialization compatible. Moreover, in the explicit case, if Ct /At and B are presented, then I(t), 0 /A0ti , i ∈ I(t), and D/B 0 can be effectively computed. Cti Proof: Let t ∈ T . Apply Proposition 19.7.3 to find a stratification of At ∩ S π −1 (B) into a disjoint union · i∈J(t) Ati of normal K-basic sets. In particular, π(Ati ) ⊆ B, for i ∈ J(t). Let I(t) = {j ∈ J(t) | dim(π(Atj )) = dim(B)} and I 0 (t) = J(t) r I(t). Then [ [ [  (B r π(Atj )) ∪ π(Atj ) B0 = t∈T

j∈I(t)

j∈I 0 (t)

is of dimension smaller than B. Find a polynomial f ∈ K[X1 , . . . , Xn ] that vanishes on B0 but not on B. Note that for each i ∈ I(t) either dim(Ati ) = dim(B) or dim(Ati ) = dim(B) + 1. Let Cti be the ring cover of Ati induced by Ct . If dim(Ati ) = dim(B), let Fti be the separable closure of K(B) in K(Cti ). If dim(Ati ) = dim(B) + 1, let Fti be the algebraic closure of K(B) in K(Cti ). Choose a finite Galois extension P of K(B) containing all Fti , i ∈ I(t), t ∈ T . Choose a generic point x for B. Lemmas 19.7.2 and 30.2.1 produce a multiple g ∈ K[X1 , . . . , Xn ] of f that vanishes on B0 but not on B such that with 0 = Cti [g(x)−1 ] the following conditions hold: A0ti = Ati r V (g) and Cti (1a) B 0 = B r V (g) has a Galois ring cover D with K(D) = P . (1b) If dim(Ati ) = dim(B), then K[A0ti ] is integral over K[B 0 ]. 0 /A0ti , D/B 0 ) of Galois (1c) If dim(Ati ) = dim(B) + 1, then the pair (Cti set/ring covers is specialization compatible. These conditions imply the conditions of the lemma.  Let n ≥ 0 and let A be a K-constructible set in An . A normal stratification, A = hA, Ci /Ai ii∈I , S of A over K is a partition A = · i∈I Ai of A as a finite union of disjoint normal equipped with a Galois ring/set cover Ci . Denote the K-basic sets Ai , each S collections of groups, i∈I Subgr(Gal(Ci /Ai )), by Subgr(A).

30.4 Elimination of One Variable

717

Lemma 30.3.2: Let n ≥ 0 and let {Ct /At | t ∈ T } be a finite collection of Galois ring/set covers over K with At ⊆ An+1 for each t ∈ T . Let B ⊆ An be a normal K-basic set such that B ⊆ π(At ) for each t ∈ T . Then there exist normal stratifications B = hB, Dj /Bj ij∈J and A = hπ −1 (B), Cjk /Ajk ij∈J, k∈K(j) satisfying the following conditions: S (a) π(Ajk ) = Bj for all j ∈ J and k ∈ K(j) and π −1 (Bj ) = · k∈K(j) Ajk . (b) Each Ajk is contained in a unique At and Cjk /Ajk is induced by Ct /At . (c) If dim(Ajk ) = dim(Bj ), then K[Ajk ] is integral over K[Bj ] and K(Dj ) contains the maximal separable extension of K(Bj ) in K(Cjk ). (d) If dim(Ajk ) = dim(Bj ) + 1, then K(Dj ) contains the algebraic closure of K(Bj ) in K(Cjk ) and the pair (Cjk /Ajk , Dj /Bj ) is specialization compatible. Proof: First suppose B is a normal K-basic set. Let B 0 be the nonempty open subset of B which Lemma 30.3.1 supplies. Then, in the notation of that 0 /A0ti it∈T, i∈I(t) and hB 0 , D/B 0 i lemma, the normal stratifications hπ −1 (B 0 ), Cti satisfy Conditions (a)-(d) of our lemma. The general case follows now by the stratification lemma (Lemma 19.6.6). 

30.4 Elimination of One Variable This section defines Galois stratification and establishes basic lemmas to eliminate quantifiers from the Galois formulas of Section 30.5. Let n ≥ 0, A a K-constructible set in An , and A = hA, Ci /Ai ii∈I a normal stratification, of A over K. If H is a family of finite groups, then A may be augmented to a Galois stratification (with respect to H): A(H) = hA, Ci /Ai , Con(Ai , H)ii∈I where, in addition to the above conditions, Con(Ai , H) is a conjugacy domain of subgroups of Gal(Ci /Ai ) belonging to H, i ∈ I. Then A is said to be the underlying normal stratification of A(H). Let M be a field containing K and a ∈ A(M ). Write Ar(A, M, a) ⊆ Con(A(H)) if Ar(Ai , M, a) ⊆ Con(Ai , H) for the unique i ∈ I with a ∈ Ai . If H0 is another family of finite groups with H ∩ Subgr(A) ⊆ H0 , then we freely rename Con(Ai , H) to be Con(Ai , H0 ), i ∈ I. Then A(H0 ) = hA, Ci /Ai , Con(Ai , H0 )ii∈I is a Galois stratification with respect to H0 . For every field M containing K and each a ∈ A(M ), clearly Ar(A, M, a) ⊆ Con(A(H)) if and only if Ar(A, M, a) ⊆ Con(A(H0 )). Suppose A0 = hA, Cj0 /A0j ij∈J is another normal stratification of A. Call 0 A a refinement of A if for each j ∈ J there exists a unique i ∈ I such that A0j ⊆ Ai and the cover Cj0 /A0j is induced by the cover Ci /Ai . If A0 (H) = hA, Cj0 /A0j , Con(A0j , H)ij∈J is an extension of A0 to a Galois stratification,

718

Chapter 30. Galois Stratification

then A0 (H) is said to be a refinement of A(H) if in addition Con(A0j , H) is induced by Con(Ai , H) (Section 30.1) whenever A0j ⊆ Ai . In this case we have for a ∈ A(M ) that Ar(A, M, a) ⊆ Con(A(H)) if and only if Ar(A0 , M, a) ⊆ Con(A0 (H)). The next two lemmas are based on Lemma 30.2.6. They will allow us in Section 30.5 to eliminate, respectively, one existential or universal quantifier from a given Galois formula. Lemma 30.4.1 (Existential Elimination Lemma): Let n ≥ 0 and let A(H) = hAn+1 , Ci /Ai , Con(Ai , H)ii∈I be a Galois stratification of An+1 over K with respect to a family H of finite groups. Then there exists a Galois stratification B(H) = hAn , Dj /Bj , Con(Bj , H)ij∈J with the following property: Let I = Subgr(A) ∪ Subgr(B). Suppose M is a perfect Frobenius field that contains K and satisfies (1)

Im(Gal(M )) ∩ I = H ∩ I.

Then, for each b ∈ An (M ), (2) Ar(B, M, b) ⊆ Con(B(H)) if and only if there exists a ∈ An+1 (M ) such that π(a) = b and Ar(A, M, a) ⊆ Con(A(H)). Moreover, the underlying normal stratification B = hAn , Dj /Bj ij∈J of B(H) depends only on the normal stratification A = hAn+1 , Ci /Ai ii∈I . In the explicit case, if H is primitive recursive and A(H) presented, then B(H) can be effectively computed. Proof: The union of the constructible sets π(Ai ) is equal to An . Apply Lemma 19.6.6 to stratify An into a union of disjoint normal K-basic sets Us , s ∈ S, such that for all i ∈ I and s ∈ S, either Us ⊆ π(Ai ) or Us ∩ π(Ai ) = ∅. Lemma 30.3.2 allows us to stratify Us and π −1 (Us ) for each s ∈ S separately and then S combine the separate stratifications into basic normal stratS S ifications An = · j∈J Bj and An+1 = · j∈J · k∈K(j) Ajk with the following properties: (3a) Each Ajk is contained in a unique Ai and has a Galois ring cover Cjk which is induced by Ci /Ai . −1 (3b) π(A S jk ) = Bj for each j ∈ J and each k ∈ K(j) and π (Bj ) = · k∈K(j) Ajk . (3c) Each Bj is equipped with a Galois ring cover Dj . (3d) If dim(Ajk ) = dim(Bj ), then K[Ajk ] is an integral extension of K[Bj ] and K(Dj ) contains the maximal separable extension of K(Bj ) in K(Cjk ). (3e) If dim(Ajk ) = dim(Bj ) + 1, then K(Dj ) contains the maximal algebraic extension of K(Bj ) in K(Cjk ) and the pair (Cjk /Ajk , Dj /Bj ) of Galois ring/set covers is specialization compatible. Following (3a), let Con(Ajk , H) be the conjugacy domain of subgroups of Gal(Cjk /Ajk ) induced by Con(Ai , H). Use Lemma 30.2.5 (resp. Lemma

30.5 The Complete Elimination Procedure

719

30.2.3), to define a conjugacy domain Conk (Bj , H) of subgroups of Gal(Dj /Bj ) from Con(Ajk , H) in case (3d) (resp. (3e)). The stratification A0 (H) = hAn+1 , Cjk /Ajk , Con(Ajk , H)ij∈J, k∈K(j) reS fines A(H). For each j ∈ J define Con(Bj , H) to be k∈K(j) Conk (Bj , H). Then B(H) = hAn , Dj /Bj , Con(Bj , H)ij∈J is a Galois stratification of An . Moreover, the conclusion of the lemma follows from Lemmas 30.2.3 and 30.2.5: Ar(B, M, b) ⊆ Con(B(H)) if and only if there exists a ∈ An+1 (M ) such that π(a) = b and Ar(A0 , M, a) ⊆ Con(A0 (H)) (i.e. Ar(A, M, a) ⊆ Con(A(H))).  Lemma 30.4.2 (Universal Elimination Lemma): Let A(H) = hAn+1 , Ci /Ai , Con(Ai , H)ii∈I be a Galois stratification of An+1 over K with respect to a family H of finite groups. Then there exists a Galois stratification B(H) = hAn , Dj /Aj , Con(Bj , H)ij∈J such that for each perfect Frobenius field M that contains K and satisfies (1) the following holds: for each b ∈ An (M ). (4) Ar(B, M, b) ⊆ Con(B(H)) if and only if Ar(A, M, a) ⊆ Con(A(H)) for each a ∈ An+1 (M ) with π(a) = b. Moreover, the underlying normal stratification B = hAn , Dj /Aj ij∈J of B(H) depends only on the normal stratification A = hAn+1 , Ci /Ai ii∈I . In the explicit case, if H is primitive recursive and A(H) is presented, then B(H) can be effectively computed. Proof: Let Ac (H) = hAn+1 , Ci /Ai , Conc (Ai , H)ii∈I be the complementary Galois stratification to A(H) of An+1 , where / Con(Ai , H)}. Conc (Ai , H) = {H ≤ Gal(Ci /Ai ) | H ∈ H and H ∈ Apply Lemma 30.4.1 to find a Galois stratification B c (H) = hAn , Dj /Bj , Conc (Bj , H)ij∈J over K such that for each M and b as in the lemma Ar(B c , M, b) ⊆ Con(Bc (H)) if and only if there exists a ∈ An+1 (M ) such that π(a) = b and Ar(Ac , M, a) ⊆ Con(Ac (H)). The complementary Galois stratification to B c (H) of An satisfies the conclusion of the lemma. 

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Chapter 30. Galois Stratification

30.5 The Complete Elimination Procedure Lemma 30.4.1 and 30.4.2 lead to an elimination of quantifier procedure for “Galois formulas” (Proposition 30.5.2). For a “Galois sentence” θ this procedure produces a finite Galois extension L of K and a conjugation domain Con(H) of subgroups of Gal(L/K) such that θ holds in a perfect Frobenius field M with a certain restriction on Im(Gal(M )) if and only if Gal(L/L ∩ M ) ∈ Con(H). Let m, n ≥ 0 be integers, Q1 , . . . , Qm quantifiers, and A(H) = hAm+n , Ci /Ai , Con(Ai , H)ii∈I a Galois stratification of Am+n over K with respect to a family H of finite groups. Then the expression (1)

(Q1 X1 ) · · · (Qm Xm )[Ar(X, Y) ⊆ Con(A(H))]

with X = (X1 , . . . , Xm ) and Y = (Y1 , . . . , Yn ) is said to be a Galois formula (with respect to K and A(H)) in the free variables Y. Denote it by θ = θ(Y). For a field M that contains K and for b1 , . . . , bn ∈ M , write M |= θ(b) if for Q1 a1 ∈ M, . . . , Qm am ∈ M , Ar(A, M, (a, b)) ⊆ Con(A(H)). Here read “Qi ai ∈ M ” as “there exists an ai in M ” if Qi is ∃, and as “for each ai in M ” if Qi is ∀. In the case that n = 0, θ has no free variables and is called a Galois sentence. Remark 30.5.1: Each formula ϕ(Y1 , . . . , Yn ) in the language L(ring, K) can be written (effectively, in the explicit case) in prenex normal form  (Q1 X1 ) · · · (Qm Xm ) 

k ^ l _

 fij (X, Y) = 0 ∧ gij (X, Y) 6= 0

i=1 j=1

with fij , gij ∈ K[X, Y]. The formula in the brackets defines a K-construconly the trivial group. tible set A ⊆ Am+n . Let H be the family containing S Construct a K-normal stratification Am+n = · i∈I Ai such that for each i ∈ I either Ai ⊆ A or Ai ⊆ Am+n r A. In the first case let Ci be K[Ai ] and define Con(Ai , H) to be H. In the second case let Ci = K[Ai ] and define Con(Ai , H) to be the empty collection. The corresponding Galois stratification A(H) defines a Galois formula θ as in (1). If a field M contains K and if b1 , . . . , bn ∈ M , then M |= θ(b) if and only if M |= ϕ(b). Thus, each formula in L(ring, K) is equivalent to a Galois formula over K.  Proposition 30.5.2: Let C be a family of finite groups. Consider a Galois formula θ(Y1 , . . . , Yn ) (Q1 X1 ) · · · (Qm Xm )[Ar(X, Y) ⊆ Con(A(C))],

30.5 The Complete Elimination Procedure

721

with respect to a Galois stratification A(C) of Am+n over K. Then there exists a finite family S of finite groups and a normal stratification B of An over K with Subgr(A)∪Subgr(B) ⊆ S. Moreover, for every family H of finite groups which satisfies Con(A(C)) ⊆ H, the normal stratification B can be expanded to a Galois stratification B(H) with respect to H such that for each perfect Frobenius field M that contains K and satisfies Im(Gal(M )) ∩ S = H ∩ S and for each b ∈ An (M ) the following holds: (2)

M |= θ(b) ⇐⇒ Ar(B, M, b) ⊆ Con(B(H)).

In the explicit case, if θ(Y) is presented, then S and B can be effectively computed. In this case, for each primitive recursive H an effective computation gives B(H) with the above properties. Proof: Lemmas 30.4.1 and 30.4.2 give a normal stratification Am−1 of Am−1+n such that for each family H of finite groups which satisfies Con(A(C)) ⊆ H, the normal stratification Am−1 can be expanded to a Galois stratification Am−1 (H) (depending on Qm ) with the following property: For each perfect Frobenius field M that contains K and satisfies Im(Gal(M )) ∩ (Subgr(A) ∪ Subgr(Am−1 )) = H ∩ (Subgr(A) ∪ Subgr(Am−1 )) and for each (a1 , . . . , am−1 , b) ∈ Am−1+n (M ), Ar(Am−1 , M, (a1 , . . . , am−1 , b)) ⊆ Con(Am−1 (H)) if and only if Qm am ∈ M such that Ar(A, M, (a, b)) ⊆ Con(A(C)). This eliminates Qm from θ. Continue to eliminate Qm−1 , . . . , Q1 in order by constructing the corresponding normal stratifications Am−2 , . . . , A0 . Let Am = A, B = A0 , and S = Subgr(Am ) ∪ · · · ∪ Subgr(A0 ). The proposition follows.  The case n = 0 is of particular interest: θ is a Galois sentence, the normal stratification B of A0 is trivial, and Con(B(H)) is a conjugacy domain Con(H) of Gal(L/K) with L a finite Galois extension of K. The condition Ar(B, M, b) ⊆ Con(B(H)) simplifies to Gal(L/L ∩ M ) ∈ Con(H). Proposition 30.5.3: Let C be the family of finite groups and let θ be a Galois sentence, (Q1 X1 ) . . . (Qm Xm )[Ar(X) ⊆ Con(A(C))], with respect to a Galois stratification A(C) of Am over K. Then there exists a finite family S of finite groups and a finite Galois extension L of K with Subgr(A) ∪ Subgr(Gal(L/K)) ⊆ S. Moreover, for each family H of finite groups which satisfies Con(A(C)) ⊆ H there exists a conjugacy domain Con(H) of Gal(L/K) consisting of groups belonging to H with the following property:

722

Chapter 30. Galois Stratification

For each perfect Frobenius field M that contains K and satisfies Im(Gal(M )) ∩ S = H ∩ S, (3)

M |= θ ⇐⇒ Gal(L/L ∩ M ) ∈ Con(H).

In the explicit case, if θ is presented, then an effective computation gives S and L, and for each primitive recursive H, Con(H) satisfying (3) can be found effectively.

30.6 Model-Theoretic Applications The decidability of the theory of Frobenius fields is now a consequence of the elimination of quantifiers procedure of Section 30.5. Let K be a field and C a full formation of finite groups. Write Frob(K) for the class of all perfect Frobenius fields that contain K. Put Frob(K, pro-C) = {M ∈ Frob(K) | Gal(M ) is a pro-C-group}. Then let Th(Frob(K, pro-C)) be the theory of all sentences θ ∈ L(ring, K) which are true in all fields M ∈ Frob(K, pro-C). Theorem 30.6.1: Let K be a presented field with elimination theory and C a primitive recursive full formation of finite groups. Then Th(Frob(K, pro-C)) is primitive recursive. Proof: Let θ0 be a sentence of L(ring, K). Use Remark 30.5.1 to find a Galois stratification A over K with Con(Ai ) either empty or consisting only of the trivial group for each Ai in the normal stratification underlying A. Then let θ be the associated Galois sentence which is equivalent to θ0 . Proposition 30.5.3 gives a finite family S of finite groups and a finite Galois extension L of K such that Subgr(A)∪Subgr(Gal(L/K)) ⊆ S. List all subfamilies, H1 , . . . , Hs , of C ∩ S which contain Con(A). For each i, Corollary 24.5.3 gives a check if there exists a superprojective pro-C-group G such that Im(G) ∩ S = Hi . Find r, 0 ≤ r ≤ s, and reorder H1 , . . . , Hs so that Im(G) ∩ S = Hi for some superprojective pro-C group G, if and only if i ∈ {1, 2, . . . , r}. For each i, 0 ≤ i ≤ r, Proposition 30.5.3 gives a conjugacy domain Con(Hi ) of subgroups of Gal(L/K) belonging to Hi with the following property: for each perfect Frobenius field M that contains K and satisfies Im(Gal(M )) ∩ S = Hi , (1)

M |= θ ⇐⇒ Gal(L/L ∩ M ) ∈ Con(Hi ).

