Local Analytic Geometry: Basic Theory and Applications

i

ii

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v

Voor Jeannette, Nils en Noah F¨ ur Marlis

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Preface Algebraic geometry is, loosely speaking, concerned with the study of zero sets of polynomials (over an algebraically closed field). As one often reads in prefaces of introductory books on algebraic geometry, it is not so easy to develop the basics of algebraic geometry without a proper knowledge of commutative algebra. On the other hand, the commutative algebra one needs is quite difficult to understand without the geometric motivation from which it has often developed. Local analytic geometry is concerned with germs of zero sets of analytic functions, that is, the study of such sets in the neighborhood of a point. It is not too big a surprise that the basic theory of local analytic geometry is, in many respects, similar to the basic theory of algebraic geometry. It would, therefore, appear to be a sensible idea to develop the two theories simultaneously. This, in fact, is not what we will do in this book, as the “commutative algebra” one needs in local analytic geometry is somewhat more difficult: one has to cope with convergence questions. The most prominent and important example is the substitution of division with remainder. Its substitution in local analytic geometry is called the Weierstraß Division Theorem. The above remarks motivated us to organize the first four chapters of this book as follows. In Chapter 1 we discuss the algebra we need. Here, we assume the reader attended courses on linear algebra and abstract algebra, including some Galois theory. Probably the reader could just start with Chapter 2, referring to Chapter 1 only when needed. Chapter 2 deals with the basics of affine algebraic geometry, up to say, Hilbert’s Nullstellensatz and decomposition into irreducible components. In Chapter 3 we tackle the corresponding basics for local analytic geometry. Here it is assumed that the reader has knowledge of the theory of holomorphic functions in one variable. Although Chapter 4 is written in the language of local analytic geometry, most statements and proofs make sense for affine algebraic geometry, too. As an application of the general theory, in Chapter 5 we study the “simplest” germs of local analytic spaces: plane curve singularities. Topics here are Puiseux expansion, semigroups of curves, and resolutions of plane curve singularities. Many of the topics in the rest of the book cannot usually be found in books on local analytic geometry. The principle of conservation of number in Chapter 6 is such an example. This is about interpretation of invariants in “families”. The simplest example of this is the intersection number of two plane curve singularities. This intersection number can be interpreted as the number of intersection points appearing after slightly “perturbing” the singularities. To put the proofs in their proper context, we discuss and prove the basic coherence theorems of Oka and Cartan, and the finite mapping theorem of Grauert and Remmert. Next in line is the theory of standard bases in a power series ring, due to Hironaka and Grauert. The corresponding notion in a polynomial ring is called Gr¨obner basis, and has received much attention lately because of its applications in computer algebra. We refer to the computer algebra system Singular [Singular 2000].1 One of the main ideas is that standard bases allow a well-defined representative of an element f modulo an ideal I, called the normal form of f . In particular, this normal form is zero 1

Our advise to the reader is to compute many examples of the theory in this book by means of a computer algebra system, which can calculate in localizations of polynomials rings.

viii

Preface

if and only if f is an element of I. Chapter 8 is devoted to approximation theorems. Most famous, and the easiest to prove, is the Artin Approximation Theorem. It states that if one has a formal solution (that is, without regarding the convergence) of a set of analytic equations, then there exists an analytic, that is, convergent solution to these equations. Further, a formal solution can be arbitrarily approximated by an analytic one. For applications it is sometimes necessary that some solutions do not depend on some of the variables. This so-called nested approximation theorem does not hold in general, however. Nevertheless, it does hold under stronger assumptions. This is called Grauert’s Approximation Theorem. In Chapter 9 we give the classification of simple hypersurface singularities in all its details. One of the important tools here is the finite determinacy theorem. The fact that an isolated hypersurface singularity is finitely determined is quite easy to prove: one can prove it by an application of Newton’s Lemma, which is a “souped-up” version of the implicit function theorem. For a sharper statement we use the Artin Approximation Theorem. Finally, in Chapter 10, we give, as application of Grauert’s Approximation Theorem, a proof of the existence of a semi-universal deformation of an isolated singularity. Some proofs in this book are new, or at least we have not seen them in this form in the literature. We mention: • The proof of the convergence of the Puiseux expansion by using Newton’s Lemma. • The proof of the inversion theorem for Puiseux pairs. • Direct construction of factors of a power series in two variables, without using the Puiseux expansion. • Certain parts of the proof of Grauert’s Approximation Theorem, in particular the proof of Cartan’s Lemma by using standard bases. • The fact that the resolution graph determines the semigroup of the curve. • Parts of the proof of the Mather-Yau Theorem. Assuming the reader has the knowledge mentioned above, it was our intention to give either full proofs of all the statements in the book, or to put them into exercises with sufficient hints. Except for the descriptive third section of Chapter 8, we hope to have succeeded in this goal. Acknowledgment: Various people read parts of the manuscript and gave us useful comments. We mention Holger Cr¨oni, Anne Fr¨ uhbis Kr¨ uger, Tobias Hirsch, Rainer Kaenders and Dorin Popescu. We thank Thomas Keilen, who also made some of the pictures for us. We thank Pauline Bitsch for typing a large part of the manuscript. In particular we thank Olaf Bachmann for helping us with LATEX, and explaining us the use of the cvs system. Last but not least, we thank our wives Jeannette and Marlis for being so patient with us during the three years we were writing this book.

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Introductory Guide 1

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8.2 8.3

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Contents Preface

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Introductory Guide

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Contents

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1 Algebra 1.1 Noetherian Rings . . . . . . . . 1.2 Modules . . . . . . . . . . . . . 1.3 Local Rings and Localization . 1.4 Primary Decomposition . . . . 1.5 Finite and Integral Extensions

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1 1 6 17 22 34

2 Affine Algebraic Geometry 46 2.1 Affine Hypersurfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 2.2 Affine Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 2.3 Maps between Algebraic Sets . . . . . . . . . . . . . . . . . . . . . . . 66 3 Basics 3.1 3.2 3.3 3.4

of Analytic Geometry Holomorphic Functions of Several Complex Variables Weierstraß Division and Preparation Theorem . . . . Applications . . . . . . . . . . . . . . . . . . . . . . . Germs of Analytic Spaces . . . . . . . . . . . . . . . .

4 Further Development of Analytic Geometry 4.1 Dimension Theory . . . . . . . . . . . . . . . . . 4.2 Hilbert-Samuel Function and Multiplicity . . . . 4.3 Regular Local Rings and the Jacobian Criterion 4.4 Normalization . . . . . . . . . . . . . . . . . . . 5 Plane Curve Singularities 5.1 Puiseux Expansion . . . . . . . . . . . 5.2 Invariants . . . . . . . . . . . . . . . . 5.3 Resolutions of Irreducible Plane Curve 5.4 Reducible Plane Curve Singularities .

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124 126 135 148 159

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169 171 183 193 208

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219 220 230 238 243 252

6 The Principle of Conservation of Number 6.1 Sheaves . . . . . . . . . . . . . . . . . . . . . 6.2 Fundamental Properties of Coherent Sheaves 6.3 The Four Basic Coherence Theorems . . . . . 6.4 The Principle of Conservation of Number . . 6.5 Cohen-Macaulay Spaces . . . . . . . . . . . .

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xi 7 Standard Bases 269 7.1 The Division Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 7.2 Characterizations and Properties of Standard Bases . . . . . . . . . . 275 7.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 8 Approximation Theorems 8.1 Artin’s Approximation Theorem . . . . . . . . . . . . . . . . . . . . . 8.2 Grauert’s Approximation Theorem . . . . . . . . . . . . . . . . . . . . 8.3 Some other Approximation Theorems . . . . . . . . . . . . . . . . . .

286 286 290 299

9 Classification of Simple Hypersurface Singularities 9.1 Finite Determinacy of Hypersurface Singularities . . . . . . . . . . . . 9.2 The A-D-E–singularities are simple. . . . . . . . . . . . . . . . . . . . 9.3 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

302 303 313 322

10 Deformations of Singularities 10.1 Deformations of Functions . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Deformations of Singularities . . . . . . . . . . . . . . . . . . . . . . . 10.3 Existence of a Semi-Universal Deformation . . . . . . . . . . . . . . .

329 330 333 346

Bibliography

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1

1

Algebra

In the first chapter of the book we collect some facts from algebra which we will use in this book. We assume that the reader is familiar with the concepts of ring, ideal in a ring, principal ideal domain, prime ideal, unique factorization domain, etc. Throughout this book, a ring will mean a commutative ring with 1. All homomorphisms of rings always take 1 to 1. The first topic we discuss in this chapter is Noetherian rings, that is, rings in which every ideal is finitely generated. The main thing to prove here is the Hilbert Basis Theorem. This says that the polynomial ring K[x1 , . . . ,xn ], with K a field, is Noetherian. Then we will discuss the basic notions of the theory of modules over a ring, a notion which is analogous to vector spaces over a field. In the third section we will discuss local rings, that is, rings which have a unique maximal ideal. They play a very important role in this book. For us, the most important example will appear in the third chapter, the power series ring in n variables C {x1 , . . . ,xn }. Localization of rings will be defined. Geometric interpretations of localization will be deferred to Chapter two. In section four we discuss primary decomposition of ideals in Noetherian rings. Geometrically, this corresponds to the decomposition of an algebraic or analytic set into irreducible components: these notions will also be discussed in chapters two and three. Algebraically, it can be seen as a generalization of unique factorization in Z, or in K[x1 , . . . ,xn ]. In the final section of this chapter we treat ring extensions. The notions of finite and algebraic extensions of fields here correspond to finite and integral extensions of rings. A reduced ring is called normal (or integrally closed), if every element of the total quotient ring Q(R) which is integral over R is already in R. The most important examples are K[x1 , . . . ,xn ], and, as we will see in Chapter three, the power series rings C {x1 , . . . ,xn }. We will prove the Cayley-Hamilton Theorem, and the very important theorem of the finiteness of integral closure, which is due to Dedekind and Kronecker. It says that, under some weak hypothesis, the normalization of a ring R is finitely generated as an R–module. This is a nontrivial statement, and it is only because of this proof that we need some Galois theory as prerequisite.

1.1

Noetherian Rings

Definition 1.1.1. A ring R is called Noetherian if every ideal in R is finitely generated. This means that for every idealP I ⊂ R there exist finitely many elements f1 , . . . ,fs s in I such that I = (f1 , . . . ,fs ) = { ν=1 aν fν : aν ∈ R}. In particular, principal ideal domains (domains in which every ideal can be generated by one element, for example fields) are Noetherian. But the Noetherian condition is much less restrictive, and almost all rings we consider in this book are Noetherian. Before giving some more examples we give the following characterizations of Noetherian rings. Lemma 1.1.2. Let R be a ring. Then the following conditions are equivalent: (1) R is Noetherian.

2

1 Algebra (2) Any chain of ideals in R I1 ⊂ I2 ⊂ · · · ⊂ Ik ⊂ · · · becomes stationary, that is, there exists an n > 0 such that Ik = In for all k ≥ n. (3) Every nonempty set of ideals in R has a maximal element with respect to inclusion.

Proof. (1) =⇒ (2) It is easily checked that the union ∪∞ k=1 Ik is an ideal again, which by assumption is finitely generated, say by a1 , . . . ,as . For n big enough, one has ai ∈ In for i = 1, . . . ,s. Then Ik = In for all k ≥ n. (2) =⇒ (1) Suppose that there exists an ideal I in R which is not finitely generated. Then we can inductively define elements as follows: • Take a1 ∈ I. • Given a1 , . . . ,ak−1 take an element ak ∈ I \ (a1 , . . . ,ak−1 ). But then the chain (a1 ) ⊂ (a1 ,a2 ) ⊂ · · · is not stationary, which is a contradiction. (2) =⇒ (3) If (3) were false, then there would exist a set of ideals such that for any ideal Ik in this set we can find an ideal Ik+1 with Ik $ Ik+1 . This would lead to a chain of ideals which is not stationary, a contradiction. (3) =⇒ (2) Apply (3) to the set of ideals Ik for k ≥ 1. If In is a maximal element, then Ik = In for all k ≥ n. Theorem 1.1.3 (Hilbert’s Basis Theorem). Let R be a Noetherian ring. Then the polynomial ring R[x] is also Noetherian. Proof. By contradiction. Let I ⊂ R[x] be an ideal in R[x] which is not finitely generated. We then inductively define a sequence of polynomials f1 ,f2 , . . . in I by: • Let f1 ∈ I be a polynomial of minimal degree. • Given f1 , . . . ,fk−1 , let fk be a polynomial of minimal degree in I \ (f1 , . . . ,fk−1 ). Let ak be the leading coefficient of fk . Then we claim that for all k ≥ 2: (a1 , . . . ,ak−1 ) $ (a1 , . . . ,ak ) which would give us a chain of ideals in R which does not stabilize, in contradiction to the assumption that R is Noetherian. Suppose on the contrary that ak ∈ (a1 , . . . ,ak−1 ) for some k. We then can write ak = b1 a1 + . . . + bk−1 ak−1 for some bi ∈ R. The polynomial fk − b1 xdeg(fk )−deg(f1 ) f1 − · · · − bk−1 xdeg(fk )−deg(fk−1 ) fk−1 is in I \ (f1 , . . . ,fk−1 ), but has degree smaller than fk , in contradiction to the choice of fk . The following corollary gives us the first class of examples of Noetherian rings using the fact that fields are Noetherian.

1.1 Noetherian Rings

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Corollary 1.1.4. Let R be a Noetherian ring, and I ⊂ R[x1 , . . . ,xn ] be an ideal. Then the quotient ring R[x1 , . . . ,xn ]/I is Noetherian. Proof. Using induction and Theorem 1.1.3 we obtain that R[x1 , . . . ,xn ] is Noetherian. The rest is given as Exercise 1.1.20. Definition 1.1.5. (1) Let R be a ring. An R–algebra S is a ring S together with a ring homomorphism ϕ : R −→ S. Via this ring homomorphism we can define multiplication with scalars r · s := ϕ(r)s. Alternatively, one can define an R–algebra S to be a ring S which is at the same time an R–module, with the condition that r · (st) = (rs) · t for r,s ∈ R and t ∈ S. (2) Let R be a ring and S = R[x1 , . . . ,xn ]/I be the quotient of the polynomial ring by an ideal I. Then the R–algebra S is called a ring of finite type over R or, alternatively, an affine algebra over R. From Corollary 1.1.4 we obtain that affine algebras over a Noetherian ring (or in another language, rings of finite type over a Noetherian ring) are Noetherian. Our most important case is the case of affine algebras over a field (which is Noetherian). Now we will investigate another special class of Noetherian rings. Definition 1.1.6. Let K be a field and R be a K–algebra. R is called an Artinian K–algebra if R, considered as K–vector space, has dimK (R) < ∞. Lemma 1.1.7. Let R be an Artinian K–algebra. Then R is Noetherian. Proof. Let I ⊂ R be an ideal, then I is a finite-dimensional K–vector space because R is a finite-dimensional K–vector space. Any K–basis of I, in particular, generates I as ideal and, therefore, I is finitely generated. Lemma 1.1.8. Let R be an Artinian K–algebra. Then any chain of ideals in R I1 ⊃ I2 ⊃ . . . ⊃ Ik ⊃ . . . becomes stationary, that is, there exists an n > 0 such that Ik = In for all k ≥ n.1 Proof. A chain of ideals is especially a chain of K–vector spaces. Such a chain becomes stationary because R is a finite-dimensional K–vector space. Examples 1.1.9. Let K be a field. (1) K[x]/(xn ), n ≥ 1, is an Artinian K–algebra. (2) Let A be an Artinian K–algebra and I ⊂ A be an ideal. Then A/I is an Artinian K–algebra. At the end of this section, we introduce the notion of graded rings and prove some other useful tools. 1

This is the usual way to define an Artinian ring. We need later on only Artinian K–algebras in the sense of 1.1.6 which makes proofs a little simpler. Note that C [x]/(x2 ) is a Q–algebra which is an Artinian ring, but not an Artinian Q–algebra in the sense of our definition.

4

1 Algebra

Definition 1.1.10. Let G be an abelian group. A G–graded ring is a ring R together with a direct sum decomposition M (1.1) R= Rg , g∈G

such that Rg · Rh ⊂ Rg+h for all g,h ∈ G.

(1.2)

The elements of Rg are called homogeneous of degree g. A Z–graded ring is simply called a graded ring. Examples 1.1.11. (1) Let S be a ring (for example a field), and R := S[x1 , . . . ,xn ] be the polynomial ring in n variables with coefficients in S. Then R is a graded ring, by defining ( ) X ν1 νn Rd := cν1 ,...,νn x1 . . . xn ; cν1 ,...,νn ∈ S , and Rd = (0) for d < 0. ν1 +···+νn =d

An element of Rd is called a homogeneous polynomial of degree d. Consider f ∈ R. Then we can write f = f0 + f1 + . . . + fd , fi ∈ Ri , fd 6= 0 in a unique way. We call d =: deg(f ) the degree of the polynomial f . (2) More generally, consider the polynomial ring R := S[x1 , . . . ,xn ] as above and let λ = (λ1 , . . . ,λn ) ∈ Nn . We call f ∈ S[x1 , . . . ,xn ] quasi-homogeneous of degree d with respect to the weight λ if f can be written as X f= cν1 ,...,νn xν11 . . . xνnn . ν1 λ1 +...+νn λn =d

Again, R = ⊕d≥0 Rd , where Rd consists of all quasi-homogeneous polynomials of degree d. Let f ∈ R, f = f0 + · · · + fd , fi ∈ Ri , fd 6= 0. If we wish to distinguish between the usual degree and the quasi-homogeneous degree, we call d =: w-deg(f ) (the weighted degree). Theorem 1.1.12 (Chinese Remainder Theorem). Let R be a ring and I1 , . . . ,Im be ideals in R. Assume that m

(1) ∩ Iν = (0), ν=1

(2) Iν + Iµ = R for ν 6= µ. Then the canonical map R −→ x is an isomorphism of rings.

7→

m M

R/Iν ,

ν=1

(x + I1 , . . . ,x + Im )

1.1 Noetherian Rings

5

Proof. The first condition implies that the map is injective. To prove the surjectivity it is enough to see that for all j there exists an x ∈ R such that x ∈ Iν for all ν 6= j and x ≡ 1 mod Ij . Without loss of generality, it suffices to prove this for j = 1. By assumption we have Iν + I1 = R for all ν 6= 1. In particular we can find xν ∈ Iν and yν ∈ I1 for ν = 2, . . . ,m such that xν + yν = 1. We define x :=

m Y

ν=2

xν =

m Y

ν=2

(1 − yν ).

As xν ∈ Iν for ν = 2, . . . ,m, it follows that x ∈ Iν for ν = 2, . . . ,m. As yν ∈ I1 , it follows that x ≡ 1 mod I1 . Finally, we prove a useful Lemma which will be needed in Chapters 4 and 6. Lemma 1.1.13 (Prime Avoidance). Let R be a Noetherian ring and p1 , . . . ,pn be ideals in R. Suppose that p1 , . . . ,pn−1 are prime ideals. Let I ⊂ ∪ni=1 pi be an ideal. Then there exists k such that I ⊂ pk . To put it differently, let I be an ideal such that I 6⊂ pi for all i. Then I 6⊂ ∪ni=1 pi . This means that we can find f ∈ I with f ∈ / pi for all i, whence the name prime avoidance. Proof. The proof is done by induction on n: the case n = 1 is obvious. So it suffices to show that I ⊂ ∪j6=i pj for some i. This we will show by contradiction. So suppose that I 6⊂ ∪j6=i pj for all i. Therefore, we can choose x1 , . . . ,xn ∈ I such that xi 6∈ ∪j6=i pj . Because of the fact that xi ∈ I ⊂ ∪nj=1 pj we get xi ∈ pi . Now we consider x1 + x2 . . . xn ∈ I. In particular, there exists a k such that x1 + x2 . . . xn ∈ pk . If k = 1 then x1 ∈ p1 and, therefore, x2 . . . xn ∈ p1 . This implies xν ∈ p1 for some ν > 1 which is not possible by the choice of xν . If k > 1 then x2 . . . xn ∈ pk because xk ∈ pk and, therefore, x1 ∈ pk which is again not possible. Altogether we obtain a contradiction to the assumption that I 6⊂ ∪j6=i pj for all i. This shows the Lemma.

Exercises 1.1.14. Let I, J ⊂ R be ideals. The product I ·J ⊂ R is the ideal in R generated by the elements x · y, where x runs through I and y runs through J. Show that I · J ⊂ I ∩ J. 1.1.15. (1) Let I, J ⊂ R be ideals. Show that I : J := {x ∈ R : xJ ⊂ I} is an ideal in R, and that I ⊂ I : J.

(2) Let a ∈ R. Prove that (I ∩ J) : a = (I : a) ∩ (J : a).

(3) Let I : f ∞ := {x ∈ R : xf s ∈ I for some s}. If R is Noetherian, show that I : f ∞ = I : f k for some k ∈ N.

(4) Let I ⊂ R be an ideal, and x,y ∈ R. Prove that (I : x) : y = I : xy. √ 1.1.16. Let I ⊂ R be an ideal. The radical I of I is defined by: √ I = {f ∈ R : f s ∈ I for some s ∈ N}. √ An ideal is called radical if and only if I = I. √ (1) Prove that I is an ideal. p√ √ √ (2) Prove that I + J = I + J.

6

1 Algebra √ √ √ (3) Find ideals I,J for which I + J % I + J . √ √ √ (4) Prove that I ∩ J = I ∩ J. Generalize to finitely many ideals.

1.1.17. Let R be a ring. An element x ∈ R is called nilpotent if there exists an n ∈ N with xn = 0. Show that the set of nilpotent elements is an ideal N . The ring Rred := R/N is called the reduction of R. A ring is called reduced if it has no nonzero nilpotent elements. Show that Rred is reduced. P 1.1.18. Let R be a G–graded ring. An ideal I ⊂ R is called homogeneous if I = g∈G (I ∩ Rg ). Prove the following. (1) I is homogeneous if and only if it can be generated by homogeneous elements. (2) Sums, products and intersections of homogeneous ideals are homogeneous. 1.1.19. An abelian group G is called ordered, if there is a total ordering < on G satisfying: g < h =⇒ a + g < a + h for all a ∈ G. For example, the group Z with its natural ordering is an ordered group. Let G be an ordered group and R be a G–graded ring. (1) Prove that a homogeneous ideal I is prime if and only if for all homogeneous elements a,b ∈ R with ab ∈ I it follows that a ∈ I or b ∈ I. (2) Prove that the radical of a homogeneous ideal is homogeneous.

1.1.20. Let R be a Noetherian ring and I ⊂ R be an ideal. Prove that R/I is Noetherian. 1.1.21. Consider two ideals I ⊃ J in a Noetherian ring R. Suppose that for all f ∈ I there exists a k (depending on f) with f k ∈ J. Prove that some power of I is contained in J. In √ k particular I ⊂ I for some k.

1.2

Modules

The notion of a module over a ring is the analog of the notion of vector space over a field, and is obtained by simply copying the definition. Here it is. Definition 1.2.1. Let R be a ring. An R–module M is a set M with two operations: (1) There is an addition + : M −→ M : x,y ∈ M 7→ x + y ∈ M (2) We have multiplication with scalars ·R × M −→ M : x ∈ M, a ∈ R 7→ a · x = ax ∈ M. These should satisfy the following conditions. • The set M is an abelian group with respect to the addition +. • a(x + y) = ax + ay (ab)x = a(bx)

(a + b)x = ax + bx 1·x =x

This holds for all x,y ∈ M and a,b ∈ R.

1.2 Modules

7

As in the theory of rings, groups, and vector spaces, we want to consider modules up to isomorphism. The definition should not surprise the reader: Definition 1.2.2. Let M and N be R–modules, ϕ : M → N a map. • ϕ is called a homomorphism of R–modules (or simply homomorphism) if – ϕ(x + y) = ϕ(x) + ϕ(y) – ϕ(ax) = aϕ(x) for all x,y ∈ M and all a ∈ R. • ϕ is called an isomorphism if and only if ϕ is a bijective homomorphism.

• Two modules M and N are called isomorphic, notation M ∼ = N , if and only if there exists an isomorphism ϕ : M → N .

From the definition it is not immediate that isomorphy is an equivalence relation. Of course, the fact that M is isomorphic to itself is trivial, as is the fact that the isomorphism property is transitive. To prove that it is symmetric too, one has to show that the inverse ϕ−1 : N → M of a bijective homomorphism ϕ : M → N is a homomorphism again. This will be left to the reader as an exercise. This property distinguishes algebra from for example topology. A bijective continuous map between topological space need not be a homeomorphism. To give a very trivial example, the identity map from a set with more than one element to itself, the first equipped with the discrete topology, and the second equipped with the trivial topology, is a bijective continuous map which is not a homeomorphism. We now give some examples of modules. Examples 1.2.3. (1) A K–module, for a field K, is just a K–vector space. (2) A Z–module is an abelian group. (3) An ideal I of R is an R–module. In particular, R itself is an R–module. (4) Consider the polynomial ring K[X], and let M be a K[X]–module. Then in particular M is a K–module, that is, a K–vector space. We also get a map (multiplication with X): X : M −→ M which one checks to be K–linear. In fact, the data M is a K[X]–module, and M is a K–vector space together with a linear map, are equivalent. Classifying K[X]– modules up to isomorphism is the same as classifying endomorphisms of K–vector spaces up to conjugacy, see Exercise 1.2.30. We collect some more definitions. Definition 1.2.4. Let R be a ring, M and N be R–modules.

8

1 Algebra (1) Let ∅ 6= A ⊂ M . Then A is called a submodule of M if for all x,y ∈ A and a ∈ R we have x + y and ax ∈ A. Then A is in a natural way an R–module, where addition and scalar multiplication are induced by those in M . One shows without difficulty that the image ϕ(M ) of a homomorphism ϕ : M −→ N is in a natural way a submodule of N . Let N ⊂ M be a submodule. Then the quotient module M/N is the set of equivalence classes x := x + N where x runs through M . It is in a natural way an R–module via • (x + N ) + (y + N ) := (x + y) + N , • a(x + N ) := ax + N . Here x,y ∈ M and a ∈ R. One checks that addition and scalar multiplication are well-defined. (2) Let ϕ : M −→ N be a homomorphism of R–modules. We define • the kernel of ϕ, notation Ker(ϕ), as Ker(ϕ) := {x ∈ M : ϕ(x) = 0}. One checks, exactly as in linear algebra, that Ker(ϕ) = {0} if and only if ϕ is injective. Moreover, Ker(ϕ) is a submodule of M , • the cokernel of ϕ, notation Coker(ϕ), to be the quotient module N/ϕ(M ). (3) The direct sum M ⊕ N of M and N is defined to be the direct sum of abelian groups M ⊕ N = {(m,n),m ∈ M,n ∈ N }, with obvious addition, and with scalar multiplication defined by a(m,n) := (am,an). It is sometimes also called the direct product M × N . Similarly, one defines the direct sum ⊕i∈I Mi of R–modules. Here I is any index set. As a set: M Mi = {(mi )i∈I with only finitely many mi 6= 0} i∈I

and we have the obvious addition and scalar multiplication. One also has the direct product Y Mi = {(mi )i∈I }. i∈I

with obvious addition and scalar multiplication. Here it is allowed that infinitely many mi are nonzero.

(4) An R–module is called a free module if it is isomorphic to a direct sum ⊕i∈I R of copies of R. Often we will have the case that I is a finite set of cardinality t. This free module is called the free module of rank t, denoted by Rt . The fact that the rank is well-defined is not trivial and will be shown in the exercises. (5) Let M be an R–module, m1 , . . . ,mt be elements of M . The submodule (m1 , . . . ,mt ) generated by m1 , . . . ,mt is by definition the submodule of M given by {a1 m1 + . . . + at mt : a1 , . . . ,at ∈ R}. Similarly one defines the submodule generated by an arbitrary number of elements of M .

1.2 Modules

9

(6) An R–module M is called finitely generated if there exist finitely many elements m1 , . . . ,mt of M such that M = (m1 , . . . ,mt ). If this is the case, the elements m1 , . . . ,mt are called generators of M . Finite generation of a module is equivalent to having a surjective homomorphism

ϕ : Rt −→ M,

(a1 , . . . ,at ) 7→ a1 m1 + . . . + at mt . (7) A set {m1 , . . . ,mt } is called a basis of M if every m ∈ M can be written as an R– linear combination of the elements m1 , . . . ,mt in a unique way, that is, for all m ∈ M there exist uniquely defined a1 , . . . ,at ∈ R such that m = a1 m1 + . . . + at mt . One can show (see the exercises) that a module has a basis if and only if it is isomorphic to a finitely generated free module. Similarly an infinite set is called a basis of M if every element of M can be written as an R–linear combination of finitely many elements of the set in a unique way. (8) The torsion module of M is defined by: Tors(M ) := {m ∈ M : there exists a nonzerodivisor a ∈ R with a · m = 0}. So in case R is an integral domain, we have to find a nonzero element a ∈ R with am = 0. A module M is called torsion free if Tors(M ) = 0. M is called a torsion module if Tors(M ) = M . (9) The set HomR (M,N ) of all homomorphisms from M to N is an R–module via the operations (ϕ + ψ)(m) := ϕ(m) + ψ(m). (aϕ)(m) := aϕ(m), for all m ∈ M and a ∈ R. (10) Let m ∈ M . The annihilator of m is defined by Ann(m) := {r ∈ R : r · m = 0}. The annihilator of M by definition is Ann(M ) := {r ∈ R : r · m = 0 for all m ∈ M }. The annihilator is obviously an ideal. Despite the obvious similarity with vector spaces, the general theory of modules over a ring is much more difficult and much richer. The inability of to divide is just too much an obstacle. For example, we know from the theory of vector spaces, that any finitely generated vector space over a field K is isomorphic to K t for some t. This is certainly not true for rings, as the following trivial example shows: Example 1.2.5. (1) The Z–module (that is, abelian group) Z/(2) is not free. Indeed a free group either has 1 or an infinite number of elements. (2) More generally, any module M whose torsion submodule Tors(M ) is nontrivial, is not a free module. Indeed, suppose the converse, that is, let {mi }i∈I be a basis of M . Let m ∈ Tors(M ), m 6= 0 and assume

10

1 Algebra m = a1 m 1 + . . . + at m t . As m 6= 0 we have ai 6= 0 for some i. Because m ∈ Tors(M ) there exists a nonzerodivisor a ∈ R with am = 0. We obtain a · m = aa1 m1 + . . . + aat mt = 0. By definition of basis aai = 0, which is a contradiction to the fact that a is a nonzerodivisor.

One could ask the question whether at least finitely generated torsion free modules are free. But this is also much too much to ask for, as the following simple lemma shows: Lemma 1.2.6. Let I ⊂ R be an ideal, with R an integral domain. Then I is a free module if and only if I is principal. Proof. The case where I = (0) is trivial. Otherwise, suppose that I is principal, that is, I = (f ) for some f ∈ R. Then the map: R −→ I given by a 7→ af gives an isomorphism of R–modules. It is injective because R is supposed to be an integral domain. To prove the converse, let I be a free R-module. If the rank is one, let ϕ : R −→ I be an isomorphism. It is then immediate that I is the ideal generated by ϕ(1). We are finished if we show that an ideal cannot be isomorphic to a free module of rank at least two. For this, it suffices to show that there does not exist an injective map α : Rn −→ I, as soon as n ≥ 2. Suppose the converse. Without loss of generality we may assume n = 2. Let f = α(1,0), and g = α(0,1). Since α is injective, both f and g are nonzero. But then α(g, − f ) = gf − f g = 0, a contradiction. So for example the ideal (x,y) in the polynomial ring K[x,y] is not a free K[x,y]– module. From this lemma one might hope that at least for principal ideal domains (that is, domains where every ideal can be generated by one element), a finitely generated torsion free module is free. This is indeed the case, in fact one has the following classification theorem: Theorem 1.2.7. Let R be a principal ideal domain, and M be a finitely generated R– module. Then M∼ = Rs ⊕ Tors(M ). The torsion module has the following form: Tors(M ) ∼ = R/(a1 ) ⊕ · · · ⊕ R/(at ) with (a1 ) ⊃ (a2 ) ⊃ . . . ⊃ (at ) 6= (0). For a proof, we refer to Lang, [Lang 1965], Chapter XV, §2. At one point in the book we will need the statement that a finitely generated torsion free module over the power series ring in one variable is free. This is a special case of the theorem, for which an easier proof is available. We refer to 1.3.9. Of course, the finitely generated free modules Rp with p ∈ N are the simplest R– modules. Just as in linear algebra, maps between such modules can be given by matrices. Consider Rp with basis {e1 , . . . ,ep }, and Rq with basis {f1 , . . . ,fq }. Let ϕ : Rp −→ Rq be an R–module homomorphism. The matrix of ϕ (with Pq respect to the chosen bases) is the matrix (aij ) with entries in R given by ϕ(ei ) = j=1 aji fj . All the results in linear algebra which do not use division in their proofs carry over to the general case of rings. In particular, the determinant of a map between free modules of the same rank is defined. On the other hand, because in a general ring we cannot divide, Cramer’s rule does not carry over. But at least we can save the following statement.