Then Con(Hi ) ⊆ Hi ∩ Subgr(Gal(L/K)) for each i, 1 ≤ i ≤ r. Suppose first that (2)

Con(Hi ) = Hi ∩ Subgr(Gal(L/K)),

i = 1, . . . , r.

30.6 Model-Theoretic Applications

723

Then θ is true in each M ∈ Frob(K, pro-C). Indeed, for each such M , the pro-C-group Gal(M ) is superprojective (Proposition 24.1.5). Hence, there exists an i, 1 ≤ i ≤ r such that Im(Gal(M )) ∩ S = Hi . Then Gal(L/L ∩ M ) ∈ Im(Gal(M )) ∩ S ∩ Subgr(Gal(L/K)) = Con(Hi ). Thus, by (1), M |= θ. Suppose on the other hand, that there exists i, 1 ≤ i ≤ r, and there exists a subgroup H of Gal(L/K) which belongs to Hi but not to Con(Hi ). By assumption, there exists a superprojective pro-C-group G such that Im(G) ∩ S = Hi . In particular, H ∈ Im(G). Lemma 24.1.6 produces a field M ∈ Frob(K) such that Gal(M ) ∼ = G and Gal(L/L ∩ M ) = H. In particular, Gal(L/L ∩ M ) ∈ / Con(Hi ) but Im(Gal(M )) ∩ S = Hi . We conclude, by (1) that M 6|= θ. Thus, condition (2) is equivalent to the validity of θ in each field M ∈ Frob(K, pro-C). A check of (2) therefore decides whether or not θ is true in each M ∈ Frob(K, pro-C).  We now specialize the previous theory. Consider a fixed superprojective group G. Let Frob(K, G) be the class of all fields M ∈ Frob(K) with Im(G(M )) = Im(G). A simplification of the proof of Theorem 30.6.1 yields a primitive recursive decision procedure for this theory, provided Im(G) is primitive recursive. Indeed, let θ be a sentence in L(ring, K) or, more generally, a Galois sentence with respect to the family Im(G). Take H in Lemmas 30.4.1 and 30.4.2 to be Im(G). Then Condition (1) of Lemma 30.4.1 is satisfied for each field M ∈ Frob(K, G). Use these lemmas to eliminate the quantifiers of θ one by one. The procedure produces in its final stage a finite Galois extension L of K and a conjugacy domain Con of subgroups of Gal(L/K) which belong to Im(G) with this property: For each M ∈ Frob(K, G), (3)

M |= θ ⇐⇒ Gal(L/L ∩ M ) ∈ Con.

Therefore, M |= θ for each such M if and only if Con = Im(G) ∩ Subgr(Gal(L/K)). This is, again, a checkable condition and this establishes the procedure. These arguments prove the following result: Theorem 30.6.2: Let K be a presented field with elimination theory and G a superprojective group such that Im(G) is primitive recursive. Then there exists a primitive recursive decision procedure for the theory of Frob(K, G). Similarly, a perfect Frobenius field M is determined up to elementarily equivalence by Im(Gal(M )) and by its algebraic part:

724

Chapter 30. Galois Stratification

Theorem 30.6.3: Let M and M 0 be perfect Frobenius fields containing a field K. Then M and M 0 are elementarily equivalent with respect to L(ring, K) if and only if Im(Gal(M )) = Im(Gal(M 0 )) and Ks ∩M ∼ = Ks ∩M 0 . 0 In particular, if M is algebraically closed in M and Im(Gal(M )) = Im(Gal(M 0 )), then M is an elementary subfield of M 0 . Proof: Suppose Im(Gal(M )) = Im(Gal(M 0 ) and Ks ∩ M ∼ =K Ks ∩ M 0 . 0 The elementary equivalence of M and M follows by taking Gal(M 0 ) = G in Theorem 30.6.2: the right hand side of (3) is simultaneously fulfilled or not fulfilled for both M and M 0 . Hence, so is the left hand side of (3). Conversely, suppose that M ≡ M 0 . Since the statement “the group G is realizable over M ” is elementary (Remark 20.4.5(d)), Im(Gal(M )) =  Im(Gal(M 0 )). Lemma 20.6.3 shows that Ks ∩ M ∼ =K Ks ∩ M 0 . Remark 30.6.4: The converse of Remark 30.5.1 also holds: Each Galois sentence over K is equivalent in a certain sense to a sentence of L(ring, K). We explain in detail: Consider a fixed superprojective group G and let θ be a Galois sentence with respect to the family H = Im(G). Let L and Con(H) be as in the discussion preceding Theorem 30.6.2. Choose representatives H1 , . . . , Hr for the conjugacy classes of Con(H). For each i, 1 ≤ i ≤ r, let xi0 , xi1 , . . . , xi,s(i) be elements of L such that K(xi0 ) = L(Hi ) and K(xi1 ), . . . , K(xi,s(i) ) is a list of all proper extensions of L(Hi ) in L. For fij = irr(xij , K), consider the sentence λ of L(ring, K): (4)

r _ 

s(i)

i=1

j=1

(∃X)[fi0 (X) = 0] ∧

^

 ¬(∃X)[fij (X) = 0] .

Obviously, if a field M contains K, then Gal(L/L ∩ M ) ∈ Con(H) if and only if M |= λ. From (3), for each M ∈ Frob(K, G), (5)

M |= θ ⇐⇒ M |= λ.

Note that λ is a test sentence in the sense of (2) of Section 20.6. Thus, (5) generalizes the case G = Fˆe of Proposition 20.6.6(b). When θ is a sentence of L(ring, K), note the difference between the methods by which λ is obtained from θ. In Proposition 20.6.6, λ is found by examining all proofs from the axioms (a recursive procedure). In this chapter, however, λ is achieved by an explicit step by step elaboration, using an algebraic-geometric technique (and a primitive recursive procedure). In the case where C is the family of all finite groups and θ is a Galois sentence over K with respect to C we may assert only a weaker statement than that θ is equivalent to one sentence of L(ring, K). In the notation of the proof of Theorem 30.6.1, as above we find for each family Hi a test sentence λi such that for each M ∈ Frob(K) with Im(Gal(M )) ∩ S = Hi , M |= θ ⇐⇒ M |= λi .

30.7 A Limit of Theories

725

Thus, in this case, we say that θ is equivalent to a sequence λ1 , . . . , λr of sentence of L(ring, K) indexed by cases. 

30.7 A Limit of Theories In Section 20.5, K is a countable Hilbertian field and Almost(K, e) is the ˜ theory of all sentences θ of L(ring, K) which are true in K(σ), for almost all e σ ∈ Gal(K) . We now investigate the limit of these theories: namely, the theory of all sentences θ of L(ring, K) which belong to Almost(K, e) for all large e. Let C be a family of finite groups. Put Frob(K, C) = {M ∈ Frob(K) | Im(Gal(M )) = C}. Denote the theory of all sentences θ in L(ring, K) which are true in all M ∈ Frob(K, C) by Th(Frob(K, C)). For each integer e ≥ 0 let Ce = {G ∈ C | rank(G) ≤ e}. We concentrate on the case where C is a formation of finite groups. In this case if M is a field, then Im(Gal(M )) = Ce if and only if Gal(M ) ∼ = Fˆe (C) (Lemma 17.7.1). If M is a countable Frobenius field, then Im(Gal(M )) = C if and only if Gal(M ) ∼ = Fˆω (C) (Theorem 24.8.1). Moreover, if C is full, ˆ ˆ then Fe (C) and Fω (C) are projective (Corollary 22.4.5). Since, in addition, they have the embedding property (Lemma 24.3.3), they are superprojective. Thus, Lemma 24.1.6 shows that the classes Frob(K, Ce ) and Frob(K, C) are nonempty. The following theorem may be regarded as an interpretation of the statement that Th(Frob(K, C)) is a limit of the theories Th(Frob(K, Ce )) as e approaches infinity: Theorem 30.7.1: Let C be a full family of finite groups and θ a sentence of L(ring, K). Then θ ∈ Th(Frob(K, C)) if and only if there exists an integer e0 ≥ 0 such that θ ∈ Th(Frob(K, Ce )) for each e ≥ e0 . In the explicit case, if C is primitive recursive, then the theories Th(Frob(K, Ce )) and Th(Frob(K, C)) are primitive recursive. Moreover, the function e0 (θ) = min(e00 | θ ∈ Th(Frob(K, CeT) for each e ≥ e00 ) is primitive ∞ recursive. Therefore, the intersection T = e=1 Th(Frob(K, Ce )) is also a primitive recursive theory. Proof: Remark 30.5.1 allows us to assume that θ is a Galois sentence. Proposition 30.5.3 gives a finite family S of finite groups, a finite Galois extension L of K, and a conjugacy domain Con of Gal(L/K) consisting of groups of H = C∩S such that Subgr(Gal(L/K)) ⊆ S. Moreover, for each M ∈ Frob(K) with Im(Gal(M )) ∩ S = H, (1)

M |= θ ⇐⇒ Gal(L/L ∩ M ) ∈ Con.

Let e0 = max{rank(G) | G ∈ S}. Then Ce ∩ S = H for each e ≥ e0 .

726

Chapter 30. Galois Stratification

Suppose θ ∈ Th(Frob(K, C)), e ≥ e0 and M ∈ Frob(K, Ce ). Then Gal(M ) ∼ = Fˆe (C), so Gal(L/L∩M ) is a quotient of Fˆω (C). Lemma 24.1.6 gives a perfect Frobenius field M 0 which contains K satisfying L∩M 0 = L∩M and Gal(M 0 ) ∼ = Fˆω (C). In particular, M 0 ∈ Frob(K, C). By assumption, M 0 |= θ. Hence, by (1), Gal(L/L ∩ M ) = Gal(L/L ∩ M 0 ) ∈ Con, so M |= θ. Therefore, θ ∈ Th(Frob(K, Ce )). Conversely, suppose θ ∈ Th(Frob(K, Ce0 )). Let M ∈ Frob(K, C). Then Gal(M ) ∼ = Fˆω (C) and Gal(L/L ∩ M ) ∈ S. Hence, Gal(L/L ∩ M ) is a pro-C group of rank at most e0 . Lemma 24.1.6 gives a field M 0 ∈ Frob(K, Ce0 ) with L ∩ M 0 = L ∩ M . By assumption, M 0 |= θ. Hence, by (1), Gal(L/L ∩ M ) = Gal(L/L ∩ M 0 ) ∈ Con, so M |= θ. Consequently, θ ∈ Th(Frob(K, C)). Suppose θ ∈ Th(Frob(K, C)). Then θ ∈ Th(Frob(K, Ce0 )). Apply Theorem 30.6.2 to check if θ ∈ Th(Frob(K, Ce )) for each 0 ≤ e < e0 . Thus, decide  if θ belongs to each of the theories Th(Frob(K, Ce )). Finally, suppose C is the family of all finite groups and (K, e) is a Hilbertian pair (Section 20.5). In this case Frob(K, Ce ) is the class of all perfect e-free PAC fields that contain K. If L is a finite Galois extension of K and H is a subgroup of Gal(L/K) with rank(H) ≤ e, then the set of all σ ∈ Gal(K)e such that hresL σi = H is of positive rational (and effectively computable) measure. For each sentence θ of L(ring, K) let ˜ S(K, e, θ) = {σ ∈ Gal(K)e | K(σ) |= θ}. By theorem 20.6.7, the function µ(θ(K, e, θ) is recursive. The arguments of the preceding paragraph strengthens this result: Theorem 30.7.2: Let (K, e) be a Hilbertian pair. Suppose K has elimination theory. Then the function µ(S(K, e, θ)) from sentences of L(ring, K) to rational numbers is primitive recursive. The theory Almost(K, e) is primitive recursive. Similarly, Frob(K, C) in this case is the family of all ω-free perfect PAC fields which contain K. Its theory is the limit of Almost(K, e) as e approaches infinity. Theorem 30.7.3: Let K be a Hilbertian field with elimination theory. Then the theory of all ω-free perfect PAC fields that contain K is primitive recursive.

Exercises 1. This exercise traces through the concepts of Section 30.1. Let A be the affine space An over a field K, so that K[A] = K[x], with x = (x1 , . . . , xn )

Exercises

727

an n-tuple of algebraically independent elements over K. Let z1 , . . . , zn be the roots of the general polynomial of degree n f (x, Z) =

n Y

(Z − zi ) = Z n +

i=1

n X

(−1)i xi Z n−i

i=1

so that xi is the symmetric polynomial of degree n in z1 , . . . , zn , i = 1, . . . , n. Then C = K[z] is an integrally closed ring extension of K[A]. Use the notation of Section 30.1. Throughout assume that n > 2. (a) Show that the field extension K(C)/K(A) is Galois with group Sn . n(n−1) Q The discriminant of f (x, Z) is d(f ) = (−1) 2 i6=j (zi − zj ). For each Q P n c1 , . . . , cn ∈ K let D(c) = i6=j (ci − cj ) and zc = i=1 ci zi . Show that if D(c) 6= 0, then K(C) = K(x, zc ). (b) Let A1 = An r V (d(f )) and C1 = K[z, d(f )−1 ]. Show that if fc = irr(zc , K(A)) with D(c) 6= 0, then d(fc ) is a unit in K[x, zc , d(f )−1 ]. (c) Let K be a finite field and let c ∈ An (K) such that D(c) 6= 0. With Cc = K[x, zc , d(fc )−1 ] and Ac = An r V (d(fc )) consider the Galois ring/set cover Cc /Ac . Let a = (a1 , . . . , an ) ∈ Ac (K). Explicitly describe Ar(Ac , K, a) in terms of the degrees of the irreducible factors of g(a, Z). (d) Show directly, for each σ ∈ Sn , that there exists a ∈ Ac (K) (in (c)) such that Ar(Ac , K, a) consists of the conjugates of σ in Sn , if |K| is suitably large. 2. This exercise expands on Exercise 1 to give some practice with the two essential cases that appear in Section 30.2. Here K is a field of characteristic zero. (a) let A1 be the Zariski closed subset V (f ) of An+1 where f (X1 , . . . , Xn , Z) = Z n +

n X

(−1)i Xi Z n−i .

i=1

Let C1 /A1 be the trivial Galois ring/set cover (i.e. C1 = K[A1 ]). Take H to be the collection of all finite cyclic groups and let Con(A1 , H) consists of only the trivial group. Consider B1 = π(A1 ) where π: An+1 → An is projection onto the first n coordinates. Explicitly find h ∈ K[X1 , . . . , Xn ] not vanishing on B1 and a ring D10 such that for C10 = C[h−1 ], A01 = A1 r V (h), B10 = B1 r V (h) the pair (C10 /A01 , D10 /B10 ) consists of Galois ring/set covers for which K(C10 ) ⊆ K(D10 ). Also find a conjugacy domain Con(B10 , H) consisting of cyclic subgroups of Gal(D10 /B10 ) such that condition (7) of Lemma 30.2.5 holds. (b) With D10 /B10 replacing C1 /A1 in part (a), Con(B10 , H) replacing Con(A1 , H) and (4) of Lemma 30.2.3 replacing (7) of Lemma 30.2.5, do part (a) applied to the projection of An onto An−1 . Note that, in this case, dim(B10 ) = dim(π(B10 )) + 1 in contrast to part (a) where dim(A1 ) = dim(π(A1 )).

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Chapter 30. Galois Stratification

3. As in Lemma 30.2.5 let C/A and D/B be Galois ring/set covers over K with K[A] integral over K[B] such that π(A) = B. Denote the maximal separable extension of K(B) in K(A) (resp. K(C)) by E (resp. F ) and assume that F ⊆ K(D). Suppose G = Gal(C/A) ∼ = Z/3Z and that the group S3 acts on (Z/3Z)3 by permutation of the coordinates. The subgroup U = {(z1 , z2 , z3 ) ∈ (Z/3Z)3 | z1 + z2 + z3 = 0} is invariant under this action. Let H = S3 n U be the corresponding semidirect product. Suppose there is an isomorphism θ: Gal(D/B) → H which maps Gal(K(D)/E) onto U . Let π1 : U → Z/3Z be the projection on the first coordinate. Suppose θ induces ¯ ¯ Gal(F/E) → Z/3Z such that π1 (θ(σ)) = θ(res an isomorphism θ: F σ) for each σ ∈ Gal(K(D)/E). Now let H consists of all finite cyclic groups and suppose that Con(A, H) = {G}. Compute Con(B, H) as in Lemma 30.2.5. Compute also Conc (A, H) and Conc (B, H) as in the proof of Lemma 30.4.2. 4. This is an analog of Exercise 3 for the case that dim(A) = dim(B)+1. Use the notation of Lemma 30.2.3. Suppose the conjugacy domain Con(A, H) of subgroups of Gal(C/A) is nonempty. For which of the following possibilities for H, and all possible pairs (C/A, D/B) satisfying the above, is Con(B, H) nonempty: (a) H is all finite groups of rank at most 2; (b) H is all finite p-groups for some prime p; (c) H is all finite p-groups of rank at most 2; or (d) H is all Abelian groups. 5. Apply the procedure of Sections 30.4 and 30.5 to the case that K = Q, H is the collection of finite cyclic groups and θ is the sentence (∀X)(∃Y )[Tn (Y ) = X], where n is an odd positive integer and Tn is the Chebyshev polynomial in Q[Y ] satisfying Tn (Z + Z −1 ) = Z n + Z −n (Section 21.8) by following this outline: (a) Eliminate Y (after applying Remark 30.5.1) by noting that the splitting field of Tn (Y ) − x over Q(x) is Q(ζn , z), where x is transcendental over Q, z satisfies Tn (z + z −1 ) = x, and ζn is a primitive nth root of 1. Moreover, Gal(Q(ζn , z)/Q(ζn , x)) is generated by σ and τ , where σ(z) = ζn z and τ (z) = z −1 . (b) Apply (a) and Lemma 30.4.1 to produce B(H) = hA1 , Dj /Bj , Con(Bj , H) | j = 1, 2, 3i satisfying the conclusion (2) of Lemma 25.6. Indeed, take B1 = {2}, B2 = {−2} and B3 = A1 r{2, −2}. Note that 2 and -2 are the values of X for which Tn (Y ) − X has multiples zeros. (c) Now apply (b) and Lemma 30.4.1 to eliminate X from (∀X)[Ar(X) ⊆ Con(B(H))] and finally arrive at a finite Galois extension L/Q and a set of

Notes

729

conjugacy classes Con of Gal(L/Q) the following holds: Suppose M is a Frobenius field of characteristic 0 and Im(G(M )) ∩ Subgr(Gal(Q(ζn , z))/Q(x))) consists of exactly the cyclic subgroups of Gal(Q(ζn , z)/Q(x)). Then M |= θ if and only if Gal(L/L ∩ M ) ∈ Con. (d) Your answer to (c) should be explicit enough that you will be able to ˜ conclude that if gcd(n, 6) = 1, then the set of σ ∈ G(Q) such that Q(σ) |= θ has positive measure.