1.2 Modules

11

Theorem 1.2.8 (Cramer’s Rule). Let a ring R, an n × n matrix A = (aij ) with coefficients in R and vectors (x1 , . . . ,xn ) and (b1 , . . . ,bn ) with entries in R be given. For all j let Aj be the matrix obtained from A by replacing its j–th column2 by (b1 , . . . ,bn )t . Consider the system of equations n X

aij xj = bi

i = 1, . . . ,n.

j=1

Then det(A)xj = det(Aj )

for j = 1, . . . ,n..

For a proof, look in any textbook on linear algebra, and forget to do the division at the end of the proof, or do the proof yourself, Exercise 1.2.28. Definition 1.2.9. Let A be an n × n matrix with entries in R. The adjoint matrix Aad = (aad ij ) is defined by i+j aad det(Aij ) ij = (−1)

where Aij is the matrix obtained from A by deleting the i–th column and j–th row. For a matrix A with adjoint matrix Aad we have the following theorem, which follows from Cramer’s rule: Theorem 1.2.10. Aad A = AAad = det(A) · Idn , where Idn denotes the n × n unit matrix. So, up to a factor (the determinant), the adjoint matrix describes “the inverse matrix”. Definition 1.2.11. A (finite or infinite) sequence of R–modules and homomorphisms αk−1

α

k · · · −→ Mk−1 −−−→ Mk −−→ Mk+1 −→ · · ·

is called a complex if Ker(αk ) ⊃ Im(αk−1 ). It is called exact at Mk if Ker(αk ) = Im(αk−1 ). It is called exact if it is exact at all Mk . In particular we have: α

• 0 −→ M −→ N is exact ⇐⇒ α is injective, α

• M −→ N −→ 0 is exact ⇐⇒ α is surjective, α

• 0 −→ M −→ N −→ 0 is exact ⇐⇒ α is an isomorphism. An exact sequence of type α

β

0 −→ M1 −→ M2 −→ M3 −→ 0 is called a short exact sequence. This is equivalent to saying that α is injective, β is surjective, and β induces an isomorphism M2 / Im(α) ∼ = M3 . 2

Given a matrix A we denote by At the transposed matrix.

12

1 Algebra

Lemma 1.2.12 (Snake Lemma). Consider a commutative diagram3 f2

f1

M1 −→ M2 −→ M3 −→ 0 h1 ↓ h2 ↓ h3 ↓ g2 g1 0 −→ N1 −→ N2 −→ N3 −→ 0

0 −→

of R–modules, with exact rows. Then the sequence f′

f′

δ

g

g

1 2 1 2 Coker(h2 ) −→ Coker(h3 ) → 0 0 → Ker(h1 ) −→ Ker(h2 ) −→ Ker(h3 ) −→ Coker(h1 ) −→

is exact. In this lemma f1′ and f2′ are restrictions of f1 and f2 , and g 1 and g2 are induced by g1 and g2 . The map δ is defined as follows. Let x ∈ Ker(h3 ), that is, x ∈ M3 with h3 (x) = 0. As the upper row is exact, it follows that f2 is surjective, hence we can take y ∈ M2 with f2 (y) = x. Because of the commutativity of the diagram we have g2 h2 (y) = h3 f2 (y) = h3 (x) = 0. Therefore, h2 (y) ∈ Ker(g2 ). Since the lower row is exact, we have Ker(g2 ) = Im(g1 ). Hence we can find w ∈ N1 with g1 (w) = h2 (y). Now we put δ(x) to be the class of w in Coker(h1 ). We have to prove that δ is well-defined and that this sequence is exact. The proof of the Snake Lemma is left as Exercise 1.2.27. Definition 1.2.13. Let R be a ring, and M be an R–module. Then M is called Noetherian if every submodule N ⊂ M is finitely generated. This just means that we can find finitely many elements m1 , . . . ,mt ∈ N such that N = R · m1 + . . . + R · mt = (m1 , . . . ,mt ). Note that a ring R is Noetherian if and only if R, considered as an R–module, is Noetherian. The following lemma is left as Exercise 1.2.29. Lemma 1.2.14. • Submodules and quotient modules of Noetherian modules are Noetherian. • Suppose that N ⊂ M are R–modules. Then M is Noetherian ⇐⇒ N and M/N are Noetherian. We could rephrase this by saying that, in a short exact sequence of modules, if any two are Noetherian, then the third is Noetherian. Noetherian modules over a Noetherian ring are easy to construct: Lemma 1.2.15. Let R be a Noetherian ring, and M be a finitely generated R–module. Then M is a Noetherian R–module. 3

A diagram

M1 h1 ↓

f

− →

M2 h2 ↓

is called commutative if h2 f = gh1 , that is it does not matter which route one − → N2 takes along the diagram. N1

g

1.2 Modules

13

Proof. As M is a quotient of a free module, we only have to prove the lemma for M = Rt , by the previous lemma. This will be done by induction on t. The case t = 1 is the assumption. Apply the previous lemma to the exact sequence 0 1 1 B0C B.C @.A

0 0 1 0 ... 0 1 .. A @ .. .. ..

. .. . . 0 0 0 0 ... 1 −−−−−−→ Rt−1 → 0. 0 → R −−−−→ Rt −−− Definition 1.2.16. (1) The tensor product M ⊗R N is an R–module with the following universal property: There exists a canonical bilinear map λ : M × N −→ M ⊗R N . If ϕ : M × N −→ P is any bilinear map, then there exists a unique homomorphism h : M ⊗R N −→ P such that h ◦ λ = ϕ. λ / M ⊗R N M × NM MMM MMM h ϕ MMM M&  P. One can give a direct construction of the tensor product as follows. M ⊗R N is the quotient of the free R–module generated by {m ⊗ n : m ∈ M, n ∈ N } divided out by the submodule of the tensor product relations generated by {(am + bn) ⊗ (cp + dq) − ac · m ⊗ p − ad · m ⊗ q − bc · n ⊗ p − bd · n ⊗ q | a,b,c,d ∈ R, m,n ∈ M, p,q ∈ N }. The bilinear form λ : M × N −→ M ⊗R N is defined by λ(m,n) = m ⊗ n and has the property mentioned above (Exercise 1.2.45). (2) Let S be an R–algebra, then M ⊗R S is an S–module with the multiplication s · m ⊗ s′ := m ⊗ (ss′ ) (Exercise 1.2.44). For example, if V is a real vector space, then V ⊗R C is a C –vector space, called the complexification of V . (3) Let S,T be R–algebras, then S ⊗R T is an R–algebra with the multiplication (s ⊗ t) · (s′ ⊗ t′ ) := (ss′ ) ⊗ (tt′ ). The simplest example here is the case that R = k is a field, S = k[X] and T = k[Y ]. Then S ⊗R T = k[X,Y ]. Let i1 : S −→ S ⊗R T , i2 : T −→ S ⊗R T be defined by i1 (s) = s ⊗ 1, i2 (t) = 1 ⊗ t, then i1 S ⊗O R T o SO i2

T o

R

is commutative, and universal (cf. Exercise 1.2.45), that is: let A be an R–algebra and j1 : S −→ A, j2 : T −→ A be homomorphisms such that AO o

j1

SO

j2

T o

R

14

1 Algebra

is commutative, then there exists a unique homomorphism h : S ⊗R T −→ A such that h ◦ iν = jν , ν = 1,2. Definition 1.2.17. Let R be a G–graded ring. An R–module M is called G–graded if there is a decomposition M = ⊕g∈G Mg such that Rg Mh ⊂ Mg+h for all g,h ∈ G. In particular a G–graded ring is a G–graded module over itself. A Z– graded module is called a graded module. A homomorphism ϕ : M −→ N is called homogeneous if ϕ(Mg ) ⊂ Ng for all g ∈ G. More generally, one says that ϕ : M −→ N is homogeneous of degree h if ϕ(Mg ) ⊂ Ng+h for all g ∈ G. For a G–graded module M and g ∈ G we define the G–graded module M (g). It has M as the underlying module, but has different grading: M (g)h := Mg+h . So a homogeneous homomorphism ϕ : M −→ N of degree g induces a homogeneous homomorphism ϕ : M −→ N (g) (or ϕ : M (−g) −→ N ). Definition 1.2.18. Let R be a ring and ϕ : Rn −→ Rm be an R–module homomorphism. Assume that ϕ is defined by the matrix A for some choice of bases in Rn and Rm . Then for all 1 ≤ t ≤ min{n,m} the ideal generated by the t–minors of A does not depend on A and is called the t–th Fitting ideal It (ϕ) of ϕ. For convenience we define It (ϕ) = R if t ≤ 0. Remark 1.2.19. (1) The proof that It (ϕ) is independent on the choice of the bases is left as Exercise 1.2.47. ϕ

ψ

→ Rt −→ M −→ 0 be exact. Then Im−ν (ϕ) = → Rm −→ M −→ 0 and Rs − (2) Let Rn − It−ν (ψ) for all ν. The proof uses some facts on so-called local rings, to be treated in the next section. The proof is left as Exercise 1.3.28. (3) If S is an R–algebra, then Iν (ϕ ⊗ idS ) = Iν (ϕ)S. This is left as Exercise 1.2.47.

Exercises 1.2.20. Prove that the ring of integers Z and the polynomial ring in one variable K[x] over a field K are principal ideal domains. (Hint: Use division with remainder.) 1.2.21. (1) Let ϕ : Rp → Rs be an isomorphism between free R–modules. Show that p = s. (Hint: Suppose not, for example p > s. Let ψ : Rs → Rp be the inverse of ϕ. Extend ϕ resp. ψ to maps ϕ′ : Rp → Rs ⊕ Rp−s resp.

ψ ′ : Rs ⊕ Rp−s → Rp

in the obvious way. Derive a contradiction by looking at determinants.) (2) Conclude that the rank of a free module is well-defined. 1.2.22. Let M be an R–module, and I = Ann(M ) = {x ∈ R : x · M = 0}. Then M is in a natural way an R/I–module. Prove this.

1.2 Modules

15

1.2.23. Let R be a subring of S. Show that one can, in a natural way, view S as an R–module. 1.2.24. Prove that Tors(M ) is a submodule of M . 1.2.25. Consider an exact sequence αk−1

α

k · · · −→ Mk−1 −→ Mk −→ Mk+1 −→ · · · .

Put Nk = Im(αk ) = Ker(αk+1 ). Show that for all k one has short exact sequences: 0 −→ Nk−1 −→ Mk −→ Nk −→ 0. (So any exact sequence can be split into short exact sequences.) 1.2.26. Let K be a field, and consider an exact sequence 0 −→ M1 −→ M2 −→ · · · −→ Ms −→ 0 P of finite-dimensional K–vector spaces. Prove that sk=1 (−1)k dimK (Mk ) = 0.

1.2.27. Prove the Snake Lemma 1.2.12. 1.2.28. Prove Cramer’s rule 1.2.8. 1.2.29. Prove Lemma 1.2.14. 1.2.30.

(1) Let M be a K[X]–module, which is finite-dimensional as a K–vector space. Show that M is a torsion module. (2) Prove the theorem on Jordan normal form of matrices over an algebraically closed field, using the classification of finitely generated modules over a principal ideal domain, see Theorem 1.2.7. 1.2.31. Let K be a field, S be a K–algebra, and M be an S–module. Suppose that Ann(M ) = (0) and that dimK (M ) < ∞. Prove that dimK (S) < ∞. (Hint: Show that one can embed S into HomS (M,M ).) 1.2.32. Let N be an R–module. Prove that the map φ : HomR (R,N ) −→ N , φ(ϕ) = ϕ(1), is an isomorphism. 1.2.33. Let M,M ′ ,N,N ′ be R–modules and α : M ′ −→ M , β : N −→ N ′ be homomorphisms. Prove that ϕ 7→ β ◦ ϕ ◦ α defines a homomorphism HomR (M,N ) −→ HomR (M ′ ,N ′ ). n

1.2.34. Let M,N be R–modules and M = ⊕ Mi . Prove that i=1

n

Hom(M,N ) ∼ = ⊕ Hom(Mi ,N ) i=1 n

Hom(N,M ) ∼ = ⊕ Hom(N,Mi ). i=1

1.2.35. Consider an exact sequence · · · −→ Mk+1 −→ Mk −→ Mk−1 −→ · · · of R–modules. Let x ∈ R be a nonzero divisor of Mk for all k. Prove that the induced sequence · · · Mk+1 /xMk+1 −→ Mk /xMk −→ Mk−1 /xMk−1 −→ · · · is also exact. (Hint: Split the exact sequence in many short exact sequences and use the Snake Lemma.)

16

1 Algebra

1.2.36. Let 0 −→ M −→ N −→ L be a complex of R–modules. Prove that the complex is exact if and only if for all R–modules A 0 −→ HomR (A,M ) −→ HomR (A,N ) −→ HomR (A,L) is exact. 1.2.37. Let M −→ N −→ L −→ 0 be a complex of R–modules. Prove that the complex is exact if and only if for all R–modules A 0 −→ HomR (L,A) −→ HomR (N,A) −→ HomR (M,A) is exact. 1.2.38. Let M,M ′ ,N,N ′ be R–modules and α : M ′ −→ M , β : N ′ −→ N be homomorphisms. Prove that m′ ⊗ n′ 7→ α(m′ ) ⊗ β(n′ ) defines a homomorphism M ′ ⊗R N ′ −→ M ⊗R N . 1.2.39. Prove that M ⊗R N ∼ = N ⊗R M . 1.2.40. Prove that the canonical map φ : HomR (M ⊗R N,P ) −→ HomR (M, HomR (N,P )) defined by φ(λ)(m)(n) := λ(m ⊗ n) is an isomorphism. n

n

i=1

i=1

1.2.41. Let M,N be R–modules and M = ⊕ Mi . Prove that M ⊗R N ∼ = ⊕ (Mi ⊗R N ). 1.2.42. Prove that M ⊗R R ∼ = M. 1.2.43. Let M −→ N −→ L −→ 0 be an exact sequence of R–modules and A be an R–module. Prove that M ⊗R A −→ N ⊗R A −→ L ⊗R A −→ 0 is exact. (Hint: Use Exercises 1.2.37 and 1.2.40.)

1.2.44. Let M be an R–module and S be an R–algebra. Prove that with the multiplication s · m ⊗ s′ = m ⊗ ss′ , M ⊗R S is an S–module. 1.2.45. Prove the universal properties mentioned in Definition 1.2.16. P 1.2.46. Let V and W be vector spaces over K. For ϕ ∈ V ∗ and vi ⊗ wi ∈ V ⊗K W , consider P ϕ(vi )wi ∈ W . Show that we get a well defined map V ∗ −→ W . Show that V ⊗K W ∼ = Hom(V ∗ ,W ). 1.2.47. Prove (1) and (3) of Remark 1.2.19. (Hint: It is enough to consider the case of two matrices A and B, B obtained from A by one column operation, and to prove that It (A) = It (B).) ϕ

ψ

1.2.48. Let M,N be R–modules and 0 −→ M −→ N −→ Rs −→ 0 be an exact sequence. Prove that N ∼ = M ⊕ Rs , more precisely 0

−→

0

−→

M || M

ϕ

−→ −→ i

N ↑ M ⊕ Rs

ψ

−→ −→ π

Rs || Rs

−→

0

−→

0

commutes, i,π the canonical injection resp. projection. In this case one says that M is a direct summand of N .

1.3 Local Rings and Localization

1.3

17

Local Rings and Localization

Definition 1.3.1. A ring R is called local if it has a unique maximal ideal m. One often says that (R,m) is a local ring, to indicate that m is its unique maximal ideal. Later on in this section we will see how to construct local rings from general rings by means of “localization”. In this book, the local rings we are dealing with most frequently are power series rings and their quotients, see Section 3.1. Here we define those rings for the case of one variable. Definition P∞ 1.3.2. The formal power series ring C [[x]] in one variable consists of elements k=0 ak xk , with ak ∈ C . We have the obvious addition and multiplication. The P∞ convergent power series ring C {x} in one variable consists of elements k=0 ak xk , with ak ∈ C , that converge in some open neighborhood of 0 in C . Note that in the formal power series ring as well as in the convergent power series ring, the elements which do not lie in (x), that is, those with nonzero constant term, are units. Therefore, the fact that the power series rings are local rings follows from the following lemma. Lemma 1.3.3. Let R be a ring, and m ⊂ R be an ideal. Then R is local, with maximal ideal m, if and only if R \ m is the set of units of R. Proof. Suppose R is local with maximal ideal m. Take an element a ∈ R. If (a) 6= R, then (a) is contained in a maximal ideal, which by definition of a local ring must be m. Therefore, if a ∈ / m, then (a) = R, that is, a is a unit. Moreover, if a ∈ m, then a is not a unit, as otherwise (a) = R. Suppose, on the other hand, that R \ m is the set of units. Take an ideal I 6= R. Then the ideal I does not contain units, and therefore must be contained in m. This shows that m is the unique maximal ideal of R. Theorem 1.3.4 (Nakayama’s Lemma). Let (R,m) be a local ring and M be a finitely generated R–module with m · M = M . Then M = 0. Proof. By contradiction. Let m1 , . . . ,mt be a minimal number of generators for M . As M = m · M there exist a1 , . . . ,at ∈ m such that mt = a1 m1 + . . . + at mt . Therefore (1 − at )mt = a1 m1 + . . . + at−1 mt−1 . Now 1 − at ∈ / m, and as R is a local ring it follows that 1 − at is a unit. Therefore mt ∈ (m1 , . . . ,mt−1 ), in contradiction to the minimality of t. For more a general version of Nakayama’s Lemma, we refer to Exercise 1.3.17. A corollary of this important lemma is the following corollary. Corollary 1.3.5 (Krull’s Intersection Theorem). Let (R,m) be a Noetherian local ring, and M be a finitely generated R–module. Then ∩k∈N mk M = (0). Proof. We will prove this theorem for the case M = R, the general case is similar, see Exercise 1.3.24. Put I = ∩k∈N mk . Then I is an ideal (check this), finitely generated because R is Noetherian. We will show that m · I = I, so that I = (0) by Nakayama’s Lemma. For this purpose consider the following set of ideals

18

1 Algebra A := {J ideal in R : J ∩ I = mI}.

Our ring R is Noetherian, so that this set has a maximal element which we call J. Obviously mI ⊂ J. Note that it suffices to prove that mk ⊂ J for certain k. Indeed, as obviously I ⊂ mk , it follows I ⊂ mk ⊂ J, hence I ⊂ J ∩ I = mI, and we can apply Nakayama. As m is finitely generated, it suffices to find for each f ∈ m an α with f α ∈ J. This we will do by showing J + (f α ) = J by using the maximality property of J. Thus it remains to show that (J +(f α ))∩I = mI. The inclusion ⊃ is obvious. As R is Noetherian, the chain J : f ⊂ J : f 2 ⊂ · · · stabilizes. Hence there exists an α with J : f α = J : f α+1 , and this is the α we need. Now let x ∈ (J + (f α )) ∩ I, say x = y + af α ∈ I for y ∈ J. It follows that f α+1 a = f x − f y ∈ mI + J = J, as mI ⊂ J. Hence a ∈ J : f α = J : f α+1 . Thus f α a ∈ J, so that x ∈ J follows. As x ∈ I it follows x ∈ J ∩ I = mI. This holds for all x ∈ (J + (f α )) ∩ I so that (J + (f α )) ∩ I ⊂ mI follows. This is what we had to show. Another corollary is the following lemma on the number of generators of a module over a local ring. The proof is left as Exercise 1.3.18. Corollary 1.3.6. Let M be a finitely generated module over a local ring (R,m). (1) Let f1 , . . . ,fs ∈ M such that the classes of the fi generate M/mM as R/m–vector space. Then f1 , . . . ,fs generate M . (2) The minimal number of generators of M is equal to dimR/m (M/mM ). Definition 1.3.7. Let R be a ring, and M be an R–module. A presentation of M is a short exact sequence (1.3)

α

β

G −→ F −→ M −→ 0

of R–modules, in which F and G are free R–modules. In case R is a local ring, and M is finitely generated, a presentation as in 1.3 is called minimal if the rank of F is equal to the minimal number of generators of M , that is, equal to dimR/m (M/mM ), see 1.3.6. This is equivalent to the statement Im(α) ⊂ m · F . ϕ

Proposition 1.3.8. Let R be a local ring and Rn −→ Rm −→ M −→ 0 be a presentation of M . The following conditions are equivalent: (1) M is a free module of rank m − r, (2) Im(ϕ) is a free module of rank r and a direct summand of Rm , (3) Ir (ϕ) = R and Ir+1 (ϕ) = (0). Proof. (3) ⇒ (2) The fact that Ir (ϕ) = R and R is local implies that for a matrix A representing ϕ there is an r–minor which is not in the maximal ideal of R, hence a unit. By renumbering we may assume that this minor is obtained by taking the determinant of the first r rows ad 0 gives and columns. We write A = ( B∗ ∗∗ ) with B an r × r–matrix. Then det1 B B0 Idn−r n an automorphism of R and we multiply A from the right with this matrix. So we may assume that A is of type ( Id∗r ∗∗ ). As in linear algebra, we can now do row and column operations, corresponding
to a change of basis in target and source. We may then assume that A
is of type Id0r 0∗ . As by assumption Ir+1 (ϕ) = 0, it then follows that A is of type Idr 0 . This immediately implies (2). 0 0

1.3 Local Rings and Localization

19

(2) ⇒ (1) is trivial, using Exercise 1.3.26.

(1) ⇒ (3) By Exercise 1.2.48 and Exercise 1.3.26 we obtain that Rm ∼ = M ⊕ Im(ϕ), and Im(ϕ) is free of rank r. Similarly, Rn ∼ = Ker(ϕ) ⊕ Im(ϕ)
and Ker(ϕ) is free of rank n − r. So we may assume that ϕ is given by the matrix Id0r 00 , see Remark 1.2.19. This implies (3). The following theorem is a special case of the classification of finitely generated modules over principal ideal domains. Theorem 1.3.9. Let M be a finitely generated torsion free C {x}–module. Then M is free. As a finitely generated vector space is free, the theorem immediately follows from the following lemma. Lemma 1.3.10. Let (R,m) be a local ring, M a finitely generated R–module, and x ∈ m a nonzero divisor of M . Then M is a free R–module ⇐⇒ M/xM is a free R/(x)–module. Proof. The implication =⇒ is trivial. Suppose, therefore, that M/xM is a free R/(x)module. Take a map α : Rn = F → M , sending each basis element to a generator of M . We might take n to be the minimal number of generators of M . It follows that the map n F/xF → M/xM is also surjective. By 1.3.6 M/mM ∼ = (R/m) . As we supposed that M/xM is a free R/(x)–module the map α induces an isomorphism (R/(x))n ∼ = M/xM . Let K be the kernel of the map α. We obtain a commutative diagram: 0 /

K /

F

/

 F

·x

0 /

 K

M

/

 M

·x

/

/

0

/

0

·x

where ·x denotes multiplication by x. By assumption, x is not a zero divisor of M , so that the kernel of ·x : M → M is zero. It follows from the Snake Lemma that we have an exact sequence: 0 −→ K/xK −→ F/xF −→ M/xM −→ 0. From the remark above, we have that F/xF → M/xM is an isomorphism. It follows that K/xK = 0. From Nakayama, in the version of Exercise 1.3.17 it follows that K = 0 or, equivalently, that M ∼ = F is free. There is a procedure to construct local rings from a given ring, called localization. This is a special case of forming rings (and modules) of fractions, which is itself a generalization of defining the quotient field of an integral domain. In order to define rings of fractions, we have to introduce the concept of a multiplicatively closed set. Definition 1.3.11. Let R be a ring. A set S ⊂ R is called multiplicatively closed if and only if it satisfies the following conditions. • 1 ∈ S, • If f,g ∈ S, then f · g ∈ S.

20

1 Algebra

Definition 1.3.12. Let R be a ring and S be a multiplicatively closed set. We will define the ring of fractions RS , also denoted by S −1 R. As a set it is the following: RS = { sr : r ∈ R, s ∈ S} and satisfies the following properties: •

r1 r2 = ∈ RS s1 s2 if and only if there exists an s ∈ S such that s(r1 s2 − r2 s1 ) = 0.

• Multiplication: • Addition:

r1 r2 r1 r2 · := . s1 s2 s1 s2 r1 r2 r1 s2 + r2 s1 + := . s1 s2 s1 s2

Similarly, for any R–module M , and any multiplicatively closed set S ⊂ R, one can define the RS –module MS , also denoted S −1 M . In particular this can be done for an ideal I ⊂ R. Its definition is straightforward, and is left to the reader. Examples 1.3.13. (1) Let R be an integral domain, and S = R \ {0}. Then RS is the quotient field Q(R) of R. It is indeed a field, as Q(R) = { ab : a,b ∈ R, b 6= 0}. So if ab 6= 0, then a 6= 0, and ab ∈ Q(R). (2) More generally, for a ring R, one can take as multiplicatively closed set S the set of all nonzero divisors. The resulting ring RS is called the total quotient ring, or total ring of fractions of R, notation Q(R). We will prove in Proposition 1.4.27 that for a reduced Noetherian ring Q(R) is a direct sum of fields. (3) Consider the multiplicatively closed set S = {1,f,f 2 ,f 3 , . . .} for f ∈ R. In this case one usually writes Rf instead of RS . (4) Let p be a prime ideal of R. According to the definition of a prime ideal the set R \ p is a multiplicatively closed set. The resulting ring of fractions is denoted by Rp (instead of the “correct” RR\p ), and is called the localization of R in p. So Rp = { ab : a ∈ R, b ∈ / p}. At first it might be surprising that the following lemma holds generally for prime ideals, and not only for maximal ideals. Nevertheless, the proof is easy! Lemma 1.3.14. Let R be a ring and p ⊂ R be a prime ideal. Then Rp is a local ring, / p}. with maximal ideal pRp = { ab : a ∈ p, b ∈ Proof. By 1.3.3, it suffices to show that the elements of pRp are the nonunits of Rp . An element in Rp \ pRp is of type ab , where a,b ∈ / p. But then ab ∈ Rp , so that ab is a unit. Of a course, every element b , with a ∈ p is not a unit in Rp .

1.3 Local Rings and Localization

21

Theorem 1.3.15. Let R be a ring, S a multiplicatively closed set. There is a one-one correspondence: prime ideals p ⊂ R with p ∩ S = ∅ ←→

prime ideals pS ⊂ RS .

The correspondence is given by the following rules. Given a prime ideal pS ⊂ RS , the ideal p := pS ∩ R is a prime ideal in R. Given a prime ideal p in R whose intersection with S is empty, its localization pS is a prime ideal in RS . Proof. Suppose p is a prime ideal in R, with p ∩ S = ∅. Then the inclusion pS $ RS is strict. This is because otherwise 1 = as , with a ∈ p, s ∈ S. Hence there would exist a t ∈ S such that t(s − a) = 0, that is ta = ts. But ts ∈ S, and ta ∈ p, which by assumption cannot be the case. Now suppose as bt ∈ pS . It follows that ab ∈ p, and therefore either a ∈ p or b ∈ p. Therefore either as ∈ pS or bt ∈ pS . The converse is left to the reader.

Exercises 1.3.16. Prove that IRf ∩ R = I : f ∞ (see 1.1.15). 1.3.17 (Nakayama’s Lemma). Let R be a ring, I ⊂ R be an ideal which is contained in all the maximal ideals of R, and M be a finitely generated R–module. (1) Suppose that M = I · M . Prove that M = 0.

(2) Let N be a submodule of M such that M = N + I · M . Prove that M = N . 1.3.18. Prove Corollary 1.3.6. (Hint: Let N = (f1 , . . . ,fs ) and apply Nakayama’s Lemma.) 1.3.19. Let (R,m) be a Noetherian local ring, and J an ideal in R. (1) Suppose mJ = (0). (a) Prove that J is a finite-dimensional R/m–vector space. (b) Prove that there exists an n ∈ N with J ∩ mn = (0).

(2) Show that that there exists an n with J ∩ mn ⊂ mJ. (3) Prove that for such an n we have J/mJ ∼ = J + mn /mJ + mn . 1.3.20. Let K be a field, and R be a local K–algebra, with maximal ideal m. (1) Suppose that dimK R = n < ∞. Show that mn = 0. (Hint: Show that the sequence of ideals m ⊃ m2 ⊃ · · · stabilizes and use Nakayama’s Lemma.) (2) Let I ⊂ R be an ideal such that dimK (R/I) = n < ∞. Show that mn ⊂ I. 1.3.21. Let M be an R–module. Prove that the following conditions are equivalent. (1) M = 0. (2) Mp = 0 for all prime ideals p ⊂ R.

(3) Mm = 0 for all maximal ideals m ⊂ R.

(Hint: For (3) =⇒ (1) show that for all m ∈ M the ideal Ann(m) is not contained in any maximal ideal of R.) 1.3.22. Let R be a ring. (1) Let S ⊂ R a multiplicatively closed subset of R and M −→ N −→ P be an exact sequence of R–modules. Prove that the induced sequence MS −→ NS −→ PS is exact. One phrases this by saying that localization is an exact functor.

22

1 Algebra

(2) Let · · · −→ Mi−1 −→ Mi −→ Mi+1 −→ · · · be a complex. Prove that the complex is exact if and only if for all prime ideals p ⊂ R the induced complex · · · −→ Mi−1p −→ Mip −→ Mi+1p −→ · · · is exact. ∼ MS . (3) Show that RS ⊗R M =

(4) Show that (M ⊗R N )S ∼ = MS ⊗RS NS .

1.3.23. Let R be a ring, and p ⊂ R be a prime ideal. We define the n–th symbolic power p(n) of p by p(n) := pn Rp ∩ R. We obviously have pn ⊂ p(n) .

(1) Let R = k[x,y,z]/(z 2 − xy), and p = (x,z). Show that p is prime, and that p(2) = (z 2 ,x). In particular p2 6= p(2) . (2) Let f,g ∈ R be given such that f ∈ / p and f · g ∈ p(n) . Show that g ∈ p(n) .

1.3.24. Prove Krull’s Intersection Theorem 1.3.5 for modules. 1.3.25. Let R be a ring. The Jacobson radical J(R) of R by definition is the intersection of all maximal ideals of R. Show that x ∈ J(R) implies that 1 + x is a unit. 1.3.26. Let R be a local ring and Rn = Rs ⊕ N . Prove that N is free of rank n − s. (Hint: Use 1.3.6.) This statement does not hold for general (nonlocal) rings. 1.3.27. Let R be a ring and p1 , . . . ,pm be prime ideals. Let (0) 6= M be a finitely generated R–module such that Mp j 6= (0) for all j. Prove that there exists an x ∈ M such that x 6∈ pj Mp j for all j. (Hint: Use the idea of the proof of prime avoidance, see 1.1.13: Use induction on m and assume, therefore, y 6∈ pj Mp j for j = 1, . . . ,m − 1 and a suitable y. Assume the worst case, that is, m−1 y ∈ pm Mp m . You may also assume that ∩m−1 j=1 pj 6⊂ pm and choose r ∈ ∩j=1 pj \ pm . Let ′ ′ M = (y,y2 , . . . ,yn ) and M = (y,ry2 , . . . ,ryn ) then Mp m = Mp m and y + ryi 6∈ pj Mp j for i ≥ 2 and j = 1, . . . ,m − 1. One of these elements is the element we are looking for because, otherwise, y,y2 , . . . ,yn ∈ pm Mp m , that is, Mp m = (0).) 1.3.28. Prove (2) of Remark 1.2.19. (Hint: Localize by prime ideals to reduce the assertion to the case of local rings. Let m be the ϕ1 ϕ2 maximal ideal of R and Rn1 −→ Rm1 −→ M −→ 0, Rn2 −→ Rm2 −→ M −→ 0 be two presentations of M . Show that we may assume that Ker(ϕ1 ) ⊂ mRn1 and Im(ϕ1 ) ⊂ mRm1 . Suppose first that Ker(ϕi ) ⊂ mRni and Im(ϕi ) ⊂ mRmi . Prove that there exist isomorphisms λ,µ such that Rn1

ϕ1

/ Rm1

/M

µ

λ



Rn2

ϕ2



/ Rm2

/M

commutes. Use Remark 1.2.19 (1) for this` case. Now show that for a suitable choice of basis in ´ 0 0 Rn2 and Rm2 the map ϕ2 has a matrix ϕ02 Id ϕ2 : Rn1 −→ Rm1 and Id the identity with 0 matrix.)