Notes The notion of Galois stratification is developed for the first time in [FriedSacerdote] in order to establish an explicit decision procedure for the theory of finite fields. In our opinion the real progress made by introducing Galois stratification is not so much the primitive recursiveness of this algorithm, but the precise algebraic-geometric indication of how to eliminate quantifiers. No notion from recursion theory is required to appreciate its importance. Chapter 30 is an elaboration of [Fried-Haran-Jarden, Section 3]. Since, however, [Fried-Haran-Jarden] treats only the special case where the Frobenius fields considered have a fixed superprojective group G as their absolute Galois group, this has required additions to [Fried-Haran-Jarden]. A discussion of the theory of Frobenius fields whose absolute Galois groups are pro-C-groups in the case where C is an arbitrary fixed full family of finite groups appears in [Jarden13]. A recursive variant of Theorem 30.7.3 (and the remark on limits of theories) appears in [Jarden5]. The language L(ring) is extended in [Ershov5] so that Galois formulas become formulas in the usual sense. Galois stratification then yield elimination of quantifier procedures. Galois stratification is used in [Jarden15] in order to establish an algebraic dimension for definable sets over Frobenius fields. [Haran-Lauwers] establishes Galois stratification over e-fold ordered Frobenius fields. [Denef-Loeser] uses Galois stratification to associate to each reduced and separated scheme of finite type over Z a canonical rational function which specializes for almost every p to the p-adic Poincar´e series of the scheme.

Chapter 31. Galois Stratification over Finite Fields Chapter 30 establishes the Galois stratification procedure over a fixed field K with elimination theory. The outcome is an explicit decision procedure for the theory of Frobenius fields that contain K. Section 31.1 modifies this to replace K by localizations Z[k −1 ] of Z. This, in particular, extends the above result to establish an explicit decision procedure for the theory of all Frobenius fields. The next two sections return to the opening subject of the book - finite fields. Although no finite field is Frobenius, as a collection they behave like Frobenius fields with respect to any given Galois sentence over Z[k −1 ], provided we forget about finitely many (exceptional) primes. As a consequence this establishes an explicit decision procedure for the theory of finite fields (Theorem 31.2.4). Section 31.3 applies the elimination of quantifiers procedure for finite fields and a theorem of Artin in representation theory to prove that the zeta function of a Galois formula over a finite field is essentially the radical of a rational function by reducing this to a corollary of Dwork’s theorem that the zeta function of a variety over a finite field is a rational function. The explicit version that we use is a result of Bombieri. The last section of this chapter gives a survey on some generalizations of the above zeta functions to p-adic fields.

31.1 The Elementary Theory of Frobenius Fields Let K be a presented field with elimination theory. Theorem 30.6.1 states that the theory of all perfect Frobenius fields M that contains the field K is primitive recursive. We now extend this result to ask about the theory of all perfect Frobenius fields in the language L(ring). This requires a slight change in basic definitions and a minor amendment to the stratification procedure. Instead of giving Galois statements over Q, we attach to each Galois stratification A a presented ring of the form Z[k−1 ] = R(A) = R with k = k(A) ∈ N and with this property: each algebraic set that appears in A is defined by polynomials with coefficients in R, and the coordinate ring of each algebraic set is finitely generated over R. The stratification procedure, in this case, starts with (A, R(A)) with A a stratification of An . The inductive step produces (B, R(B)) with B a Galois stratification of An−1 and R(B) = Z[k(B)−1 ] where k(B) is a multiple of k(A). The next lemma is an addition to Lemma 19.7.2 that allows us to construct appropriate normal Q-basic sets in this new context. Recall that a polynomial h ∈ Z[X1 , . . . , Xr ] is said to be primitive if the greatest common divisor of its coefficients is 1.

31.1 The Elementary Theory of Frobenius Fields

731

Lemma 31.1.1: Let (x1 , . . . , xn ) be a presented n-tuple over Q. Then we can effectively find a positive integer k0 and h ∈ Z[X1 , . . . , Xn ] such that h(x) 6= 0 and for each multiple k of k0 , the ring Z[k −1 , x, h(x)−1 ] is presented in Q(x) (Definition 19.1.1). Moreover, if x1 , . . . , xr is a transcendence basis for Q(x)/Q, then h can be chosen in Z[X1 , . . . , Xr ]. Proof: Without loss assume that x1 , . . . , xr is a transcendence basis for Q(x)/Q. For each i, r < i ≤ n use Q(x) = Q(x1 , . . . , xr )[xr+1 , . . . , xn ] to write fi (x1 , . . . , xi−1 , Xi ) irr(xi , Q(x1 , . . . , xi−1 )) = h(x1 , . . . , xr ) with fi ∈ Z[k0−1 , X1 , . . . , Xi ] with k0 ∈ N and h ∈ Z[X1 , . . . , Xr ] primitive. Let k be a multiple of k0 . Each element of S = Z[k−1 , x, h(x)−1 ] has a unique presentation X gi (x1 , . . . , xr ) i · x r+1 · · · xinn , h(x1 , . . . , xr )d(i) r+1 where ir+1 , . . . , in range from 0 to degXr+1 (fr+1 ) − 1, . . . , degXn (fn ) − 1, respectively, and for each i = (ir+1 , . . . , in ) the polynomial gi (x1 , . . . , xr ) of Z[k −1 , x1 , . . . , xr ] is not divisible in Q[x1 , . . . , xr ] by h(x1 , . . . , xr ). An arbitrary element u of Q(x) can be presented in the form (1)

X gi (x1 , . . . , xr ) i · x r+1 · · · xinn hi (x1 , . . . , xr ) r+1

where the exponents ir+1 , . . . , in satisfy the same condition as above and gi (x1 , . . . , xr ), hi (x1 , . . . , xr ) are polynomials in Z[x1 , . . . , xr ] which are relatively prime in Q[x1 , . . . , xr ]. A necessary condition for u to belong to Q[x, h(x)−1 ] is that each hi (x1 , . . . , xr ) divides a power of h(x1 , . . . , xr ) in pi (x1 ,...,xr ) Q[x1 , . . . , xr ]. In this case rewrite the ith coefficient in (1) as ci h(x e(i) 1 ,...,xr ) with pi ∈ Z[X1 , . . . , Xr ] and ci ∈ N. Then u belongs to Z[k −1 , x, h(x)−1 ] if −1 , x1 , . . . , xr ] for each i. This check is effective and and only if c−1 i pi ∈ Z[k it concludes a procedure to decide whether u belongs to Z[k −1 , x, h(x)−1 ].  Thus, Z[k −1 , x, h(x)−1 ] is presented. Lemma 31.1.2: Let (x1 , . . . , xn , z) be a presented (n + 1)-tuple over Q with z algebraic over Q(x). Then we can effectively find a polynomial g ∈ Q[X] and a common multiple k0 of the denominators of the coefficients of g such that for each multiple k of k0 (a) g(x) 6= 0, and the ring Z[k −1 , x, g(x)−1 ] is integrally closed and presented in Q(x); and (b) the ring Z[k−1 , x, g(x)−1 , z] is presented in Q(x, z) and is a cover of Z[k −1 , x, g(x)−1 ] with primitive element z. Proof: We follow the proof of Lemma 19.7.2:

732

Chapter 31. Galois Stratification over Finite Fields

Part A: (2a) gives rise to (2b). Suppose g and k0 are such that for each and disc(z) = dd10 (x) multiple k of k0 (2a) holds. Write irr(z, Q(x)) = f1f(x,Z) (x) 0 (x) with f0 , f1 , d0 , d1 polynomials with coefficients in Z. Lemma 31.1.1 produces a multiple k2 of k0 and a multiple h(x) of d0 (x)d1 (x)f0 (x)g(x) such that for each multiple k of k1 the ring Z[k −1 , x, z, g1 (x)−1 ] is presented in Q(x, z). By Definition 6.1.3, the latter ring is a cover of Z[k −1 , x, g1 (x)−1 ], with primitive element z. Part B: Proof of (2a). The case where n = 0 reduces to the fact that Z is integrally closed. Suppose n ≥ 1. An induction on n effectively gives a polynomial g0 ∈ Q[X1 , . . . , Xn−1 ] and a common multiple k0 of the denominators of the coefficients of g0 such that g0 (x1 , . . . , xn−1 ) 6= 0, and for each multiple k of k0 , the ring Z[k −1 , x1 , . . . , xn−1 , g0 (x1 , . . . , xn−1 )−1 ] is integrally closed and presented in Q(x1 , . . . , xn−1 ). If xn is transcendental over Q(x1 , . . . , xn−1 ), then Z[k −1 , x1 , . . . , xn , g0 (x1 , . . . , xn−1 )−1 ] is integrally closed [Zariski-Samuel2, p. 85 or p. 126] and presented in Q(x1 , . . . , xn ). Otherwise, use Part A with z = xn to effectively find a multiple g ∈ Q[X] of g0 and a common multiple k1 of k0 and the denominators of the coefficients of g such that g(x) 6= 0 and for each multiple k of k1 the ring Z[k −1 , x, g(x)−1 ] is presented in Q(x) and is a cover of Z[k −1 , x1 , . . . , xn−1 , g(x)−1 ]. In particular, Z[k −1 , x, g(x)−1 ] is integrally closed.  Let A = V (f1 , . . . , fm ) r V (g) be a presented normal Q-basic set with a generic point x = (x1 , . . . , xn ), and let Q[x, g(x)−1 , z] be a presented Galois ring cover of A over Q. Apply Lemma 31.1.2 to find a multiple h ∈ Q[X] of g and a positive integer k such that the coefficients of f1 , . . . , fm , g, h lie in Z[k−1 ], h(x) 6= 0, Z[k −1 , x, h(x)−1 ] is an integrally closed ring, presented in Q(x), and Z[k −1 , x, h(x)−1 , z]/Z[k −1 , x, h(x)−1 ] is a ring cover. Then A0 = A r V (h) is a presented normal Q-basic open subset of A. Call Z[k−1 , A0 ] = Z[k −1 , x, h(x)−1 ] the coordinate ring of A0 over Z[k−1 ]. We say that A0 is normal over Z[k−1 ]. Also, with C = Z[k−1 , x, h(x)−1 , z] call C/A0 a Galois ring /set cover over Z[k −1 ]. Let M be a field of characteristic p (which might be 0). Suppose p - k. Then reduction modulo p of Z uniquely extends to Z[k −1 ]. In this case, with A0 as above, let A0 (M ) = {a ∈ M n | f1 (a) = · · · = fm (a) = 0 and h(a) 6= 0} where the coefficients of f1 , . . . , fm and h are interpreted as being in Fp (or in Q if p = 0). For each a ∈ A0 (M ) there is a unique homomorphism ϕ0 : Z[k −1 , A0 ] → M with ϕ0 (x) = a. As usual the homomorphism ϕ0 extends to a homomorphism ϕ of C into a finite Galois extension N = M (ϕ(z)) of M . We may define the decomposition group of ϕ with respect to M as in Section 30.1: DM (ϕ) = {σ ∈ Gal(C/A0 ) | (∀u ∈ C)[ϕ(u) ∈ M =⇒ ϕ(σ(u)) = ϕ(u)]}.

31.1 The Elementary Theory of Frobenius Fields

733

It is a subgroup of Gal(C/A0 ) isomorphic to Gal(N/M ) (Lemma 6.1.4). Then we define Ar(C/A0 , M, a) (or Ar(A0 , M, a), if there can be no confusion) to be the conjugacy class of subgroups of Gal(C/A0 ) generated by DM (ϕ). Let n ≥ 0, A a Q-constructible set in An , and k be a nonzero integer. A normal stratification A = hA, Ci /Ai ii∈I S of A over Z[k−1 ] is a decomposition A = · i∈I Ai of A as a finite union with a Galois ring of disjoint normal Z[k−1 ]-basic sets Ai , each equipped S cover Ci over Z[k −1 ]. Assume also that A(M ) = · i∈I Ai (M ) for each field M with char(M ) = p and p - k. In this case for each family H of finite groups, A can be augmented to a Galois stratification A(H) with respect to Z[k −1 ] by adding conjugacy domains Con(Ai , H), as in Section 30.3. For each a ∈ A(M ), write Ar(A, M, a) ⊆ Con(A(H)) if Ar(Ai , M, a) ⊆ Con(Ai , H) for the unique i ∈ I such that a ∈ Ai (M ). Here, as above, we interpret the polynomials defining A and Ai as having coefficients in Fp , so that A(M ) and Ai (M ) are well defined. Note, however, that the reduction of Z[k−1 , Ai ] modulo p is not assumed to be integrally closed. Thus, the reduction of A modulo p need not be a normal stratification of A over Fp . In case n = 0, the affine space A0 consists of the origin O only. Let C/Z[k −1 ] be the corresponding Galois ring cover and let L = Q(C). Notationally simplify Ar(A, M, O) to Ar(L/Q, M ) and Con(A(H)) to Con(H). If char(M ) = 0, then Ar(L/Q, M ) ⊆ Con(H) means that Gal(L/L ∩ M ) ∈ H. If M = Fp , and p - k, then Ar(L/Q, M ) is the conjugacy class of subgroups
of Gal(L/Q) generated by the elements of the classical Artin symbol L/Q p . Now define Galois formulas and Galois sentences over Z[k −1 ] as in Section 30.4. In particular, the stratification lemma (Lemma 19.7.3) and Lemma 31.1.2 imply, as in Remark 30.5.1, that each formula ϕ(X) of L(ring) is equivalent to a Galois formula θ(X) over Z[k −1 ] for a suitable k which can be effectively computed. Thus, if M is a field with char(M ) - k and b1 , . . . , bn ∈ M , then M |= θ(b) if and only if M |= ϕ(b). Proposition 31.1.3: Let θ be a sentence of L(ring). Then we can effectively find a finite family S of finite groups, a finite Galois extension L of Q with Subgr(Gal(L/Q)) ⊆ S, and a positive integer k such that the following holds: For each subfamily H of S that contains the trivial group we can effectively find a conjugacy class Con(H) ⊆ Subgr(Gal(L/K)) ∩ H such that for each perfect Frobenius field M with char(M ) - k and Im(Gal(M )) ∩ S = H (2)

M |= θ ⇐⇒ Ar(L/Q, M ) ∈ Con(H).

Sketch of proof: Write θ in prenex normal form with m quantifier Q1 , . . . , Qm . Then apply the stratification lemma (Lemma 19.7.3) and Lemma 31.1.2 to find a positive integer k0 such that θ is defined over Z[k0−1 ], and a Galois stratification Am of Am over Z[k0−1 ] such that for each basic set Ai in

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Chapter 31. Galois Stratification over Finite Fields

Am , Con(Ai ) is either empty or consists of the trivial group, and the Galois sentence θ0 , (Q1 X1 ) · · · (Qm Xm )[Ar(X) ⊆ Con(Am )], is equivalent to θ over each perfect Frobenius field M with char(M ) - k. As in the proof of Proposition30.4.2 apply Lemmas 31.1.2 and 19.7.3 to proceed inductively for i, 0 ≤ i ≤ m: find ki , with ki+1 a multiple of ki , and a normal stratification Am−i of Am−i over Z[ki−1 ] which eliminates Qi from θ0 with respect to each perfect Frobenius Smfield M with char(M ) - ki . Now take L as the ring cover of Q in A0 , S = i=0 Subgr(Ai ), and the final integer k to be km . Then the conclusion of the proposition holds. For the verification one must replace each occurrence of the phrase “a field M that contains K” in the appropriate lemmas of Chapter 30 that contribute to Propositions 30.5.2 and 30.5.3, by the phrase “a field M of characteristic not dividing k.” The stratification procedure is carried out in characteristic 0, over the rings Z[ki−1 ]. The decomposition groups with respect to points over Q behave as in the case of Chapter 30. In particular, they are isomorphic to the Galois groups of the corresponding residue fields. Thus, the conclusion of Propositions 30.5.2 and 30.5.3 hold.  The next theorem generalizes Theorem 30.5.3 to include all perfect Frobenius fields: Theorem 31.1.4: Let C be a primitive recursive full family of finite groups. (a) For each given sentence θ of L(ring) we can effectively find a nonzero integer k with the following property: If θ is true in all fields M ∈ Frob(Q, pro-C), then θ is true in all fields M ∈ Frob(Fp , pro-C) such that p - k. (b) The theory of perfect Frobenius fields M such that Gal(M ) is a pro-Cgroup is primitive recursive. Proof of (a): Find S, L and k as in Proposition 31.1.3. Suppose θ is true in all fields M ∈ Frob(Q, pro-C). Let H1 , . . . , Hr be all subfamilies of C ∩ S such that there exists a superprojective group Gi with Im(Gi ) ∩ S = Hi , i = 1, . . . , r. Consider a group H ∈ Hi ∩ Subgr(Gal(L/Q)). Lemma 24.1.6 produces a Frobenius field M of characteristic 0 such that Gal(M ) ∼ = Gi and Gal(L/L ∩ M ) = H. Since M |= θ, Proposition 31.1.3 implies that H ∈ Con(Hi ). Therefore, (3)

Con(Hi ) = Hi ∩ Subgr(Gal(L/Q)),

i = 1, . . . , r.

Now consider M ∈ Frob(Fp , pro-C) for p - k. Then Im(Gal(M )) ∩ S = Hi for some i, 1 ≤ i ≤ r. Since Subgr(Gal(L/Q) ⊆ S, (3) gives Ar(L/Q, M ) ∈ Con(Hi ). By Proposition 31.1.3, M |= θ. This proves (a). Proof of (b): The proof of Theorem 30.6.1 gives an effective check for condition (3). If (3) holds, then Theorem 30.6.1 gives a check to decide for each  prime p|k if θ is true in each field M ∈ Frob(Fp , C). This proves (b).

31.2 The Elementary Theory of Finite Fields

735

In particular, we may take C to be the family of all finite groups.

Corollary 31.1.5: The theory of perfect Frobenius fields is primitive recursive.

We leave to the reader the formulation and proof of the appropriate analog of Theorem 30.6.3.