1.4

Primary Decomposition

It is well-known that in Z every number is a product of prime numbers. More precisely, let a ∈ Z and a 6= ±1,0, then

1.4 Primary Decomposition

23 a = sign(a)

Y

pν11 · · · pνrr

with pairwise different prime numbers pi . The pi and the νi are uniquely determined (up to permutation). It is easily shown that equivalently (a) = (p1 )ν1 ∩ · · · ∩ (pr )νr holds. This property can be generalized in the following manner: Definition 1.4.1. Let R be an integral domain. (1) An element 0 6= p ∈ R with (p) 6= R is called prime if p | gh implies that p | g or p | h. An element f in R is called irreducible if f = gh implies that g or h is a unit. (2) R is called a unique factorization domain, or a factorial ring, if every element different from zero and not being a unit is a product of finitely many primes. (3) Let R be a unique factorization domain and a1 , . . . ,an ∈ R elements different from zero. An element d ∈ R is called a common divisor of a1 , . . . ,an if d | ai for i = 1, . . . ,n. An element d ∈ R is called a greatest common divisor , d = gcd(a1 , . . . ,an ), if d is a common divisor of a1 , . . . ,an and every common divisor of a1 , . . . ,an divides d. Remarks 1.4.2. (1) It follows directly from the definition that a prime element √ is irreducible. The converse is not true in general. For example, in the ring Z[ √ −5] the element 2 is √ −5)(1 − −5). However, it can irreducible, but not prime as 2 | 6, and 6 = (1 + √ √ be shown that 2 does not divide 1 + −5 and 1 − −5. To give a different example, consider the ring C [x,y,z]/(z 2 − xy). Then we have z | xy, but z does not divide x or y. The converse does hold however in a unique factorization domain. (2) One could, alternatively, define a unique factorization domain as a domain in which each nonzero element which is not a unit has (up to units and renumbering) a unique decomposition into irreducibles. This will be shown in the exercises. (3) The existence of a greatest common divisor will be proved in Exercise 1.4.37. Example 1.4.3. Any principal ideal domain, especially Z and the polynomial ring K[x] over a field K in one variable, is a unique factorization domain. This will be proved in Exercise 1.4.35. Moreover, it follows from the Lemma of Gauß, to be proved next, that the polynomial ring K[x1 , . . . ,xn ] over any field K is a unique factorization domain. Lemma 1.4.4 (Lemma of Gauß). Let R be a unique factorization domain. Then R[x] is a unique factorization domain. One of the main steps in the proof of this Theorem is the following lemma. Lemma 1.4.5. Let R be a unique factorization domain, K = Q(R) its quotient field, and f ∈ R[x] irreducible. Then f is also irreducible considered as element of K[x].

24

1 Algebra

Proof. Suppose the converse. Let f = g · h, g,h ∈ K[x] and deg(h) < deg(f ), deg(g) < deg(f ). We choose c ∈ R such that cg =: g, ch =: h are in R[x] and obtain c2 f = gh. Now c2 is not a unit in R. This is because f is by assumption irreducible in R[x]. So we can write c2 = p1 · · · ps , pi ∈ R prime. Exercise 1.4.39 implies p1 |g or p1 |h. So we can find g ′ and h′ in R[x] with p2 · · · ps f = g ′ h′ . Repeating this argument, shows that f = g0 · h0 , deg(g0 ) = deg(g), deg(h0 ) = deg(h). This is a contradiction to the fact that f is irreducible in R[x]. Proof of the Lemma of Gauß 1.4.4. Let K = Q(R) be the field of fractions ofP R. Let f ∈ m k R[x]. We have to prove that f = f1 · · · fn , fP ∈ R[x] prime. We can write f = i k=0 ak x . m ak k Define d = gcd(a0 , . . . ,am ). Then f = d · k=0 d x . We know, because R is a unique factorization domain, that d is a unit or a product of finitely many primes. Because of Exercise 1.4.39 they are also prime in R[x]. Therefore, we may assume that d = 1. Let f = f1 · · · fn , fi ∈ R[x] be irreducible. Such a decomposition into finitely many P polynomials exists because deg(f ) = deg(fi ). Therefore, we may assume that f is irreducible and we have to prove that f is prime. From Lemma 1.4.5 it follows that f is irreducible in K[x]. From Exercise 1.4.33 it follows that f is prime in K[x]. Now we are ready to prove the theorem. Let f | g · h, g,h ∈ R[x]. Then f | g or f | h in K[x]. Without loss of generality, suppose that f | h. Then it follows that f · b = h for a suitable b ∈ K[x]. Pm Then we choose c ∈ R such that c · b =: b ∈ R[x] and obtain f b = c · h. Now f = k=0 ak xk and gcd(a0 , . . . ,am ) = 1 implies, using 1.4.39 that c | b, that is, b ∈ R[x]. Corollary 1.4.6. Let R be a field or Z. Then R[x1 , . . . ,xn ] is a unique factorization domain. In the rest of the section, we discuss a generalization of this result for an arbitrary Noetherian ring. From now on we fix a Noetherian ring R. First we need to find a substitute of the notion of prime power for a general ring. Definition 1.4.7. An ideal q $ R is called a primary ideal if ab ∈ q, a 6∈ q implies bn ∈ q for some n ∈ N. An ideal p is prime, if R/p is a domain, that is, has no zero divisors. Similarly, an ideal I is primary, if every zerodivisor of R/I is nilpotent. An ideal I is radical , exactly if R/I does not have any nilpotent elements, except for the zero element. This means exactly that R/I is a reduced ring. Lemma 1.4.8. (1) Let m be a maximal ideal of R. Then m is a prime ideal. √ (2) Let q be an ideal in R such that m := q is a maximal ideal. Then q is a primary ideal. √ (3) Let q ⊂ R be a primary ideal. Then q is prime. (4) Let q ⊂ R be a primary ideal and d ∈ / q be an element of R. Then q : d is a primary ideal.

1.4 Primary Decomposition

25

Proof. (1) Let ab ∈ m and a 6∈ m. Then (a) + m = R because of the maximality of m. Let 1 = r · a + m for some r ∈ R, m ∈ m, then b = r · a · b + mb ∈ m because a · b and m · b ∈ m. This implies that m is a prime ideal. (2) Now let a · b ∈ q with a 6∈ q. If bs 6∈ q for all s ∈ N, that is, b 6∈ m, then again (b) + m = R. Thus 1 = rb + m for some r ∈ R and m ∈ m. Hence a = rba + m · a. Iterating this we obtain a = r(1 + m + m2 + · · · + mk )ba + mk+1 a for all k > 0. As √ q = m, we have mk+1 ∈ q for some k (R is Noetherian). This is in contradiction to a∈ / q. This implies b ∈ m and thus bs ∈ q for some s. √ √ (3) Let ab ∈ q and a 6∈ q. Then (ab)m ∈ q for some m but am 6∈ q. This implies that √ m s for some s, (b ) ∈ q and, therefore, b ∈ q. (4) Exercise 1.4.42. Remarks 1.4.9. (1) It is not true that any power of a prime ideal is primary. For example, the ideal p = (x,z) is prime in R := K[x,y,z]/(xy − z 2 ), because R/p ∼ = K[y] does not have zero divisors. However, p2 is not primary. To see this, note that z 2 = xy ∈ p2 , x 6∈ p2 and no power of y is in p2 . (2) It is always true from (f ) is a prime ideal it follows that f is an irreducible element. The converse does not always hold, but it holds if R is a unique factorization domain. Furthermore (f m ) is primary in this case. This applies for example for rings like K[x1 , . . . ,xn ] or Z[x1 , . . . ,xn ] by the Lemma of Gauß. One of the main goals in this section is to show that any ideal in a Noetherian ring is a finite intersection of primary ideals. Before proving this, we need the following lemma. Lemma 1.4.10. Let I ⊂ R be an ideal, and f ∈ R such that I : f = I : f 2 . Then (I : f ) ∩ (I,f ) = I. Proof. The inclusion I ⊂ (I : f ) ∩ (I,f ) is obvious. For the other inclusion, let g ∈ (I : f ) ∩ (I,f ). We have to prove g ∈ I. Let g = x + y · f, x ∈ I, y ∈ R. As g ∈ I : f it follows that f g = xf + y · f 2 ∈ I. Hence y · f 2 ∈ I, so that y ∈ I : f 2 = I : f . It follows that yf ∈ I, and therefore g ∈ I. Theorem 1.4.11 (Primary Decomposition). Let R be a Noetherian ring and I $ R be an ideal. Then there exist finitely many primary ideals q1 , . . . ,qr in R such that I = q1 ∩ . . . ∩ qr . Proof. Suppose the converse. Then the set A of ideals for which the theorem is not true, that is, A := {I ideal in R : I is not an intersection of finitely many primary ideals}, is nonempty. The ring R is Noetherian by assumption, and therefore by 1.1.2 the set A has a maximal element with respect to inclusion. Let I be such a maximal element. In particular, I is not primary! Therefore, there exist a,b ∈ R such that a · b ∈ I, a 6∈ I and bn 6∈ I for all n. We have a chain of ideals I : b ⊂ I : b2 ⊂ · · · . As our ring R is Noetherian, we get by 1.1.2 that this chain is stationary. Hence, there exists an n with the property I : bn = I : bn+1 = I : bn+2 = · · · = I : b2n . We use Lemma 1.4.10 to obtain I = (I : bn ) ∩ (I,bn ).

26

1 Algebra

Now, since bn ∈ / I, we have I $ (I,bn ). Moreover as a ∈ / I and abn ∈ I we have I $ (I : bn ). As I is maximal in A it follows that both (I : bn ),(I,bn ) ∈ / A . Therefore, both ideals are intersections of finitely many primary ideals. But as I = (I : bn )∩(I,bn ), it would follow that I is the intersection of finitely many primary ideals too, in contradiction to our assumption. Examples 1.4.12. (1) (xy,xz) = (x) ∩ (y,z) ⊂ K[x,y,z]; (2) (xy − s,xz − s,yz − s) = (x,y,s) ∩ (x,z,s) ∩ (y,z,s) ∩ (x − y,y − z,x2 − s) ⊂ K[x,y,z,s]. We now turn our attention to the problem of uniqueness of the primary decomposition. Definition 1.4.13. √ • Let q be a primary ideal. The prime ideal p = q is called the associated prime of q. In this case one says that q is a p–primary ideal. • A primary decomposition I = q1 ∩ . . . ∩ qr is called irredundant4 if the following two conditions are satisfied: (1) None of the qi can be removed from the primary decomposition, that is for all i, the inclusion I ⊂ ∩j6=i qj is strict. √ (2) The associated primes pi := qi are pairwise different. Lemma 1.4.14. (1) Let q and q′ ⊂ R be p–primary ideals. Then q ∩ q′ is a p–primary ideal. (2) Let R be Noetherian and I $ R be an ideal in R. Then there exists an irredundant primary decomposition of I. Proof. √ (1) From 1.1.16, parts 2 and 4, we have q ∩ q′ = p. We have to prove that q ∩ q′ is primary. So let ab ∈ q ∩ q′ , and a 6∈ q ∩ q′ . We may assume √ without restriction of √ generality that a 6∈ q. Then it follows that b ∈ q = p = q ∩ q′ , which shows the lemma for q ∩ q′ . (2) This follows from the first part of the lemma, as we can combine different p–primary ideals into one p–primary ideal, for every prime ideal p. It is important to note that an irredundant primary decomposition might not be unique. The simplest example is: Example 1.4.15. (x2 ,xy) = (x) ∩ (x,y)2 = (x) ∩ (x2 ,y). But at least the associated prime ideals are the same, that is, (x,y) and (x), in both cases. This turns out to be the case in general. 4

In the literature this is often also called minimal, normal or reduced. Some authors, for example Eisenbud, [Eisenbud 1995], reserve the notion irredundant primary decomposition for a primary decomposition in which none of the terms can be dropped.

1.4 Primary Decomposition

27

Theorem 1.4.16 (First Uniqueness Theorem of Primary Decomposition). Let R be a Noetherian ring and I = q1 ∩ . . . ∩ qr be an irredundant primary decomposi√ √ tion of I. Then the prime ideals p1 = q1 , . . . ,pr = qr are uniquely determined by I, up to permutation. For the proof, we need the following lemma. Lemma 1.4.17. Let R be a Noetherian ring and q be a primary ideal with associated prime p and I = J ∩q for some ideal J 6⊂ q. Then there exists a d ∈ R such that I : d = p. Proof. Step 1. We first reduce to the case that J = R, that is, I = q. We choose d ∈ J and d 6∈ q. Then it follows that I : d = (J ∩ q) : d = (J : d) ∩ (q : d) = q : d. As q : d is also primary by Exercise 1.4.47, and because of the property (I : x) : y = I : xy (Exercise 1.1.15) we reduced the problem to the case J = R. Step 2. The crucial thing to note is that for any g ∈ p \ q it follows that q $ q : g. Indeed, g ∈ p \ q implies g m ∈ q for some m, and g 6∈ q implies m ≥ 2. If we choose m minimal such that g m ∈ q, then g m−1 ∈ q : g, but g m−1 6∈ q. Note, moreover, that by 1.4.47 q : g is p–primary. In particular q : g ⊂ p. Step 3. The proof of the lemma is by contradiction. So suppose that the set A := {q ⊂ R p–primary : p 6= q : d for all d ∈ R} is nonempty. As R is Noetherian, we can take a maximal element q in A. In particular, p 6= q, so that we can take g ∈ p \ q with q $ q : g. Now q : g is p–primary, and by maximality of q there exists a d with (q : g) : d = p. Hence q : (gd) = p, which is a contradiction. The prime ideals one can get by taking ideal quotients are called the associated primes of I. More precisely: Definition 1.4.18. Let I $ R be an ideal. A prime ideal p of R is called an associated prime of I if it can be written as p=I :d for some d ∈ R. The set of associated primes of I is denoted by Ass(I). Remarks 1.4.19. (1) For a primary ideal q, we have now two definitions of an associated prime of q. We will prove in Theorem 1.4.20 that the two definitions coincide. (2) Alternatively, but very similarly, one can define the set of associated primes by Ass(I) := {p : p prime, p = Ann(b), b ∈ R/I.} The First Uniqueness Theorem 1.4.16 follows directly from the following theorem.

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Theorem 1.4.20. Let I = q1 ∩ · · · ∩ qr be an irredundant primary decomposition of I, √ and pi = qi . Then Ass(I) = {p1 , . . . ,pr }, that is, the set of associated primes is equal to the set of prime ideals which are associated to the primary ideals occurring in an irredundant primary decomposition of I. Proof. We apply 1.4.17 with J = ∩j6=i qj , and q = qi . We conclude that there exist di such that I : di = pi for i = 1, . . . ,r. This gives the inclusion Ass(I) ⊃ {p1 , . . . ,pr }. To show the other inclusion, let p = I : d ∈ Ass(I). Then p = I : d = (q1 : d) ∩ · · · ∩ (qr : d). /I= Let {i1 , . . . ,it } be the set of indices with d ∈ / qij . This set is not empty because d ∈ ∩qi . Then p = (qi1 : d) ∩ · · · ∩ (qit : d). p √ By Exercise 1.4.47, qik : d is pik –primary. In particular qik : d = qik = pik . By Exercise 1.1.16 part 4 it follows that: p p √ √ √ p = p = qi1 : d ∩ · · · ∩ qit : d = qi1 ∩ · · · ∩ qit = pi1 ∩ · · · ∩ pit

As p is a prime ideal it follows that p ⊃ pik for some k. As the other inclusion is clear it follows that p = pik . There is a second uniqueness theorem which we now want to formulate. Definition 1.4.21. (1) Let p ∈ Ass(I). Then we define M (I,p) = {q : q ∈ Ass(I), q ⊂ p}. (2) Suppose that p ∈ Ass(I) with M (I,p) = {p}. Then p is called a minimal associated prime of I. This means that there does not exist an associated prime q of I with q $ p. (3) If p,q ∈ Ass(I), p $ q, then q is called an embedded prime ideal of I. This exactly means that M (I,q) 6= {q}. Example 1.4.22. Let
I = (x2 ,xy) ⊂ K[x,y]. Then I : x = (x,y), I : y = (x). Ass(I) = {(x),(x,y)}, M I,(x,y) = Ass(I). The ideal (x,y) is an embedded prime ideal.

Theorem 1.4.23 (Second Uniqueness Theorem of Primary Decomposition). Let R be a Noetherian ring and I ⊂ R be an ideal, I = q1 ∩ · · · ∩ qr an irredundant primary decomposition of I. Let p be an associated prime ideal of I, and put √ √ M (I,p) = { qi1 , . . . , qis }.

Then qi1 ∩ · · · ∩ qis is uniquely determined (that is, independent of the irredundant decomposition). In fact: qi1 ∩ · · · ∩ qis = Ip ∩ R. In particular, the primary ideals belonging to minimal associated primes are uniquely determined. For such a minimal prime ideal pi one, therefore, has qi = Ipi ∩ R.

1.4 Primary Decomposition

29

Proof. There is the following obvious generalization of 1.3.15: primary ideals q ⊂ R with q ∩ S = ∅

←→

primary ideals qS ⊂ RS .

The correspondence is that to a primary ideal q we assign qS . The ideal qS is primary if q ∩ S = ∅, it is equal to RS if q ∩ S 6= ∅. On the other hand, for a primary ideal qS in RS , we get a primary ideal q := qS ∩ R in R. For a proof, see Exercise 1.4.49. Using Exercise 1.4.50 we get Ip = qi1 p ∩ . . . ∩ qis p and Ip ∩ R = qi1 ∩ · · · ∩ qis . This is what we had to show.
Theorem 1.4.24. Let R be a Noetherian ring and Ass (0) = {p1 , . . . ,pk }. Then (1)

p (0) = p1 ∩ · · · ∩ pk . In particular this means that the intersection of the minimal prime ideals is the set of nilpotent elements.

(2) Let p ⊂ R be a prime ideal. Then pi ⊂ p for some i. k

(3) ∪ pi \ {0} is the set of zerodivisors of R. i=1

(4) If R is reduced, then the set of zerodivisors of R is equal to the union of the minimal prime ideals of R.5 p Proof. Let (0) = q1 ∩ · · · ∩ qk be an irredundant primary decomposition. Then (0) = pp1 ∩· · ·∩pk follows, which is (1). For (2), as p is a prime ideal, it follows that p1 ∩· · ·∩pk = (0) ⊂ p, so (2) follows. To prove (3), let x be a zero divisor, that is, there is a y ∈ R, y 6= 0 and xy = 0. Let √ (0) = q1 ∩ · · · ∩ qk be an irredundant primary decomposition and qi = pi . Since y 6= 0 there exists an i such that y 6∈ qi . This implies x ∈ pi . To prove the other direction, let x ∈ pi for some i. If x is nilpotent, we are done. Otherwise xm ∈ qi for some m and xm 6= 0. In particular, k ≥ 2. But qi ·(q1 ∩· · ·∩qi−1 ∩qi+1 ∩· · · ∩qk ) ⊂ q1 ∩· · ·∩qk = (0) implies that there exists a y 6= 0 such that xm y = 0. Let n ≥ 1 be minimal such that xn y = 0. Then x · (xn−1 y) = 0 and, consequently, x is a zero divisor. The final statement follows from the third. As an application, we first want to study the structure of Artinian K–algebras (cf. 1.1.6). Lemma 1.4.25. Let R be an Artinian K–algebra and p ⊂ R be a prime ideal. Then p is a maximal ideal.6 Proof. R/p is an Artinian K–algebra, see 1.1.9, and an integral domain. We have to prove that R/p is a field. Let x ∈ R/p, x 6= 0, then Lemma 1.1.8 implies that (xn ) = (xn+1 ) for some n. Hence, xn = xn+1 y for some y ∈ R/p. Since R/p is an integral domain, this implies 1 = xy. So x is invertible, and R/p is a field. Theorem 1.4.26. Let R be an Artinian K–algebra. Then R is uniquely (up to isomorphism) a finite direct sum of local Artinian K–algebras. 5 6

Note that minimal prime ideals of R are the minimal associated prime ideals of (0). This is a very special property, and says that an Artinian K–algebra has Krull dimension zero. For a discussion on these matters, see Section 4.1.

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Proof. Step 1. We first prove the existence. Let (0) = q1 ∩ · · · ∩ qn be an irredundant primary √ decomposition of the zero ideal. Then all the associated primes mi := qi are maximal (1.4.25). By the Second Uniqueness Theorem 1.4.23 the qi are uniquely determined. We claim that furthermore qi + qj = R if i 6= j. Indeed, otherwise m ⊃ qi + qj for some √ √ maximal ideal m. But then m ⊃ qi = mi and m ⊃ qj = mj . This implies m = mi = mj and, therefore, i = j (the decomposition was irredundant). Therefore, the conditions to apply the Chinese Remainder Theorem 1.1.12 are fulfilled and the canonical map n M R/qi R −→ i=1

is an isomorphism. Because of 1.1.9 the R/qi are Artinian K–algebras. They are local because qi 6= qj for i 6= j, again because the primary decomposition was irredundant.

Step 2. We now prove the uniqueness of the decomposition. So consider R ∼ = ⊕m i=1 Ri with Ri local. In step 1 we proved that an irredundant primary decomposition of the zero ideal in an Artinian K–algebra is uniquely determined (up to permutation). We will show that if qi := Ker(R −→ Ri ), then q1 ∩ . . . ∩ qm is an irredundant primary decomposition of the zero ideal in R. This suffices to show the uniqueness, as Ri ∼ = R/qi . (1) Consider the zero ideal in Ri . As Ri has only one maximal ideal, there is only one primary ideal in an irredundant primary decomposition. This ideal has to be (0), hence (0) is a primary ideal in Ri . As Ri = R/qi , this shows that the ideals qi are primary. ∼ m (2) ∩m i=1 qi = (0) because R = ⊕i=1 Ri . Thus q1 ∩ · · · ∩ qm = (0) is a primary decomposition. (3) To prove that the primary decomposition is irredundant, it suffices to show that √ √ qi 6= qj for i 6= j. For this it suffices to show that qi + qj = R for i 6= j. Now, because R ∼ = ⊕m i=1 Ri we find x ∈ R corresponding to (0, . . . ,0,1,0, . . . ,0). So x − 1 ∈ qi and x ∈ qj . This implies 1 = x + (1 − x) ∈ qi + qj . Next we study the structure of quotient rings, as promised in the Examples 1.3.13.
Proposition 1.4.27. Let R = Q(R) be a reduced Noetherian ring and Ass (0) = {p1 , . . . ,ps }. Then R ∼ = ⊕si=1 R/pi . Proof. Note that p1 , . . . ,ps are minimal, because R is reduced. By prime avoidance, Lemma 1.1.13, pi + pj 6⊂ ∪sℓ=1 pℓ if i 6= j. By Theorem 1.4.24 there is a nonzerodivisor in pi + pj . As R = Q(R), all nonzerodivisors are units, so that pi +pj = R. As R is, moreover, reduced, we have (0) = p1 ∩· · ·∩ps by 1.4.24. The conditions of the Chinese Remainder Theorem (1.1.12) are satisfied and we get R ∼ = ⊕si=1 R/pi . Finally, we wish to study the rank of a module. Definition 1.4.28. Let R be a Noetherian ring and M be a finitely generated R–module. One says that M has rank r if M ⊗R Q(R) is a free Q(R)–module of rank r. We write rank(M ) = r.

1.4 Primary Decomposition

31

Lemma 1.4.29. Let R be a Noetherian ring and M be a finitely generated R–module.
M has rank r if and only if Mp is a free Rp –module of rank r for all p ∈ Ass (0) .

Proof. Since Rp = Q(R)pQ(R) , see Exercise 1.4.51, we may assume that R = Q(R). One implication is clear. Assume that Mp is free of rank r for all prime ideals p. We use induction on r. The case r = 0 is trivial. Suppose r > 0. Now we choose x ∈ M such that
x 6∈ pMp for all prime ideals p in Ass (0) . Such an x exists because of Exercise 1.3.27. Now x is an element of a minimal system of generators of Mp for all p, by Nakayama, that is, an element of a basis of the free module Mp for all p. This implies that (M/xR)p is free of rank r − 1 for all p. Using induction, we obtain that M/xR is free of rank r − 1 and, therefore, by Exercise 1.2.48, M ∼ = xR ⊕ M/xR is free of rank r because xR ∼ =R ∼ (xRp = Rp for all p). Remark 1.4.30. If R is an integral domain, then every finitely generated R–module M has a rank, namely rank(M ) = dimQ(R) M ⊗R Q(R). In general this is not true: let R = C [x,y]/(xy) and M = (x). Then M(x) = (0) and M(y) = R(y) . i

π

Proposition 1.4.31. Let R be a Noetherian ring and 0 −→ M −→ N −→ P −→ 0 be an exact sequence of R–modules. If two of the modules M,N,P have a rank then the third one has a rank and rank(M )+ rank(P ) = rank(N ). Proof. Because of 1.4.29, we may assume that R is local, and that m ∈ Ass((0)). Because of Exercise 1.2.48 we may assume that N is always free. If N and P are free, we can use 1.3.26 to deduce that M is free. Now let N and M be free. Suppose M ⊂ mN , then M = (0). Indeed, as m ∈ Ass((0)), there exists a d ∈ R with m = (0) : d. It follows that dM = dmN = (0). As M is free, this implies M = (0). If M 6⊂ mN we use induction on the rank of M . If the rank of M is 1 and m is a free generator then i(m) 6∈ mN implies that i(m) can be extended to a set of free generators of N . This implies that N ∼ = M ⊕ P and P is free because of 1.3.26. If the rank of M is bigger than 1 we can, for the same reason, write M = M ′ ⊕ R, N = N ′ ⊕ R such that i(M ′ ) ⊂ N ′ and the induced sequence 0 −→ M ′ −→ N ′ −→ P −→ 0 is still exact. Then P is free by induction. Proposition 1.4.32. Let R be a Noetherian ring and F, G be free R–modules of finite rank. Let ϕ : F −→ G be a homomorphism. Then the following conditions are equivalent: (1) Im(ϕ) has rank r, (2) Ir (ϕ) contains a nonzerodivisor and Ir+1 (ϕ) = 0. Proof. (1) ⇒ (2).
Assume that all elements of Ir (ϕ) are zerodivisors. Then Ir (ϕ) ⊂ p for some p ∈ Ass (0) by prime avoidance. But Im(ϕ) ⊗R Q(R) is free of rank r, especially Im(ϕ) ⊗R Rp is free of rank r. This implies that Ir (ϕ)Rp = Rp by Proposition1.3.8, and is a contradiction to Ir (ϕ) ⊂ p. Now assume Ir+1 (ϕ) 6= 0. As Im(ϕ) has rank r, Ir+1 (ϕ) does not contain a nonzerodivisor. (Otherwise this would imply Ir+1 (ϕ)Q(R) = Q(R) and by Exercise 1.3.8 give that
Im(ϕ) has rank at least r + 1.) By prime avoidance and 1.4.24 there exists a p ∈ Ass (0) with Ir+1 (ϕ)Rp ⊂ pRp . But we just proved that Ir (ϕ)Rp = Rp and, therefore, for a suitable choice
of bases in F ⊗R Rp and G ⊗R Rp the matrix corresponding to ϕ : Fp −→ Gp is Id0r H0 with H 6= 0 because Ir+1 (ϕ) 6= 0. Now we can choose x ∈ p \ {0} such that xp = (0), then xH = 0. This implies that Im(ϕ)⊗R Rp has torsion elements, and therefore is not free which is a contradiction.

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For (2) ⇒ (1), the existence of a nonzerodivisor, implies Ir (ϕ)Q(R) = Q(R). Together with Proposition 1.3.8 we get that Im(ϕ) has rank r.

Exercises 1.4.33. Let R be a principal ideal domain. Prove that f ∈ R is prime if and only if it is irreducible. (Hint: Irreducible elements in a principal ideal domain generate maximal ideals. Maximal ideals are prime ideals.) 1.4.34. Prove that y 2 − x2 (x + 1) ∈ C [x,y] is irreducible. 1.4.35. Use Exercise 1.4.33 to prove that principal ideal domains are unique factorization domains. ( Hint: Use the property that a principal ideal domain is Noetherian to prove that every f = 6 0 not being a unit is a finite product of irreducible elements.) 1.4.36. Let R be a unique factorization domain. Prove that R has the unique factorization n m Q Q property: let p1 , . . . ,pn , q1 , . . . ,qm be primes in R and pi = qi . Prove:

i=1

i=1

(1) n = m. (2) There exist a permutation π ∈ Sn and units u1 , . . . ,un ∈ R such that pi = ui qπ(i) . 1.4.37. Let R be a unique factorization domain and a1 , . . . ,an ∈ R be elements different from zero. Prove that a1 , . . . ,an have a greatest common divisor. 1.4.38. Recall the Euclidean algorithm and prove that it computes in Z and K[x] the greatest common divisor of two elements. 1.4.39. Let R be a unique factorization domain and p ∈ R prime. Let f,g ∈ R[x] be two polynomials and assume p|f · g. Prove that p|f or p|g, that is, p is also prime in R[x]. 1.4.40. Let a,b ∈ Z. Prove that (a) · (b) = (a) ∩ (b) iff gcd(a,b) = 1. 1.4.41. Let R be a unique factorization domain and f ∈ R be an irreducible element. Prove that (f ) is a prime ideal and (f )m is primary for m ≥ 1. 1.4.42. Prove (4) of Lemma 1.4.8. 1.4.43. Let I ⊂ R be an ideal and q ⊃ I be a prime (respectively primary) ideal. Prove that q = {f : f ∈ q} ⊂ R/I is prime (respectively primary). 1.4.44. Prove that (xi1 , . . . ,xik ) ⊂ K[x1 , . . . ,xn ] is a prime ideal. 1.4.45. Let q be a prime (respectively primary) ideal in a Noetherian ring R. Let I · J ⊂ q and I 6⊂ q for two ideals I,J ⊂ R. Prove that J (respectively J s for some s) is contained in q. 1.4.46. Let I = (x2 yz 3 ,xyw,zw2 ) ⊂ K[x,y,z,w]. Compute the associated primes of I. 1.4.47. Let q be a primary ideal and p its associated prime ideal. Let x ∈ R, then 8 > :primary with associated prime p if x 6∈ q.

1.4.48. Prove that p p p (I,f · g) = (I,f ) ∩ (I,g). (1)

1.4 Primary Decomposition (2)

√

I=

p

IRf ∩ R ∩

p

33

(I,f ).

1.4.49. Formulate and prove a generalization of 1.3.15 to the case of primary ideals. 1.4.50. Let R be a Noetherian ring, p ⊂ R be a prime ideal. (1) Let I,J be ideals in R. Prove that (I ∩ J)Rp = IRp ∩ JRp . (2) Let q be a primary ideal. Show that (a) qRp = Rp ⇐⇒ q 6⊂ p. (b) Suppose that q ⊂ p. Then qRp is primary, and qRp ∩ R = q.