31.2 The Elementary Theory of Finite Fields A special case of Theorem 30.7.2 implies that the theory Almost(Q, 1) of all ˜ sentences θ of L(ring) which are true in Q(σ) for almost all σ ∈ Gal(Q), is primitive recursive. By the transfer theorem (Theorem 20.9.3), each such θ is true in Fp , for almost all primes p. Thus, we have a primitive recursive procedure to decide the truth of a given sentence θ of L(ring) in Fp for almost all primes p. Still, even if θ is true in Fp for almost all p, this does not yet include a procedure to compute the finite set of exceptional primes p for which θ is not true in Fp . This section modifies the Galois stratification procedure to fill this gap. The basic idea is to replace the Frobenius fields in the procedure of Section 31.1 by finite fields. Since the procedure depends on the Frobenius property, which does not hold for finite fields, we must show that, relative to a given sentence θ, the finite fields behave like Frobenius fields provided they are sufficiently large. Exactly how large depends on the following quantitative variant of Proposition 6.4.8:

Lemma 31.2.1 ([Haran-Jarden1, p. 15]): Let d be a positive integer, M a finite field, g ∈ M [Y ], f ∈ M [Y, Z] polynomials with g 6= 0, y an indetermi^ ˜ ∩M (y, z). nate, and z ∈ M (y). Put R = M [y, g(y)−1 ], C = R[z], and N = M Suppose q = |M | ≥ d4 , deg(g) < d, f (y, Z) = irr(z, M (y)), deg(f ) ≤ d, and C/R is a Galois ring cover with primitive element z. Consider τ ∈ Gal(C/R) satisfying hresN τ i = Gal(N/M ). Then there exists an M -homomorphism ˜ with ψ(R) = M and DM (ψ) = hτ i. ψ: C → M ˆ ˜ (y, z). Since Gal(M ) ∼ Proof: Let E = M (y), F = M (y, z) and P = M = Z, resN τ extends to a generator τ0 of Gal(M ). Then τ extends to an element τ˜ ∈ ˜ is M . τ ) with M Gal(P/E) such that resM˜ τ˜ = τ0 . The intersection of D = P (˜ In particular, D is a regular extension of M and the map resM˜ : Gal(P/D) → ˆ res ˜ is bijective (Proposition Gal(M ) is surjective. Since Gal(M ) ∼ = Z, M ˜ D = P , so [D : E] = [P : M ˜ E] ≤ [F : E] ≤ d. Thus, D is 16.10.6). Hence, M

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Chapter 31. Galois Stratification over Finite Fields

a function field of one variable over M . ˜ M

N

M

˜E M

P x xx x F xx xx NE ND

E

D

Let n be the number of prime divisors of D/M of degree 1. By the Riemann hypothesis (Theorem 4.5.2) √ |n − (q + 1)| ≤ 2gD q, where gD = genus(D/M ). Since P is a separable constant field extension of both D and F , gD = gF (Proposition 3.4.2). Corollary 5.3.5 now gives gF ≤ By assumption,

√

1 1 (deg(f ) − 1)(deg(f ) − 2) ≤ (d − 1)2 . 2 2

q ≥ d2 ≥ (d − 1)2 + 1. Hence

√ √ n ≥ (q + 1) − 2gF q ≥ (q + 1) − (d − 1)2 q √ √ √ = 1 + q( q − (d − 1)2 ) ≥ 1 + q ≥ 1 + d2 . Thus, D/M has at least d2 + 1 prime divisors of degree 1. Also, E/M has at most (1 + deg(g)) ≤ d prime divisors which are not finite on R. Each of them has at most [D : E] ≤ d extensions to D. Using (1 + d2 ) − d2 = 1, we find an M -rational place ψ0 : D → M ∪{∞} finite on R such that ψ0 (R) = M . ˜ P →M ˜ ˜ is the identity map. ˜ ∪{∞} such that ψ| Extend ψ0 to a place ψ: M ˜ Put ψ = ψ|C . Then ψ: C → M is an M -homomorphism and ψ(R) = M . ˜ S is the integral closure Let S be the integral closure of R in D. Then M ˜ ˜ of R in P (Lemma 2.5.8). From the definitions, ψ(˜ τ x) = τ0 ψ(x) for each x ∈ ˜ ˜ M and each x ∈ S. Since C ⊆ M D0 , this holds also for each x ∈ C. In particular, ψ(τ x) = τ0 ψ(x) for each x ∈ C. Let ψ ∗ : Gal(ψ(C)/M ) → Gal(F/E) be the embedding induced by ψ. Then ψ ∗ (resψ(C) τ0 ) = τ . Consequently,  DM (ψ) = hτ i. Finite fields compensate for their being only “approximately” Frobenius fields, by the simplicity of their absolute Galois groups. Using this, there is a simple finite field analog to Proposition 31.1.3 which costs us a few more “exceptional primes.”

31.2 The Elementary Theory of Finite Fields

737

Proposition 31.2.2: Let k be a positive integer and let θ(Y) be the Galois formula (Q1 X1 ) · · · (Qm Xm )[Ar(X, Y) ⊆ Con(A(C))] with respect to a Galois stratification A(C) of Am+n over Z[k −1 ], where C is the family of all finite cyclic groups. Then we can effectively find a multiple l of k and a Galois stratification B(C) of An over Z[l−1 ] such that the following holds: (1) for each finite field M of characteristic not dividing l, and for each b ∈ An (M ), M |= θ(b) ⇐⇒ Ar(B, M, b) ⊆ Con(B(C)). Sketch of Proof: As in the proof of Proposition 30.5.2, we may use the stratification lemma and Lemma 31.1.2 to find a multiple l0 of k and to eliminate the quantifiers Qm , Qm−1 , . . . , Q1 from θ(Y) with respect to C over Z[l0−1 ]. From this we may successively obtain Galois stratifications Bm−1 (C),Bm−2 (C), . . . , B0 (C) of Am−1+n , Am−2+n , . . . , An , respectively, over Z[l0−1 ]. Denote A(C) by Bm (C). Enlarge l0 if necessary to assume that for each i, 0 ≤ i ≤ m, and for each prime p - l0 the reduction of Bi modulo p gives a stratification of Ai+n in characteristic p (Theorem 9.3.1). Let C/A be a Galois ring/set cover involved in the normal stratification Bi+1 , 0 ≤ i ≤ m − 1. Suppose the projection π: Ai+1+n → Ai+n maps A onto a basic set B (of Bi ) and (2)

dim(A) = dim(B) + 1.

Find a generic point x = (x1 , . . . , xn+i ) of B, a transcendental element y over Q(B) with Q(A) = Q(B)(y), a polynomial g ∈ Z[l0−1 , X1 , . . . , Xn+i , Y ] with Z[l0−1 , A] = Z[l0−1 , x, y, g(x, y)−1 ], a primitive element z for the ring cover C/Z[l0−1 , A] and a polynomial f ∈ Z[k −1 , X1 , . . . , Xn+i , Y, Z] so that f (x, y, z) = 0 and the coefficient of the highest power of Z in f divides g(X1 , . . . , Xn+i , Y ). Define l(C/A) to be the product of l0 with all primes which do not exceed max((deg(Y,Z) f )4 , (degY g)4 ). Let M be a finite field with p = char(M ) - l(C/A). Then Bi+1 and Bi stratify Ai+1+n and Ai+n , respectively. Therefore, π maps the reduction of A modulo p onto the reduction of B modulo p. Let y 0 be a transcendental element over M . If b ∈ B(M ) this implies that g(b, y 0 ) 6= 0. Hence, deg(g(b, Y )) = degY g(X, Y ) and deg(f (b, Y, Z)) = deg(Y,Z) f . Also, the specialization compatibility of the Galois covers of A and B implies that f (b, y 0 , Z) is irreducible over M (y 0 ). Thus, we may replace the use of Proposition 24.1.4 (the Frobenius property) in Part B of the proof of Lemma 30.2.3 by Lemma 31.2.1. By Remark 30.2.4, Part B of the proof of Lemma 30.2.3 is the only point where the Frobenius property of M is used. Therefore, take l to be the least common multiple of all l(C/A) as i ranges from 0 to m − 1 and C/A runs over all covers involved in the normal stratification Bi+1 for which (2) holds.

738

Chapter 31. Galois Stratification over Finite Fields

With this value of l and with B(C) = B0 (C) the conclusion of the proposition holds.  In the next result we replace Z[k −1 ] by a finite field: Proposition 31.2.3: Let K be a finite field, C a family of finite cyclic groups, and θ(Y) the Galois formula (Q1 X1 ) · · · (Qm Xm )[Ar(X, Y) ⊆ Con(A(C))] with respect to a Galois stratification A(C) of Am+n over K. Then we can effectively find a positive integer c and a Galois stratification B(C) of An over K such that the following holds: (3) for each finite field M of cardinality exceeding c that contains K and for each b ∈ An (M ), M |= θ(b) ⇐⇒ Ar(B, M, b) ⊆ Con(B(C)). Proof: Apply the Galois stratification procedure to A(C) over K. For each Galois ring/set cover C/A that appears in the Galois stratification procedure for which dim(A) = dim(π(A)) + 1 we may assume that k[A] = K[x1 , . . . , xn+i , y, g(x1 , . . . , xn+i , y)−1 ] and C = K[A][z] as in the proof of Proposition 31.2.2. Let f (x, y, Z) = irr(z, K(x, y)). Then choose c to exceed max((deg(Y,Z) f )4 , (degY g)4 ), for all C/A. Use Lemma 31.2.1 and Remark 30.2.4 to conclude Condition (3).  We are now ready to prove the primitive recursive version of Theorem 20.10.6. Theorem 31.2.4 ([Fried-Sacerdote]): Let θ be a sentence of L(ring). (a) There is a primitive recursive procedure to decide if θ is true in almost every field Fp and to compute the set of exceptional primes in case θ is true in almost every field. (b) The theory of sentences θ of L(ring) which are true in each field Fp is primitive recursive. (c) The theory of sentences θ of L(ring) which hold in almost every finite field (equivalently, in every pseudo-finite field, Proposition 20.10.2) is primitive recursive. (d) The theory of finite fields is primitive recursive. Proof: Apply Proposition 31.2.2, in the case n = 0, to θ. This gives a nonzero integer l, a finite Galois extension L of Q, and a conjugacy class Con of cyclic subgroups of Gal(L/Q) satisfying this: For each finite field M with char(M ) - l, (4)

M |= θ ⇐⇒ Ar(L/Q, M ) ⊆ Con.

Suppose there exists σ ∈ Gal(L/Q) with hσi ∈ / Con. By the Chebotarev density theorem (Theorem 6.3.1), there exist infinitely many primes p not

31.3 Near Rationality of the Zeta Function of a Galois Formula

739


dividing l such that L/Q = Ar(L/Q, Fp ) is the conjugacy class of groups p generated by hσi. Thus, Ar(L/Q, Fp ) 6⊆ Con. By (4), θ is false in Fp . If, on the other hand, Con contains each cyclic subgroup of Gal(L/Q), then the right hand side of (4), and therefore its left hand side, is true for each finite field M with char(M ) - l. In particular, Fp |= θ for each prime p - l. Assume that Fp |= θ for each p - l. Check directly, for each prime p | l, if Fp |= θ. Determine in this way the finite set of primes p such that Fp 6|= θ. This concludes the proofs of (a) and (b). The argument above actually shows that we can decide the truth of θ in M for each finite field M of characteristic not dividing l. It remains to check the truth of θ in each field of characteristic p where p|l. Apply Proposition 31.1.3 to effectively find a positive integer mp for which we can decide the truth of θ in all finite fields of characteristic p and of cardinality exceeding mp . If θ is true in all these fields with p running over the divisors of l, then we can check directly the truth of θ in M as M runs over the finite list of finite fields with the property that char(M ) = p | l and |M | ≤ mp . This proves (d) and (e). 

31.3 Near Rationality of the Zeta Function of a Galois Formula Let K be the finite field with q elements. Denote the unique extension of K of degree k by Kk . Throughout this section C is the family of finite cyclic groups. Consider a Galois formula θ(Y1 , . . . , Yn ) over K with respect to a Galois stratification A(C), A(C) = hAm+n , Ci /Ai , Con(Ai , C)ii∈I . For each positive integer k let ν(θ, k) = #{b ∈ Kkn | Kk |= θ(b)}. Define the Poincar´ e series of θ to be the formal power series (1)

P (θ, t) =

∞ X

ν(θ, k)tk .

k=1

The zeta function, Z(θ, t), is then defined by the formulas (2) where

(3)

P (θ, t) = t d dt

d (log Z(θ, t)) dt

and Z(θ, 0) = 1,

indicates the operation of taking the formal derivative. Thus,

Z(θ, t) = exp

∞ X ν(θ, k) k=1

k

! t

k

,

740

Chapter 31. Galois Stratification over Finite Fields

P∞ n where exp(t) = n=0 tn! . The remainder of this section consists of the proof that P (θ, t) is rational in t, that Z(θ, t) is “nearly” rational in t and that the polynomial expressions involved in these statements are given by a primitive recursive procedure. We give a brief account of the definitions and results of the representation theory of finite groups (see, e.g. [Serre4], for details) as these will be used in the proof. A representation of a finite group G is a homomorphism ρ: G → GL(n, C). The character χ of ρ is the complex valued function on G obtained by composing ρ with the trace: χ(σ) = trace(ρ(σ)). If L is a subfield of C with ρ(σ) ∈ GL(n, L) for all σ ∈ G, then ρ and χ are said to be rational over L. In this case χ(σ) ∈ L for each σ ∈ G (but the converse does not hold [Serre4, Section 12.2]). Call a character χ of G irreducible over L if it is rational over L and it is not the sum χ = χ1 + χ2 of two rational characters over L. There are only finitely many irreducible characters over L and they are linearly independent. Note (as a property of taking trace) that a character χ of G is constant on conjugacy classes of elements: (4)

χ(τ −1 στ ) = χ(σ) for all σ, τ ∈ G.

Any function χ: G → C that satisfies (4) is said to be central. We record here a criterion for a central function to belong to the L-linear space generated by the L-irreducible characters of G [Serre4, Section 12.4] in the case where L = Q. Lemma 31.3.1: Consider a function χ: G → Q from a finite group G. The following conditions are equivalent: (a) χ is a linear combination over Q of the Q-irreducible characters of G over Q. (b) χ is central and χ(σ) = χ(σ i ) for each σ ∈ G and for each i ∈ Z, with gcd(i, ord(σ)) = 1. (c) χ(σ) = χ(σ 0 ) for all σ, σ 0 ∈ G such that hσi is conjugate to hσ 0 i. Definition 31.3.2: Refer to a central function χ: G → Q as Q-central if Condition (a) (or (b) or (c)) of Lemma 31.3.1 holds.  Consider a subgroup H of G. Let ρ0 : H → GL(m, C) be a representation of H. The group GL(m, C) acts faithfully on the vector space W = Cm by multiplication from the left. We may therefore consider ρ0 as a homomorphism from H into Aut(W ). Let V = IndG H W be the vector space of all functions f : G → W satisfying f (ησ) = ρ0 (η)f (σ) for all η ∈ H and σ ∈ G. Define a homomorphism ρ: G → Aut(V ) by (ρ(τ )f )(σ) = f (στ ),

σ, τ ∈ G.

Now choose a basis for V to consider ρ as a representation ρ: G → GL(n, C), with n = dim(V ) = (G : H)m. This is the induced representation of ρ0

31.3 Near Rationality of the Zeta Function of a Galois Formula

741

G from H to G: ρ = indG H ρ0 . The corresponding characters χ0 and χ = indH χ0 are related by the formula

χ(σ) =

1 X χ0 (τ στ −1 ), |H|

where τ ranges over all elements of G satisfying τ στ −1 ∈ H. In particular, let 1H be the trivial character of H, 1H (η) = 1 for each G η ∈ H. Then 1G H = indH 1H is given by the formula 1G H (σ) =

(5)

1 #{τ ∈ G | τ στ −1 ∈ H}. |H|

P Lemma 31.3.3: Let χ: G → Q be a Q-central function. Then χ = H rH 1G H, where H ranges over all cyclic subgroups of G and rH are rational numbers which can be effectively computed from G and χ. Proof: The existence of the rH ’s is a well known result of a theorem of Artin [Serre4, p. 103]. Now suppose that the multiplication table of G is known and that χ is presented. For each cyclic subgroup H of G compute the induced character 1G H from (5). Then solve the system of linear equations χ(σ) =

X

r H 1G H (σ),

σ∈G

H

in the unknowns rH , where H ranges over all cyclic subgroups of G.



Now suppose that C/A is a Galois ring/set cover over K with G = Gal(C/A). Let χ: G → Q be a Q-central function. For all k ∈ N and a ∈ A(Kk ) choose an element σ ∈ G with hσi ∈ Ar(C/A, Kk , a) and define χ(k, a) to be χ(σ). If σ 0 is another element of G such that hσ 0 i ∈ Ar(C/A, Kk , a), then hσi is conjugate to hσ 0 i. Hence, χ(σ 0 ) = χ(σ) and consequently χ(k, a) is well defined. Define the L-series of χ to be   ∞ X X 1 (6) L(C/A, χ, t) = exp  χ(k, a)tk  k k=1

a∈A(Kk )

This definition agrees with other definitions of the L-series (e.g. as in [Serre2, p. 87] or [Dwork1, p. 44]). Indeed, for a ∈ A(Kk ) let deg(a) = [K(a) : K] be the number of points of A conjugate to a over K. Let N (a) = q deg(a) = ˜ be a homo|K(a)|. Choose a generic point x of A over K and let ϕ: C → K morphism which extends the specialization x → a. Then Ar(C/A, Kk , a) is the class of subgroups of G conjugate to DKk (ϕ) (Section 31.1). The group DK(a) (ϕ) is generated by an element σ that corresponds to the automorphism deg(a)

x 7→ xq of ϕ(C) over K(a) while DKk (ϕ) is generated by σ k/ deg(a) . Denote the prime ideal of K[A] corresponding to the K-specialization x → a

742

Chapter 31. Galois Stratification over Finite Fields

(and also x → a0 , for each conjugate a0 of a over K) by p and let N (p) = N (a). Then, in the notation of [Serre2, p. 87], χ(k, a) = χ(pk/ deg(a) ). Put t = q −s and m = k/ deg(a). Then X χ(k, a0 )tk a0

k

=

χ(pm )N (p)−ms , m

where a0 ranges on the points conjugate to a. Thus ∞ XX χ(pm )N (p)−ms L(C/A, χ, t) = exp m p m=1

! ,

where p ranges over all primes of A. This is exactly (8) of [Serre2, p. 87]. In the special case where χ = 1G the L-series becomes the zeta function of A: ! ∞ X 1 k |A(Kk )|t . (7) L(C/A, 1G , t) = Z(A, t) = exp k k=1

If χ1 and χ2 are Q-central functions of G, then (6) implies that L(C/A, χ1 + χ2 , t) = L(C/A, χ1 , t) · L(C/A, χ2 , t). Hence, (8)

L(C/A, rχ, t) = L(C/A, χ, t)r

for each Q-central function χ of G and every r ∈ Q. If H is a subgroup of G let B be a K-variety such that K[B] is the integral closure of K[A] in the fixed field of H in K(C). Then C/B is a Galois ring/set cover and for each character χ of H, rational over Q (9)

L(C/B, χ, t) = L(C/A, indG H χ, t)

(see [Serre2, p. 88] and also [Cassels-Fr¨ ohlich, p. 222] for a proof which is valid in our more general situation). Definition 31.3.4: The total degree of a rational function r ∈ C(t) is the sum deg(f ) + deg(g), where f, g ∈ C[t] are relatively prime polynomials with r = fg . Alternatively, the total degree of r is the total number of zeros and poles of r counted with multiplicity.  In the rest of this section, whenever we speak about a “rational function”, we mean “a rational function with coefficients in C ”.

31.3 Near Rationality of the Zeta Function of a Galois Formula

743

Lemma 31.3.5 (Dwork–Bombieri): Let V = V (f1 , . . . , fm ) be the K-closed subset of An defined by f1 , . . . , fm ∈ K[X1 , . . . , Xn ]. Then Z(V, t) is a rational function of total degree at most (5 + 4 max1≤i≤m (1 + deg(fi )))m+n . Proof: The rationality of Z(V, t) is proved in [Dwork1]. In [Dwork2, lemma 14.1, p. 489] Dwork gives a bound on certain vector spaces which leads to a bound on the total degree of Z(V, t). Here we follow [Bombieri2]: Pak−1 i Let tracek : Kk → Fp be the trace from Kk to Fp , tracek (x) = i=0 xp a 2πi·tracek (x)/p where q = p . Then ψk : K → C defined by ψk (x) = e is a nontrivial character of the additive group of Kk . Thus, ψk (x + y) = ψk (x)ψk (y) for all x, y ∈ Kk , ψk (0) = 1, and there is a c ∈ Kk with ψk (c) 6= 1. For f ∈ K[X1 , . . . , Xn ] use the exponential sum X

Sk =

ψk (f (x))

x∈Kkn

to define an L-series L(An , f, t) = exp

∞  X 1 S k tk . k k=1

By [Bombierie2, Theorem 2], L(An , f, t) is a rational function and (5 + apply Bombieri’s result to Z(V, t) 4 deg(f ))n is a bound on its total degree. To Pm replace n by n + m and take f (X, Y) = i=1 fi (X)Yi . If x ∈ Kkn r V (Kk ), then there is P an i between 1 and m with fi (x) 6= 0. Hence, there exists m b ∈ Kkm with i=1 fi (x)bi = c. Thus, X

ψk (f (x, y)) =

y∈Kk

X

ψk (f (x, y + b)) =

=




ψk f (x, y) ψk (c)

y∈Kkm

so

P

y∈Kk

ψk (f (x, y)) = 0. If x ∈ V (Kk ), then f (x, y) = 0. Therefore, Sk =

X

X

ψk (f (x, y)) = q km |V (Kk )|.

x∈V (Kk ) y∈Kkm

Consequently, Z(V, q m t) = exp

∞  X 1 |V (Kk )|(q m t)k k k=1

∞ X  1 = exp Sk tk = L(An+m , f, t). k k=1




ψk f (x, y) + f (x, b)

y∈Kkm

y∈Kk

X

X

744

Chapter 31. Galois Stratification over Finite Fields

Since deg(f ) ≤ max1≤i≤m (1 + deg(fi )), the expression (5 + 4 max (1 + deg(fi )))m+n 1≤i≤m

bounds the total degree of Z(V, t).



The coefficients of a monic polynomial with P the roots x1 , . . . , xn can be n effectively expressed as a polynomials in the sums i=1 xki , k = 1, . . . , n, with integral coefficients [Bˆ ocher, p. 241]. To generalize this to rational functions we introduce the following polynomials sk (X, Y) =

n X

Yjk

−

j=1

m X

Xik ,

k = 1, 2, 3, . . . .

i=1

Lemma 31.3.6: Let F be a field of characteristic 0, x1 , . . . , xm , y1 , . . . , yn ∈ F , t an indeterminate, andQr a positive integer with m + n ≤ r. Put Qm n g(t) = i=1 (1−xi t), h(t) = j=1 (1−yj t), and sk = sk (x, y), k = 1, 2, 3, . . . . Suppose (s1 , . . . , sr ) is presented. Then we can effectively compute polynomig(t) als g1 (t), h1 (t) ∈ Q(s1 , . . . , sr )[t] such that g1 (0) = h1 (0) = 1 and hg11(t) (t) = h(t) . Proof: The proof splits into two parts. Part A: A system of linear equations for the coefficients of g(t) and h(t). Let g(t) = 1+u1 t+· · ·+um tm and h(t) = 1+v1 t+· · ·+vn tn . A straightforward computation in the field F ((t)) of formal power series shows that ∞

∞

m

n

XX XX d(log(g(t))) d(log(h(t))) =− =− xk+1 tk and yjk+1 tk . i dt dt i=1 j=1 k=0

k=0

Hence, ∞

d g(t)  X log = sk+1 tk . dt h(t) k=0

Therefore, 



g(t) log h(t)

(10)

=

∞ X sk k=1

k

tk .

Apply exp on both sides of (10): ∞

∞

∞

k=1

k=1

k=1

X sk g(t) 1  X sk k 2 1  X sk k 3 =1+ tk + t t + + ··· . h(t) k 2! k 3! k

31.3 Near Rationality of the Zeta Function of a Galois Formula

745

Thus, ∞

X g(t) =1+ d k tk , h(t)

(11)

k=1

where dk =

(12)

1 sk + hk (s1 , . . . , sk−1 ) k

and hk is a polynomial with rational coefficients which does not depend on s1 , . . . , sk−1 and which can be effectively computed, k = 1, 2, 3, . . . . Multiply (11) by h(t): (13)

1 + u1 t + · · · + um tm = (1 +

∞ X

dk tk )(1 + v1 t + · · · + vn tn ).

k=1

Now compare the coefficients of tk , k = 1, 2, . . . , r on both sides of (13) to obtain r equalities, (14)

m X

ekj uj +

j=1

n X

ek,m+j vj = dk ,

k = 1, 2, . . . , r,

j=1

with computable ekj in {0, ±1, ±d1 , . . . , ±dk−1 }, k = 1, 2, . . . , r. Part B: Computation of g1 (t) and h1 (t). Put E = Q(s1 , . . . , sr ). Consider (14) as a system of linear equations in the variables u1 , . . . , um , v1 , . . . , vn with coefficients ekj ∈ E. Since the system has a solution in F m+n , we may a solution (u01 , . . . , u0m , v10 , . . . , vn0 ) ∈ E m+n . Thus, Pcompute Pm effectively n 0 0 j=1 ekj uj + j=m+1 ek,m+j vj = dk , k = 1, . . . , r, so (15)

1 + u01 t + · · · + u0m tm = (1 +

∞ X

dk tk )(1 + v10 t + · · · + vn0 tn ).

k=1

On the other hand let g1 (t) = 1 + u01 t + · · · + u0m tm

and h1 (t) = 1 + v10 t + · · · + vn0 tn .

Consider a factorization of g1 and h1 in F˜ [t]: g1 (t) =

m Y i=1

(1 − x0i t)

and h1 (t) =

n Y

(1 − yj0 t),

j=1

746

Chapter 31. Galois Stratification over Finite Fields

Then put s0k = sk (x0 , y0 ), k = 1, 2, 3, . . . . Replacing ui , vj , sk in Part A by u0i , vj , s0k , respectively, gives 1+

∞ X s0

∞ ∞ 1 X s0k k
2 1 X s0k k
3 1 + u01 t + · · · + u0m tm t + t t + +··· = , k 2! k 3! k 1 + v10 t + · · · + vn0 tn k k

k=1

so, by (15), dk =

k=1

1 0 k sk

k=1

+ hk (s01 , . . . , s0k−1 ) for k = 1, . . . , r. Hence, by (12),

s0k = sk for k = 1, . . . , r. Thus, an application of (10) to  log

g(t)h1 (t) h(t)g1 (t)

 =

g1 (t) h1 (t)

gives

∞ X sk − s0k k t . k

k=r+1

Hence, ∞ X g(t)h1 (t) =1+ ck tk h(t)g1 (t) k=r+1

with ck ∈ F˜ , k = r + 1, r + 2, . . . . Therefore, (16)

∞ X

g(t)h1 (t) = h(t)g1 (t) +

bk t k ,

k=r+1

with bk ∈ F˜ , k = r + 1, r + 2, . . . . Since deg(gh1 ), deg(hg1 ) ≤ m + n ≤ r, g(t) = hg11(t)  (16) implies that g(t)h1 (t) = h(t)g1 (t). Consequently, h(t) (t) . Theorem 31.3.7: Suppose the Galois formula θ(Y1 , . . . , Yn ) of the beginning of this section is presented. Then (a) we can effectively compute polynomials f, g, h ∈ Q[t] and a positive integer l such that (17)


 g(t) 1/l h(t)

Z(θ, t) = exp f (t)

(b) and we can effectively compute p, q ∈ Q[t] such that P (θ, t) =

p(t) q(t) .

Proof: The assertion about P (θ, t) is an immediate consequence of (2) and of (a). The proof of (a) applies the Galois stratification procedure. It splits into three parts: Part A: Elimination of quantifiers. By Proposition 31.2.3, we can effectively find a quantifier free Galois formula θ0 (Y) and a positive integer c such that the following holds for each k > c and for each a ∈ (Kk )n : (18)

Kk |= θ(a) ⇐⇒ Kk |= θ0 (a).

31.3 Near Rationality of the Zeta Function of a Galois Formula

747

In particular, ν(θ, K) = ν(θ0 , k) for k > c. Hence Z(θ, t) = exp

c X ν(θ, k) − ν(θ0 , k)

k

k=1

! k

t

Z(θ0 , t).

Compute ν(θ, k) and ν(θ0 , k), k = 1, . . . , c, to obtain f (t) =

c X ν(θ, k) − µ(θ0 , k) k=1

tk .

k

g(t)
1/l We have therefore only to compute g, h, and l such that Z(θ0 , t) = h(t) . Suppose we have computed l, r ∈ N and proved the existence of polynomials g(t), h(t) ∈ C[t] with g(0) = h(0) = 1, deg(g) + deg(h) ≤ r, and g(t) Z(θ0 , t)l = h(t) . Then consider a factorization of g and h over C:

g(t) =

m Y

(1 − xi t),

h(t) =

i=1

n Y

(1 − yi t),

j=1

where m = deg(g) and n = deg(h). Put sk = s(x, y), k = 1, 2, 3, . . . . By P∞ 1 k 0 (10), log g(h) k=1 k sk t . Now compute ν(θ , k) for k = 1, . . . , r. By (3), h(t) = 0 P ∞ g(t) log h(t) = k=1 lν(θk ,k) tk . Hence, sk = lν(θ0 , k) ∈ Z for all k. By Lemma 31.3.6 we may effectively compute g1 , h1 ∈ Q[t] such that Z(θ0 , t)l = hg11(t) (t) . In the remaining two parts of the proof we reduce the computation of l and r to the effective rationality of the zeta function of a variety. Part B: The zeta function of a Galois cover. Let A(C) = hAn , Ci0 /A0i , Con(A0i , C)ii∈I be the Galois stratification underlying θ0 . For each i ∈ I and for each positive integer k let ν(Ci0 /A0i , Con(A0i , C), k) = #{b ∈ A0i (Kk ) | Ar(Ci0 /A0i , Kk , b) ⊆ Con(A0i , C)}. Define Z(Ci0 /A0i , Con(A0i , C), t)

∞ X ν(Ci0 /A0i , Con(A0i , C), k) k  t . = exp k k=1

Then ν(θ0 , k) =

P

i∈I

ν(Ci0 /A0i , Con(A0i , C), k). Hence,

Z(θ0 , t) =

Y i∈I

Z(Ci0 /A0i , Con(A0i , C), t).

748

Chapter 31. Galois Stratification over Finite Fields

It therefore suffices to consider a Galois set/ring cover C/A over K and a conjugacy domain Con of cyclic subgroups of G = Gal(C/A), and to compute l(C/A, Con) ∈ N such that Z(C/A, Con, t)l(C/A,Con) is a rational function with an explicit bound on its total degree. Then with l(θ0 ) = 0 lcmi∈I l(Ci0 /A0i , Con(A0i , C)), the function Z(θ0 , t)l(θ ) is rational with an explicit bound on its total degree. Part C: L-series.

Consider the following central function on G:  χ(σ) =

1 0

if hσi ∈ Con if hσi ∈ / Con.

For all k ∈ N and a ∈ A(Kk ) let χ(k, a) = χ(σ) if hσi ∈ Ar(C/A, Kk , a). Then X χ(k, a). (19) ν(C/A, Con, k) = a∈A(Kk )

By (6), Z(C/A, Con, t) = L(C/A,P χ, t). Compute rH ∈ Q for each cyclic subgroup H of G such that χ = H rH 1G H (Lemma 31.3.3). In addition, for each such H compute a normal K-basic set BH such that K[BH ] is the integral closure of K[A] in the fixed field of H in K(C) (Lemma 19.3.2 and Lemma 19.7.2). By (7), (8), and (9), L(C/A, χ, t) =

Y

rH L(C/A, 1G = H , t)

Y

H

Z(BH , t)rH .

H

Now Lemma 31.3.5 states that each Z(BH , t) is a rational function and gives a bound on its total degree. Taking l(C/A, Con) to be a common denominator of all rH ’s, we conclude that L(C/A, χ, t)l(C/A,Con) can be presented as a rational function with an effective bound on its total degree.  Example 31.3.8: Necessity of the exponential factor in Theorem 31.3.7. Let λ(Y ) be the following formula of L(ring, Fq ): (∀X)[X q = X ∧ Y = 1] Obviously ν(λ, 1) = 1 and ν(λ, k) = 0 for each k ≥ 2. Hence, Z(λ, t) = et . Thus, the exponential factor in (17) is necessary. 

Exercises 1. For a positive integer k, let A1 = V (1 − kX) and A2 = A1 r A1 . Show that A1 = A1 ∪· A2 is a stratification of A1 into a disjoint union of normal Z-basic sets.

Exercises

749

2. Let A1 = V (X 2 + 3) and A2 = A1 r A1 . Prove that 2 is the minimal integer k such that Ai is Z[k−1 ]-normal, i = 1, 2. 3. Let A1 = V (X − 1) and A2 = A1 r V (2X − 2). Show that A1 = A1 ∪· A2 is a stratification of A1 into disjoint union of normal Z-basic sets. Observe ˜ 2 ) = {1} and A2 (F ˜ 2 ) = ∅. that A1 (F 4. (a) Let A = V (X 2 + Y 3 − pY ) where p is an odd prime. Show that A is normal over Z, but V (X 2 + Y 3 ) is not normal over Fp . Hint: Let (x, y) be a generic point of A over Fp . Show that xy is not
2 is. Also, note that A is normal over Q because it is in Fp [x, y] but xy nonsingular (Lemma 5.2.3). 1 , z] with z 2 = y − 1, is a Galois ring cover (b) Show that Z[A, 2(y−1) 1 1 ]. Note, however, that Fp [A, 2(y−1) , z] is not a ring cover of of Z[Z, 2(y−1) 1 ] because the latter ring is not integrally closed. Fp [A, 2(y−1)

5. Let K be a finite field and let Kk be the unique finite extension of K of degree k. For a sentence θ of L(ring) let ν(θ, k) = 1 if θ is true in Kk and ν(θ, k) = 0 otherwise. For each positive integer m let χm (k) = 1 if m|k and χm (k) = 0 otherwise. Pe (a) Suppose ν(θ, k) = i=1 srii mi χmi (k), k = 1, 2, 3, . . ., where ri , si are relatively prime integers and mi is a positive integer, i = 1, . . . , e. Prove that Z(θ, t) =

∞ X ν(θ, k) k=1

k

tk =

e Y

(tmi − 1)si /ri .

i=1

Thus, l = lcm(r1 , . . . , re ) is the minimal integer such that Z(θ, t)l is a rational function. (b) Let θ be the sentence (∃X1 )[f (X1 ) = 0] ↔ (∃X2 )[g(X2 ) = 0], where f = f1 . . . fr and g = g1 . . . gs with fi , gi irreducible polynomials in K[X]. Let ai = deg(fi ), i = 1, . . . , r and bj = deg(gj ), j = 1, . . . , s. Prove that ν(θ, k) = 1 if and only if the following statement holds: there exists i such that ai | k ⇐⇒ there exists j such that bj | k. Conclude that ν(θ, k) is fixed on congruence classes modulo n = lcm(a1 , . . . , ar , b1 , . . . , bs ). (c) Let θ be as in (b) and write ν(θ, k) in the form of (a). Hint: Consider the functions fa (k) =

r Y i=1

(1 − χai (k))

and

fb (k) =

s Y

(1 − χbj (k)).

j=1

Then show that ν(θ, k) = (1 − fa (k))(1 − fb (k)) + fa (k)fb (k). Also, note that χc (k)χd (k) = χlcm(c,d) (k).

750

Chapter 31. Galois Stratification over Finite Fields

6. Let A = hA2 , Ci /Ai , Con(Ai , C)ii=1,2 be a Galois stratification over Fq where A1 = V (X q − X − Y ), A2 = A2 r A1 , Ci = Fq [Ai ], i = 1, 2, C is the family of finite cyclic groups, Con(A1 ) consists of the trivial group and Con(A2 ) is empty. Let θ1 be (∀X)[Ar(X, Y ) ⊆ Con(A)] and let θ2 be (∃X)[Ar(X, Y ) ⊆ Con(A)]. Thus, in Section 31.3 take m = 1 and n = 1. (a) Show that Z(θ1 , t) = et and Z(θ2 , t) = (1 − qt)−1/q . (b) As in Proposition 31.2.3, compute a minimal positive integer li and a Galois stratification Bi of A1 such that for each k ≥ li and for each b ∈ Fqk , Fqk |= θi (b) if and only if Ar(Bi , Fqk , b) ⊆ Con(Bi (C)), i = 1, 2. Let θi0 be the Galois formula attached to Bi , i = 1, 2. Prove that Z(θ1 , t) = et Z(θ10 , t) while Z(θ2 , t) = Z(θ20 , t). 7. Let Γ be a smooth curve which is defined over a finite field K. Use the notation ofPSection 31.3 and define the Zeta function of Γ over K to be ∞ Nr r
t . Now denote the function field of Γ over K by F ZΓ (t) = exp k=1 r and consider the zeta function ZF/K (t) as defined in Section 4.2. Prove that P2g ZΓ (t) = ZF/K (t). Hint: Use the identity − i=1 ωik = Nk − (q k + 1) ((5) of Section 4.5).

Notes Kiefe proves Part (b) of Theorem 31.3.7 for a formula θ(Y) of L(ring, Fq ) [Kiefe, p. 52]. The elimination of quantifiers step is based on the recursive methods which are introduced in Chapter 19, rather than on Galois stratification. In addition, the representation theory arguments that we use are replaced by combinatorial arguments. One finds a remark on page 58 of [Kiefe] that the zeta function of a formula is a radical of a rational function. This is inaccurate as Example 31.3.8 shows.