(3) Let q,q′ ⊂ p be two primary ideals with different associated primes. Prove that the associated primes of qRp and q′ Rp are different. (4) Let I = q1 ∩ . . . ∩ qs ∩ qs+1 ∩ . . . ∩ qt be an irredundant primary decomposition of I. Suppose q1 , . . . ,qs ⊂ p, and qs+1 , . . . ,qt 6⊂ p Show that (q1 ∩ . . . ∩ qs−1 )Rp 6⊂ qs Rp . Deduce that IRp = q1 Rp ∩ . . . ∩ qs Rp is an irredundant primary decomposition of IRp . (5) With the assumptions of (4), let q ∈ Ass(I) and q ⊂ p be a prime ideal. Suppose that IRp is prime and prove IRp = qRp . ∼ Q(R)pQ(R) . 1.4.51. Let R be a Noetherian ring and p ∈ Ass((0)). Prove that Rp = 1.4.52. Let R be a√Noetherian ring and M be a finitely generated R–module, N ⊂ M a submodule. Define M N = {x ∈ R : xq M ⊂ N for some q}, N : M = {x ∈ R : xM ⊂ N } and Ann(M/N ) = {x ∈ R : xM/N = 0}. An element x ∈ R is called a zerodivisor in M/N if there exists a nonzero m ∈ M/N such that xm = 0. N ⊂ M is primary in M if N 6= M and every zerodivisor in M/N is nilpotent, that is, x being a zerodivisor in M/N implies that xρ ∈ Ann(M/N ) for some ρ. Prove the following statements. p √ (1) M N = Ann(M/N ). √ (2) N being primary in M implies that N : M is a primary ideal (call N : M the associated prime ideal to N ). (3) N has an irredundant primary decomposition and the associated prime ideals are uniquely determined. (4) If p is an associated prime to N , then p = N : (m) for some m ∈ M . (5) Let p1 , . . . ,ps be the set of associated primes of N , then the zerodivisors of M/N are s

∪ pi \ {0}.

i=1

1.4.53. (a) In 1.4.24 we proved that the set of nilpotent elements in a Noetherian ring is the intersection of all (minimal) prime ideals in the ring. Prove that this holds in fact for all rings (Noetherian or not) by an application of Zorn’s Lemma. √ (b) Let R be a ring and I ⊂ R be an ideal. Prove that I = ∩ p⊃I p. p prime

1.4.54. Let (R,m) be a Noetherian local ring and M be an R–module. Suppose that for all prime ideals p 6= m the localization Mp is the zero module. Show that there exists a k such the mk · M = 0. (Hint: Look at a primary decomposition of Ann(M ).) 1.4.55. Let (R,m) be a Noetherian local ring. Suppose that all x ∈ m \ m2 are zerodivisors. Show that all x ∈ m are zerodivisors. (Hint: Use 1.4.24.)

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1.5

Finite and Integral Extensions

Finite and integral extensions of rings are the ring counterparts of finite and algebraic extensions of fields, which you (hopefully) know from a course in Galois theory. First we give the relevant definitions. Definition 1.5.1. Let R ⊂ S be rings.

• R ⊂ S is called a finite extension if S, considered as an R–module, is finitely generated. • An element α ∈ S is called integral over R if and only if α satisfies an equation: αn + a1 αn−1 + . . . + an−1 α + an = 0

with all ai ∈ R. Such an equation is called an integral equation for α. The extension R ⊂ S is called integral if every element x ∈ S is integral over R. • On the other hand, R is called integrally closed in S, if every element in S which is integral over R already belongs to R. • R is called integrally closed if R is integrally closed in Q(R), the total quotient ring of R. • R is called normal if R is reduced and integrally closed.

We obtain the first lemma by copying the proofs of the corresponding statements in field theory. As usual we denote by R[α] ⊂ S the smallest ring in S containing R and α. Lemma 1.5.2. Let R ⊂ S and α ∈ S.

(1) Let R ⊂ S ⊂ T be ring extensions, with R ⊂ S and S ⊂ T finite. Then R ⊂ T is finite. (2) α is integral over R ⇐⇒ R[α] is a finitely generated R–module.

In particular, every finite extension R ⊂ S is an integral extension.

(3) Let R ⊂ S ⊂ T be ring extensions, with R ⊂ S and S ⊂ T integral. Then R ⊂ T is an integral extension. (4) Let S be a finitely generated R–algebra. Then S is finitely generated as an R–module if and only if every element of S is integral over R. Proof. (1) If u1 , . . . ,us are generators of S as R–module and v1 , . . . ,vt are generators of T as S–module, then one proves directly that the products ui vj generate T as an R−module. (2) Let α be integral, with integral equation p(α) = 0, for some monic7 polynomial p ∈ R[X] of degree n, and g(α) ∈ R[α] for some polynomial g ∈ R[X]. Then we can perform division with remainder to write g = qp + r, where deg(r) < n (cf. Theorem 2.1.5). Plugging in α, we see that g(α) = r(α). This shows that the elements 1,α, . . . ,αn−1 generate R[α] as R−module. On the other hand, if we have finitely many generators q1 , . . . ,qt of R[α] as R−module, and if we put n − 1 to be maximum of the degrees with respect to α occurring in the qi , it follows that the element αn is an R−linear combination of the qi . This gives an integral equation for α. 7

This means that the leading coefficient of p is 1.

1.5 Finite and Integral Extensions

35

(3) We give here a proof for the case that R is Noetherian. For the general case, see Exercise 1.5.28. Let α ∈ T . By hypothesis, it satisfies an integral equation: αn + a1 αn−1 + . . . + an−1 α + an = 0 with all ai ∈ S. So in fact α is integral over R[a1 , . . . ,an ]. Therefore R[a1 , . . . ,an ][α] is a finitely generated R[a1 , . . . ,an ]–module, by part (2). As all elements ai are integral over R by hypothesis, it follows from the second part and induction that R[a1 , . . . ,an ] is a finitely generated R–module. Therefore, R[α] is a submodule of R[a1 , . . . ,an ,α] which is also a finitely generated R–module by the first part. As R is Noetherian, it follows that R[α] is finitely generated. Now use (2) again, to conclude that α is integral. (4) This follows from part (2). Definition 1.5.3. The integral closure of R in S is the set of all elements in S which are integral over R. In the case that S is the total quotient ring of R, and R is reduced, e we simply call it the normalization of R. We denote the normalization by R.

To use the standard proof from field theory that sum, difference and products of integral elements are integral again we need the assumption that R is Noetherian. The statement holds without this condition however, see Exercise 1.5.29. Lemma 1.5.4. Let R ⊂ S be rings. Then the integral closure of R in S is a ring. The integral closure of R in S is integrally closed in S. In particular, the normalization of a ring is normal. Proof. We give the proof here for the case that R is Noetherian. We have to show that if α and β are integral, then α − β and αβ are integral. But those elements are in R[α,β], which is a finitely generated R–module by 1.5.2. As we supposed R to be Noetherian, it follows that R[αβ] and R[α − β] are finitely generated R−modules. By the second part of 1.5.2 it follows that α − β and αβ are integral. The other statement follows from the third part of 1.5.2. To give examples of normal rings, we prove the following theorem: Theorem 1.5.5. Let R be a unique factorization domain. Then R is normal. Proof. Suppose equation:

r s

∈ Q(R) is in lowest terms, and is integral over R. So
r n s

+ a1


r n−1 s

r s

satisfies an

+ . . . + an = 0

with ai ∈ R. We multiply the equation with sn and conclude: rn = −s(a1 rn−1 + . . . + an sn−1 ) Hence s divides rn , and thus s divides r. As we assumed that they have no common factor, it follows that s is a unit in R, and so rs ∈ R. In particular, polynomial rings over fields are normal! The converse of the theorem does not hold, see Exercise 1.5.33. Examples 1.5.6.

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(1) Consider the ring R := C [x,y]/(y 2 −x3 ). This ring is not normal. Indeed the element 2 3 t := xy satisfies t2 = xy 2 = xx2 = x. Therefore, t is integral, as it satisfies the integral e is in fact C [x,y,t]/(y 2 −x3 ,x−t2 ,y−t3 ) ∼ equation X 2 −x = 0. The normalization R = C [t]. To show that t ∈ / R, consider the map ϕ : C [x,y] −→ C [t], which sends x to t2 and y to t3 . Obviously y 2 − x3 is in the kernel of ϕ, and in fact the kernel is generated by y 2 − x3 . Indeed, let g ∈ Ker(ϕ). Then write g = q(y 2 − x3 ) + r, where r = r0 + r1 y, ri ∈ C [x]. Then g(t2 ,t3 ) = r0 (t2 ) + r1 (t2 ) · t3 = 0. The first term only has terms in even degree in t, whereas the second has only odd terms in t. From this it follows that r1 = r2 = 0. Hence C [x,y]/(y 2 − x3 ) = C [t2 ,t3 ], and t = xy is not an element of R. (2) Consider the ring R := C [x,y]/(x · y). This ring is not normal. Namely, x + y is not x . a zerodivisor in R because x + y 6∈ (x) ∪ (y) (Theorem 1.4.24). Consider u = x+y 2 2 3 2 2 2 Then u = u because x (x + y) = x = x(x + y ) = x(x + y) in R. But it can easily be checked that u ∈ / R. (3) Consider the ring R := C [x]/(x2 ). Then R = Q(R) and therefore R is integrally closed. However, R is not normal, as R is not reduced. In the second example we used quite essentially that R is not an integral domain. In fact, the following theorem holds. Theorem 1.5.7. Let R be a Noetherian normal local ring. Then R is an integral domain. Proof. R is normal and, therefore, reduced. Let (0) = p1 ∩· · ·∩pr be an irredundant prime decomposition. We have to prove that r = 1. Assuming r ≥ 2, we can find g ∈ p2 ∩· · ·∩pr , g 6∈ p1 (the decomposition is irredundant) and f ∈ p1 , f 6∈ p2 ∪ · · · ∪ pr (Prime avoidance 1.1.13). By the choice of f and g we have that f + g 6∈ pi for i = 1, . . . ,r. This implies that f + g is not a zerodivisor (Theorem 1.4.24). On the other hand, f · g = 0 because f , then u2 = u because f · g ∈ p1 (p2 ∩ · · · ∩ pr ) ⊂ p1 ∩ · · · ∩ pr = (0). Consider u := f +g 2 3 2 2 2 f (f + g) = f = f (f + g ) = f (f + g) . This implies u ∈ R because R is normal. Note f that u 6= 1, as otherwise f +g = 1, and therefore g = 0. Now u(1 − u) = 0, hence u is a zerodivisor in R. The element u is therefore contained in the maximal ideal of R. Hence 1 − u is a unit because R is local. This implies u = 0 and, therefore, f = 0 which is a contradiction to the choice of f . The assumption on R to be local in the theorem can be weakened to the assumption that R has no idempotents different from 1.8 The last part of the proof just showed that a local ring does not contain idempotents different from 1. For more subtle arguments concerning integral elements given later, we need the following important theorem: Theorem 1.5.8 (Cayley-Hamilton). Let R ⊂ S, α ∈ S. Let M be an R[α]-module, which is finitely generated as an R−module. Suppose moreover that M is a faithful9 R[α]–module. Then α is integral over R. 8 9

An element x in a ring is called idempotent if x2 = x. An R–module M is called faithful if given a ∈ R such that a · m = 0 for all m ∈ M it follows that a = 0.

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37

Proof. Let u1 , . . . ,ut be generators of M as an R−module. Then for each i the element αui is defined and is an P element of M . As the ui ′ s generate M there exist elements t ϕij ∈ R, such that αui = j=1 ϕij uj . In matrix notation:     αId − (ϕij ) 

 u1 ..  = 0, Id the unit matrix. .  ut

We multiply on
the left with the adjoint (see 1.2.9) of αId − (ϕij ) to conclude that det(αId − ϕij ) uk = 0 for k = 1, . . . ,t. This is Cramer’s rule, see 1.2.8. As we assumed
M to be faithful, it follows that det αId − (ϕij ) = 0. Expand the determinant to find an integral equation for α. Remarks 1.5.9. (1) Note that the ring S is “dummy” in some sense. We just need that our element lives in some ring. To see why this theorem is called the Cayley-Hamilton Theorem, we refer to Exercise 1.5.30. (2) If moreover α sends M into I · M for some ideal I, then α satisfies an integral equation of type αn + a1 αn−1 + · · · + an = 0, aj ∈ I. This motivates the following definition. Definition 1.5.10. Let R ⊂ S be a ring extension, and I ⊂ R be an ideal. An element x is called integral over I if it satisfies an equation of type10 xn + a1 xn−1 + . . . + an = 0, aj ∈ I. Lemma 1.5.11. Let R ⊂ S be rings, I ⊂ R be an ideal, and x ∈ S. The following conditions are equivalent. (1) x is integral over I. (2) The ring S ′ := R[x] is a finitely generated R–module, and x ∈

√ I · S ′.

(3) R[x] is contained in a subring S ′ of S, where S ′ is a finitely generated R–module, √ ′ and x ∈ I · S . Proof. Of course, (2) =⇒ (3) is trivial. The equivalence (1) ⇐⇒ (2) can be proved as in 1.5.2. The implication (3) =⇒ (1) uses the Cayley-Hamilton Theorem. Indeed, take an m ∈ N such that xm ∈ I · S ′ . Therefore, multiplication with xm sends S ′ to I · S ′ . It follows from the proof of the Cayley-Hamilton Theorem that xm satisfies an integral equation with coefficients in I (S ′ is a faithful R[xm ]–module because 1 ∈ S ′ ). Corollary √ 1.5.12. Let R ⊂ S be rings, and I ⊂ R. Let R be the integral closure of R in S. Then IR is the set of all elements of S which are integral over I. 10

There is a related notion of integrally dependence on I by using the condition aj ∈ I j . This notion will not be used in this book.

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Proof. If x p ∈ S and is√integral over I, then by√the second part of the previous lemma we get x ∈ IR[x] ⊂ IR. Conversely, if x ∈ IR then xm ∈ IR for certain m. Hence xm ∈ I · R[x1 , . . . ,xn ] for certain xi ∈ R, so that from (3) =⇒ (1) of the previous lemma it follows that x is integral over I. Next we will give a criterion for normality. Theorem 1.5.13 (Criterion for Normality). Let R be a Noetherian reduced ring. Let √ I = I ⊂ R be a radical ideal with the following properties: (1) I contains a nonzerodivisor of R. (2) Let p be a prime ideal of R such that Rp is not normal. Then p ⊃ I. Then R is normal if and only if the canonical inclusion R ⊂ HomR (I,I) is an equality. To prove the theorem we need the following lemma. e its normalization. Let I ⊂ R be Lemma 1.5.14. Let R be a Noetherian reduced ring, R an ideal containing a nonzerodivisor of R. Then e (1) R ⊂ HomR (I,I) ⊂ R. √ e ∩ HomR (I,R). (2) If, moreover, I = I, that is, I is radical, then HomR (I,I) = R

Proof. (1) The canonical map R −→ HomR (I,I) is defined by the multiplication: to y ∈ R we associate the map ϕy : I −→ I defined by ϕy (x) = yx. The injectivity of the canonical map follows from the assumption that I contains a nonzerodivisor. Now let a ∈ I be a nonzerodivisor and ϕ ∈ HomR (I,I). Then for every b ∈ I we have ϕ(b) = b · ϕ(a) because aϕ(b) = ϕ(ab) = bϕ(a). Especially, ϕ(a) ∈ Q(R) is a a independent of the choice of the nonzerodivisor. The map ϕ 7→ ϕ(a) gives the inclusion a HomR (I,I) ⊂ Q(R). In particular, we obtain an identification HomR (I,I) = {x ∈ Q(R) : xI ⊂ I}. e of R. Let x ∈ Q(R) We claim that this module is already contained in the normalization R and xI ⊂ I then I can be considered as an R[x]–module. As an ideal containing a nonzerodivisor it is a faithful R[x]–module and we can apply the Cayley-Hamilton Theorem 1.5.8 to obtain that x is integral over R. This implies that e HomR (I,I) ⊂ R.

e Moreover, the inclusion HomR (I,I) ⊂ HomR (I,R) (2) We just proved that HomR (I,I) ⊂ R. e is obvious. So we get HomR (I,I) ⊂ R ∩ HomR (I,R). As above, note that HomR (I,R) = {x ∈ Q(R) : xI ⊂ R}. e ∩ HomR (I,R). We have to prove that To prove the converse inclusion, let z ∈ R n n−1 zI ⊂ I. Let z + an−1 z + · · · + a0 = 0 for suitable a0 , . . . ,an−1 ∈ R. Let f ∈ I, then by multiplying the above equation with f n we obtain (zf )n + an−1 f (zf )n−1 + · · · + a0 f n = 0. √ This implies (zf )n ∈ I because zf ∈ R and f ∈ I. But I = I and, therefore, zf ∈ I. This implies zI ⊂ I and proves the second part of the lemma.

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e Then Lemma 1.5.14 (1) Proof of Theorem 1.5.13. Let R be normal, that is, R = R. implies that R = HomR (I,I). To prove the converse, assume that R = HomR (I,I). Suppose that the theorem is e Let h ∈ R e \ R. Consider the ideal wrong, that is, R 6= R.

J := {u ∈ R : hu ∈ R} √ √ in R. We claim that I ⊂ J. As J = ∩p⊃J p is the intersection of all prime ideals containing J, by Exercise 1.4.53, it suffices to show that p ⊃ I for all p ⊃ J. So let p be a prime ideal with p ⊃ J. Now h ∈ / Rp . Indeed, otherwise ah ∈ R for some a ∈ / p. But then also a ∈ / J, in contradiction to the definition of J. So h gives an element of Q(Rp ) which is integral over Rp (use the same integral equation) and, therefore, Rp is not normal. This implies p ⊃ I by the second assumption, and proves the claim. As our ring is Noetherian, we can now choose d ∈ N such that I d ⊂ J. This implies d hI ⊂ hJ ⊂ R. As h ∈ / R, we have d > 0. Let d be minimal with hI d ⊂ R. Thus d−1 e and hxI ⊂ hI d ⊂ R implies there exists an x ∈ I such that hx 6∈ R. But hx ∈ R e hx ∈ R ∩ HomR (I,R) = HomR (I,I) = R. This is a contradiction to the choice of x. Therefore, our original assumption h ∈ / R is wrong. This proves the theorem.

Remark 1.5.15. We will see later that for rings of finite type over a field (cf. Chapter 2) or analytic algebras (cf. Chapter 3) the ideal I describing the singular locus (which can be computed using the Jacobian Criterion, cf. Chapter 4) has the properties required in Theorem 1.5.13. Example 1.5.16 (Continuation of Example 1.5.6). (1) Let R = C [t2 ,t3 ], I = (t2 ,t3 ). Then HomR (I,I) = {x ∈ Q(C [t2 ,t3 ]) = Q(C [t]) : xI ⊂ I} = C [t].

x (2) Let R = C [x,y]/(x · y), I = (x,y) and u = x+y then R[u] = HomR (I,I) because R[u] ⊂ Hom(I,I) and R[u] = R[u]/(u) × R[u]/(1 − u) is the product of the two normal rings R[u]/(u) ∼ = C [y] and R[u]/(1 − u) = C [x].

We now come to the very important theorem of the finiteness of the normalization. This theorem, due to Kronecker and Noether, is of utmost importance for the development of the theory in this book, and will be used at several points. For example, it is used in the proof of the Nullstellensatz in local analytic geometry, see 3.4.4, and in the proof of the Local Parametrization Theorem, see 3.4.14, which is also very important in our development of the theory of local analytic geometry. It is also used in Theorem 4.4.8, in order to prove that the “normalization of a space” always exists. Before stating the theorem of the finiteness of normalization, we recall some facts concerning the discriminant of a polynomial in one variable. Definition 1.5.17. Let k be a field (not necessarily algebraically closed), and f = xr + a1 xr−1 + . . . + ar be a polynomial with coefficients in k. Consider a field extension k ⊂ K in which f splits: f = (x − α1 ) · · · (x − αr ).

Then the discriminant ∆ = ∆x (f ) with respect to the variable x is defined by: Y ∆= (αi − αj ). i6=j

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The discriminant therefore is zero exactly if f has a multiple root. It follows from the Theorem on Symmetric Functions, and is a standard fact in field theory, that the discriminant in fact is a polynomial in the coefficients a1 , . . . ,ar of f . Lemma 1.5.18. Let R be an normal integral domain, K be its quotients field, and L ⊃ K be a field extension. Let x ∈ L be integral over R. In particular, x is algebraic over K. Then the minimal polynomial f (X) ∈ K[X] of x has coefficients in R. Proof. Consider a complete set of conjugates xi of x in an algebraic closure of L. This means that we can write Y f (X) = (X − xi ).

(Each xi is counted pe times, where pe is the degree of inseparability. In particular xi is counted once, if the characteristic is zero.) By a standard result in field theory, there exists a K–isomorphism σi : K(x) ∼ = K(xi ), with σi (x) = xi for all i. It follows that also xi is integral over R. Indeed, apply the isomorphism σi to an integral equation of x xn + a1 xn−1 + . . . , + an = 0 to obtain the integral equation xni + a1 xin−1 + . . . , + an = 0. The coefficients of f (X) are the elementary symmetric functions of the xi . As sums and products of integral elements are integral, it follows that the coefficients of f are integral over R. As they are moreover in K, and R is normal by assumption, it follows that the coefficients are in R. Theorem 1.5.19 (Finiteness of the Normalization). We consider: • a finite ring extension R ⊂ S of integral domains with R a normal ring; • K, the quotient field of R and L the quotient field of S; • α ∈ S a primitive element of the field extension K ⊂ L, that is L = K(α). We suppose that α is separable; 11 • P (X) ∈ K[X] the minimal polynomial of α; • 0 6= ∆ the discriminant of P (X). Then we have the following statements. (1) For all β ∈ S, the minimal polynomial of β has coefficients in R. In particular P (X) ∈ R[X] and ∆ ∈ R. (2) Let a polynomial q ∈ R[X] with q(α) = 0 be given. Then q ∈ (P ). (3) Every element f in S is in the R–module generated by αr−1 1 ,..., . ∆ ∆ To put it in another way, for any f ∈ S there exists a polynomial q ∈ R[X] with deg(q) < deg(P ) such that ∆f = q(α). 11

Recall that separable means that a minimal polynomial for α has no multiple roots.

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(4) Let Se be the normalization of S. Then Se ⊂ generated S–module.

1 ∆ R[α].

(5) S∆ = R∆ [α] is normal.

In particular Se is a finitely

Proof. For the proof we need some Galois theory. Let r be the degree of P . (1) The first statement is a consequence of Lemma 1.5.18. (2) Note that a minimal polynomial P is monic. We perform division with remainder. Thus q = HP + B, with degree of B smaller than r. If B is nonzero, then we divide B by its leading coefficient. We get a monic polynomial B ′ in K[X] with B ′ (α) = 0. This is a contradiction to the fact that P is the minimal polynomial of α. Thus B = 0, and q ∈ (P ). (3) We consider the splitting field M of P . The element α has r (the degree of P ) conjugates α1 , . . . ,αr in M , that is, the α = α1 ,α2 , . . . ,αr are the (pairwise different) zeros of P in M . As P (αi ) = 0, it follows that the αi are integral over R. Now consider an element f ∈ S ⊂ L. It is an easy fact from field theory that we can write: f = q0 · 1 + q1 · α + . . . + qr−1 · αr−1

(1.4)

with qi ∈ K for all12 i. We have to prove that ∆ · qi ∈ R for all i. Now define f1 := q0 · 1 + q1 · α1 + . . . + qr−1 · αr−1 1 .. .

fr := q0 · 1 + q1 · αr + . . . + qr−1 · αr−1 . r We need that the fi are also integral over R. For this we need some Galois theory. There exist automorphisms σi of M over K mapping α to αi . If we let σi operate on equation (1.4) it follows that fi = σi (f ). If we now let σi operate on an integral equation for f , it follows that the fi are also integral over R. We can view the previous equations as a system of linear equations:       f1 q0 1 α1 . . . αr−1 1      .. ..  , A =  ... A  ...  =  ...  ; . .  qr−1

fr

1 αr

. . . αr−1 r

to which we apply our knowledge of linear algebra. First of all, the determinant D := det(A), as a sum and product of integral elements itself is integral. But we know more: the matrix A isQ the well-known Vandermonde matrix and its determinant can be computed to be D = i>j (αi − αj ). Furthermore, by definition of the discriminant ∆ = ±D2 . It follows from Cramer’s rule 1.2.8 that D · qν = Pν (fi ,αj )

for all ν for some polynomial Pν with integer coefficients. It follows that Dqν is integral over R. As D also was integral we conclude that ∆qν = DDqν is integral over R. Now ∆ ∈ R ⊂ K, and the qν are in K, so that ∆ · qν is an element of K, the quotient ring of R. We can now use our assumption that R is normal to conclude that ∆ · qν ∈ R for all ν. This is what we had to prove. 12

Therefore, it is trivially true that for all f there exists a ∆f with ∆f · f ∈ R[α]. The point of the theorem is, that we can take a universal ∆ which works for all f .

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e Then S[β] is a finitely generated R–module with quotient field L and, (4) Let β ∈ S. therefore, by (3) of the theorem (applied to R ⊂ S[β]) included in R∆ [α]. This implies that Se ⊂ R∆ [α] and is, therefore, finitely generated. (5) Note that S∆ = R∆ [α] and, because of (4), Se∆ = R∆ [α].

Theorem 1.5.20 (Splitting of Normalization). Let R be a reduced Noetherian ring and (0) = p1 ∩ · · · ∩ ps be an irredundant primary decomposition. Then there is a canonical ]i : e and the direct sum of the normalizations R/p isomorphism between the normalization R e∼ R =

s M

]i . R/p

i=1

Proof. We have the following statements that are easy to prove, see Exercise 1.5.27. (1) pi Q(R) is a prime ideal in Q(R). (2) ∩si=1 pi Q(R) = (0). (3) Q(R)/pi Q(R) = Q(R/pi ). Now we use Proposition 1.4.27 and obtain Q(R) ∼ =

s s M M ∼ Q(R/pi ). Q(R)/pi Q(R) = i=1

i=1

The second isomorphism is the third statement from above. Let ui ∈ Q(R) correspond to (0, . . . ,0,1,0, . . . ,0), with 1 at the i–th place. It follows immediately from the isomorphism e Note, moreover, that ui ∈ ∩j6=i pj Q(R), ui − 1 ∈ pi Q(R) for that u2i = ui . So ui ∈ R. e and ui − 1 ∈ pi Q(R) ∩ R. e The conditions i = 1, . . . ,s. Thus we get ui ∈ ∩j6=i (pj Q(R) ∩ R) of the Chinese Remainder Theorem are satisfied. We get an isomorphism e∼ R =

s M e i Q(R) ∩ R). e R/(p i=1

The theorem now follows because the canonical map

e i Q(R) ∩ R) e R/pi −→ R/(p

]i ∼ e i Q(R) ∩ R), e see Exercise 1.5.27. induces an isomorphism R/p = R/(p Remark 1.5.21.

(1) Note that we reproved that a Noetherian normal local ring is an integral domain. (2) Let K be a field, and S = K[x1 , . . . ,xn ]/p with p a prime ideal. We will prove in 2.2.9 (Noether normalization) that there exists R ⊂ S satisfying the assumptions of the Theorem 1.5.19. This implies that the normalization Se is a finitely generated S–module. In general this is not necessarily true. A similar statement holds for so-called analytic algebras, which we will consider in the third chapter.

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Theorem 1.5.22 (Lying-Over Theorem). Let R be a ring, R ⊂ S be an integral extension, and p ⊂ R be a prime ideal. Then there exists a prime ideal q in S with q ∩ R = p. For the proof, we need the following two lemmas. Lemma 1.5.23. Let R ⊂ S be an integral extension of rings, and p ⊂ R be a prime ideal. Then pS ∩ R = p. In particular pS 6= S, so that there is an inclusion R/p ֒→ S/pS. Proof. Let x ∈ pS ∩ R. Let S ′ ⊂ S be a subring such that RP⊂ S ′ is a finite extension and x ∈ pS ′ . We can find such an S ′ as follows. Write x = ni=1 xi si with si ∈ S and xi ∈ p. Then S ′ := R[s1 , . . . ,sn ]. It follows from 1.5.11 that x is integral over p. Let xn + a1 xn−1 + . . . + an = 0, with ai ∈ p. Now from x ∈ R it then follows xn ∈ p. The ideal p is prime, so that x ∈ p. Thus pS ∩ R = p. The following lemma is a generalization of the fact that every ideal is contained in a maximal ideal, corresponding to the case S = {1} in the lemma. Lemma 1.5.24. Let R be a ring, and S be a multiplicatively closed subset of R. Let I be an ideal of R, with I ∩ S = ∅. Then there exists a prime ideal p of R with I ⊂ p and p ∩ S = ∅. Proof. As I ∩ S = ∅, we get IRS 6= RS . Let m ⊂ RS be a maximal ideal with IRS ⊂ m. Then I ⊂ m ∩ R =: p. Proof of the Lying-Over Theorem 1.5.22. By 1.5.23 we get pS ∩ N = ∅, for the multiplicatively closed subset N = R \ p. By 1.5.24 there exists a prime ideal P in S with pS ⊂ P and P ∩ N = ∅. Therefore, P ∩ R ⊃ p follows from the first, and P ∩ R ⊂ p follows from the second condition. This shows that P ∩ R = p. Corollary 1.5.25 (Going-Up Theorem). Let R be a ring, R ⊂ S be an integral extension, and q ⊂ p ⊂ R be prime ideals. Let Q be a prime ideal in S lying over q, that is Q∩R = q. Then there exists a prime ideal P in S with P ∩ R = p and Q ⊂ P. Proof. Apply the Lying-Over Theorem to the integral extension R/q ⊂ S/Q. Theorem 1.5.26 (Going-Down Theorem). Let R ⊂ S be an integral extension of integral domains, and suppose that R is normal. Let p ⊂ q be two prime ideals in R. Let Q be a prime ideal in S with Q ∩ R = q. Then there exists a prime ideal P in S with P ⊂ Q and P ∩ R = p. Proof. We have multiplicatively closed sets N ′ := R \ p and N ′′ := S \ Q. Therefore, N := N ′ · N ′′ is a multiplicatively closed subset of S. Claim: We have pS ∩ N = ∅. Otherwise, we have an element x ∈ pS ∩ N . By 1.5.11 x is integral over p. We will show that the minimal polynomial P (X) = X n + a1 X n−1 + . . . + an of x over Q(R) has coefficients in p. By Lemma 1.5.18 the ai are elements of R. Let x1 , . . . ,xn be the zeros of P (X). Then x1 , . . . ,xn are also integral over p, as P (X) divides any polynomial which gives an integral equation for x over p. By Corollary 1.5.12 it follows that the elementary symmetric functions a1 , . . . ,an of the x1 , . . . ,xn are integral over p. As p is prime, it follows that a1 , . . . ,an ∈ p.

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Moreover, as x ∈ N , we can write x = u · v, with u ∈ N ′ , and v ∈ N ′′ . Consider the element v = ux . It is in S, and satisfies the equation vn +

a1 n−1 an v + . . . + n = 0, u u

which is a minimal equation for v. By Lemma 1.5.18, the uaii are elements of R. So we can write ai = bi , ui or, what is the same ai = bi ui . Now ai ∈ p, and, therefore, as ui ∈ /√ p, it follows that √ bi ∈ p. This shows that the element v is integral over p. Therefore, v ∈ pS ⊂ qS ⊂ Q. This is a contradiction to v ∈ N ′′ = S \ Q, and therefore proves the claim. By 1.5.24 there exists a prime ideal P ⊂ S with P∩N = ∅ and pS ⊂ P. In particular P ∩ N ′′ = ∅, so that P ⊂ Q. As P ∩ N ′ = ∅, it follows that P ∩ R ⊂ p, and therefore p = P ∩ R. The condition that R is normal in the Going-Down Theorem is necessary. See Example 2.3.16.

Exercises 1.5.27. Show the unproved statements in the proof of Theorem 1.5.20. ]i ∼ e i Q(R) ∩ R) e consider the canonical inclusions (Hint: To prove that R/p = R/(p M M M e i Q(R) ∩ R) e ⊂ R⊂ R/pi ⊂ R/(p Q(R/pi ).)

1.5.28. (1) Let R ⊂ S be a ring extension, and α ∈ S. Here R is not necessarily Noetherian. Use the Cayley-Hamilton Theorem to prove that the following conditions are equivalent. (a) α is integral over R. (b) R[α] is a finitely generated R–module. (c) R[α] is contained in a ring T ⊂ S, such that T is a finitely generated R–module.

(d) There exists a faithful R[α]–module M which is finitely generated as an R–module. (2) Prove that 1.5.2 (3) holds without the hypothesis that R is Noetherian. 1.5.29. Use the Cayley-Hamilton Theorem to prove that the Noetherian hypothesis in 1.5.4 is superfluous. 1.5.30. (1) Let K be a field, V be a finite-dimensional vector space, and ϕ : V → V be a linear map. Interprete 1.5.8 as the Theorem of Cayley-Hamilton you know from linear algebra. Note that we could just as well take any ring R instead of the coefficient field K, and any finitely generated module M instead of V . (2) Let R be a ring, M be a finitely generated R−module, and ϕ : M → M be a surjective R−module homomorphism. Prove that ϕ is an isomorphism. (Hint: Consider M as R[X]–module, via X · m := ϕ(m), and extend by linearity. Then (X) · M = M. Consider M as R[X,Y ]–module, via Y · m = Id(m) = m. Then apply Remark 1.5.9 (2) to deduce that Y integral over X.) fS = R eS . 1.5.31. Let R be a reduced ring, and S ⊂ R be a multiplicatively closed set. Show that R Informally speaking, one says that taking integral closure commutes with localization.