Chapter 32. Problems of Field Arithmetic

32.1 Open Problems of the First Edition The first edition of “Field Arithmetic” listed 22 open problems. Since the first edition was published fifteen of these problems were solved or partially solved. Here we list those problems and comment on the solutions whenever applicable. Problem 1: (a) Is Qsolv a PAC field? ; (b) Is there a field K which is neither formally real nor PAC all of whose Henselian hulls are separably closed? Comment: A finite field K has no nontrivial valuations. Hence, the condition “all Henselians hulls of K are separably closed” is trivially fulfilled. In addition, K is neither formally real nor PAC (Proposition 11.1.1). Thus, K (trivially) satisfies the conditions of Problem 1(b). Hrushovski’s example (Proposition 11.7.8) supplies a nontrivial example for a field K that satisfies the conditions of Problem 1(b). Indeed, in that example K is a non-PAC field of positive characteristic such that Kins is PAC. In particular, K is neither finite nor formally real. Let v be a valuation of K. Extend v to Kins in the unique possible way. Then Kins Kv is the Henselian closure of Kins at v. Hence, since Kins is PAC, Kv Kins is separably closed ˜ Consequently, Kv is separably (Corollary 11.5.5). In fact, Kins Kv = K. closed. Finally, [Geyer-Jarden5, Thm. D and Remark 2.7(a)] constructs for each finite or global field K0 an infinite regular extension K which is not formally real, each Henselian closure of K is separably closed, and neither K nor Kins are PAC. See also Remark 11.5.11.  ˜ Problem 2: Let σ ∈ Gal(Q) for which Q(σ) is neither PAC nor formally real. ˜ Does Q(σ) admit a valuation with a non-algebraically closed completion? For a field K and e ∈ N let Se (K) = {σ ∈ Gal(K)e | Ks (σ) is PAC }. Problem 3: Let E(t) be a rational function field of one variable over an uncountable field E. Is Se (E(t)) nonmeasurable? Comment: Proposition 18.8.8(b) gives an affirmative answer to Problem 3 when E is algebraically closed or E = K0 (T ), K0 is an arbitrary field, and T is an uncountable set of indeterminates.  A field K has the density property with respect to a valuation w of ˜ if for each variety V defined over K, V (K) is w-dense in V (K). ˜ K

752

Chapter 32. Problems of Field Arithmetic

Problem 4: Does each PAC field K have the density property with respect ˜ to each valuation w of K? ˜ Comment: By Proposition 11.5.3, K is w-dense for every valuation w of K. n  This proves a restricted density property for the affine spaces A . Problem 5: Give an example of a PAC field which contains no proper PAC subfield. Comment: Example 11.2.6 gives infinite extensions of Fp which contain no proper PAC subfields, yet these extensions are themselves PAC.  Problem 6: Is every perfect PAC field C1 ? Comment:

See Sections 21.2 and 21.3 for discussion on Ci -fields.



Call a field K ω-free if each finite embedding problem over K is solvable. Problem 7: Is every PAC Hilbertian field K necessarily ω-free? Comment: Problem 7 was one of the main open problems of Field Arithmetic. It was affirmatively solved for the first time by Fried-V¨olklein in characteristic 0 and then by Pop in general. The general case was also solved by Haran-Jarden. See Example 24.8.5(b) for references.  Problem 8: Let L and M be proper Galois extensions of a Hilbertian field K such that L ∩ M = K. Is LM Hilbertian? Comment: By Corollary 13.8.4, if L 6⊆ M and M 6⊆ L, then LM is Hilbertian.  Problem 9: Let K be a Hilbertian field. Prove or disprove: There exist no proper Galois extensions L and M of K such that L∩M = K and LM = Ks . Comment:

Corollary 13.8.4 affirms the statement of the problem.



Problem 10: Let K be a field equipped with an infinite set S of inequivalent discrete valuations. Suppose that for each a ∈ K, a 6= 0 the set {v ∈ S | v(a) 6= 0} is finite. Is K Hilbertian? Problem 11: Let R be a unique factorization domain with infinitely many primes. Is the quotient field of R Hilbertian? Comment: Example 15.5.8, initiated by Corvaja-Zannier, gives a unique factorization domain R with infinitely many primes such that Quot(R) is non-Hilbertian. This answers both Problem 10 and Problem 11 negatively.  Problem 12 (Conjecture of Geyer-Jarden): Let K be a field which is finitely generated over its prime field. Then for almost all σ ∈ Gal(K)e each Abelian ˜ variety A of dimension at least 2 defined over K(σ) has these properties: ˜ (a) If e = 1, then there exist infinitely many primes l such that A(K(σ)) has a point of order l.

32.1 Open Problems of the First Edition

753

˜ (b) If e ≥ 2, then for only finitely many primes l, A(K(σ)) has a point of order l. ˜ (c) If e = 1 and l is a prime, then A(K(σ)) has only finitely many points of an l-power order. Comment: Problem 12 was stated as Conjecture 16.50 in [Fried-Jarden3]. Part (b) in characteristic 0 and Part (c) in general is proved in [JacobsonJarden2]. A weaker version of Part (a) is proved in [Geyer-Jarden6] when K is a number field. See Remark 18.11.3 for more details.  Extend the language L(ring) by a unary predicate symbol Σ to a lan˜ σi be a system for L(ring, Σ) guage L(ring, Σ). For each σ ∈ Gal(K) let hQ, ˜ and with σ interpreting Σ. with a domain Q Problem 13: Is the theory of all sentences θ of L(ring, Σ) which are true in ˜ σi for almost all σ ∈ G(Q) decidable? hQ, Comment: The analogous problem where Σ is replaced by e unary predicate symbols Σ1 , . . . , Σe with e ≥ 2 has a negative solution, that is, the corresponding theory is undecidable (Theorem 29.3.1). Problem 14: Let f > e ≥ 2. What are the Haar measures of the following sets: {(x1 , . . . , xe ) ∈ (Fˆe )e | hx1 , . . . , xe i ∼ (1a) = Fˆe }; and f (1b) {(x1 , . . . , xf ) ∈ (Fˆe ) | (Fˆe : hx1 , . . . , xf i) < ∞}? Comment: By Proposition 26.1.7, the Haar measure of the set in (1a) is 1 (Lubotzky) and the Haar measure of the set in (1b) is zero (KantorLubotzky).  Problem 15: Let G be a profinite group of rank ≤ ℵ0 . Prove or S disprove: For each proper closed subgroup H of G of infinite index the set g∈G H g contains no open neighborhood of 1. Problem 16: Describe the universal Frattini cover of a non-Abelian finite simple group G. Problem 17: Denote the universal Frattini cover of a profinite group G by ˜ and let E(G) be a smallest embedding cover of G. G (a) Is E(G) projective whenever G is? ˜ isomorphic to the universal Frattini cover of E(G)? (b) Is E(G) Comment: Proposition 24.7.3 gives an Example of Chatzidakis for a projective group G for which E(G) is not projective. This refutes both (a) and (b).  Algebraic extensions L and L0 of a global field K are Kronecker equivalent if for almost all primes p of K, p has a prime divisor of relative degree 1 in L if and only if p has a prime divisor of relative degree 1 in L0 . Denote the set of all algebraic extensions L0 of K which are Kronecker equivalent to L by K(L/K).

754

Chapter 32. Problems of Field Arithmetic

Problem 18 (Conjecture of Jehne): If L is a quadratic extension of a number field K, then L is the unique element of the class K(L/K). Comment:

The conjecture was proved by Saxl. See Remark 21.5.7(c).



Problem 19 (Problem of Jehne): Do there exist fields K ⊆ L ⊆ M with K global, L/K finite separable, and M/K infinite separable such that M is Kronecker equivalent to L over K? Comment: Problem 19 is related to problem 18. See Notes to Section 21 for details.  Problem 20 (Problem of Davenport): Let f, g ∈ Z[X] be nonconstant polynomials. Suppose that for almost all primes p {f (x) | x ∈ Fp } = {g(x) | x ∈ Fp }. Are f and g necessarily strictly linearly related (i.e. there exist a, b ∈ Q, a 6= 0 such that g(X) = f (aX + b))? Comment: Remark 21.6.1 supplies counter examples to Davenport’s problem.  For the concepts involved in the last two problems see Section 26.4 of [Fried-Jarden3]. Problem 21: For a formula θ of L(ring) give an effective computation of the value k0 such that Hk (θ, t) is invariant for k ≥ k0 . Problem 22: Is Hk∗ (θ, t) invariant? If so, give an effective computation of the value k0 such that Hk∗ (θ, t) is invariant for k ≥ k0 . Comment: 

Both Problems 21 and 22 were solved affirmatively in [Pas].

32.2 Open Problems of the Second Edition The second edition of “Field Arithmetic” listed 34 open problems. Since that edition was published in 2005 five of these problems were solved or partially solved. Here we list those problems and comment on the solutions whenever applicable. 1. Does there exists a minimal PAC field which is not an algebraic extension of a finite field (Problem 11.2.7)? 2. Prove or disprove: For each non-PAC field K there exists a plane projective curve without K-rational points (Problem 11.2.10). Comment: The statement is true. Thus, for a field K to be PAC it is necessary and sufficient that each plane projective curve defined over K has a K-rational point (Remark 11.2.10). 

32.2 Open Problems of the Second Edition

755

3. Let K be a PAC field, w a valuation of K, and V a variety over K. Is ˜ (Problem 11.5.4). V (K) w-dense in V (K)? Comment:

The question has an affirmative answer (Remark 11.5.4).



4. (a) Is Qsolv a PAC field? (b) Does there exist an infinite field K of a finite transcendence degree over its prime field which is neither finite nor formally real nor PAC all of its Henselian closures are separably closed (Problem 11.5.9)? 5. Does there exist a finitely generated field extension E of Fp such that E is not PAC but Eins is PAC (Problem 11.7.9)? 6. Let K be an infinite field and H a separable Hilbertian subset of K r . Does there exist an absolutely irreducible polynomial f ∈ K[T1 , . . . , Tr , X] which is monic and separable in X such that HK (f ) ⊆ H (Problem 13.1.5)? 7. Let A be a generalized Krull domain which is not a field. (a) Is A[[X]] a generalized Krull domain? (b) Is Quot(A[[X]]) Hilbertian (Problem 15.5.9)? 8. Let K be a finite field and G a finite group. Suppose G is K-regular. Prove or disprove: There is an X-stable polynomial h ∈ K[T, X] with Gal(h(T, X), K(T )) ∼ = G (Problem 16.2.9)? 9.

Is every finite p-group regular over Q (Remark 16.4.6)?

10. Does every finite p-group occurs as a Galois group over every Hilbertian field of characteristic 0 (Remark 16.4.6)? By Lemmas 13.1.1 and 16.2.1, an affirmative answer to Problem 9 implies an affirmative answer to Problem 10. 11.

Prove that A6 is GAR over Q (Remark 16.9.5).

12. Prove that An , with n 6= 2, 3, 6, is GAR over every field K of characteristic 2 (Remark 16.9.5). 13. (a) (b) (c)

Following Remark 16.9.5 we ask the following questions: Is every finite non-Abelian simple group GAR over Q? Is every finite non-Abelian simple group GAR over Qsolv ? Is every finite non-Abelian simple group GAR over every field K contain˜ ing Q?

By Definition 16.8.1, if a finite group G is GAR over a field K, then G is GAR over every separable algebraic extension of K. Thus, Problem 13(b) is easier than Problem 13(a). 14. Let K be a number field and S a finite number of prime ideals of OK . Denote the compositum of all finite Galois extensions of K unramified outside S by KS . Is the Galois group Gal(KS /K) finitely generated (Example 16.10.9(c))?

756

Chapter 32. Problems of Field Arithmetic

˜ 15. Let σ ∈ Gal(Q) for which Q(σ) is neither PAC nor formally real. Does ˜ Q(σ) admit a valuation with a non-algebraically closed completion? (Problem 18.6.3) 16. Is the following generalization of the bottom theorem true: Let K be a Hilbertian field and e a positive integer. Then for almost all σ ∈ Gal(K)e , Ks (σ) is a finite extension of no proper subfield (Problem 18.7.8). For a field K and e ∈ N let Se (K) = {σ ∈ Gal(K)e | Ks (σ) is PAC}. 17. Let E(t) be a rational function field of one variable over an uncountable field E. Is Se (E(t)) nonmeasurable (Problem 18.8.10)? For Problems 18–21 see Conjecture 18.11.2 and Remark 18.11.3: 18. Let K be an infinite finitely generated field and A an Abelian variety. Prove that for almost all σ ∈ Gal(K) there exist infinitely many prime ˜ numbers l such that A(K(σ)) has a point of order l (Geyer-Jarden). The following weaker version of Problem 18 is proved when K is a number field in [Geyer-Jarden6]: 19. Let K be an infinite finitely generated field and A an Abelian variety. Then K has a finite Galois extension L with the following property: for almost all σ ∈ Gal(L) there exist infinitely many prime numbers l such that ˜ A(L(σ)) has a point of order l. The analog of the following problem for characteristic 0 is proved in [Jacobson-Jarden2]: 20. Let K be an infinite finitely generated field extension of Fp , A an Abelian variety over K, and e ≥ 2. Then, for almost all σ ∈ Gal(K)e there exist only ˜ finitely many prime numbers l such that A(K(σ)) has a point of order l. 21.

Is every perfect PAC field C1 (Problem 21.2.5)?

Comment: The problem has an affirmative solution in characteristic 0 (Remark 21.3.7).  22. Are there fields K ⊂ L ⊂ M with K global, L/K finite separable, and M/K infinite separable such that M0 ∈ K(L/K) for every intermediate field L ⊂ M0 ⊂ M of finite degree over K (Problem 21.5.8). For the concepts appearing in the following problem see Section 21.6: 23. (a) Are all exceptional pairs in Q[X] of the form (h(X 8 ), h(16X 8 )) with h ∈ Q[X]? (b) For a given global field K find all exceptional pairs (f, g) with f, g ∈ K[X] (Problem 21.6.2).

32.2 Open Problems of the Second Edition

757

24. Characterize the class of absolute Galois groups among all profinite groups by means of group theoretic and topological properties (Remark 23.1.5). 25. Describe the universal p-Frattini cover of PSL(2, Zp ) (hence of PSL(2, Fp )) (Problem 22.14.6). 26. Let L/K be a finite Galois extension of fields with Galois group G and t an indeterminate. Does K(t) has a Galois extension F with the following properties: (a) Gal(F/K(t)) ∼ = G. (b) F/K is a regular extension. (c) There is a prime divisor p of F/K with decomposition field K(t) and residue field L (Section 24.2). 27. Prove or disprove: Every field K with an affirmative solution for the regular inverse Galois problem has an affirmative solution for the BeckmannBlack problem (Problem 24.2.3). 28.

Prove that Gal(Qsolv ) ∼ = Fˆω (Example 24.8.5)

29. Let K be a Hilbertian field with Gal(K) projective. Prove that each finite embedding problem for Gal(K) is solvable (Conjecture 24.8.6). 30. Give a pure group theoretic proof for the uniqueness of the smallest embedding cover of a profinite group (Notes to Chapter 24). 31. Let C be a Melnikov formation of finite groups, e ≥ 2 an integer, and M, M1 , M2 closed subgroups of Fˆe (C). Suppose M is pro-C of infinite index, M1 , M2 / Fˆe (C), M1 ∩ M2 ≤ M , but M1 , M2 6≤ M . Is M ∼ = Fˆω (C) (Problem 25.4.9). Comment:

The problem has an affirmative solution (Remark 25.4.10).



32. For i = 1, 2 let ϕi : Fˆω → Fˆω be a Melnikov cover. Does there exist an isomorphism θ: Fˆω → Fˆω such that ϕ2 ◦ θ = ϕ1 (Remark 25.9.13)? 33. Let I be the set of all sentences in the language of graphs which hold in infinitely many finite graphs. Let FG be the theory of finite graphs. Does there exist a recursive set of sentences in the language of graphs which lie between FG and I (Remark 28.5.5)? Comment: Remark 28.5.5 settles Problem 33 by giving a recursive set of sentences in the language of graphs between FG and I.  34. Let T (Q, 1) be the set of all sentences in L(ring, Q, Σ) which hold in ˜ σ) for almost all σ ∈ Gal(Q). Is T (Q, 1) decidable (Problem 29.3.2)? (Q,

758

Chapter 32. Problems of Field Arithmetic

32.3 Open Problems We list all open problems of the present edition. Some are left over from the second edition, others are reformulation of solved problems of the second edition, and still others are completely new: 1. Does there exists a minimal PAC field which is not an algebraic extension of a finite field (Problem 11.2.7)? 2. (a) Is Qsolv a PAC field? (b) Does there exist an infinite field K of a finite transcendence degree over its prime field which is neither finite nor formally real nor PAC all of its Henselian closures are separably closed (Problem 11.5.9)? 3. Does there exist a finitely generated field extension E of Fp such that E is not PAC but Eins is PAC (Problem 11.7.9)? 4. Let K be an infinite field and H a separable Hilbertian subset of K r . Does there exist an absolutely irreducible polynomial f ∈ K[T1 , . . . , Tr , X] which is monic and separable in X such that HK (f ) ⊆ H (D`ebes and Haran — Problem 13.1.5)? 5. Let A be a generalized Krull domain which is not a field. (a) Is A[[X]] a generalized Krull domain? (b) Is Quot(A[[X]]) Hilbertian (Problem 15.5.9)? 6. Let K be a finite field and G a finite group. Suppose G is K-regular. Prove or disprove: There is an X-stable polynomial h ∈ K[T, X] with Gal(h(T, X), K(T )) ∼ = G (Problem 16.2.9)? 7.

Is every finite p-group regular over Q (Remark 16.4.6)?

8. Does every finite p-group occurs as a Galois group over every Hilbertian field of characteristic 0 (Remark 16.4.6)? By Lemmas 13.1.1 and 16.2.1, an affirmative answer to Problem 7 implies an affirmative answer to Problem 8. 9.

Prove that A6 is GAR over Q (Remark 16.9.5).

10. Prove that An , with n 6= 2, 3, 6, is GAR over every field K of characteristic 2 (Remark 16.9.5). 11. (a) (b) (c)

Following Remark 16.9.5 we ask the following questions: Is every finite non-Abelian simple group GAR over Q? Is every finite non-Abelian simple group GAR over Qsolv ? Is every finite non-Abelian simple group GAR over every field K contain˜ ing Q?

By Definition 16.8.1, if a finite group G is GAR over a field K, then G is GAR over every separable algebraic extension of K. Thus, Problem 11(b) is easier than Problem 11(a).