1.5 Finite and Integral Extensions

45

1.5.32. Let R be a reduced ring. Prove that the following conditions are equivalent. (1) R is normal. (2) Rp is normal for all prime ideals p in R. (3) Rm is normal for all maximal ideals m in R. 1.5.33. Prove that the ring R = k[x,y,z]/(z 2 − xy) is normal. (Hint: Localize in the prime ideals (x,z) and in (y,z), and show that those localizations are normal.) Show however, that R is not a unique factorization domain. This shows that the converse of 1.5.5 in general does not hold. 1.5.34. Prove Nakayama’s Lemma using 1.5.8. 1.5.35. Let f = xsn + a1 xs−1 + a2 xs−2 + . . . + as with ai ∈ K[x1 , . . . ,xn−1 ]. Consider the n n discriminant ∆ ∈ K[x1 , . . . ,xn−1 ] of f with respect to xn . Suppose that f , viewed as an element of K[x1 , . . . ,xn ], is square-free.13 Prove that ∆ 6= 0. 1.5.36. To illustrate the Going-Up Theorem, let R = K[x,y] ⊂ S = K[x,y,z]/(z 2 − xy). (1) Show that S is an integral extension of R.

(2) Find a prime ideal q in S with q ∩ R = (x). The ideal (x) is of course prime in R. (3) Find a prime ideal q in S with q ∩ R = (y 2 − x3 ) ⊂ K[x,y].

13

A polynomial is by definition called square-free, if it does not have multiple factors.

46

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Affine Algebraic Geometry

The purpose of this chapter is to introduce the reader to the basic definitions, ideas and notations of affine algebraic geometry. Loosely speaking, an affine algebraic set is the zero set of finitely many polynomials in an affine space K n , where K is an algebraically closed field.1 Some of the results of this chapter are true and proved over arbitrary algebraically closed fields. This is the reason why we do not consider only the case K = C as we will do from Chapter 3 onwards. Admittedly, our main interest in this book is the behavior of such zero sets in the neighborhood of a point. We will discuss and study the algebraic case first, however, simply because the proofs of the statements in this case usually are easier. This is because one does not have to bother one-selves with convergence questions. Having grasped the main geometric and algebraic ideas in the affine case, the reader is hopefully ready to tackle the local case in Chapter 3. A second reason for doing the affine case, too, is that we need to do some algebraic geometry, in order to classify hypersurface singularities, see Chapter 9. Again for reasons of simplicity, we first study the zero set V (f ) of a single polynomial f . These zero sets are called affine hypersurfaces. The main result in this section is Study’s Lemma. It says that if f and g are polynomials, with f reduced and V (f ) a subset of V (g), then f divides g. We give two proofs, whose ideas permeate large parts of the first part of the book. Namely, we consider a “general” linear projection of the hypersurface V (f ) on a hyperplane. This is called Noether normalization. One shows that the projection is surjective (this is quite trivial), and the proof is completed by using some elementary algebra, in particular, division with remainder. A second proof of Study’s Lemma is given. Here it is not only proved that the projection is surjective, but also the number of points in the fiber are studied. Study’s Lemma allows us to interpret certain algebraic operations and statements geometrically. We mention two of them. The first is the decomposition of a hypersurface in irreducible components. A hypersurface is called reducible, if it can be written in a nontrivial way as the union of two (or more) hypersurfaces. Now, if f and g are polynomials, the zero set of f · g, obviously, is the union of the zero sets of f and g. Study’s Lemma allows us to conclude that for the zero set of an irreducible polynomial f such a decomposition is not possible. The second is the interpretation of the K–algebra K[x1 , . . . ,xn ]/(f ). On a hypersurface V = V (f ), where f is a polynomial, it is natural to consider functions which are restrictions of polynomials on the ambient space K n . This set of functions is in a natural way a K–algebra. If f is reduced, this algebra can be identified with the quotient K[x1 , . . . ,xn ]/(f ) := K[V ]. In the proof, essential use of Study’s Lemma is made. In the second section, we will study affine algebraic sets in general. That is, we study zero sets V (I) := {a ∈ K n : f (a) = 0 for all f ∈ I} of ideals I in the polynomial ring K[x1 , . . . ,xn ]. Hilbert’s Basis Theorem allows us to consider zero sets of only finitely many polynomials. The main discussion is on a generalization of Study’s Lemma, called Hilbert’s Nullstellensatz. A weaker form is given by the Weak Nullstellensatz. It says that 1

Just keep in mind the case K = C .

2.1 Affine Hypersurfaces

47

if I is a proper ideal, then V (I) 6= ∅. This statement is in fact equivalent to the structure of maximal ideals in the polynomial ring: every maximal ideal in K[x1 , . . . ,xn ] is of type (x1 − a1 , . . . ,xn − an ), for some (a1 , . . . ,an ) ∈ K n . So we have a one-one correspondence between maximal ideals and points. The proof of the Weak Nullstellensatz is a consequence of the Projection Theorem, which is a statement on how one can describe, under certain circumstances, the image of an affine space under projection. To prove Hilbert’s Nullstellensatz, we again use the Projection Theorem and Noether normalization. This means that we study a general projection π : X = V (I) −→ K r , where r is determined by X. Morally speaking, r is equal to the dimension of X, a property of X which is not defined at this point of our study. The proof of the fact that the map π for a Noether normalization then is surjective, is done by means of the Projection Theorem. Hilbert’s Nullstellensatz itself is an easy consequence of the surjectivity of π, and is an immediate generalization of the second part of the proof of Study’s Lemma. A shorter proof of the Nullstellensatz out of the weak Nullstellensatz is by using the so-called Rabinowitch trick. This is done in the exercises. We presented the other proof however, as a substitute for the Rabinowitch trick in the local case does not, to our knowledge, exist. We want to have similar proofs for both the affine and the analytic case. In Section 2.3, we consider regular maps between affine algebraic sets. These maps are restrictions of polynomial maps. In particular, we define isomorphisms between affine algebraic sets. If ϕ : V −→ W is a regular map between affine algebraic varieties, we get, by composition, a K–algebra homomorphism ϕ∗ : K[W ] −→ K[V ]. (Arrows reverse!) Again geometric statements about maps between V and W are interpreted as algebraic statements on the map ϕ∗ : K[W ] −→ K[V ]. For example, it is shown that ϕ is injective, if and only if ϕ∗ is surjective.

2.1

Affine Hypersurfaces

Definition 2.1.1. Let K be a field, f ∈ K[x1 , . . . ,xn ] be a polynomial. The zero set V (f ) ⊂ K n of f is defined by: V (f ) := {(a1 , . . . ,an ) ∈ K n : f (a1 , . . . ,an ) = 0}.

A set V ⊂ K n is called an affine hypersurface if there exists a nonzero f ∈ K[x1 , . . . ,xn ], such that V = V (f ). We give some examples. For the examples, we use K = R for two reasons: real pictures are easier to draw; R is not algebraically closed and almost everything which fails over a field which is not algebraically closed fails already over R. Examples 2.1.2. (1) f = x2 + y 2 − 1 ∈ R[x,y]. The set V (f ) is a circle in the plane.

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(2) f = x2 − yz ∈ R[x,y,z]. The set V (f ) is a cone in three-dimensional space.

As is well-known, see 1.4.4, the polynomial ring K[x1 , . . . ,xn ] is factorial. It follows easily from the definition, see Exercise 2.1.18, that f |g (or what is the same (f ) ⊃ (g)) implies that there is an inclusion of zero sets V (f ) ⊂ V (g). The converse does not hold, as the following simple examples shows. Examples 2.1.3. (1) Take f = x2 + 1 ∈ R[x], g = x ∈ R[x]. Then ∅ = V (f ) ⊂ V (g) = {0}. But as the degree of f is bigger than the degree of g, f cannot divide g. (2) Let f be a nonconstant polynomial. Then V (f 2 ) = V (f ), so in particular V (f 2 ) ⊂ V (f ). But obviously f 2 does not divide f . The phenomena occurring in the first example is typical for fields which are not algebraically closed. In fact, one has the following theorem. Theorem 2.1.4 (Study’s Lemma). Let K be an algebraically closed field, f,g ∈ K[x1 , . . . ,xn ]. Suppose that f is irreducible, and that V (f ) ⊂ V (g). Then f |g. There is a proof of Study’s Lemma using so-called resultants, see for example, [Brieskorn-Kn¨ orrer 1986], page 192. We present two other proofs of Study’s Lemma here. Both proofs are based on the following two facts. • Division with remainder for polynomials. • Noether normalization for hypersurfaces. We will first discuss the well-known Division with Remainder Theorem. Theorem 2.1.5 (Division with Remainder). Let R be a ring and f ∈ R[x] be a polynomial. Suppose that the leading coefficient of f is a unit in R. Then for all g ∈ R[x] there exist q,r ∈ R[x], with deg(r) < deg(f ) such that: g = qf + r. Moreover, q and r are uniquely determined by g and f . Proof. Without loss of generality, we may assume that f is monic, that is, the leading coefficient of f is equal to 1. Step 1: Existence: The proof is by induction on deg(g). If deg(g) < deg(f ) then one can take q = 0 and r = g. Otherwise, write

2.1 Affine Hypersurfaces

49

f = xd + bd−1 xd−1 + . . . + b0 ;

g = am xm + am−1 xm−1 + . . . + a0 ,

with m ≥ d. Consider g = g − am xm−d f = (am−1 − am bd−1 )xm−1 + . . . e

The degree of ge is smaller than the degree of g. By induction e g = qef + r with deg(r) < deg(f ). With q = qe + am xm−d the equality g = qf + r holds.

Step 2: Uniqueness. Suppose g = q1 f + r1 = q2 f + r2 with deg(r1 ), deg(r2 ) < deg(f ). It suffices to show that q1 = q2 , as r1 = r2 follows immediately from it. Subtracting we get: (q1 − q2 )f = (r2 − r1 )

As the degree of r1 and r2 is smaller than the degree of f , all the terms on the left hand side of degree at least d have to vanish. Because f is monic, and looking at the term of highest degree, it follows that q1 and q2 have the same degree and the same leading coefficient. Subtraction of this highest degree term from both q1 and q2 gives us q1′ and q2′ with (q1′ − q2′ )f = (r2 − r1 ). By (descending) induction we have that q1′ = q2′ . Hence q1 = q2 . Lemma 2.1.6 (Noether Normalization for Hypersurfaces). Let K be a field with an infinite number of elements, and let f ∈ K[x1 , . . . ,xn ] be a polynomial of degree s. Then after a general linear change of coordinates, one has: (2.1)

f = c(xsn + a1 xns−1 + . . . + as )

with c ∈ K a nonzero constant, and the ai are elements of K[x1 , . . . ,xn−1 ]. Proof. We write f = fs + fs−1 + . . . + f0 where each fi is homogeneous of degree i. The polynomial fs is nonzero, so by Exercise 2.1.19 for general (b1 , . . . ,bn ) ∈ K n the value c := fs (b1 , . . . ,bn ) 6= 0. After renumbering the coordinates, we may furthermore assume that bn 6= 0. Take new coordinates y1 , . . . ,yn related to x1 , . . . ,xn by the formula’s: xi = bi yn + yi for 1 ≤ i ≤ n − 1, xn = bn yn Then g(y1 , . . . ,yn ) := f (x1 (y1 , . . . ,yn ), . . . ,xn (y1 , . . . ,yn )) is such that g(0, . . . ,0,yn ) = fs (b1 , . . . ,bn )yns + fs−1 (b1 , . . . ,bn )yns−1 + . . ., with fs (b1 , . . . ,bn ) = c 6= 0. Hence g = c(yns + a1 yns−1 + . . . + as ), for certain ai ∈ K[y1 , . . . ,yn−1 ]. This shows the lemma. The Noether normalization for hypersurfaces can be reformulated in a more algebraic manner. The algebraic formulation is the one which is going to be generalized later. Corollary 2.1.7 (Noether Normalization for Hypersurfaces). For a general coordinate system, K[x1 , . . . ,xn ]/(f ) is a finitely generated K[x1 , . . . ,xn−1 ]–module.2 2

In this case it is even a free module.

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Proof. After a general change of coordinates we may assume that f is of type (2.1). We claim that K[x1 , . . . ,xn ]/(f ) is as K[x1 , . . . ,xn−1 ]–module generated by {1,xn , . . . ,xns−1 }. Consider a g ∈ K[x1 , . . . ,xn ]. We do division with remainder g = qf + r where r is a polynomial in xn of degree less than s with coefficients in K[x1 , . . . ,xn−1 ]. Hence, modulo f , the element g is indeed in the module generated by {1,xn , . . . ,xns−1 }. Examples 2.1.8. (1) Consider f = xy. Then obviously C [x] ⊂ C [x,y]/(xy), but C [x,y]/(xy) is not a finitely generated C [x]–module. Indeed, if it was finitely generated, say by the elements g1 , . . . ,gs ∈ C [x,y], then let d be the maximum of the degree in y in the polynomials gi . Then y d+1 is not in the module generated by the g1 , . . . ,gs . The module C [x,y]/(xy) is as C [x]–module generated by {1,y,y 2 , . . .}.

f

(2) Consider f = xy − 1. Then C [x] ⊂ C [x,y]/(xy − 1). But again C [x,y]/(xy − 1) = C [x,x−1 ] is not a finitely generated C [x]–module. It is generated by {1,y,y 2 , . . .} = {1,x−1 ,x−2 , . . .}.

f

2.1 Affine Hypersurfaces

51

These two examples are quite typical. In both cases, we considered the inclusion of rings C [x] ⊂ C [x]/(f ). Consider the projection π : X = V (f ) −→ C which sends (a,b) to a. In the first example, this map is surjective, but the fiber π −1 (0) consists of infinitely many points, in fact the whole y–axis. In the second example, the map π is not surjective: π −1 (0) = ∅. But, if f is as in formula (2.1), it turns out that the projection π : V (f ) −→ C n−1 on the first n − 1 coordinates is surjective, and has only finitely many points in each fiber. This fact, which is very easy to prove, is crucial for our first proof of Study’s Lemma. Lemma 2.1.9. Let K be an algebraically closed field, and f = xsn + a1 xns−1 + . . . + as , with ai ∈ K[x1 , . . . ,xn−1 ]. Consider the projection π : V (f ) −→ K n−1 ;

(p1 , . . . ,pn ) 7→ (p1 , . . . ,pn−1 ).

Then π is surjective. Proof. This is quite obvious. For, let p = (p1 , . . . ,pn−1 ) ∈ K n−1 . Because K is algebraically closed, the equation X s + a1 (p)X s−1 + . . . + as (p) = 0, has a solution pn . Then (p1 , . . . ,pn ) ∈ V (f ). First proof of Study’s Lemma 2.1.4. Let g ∈ K[x1 , . . . ,xn ] with V (f ) ⊂ V (g). We have to show that f | g. By the Noether Normalization Lemma for Hypersurfaces, we may suppose that f is of type f = xsn + a1 xns−1 + . . . + as . By 2.1.7 K[x1 , . . . ,xn ]/(f ) is a finitely generated K[x1 , . . . ,xn−1 ]–module. Consider the class g of g in K[x1 , . . . ,xn ]/(f ). Hence, by 1.5.2, g is integral over K[x1 , . . . ,xn−1 ], that is, it satisfies an equation of type gk + b1 g k−1 + . . . + bk = 0, with bi ∈ K[x1 , . . . ,xn−1 ]. This equation is to be considered in the ring K[x1 , . . . ,xn ]/(f ). Therefore g k + b1 g k−1 + . . . + bk ∈ (f ). We take k minimal with this property. Take a point p ∈ V (f ), so that f (p) = 0. From V (f ) ⊂ V (g) it follows that g(p) = 0. Therefore bk (p) = 0. But bk only depends on the first n − 1 coordinates. As π : V (f ) −→ K n−1 is surjective by the previous lemma, it follows that bk (p1 , . . . ,pn−1 ) = 0 for all (p1 , . . . ,pn−1 ) ∈ K n−1 . Hence bk is the zero polynomial. Therefore g · (g k−1 + b1 g k−2 + . . . + bk−1 ) ∈ (f ). As f is irreducible, either f | g or f | g k−1 +b1 g k−2 +. . .+bk−1 . But the second possibility is excluded by our choice of k to be minimal. The second proof of Study’s Lemma we present looks somewhat more complicated. It gives, however, also some more information on the fibers of the map π : V (f ) −→ K n−1 . Moreover, the condition irreducible in this proof can be somewhat weakened: we only need f to be square-free.

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Lemma 2.1.10. Let K be an algebraically closed field. Consider a square-free polynomial f ∈ K[x1 . . . ,xn ] of the form f = xsn + a1 xns−1 + a2 xns−2 + . . . + as ;

ai ∈ K[x1 , . . . ,xn−1 ].

Let ∆ ∈ K[x1 , . . . ,xn−1 ] be the discriminant of f (with respect to xn ), see 1.5.17. Consider the projection π : V (f ) −→ K n−1 , (p1 , . . . ,pn ) 7→ (p1 , . . . ,pn−1 ) =: q.

Then π −1 (q) has s, the degree of f , elements exactly when ∆(q) 6= 0. Proof. Consider q = (p1 , . . . ,pn−1 ) ∈ K n−1 . Look at the polynomial (2.2)

xsn + a1 (q)xns−1 + a2 (q)xns−2 + . . . + as (q) ∈ K[xn ].

Because K is algebraically closed, this polynomial has s (distinct) roots exactly when its discriminant, which is ∆(q), is nonzero. If b1 , . . . ,bt , t ≤ s are the roots of the polynomial (2.2), then the points in the fiber π −1 (q) are (p1 , . . . ,pn−1 ,b1 ), . . . ,(p1 , . . . ,pn−1 ,bt ). Example 2.1.11. Consider the curve C given by the equation y 2 − x2 (x + 1) = 0. The discriminant is given by 4x2 (x + 1), so is zero for x = 0 and x = −1. Consider the projection π from C to the x–axis.

y

x

We see indeed from the picture (over C !) that π −1 (p) consists of two points for p 6= 0, − 1, whereas for p = 0 or −1 it consists of one point. For other examples see Exercise 2.1.21. Second proof of Study’s Lemma. After, if necessary, performing a linear change of coordinates, we may assume, by Noether normalization for hypersurfaces, that f = xsn + a1 xns−1 + a2 xns−2 + . . . + as with ai ∈ K[x1 , . . . ,xn−1 ]. As f is square-free, by Exercise 2.1.20 the discriminant ∆ of f is nonzero. We consider the projection π : V (f ) −→ K n−1 as in Lemma 2.1.10. Let D = {q ∈ K n−1 : ∆(q) = 0}. We have shown that the inverse image of q 6∈ D under π has exactly s points. Now do division with remainder for polynomials in the variable xn : g = qf + r r = r0 + r1 xn + . . . + rs−1 xns−1 , ri ∈ K[x1 , . . . ,xn−1 ]. Let (p1 , . . . ,pn−1 ,pn ) ∈ V (f ). As by assumption V (f ) ⊂ V (g), it follows that

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53

g(p1 , . . . ,pn−1 ,pn ) = 0. As g = qf + r, it follows that r(p1 , . . . ,pn−1 ,pn ) = 0. Hence, if q = (p1 . . . ,pn−1 ) 6∈ D the polynomial r0 (q) + r1 (q)xn + . . . + rs−1 (q)xns−1 has s roots. But the polynomial r(p1 , . . . ,pn−1 ) has degree s − 1, has s roots, so must be the zero polynomial. Therefore, its coefficients are zero. Hence r0 (p1 , . . . ,pn−1 ) = · · · = rs−1 (p1 , . . . ,pn−1 ) = 0. As we can take q = (p1 , . . . ,pn−1 ) general, it follows from Exercise 2.1.19 that the polynomials, r0 , . . . ,rs−1 ∈ K[x1 , . . . ,xn−1 ] are zero. Hence r = 0. Therefore, g = qf , that is, f | g. Definition 2.1.12. An affine hypersurface C ⊂ K n is called irreducible if from C = C1 ∪ C2 , with C1 and C2 affine hypersurfaces, it follows that either C1 = C or C2 = C. A hypersurface is called reducible if it is not irreducible. An irreducible component of an affine hypersurface C is an irreducible affine hypersurface contained in C. A consequence of Study’s Lemma is that one can characterize irreducible hypersurfaces by its defining equation. Theorem 2.1.13. Let K be an algebraically closed field. Let C = V (f ) be an affine hypersurface. Then the following conditions are equivalent. (1) C is irreducible. (2) There exists an irreducible g ∈ K[x1 , . . . ,xn ], and a k ∈ N with f = g k . It follows that that for all affine hypersurfaces C we have a decomposition C = C1 ∪ · · · ∪ Cs , where Ci are the irreducible components of C, and Ci 6= Cj for i 6= j. Proof. Suppose first that C is irreducible. Suppose f = f1n1 · · · fsns . Then V (f ) = V (f1 · · · fs ). So we may assume that f is square-free. We have to prove that f is irreducible under this assumption. Suppose the converse. Then we can write f = f1 · f2 , where f1 and f2 do not have common factors and are not constant. In particular f1 ∤ f2 and f2 ∤ f1 . From Study’s Lemma it follows that V (f1 ) 6⊂ V (f2 ) and V (f2 ) 6⊂ V (f1 ). Therefore C 6= V (f1 ) and C 6= V (f2 ). But as f = f1 ·f2 it follows that C = V (f1 )∪V (f2 ). This is a contradiction to the assumption that C is irreducible. To prove the converse, suppose C = C1 ∪ C2 , C = V (f ), C1 = V (f1 ) and C2 = V (f2 ), and both C1 and C2 are not equal to C. Let h be irreducible and h | f1 . Then V (h) ⊂ V (f1 ) $ V (f ) (Exercise 2.1.18). Study’s Lemma implies h | f . and thus h | g k . As the polynomial ring is a unique factorization domain, and g is irreducible, it follows that h = g · c for some c ∈ K. This implies V (h) = V (g) = V (f ) which is in contradiction to the assumption V (h) $ V (f ).

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Examples 2.1.14. (1) f (x,y) = x · y. Then V (f ) has two components, given by x = 0 and y = 0. x=0

y=0

(2) As f (x,y) = y 2 − x2 (x + 1), see 2.1.11, is an irreducible polynomial, see Exercise 1.4.34, the affine hypersurface V (f ) is irreducible. We now consider polynomial functions on affine hypersurfaces. Definition 2.1.15. Let C = V (f ) ⊂ K n be an affine hypersurface. A function g : C −→ K is called regular, if g is the restriction of a polynomial map g : K n −→ K. With the obvious addition and multiplication the regular functions on C form a ring . It is in fact a K–algebra. The ring of regular functions of C we denote by K[C]. A polynomial map g : K n −→ K is just given by a polynomial g ∈ K[x1 , . . . ,xn ]. Let such a g be given. The restrictions g and g + αf for any α ∈ K[x1 , . . . ,xn ] define the same function on C = V (f ). Using Study’s Lemma, we have the converse, as soon as we assume that the hypersurface C is defined by a square-free f . Theorem 2.1.16. Let K be an algebraically closed field. Let f ∈ K[x1 , . . . xn ] be squarefree. Then the ring of regular functions on C = V (f ) is equal to K[x1 , . . . ,xn ]/(f ). Proof. Let g1 ,g2 ∈ K[x1 , . . . xn ] be polynomial functions such that the functions g1|C and g2|C are equal. We have to show that the classes of g1 and g2 in K[x1 , . . . ,xn ]/(f ) are equal. By assumption g1 − g2 vanish on C, which means that V (f ) ⊂ V (g1 − g2 ). From Study’s Lemma it follows that f |g1 − g2 , or, what is the same g1 − g2 ∈ (f ). This shows the theorem. The assumption that f is square-free is essential. For example, let f = x2 ∈ K[x,y], and C = V (f ). Then x and 0 define the same function on V (f ), but x 6= 0 ∈ K[x,y]/(x2 ).

Exercises In these exercises, K is an algebraically closed field. 2.1.17. Deduce Study’s Lemma for square-free f from Study’s Lemma for irreducible f . 2.1.18. Let f,g ∈ K[x1 , . . . ,xn ] with f | g. Show that V (f ) ⊂ V (g). 2.1.19. Let K be a field with an infinite number of elements. Let f ∈ K[x1 , . . . ,xn ] be a nonconstant polynomial. Prove, by induction on n, that for “general” points (a1 , . . . ,an ) ∈ K n we have f (a1 , . . . ,an ) 6= 0. Specify what we mean by general.

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55

2.1.20. Let f = xsn + a1 xs−1 + a2 xs−2 + . . . + as , with ai ∈ K[x1 , . . . ,xn−1 ]. Suppose that f is n n square-free. Prove that the discriminant of f is nonzero. 2.1.21. (1) Consider the curve defined by y 2 − xy + x2 = 0, and the projection π on the x–axis. Compute for which points on the x–axis the inverse image of π consists of less than two points. (2) Consider the cone x2 − yz, see example 2.1.2., and the projection π on the y,z–plane. Compute for which points on the y,z–plane the inverse image of π consists of less than two points. What about projection on the x,y–plane? 2.1.22. Prove, using only the statement of Study’s Lemma, that V (f ) 6= ∅ for any nonunit f ∈ K[x1 , . . . ,xn ].

2.2

Affine Varieties

In this rest of this chapter we assume, unless said otherwise explicitly that K is algebraically closed. In this section, we will, more generally, consider zero sets of any set of elements in K[x1 , . . . ,xn ]. Definition 2.2.1. Let F ⊂ K[x1 , . . . ,xn ] be a subset. We put V (F ) := {p ∈ K n : f (p) = 0 for all f ∈ F }. and call it the zero set of F . A subset V of K n is called an (affine) algebraic set, or an affine variety if there exists a subset F ⊂ K[x1 , . . . ,xn ] with V = V (F ). Conversely, for any set A ⊂ K n we define I (A) = {f ∈ K[x1 , . . . ,xn ] : f (p) = 0 for all p ∈ A}, and call it the ideal of (functions vanishing on) A. Elementary properties of these notions are gathered in the following lemma. Lemma 2.2.2. Let V,W be algebraic subsets of K n , A be a subset of K n , I,J,Iα be ideals of K[x1 , . . . ,xn ], F be a subset of K[x1 , . . . ,xn ], and B be an index set. (1) I (A) is an ideal. Even more, I (A) is a radical ideal. (2) V ⊂ W =⇒ I (V ) ⊃ I (W ). (3) I ⊂ J =⇒ V (I) ⊃ V (J). (4) Let I be the ideal generated by F ⊂ K[x1 , . . . ,xn ]. Then V (F ) = V (I). (5) ∅ and K n are algebraic sets. (6) V (I · J) = V (I ∩ J) = V (I) ∪ V (J). In particular, (induction) finite unions of algebraic sets are algebraic. P (7) V ( α∈B Iα ) = ∩α∈B V (Iα ). In particular, arbitrary intersections of algebraic sets are algebraic sets.

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(8) I (V ∪ W ) = I (V ) ∩ I (W ). √ (9) I ⊂ I (V (I)). Proof. (1) The fact that I (A) is an ideal is easy. Moreover, if f m ∈ I (A), then f m (p) = 0 for all p ∈ A. Hence f (p) = 0 for all p ∈ A, that is f ∈ I (A). (2) and (3) are trivial. (4) As F ⊂ I, Pnit follows that V (F ) ⊃ V (I). Suppose on the other hand that a ∈ V (F ), and let f = i=1 ai fi , fi ∈ F be an element of I. As a ∈ V (F ), we have that fi (a) = 0 for all i. Hence f (a) = 0. As f is arbitrary it follows that a ∈ V (I). (5) ∅ = V ({1}), hence is algebraic. Furthermore, K n = V ({0}), hence is algebraic too. (6) We prove three inclusions. (a) V (I) ∪ V (J) ⊂ V (I ∩ J): let a in V (I) ∪ V (J). Therefore, either f (a) = 0 for all f ∈ I or f (a) = 0 for all f ∈ J. In particular, if f ∈ I ∩ J, it follows that f (a) = 0. (b) V (I ∩ J) ⊂ V (I · J): this follows from the inclusion of ideals I · J ⊂ I ∩ J. (c) V (I · J) ⊂ V (I) ∪ V (J): let a ∈ V (I · J). In particular f (a) · g(a) = 0 for all f ∈ I and g ∈ J. Suppose a ∈ / V (J). Then we must show a ∈ V (I). As a ∈ / V (J) there exists a g ∈ J with g(a) 6= 0. But as f (a)g(a) = 0 for all f ∈ I, it follows that f (a) = 0 for all f ∈ I. This gives a ∈ V (I). P (7)P As the ideal generated by ∪α∈B Iα is exactly α∈B Iα , it follows that V (∪α∈B Iα ) = V ( α∈B Iα ). The equality V (∪α∈B Iα ) = ∩α∈B V (Iα ) is easy, and therefore left to the reader. (8) If f vanishes on V ∪ W , it must vanish on both. The inclusion I (V ∪ W ) ⊂ I (V ) ∩ I (W ) follows. Suppose on the other hand that f ∈ I (V ) ∩ I (W ), and let a ∈ V ∪ W . Then either a ∈ V or a ∈ W . In any case, for f ∈ I (V ) ∩ I (W ) it follows that f (a) = 0.

(9) The inclusion I ⊂ I (V (I)) is a√tautology. As I (V (I)) is a radical ideal by the first part of the lemma, it follows that I ⊂ I (V (I)). Examples 2.2.3. (1) Consider I = (xy,xz,yz). It is not too difficult to show, see Exercise 2.2.30, that I = (x,y) ∩ (x,z) ∩ (y,z). Hence X := V (I) consist of the union of the coordinate axes. On the other hand, X is also the zero set of the ideal I ′ := (x,y) · (x,z) · (y,z) = (x2 y,x2 z,xyz,xz 2 ,xy 2 ,yz 2 ,y 2 z). This ideal is smaller than I and, therefore, is not radical. This can be seen immediately. For example, xy ∈ / I ′ but (xy)2 = y · x2 y is ′ in I . This is quite typical. Products of radical ideals in general are not radical.

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(2) Consider I1 = (z,y 2 − x3 ), and I2 = (x,y), and X1 = V (I1 ),X2 = V (I2 ). Then the union X
= X1 ∪ X2 is the zero set of the ideal I = I1 · I2 = xz,yz,x(y 2 − x3 ),y(y 2 − x3 ) . Again this ideal is not radical. Indeed, y 2 − x3 ∈ / I, but (y 2 − x3 )2 = √ 2 3 2 2 3 2 3 y · y(y − x ) − x · x(y − x ) ∈ I, so that y − x ∈ I.

Definition 2.2.4. We define a topology on K n by taking the closed subsets to be the algebraic subsets. That this is indeed a topology follows from Lemma 2.2.2 (5), (6) and (7). This topology is called the Zariski topology of K n . More generally, for any algebraic set V ⊂ K n , we define the Zariski topology on V to be the topology on V induced by the Zariski topology on K n . So a closed subset in V is of type V (I) ∩ V for some ideal I ⊂ K[x1 , . . . ,xn ]. The Zariski topology of C n is much coarser than the usual (Euclidean) topology of C . For example, for n = 1, a subset of C is closed in the Zariski topology when this subset is finite or equal to C . The Zariski topology does not behave well under products, see Exercise 2.2.34. n

Lemma 2.2.5. Let A ⊂ K n be a subset. Then A = V (I (A)). Proof. The inclusion A ⊂ V (I (A)) is a tautology. Moreover, V (I (A)) is the smallest closed subset containing A. Indeed, let B be a closed subset with A ⊂ B. As B is closed, there exists an ideal J with B = V (J). Then J ⊂ I (V (J)), hence B = V (J) ⊃ V (I (V (J))) = V (I (B)) ⊃ V (I (A)). The last inclusion is by combining (2) and (3) of the Lemma 2.2.2. As A is the smallest closed subset containing A, the lemma follows. Theorem 2.2.6 (Hilbert’s Nullstellensatz). Let I ⊂ K[x1 , . . . ,xn ] be an ideal. Then

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I = I (V (I)).

We will give several proofs of the Nullstellensatz in this book. For still other proofs, we refer to the book of Eisenbud [Eisenbud 1995]. Hilbert’s Nullstellensatz is a generalization of Study’s Lemma, see Exercise 2.2.31. Why is it called the Nullstellensatz (German for Theorem of Zeros)? Well, let I $ K[x1 , . .
. ,xn ]. Then it follows from the Nullstellensatz that V (I) 6= ∅: indeed, otherwise I V (I) = I (∅) = K[x1 , . . . ,xn ], hence √ I = (1). Therefore for some n we have 1 = 1n ∈ I and, therefore, I = K[x1 , . . . ,xn ], a contradiction. Hence from the Nullstellensatz it follows: Theorem 2.2.7 (Weak Nullstellensatz). Let I $ K[x1 , . . . ,xn ] be an ideal. Then V (I) 6= ∅. In fact, one is able to show via the so-called Rabinowitch trick, see Exercise 2.2.32, that conversely, the weak Nullstellensatz implies the Nullstellensatz. In this reduction, both the number of variables and the number of equations grow by one! In particular we get an alternative proof of Study’s Lemma, by following the lines in this proof of the Nullstellensatz, in which one would have to prove the weak Nullstellensatz for two functions. We will prove the Weak Nullstellensatz by using the Projection Theorem, which we treat now. Theorem 2.2.8 (Projection Theorem). Consider a nonzero ideal I ⊂ K[x1 , . . . ,xn ]. Suppose that I contains an element f0 which is regular in xn , that is f0 = xdn + a1 xd−1 + . . . + a0 n for certain elements ai ∈ K[x1 , . . . ,xn−1 ]. Let p : K n −→ K n−1 be the map which sends (x1 , . . . ,xn ) to (x1 , . . . ,xn−1 ). Define moreover I ′ := I ∩ K[x1 , . . . ,xn−1 ], and put X ′ = V (I ′ ). Then p(X) = X ′ . Proof. The inclusion p(X) ⊂ X ′ is the easy inclusion. Indeed if a ∈ p(X), then there exists a b ∈ K with (a,b) ∈ X. Let f ∈ I ′ . Then f ∈ I and thus f (a,b) = 0. But f does not depend on xn so that f (a) = 0. For the inclusion X ′ ⊂ p(X) we will use the Nullstellensatz for the case n = 1. This case is easy, as K[x] is a principal ideal domain, and every f has a root, since by assumption K is algebraically closed. Assume that a ∈ / p(X). We will show that a ∈ / X ′. ′ To prove this, it suffices to find an g ∈ I with g(a) 6= 0. This g is constructed in two steps. Step 1. First we will show that for all f ∈ K[x1 , . . . ,xn ] there exists a representation (2.3)

f = g0 + g1 xn + . . . + gn−1 xd−1 + hf n

with hf ∈ I, gi ∈ K[x1 , . . . ,xn−1 ] and gi (a) = 0. To show this we look at the map ϕ : K[x1 , . . . ,xn ] −→ K[xn ], f 7→ f (a,xn ). Now ∃b : b ∈ V (ϕ(I)) ⇐⇒ ∃b : f (a,b) = 0 for all f ∈ I ⇐⇒ a ∈ p(X),

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By assumption a ∈ / p(X), so that V (ϕ(I)) = ∅. From the Nullstellensatz in one variable we get ϕ(I) = K[xn ]. In particular for all f we can find an h′f ∈ I with ϕ(h′f ) = ϕ(f ). With gf := f − h′f we get a decomposition f = gf + h′f , such that h′f ∈ I and gf (a,xn ) = 0. We divide gf through f0 . We get gf = qf0 +

d−1 X

gj xjn .

j=0

Pd−1 If we plug in (a,xn ) we get 0 = q(a,xn )f0 (a,xn ) + j=0 gj (a)xjn . By assumption f0 (a,xn ) is a polynomial in xn of degree d. By the uniqueness statement in division with remainder it follows that gj (a) = 0 for all j. Now we put hf := h′f +qf0 and we get the representation (2.3). Step 2. We apply (2.3) to the functions 1,xn , . . . ,xd−1 n . We get 1 xd−1 n

= .. .

g0,0

= gd−1,0

+...+ .. .

g0,d−1 xd−1 n

+ . . . + gd−1,d−1xd−1 n

+ .. .

h0

+ hd−1

with gij (a) = 0 for all i,j and hj ∈ I for all j. In matrix notation     1 h0     (Id − gij )  ...  =  ...  . xd−1 n

Let B be the adjoint matrix of (Id − gij ). follows from Cramer’s rule that  1  .. (2.4) det (Id − gij )  .

xd−1 n

hd−1

We multiply with B from the left, and it 



  =B

h0 .. . hd−1



 .

We define g := det (Id − gij ) ∈ K[x1 , . . . ,xn−1 ]. From the first row of (2.4) we see that g ∈ I, so g ∈ I ′ . As moreover gij (a) = 0, it follows that g(a) = 1 6= 0. This was our goal. Proof of the weak Nullstellensatz. The proof is by induction on n. The case I = (0) is obvious. Otherwise there exists an f0 ∈ I, for which we may, by the Noether normalization for hypersurfaces 2.1.6, assume to fulfill the conditions of the Projection Theorem. Furthermore I ′ 6= (1), as otherwise I = (1). By induction V (I ′ ) 6= ∅, and thus by the Projection Theorem p(X), and in particular X = V (I) is nonempty. For the following theorem we do not need that K is algebraically closed. The statement of the following so-called Noether normalization is a direct generalization of the Noether normalization for hypersurfaces, see 2.1.7.

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Theorem 2.2.9 (Noether Normalization). Let K be a field with infinitely many elements, A = K[x1 , . . . ,xn ]/I a finitely generated K–algebra. Then after a general linear coordinate change, there exists a number r ≤ n, and an inclusion K[x1 , . . . ,xr ] ⊂ A, such that A is a finitely generated K[x1 , . . . ,xr ]–module. If moreover I 6= (0), then r < n. If V = V (I), we also say that the projection on the first r coordinates π : V −→ K r is a Noether normalization of V . Proof. If I = (0), we can take x1 , . . . ,xr to be x1 , . . . ,xn . Otherwise the proof goes by induction on n, the case n = 0 being trivial. Let g ∈ I with g 6= 0. By Noether normalization for hypersurfaces see 2.1.7, after a general (linear) change of coordinates, we may assume that K[x1 , . . . ,xn ]/(g) is a finitely generated K[x1 , . . . ,xn−1 ]–module. Because g ∈ I, then certainly K[x1 , . . . ,xn ]/I is a finitely generated K[x1 , . . . ,xn−1 ]/(I ∩ K[x1 , . . . ,xn−1 ])–module. We apply the induction hypotheses to I ∩K[x1 , . . . ,xn−1 ]. After a general linear coordinate change in x1 , . . . ,xn−1 we have that
K[x1 , . . . ,xr ] ⊂ K[x1 , . . . ,xn−1 ]/ I ∩ K[x1 , . . . ,xn−1 ] is finitely generated. Composing with the inclusion


K[x1 , . . . ,xn−1 ]/ I ∩ K[x1 , . . . ,xn−1 ] ⊂ K[x1 , . . . ,xn ]/I,

allows us to deduce that K[x1 , . . . ,xr ] ⊂ K[x1 , . . . ,xn ]/I is finitely generated. This proves the theorem. Corollary 2.2.10. Let π : V −→ K r be a Noether normalization. Then π is surjective. Proof. This follows by induction from the Projection Theorem 2.2.8. Corollary 2.2.11. Hilbert’s Nullstellensatz holds for prime ideals p ⊂ K[x1 , . . . ,xn ]. Proof. As a prime ideal is radical, the Nullstellensatz says that p = I (V (p)) for a prime √ ideal p. The inclusion ( p = p ⊂ I (V (p)) is easy, see Lemma 2.2.2 (9). For the other inclusion, consider a Noether normalization K[x1 , . . . ,xr ] ֒→ K[x1 , . . . ,xn ]/p. Now let f ∈ I (V (p)) be given. By looking at its class in K[x1 , . . . ,xn ]/p, we see that there exists an integral equation f s + a1 f s−1 + . . . + as = 0 ∈ K[x1 , . . . ,xn ]/p, ai ∈ K[x1 , . . . ,xr ]. We take s minimal with this property. From p ⊂ I (V (p)), it follows as ∈ I (V (p)). This implies that as (p) = 0 for all p ∈ V (p). If p = (p1 , . . . ,pn ) then as (p) = as (p1 , . . . ,pr ). Using Corollary 2.2.10 we know that V (p) −→ K r is surjective and, consequently, for all (p1 , . . . ,pr ) ∈ K r there exists a p = (p1 , . . . ,pr ,pr+1 , . . . ,pn ) ∈ V (p) which implies as (p1 , . . . ,pr ) = 0 for all (p1 , . . . ,pr ) ∈ K r . This implies as ∈ I (K r ) = 0. Hence f (f s−1 + a1 f s−2 + . . . + as−1 ) ∈ p.

Because p is a prime ideal, and s was to be taken minimal, it follows that f ∈ p. This is what we had to show.

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61

To complete our proof of the Nullstellensatz it remains to show the following lemma. Lemma 2.2.12. Suppose the Nullstellensatz holds for all prime ideals p ⊂ K[x1 , . . . ,xn ]. Then the Nullstellensatz holds for all ideals I ⊂ K[x1 , . . . ,xn ]. √ Proof. Let I = q1 ∩ . . . ∩ qr be a primary decomposition of I. Let pi = qi for all i. In particular V (pi ) = V (qi ). Then V (I) = V (p1 ∩ . . . ∩ pr ) = V (p1 ) ∪ . . . ∪ V (pr ) by Lemma 2.2.2 (6). Hence by (8) of the same lemma

Thus

√

I (V (I)) = I (V (p1 )) ∩ . . . ∩ I (V (pr )). p r √ I= ∩i=1 qi = ∩ri=1 qi = ∩ri=1 pi = ∩ri=1 I (V (pi )) = I (V (I)).

Hilbert’s Nullstellensatz can be used to give a geometric interpretation of primary decomposition for radical ideals in a polynomial ring over an algebraically closed field. For this, we need the following definition. Definition 2.2.13. Let X be a topological space. Then X is called irreducible if from X = X1 ∪ X2 with X1 and X2 closed subsets of X it follows that either X1 = X or X2 = X. The space X is called reducible if it is not irreducible. An irreducible decomposition of a topological space is a decomposition X = X1 ∪ . . . ∪ Xr , where the Xi are irreducible, and for all i 6= j, Xi is not contained in Xj . We will prove that irreducible algebraic sets with the Zariski topology correspond to zero sets of prime ideals. By an application of the Primary Decomposition Theorem it will follow that an affine algebraic set X always has an irreducible decomposition. Example 2.2.14. Let X = V (zx,zy) ⊂ K 2 , with the Zariski topology.

Then X = X1 ∪ X2 , where X1 = V (x,y) and X2 = V (z). Moreover, none of the Xi , for i = 1,2 is contained in the other. Theorem 2.2.15. (1) Let X ⊂ K n be an irreducible algebraic set. Then I (X) is a prime ideal. (2) Conversely let p be a prime ideal in K[x1 , . . . ,xn ]. Then V (p) is irreducible.

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Proof.
(1) Suppose f · g ∈ I (X). Then V (f ) ∪ V (g) = V (f · g) ⊃ V I (X) = X. The last equality is by 2.2.5. Hence

X = X ∩ V (f ) ∪ X ∩ V (g) ,

so that by assumed irreducibility of X either X = X ∩ V (f ), or X = X ∩ V (g). So either X ⊂ V (f ), of X ⊂ V (g), which means that either f ∈ I (X) or g ∈ I (X). This is what we had to show.

(2) Suppose V (p) = X1 ∪ X2 with X1 and X2 closed. Then by the Nullstellensatz p =
I V (p) = I (X1 ) ∩ I (X2 ). As p is prime, either p = I (X1 ) or p = I (X2 ). Therefore, either V (p) = X1 or V (p) = X2 . Theorem 2.2.16. Let I be a radical ideal and let I = p1 ∩ . . . ∩ pr be an irredundant primary decomposition of I. Let Xi = V (pi ). Then X = X1 ∪ . . . ∪ Xr is an irreducible decomposition of X. An irreducible decomposition of X is, up to permutation of the Xi , uniquely determined. Proof. For the first statement, we have X = V (I) = V (p1 ) ∪ . . . ∪ V (pr ). The V (pi ) are irreducible by Theorem 2.2.15. Suppose V (pi ) ⊂ V (pj ). Then by the Nullstellensatz pi = I (V (pi )) ⊃ I (V (pj )) = pj . As the primary decomposition of I is irredundant, it follows that i = j. This shows that the Xi are pairwise not contained in each other. Hence, X = X1 ∪ . . . ∪ Xr is a decomposition of X into irreducible components. To show that an irreducible decomposition is uniquely determined, consider an irreducible decomposition X = X1 ∪ . . . ∪ Xr . Let pi = I (Xi ). Then the pi are prime ideals and I (X) = p1 ∩ . . . ∩ pr . The irredundancy of the primary decomposition follows from pj 6⊂ pi for i 6= j, which follows from Xi 6⊂ Xj . By the Uniqueness Theorems of a Primary Decomposition, a primary decomposition of a radical ideal is uniquely determined. Now Xi = V (I (Xi )) and it follows that the irreducible decomposition is uniquely determined. Lemma 2.2.17. Let X ⊂ K n be an irreducible algebraic set, U ⊂ X an open nonempty subset. Then U = X. In particular, let h ∈ K[x1 , . . . ,xn ] be a polynomial with h 6∈ I (X). Then X \ V (h) = X. Proof. As X is closed the inclusion U ⊂ X is trivial. Suppose the inclusion is proper. Then U is closed, and we can write X = U ∪X \ U. This gives a nontrivial decomposition of X into closed subspaces, a contradiction. Hence the inclusion is an equality. Despite the Nullstellensatz, it is often important to consider ideals which are not radical. Example 2.2.18. Let I = (y 2 ,xy) ∈ K[x,y]. Then, see 1.4.15, I = (y) ∩ (y 2 ,x) is a primary decomposition. Hence V (I) = V (y) ∪ V (x2 ,y). Hence V (I) = {(x,y) : y = 0} ∪ {(0,0)}. One tries to visualize this by regarding the point (0,0) to be “embedded” in the line {(x,y) : y = 0} as the following picture suggests. This is the reason why, for an ideal I in a Noetherian ring R, the associated primes of I which are not minimal are called embedded primes.

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63

The next thing we do is to give a geometric interpretation of the the ideal quotient. Theorem 2.2.19. Consider an ideal I ⊂ K[x1 , . . . ,xn ], and let d be an element in K[x1 , . . . ,xn ]. Let I = q1 ∩ . . . ∩ qr be an irredundant primary decomposition of I. Then [ V (qi ). V (I : d) = i:d6∈qi

Proof. From (the easy) Exercise 1.1.15 it follows that I : d = (q1 : d)∩. . .∩(qr : d) so that V (I : d) = V (q1 : d) ∪ . . . ∪ V (qr : d). Now if d ∈ qi , then qi : d is the whole polynomial ring. In√particular V (qi : d) is empty. Suppose, on the other hand, that d ∈ / qi . Then √ qi = qi : d by 1.4.47, so it follows that V (qi : d) = V (qi ). From this the theorem follows. For radical ideals I the theorem means that V (I : d) picks out the irreducible components of V (I) on which the function d does not vanish identically. In mathematical terms, we can write V (I : d) = V (I) \ V (d). Here the bar means taking the closure in the Zariski topology. This will be slightly generalized in Exercise 2.2.38. Examples 2.2.20. (1) Let I = (xy,xz,yz) = (x,y)∩(x,z)∩(y,z). Then I is a radical ideal. Let d = x. Then d vanishes on the y and z–axis so that V (I : d) is the x–axis. Indeed I : d = (y,z). (2) Consider the ideal I = (x2 ,xy). Let d = x. Then I : d = (x,y). So V (I : d) does consist of the origin. So the statement V (I : d) = V (I) \ V (d) does not generalize to nonradical ideals. The final topic in this section is the discussion of localizations. First a definition. Definition 2.2.21. Let V ⊂ K n be an algebraic set (1) A function f : V −→ K is called regular, if f is the restriction of a polynomial map f : K n −→ K. The regular functions on V form a K–algebra, which we denote by K[V ]. It is easy to see, see Exercise 2.2.27, that K[V ] is isomorphic to K[x1 , . . . ,xn ]/I (V ). The ring K[V ] is called the coordinate ring, or ring of regular functions of V . (2) Let V be an irreducible algebraic set. A rational function on V is an element of K(V ) := Q(K[V ]), the quotient field of K[V ]. Note that a rational function on V in general is not a genuine function on V : if f = hg , then, in general, the value of f at a point p with h(p) = 0 is not defined. Lemma 2.2.22. (1) Let V be an irreducible algebraic set, and p a point of V . Let mp be the maximal ideal of K[V ] of regular functions of V vanishing at the point p. Then K[V ]mp are those rational functions f on V which have a well-defined value at p. (2) More generally, let W ⊂ V be an irreducible subset of V , and p ⊂ K[V ] induced by I (W ). Elements of K[V ]p are those rational functions on V which have a well defined value on a nonempty Zariski open subset of W .

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(3) Let f 6= 0 be an element of K[V ]. Then elements of K[V ]f have a well defined value on V \ V (f ). Proof. Elements f in K[V ]mp are of type f = hg with h ∈ / mp . In particular h(p) 6= 0. g(p) So f (p) = h(p) is a well defined element of K. The other two statements are proved in a similar way. Now about localizations for not necessarily irreducible V . First an example. Example 2.2.23. Consider the coordinate axes V (x · y), so K[V ] = K[x,y]/(xy). Let p = (x). Then all elements which are not divisible by x are units in K[V ]p . In particular y is a unit. Hence x = 0 in K[V ]p , so that K[V ]p consists of the rational functions on the y–axis. This was an example for the following theorem, whose proof is left as an exercise. Theorem 2.2.24. Let X = X1 ∪ . . . ∪ Xr be an irreducible decomposition of an algebraic set X, and Y be an irreducible subset of X, defined by the prime ideal p. Then K[X]p ∼ = ⊕Y ⊂Xi K[Xi ]p . In particular, if pi is the ideal of Xi then K[X]pi ∼ = K(Xi ), the ring of rational functions on Xi . Remark 2.2.25. Let V be an algebraic set, with coordinate ring K[V ]. Then for any set F ⊂ K[V ] we can define the zero set V (F ) which is a subset of V , and for any subset A ⊂ V we can define the ideal of functions vanishing on A. All the properties of 2.2.2 hold in this more general case, and the proofs are literally the same. Furthermore, the Nullstellensatz holds in this more general case, and follows from the ordinary Nullstellensatz. The proofs are left as Exercise 2.2.27.

Exercises 2.2.26. (1) Let (a1 , . . . ,an ) ∈ K n be a point. Show that (x1 − a1 , . . . ,xn − an ) is a maximal ideal of K[x1 , . . . ,xn ]. (Hint: Consider the map K[x1 , . . . ,xn ] −→ K : f 7→ f (a1 , . . . ,an )).

(2) Let m ⊂ K[x1 , . . . ,xn ] be a maximal ideal. Prove that the following two statements are equivalent. (a) V (m) 6= ∅.

(b) There exists (a1 , . . . ,an ) ∈ K n such that m = (x1 − a1 , . . . ,xn − an ). (3) Let V ⊂ K n be an algebraic set. Prove that there is a one-one correspondence, points of V ←→ maximal ideals of K[V ], where K[V ] = K[x1 , . . . ,xn ]/I (V ). The point (a1 , . . . ,an ) ∈ V corresponds to the maximal ideal generated by the classes of x1 − a1 , . . . ,xn − an . 2.2.27. (1) Show that K[V ] is isomorphic to K[x1 , . . . ,xn ]/I (V ). (2) For a subset F ⊂ K[V ], define V (F ) := {a ∈ V : f (a) = 0 for all f ∈ F }. V (F ) is called the zero set of F . Show that the properties of Lemma 2.2.2 also holds in this more general context. Show that the Nullstellensatz holds in the ring K[V ]. (3) Show that V is irreducible if and only if K[V ] is an integral domain.

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65

(4) Let a,b ∈ V . Show that there exists a regular function f ∈ K[V ] with f (a) 6= f (b). (Hint: Reduce to the case V = K n . ) 2.2.28. We proved hat V (I + J) = V (I) ∩ V (J). It is not true in general that I (V ∩ W ) = I (V ) + I (W ). Give a counterexample. 2.2.29. (1) Let I ⊂ K[x1 , . . . ,xn ] be an ideal, such that the vector space K[x1 , . . . ,xn ]/I has finite dimension s. Show that the set V (I) consists of at most s points. (Hint: Use the fact that an Artinian algebra is the direct sum of local Artinian algebras, see 1.4.26.) (2) Suppose moreover that I is radical. Show that V (I) consists of exactly s points. (3) Show that for a Noether normalization π : X → K k , with s the degree of the corresponding field extension K(x1 , . . . ,xk ) ⊂ K(X) = Q(K[X]), the number of points in a fiber is at most s. 2.2.30. (1) Prove that (x,y)∩(x,z)∩(y,z) is an irredundant primary decomposition of I = (xy,xz,yz). (2) Prove that (z,y 2 − x3 ) ∩ (x,y) = (yz,xz,y 2 − x3 ). 2.2.31. Prove that Study’s Lemma is a consequence of Hilbert’s Nullstellensatz. 2.2.32. Show that the Nullstellensatz follows from the weak Nullstellensatz by using the socalled Rabinowitch ` ´ trick which goes as follows. Take f ∈ I V (I) , and consider the polynomial ring K[x1 , . . . ,xn ,y]. Take the ideal J generated by I and yf − 1. Show that V (J) = ∅. Apply the weak Nullstellensatz to conclude that 1 ∈ J. Then use the fact that modulo J, the equality y = f1 holds. 2.2.33. Consider affine varieties, X = V (f1 , . . . ,fs ) ⊂ K n , Y = V (g1 , . . . ,gt ) ⊂ K m , where f1 , . . . ,fs ∈ K[x1 , . . . ,xn ], g1 , . . . ,gt ∈ K[y1 , . . . ,ym ]. Show that the Cartesian product X × Y is also an affine variety, as zero set of the ideal (f1 , . . . ,fs ,g1 , . . . ,gt ), where the fi and gi are interpreted as elements of K[x1 , . . . ,xn ,y1 , . . . ,ym ]. 2.2.34. Suppose m,n > 0. Prove that the Zariski topology of K n × K m is not equal to the product of the Zariski topologies of K n and K m . 2.2.35. Let V be an algebraic set. Let mk = (x1 , . . . ,xk ) ⊂ K[x1 , . . . ,xk ], and let m be a maximal ideal of K[V ]. A Noether normalization of K[V ]m , by definition is an inclusion K[x1 , . . . ,xk ]m k ⊂ K[V ]m for some k, such that K[V ]m is a finitely generated K[x1 , . . . ,xk ]m k –module. Show that after a general coordinate change, a Noether normalization of K[V ]m exists. 2.2.36. (1) Let V be an irreducible topological space, f : V −→ W be continuous. Show that f (V ) is irreducible. (2) Let V ⊂ W be topological spaces. Suppose that V is irreducible. Show that the closure V of V in W is also irreducible. 2.2.37. In this exercise we look at the Going-Up Theorem in the special case that p is a maximal ideal.

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(1) Let R ⊂ S be an integral extension of integral domains. Show that R is a field if and only if S is a field. (2) Let R ⊂ S be an integral extension of rings. Let m be a maximal ideal in R, and q ⊂ S be a prime ideal with q ∩ R = m. Prove that q is a maximal ideal.

Give geometric interpretations of these statements.

2.2.38. Let I,J be ideals in the polynomial ring K[x1 , . . . ,xn ]. Suppose I is radical. Prove that V (I : J) = V (I) \ V (J). 2.2.39. Let Y ⊂ V be algebraic sets. Show that I (Y ) = ∩p∈Y mp .

2.3

Maps between Algebraic Sets

In this section we look at the morphisms we allow between algebraic sets. As an algebraic set is defined by polynomial equations, it is a natural idea to allow between them only polynomial mappings. The simple case of regular functions, that is, regular maps V −→ K has already been considered in 2.2.21. Definition 2.3.1. Let V ⊂ K n and W ⊂ K m be affine varieties. A map ϕ : V −→ W is called a regular map between the affine varieties V and W if ϕ is the restriction of a polynomial map ϕ : K n −→ K m . Regular maps V −→ K are called regular functions. This means the following: suppose (x1 , . . . ,xn ) are coordinates on K n , and (y1 , . . . ,ym ) are coordinates on K m . Then there exist functions ϕ1 , . . . ,ϕm ∈ K[x1 , . . . ,xn ], such that the point with coordinates (x1 , . . . ,xn ) ∈ V
is sent to the point with coordinates (y1 , . . . ,ym ) = ϕ1 (x1 , . . . ,xn ), . . . ,ϕm (x1 , . . . ,xn ) ∈ W .

Definition 2.3.2. An isomorphism ϕ : V −→ W is a regular map which has a two-sided inverse. Two affine varieties V,W are called isomorphic if there exists an isomorphism ϕ : V −→ W between them. Of course, being isomorphic is an equivalence relation.

It is easy to check that the composition of two polynomial maps is a polynomial map. This in particular gives the following. Let ϕ : V −→ W be a regular map. Let f ∈ K[W ], ϕ

f

that is, f : W −→ K be a regular function. Then the composition f ◦ ϕ : V −→ W −→ K is a regular function too. Hence we get a map ϕ∗ : K[W ] −→ K[V ];

f 7→ ϕ ◦ f

which one checks to be a K–algebra homomorphism. Note that if moreover ψ : W −→ X, we have the property (ψ ◦ ϕ)∗ = ϕ∗ ◦ ψ ∗ . So we assign to each regular map a K–algebra homomorphism. There is a converse to this. Theorem 2.3.3. Suppose a K–algebra homomorphism α : K[W ] −→ K[V ] is given. Then there is a unique regular map ϕ : V −→ W with ϕ∗ = α. Proof. We write K[V ] = K[x1 , . . . ,xn ]/I (V ),

K[W ] = K[y1 , . . . ,ym ]/I (W ).

2.3 Maps between Algebraic Sets

67 α

The composition K[y1 , . . . ,ym ] −→ K[W ] −→ K[V ] we also denote by α. Consider the “coordinate” functions y1 , . . . ,ym ∈ K[y1 , . . . ,ym ]/I (W ). Then α(yi ) = ϕi (x1 , . . . ,xn ) ∈ K[V ] for some ϕi . Take lifts ϕi ∈ K[x1 , . . . ,xn ] of ϕi . Then ϕ := (ϕ1 , . . . ,ϕm ) is a polynomial map ϕ : K n −→ K m . As both α and ϕ∗ are K–algebra maps, and they agree on yi , it follows that they agree on all polynomials in the yi . Hence α = ϕ∗ : K[y1 , . . . ,ym ] −→ K[V ]. So, uniqueness is clear. As different lifts differ by elements in I (V ) the restriction of ϕ to V is well-determined. We need to show ϕ(V ) ⊂ W. Take p ∈ V , and consider ϕ(p). As W = V (I (W )) by 2.2.5 it suffices to show f (ϕ(p)) = 0 for all f ∈ I (W ). Now f (ϕ(p)) = f ◦ϕ(p) = ϕ∗ (f )(p) = α(f )(p). As α : K[W ] −→ K[V ], it maps all functions in I (W ) to zero. Hence α(f ) is the zero function, in particular α(f )(p) = 0. The following follows immediately. Corollary 2.3.4. Algebraic sets V and W are isomorphic if and only if K[V ] and K[W ] are isomorphic as K–algebras. As K–algebra homomorphisms between affine rings correspond to regular maps of the corresponding varieties, one would like to translate certain properties of K–algebra homomorphisms into properties of the corresponding regular map. The following theorem gives a first example of this principle. Theorem 2.3.5. Let algebraic sets V,W , algebraic subsets i : X ֒→ V , j : Y ֒→ W , and a regular map ϕ : V −→ W be given. Then
(1) ϕ(X) ⊂ Y ⇐⇒ ϕ∗ I (Y ) ⊂ I (X),
(2) the fiber ϕ−1 (Y ) is the zero set of the set ϕ∗ I (Y ) , (3) a regular map is continuous in the Zariski topology.

Proof. (1) As I (X) ⊃ I (V ), we get a canonical surjection i∗ : K[V ] −→ K[X] which is just restriction of regular maps. Similarly, we have a canonical surjection j ∗ : K[W ] −→ K[Y ]. Now if ϕ∗ (I (Y )) ⊂ I (X), we get an induced map ϕ∗ : K[Y ] −→ K[X], with i∗ ◦ϕ∗ = ϕ∗ ◦j ∗ By the previous theorem we get a map ϕ : X −→ Y , such that j◦ϕ = ϕ◦i. This exactly means that ϕ(X) ⊂ Y . If, on the other hand ϕ sends X to Y , that is, we have a map ϕ : X −→ Y , we get an induced map ϕ∗ : K[Y ] −→ K[X], so a commutative diagram ϕ∗ // K[V ] K[W ] j∗

i∗

 K[Y ]
It follows that ϕ∗ I (Y ) ⊂ I (X).

ϕ∗

 // K[X].

(2) We apply the first part to X = ϕ−1 (Y ). We get

ϕ∗ I (Y ) ⊂ I ϕ−1 (Y ) ,

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∗ so that, by taking zero sets, V ϕ I (Y ) ⊃ ϕ−1 (Y ). To

prove the other

inclusion, ∗ let p ∈ V ϕ I (Y ) . Then mp = I (p) ⊃ I V ϕ∗ I (Y ) ⊃ ϕ∗ I (Y ) . Therefore ϕ(p) ∈ Y follows from the first part applied to X = {p}. Hence p ∈ ϕ−1 (Y ) as was to be proved. (3) From the second part it follows that inverse images of Zariski closed sets are Zariski closed. Thus, ϕ is continuous. Example 2.3.6. Let V = V (I), where I = (y 2 − x) ⊂ K[x,y]. Consider the projection ϕ : V −→ W = K, the projection on the x–axis. Then the map ϕ∗ is just the inclusion map ϕ∗ : K[x] ⊂ K[x,y]/(y 2 −x). We take the subset Y of W = K defined by x(x−1) = 0, so consisting of the points 0 and 1. Hence the fiber is defined by the ideal (x(x−1),y 2 −x)) in K 2 so consists of three points, namely (0,0),(1,1) and (1, − 1).

0

1

Lemma 2.3.7. Let a regular map ϕ : V −→ W be given, and let ϕ∗ : K[W ] −→ K[V ] be the induced map on coordinate rings. Then ϕ is injective ⇐⇒ ϕ∗ is surjective. Proof. First suppose that ϕ∗ is surjective. Suppose that a,b ∈ V with ϕ(a) = ϕ(b). Then for all g ∈ K[W ], it follows that ϕ∗ (g)(a) = g ◦ ϕ(a) = g ◦ ϕ(b) = ϕ∗ (g)(b). Suppose a 6= b. Choose an f ∈ K[V ] with f (a) 6= f (b). This we can do by Exercise 2.2.27. Because ϕ∗ is surjective, there is a g ∈ k[W ] with f = ϕ∗ (g). This is a contradiction. For the converse, suppose ϕ is injective. Hence V can be viewed as a subset of W : V ⊂ W ⊂ K n . Hence I (V ) ⊃ I (W ), so that K[W ] = K[x1 , . . . ,xn ]/I (W ) −→ k[V ] = K[x1 , . . . ,xn ]/I (V ) is surjective. The corresponding statement with injective and surjective interchanged, is not true, by the following example which we have seen before. Example 2.3.8. Consider V = {(x,y) ∈ K 2 : xy − 1 = 0}, let W be the x–axis, and ϕ : V −→ W be the projection. Then ϕ∗ is the inclusion K[W ] = K[x] ⊂ K[V ] = K[x,y]/(xy − 1) = K[x,x−1 ]. However, ϕ is not surjective: it misses the point 0. Note that in this example, K[V ] is not a finitely generated K[W ]–module. If we insist however on this, the converse holds. The following therefore is a generalization of the fact that for a Noether normalization π : V −→ K r , the map π is surjective, and that the fibers consist of finitely many points. The proof is left as Exercise 2.3.17.