32.3 Open Problems

759

12. Let K be a number field and S a finite number of prime ideals of OK . Denote the compositum of all finite Galois extensions of K unramified outside S by KS . Is the Galois group Gal(KS /K) finitely generated (Shafarevich — Example 16.10.9(c))? ˜ 13. Let σ ∈ Gal(Q) for which Q(σ) is neither PAC nor formally real. Does ˜ Q(σ) admit a valuation with a non-algebraically closed completion? (Problem 18.6.3) 14. Is the following generalization of the bottom theorem true: Let K be a Hilbertian field and e a positive integer. Then for almost all σ ∈ Gal(K)e , Ks (σ) is a finite extension of no proper subfield (Problem 18.7.8). 15. Let K be a Hilbertian field. Is Kalt PAC (Fried-V¨olklein — Problem 18.10.6)? For a field K and e ∈ N let Se (K) = {σ ∈ Gal(K)e | Ks (σ) is PAC}. 16. Let E(t) be a rational function field of one variable over an uncountable field E. Is Se (E(t)) nonmeasurable (Problem 18.8.10)? For Problems 17–20 see Conjecture 18.11.2 and Remark 18.11.3: 17. Let K be an infinite finitely generated field and A an Abelian variety. Prove that for almost all σ ∈ Gal(K) there exist infinitely many prime ˜ numbers l such that A(K(σ)) has a point of order l (Geyer-Jarden). The following weaker version of Problem 17 is proved when K is a number field in [Geyer-Jarden6]: 18. Let K be an infinite finitely generated field and A an Abelian variety. Then K has a finite Galois extension L with the following property: for almost all σ ∈ Gal(L) there exist infinitely many prime numbers l such that ˜ A(L(σ)) has a point of order l. The analog of the following problem for characteristic 0 is proved in [Jacobson-Jarden2]: 19. Let K be an infinite finitely generated field extension of Fp , A an Abelian variety over K, and e ≥ 2. Then, for almost all σ ∈ Gal(K)e there exist only ˜ finitely many prime numbers l such that A(K(σ)) has a point of order l. 20. Is every perfect PAC field of positive characteristic C1 (Ax — Problem 21.2.5)? 21. Are there fields K ⊂ L ⊂ M with K global, L/K finite separable, and M/K infinite separable such that M0 ∈ K(L/K) for every intermediate field L ⊂ M0 ⊂ M of finite degree over K (Jehne — Problem 21.5.8). For the concepts appearing in the following problem see Section 21.6:

760

Chapter 32. Problems of Field Arithmetic

22. (a) Are all exceptional pairs in Q[X] of the form (h(X 8 ), h(16X 8 )) with h ∈ Q[X] (M¨ uller)? (b) For a given global field K find all exceptional pairs (f, g) with f, g ∈ K[X] (Davenport — Problem 21.6.2). 23. Characterize the class of absolute Galois groups among all profinite groups by means of group theoretic and topological properties (Remark 23.1.5). 24. Describe the universal p-Frattini cover of PSL(2, Zp ) (hence of PSL(2, Fp )) (Problem 22.14.6). 25. Let L/K be a finite Galois extension of fields with Galois group G and t an indeterminate. Does K(t) has a Galois extension F with the following properties: (a) Gal(F/K(t)) ∼ = G. (b) F/K is a regular extension. (c) There is a prime divisor p of F/K with decomposition field K(t) and residue field L (Beckmann-Black — Section 24.2). 26. Prove or disprove: Every field K with an affirmative solution for the regular inverse Galois problem has an affirmative solution for the BeckmannBlack problem (Problem 24.2.3). 27.

Prove that Gal(Qsolv ) ∼ = Fˆω (Iwasawa’s Conjecture — Example 24.8.5)

28. Let K be a Hilbertian field with Gal(K) projective. Prove that each finite embedding problem for Gal(K) is solvable (Fried-V¨olklein — Conjecture 24.8.6). 29. Give a pure group theoretic proof for the uniqueness of the smallest embedding cover of a profinite group (Notes to Chapter 24). 30. For i = 1, 2 let ϕi : Fˆω → Fˆω be a Melnikov cover. Does there exist an isomorphism θ: Fˆω → Fˆω such that ϕ2 ◦ θ = ϕ1 (Remark 25.9.13)? 31. Let T (Q, 1) be the set of all sentences in L(ring, Q, Σ) which hold in ˜ σ) for almost all σ ∈ Gal(Q). Is T (Q, 1) decidable (Problem 29.3.2)? (Q,

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Index

Abelian extension, 294 Abelian pro-p groups, 308 Abelian profinite, 345 Abhyankar, 70 Abhyankar’s conjecture, 70 absolute Galois group, 12 absolute norm (of a prime ideal), 112 absolute value, 238 absolutely converges (infinite product), 80, 92 absolutely irreducible (polynomial), 43 absolutely irreducible (variety), 175 abstract variety, 189 accessible, 626 Ackermann, 158 Ackermann function, 158 additive absolute value, 278 adele, 64 admissible (ideal), 123 admissible, 541 Adler-Kiefe, 670 affine n-space An , 172 affine plane, 95 affine plane curve, 95 algebraic function field of one variable, 52 algebraic set, 172 algebraically independent (fields), 40 algebraically independent (set), 40 almost all, 146 almost contained (sets), 140 almost equal (sets), 140 almost full, 542 amalgamation property, 656 ample, 335 Arason-Fein-Schacher-Sonn, 307 archimedean (absolute value), 239 Arithmetic, 706 arithmetical primes, 280 arithmetically definable, 707 arithmetic progression, 241 Artin, 41, 64, 76, 130, 494, 537, 738, Artin reciprocity law, 124 Artin-Schreier extension, 29 Artin symbol, 113, 709 atomic formula (of Ln ), 700

atomic formulas, 133 Ax, 141, 148, 162, 191, 200, 207, 209, 217, 218, 381, 401, 447, 448, 449, 450, 452, 453, 456, 524, 759 Ax-Roquette, 196 B¨ uchi, 682 Bary-Soroker, 613 basic field, 172 basic functions, 402 basic sets, 402 basic statement, 551 basis (of a free abstract group), 346 basis (of a free pro-C group), 348 Bauer, 130, 236 beat, 671 Beckmann-Black, 760 Beckmann-Black Problem, 565 Bell-Slomson, 145, 148 Bensimhoun, 493 Bertini-Noether, 169, 171, 179 Bezout ring, 276 bijective (polynomial function), 477 Binz-Neukrich-Wenzel, 362, 542 birational equivalence (varieties), 178 birationally equivalent (varieties), 178 block (of a permutation group), 474 Bˆ ocher, 744 Bombieri, 94, 743 Boolean algebra generated by a family, 140 Boolean algebra of sets, 140 Boolean polynomials, 140 Borel-Cantelli, 372 Borel field, 362 Borevich-Shafarevich, 61 bottom theorem, 385 bounded (degrees), 417 bounded (occurence of a variable), 133 bounded minimum operator, 157 Bourbaki, 24, 25, 226 branch point, 214 Brandis, 402, 525 Brink, 320, 328

Index Burnside, 206, 477, 495 Ci -field, 453 Ci,d -field, 453 C-embedding problem, 503 C-homogeneous, 615 C-homogeneous cover, 633 C-projective, 503 canonical class, 55 canonical divisor, 55 Caratheodory, 365 Carlitz’s Conjecture, 489 cartesian product, 182 cartesian square, 400 Cassels-Fr¨ ohlich, 21, 32, 62, 63, 65, 206 239, 241, 314, 363, 381, 469, 470 483, 485, 742 Cauchy sequence, 61 center (of a divisor), 97 center (of a prime divisor), 57 central (function), 740 central simple (K-algebra), 208 Chaevalley, 76 character, 740 character (of a finite abelian group), 122 characteristic (subgroup), 330 characteristic function, 156 Chatzidakis, 401, 580, 593, 595, 603, 633, 789 Chebotarev, 130 Chebotarev density theorem, 114 Chebyshev polynomials, 479 Chebyshev’s inquality, 373 Cherlin, 383 Cherlin-Jarden, 706, 707 Cherlin-v.d.Dries-Macintyre, 453, 542, 560, 690, 697 Chevalley, 24, 64, 76, 423, 455 Chevalley-Warning, 455 class number, 77 class of divisors, 53 classical Hilbertian fields, 242 coefficients (of an element in Z[H]), 475 cofinite, 143 Cohen-Fried, 490 Cohn, 188 cohomological dimension, 210 Colliot-Th´el`ene, 230, 243, 593 commutator, 309 commutator subgroup, 310

781 compactness theorem, 144 compatible (maps), 1, 2 complementary Galois stratification, 719 complementary module, 65 complement (of a closed subgroup), 524 complete (measure), 365 complete (theory), 152 complete (valued field), 61 completeness (of projective varieties), 186 completion, 61 components (of a K-closed set), 174 composition factor, 326 composition series, 688 compositum (of places), 23 computes (a function), 672 concurrent, 271 conductor (of a ring extension), 100 cone, 272 congruence test, 132 conjugacy domain of subgroups, 710 conjunction symbol, 133 consequence, 149 conservative (function field of one variable), 393 conservative (function field of several variables), 392 consistent, 152 constant, 52, 134 constant field extension, 59 constant symbols, 132 constant terms, 153 continuous set theoretic section, 9 contradiction (formula), 152 converge to 1 (subset), 338 convergent to 1 (map), 346 converges (infinite product), 80 convex hull, 280 convex, 274 coordinate ring (of a K-basic set), 424 coordinate ring (of a K-variety), 173 coordinate ring (of a curve), 96 coordinate ring (over Z[k−1 ]), 732 corank (of a field), 439 Corvaja-Zannier, 266, 289 Cossey-Kegel-Kov´ acs, 542 covers, 513 crossed homomorphism, 526 cusp, 394 cyclotomic (extension), 121

782 data for being algebraically closed, 662 Davenport (Problem), 753 Davenport, 466 D`ebes, 566, 593 D`ebes-Haran, 236, 758 de-Morgans’s laws, 164 decidable theory, 159 decision procedure, 159 decomposable, 471 decomposition factor (of a polynomial), 485 decomposition field (of a place), 110 decomposition field (of a prime ideal), 107 decomposition group, 25 decomposition group (of a place), 110 decomposition group (of a prime ideal), 107 decomposition group condition, 564 Dedekind, 63, 464 Dedekind domain, 32 Dedekind zeta function, 124 deduction theorem, 151 deductively close, 150 defined (absolutely irreducible K-variety), 175 defined over K (scheme), 189 degree (of a prime divisor), 52 degree (of a curve), 95 degree (of a divisor), 53 degree (of a field extension), 519 degree (of a polynomial function), 478 degree (of a rational map), 178 degree (of a stabilizing base), 391 Deligne, 94 Denef-Jarden-Lewis, 460 Denef-Loeser, 729 density property, 397, 751 Dentzer, 306 derivation, 48 Deuring, 53, 54, 55, 59, 60, 61, 76, 130, 210 diamond theorem, 260 Dickson polynomial, 480 different, 67 different (of a ring extension), 112 different exponent, 65 differential, 54, 64 dihedral group, 472

Index dimension (of a constructible set), 423 dimension (of a K-variety), 173 dimension (of a ring), 283 dimension theorem, 174 directed (family), 341 directed (family of subsets), 6 directed partially ordered set, 1 Dirichlet, 114, 130 Dirichlet density, 113 discrete (valuation), 21 discriminant (of an element), 109 discriminant (of a polynomial), 108 discriminant (of a ring extension), 112 distinct (valuations), 20 divides (supernatural number), 520 divisor, 52 divisor (of a differential), 55, 64 divisor of poles, 52 divisor of zeros, 52 Dixon, 636, 639 Dixon-du.Sautoy-Mann-Segal, 539 Dixon-Pyber-Seress-Shalev, 358 domain, 134 domain of definition (of a rational map), 178 dominant (rational map), 178 Douady, 338 double implication symbol, 133 double tangents, 392 doubly transitive (permutation group), 474 Dries, v. d., 426, 590 Dries-Smith, 401 Drinfeld-Vlˆ adut, 106 Duret, 190, 191, 699 Dwork, 741, 743 Dwork-Bombieri, 743 e-basic statement, 551 e-cycle, 315 e-free (field), 379 effective algorithm, 405 effective computation, 412 Efrat, 207 Efrat-V¨ olklein, 541 Eichler, 266 Eisenbud, 99 Eisenstein’s Criterion, 30, 217 elementarily equivalent, 136 elementary equivalence theorem, 435

Index elementary extension, 137 elementary statement, 136 elementary substructure, 137 elimination (of a quantifier), 164 elimination of quantifiers, 163 elimination theory, 410 embedding (of structures), 137 embedding cover, 571 embedding lemma, 431 embedding problem, 303, 502 embedding property, 564 ends with (word), 351 enlargement, 268 equality symbol, 132 equivalent (absolute values), 239 equivalent (algebras), 208 equivalent (formulas), 150 equivalent ((n + 1)-tuples), 185 equivalent (places), 20 equivalent (points), 189 equivalent (valuations), 20 Ershov, 196, 362, 402, 436, 540, 541, 542, 538, 539, 576, 670, 689, 697, 729 Ershov-Fried, 542, 593 Ershov-Lavrov-Taimanov-Taitslin, 683, 694, 697, 706 even (ordinal number), 618 eventually stationary, 672 exceptional (pair of polynomials), 467 exchange principle, 45 existential (sentence), 462 existential elimination lemma, 718 existential symbol, 132 existentially closed, 137, 656 existentially complete, 561 explicit case, 440, 708 extension (of function fields), 59 extension (of structures), 136, 656 external (object), 268 faithful (permutation group), 472 Faltings, 248 Faltings-W¨ ustohlz, 247 fiber product (of groups), 499 fiber product (of schemes), 246 field cover, 109 field crossing argument, 107, 130, 324, 431, 558, 562 field of formal power series, 287 field of p-adic numbers, 17

783 field with a product formula, 281 filter, 138 final tape, 672 finite (element at a place), 20 finite (embedding problem), 303, 502 finite (K-morphism), 180 finite (place at an element), 20 finite intersection property, 139 finite points (on a plane curve), 95 finitely generated (profinite group), 328 finitely generated (ring cover), 562 first isomorphism theorem for compact groups, 5 first order language, 132 formal power series, 62 formal proof, 150 formation, 344 formulas, 133 fractional ideal, 31 Frane, 145 Franz, 266 Frattini cover, 508 Frattini embedding problem, 511 Frattini group, 497 Frattini p-cover, 529 free (abstract group), 346 free (variable), 133 free generators theorem, 379 free occurrence of a variable in a formula, 133 free pro-C group, 346, 349 free pro-C group of rank m, 348 free product (of profinite groups), 508 free set of generators, 346, 605 free variables (of a formula), 268 Frey, 196, 205, 218, 217 Frey-Geyer, 194 Frey-Jarden, 337, 398 Frey-Prestel, 205 Fried, 130, 206, 244, 264, 266, 290, 471, 492, 494, 540 Fried-Guralnick-Saxl, 489 Fried-Haran-Jarden, 428, 564, 593, 729 Fried-Jarden, 51, 131, 196, 207, 230, 262, 266, 289, 336, 391, 392, 395, 397, 401, 402, 495, 542, 581, 593, 582, 635, 639, 753 Fried-MacCrae, 486 Fried-Sacerdote, 495, 729, 738

784 Fried-V¨ olklein, 337, 395, 397, 582, 583, 752 Fried-V¨ olklein (Conjecture), 759, 760 Frobenius automorphism, 9, 15 Frobenius automorphism (of a prime ideal), 112 Frobenius, 130 Frobenius density theorem, 130 Frobenius field, 564 Frobenius fields, 559 Fr¨ ohlich-Shepherdson, 411 from the n-tracks, 672 full formation, 344 function field, 52 function field (of a curve), 96 function field (of a K-basic set), 424 function field (of a K-variety), 173 function field (of a projective variety), 186 function field of several variables, 391 function symbol, 132 functional prime, 280 G-extesnion, 294 G-extension, 294 g-Hilbertian, 248 GA (realization), 321 Galois, 493 Galois (polynomial), 235 Galois (ring cover), 562 Galois closure, 10 Galois formula, 717 Galois ring/set cover (over Z[k−1 ]), 732 Galois ring cover, 109 Galois sentence, 720 Galois splitting field (of a Zariski closed set), 454 Galois stratification, 717 GAR (realization), 321 Gasch¨ utz, 362, 525 general polynomial of degree n, 296 generalized Abhyankar’s conjecture, 70 generalized Krull domain, 283 generated (formation), 346 generated (σ-algebra), 362 generates (a profinite group), 328 ˆ 14 generates (the group Z), generic (point), 96 generic point (of a K-variety), 173 genus (of a curve), 96

Index genus (of a function field), 54 geometric points, 188 Geyer, 51, 176, 211, 243, 288, 337, 390, 393, 402 Geyer-Jarden, 131, 206, 211, 332, 337, 391, 392, 398, 402, 706, 751, 752, 753, 756, 759 Geyer-Jarden (Conjecture), 749 Geyer-Jensen, 307, 336, 519 Gilmore-Robinson, 277, 290 Gilmore-Robinson criterion, 289 global field, 112 G¨ odel, 162 G¨ odel completeness theorem, 154 G¨ odel number, 159 G¨ odel numbering, 403 Goldstein, 122, 124 Golod-Shafarevich inequality, 647 Goppa, 106 graph conditions, 689 graph, 679 Grauert-Manin, 212 greatest common divisor (of supernatural numbers), 520 Greenleaf, 495 Grothendieck, 70, 187 group of divisor classes, 53 group ring, 310 group theoretic diamond theorem, 610 group theoretic section, 502 Gruenberg, 210, 503, 542 Guralnick-M¨ uller, 487, 496 Guralnick-Thompson, 266 Haar measure, 362 Hall, 252, 635 Halmos, 363, 366, 376, 399, 640, 645, 668 Haran, 130, 207, 385, 402, 496, 540, 542, 610, 633, 686 Haran’s diamond theorem, 265 Haran-Jarden, 262, 266, 335, 545, 583, 593, 735, 752 Haran-Lauwers, 729 Haran-Lubotzky, 542, 568, 575, 578, 593 Haran-V¨ olklein, 334 Harbater, 70, 334 Hartshorne, 187, 392 Henkin, 162 Hensel’s lemma, 62 Henselian, 203

Index Henselian closure, 203 Herfort-Ribes, 542 Hermes, 158 Hermite-Minkowski, 331 higher order set, 267 higher order structure, 267 Hilbert, 230, 231, 266, 299, 336 Hilbert’s basis theorem, 172 Hilbert’s irreducibility theorem, 219 Hilbert’s Nullstellensatz, 169 Hilbert set, 219 Hilbert subset, 219 Hilbertian field, 219 Hilbertian (integral domain), 241 Hilbertian pair, 439 Hirschfeld-Wheeler, 670 holomorphy ring, 56 holomorphy ring theorem, 57 homogeneous (pro-C-group), 595 Hrushovski, 215, 751 Huppert, 208, 326, 495, 498, 506, 526, 536, 543, 630, 637, 647, 688 hyperelliptic, 70 hyperplane, 174 hypothesis H(p, q), 700 I-cover, 568 identity axioms, 151 Ihara, 106 Ikeda, 304, 542 immediate extension, 203 imperfect degree, 45 imperfect exponent, 45 implication symbol, 133 imprimitive (permutation group), 474 Inaba, 191, 230, 266 inclusion-exclusion principle, 370 indecomposable, 471 independent (sets), 369 independent (valuations and orderings), 242 index (of a closed subgroup), 520 induced (conjugacy domain), 707 induced (cover), 710 induced (prime divisor), 281 induced representation, 740 induces (homomorphism), 430 induction on structure, 133 inductive, 656 inertia field (of an ideal), 107