2.3 Maps between Algebraic Sets

69

Theorem 2.3.9. Let ϕ : V −→ W be a regular map, with ϕ∗ : K[W ] −→ K[V ] injective, and such that K[V ] is a finitely generated K[W ]-module. Then ϕ is surjective, and the fibers consist of finitely many points. Note that although in Example 2.3.8, the map ϕ is not surjective, it is at least “almost” surjective. To make this more precise, we need to understand localizations of affine rings. Lemma 2.3.10. Let V ⊂ K n be an algebraic set, with coordinate K[V ]. Take 0 6= f ∈ K[V ]. Then V \ V (f ) ⊂ V is homeomorphic in the Zariski topology to an algebraic set X, whose coordinate ring K[X] is the localization K[V ]f . Proof. Let K[V ] = K[x1 , . . . ,xn ]/I. Take a new coordinate y, and consider the set X := V (I,y · f − 1) ⊂ K n+1 . The coordinate ring of X is K[x1 , . . . ,xn ,y]/(I,y · f − 1) which is as K–algebra, by sending y to f1 , isomorphic to K[V ]f . The projection π : X −→ K n on the first n coordinates is easily seen to be a one to one map of X onto its image V r V (f ). The map π : X −→ V r V (f ) is regular, hence continuous. To show that π is a homeomorphism, we have to show that a closed subset in X is sent to a closed subset in V r V (f ). It suffices to show this for irreducible sets Y in X, corresponding to a prime ideal pf in K[V ]f . By 1.3.15 pf ∩ K[V ] = p is a prime ideal in K[V ], and conversely a prime ideal in K[V ] with f ∈ / p extends to a prime ideal in K[V ]f . We claim that the image of V (pf ) is equal to V (p) r V (f ), which would prove the lemma. Indeed, let a ∈ V (p) r V (f ). Then for all g ∈ p, and all k ∈ N we have fgk (a) = 0, as g(a) = 0, and f (a) 6= 0. As pf consists of such elements it follows that a ∈ V (pf ). On the other hand, if a ∈ V (pf ), then fgk (a) = 0 for all g ∈ p and k ∈ N. Hence g(a) = 0, and therefore a ∈ V (p). This proves the lemma. Definition 2.3.11. Consider irreducible affine varieties, and ϕ : V −→ W a regular map. Then ϕ is called dominant if ϕ(V ) contains a nonempty Zariski open subset U . So ∅ 6= U ⊂ ϕ(V ). Theorem 2.3.12. Let V,W be irreducible affine varieties, and ϕ : V −→ W be a regular map. Then the following conditions are equivalent: (1) ϕ∗ : K[W ] −→ K[V ] is injective. (2) ϕ is dominant. (3) ϕ(V ) = W . Proof. The implication (2) =⇒ (3) is easy, see Exercise 2.2.36. (3) =⇒ (1). Take f,g ∈ K[W ]. Then we have the following two equivalences. f = g ⇐⇒ f|ϕ(V ) = g|ϕ(V ) ⇐⇒ f ◦ ϕ = g ◦ ϕ. The second equivalence is clear. For the first, the only not obvious implication is ⇐=. As regular functions are continuous, it follows that (f − g)−1 (0) is closed. By assumption, this set contains ϕ(V ), and therefore contains ϕ(V ) = W . Hence f = g on the whole of W.

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The implication (1) =⇒ (2) is the “difficult” case. We have to prove that W r V (f ) ⊂ ϕ(V ) for some 0 6= f ∈ K[W ]. The idea is to find a nonzero f ∈ K[W ] ⊂ K[V ] and an r such that there exists an inclusion K[W ]f [x1 , . . . ,xr ] ⊂ K[V ]f , such that K[V ]f is a finitely generated K[W ]f [x1 , . . . ,xr ]-module. Indeed, this inclusion gives   • a finite map, hence surjective map ϕ e : V \ V (f ) −→ W r V (f ) × K r ,

  • a projection map π : W rV (f ) ×K r −→ W rV (f ), which is obviously surjective,

such that ϕ = π ◦ ϕ. e Hence, the image of ϕ would contain W r V (f ), which is a nonempty Zariski open subset of W . Now we will find the f and r. For simplicity we write R = K[W ], and S = K[V ]. By assumption we have an inclusion R ⊂ S. Consider the following diagram: R ֒→ S ↓ ↓ E = Q(R) ֒→ SRr{0} ⊂ Q(S).

Obviously, SRr{0} is a finitely generated E–algebra, and E is a field. In this situation we can apply the Noether Normalization Theorem, see Theorem 2.2.9. There exist x1 , . . . ,xr ∈ SRr{0} such that E[x1 , . . . ,xr ] ⊂ SRr{0} , and SRr{0} is finitely generated as E[x1 , . . . ,xr ]–module. By clearing denominators, we can achieve that the xi are elements of S, and SRr{0} is still a finitely generated E[x1 , . . . ,xr ]–module. Now consider R[x1 , . . . ,xr ] ⊂ S. This need not be a finitely generated extension. But we can do the following. Take generators α1 , . . . ,αs of SRr{0} as E[x1 , . . . ,xr ]–algebra. Again we may assume that in fact αi in S. The elements αi satisfy an integral equation with coefficients in E[x1 , . . . ,xr ] = Q(R)[x1 , . . . ,xr ]. Let f be the product of all the denominators that appear as coefficients in all such equations. Therefore, over Rf [x1 , . . . ,xr ] every αi is integral. We therefore forced that the extension Rf [x1 , . . . ,xr ] ⊂ Sf is integral. This was our goal. Theorem 2.3.13. Consider an ideal I ⊂ K[x1 , . . . ,xn ,y1 , . . . ,ym ], and let V = V (I) ⊂ K n+m . Let p : K n+m −→ K n be the projection on the first n factors. Then p(V ) = V (I ∩ K[x1 , . . . ,xn ]). The closure is taken here in the Zariski topology. Proof. To show the inclusion p(V ) ⊂ V (I ∩ K[x1 , . . . ,xn ]), it suffices to show p(V ) ⊂ V (I ∩ K[x1 , . . . ,xn ]), as the right hand side is closed. So let a ∈ p(V ). Hence there exists a b ∈ K m with (a,b) ∈ V . Let f ∈ I ∩ K[x1 , . . . ,xn ] then f (a,b) = 0. But f does not depend on the variables y1 , . . . ,ym at all, and therefore f (a) = 0. This holds for all f ∈ I ∩ K[x1 , . . . ,xn ] so that the inclusion ⊂ follows.

2.3 Maps between Algebraic Sets

71

For the converse inclusion consider K[x1 , . . . ,xn ]/(I ∩ K[x1 , . . . ,xn ]) ⊂ K[x1 , . . . ,xn ,y1 , . . . ,ym ]/I, so that p(V ) is indeed given by the zero set of the ideal I ∩ K[x1 , . . . ,xn ] by 2.3.12. Remark 2.3.14. Let ϕ = (ϕ1 , . . . ,ϕm ) : V −→ W ⊂ K m be a regular map, V = V (I). Consider the graph of ϕ: Graph(ϕ) = {(x,ϕ(x)) : x ∈ V } ⊂ V × W ⊂ V × K m . Then the zero set of (y1 − ϕ1 , . . . ,ym − ϕm ) ⊂ K[V ] ⊗ K[W ] is the graph of ϕ. The graph of ϕ is canonically isomorphic to V : in the
ideal (y1 − ϕ1 , . . . ,ym − ϕm ) one eliminates all y–variables, so that K[V ] ⊗ K[W ] /(y1 − ϕ1 , . . . ,ym − ϕm ) is isomorphic to K[V ]. Note that with this identification the map of K–algebras
K[W ] −→ K[V ] ⊗ K[W ] /(y1 − ϕ1 , . . . ,ym − ϕm )

gives the map Graph(ϕ) −→ W , which maps (x,ϕ(x)) to ϕ(x). We come to the conclusion that every map ϕ : V −→ W can be viewed as the restriction of a projection mapping. This is a point of view which is sometimes useful. For example its local version, see 3.4.43 will be used in the proof of the Finite Mapping Theorem, see 6.3.5. Example 2.3.15. Consider the regular map ϕ : K −→ K 2 given by ϕ(t) = (t2 ,t3 ). Let the coordinates of K 2 be x,y. Then the graph of ϕ lies in K 3 and is the zero set of the ideal I = (x − t2 ,y − t3 ) So the closure of the image of f is the zero set of the ideal I ′ = (x − t2 ,y − t3 ) ∩ K[x,y] by 2.3.14 and 2.3.13. One shows that I ′ = (y 2 − x3 ). The proof of this is contained in Example 1.5.6. Finally, we give a counterexample to the Going-Down Theorem for nonnormal rings. Example 2.3.16. Let K be an algebraically closed field. Consider the map ϕ : K2 (t,s)

−→ K 3 7−→ (x,y,z) = (t2 − 1,t3 − t,s).

p′

p X

X′

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Then one shows that the image of ϕ is exactly given by y 2 − x2 (x + 1) = 0. Moreover, via
ϕ∗ , the module S := K[t,s] is a finitely generated R := K[x,y,z]/ y 2 −x2 (x+1) –module. Indeed, from the definition of the map it follows
that t2 − 1 − x = 0, hence t is integral. Notice, moreover, that K[x,y,z]/ y 2 − x2 (x + 1) is not a normal ring; the element t = xy is an integral element. We now show that the Going-Down Theorem 1.5.26 does not hold for the ring extension R ⊂ S. Consider the prime ideal P := (s − t) in S (in the picture X = V (P) ⊂ K 2 ), and the prime ideal p := P ∩ R in R (in the picture X ′ = V (p) ⊂ K 3 ). One easily calculates that p = (x − z 2 + 1,y − xz). Now we claim that P is the unique prime ideal in S which lies over p. This should be clear geometrically, but let us give a precise algebraic argument. Indeed, such a prime ideal P contains at least t2 − 1 − s2 + 1 = t2 − s2 , corresponding to x − z 2 + 1, the first generator of P. Hence, either t − s ∈ P, or t + s ∈ P. If t − s ∈ P, we are done. Therefore, suppose t + s ∈ P. As y − xz ∈ p, it follows that t3 − t − s(t2 − 1) = 2t3 − 2t + (1 − t2 )(s + t) ∈ P. Hence P is either (t − 1,s + 1) or (t,s) or (t + 1,s − 1). This cannot be the case because these are all maximal and P is not, see Exercise 2.2.37. Now consider the maximal ideal q = (x,y,z − 1) of R, corresponding to the point p′ = (0,0,1) in the picture. There are two points in the preimage of p′ = (0,0,1) namely (1,1) and p = (−1,1). In particular the maximal ideal Q : (t + 1,s − 1) corresponding to p is such that Q ∩ R = q. However, P 6⊂ Q, as otherwise p ∈ X. So we conclude that the statement of the Going-Down Theorem does not hold in this case.

Exercises 2.3.17. Prove Theorem 2.3.9. (Hint: Either use Theorem 2.3.5 to describe the fibers of y, and Exercise 2.2.37, or use the Graph Construction 2.3.14 and the Projection Theorem 2.2.8.) 2.3.18. (1) Consider the map f : K 2 −→ K 4 given by f (s,t) = (s,st,t2 ,t3 ) = (x,y,z,u). Show that the ideal of the image is equal to I ′ = (y 2 − zx2 ,yz − ux,z 2 x − uy,u2 − z 3 ). (Hint: Show that any g is modulo I ′ equivalent to ay + bu + c with a,c ∈ K[x,z] and b ∈ K[x].) (2) Consider the map f : K −→ K 3 , f (t) = (t3 ,t4 ,t5 ) = (x,y,z). Show that the image of f is given by the zero set of the ideal (y 2 − xz,x3 − yz,z 2 − x2 y).

(3) Use a computer algebra system, for example [Singular 2000] to check the results in (1) and (2).

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Basics of Analytic Geometry

In this chapter we start studying local analytic geometry, that is, the zero sets of analytic functions in a (small) neighborhood of a point. To see why we want to do so, we look at the affine hypersurface in C 2 defined by f (x,y) = y 2 − x2 (x + 1) = 0. In a small neighborhood U of (0,0) we see two “parts”, or components of f (x,y) = 0.

U

So we would like to say that the “singularity” consists of two components. But how small should we take U ? The answer to this question is: as small as we need to. To make a precise notion, we define the notion of a germ of a space. A germ of a topological space X at a point p by definition is an equivalence class of open neighborhoods of p. But f , considered as element of C [x,y] is irreducible (Exercise 1.4.34), but reducible if considered as an element of C {x,y}. Here C {x,y} is the ring of power series which converge in some neighborhood of 0.1 Indeed, we have the following factorization in C {x,y} √ √

f (x,y) = y − x x + 1 y + x x + 1 . √ So we see that in a neighborhood on which √ x + 1 converges, the zero set √ f = 0 decomposes into two parts, namely {(x,y) : y = x x + 1} and {(x,y) : y = −x x + 1}. As we assume that the reader heard a course on holomorphic functions in one variable, we will, in Section 3.1, treat only the basic facts on holomorphic functions in several variables. Many of the statements and the proofs are similar or direct generalizations of the one variable case. For polynomials f ∈ C [x1 , . . . ,xn ] one of the main invariants is the degree. It can be interpreted (in case f is reduced) as the number of intersection points of the set {p ∈ C n : f (p) = 0} with a general line. By applying the Noether Normalization Theorem, one may assume that this line is given by x1 = a1 , . . . ,xn−1 = an−1 . In case of convergent power series f , its replacement is the multiplicity or order of the power series, and is defined as the lowest order term of f . Assuming that f converges on the (small enough) neighborhood U of 0, the multiplicity can be interpreted as the maximal number of intersection points of f = 0 with a general line L in U , in case f is reduced. This 1

The neighborhood might depend on the power series.

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3 Basics of Analytic Geometry

statement follows from the so-called continuity of roots. So for example the element f (x,y) = y 2 − x2 (x + 1) has degree three considered as element of C [x,y], and has multiplicity two considered as element of C {x,y}. There is a substitute for the Noether normalization for the case f ∈ C {x1 , . . . ,xn }, so that we can always assume that the line L from above points in the xn direction. In terms of the power series f of multiplicity b, this is equivalent to saying that the term xbn occurs with nonzero coefficient in the power series expansion of f . In this case one says that f is regular of order b in xn . The replacement for division with remainder, which played an important role in the proof of Study’s Lemma, or more generally the Nullstellensatz, is the Weierstraß Division Theorem. The Weierstraß Division Theorem says the following: Let a convergent power series f ∈ C {x1 , . . . ,xn } be regular of order b in xn , and a further power series g ∈ C {x1 , . . . ,xn } be given. Then there exist uniquely defined q ∈ C {x1 , . . . ,xn } and r ∈ C {x1 , . . . ,xn−1 }[xn ], where r has, considered as polynomial in xn , degree less than b, such that g = q · f + r. The similarity with ordinary division with remainder is obvious. It follows from the Weierstraß Division Theorem, that the C {x1 , . . . ,xn−1 }–module C {x1 , . . . ,xn }/(f ) is finitely generated.2 As the ring C {x1 , . . . ,xn−1 } is local, it even follows by an application of Nakayama’s Lemma, that the Weierstraß Division Theorem follows from the fact that C {x1 , . . . ,xn }/(f ) is a finitely generated C {x1 , . . . ,xn−1 }–module. In Section 3.2 we will prove a weak generalization of this seemingly weaker statement. If one looks closely at the proof, we will “only” show the existence of q and r where r has degree less than or equal to b. It turns out that in order to show the convergence of q and r, it is much easier to prove this last statement. In Section 3.3, we give applications of the Weierstraß Theorems. To mention the most important ones, we show the Implicit and Inverse Function Theorems and Newton’s Lemma (a “cooked-up” version of the Implicit Function Theorem). Furthermore, we show that the (formal) power series ring is Noetherian and factorial, prove the Noether Normalization Theorem, and Hensel’s Lemma. With this knowledge at hand we can start the study of germs of analytic spaces in Section 3.4. The main theorem, and the most difficult in this section is the Nullstellensatz. In the case of germs of analytic spaces, this theorem cannot be interpreted as a statement on maximal ideals. We will prove the analogous statement to Corollary 2.2.10. That is, let a germ of an analytic space (X,x) be given for which the ideal of (X,x) is prime, and let π : (X,x) −→ (C k ,y) be a Noether normalization. Then π is surjective. As in the affine case, the Nullstellensatz follows from this surjectivity. An essential ingredient in the proof of the surjectivity of π is the Theorem of Finiteness of Normalization 1.5.19. It will be used to describe the properties of a general projection of an irreducible germ (X,x) onto a hypersurface singularity. The precise properties of this projection of X onto a hypersurface are gathered in the Local Parametrization Theorem. This is a very important theorem, as it, in many cases, allows the reduction of general statements on germs of analytic spaces to the case of hypersurfaces, which are sometimes easier to handle. The Local Parametrization Theorem will be used several times in the next chapter. 2

This module is even free.

3.1 Holomorphic Functions of Several Complex Variables

3.1

75

Holomorphic Functions of Several Complex Variables

Before going on with geometry, we need some basic facts on holomorphic functions of several complex variables. The facts we mention here are either proved, or the proof can be easily generalized from the case of a function in one variable. In this case, no proof is given. Definition 3.1.1. Let n ∈ N, and νi , for i = 1, . . . ,n be nonnegative integers, x = (x1 , . . . ,xn ) ∈ C n . We introduce the multi index notation ν := (ν1 , . . . ,νn ); xν :=

n Y

i=1

xνi i ; |ν| :=

n X

νi .

i=1

The number |ν| is called the degree of ν. Let r = (r1 , . . . ,rn ) ∈ Rn+ , ri > 0 for i = 1 . . . ,n. The set P r := {x ∈ C n : |xi | ≤ ri for i = 1, . . . ,n} is called the closed polydisc (or polycylinder) with center {0} and polyradius r. Similarly one defines open polydiscs Pr , and polydiscs with center p ∈ C n .

Definition 3.1.2. Let U ⊂ C n be a domain3 , f : U −→ C a function, p ∈ U . Then f is called complex differentiable in p, if there exists an open neighborhood V ⊂ U of p and functions ∆1 , . . . ,∆n : V −→ C which are continuous in p such that for all x ∈ V f (x) = f (p) +

n X i=1

(xi − pi )∆i (x).

f is called holomorphic in U if f is complex differentiable in p for all p ∈ U . In this case it follows without much difficulty, by reducing it to the one-dimensional ∂f (p). case, that the values ∆i (p) are uniquely determined. Those values we denote by ∂x i Furthermore, sums, products, and quotients (whenever defined) of holomorphic functions are holomorphic. See Exercise 3.1.18. It is known that one-variable functions are holomorphic if and only if they are locally represented by their Taylor series. To have the analogous statement for the multi-variable case, we need some definitions. Definition 3.1.3. (1) Let aν for ν ∈ Nn be complex numbers. The expression4 ∞ X

ν=0

aν (x − p0 )ν

is called a formal power series in the variables x1 , . . . ,xn around p0 . With the obvious addition and multiplication the set of formal power series is a ring, and even an C –algebra. The ring of formal power series around p0 = 0 is denoted by C [[x1 , . . . ,xn ]] or sometimes C [[x]] for short, and is called the formal power series ring. 3 4

This means that U is an open and P connected subset. This is an alternative notation for ν∈Nn aν (p − p0 )ν .

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3 Basics of Analytic Geometry

P∞ (2) Let p ∈ C n , and c ∈ C . One says that the formal power series ν=0 aν (x − p0 )ν converges in p with limit c if for all ε > 0 there exists a finite set Iε ⊂ Nn such that for all finite sets I with Iε ⊂ I ⊂ Nn X aν (p − p0 )ν − c < ε. ν∈I

P ν Note that P ν∈I aν (p − p0 ) makes sense because the summation is over a finite set. ∞ In case that ν=0 aν (x − p0 )ν converges in p the number c is uniquely determined, and is called the value of the formal power series at p. We, therefore, simply write P ∞ ν ν=0 aν (p − p0 ) = c. P∞ ν The P∞ formal powerν series ν=0 aν (x − p0 ) is called absolutely convergent in p if ν=0 | aν (x − p0 ) | is convergent. An absolutely convergent power series is in fact convergent.

(3) P Let U ⊂ C n , and f : U −→ C be a function. One says that the formal power series ∞ aν (x − p0 )ν converges on U with limit f if for all p in U the formal power ν=0 P ∞ series ν=0 aν (x − p0 )ν converges in p with limit f (p).

n (4) Let again P∞U ⊂ C , andν f : U −→ C be a function. One says that the formal power series ν=0 aν (x − p0 ) converges uniformly to f on U if for all ε > 0 there exists a finite set Iε ⊂ Nn such that for all finite sets I with Iε ⊂ I ⊂ Nn and for all p ∈ U X ν aν (p − p0 ) − f (p) < ε. ν∈I

The difference to ordinary convergence is that in case of ordinary convergence, the set Iε might depend on p, whereas in case of uniform convergence, the set Iε is not allowed to depend on p.

(5) Let U ⊂ C n be a domain, f : U −→ C be a function. f is called analytic if for all p0 ∈ U , there exists a neighborhood V = V (p0 ) of p0 in U , and a power series P ∞ ν ν=0 aν (x − p0 ) which converges on V to f .

Let p ∈ C n be fixed. It will turn out that the formal power series that converge on a neighborhood of p (the neighborhood might depend on the function) form a ring, called the convergent power series ring On,p at p. The ring of convergent power series at 0 is denoted by C {x} or C {x1 , . . . ,xn } or also often by On . instead of the “correct” On,0 .

(6) Let 0 6= f ∈ C [[x1 , . . . ,xn ]] be a formal power series. We can write f = fm + fm+1 + . . . where fk are homogeneous polynomials of degree k in x1 , . . . ,xn , and fm 6= 0. Then we define mult(f ) := ord(f ) := m and call it the order or multiplicity of f . One obviously has that ord(f · g) = ord(f ) + ord(g). (7) More generally, let λ = (λ1 , . . . ,λn ) ∈ Nn+ and let 0 6= f ∈ C [[x1 , . . . ,xn ]] be a formal power series. We can write f = fm + fm+1 + · · ·

3.1 Holomorphic Functions of Several Complex Variables

77

where fk are quasi-homogeneous polynomials (with respect to λ) of weighted degree k and fm 6= 0. Then we define w-ord(f ) := m and call it the weighted order of f . The first thing we want to show is that analytic functions are holomorphic. In order to show this, we need the following lemma. P∞ Lemma 3.1.4. Consider a formal power series ν=0 aν xν , which converges at p ∈ C n . Let p = (p1 , . . . ,pn ) and suppose that pi 6= P0∞for all i. Let the polyradius r be defined by ri = |pi |. Then the formal power series ν=0 aν xν converges absolutely and uniformly on all compact subsets of the open poly-disc Pr . P Proof. From the convergence of the series aν pν it follows that the set {aν pν } is bounded, that is, there exists an M ∈ R with |aν pν | = |aν |rν < M . Consider a real number q with 0 < q < 1. By definition of Pqr for all x ∈ Pqr it follows that |xν | < q |ν| rν . Hence for all x ∈ Pqr |aν xν | < |aν |q |ν| rν < M · q |ν| . But M

X

q

|ν|

=M

ν



1 1−q

n

,

see Exercise 3.1.19. The set Nn is countable, hence there exists n a bijection α : N −→ P∞ 1 n α(k) N . Put bk (x) := aα(k) x . Hence, k=0 |bk (x)| ≤ M 1−q , so that this series is P∞ increasing and bounded. It follows that k=0 bk (x) converges absolutely and uniformly on Pqr so that ∀ε > 0 there exists nε ∈ N with

∞ X

k=nε

|bk (x)| < ε for all x ∈ Pqr .

Set Iε := {α(i) : i = 0, . . . ,nε } and choose I with I ⊃ Iε . Then ∞ ∞ X X X X ν ν |aν x | − |bk (x)| − |aν x | = |bk (x)| ν=0 ν∈I k=0 k∈α−1 (I) =

X

k∈α / −1 (I)

|bk (x)| ≤

∞ X

k=nε +1

|bk (x)| < ε.

This shows that the series converges absolutely and uniformly in Pqr . We still have to show that the series converges on all compact subsets K in Pr . But Pqr for 0 < q < 1 is an open cover of K. As K is compact we have a finite subcover Pq1 r , . . . Pqs r . Let q = max{q1 , . . . ,qs }. Hence K ⊂ Pqr , and thus the series converges uniformly on K. Remarks 3.1.5. (1) Let f be analytic on a neighborhood of 0. As f is analytic, it follows from the proof of 3.1.4 that there exists a δ > 0 such that X kf kδ := |fα |δ |α| < ∞. α

78

3 Basics of Analytic Geometry The fact that f is analytic is equivalent to this property. We could rephrase this P as follows. For all δ > 0 we put a seminorm5 on C [[x1 , . . . ,xn ]]. Let f = fα xα ∈ C [[x1 , . . . ,xn ]] be a formal power series. Then we put for δ > 0 X kf kδ := |fα |δ |α| ∈ R+ ∪ {∞}. α

It follows from the proof of Lemma 3.1.4 that f is convergent exactly when there exists a δ > 0 with kf kδ < ∞. (2) Note that if kf kδ < ∞, and 0 < t < 1, then kf ktδ < ∞. If, moreover, f ∈ (x1 , . . . ,xn ) an easy estimate shows that kf ktδ ≤ t · kf kδ . P∞ k (3) Let f = k=0 fk xn be a convergent power series, with fk ∈ C [[x1 , . . . ,xn−1 ]]. From the absolute convergence of f in some domain in C n it follows that fi are absolute convergent in some domain of C n−1 , see Exercise 3.1.23. It follows that fk ∈ C {x1 , . . . ,xn−1 }. Theorem 3.1.6. Let U ⊂ C n be open, f : U −→ C be analytic. Then f is holomorphic. Proof. Without loss of generality, it suffices to show that f is holomorphic in 0. Let P ν a x be a power series that converges on an open neighborhood V of 0 to f . We ν ν write X X aν xν = a0...0 +x1 aν x1ν1 −1 xν22 · · · xνnn ν

ν1 6=0

+x2

X

ν,ν2 6=0 ν1 =0

+xn

aν x2ν2 −1 · · · xνnn

X

.. . aν xνnn −1

ν,νn 6=0 ν1 =···=νn−1 =0

= a0...0 +x1 ∆1 + . . . + xn ∆n . P Take a point p = (p1 , . . . ,pn ) for which ν aν pν is absolutely convergent. Such a point exists by Lemma 3.1.4. We P may assume that all the pi are nonzero. It follows from the absolute convergence of ν aν pν that each of the sub series pi ∆i (p) are absolute convergent. As pi 6= 0, the series ∆i (p) are convergent. Applying Lemma 3.1.4, it follows that the ∆i for i = 1, . . . ,n converge absolutely and uniformly on small neighborhoods of 0. From a standard theorem in analysis it then follows that the ∆i are continuous. Theorem 3.1.7 (Osgood’s Lemma). Let U be an open set of C n , and f : U −→ C be a continuous function. The following conditions are equivalent. (1) f is analytic. (2) f is holomorphic. 5

A seminorm is a map C [[x1 , . . . ,xn ]] → R ∪ {∞} satisfying all conditions of a norm (with the obvious interpretation a + ∞ = ∞).

3.1 Holomorphic Functions of Several Complex Variables

79

(3) f is holomorphic in each variable. Proof. We already showed that (1) implies (2) in Theorem 3.1.6. As (2) implies (3) is trivial, we only have to show (3) implies (1). Take a point p ∈ U . Without loss of generality, we may assume that p = 0. Let P r be a closed poly-disc contained in U for r = (r1 , . . . ,rn ). Take a point (ζ1 , . . . ,ζn−1 ) with |ζi | ≤ ri . The function fn (xn ) := f (ζ1 , . . . ,ζn−1 ,xn ) by assumption is holomorphic in xn . By the Cauchy integral formula in one variable we therefore have for all xn with |xn | < rn Z f (ζ1 , . . . ,ζn−1 ,ζn ) 1 dζn . f (ζ1 , . . . ,ζn−1 ,xn ) = 2πi ζn − xn |ζn |=rn

We can do the same trick with the penultimate variable, Z 1 f (ζ1 , . . . ,ζn−1 ,xn ) f (ζ1 , . . . ,ζn−2 ,xn−1 ,xn ) = dζn−1 2πi ζn−1 − x  n−1 2  Z |ζn−1 |=rn−1 Z f (ζ1 , . . . ,ζn−1 ,ζn ) dζn−1 1   dζn  =  2πi ζn − xn ζn−1 − xn−1 |ζn−1 |=rn−1

|ζn |=rn

Continuing like this we obtain



1 2πi

n

Z

|ζ1 |=r1



1   ζ1 − x1

Z

|ζ2 |=r2

f (x1 , . . . ,xn ) =   Z 1   · · ·  ζ2 − x2

|ζn |=rn







f (ζ)    dζn  dζn−1  · · ·  dζ1 . ζn − xn

For fixed (x1 , . . . ,xn ) the integrand is continuous on T := {(ζ1 , . . . ,ζn ) : |ζi | = ri for i = 1, . . . ,n}, which is a compact set. By Fubini n Z  f (ζ) 1 dζ1 · · · dζn . f (x1 , . . . ,xn ) = 2πi (ζ1 − x1 ) · · · (ζn − xn ) T

It follows from Exercise 3.1.19 that for fixed (x1 , . . . ,xn ) the series ∞

X xν1 · · · xνn 1 n 1 = (ζ1 − x1 ) · · · (ζn − xn ) ν=0 ζ1ν1 +1 · · · ζnνn +1 converges absolutely and uniformly on T . As f is continuous, it is bounded on T . Therefore, for fixed (x1 , . . . ,xn ) the function  νn ν ∞  xn f (ζ1 , . . . ,ζn ) X x1 1 f (ζ1 , . . . ,ζn ) ··· · = (ζ1 − x1 ) · · · (ζn − xn ) ζ1 . . . ζn ζ1 ζn ν=0 converges absolutely and uniformly on T . It is a standard theorem from analysis, that one may in such a case interchange summation and integration. We get

80

3 Basics of Analytic Geometry

f (x1 , . . . ,xn ) =

=



1 2πi

T

∞ X

ν=0

=

∞ X



f (ζ) dζ1 · · · dζn (ζ1 − x1 ) · · · (ζn − xn )

xν11 · · · xνnn ·



1 2πi

n Z T

f (ζ) dζ1 · · · dζn · · · ζnνn +1

ζ1ν1 +1

aν xν , with

ν=0

aν =

n Z

1 2πi

n Z T

f (ζ) dζ1 · · · dζn . · · · ζnνn +1

ζ1ν1 +1

Remarks 3.1.8. (1) It follows that the convergent power series is a ring. Indeed, if f,g ∈ C {x1 , . . . ,xn }, then choose an open neighborhood U of 0 on which f converges, and an open neighborhood V of 0 on which g converges. Then f and g are holomorphic on U ∩ V , so that f ± g and f · g are holomorphic on V , so define an analytic function. This fact could also have been proved directly, by using only Lemma 3.1.4. Note moreover that C {x1 , . . . ,xn } is a local ring with maximal ideal (x1 , . . . ,xn ). Indeed, for convergent power series f not in (x1 , . . . ,xn ) one has f (0) 6= 0. It follows that 1 f is also holomorphic in a neighborhood of 0, and therefore, by Osgood’s Lemma, again is a convergent power series. So the elements of the complement of (x1 , . . . ,xn ) are units, and it follows from 1.3.3 that the power series ring is local. (2) Let U ⊂ C n be open. A map F = (f1 , . . . ,fm ) : U −→ C n is called holomorphic if each of the fi is holomorphic. If F (U ) ⊂ W ⊂ C m , and G : W −→ C p is a holomorphic map, then the chain rule, see Exercise 3.1.22 gives that G ◦ F is holomorphic. This in particular gives the following. Let g1 , . . . ,gn ∈ C {y1 , . . . ,ym }, with gi (0) = 0. Let f ∈ C {x1 , . . . ,xn }. Then f (g1 , . . . ,gn ) ∈ C {y1 , . . . ,ym }. Theorem 3.1.9 (Identity Theorem). Let U ⊂ C n be a domain. Let f : U −→ C be a holomorphic function, such that the restriction of f to a nonempty open subset V ⊂ U is the zero map. Then f is identically zero on U . Remark 3.1.10. In the case n = 1 we have a stronger version of the Identity Theorem, as follows. Let U ⊂ C be a domain and f : U −→ C be a holomorphic function. Let {xk } be a convergent sequence, xk ∈ U and limk→∞ xk ∈ U . If f (xk ) = 0 for all k, then f is identically zero on U . Indeed, suppose lim xk = 0, and f not identically zero in a neighborhood of 0. Then f = xb · u, where u(0) 6= 0. Therefore, for p with |p| small, it follows that u(p) 6= 0, and therefore, f (p) 6= 0. The zeros of f therefore lie isolated. This is a contradiction, and therefore, f vanishes in a neighborhood of 0. One can reduce the proof of the Identity Theorem to the Identity Theorem in the one variable case. The idea is the same as in the proof of the maximum principle, which comes next. The proof of the Identity Theorem 3.1.9 is therefore left as Exercise 3.1.24.