785 inertia field (of a place), 110 inertia group, 25, 543 inertia group (of a place), 110 inertia group (of a prime ideal), 107 inference rule, 149 infinite absolute value, 281 infinite prime, 276 infinite sentences, 442 infinitesimal (element), 274 inflection points, 392 ∞-imperfect, 45 initial tape, 671 injective (polynomial function), 487 instant, 671 integral closure, 30 integral over (element), 30 integral over (ring), 31 integrally closed (ring), 30 internal (n-tuples), 269 internal (object), 267 internal function, 270 internal substitutions , 269 interpretation (of an n-adic theory), 700 inverse (of fractional ideal), 31 inverse limit, 1 inverse system, 1 invertible (fractional ideal), 32 Ireland-Rosen, 93 irreducible (character), 740 irreducible (K-closed set), 173, 185 isomorphic (covers), 513 isomorphic (projective curves), 96 isomorphic (structures), 136 isomorphism theorem for compact groups, 5 Iwasawa, 516, 578, 581, 582 Iwasawa’s Conjecture, 582, 760 Jacobson, 209 Jacobson-Jarden, 337, 398, 443, 453, 753, 756, 759 Jannsen, 331, 519 Janusz, 69, 123, 124, 130, 464 Jarden, 130, 148, 206, 211, 243, 337, 379, 380, 381, 384, 398, 401, 402, 451, 453 561, 592, 669, 670, 729 Jarden-Kiehne, 148, 431, 442, 453 Jarden-Lubotzky, 587, 589, 590, 611, 612 633, 645, 648, 653, 654 Jarden-Ritter, 516

786 Jarden-Roquette, 193, 293 Jarden-Shelah, 388, 402 Jarden’s lemma, 542 Jehne, 463, 466, 467, 753 Jehne-Saxl, 467 Jehne (Conjecture), 467, 753 Jehne (Problem), 753 Jordan, 394 K-algebraic set (projective), 185 K-basic (constructible set), 424 K-birational (varieties), 178 K-closed subsets, 172 K-components (projective), 185 K-constructible set, 423 K-curve, 173 K-derivation, 48 K-hypersurface, 174 K-isomorphic (curves), 96, 97 K-isomorphism, 180 K-morphism, 180, 186 K-normal (point), 96 K-place, 20 K-rational (point), 178 K-rational map, 178, 186 K-variety, 173 K-variety (projective), 185 Kantor-Lubotzky, 638, 639, 644, 654, 753 kernel of the (embedding) problem, 502 Kiefe, 750 Kiming, 307 Kimmerle-Lyons-Sandling-Teaque, 650 Koll´ ar, 171, 199, 205, 218, 462 Klein, 290 Klingen, 337, 438, 466, 467, 495 Knopp, 92 Kronecker, 411, 428, 461, 493 Kronecker class, 464 Kronecker conjugate, 463, 469 Kronecker equivalent, 753 Kronecker substitution, 199 Kronecker-Weber, 335 Krull domain, 286 Krull topology, 10 Kuhlmann-Pank-Roquette, 543 Kummer, 464 Kurosh, 352, 357, 503, 508 Kurosh subgroup theorem, 361 Kuyk, 333, 546, 558, 561 Kuyk-Lenstra, 336

Index L-rational point, 188 L-series, 122 Lagarias-Montegomery-Odlyzko, 131 Lang, 11, 12, 18, 22, 23, 24, 30, 37, 38, 40, 43, 51, 61, 65, 75, 76, 98, 99, 101, 107, 108, 109, 110, 111, 112, 113, 122, 123, 124, 171, 172, 173, 174, 175, 176, 183, 190, 191, 193, 196, 205, 207, 222, 230, 237, 241, 245, 248, 258, 259, 266, 282, 287, 296, 299, 300, 301, 307, 308, 309, 316, 323, 331, 384, 398, 406, 407, 416, 428, 455, 464, 471, 497, 519, 536, 540, 545, 563, 591, 643, 699 language, 132 language of n-adic quantifiers, 700 larger (cover), 513 Lavrov, 680 least common multiple (of supernatural numbers), 520 Leibniz, 266 length (of a proof), 150 Lenstra, 426, 453, 490, 492, 496 Lenstra-Lenstra-Lov´ acz, 428 Leptin, 12, 18, 43, 494 letters, 132 LeVeque, 37, 113, 309, 310, 372, 380, 644 Levi, 357 lexicographic order, 19 Lidl-Mullen-Turnwald, 496 Lidl-Niederreiter, 93, 106 Liebeck-Shalev, 639 lies over (prime divisor), 59 lies over v (valuation), 24 Lim, 540 Lindon-Schupp, 361 line at infinity, 95 line, 95 linear fractional transformation, 213 linearly disjoint (fields), 34, 35, 36 linearly equivalent (divisors), 53 linearly independent (over a ring), 23 linearly related, Liu, 334 local parameter, 52 local ring (at a point), 96 local ring (of a K-variety at a point), 173 local ring, 31 localize (a ring at a prime ideal), 107 locally finite (word), 517 logically deducible, 150

Index logically equivalent (formulas), 163 logically valid formula, 149 long multiplications, 156 long summations, 156 Loˇs, 142 lower central series, 540 Lubotzky, 357, 539, 592, 638, 641, 654, 753 Lubotzky-Melnikov-v.d.Dries, 584 Lubotzky-Segal, 646 Lubotzky-v.d.Dries, 337, 545, 586, 591 L¨ uroth’s theorem, 69, 320 lying over (prime ideal), 107 Lyndon-Schupp, 362 m-ary variable symbols, 700 m-stationary, 676 M¨ obius transformation, 213 Moret-Bailly, 593 Morley, 593 MacCluer, 130, 492 Macintyre, 171 Macintyre-McKenna-v.d.Dries, 171 Madan-Madden, 390, 393, 402 Malle-Matzat, 306, 328, 337, 336, 542 Mann, 646 Matsumura, 287, 288 Matsusaka-Zariski, 180 Matzat, 323, 325, 337, 542 maximal pro-C quotient, 345 maximal (subgroup), 497 meaningful words, 132 measurable (set), 366, 368 measurable rectangles, 376 Melnikov, 402, 542, 598, 606, 620, 622, 623, 624, 630, 631, 632, 633, 634 Melnikov (embedding problem), 615 Melnikov cover, 615 Melnikov formation, 344 Melnikov group, 613 metric (absolute value), 239 minimal (Kronecker class), 465 minimal normal subgroup, 322 minimal PAC field, 196 minimum operator, 157 Minkowski, 69 model, 135 model companion, 655 model complete, 164 model completion, 656

787 modified valuation, 280 modus ponens, 150 monotone (collection of sets), 362 Mordell-Weil, 247, 398 Morley, 590 morphism (projective), 186 morphism, 180 move commands, 671 µ-independent, 370 M¨ uller, 266, 468, 482, 496, 760 Mumford, 187, 188, 248 n-adic structure, 700 n-adic theories, 700 n-dimensional projective space, 185 n-imperfect, 45 n-track, 672 negation symbol, 132 Neukirch-Schmidt-Wingberg, 305 Neumann, 391, 394, 402 Neveu, 376 Nielsen-Schreier formula, 355 Nikolov-Segal, 516, 517 Nobusawa, 511, 542 nodes, 392 Noether’s normalization theorem, 98, 414 Noether-Grell, 32 Noetherian (integral domain), 31 Noetherian (ring), 172 non-archimedean (absolute value), 239 nonconstant curve, 212 nonelementary (classes), 549 nonsingular homogeneous linear transformation, 97 norm (of a prime divisor), 77 normal (basic set), 424 normal (basic set over Z[k −1 ]), 732 normal (projective model), 393 normal stratification, 716 normal stratification (over Z[k−1 ]), 733 number field, 112 objects of type, 267 occurs (a group over a field), 294 odd (ordinal number), 618 ω-free, 652, 752 open-closed partition, 3 operational modes, 671 opposite (algebra), 208 order (of a profinite group), 520 order of magnitude, 275

788 ordered group, 19 orthogonality relations, 122 overring, 23 p-adic integers, 13 p-adic numbers (field of), 61 p-basis, 44 p-cover, 529 p-embedding problem, 529 p-independent, 44, 45 p-independent, 44, 45 p-projective, 528 P -stratification, 424 p-Sylow group, 522 PAC Nullstellensatz, 380 PAC, 192 partial ordering, 1 partial products, 92 Pas, 754 permutation polynomial, 479 permutation representation, 292 permutes (polynomial), 479 place, 20 Poincar´e series, 739 point (in an affince space), 172 points at infinity (on a projective curve), 95 Poizat, 453 pole, 52, 278 polefinite, 278 polynomial words, 403 Pontryagin, 18 Pop, 337, 583, 749 positive diagram, 135 positively finitely generated, 646 prenex normal form, 163 presented (basic set), 422 presented (field), 404 presented in its quotient field (integral domain), 404 presented (K-basic set), 418 presented (K-constructible set), 421 presented over (n-tuple), 407 presented over (element), 407 presented over (field extension), 405 presented over (rational field), 404 presented (ring), 404 Prestel, 204, 218, 383 primary (field extension), 38 prime (of a global field), 239

Index prime divisor, 52, 278 prime element, 21 primitive (permutation group), 474 primitive element for the cover, 109 primitive polynomial, 179 primitive recursive (function), 155, 404 primitive recursive (relation), 156 primitive recursive (theory), 159 primitive recursive indexing, 403 principal divisor, 53 principal ultrafilter, 139 pro-C group, 344 pro-C embedding problem, 500 pro-l-extension, 249 pro-p groups, 345 pro-S group, 346 procyclic (group), 16 product formula, 280 product measure, 376 product (of fractional ideals), 31 product (of supernatural numbers), 520 profinite completion (of a group), 341 profinite group, 5 profinite ring, 13 profinite space, 3, 539 projective, 506 projective (profinite group), 207 projective completion (of a curve), 95 projective cover, 513 projective limit, 1 projective plane, 95 projective plane curve, 95 projective special linear group, 536 projective system, 1 projective variety, 185 pronilpotent, 205 proper (overring), 23 prosolvable extension, 294 prosolvable group, 205, 345 Pr¨ ufer group, 12, 14 pseudo algebraically closed (field), 192 pseudo finite, 448 Q-central (function), 740 quadratic subfield, 70 quantifier axioms, 150 quantifier elimination procedure, 164 quasi-p (group), 70 Rabin, 403

Index radical, 173 ramification group, 543 ramification index, 24 rank, 327 rank 1 (valuation), 21 rank (of a cover), 633 rank (of a profinite group), 339, 346 rank (of an abstract group), 356 rank (of separable algebraic extension), 386 rank (of an infinite separable algebraic extension), 387 rational (field extension), 321 rational (representation, character), 737 rational map (projective), 186 rational map, 178 Raynaud, 70 real (valuation), 21 realizable (a group over a field), 294 realization of twisted wreath products, 259 realize (twisted wreath product), 257 realize (wreath product), 257 reciprocity map, 123 recognizable (elements in a ring), 403 recursive functions, 157 recursive (theory), 159 recursively inseparable, 677 recursively separable, 676 reduced form, 209 reduced norm, 209 reduced (ring), 187 reduced (word), 351 reducible (K-closed set), 173 refinement (of Galois stratrification), 718 refinement (of normal stratification), 717 regular (a group over a field), 294 regular (field extension), 38 regular (permutation group), 472 regular (ring cover), 562 regular (ultrafilter), 140 regular inverse Galois problem, 295 regular solution, 303 regular ultraproduct, 146 regularity (of a measure), 362 regularly solvable, 303 Reichardt, 130 relation rank, 647 relation symbol, 132

789 relations of type, 267 relativizing (variable), 147 R´enyi, 373 repartition, 54 representable (recursive set), 707 representation (of a finite group), 740 residually-C, 356 residually finite, 356 residue (of an element), 20 residue field, 20 resultant system, 413 ribbon, 671 Ribenboim, 203, 381 Ribes, 18, 210, 211, 300, 507, 534, 542, 582 Ribes-Zalesski, 4, 357, 513, 519, 542, 625, 628, 630, 631, 632, 633, 634 Riemann-Hurwitz formula for tamely ramified extensions, 69 Riemann-Hurwitz genus formula, 67 Riemann-Roch theorem, 54 Riemann hypotheses (for function fields), 85 ring cover, 109 ring of p-adic integers, 13 ring of integers, 33 ring of integers (of a global field), 112 ring/set cover, 708 Ritt, 206 Robinson, 170, 266, 290, 670 Robinson’s test for model completeness, 659 Robinson-Roquette, 237, 276, 290 Roquette, 184, 277, 290, 337, 361, 665 Rosen, 697 Rosenlicht, 393, 394 rule of generalization, 150 S-rank function, 583 Sacks, 171 Samuel, 106, 212 Sansuc, 243 satisfiable, 679 saturated, 143 Saxl, 466, 467 Schinzel, 465, 492, 495 Schmidt, F. K., 82, 333 Schmidt, K., 426, 670 Schmidt, W. M., 94 Schreier, 355

790 Schreier basis, 352 Schreier construction, 351 Schreier system, 352 Schur, 475, 495 Schur’s Conjecture, 487 Schur-Zassenhaus, 524 Scott, 145 section (group theoretic), 252 Seidenberg, 102, 105 semidirect complement, 689 semidirect product (of groups), 252 semilinear rationality criterion, 322 sentence, 133 sentential axioms, 150 separable (field extension), 38 separable (rational map), 178 separable descent, 183 separable Hilbert subset, 219 separating transcendence basis, 38 Serre, 25, 33, 65, 70, 106, 130, 187, 210, 24 331, 332, 334, 336, 401, 516, 740, 741, 742 Serre’s question, 516 Serre-Stalling, 401 set of axioms, 135 set of basic test sentences, 146 set of logical consequences, 657 set theoretic section, 9 sets of type, 267 Severi-Brauer (variety), 209 Shafarevich, 332, 759 Shelah, 145, 593 Shoenfield, 707 Siegel’s theorem, 266 Siegel-Mahler, 276, 290 Siegel, 237 σ-additivity, 363 σ-algebra, 363 simple (point on an affine curve), 98 simple (point on a projective curve), 99 singular (point on a projective curve), 99 Skolem-L¨ owenheim, 138, 171 small (profinite group), 329 small (set of powers of primes), 445 small sets, 139 smallest embedding cover, 571 smooth (curve), 104 solution (of an embedding problem), 303, 503

Index solution field, 302 solution group, 596 solution (over K(t)), 302 solvable (embedding problem), 503 Sonn, 337 specialization, 173 specialization compatible, 711 Specht, 362 splits (algebra), 208 splits (embedding problem), 302, 502 splits (short exact sequence), 252 splitting algorithm, 405 Sprind¸zuk, 290 stabilizing base, 391 stable (field extension), 391 stable (field), 391 Stalling, 401 standard (object), 268 standard decomposition (of a projective variety), 186 standard function, 270 standard prime, 280 star-additive absolute value, 280 starfinite subsets, 274 starfinite summation, 273 starts with (word), 351 stationarily satisfiable, 679 stationary, 679 Stepanov, 94 Stichtenoth, 76 St¨ ohr-Voloch, 94 store (an integer), 672 stratification lemma, 424 strictly linearly related, 467 string, 132 strong approximation theorem, 56, 238 structure, 134 structure (for Ln ), 700 subnormal subgroup, 626 substitution, 134, 267 substructure, 136 supernatural number, 520 superprojective, 565 support, 475 surjective (polynomial function), 487 Suzuki, 327, 638 symmetrically stabilizing base, 391 Tamagawa, 201, 218 tamely ramified

Index (extension of function fields), 69 tamely ramified (valuation), 25 tape, 671 Tarski, 171 Tate, 218, 518 Tate’s theory, 217 tautology, 149 terms, 132 test sentence, 146, 440 T -existentially closed (model), 658 T -extension, 656 theory, 135 thin (subset), 245 Thompson, 305 Titchmarsh, 77 topological group, 4 total degree, 742 totally disconnected, 4 totally positive (element), 122 totally ramifies (valuation), 29 tower property, 35 track, 669 transcendence base, 40 transcendence degree, 40 transfer theorem, 447 transitive (permutation group), 472 translation invariance, 364 trivial (place), 30 trivial (valuation), 21 trivial block, 474 truth (of a formula), 268 truth set, 145, 440, 705 truth value, 134 Tsen, 211 Tsfasman-Vlˆ adut-Zink, 106 Turing machine, 671 Turnwald, 495 twisted wreath product (of groups), 253 type, 267, 657 Uchida, 228, 230, 336, 524 ultrafilter, 138 ultrametric (absolute value), 239 ultrapower, 143 ultraproduct, 141 underlying normal stratification of, 717 underlying set of simple groups, 346 union (of structures), 137 universal domain, 172 universal elimination lemma, 716

791 universal Frattini p-cover, 531 universal Frattini cover, 513 universal Hilbert subset property, 289 universal quantifier, 133 unramified (extension of function fields), 69 unramified (ideal), 32 unramified (prime divisor), 59 unramified (valuation), 25 unramified extension of rings, 129 v.d.Dries-Schmidt, 428 V¨ olklein, 334 valuation, 19 valuation ring, 20 value (of infinite product), 92 value group, 20 value set, 467 valued field, 19 variable symbols, 132 variety, 175 virtually free pro-C, 632 Voloch, 94 Waerden, 48, 69, 318, 392, 428 Waterhouse, 18 weak (n-adic structure), 700 weak approximation theorem, 21 weak monadic structures, 700 weak solution (of an embedding problem), 503 weakly Ci , 456 weakly Ci,d , 456 weakly normic form, 457 weakly solvable (embedding problem), 503 Weigel, 358, 536 weight, 601 Weil, 93, 94, 106, 130, 175, 176, 183, 187, 189, 191, 208, 209, 290, 363 Weil differential, 54 Weissauer, 230, 262, 266, 278, 282, 284, 287, 290, 333 Whaples, 313 Wheeler, 561 width (of a Kronecer class), 464 Wielandt, 394 witness, 152 Witt, 300 word, 350 working functions, 671

792 wreath product, 257 X-stable, 294 Zannier, 248, 266 Zariski-closed (affine set), 176 Zariski-open (affine set), 176 Zariski K-topology, 172 Zariski K-closed sets (projective), 185 Zariski K-closed subsets, 172 Zariski K-topology, 185 Zariski-open (affine set), 176 Zariski-Samuel, 62, 98, 99, 100, 287, 426, 428, 732 Zariski topology, 176 zero, 52 zero set, 364 zeta function, 79, 371, 739

Index