3.1 Holomorphic Functions of Several Complex Variables

81

Theorem 3.1.11 (Maximum Principle). Consider a domain subset U ⊂ C n . Let f : U −→ C be a nonconstant holomorphic function. Then |f | does not have a maximum on U. In other words, if f : U −→ C takes its maximum absolute value at a point p in U , then f is constant. Proof. In case of a holomorphic function in one variable, this is well-known. We will prove the general statement by using the one-variable case and the Identity Theorem. Let p ∈ U be a point on which |f | attains its maximum. Consider a complex line L through p, and consider a small open ball B around p which lies inside U .

L

U B p

Then the restriction of f to B ∩ L is a holomorphic function of one variable. B ∩ L is a disc in L. By the maximum principle for the one variable case, the restriction of f to B ∩ L is constant. By varying the lines through p, we conclude that f is constant on B. By the Identity Theorem, f is constant on U . Definition 3.1.12. (1) A set or space X ⊂ C n is called locally analytic, if for any point p ∈ X, there exists an open subset V of p in C n , and finitely many holomorphic functions f1 , . . . ,fs defined on V such that X ∩ V = {x ∈ V : f1 (x) = . . . = fs (x) = 0}. (2) Let U be an open subset of C n . A subset X ⊂ U is called an analytic subset of U , if X is locally analytic, and closed in U . (3) Let X be an analytic subset of U , U ⊂ C n open. A function f : X −→ C is called holomorphic if for all x ∈ X there exists an open neighborhood V of x in C n such that f|V ∩X is the restriction of a holomorphic function on V . Remarks 3.1.13. (1) In this definition, we have to be more careful than in the affine case. Polynomials are always globally defined. But holomorphic functions are usually only defined in small open neighborhoods, so this is the reason why we have to choose open neighborhoods V . (2) The set X := {(x,y) : y = 0, Im(x) ≥ 0} in C 2

82

3 Basics of Analytic Geometry

is not locally analytic. Indeed, let V be a connected open neighborhood of {0}, and f be a function on V with V ∩ X ⊂ {(x,y) : f (x,y) = 0.} Then we can restrict f to the set V ∩ {y = 0}. By the Identity Theorem in one variable, f must vanish identically on {y = 0} ∩ V . As this holds for all f , it follows that X is not locally analytic. (3) In the definition of X to be an analytic subset of U , we need the condition X is closed in U to avoid phenomena as shown in the following picture.

U X Morally speaking, one wants an analytic subset of U , except for discrete points, to “reach” the boundary. Even more severely, without the closedness condition, if X is the shaded part of

X U

then X would be analytic. (4) Consider a sequence p1 ,p2 , . . . of nonzero points in C that converge to the origin. Then the union X of the points is not an analytic set. Indeed, any holomorphic function vanishing on U ∩X, for any open neighborhood U of 0 is the zero function. This follows from the Identity Theorem in one variable. It is well-known (see Remark 3.1.10) that the zeros of a holomorphic function in one variable lie isolated. The following is a generalization of this fact. Theorem 3.1.14. Let X ⊂ U be an analytic subset, with U connected. Suppose X 6= U . Then the closure6 of U \ X in U is U . Proof. The main idea of the proof is already contained in the proof of the fact that the sets in 3.1.13, (2) and (3) are not analytic. The claim is that X has no inner points of ◦

U , that is, we claim that the set X of inner points of X is empty. Suppose the converse, ◦

◦

that is, X contains a nonempty open subset X of U . We claim that X is closed. From 6

In the topological sense.

3.1 Holomorphic Functions of Several Complex Variables

83

◦

this it would follow from the connectedness of U that X is equal to U , in contradiction ◦

◦

to the assumption. So let us prove X is closed. Take a point p in the boundary of X in U . As X is closed in U , the point p is in X. Hence, there exists an open, connected neighborhood V of p and holomorphic functions f1 , . . . ,fs defined on V such that X ∩ V = {x ∈ V : f1 (x) = . . . = fs (x) = 0}. ◦

◦

As p is in the boundary of X, and p is in the open set V , it follows that V ∩ X 6= ∅.

p

V U o

X

Therefore, for all i = 1, . . . ,s the function fi is identically zero on the (nonempty) open ◦

subset V ∩ X, hence fi is, by the Identity Theorem, identically zero on V . Hence V ⊂ X, ◦

◦

and p ∈ X. So we proved that X is closed.

Theorem 3.1.15 (First Riemann Extension Theorem). Let U be an open connected subset of C n , and X ⊂ U an analytic subset. Consider f : U \ X −→ C

holomorphic. Suppose f is locally bounded, that is, for all p ∈ X there exists an open neighborhood V of p such that the restriction of f to V \ (V ∩ X) is bounded. Then there exists a holomorphic extension of f to U . More precisely, there exists a holomorphic function fe : U −→ C such that fe|U\X = f .

Proof. Let p ∈ X, and Bp an open ball around p. It suffices to extend f to a holomorphic function fep on Bp . Namely, if Bp ∩ Bq 6= ∅, and if fep ,feq are extensions of f to Bp , respectively Bq , they must agree on Bp ∩ Bq , as on W := Bp ∩ Bq ∩ (U \ X) they coincide with f , and W is not empty because the closure of U \ X in U is U , see Theorem 3.1.14. So we can use the Identity Theorem 3.1.9 to deduce that fp and fq coincide on Bp ∩ Bq . Consider a line L through p. Then L ∩ X is an analytic subset of X. As X contains no inner points, there exists a line not contained in X. As L ∩ X is an analytic subset of an open subset of L ∼ = C , hence locally given by the zero-set of one analytic function, the points of L ∩ X lie isolated. By an affine change of coordinates, we may assume that the line L is the xn –axis, given by x1 = . . . = xn−1 = 0, and p is equal to 0. Therefore, there exists an rn > 0 such that X ∩ {x = (x1 , . . . ,xn ) : x1 = . . . = xn−1 = 0, |xn | ≤ rn }

consists of just one point, namely 0. As the set {x = (x1 , . . . ,xn ) : x1 = . . . = xn−1 = 0, |xn | = rn } is compact7 , and is contained in the open set U \ X, it follows without much difficulty that there exists an open neighborhood W of 0 in C n−1 such that W × {|xn | = rn } ⊂ U \ X. So 7

It is a circle.

84 (3.1)

3 Basics of Analytic Geometry X ∩ (W × {|xn | = rn }) = ∅.

L

Lζ

U X

ζ

|xn | ≤ rn

p

W We consider a point ζ = (ζ1 , . . . ,ζn−1 ) ∈ W fixed. As X is given by the vanishing of holomorphic functions, and the restriction of such functions to the line Lζ given by x1 = ζ1 , . . . ,xn−1 = ζn−1 do not all vanish identically because of (3.1), it follows that the set X∩{(ζ1 , . . . ,ζn−1 )}×{|xn | ≤ rn } has only isolated points. The function f (ζ1 , . . . ,ζn−1 ,xn ) is a holomorphic function in xn , not defined in the finitely many points of X, and it is locally bounded. Therefore, by the Riemann Extension Theorem in one variable, it has a holomorphic extension to {(ζ1 , . . . ,ζn−1 )} × {|xn | ≤ rn } and is, in fact, given by the integral Z f (ζ) dζn . (3.2) f (ζ1 , . . . ,ζn−1 ,xn ) = ζn − xn |ζn |=rn

This holds for all (ζ1 , . . . ,ζn−1 ) ∈ W . Now we define fe for all ζ1 , . . . ,ζn−1 ,xn ∈ W ×{|xn | ≤ rn } by the right hand side of equation (3.2). It follows that fe is holomorphic with respect to each variable. Therefore, by Osgood’s Lemma 3.1.7, fe is holomorphic.

Theorem 3.1.16 (Second Riemann Extension Theorem). Let U be an open subset of C n , with n ≥ 2. Let p ∈ U , and f : U \ {p} −→ C be a holomorphic function. Then f has a holomorphic extension to the whole of U .

Proof. Consider a small open ball B around p which is contained in U . We will show that f is bounded on B \ {p}, so that the First Riemann Extension Theorem gives that f can be extended to a holomorphic function on B. Consider q ∈ B \ {p}. As n ≥ 2, we can find a complex line L through q which misses p.

3.1 Holomorphic Functions of Several Complex Variables L

U

B

q

85

p

The restriction f to B ∩ L is holomorphic in one variable. By the maximum principle |f (q)| ≤ max |f|∂B∩L | ≤ max |f|∂B |. As this holds for all q ∈ B \ {p}, it follows that f is bounded, as wanted. Remark 3.1.17. For n = 1 the theorem is not true: consider the function C \ {0}.

1 x

on U =

Exercises 3.1.18. (1) Prove that the values ∆i (p) in the definition of a holomorphic function are well-determined, by reducing it to a corresponding statement in the one variable case. P ν (2) Let f = ν aν x be a convergent power series. Describe all partial derivatives of f in terms of the aν . 3.1.19. Let q = (q1 , . . . ,qn ). Prove that X

qν =

ν∈Nn

n Y

k=1

1 . 1 − qk

3.1.20. Consider the formal power series ring C [[x1 , . . . ,xn ]] and let m = (x1 , . . . ,xn ). (1) Show that m is a maximal ideal. (2) Let f,g ∈ C [[x1 , . . . ,xn ]]. Show that f = g if and only if f ≡ g mod mk for all k ∈ N.

k (3) Show that in particular ∩∞ k=0 m = 0. P k (4) Let g ∈ m, and ak ∈ C . Show that ∞ k=0 ak g is a well-defined element in C [[x1 , . . . ,xn ]].

(5) More generally, let elements g1 , . . . ,gs in m be given. Show that there is a well-defined homomorphism of C –algebras C [[y1 , . . . ,ys ]] yi

−→ 7→

C [[x1 , . . . ,xn ]] gi .

(6) Let ϕ : C [[y1 , . . . ,ys ]] −→ C[[x1 , . . . ,xn ]] be a C –algebra homomorphism. Prove that for all f we have ϕ(f ) = f (ϕ(y1 ), . . . ,ϕ(ys )). (7) Show that C [[x1 , . . . ,xn ]] is a local ring. 1 (Hint: Let g ∈ m. What is 1−g ?) 3.1.21. Prove the following statements. (1) Let f be holomorphic in p. Then f is continuous at p. (2) Sums, products and quotients (when defined) of holomorphic functions are holomorphic.

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3 Basics of Analytic Geometry

3.1.22. (1) Let f ∈ C {x1 , . . . ,xn }, and g1 , . . . ,gn ∈ C {y1 , . . . ,ym } with gi (0) = 0 for i = 1, . . . ,n. Show that f (g1 , . . . ,gm ) ∈ C {y1 , . . . ,ym } by using the seminorm kf kδ , see 3.1.5.

(2) Use Exercise 3.1.20 to prove that for a C –algebra homomorphism ϕ : C {y1 , . . . ,ys } −→ C{x1 , . . . ,xn } for all f ϕ(f ) = f (ϕ(y1 ), . . . ,ϕ(ys )) holds. P 3.1.23. Let f = fk xkn , fk ∈ C [[x1 , . . . ,xn−1 ]], and f ∈ C {x1 , . . . ,xn }. Prove that fk ∈ C {x1 , . . . ,xn−1 }. (Hint: Use 3.1.4 to show this for k = 0. Now use induction on k.) 3.1.24. Prove the Identity Theorem 3.1.9. 3.1.25. Let U be an open connected subset of C n , and O(U ) the ring of holomorphic functions on U . Prove that O(U ) is an integral domain. 3.1.26. Let f be a holomorphic function on an open subset in C n , with n ≥ 2. Then f has no zeros which lie isolated. (Hint: Look at f1 .) 3.1.27. Consider an ideal I ⊂ C [x1 , . . . ,xn ], and f ∈ C [x1 , . . . ,xn ]. Suppose that f ∈ I · C {x1 , . . . ,xn }. Show that f ∈ I. 3.1.28. Let I = (y 2 − xz,x3 − yz,z 2 − x2 y). Show that I ∩ C {x} = (0). (Hint: Let α(y 2 − xz) + β(x3 − yz) + γ(z 2 − x2 y) ∈ C {x}. Show that modulo (x) we can write α = Az,β = Ay + Cz,γ = Cy. Take α′ = Az + Cx2 ,β ′ = Ay + Cz,γ ′ = Cx + Cy, and subtract these from α,β and γ. Iterate this and show that even in the formal power series α(y 2 − xz) + β(x3 − yz) + γ(z 2 − x2 y) = 0.)

3.2

Weierstraß Division and Preparation Theorem

We now come to the famous Weierstraß Division Theorem, and its immediate corollary, the Weierstraß Preparation Theorem. Before stating it, we need some definitions. Definition 3.2.1. (1) An element f ∈ C {x1 , . . . ,xn } is called regular of order b in xn if the power series in the variable xn , defined by f (0, . . . ,0,xn ) has a zero of order b. (2) An element xbn + a1 xnb−1 + . . . + ab−1 xn + ab , ai ∈ C {x1 , . . . ,xn } is called a Weierstraß polynomial if ai (0) = 0 for all i = 1, . . . ,b. Equivalently, it is a monic polynomial of degree b in xn which is, as convergent power series, regular of order b in xn . For general coordinate systems (x1 , . . . ,xn ) an f ∈ C {x1 , . . . ,xn } is regular in xn . In fact, we have the following result, which is analogous to the Noether Normalization Theorem for Hypersurfaces, see 2.1.7. As the proof is similar it is left as Exercise 3.2.15. Lemma 3.2.2. Let f ∈ C {x1 , . . . ,xn } be of order b. Then after a general linear change of coordinates, f is regular of order b in xn .

3.2 Weierstraß Division and Preparation Theorem

87

We can now state the Weierstraß Theorems. Theorem 3.2.3 (Weierstraß Division Theorem). Let f,g ∈ C {x1 , . . . ,xn }, and suppose that f is regular of order b in xn . Then there exist uniquely determined q ∈ C {x1 , . . . ,xn }, and r ∈ C {x1 , . . . ,xn−1 }[xn ], with r of degree smaller than b in xn such that g = qf + r. Theorem 3.2.4 (Weierstraß Preparation Theorem). Let f ∈ C {x1 , . . . ,xn } be regular of order b in xn . Then there exists a unit u ∈ C {x1 , . . . ,xn }, and a Weierstraß polynomial p of degree b, such that f = up. These u and p are uniquely determined by f . Proof. (Assuming the statement of the Weierstraß Division Theorem.) Apply the Weierstraß Division Theorem to g = xbn to obtain a q ∈ C {x1 , . . . ,xn } and a polynomial r in xn of degree smaller than b such that. xbn = qf + r. Now q is a unit, as otherwise in the expression qf + r there does not appear the term xbn , which would be a contradiction to the equality xbn = qf + r. So now we can put u = q −1 and p = xbn − r and get the equality f = up. It remains to show that p is a Weierstraß polynomial. But if p were not a Weierstraß polynomial, one of the nonleading coefficients of r would be a unit, and f would be regular of order smaller than b in xn . The fact that u and p are uniquely determined follows from the corresponding fact in the Weierstraß Division Theorem, see Exercise 3.2.16. Remark 3.2.5. The Weierstraß Division and Preparation Theorem also hold in the formal power series ring C [[x1 , . . . ,xn ]]. A proof for this case is much easier, as one does not have to cope with convergence questions. A proof will be given in Exercise 3.2.18. Corollary 3.2.6. Suppose that f ∈ C {x1 , . . . ,xn } is regular of order b. Then the C {x1 , . . . ,xn−1 }–module C {x1 , . . . ,xn }/(f ) is finitely generated and free of rank b. A basis is given by 1,xn , . . . ,xb−1 n . The proof of this corollary is similar to the proof of 2.1.7 and therefore left as Exercise 3.2.17. It is in fact not so difficult to see that the Corollary is equivalent to the Weierstraß Division Theorem. We can even do something better by invoking Nakayama’s Lemma. Lemma 3.2.7. Let f ∈ C {x1 , . . . ,xn } be regular of order b. If C {x1 , . . . ,xn }/(f ) is a finitely generated C {x1 , . . . ,xn−1 }–module, then the Weierstraß Division Theorem holds for f . Proof. Put M := C {x1 , . . . ,xn }/(f ). The maximal ideal of
C {x1 , . . . ,xn−1 } is the ideal (x1 , . . . ,xn−1 ). Now M/(x1 , . . . ,xn−1 )M = C {xn }/ f (0,xn ) , hence is a finite-dimensional C –vector space of dimension b. Indeed, a basis of this vector space is given by the classes of 1,xn , . . . ,xb−1 n . By Corollary 1.3.6 (a corollary of Nakayama’s Lemma), it follows that 1,xn , . . . ,xb−1 generate M as C {x1 , . . . ,xn−1 }–module. This exactly means that for all n g ∈ C {x1 , . . . ,xn } there exist q and r such that g = qf + r

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3 Basics of Analytic Geometry

where r is a polynomial in xn of degree less that b with coefficients in C {x1 , . . . ,xn−1 }. To finish the proof, we need to show that q and r are uniquely determined. So assume that g = qf + r = q ′ f + r′ , with the degree of both r and r′ smaller than b. Then (q − q ′ )f + r − r′ = 0. It therefore suffices to show that from qf + r = 0 with degree of r less than b, it follows that q = r = 0. For this it obviously suffices to show that q = 0. We now write b−1 X X X ri xin , f= fi xin ; q= qi xin ; r= i=0

with fi , qi and ri in C {x1 , . . . ,xn−1 }. As f is regular of order b in xn , it follows that ord(fb ) = 0. If q 6= 0, we can find a minimal k such that ord(qk ) is minimal. This means that for j < k we have ord(qj ) > ord(qk ), and for j > k we have ord(qj ) ≥ ord(qk ). As f is regular of order b it follows that ord(fi ) > 0 for i = 0, . . . ,b − 1. We now look at the coefficient of xb+k of the power series q · f . As the term xb+k does not appear in r, and n n qf + r = 0, this coefficient is equal to (3.3)

fb+k q0 + · · · + fb+1 qk−1 + fb qk + fb−1 qk+1 + · · · + f0 qb+k = 0.

We know the following • ord(fb qk ) = ord(qk ), • ord(fb+j qk−j ) ≥ ord(qk−j ) > ord(qk ) for j > 0, • ord(fb−j qk+j ) = ord(fb−j ) + ord(qk+j ) > ord(qk ) for j > 0. The only terms of degree ord(qk ) in formula (3.3) come from fb qk , so the expression (3.3) cannot be equal to zero. This is a contradiction. Hence q = 0 and the uniqueness is proved. So all we need to show now, in order to prove the Weierstraß Division Theorem, is that C {x1 , . . . ,xn }/(f ) is a finitely generated C {x1 , . . . ,xn−1 }–module. As it is not more difficult, we will in fact prove a more general version. This more general version will play an important role in many arguments in this book. First we need a definition and a simple lemma. Definition 3.2.8. (1) An analytic algebra (also called analytic C –algebra) is a C –algebra of type C {x1 , . . . ,xn }/I, where I is an ideal in C {x1 , . . . ,xn }. (2) A formal C –algebra is a C –algebra of type C [[x1 , . . . ,xn ]]/I, where I is an ideal in C [[x1 , . . . ,xn ]]. Lemma 3.2.9. Let (R,mR ), (S,mS ) be analytic (or formal) C –algebras, ϕ : R → S a morphism of C –algebras. Then ϕ(mR ) ⊂ mS . Proof. Suppose the converse. Then there exists an f ∈ mR with ϕ(f ) ∈ / mS . Thus ϕ(f )(0) = c 6= 0. Hence ϕ(f ) − c ∈ mS is not a unit. As ϕ is a ring homomorphism, it sends 1 to 1. As a ring homomorphism, it therefore sends units in R to units in S. However, as ϕ is a homomorphism of C –vector spaces, ϕ(f ) − c = ϕ(f − c). Now f − c is a unit in R, and, therefore, ϕ(f ) − c is a unit in S. This is a contradiction, proving the lemma.

3.2 Weierstraß Division and Preparation Theorem

89

Theorem 3.2.10 (General Weierstraß Division Theorem). Let R, S be analytic C – algebras, ϕ : R → S a morphism of C –algebras. One can, via ϕ, view S as an R–module. Then the following conditions are equivalent: (1) S is a finitely generated R–module. (2) dimC S/mR S < ∞. Analogous statements hold for formal C –algebra’s. Before giving the proof of Theorem 3.2.10, we look at two examples. Examples 3.2.11. (1) We show that the Weierstraß Division Theorem follows from the General Weierstraß Division Theorem by using Lemma 3.2.7. So for f ∈ C {x1 , . . . ,xn } which is regular of order b in xn we have to show that C {x1 , . . . ,xn }/(f ) is a finitely generated C {x1 , . . . ,xn−1 }–module. We take R = C {x1 , . . . ,xn−1 }, and S = C {x1 , . . . ,xn }/(f ). The map ϕ sends xi to xi for i = 1, . . . ,n−1. Then S/mR S = C {x1 , . . . ,xn }/(x1 , . . . ,xn−1 ,f ) ∼ = C {xn }/(f (0, . . . ,0,xn ) ∼ = C b . So the statement follows from the General Weierstraß Division Theorem. (2) We consider R = C {x}, S = C {t}, and the map ϕ : R −→ S be given by ϕ(x) = t2 . Then S/mR S = C {t}/(t2 ) is a two-dimensional vector space. We claim that C {t} P∞ k is as C {x}–module generated by 1 and t. Now let f = k=0 fk t ∈ C {t} be arbitrary, with fk ∈ C . Now define two power series a0 =

∞ X

f2k xk ; a1 =

k=0

Then ϕ(a0 ) =

P∞

k=0

∞ X

f2k+1 xk .

k=0

f2k t2k and ϕ(a1 ) =

P∞

k=0

f2k+1 t2k , so that

f = ϕ(a0 ) · 1 + ϕ(a1 ) · t.

We still have to show that a0 and a1 are convergent power series. In this case this follows easily from the convergence of f , by using 3.1.4. Proof of the General Weierstraß Division Theorem 3.2.10. Step 1. We first show that (1) implies (2). This is quite obvious. Any set of generators of S as an R–module will, by taking classes, be a set of generators for S/mR S as R/mR = C – module. Hence, S/mR S is a finite-dimensional vector space. Step 2. We first prove the implication (2) =⇒ (1) for the case that R = C {x1 , . . . ,xn }, S = C {y1 , . . . ,ym }. We put gi = ϕ(xi ) for i = 1, . . . ,n. The assumption then is that C {y1 , . . . ,ym }/(g1 , . . . ,gn ) is a finite-dimensional vector space. So S/(g1 , . . . ,gn ) is an Artinian C –algebra. Writing m = mS for short, it follows from Exercise 1.3.20 that there exists a k ∈ N such that mk ⊂ (g1 , . . . ,gn ). Multiplying with m, and putting r = k+1, it follows that mr ⊂ m(g1 , . . . ,gn ). In particular, for any monomial y α of degree r there exist elements pα,i ∈ m such that (3.4)

yα =

n X i=1

pα,i gi .

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3 Basics of Analytic Geometry

The claim is that the monomials {y α : |α| < r} generate S as an R-module. This would prove the finite generatedness of S as R–module in this case. Step 2a. We first prove this “formally”. Let f ∈ S. We can write X e hα (f )y α . (3.5) f = a(f ) + α:|α|=r

such that e hα (f )y α and e hβ (f )y β have no common monomial if α 6= β. Moreover a(f ) is a polynomial in the y ′ s of degree smaller than r. Plugging in (3.4) we get (3.6)

f = a(f ) +

n X

hi (f )gi ,

i=1

where hi (f ) =

P

α:|α|=r

we define inductively

hα (f ).We now iterate (3.6). For a multi-index β = (β1 , . . . ,βn ) pα,i e

(1) aβ (f ) = a(f ) for β = (0, . . . ,0), (2) hβ (f ) = f for β = (0, . . . ,0), (3) hβ (f ) = 0 if some entry in β is negative, (4) aβ (f ) = a(hβ (f )),
Pn (5) hβ (f ) = i=1 hi hβ−ei (f ) . Here e1 = (1,0, . . . ,0) etc.

Note that the aβ (f ) are polynomials in the y1 , . . . ,ym of degree smaller than r. We claim that for all k ∈ N X X hβ (f )g β . aβ (f )g β + (3.7) f= |β|=k

|β|
For k = 1 this is justP formula (3.6). We now apply (3.6) to the coefficients hβ (f ). We get hβ (f ) = a(hβ (f )) + ni=1 hi (hβ (f ))gi . Plugging this in (3.7) and using the definition of aβ (f ) and hβ (f ) gives the induction step. As (g1 , . . . ,gn ) ∈ m, it follows that for all k ∈ N X f≡ aβ (f )g β mod mk . β

So in the formal power series ring we have the equality X f= aβ (f )g β . β

The aβ (f ) are polynomials in the y ′ s of degree smaller than r. So we write X aβ,α (f )y α , aβ,α (f ) ∈ C . (3.8) aβ (f ) = α:|α|
Now for all α with |α| < r look at the formal power series

3.2 Weierstraß Division and Preparation Theorem (3.9)

Aα (f ) :=

X

91

aβ,α (f )xβ .

β

So if we look at the map ϕ : C [[x1 , . . . ,xn ]] −→ C [[y1 , . . . ,ym ]] which sends xi to gi it follows that X ϕ(Aα (f )) · y α , f= α:|α|
showing the theorem in the formal case. Step 2b. We P of Aα (f ). We use the seminorm of 3.1.5. That Phave to show the convergence is, for f = α fα xα we have kf kδ = α |fα |δ |α| . Now choose a δ > 0 so small that (1) kf kδ is finite.

(2) Let C be the number of monomials of degree r. Then for the pα,i of formula (3.4) kpα,i kδ ≤

1 C.

We can find such a δ, because the elements pα,i are in the maximal ideal, see 3.1.5 (2). We will show for all α with |α| < r r

Aα (f ) < ∞ for ε = δ . ε 2n2

(3.10)

This will show the convergence of Aα (f ). We have the following estimates for the terms in formula (3.5) and (3.6). (1) ka(f )kδ ≤ kf kδ , (2) ke hα (f )kδ ≤ δ −r kf kδ , (3) khi (f )kδ ≤ δ −r kf kδ .

The first two estimates follow from the definition of the seminorm, and the definition of a(f ) and e hα (f ). The final estimate follows because we have chosen δ so that kpα,i kδ ≤ C1 , C being the number of monomials of degree r. Inductively we get the estimates khβ (f )kδ ≤ n|β|−1 · δ −|β|r · kf kδ ,

(3.11)

(3.12) Recall that aβ (f ) =

P

kaβ (f )kδ ≤ n|β|−1 · δ −|β|r · kf kδ .

α:|α|
aβ,α (f )y α , for some aβ,α (f ) ∈ C .

Obviously |aβ,α (f )| ≤ kaβ (f )kδ δ −|α| . From (3.12) it follows with Cα = |aβ,α (f )| ≤ Cα nδ −r With ε =

δr 2n2

we get


|β|

δ −|α| n kf kδ

.

X 1 X X n |β|

X

ε|β| = Cα . aβ,α (f )xβ ε = |aβ,α (f )|ε|β| ≤ Cα r δ (2n)|β| β

β

β

that

β

As the number of β with |β| = k is at most nk this can be estimated as Cα

P∞

1 k=0 2k

< ∞.

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3 Basics of Analytic Geometry

Step 3. The implication (2) =⇒ (1) will be proved for the case that S = C {y1 , . . . ,ym }/J, where J = (f1 , . . . ,fp ) is a finitely generated ideal. This would be always the case if we knew that C {y1 , . . . ,ym } is Noetherian. Now, as the Weierstraß Division Theorem is deduced from the General Weierstraß Division Theorem by using J = (f ), see Example 3.2.11 (1), and only the Weierstraß Division Theorem is used in the proof of the fact that C {x1 , . . . ,xn } is Noetherian, see Lemma 3.2.7, this restriction is not essential. Step 4. We now prove the general case, for J finitely generated. Write R = C {x1 , . . . ,xn }/I,S = C {y1 , . . . ,ym }/J, where J = (f1 , . . . ,fp ). Define new rings: R′ := C {x1 , . . . ,xn ,xn+1 , . . . ,xn+p };

S ′ := C {y1 , . . . ,ym }.

Take representatives g1 , . . . ,gn ∈ S ′ = C {y1 , . . . ,ym } of ϕ(x1 ), . . . ,ϕ(xn ). The map ϕ′ is then defined on generators by: • ϕ′ (xi ) = gi for i = 1, . . . ,n, • ϕ′ (xn+j ) = fj for j = 1, . . . ,p. It is immediately clear that S ′ /mR′ S ′ = C {y1 , . . . ,ym }/(g1 , . . . ,gn ,f1 , . . . ,fp ) = S/mR S, which by assumption is finite-dimensional. Therefore, by Step 2 we may conclude that S ′ is a finitely generated R′ –module. Hence, there exist v1 , . . . ,vs ∈ S ′ = C {y1 , . . . ,ym }, such that any h ∈ S ′ can be written as: h = α1 v1 + . . . + αs vs

αi ∈ ϕ′ (R′ ).

Reading this equation modulo (f1 , . . . ,fp ), we see that S is an R–module generated by v1 , . . . ,vs , so, in particular, finitely generated. Now we know (under the assumptions of the theorem) that S is a finitely generated R–module, we can use Nakayama’s Lemma to get the following somewhat stronger form of the theorem. Corollary 3.2.12. Keep the notations of the previous Theorem 3.2.10. Suppose we have elements v1 , . . . ,vs ∈ S whose classes generate S/mR S as a C –vector space. Then v1 , . . . ,vs generate S as an R–module. Corollary 3.2.13. Again keep the notation of Theorem 3.2.10. (1) Suppose mR S = mS . Then the map ϕ is surjective. (2) Suppose S is an analytic algebra such that mS has s generators. Then S is a quotient of the power series ring C {x1 , . . . ,xs }. Proof. The first statement follows from 3.2.12: indeed, it follows that 1 generates S as an R–module, and thus in particular, ϕ is surjective. For the second statement, suppose that f1 , . . . ,fs generate mS . Consider the map ϕ : C {x1 , . . . ,xs } −→ S, which sends xi to fi . It follows from the first part that ϕ is surjective.

3.2 Weierstraß Division and Preparation Theorem

93

There is a “direct” proof of the second part of this corollary, using only the statement of the Weierstraß Division Theorem. See Exercise 3.2.21. Corollary 3.2.14. Let R,S be analytic C –algebras, ϕ : R −→ S a morphism of C – algebras. Let M be a finitely generated S–module such that dimC M/mR M < ∞. Then M is a finitely generated R–module. Proof. As M is an S/ AnnS (M )–module, we may assume that AnnS (M ) = (0). Exercise 1.2.31 applied to M/mR M implies that dimC S/mR S < ∞. By the previous theorem S is a finitely generated R–module and, therefore, M is a finitely generated R–module.

Exercises 3.2.15. Prove Lemma 3.2.2. 3.2.16. Prove that u and p in the Weierstraß Preparation Theorem are uniquely determined. 3.2.17. Prove that the Weierstraß Division Theorem is equivalent to the statement in Corollary 3.2.6. 3.2.18. Prove the Weierstraß Division Theorem for the formal power series ring. (Hint: Consider a formal power series f ∈ C [[x1 , . . . ,xn ]] which is regular of order b in xn . Write (3.13)

f = f0 + xbn uf .

As f is regular of order b in xn it follows that • f0 is a polynomial of degree ≤ b − 1 in xn with coefficients in (x1 , . . . ,xn−1 ). • uf is a unit.

From (3.13) it follows that xbn = ri ,qi ,gi with the following properties.

f uf

−

f0 uf

with

f0 uf

∈ (x1 , . . . ,xn−1 ). Construct inductively

(1) g = qi · f + ri + gi ,

(2) gi ,qi+1 − qi ,ri+1 − ri ∈ (x1 , . . . ,xn−1 )i ,

(3) ri is a polynomial of degree ≤ b − 1 in xn .

Use as initialization q0 = r0 = 0, and g0 = g, and divide gi by xbn . )

3.2.19. Let f be a Weierstraß polynomial in xn of degree b. Suppose that ord(f ) = b. Let g be a (formal) power series of order t. Apply the Weierstraß Division Theorem, so write g = q · f + r, for r a polynomial of degree smaller than b in xn . Prove that the order of r is at least t. (Hint: Follow the steps in the proof of the formal Weierstraß Division Theorem, see Exercise 3.2.18.) 3.2.20. (1) Let f ∈ C {x1 , . . . ,xn } be regular of order b in xn , assume that b = ord(f ) and let g ∈ (x1 , . . . ,xn )c C {x1 , . . . ,xn }. Let g = qf + r as in 3.2.3. Prove that q ∈ (x1 , . . . ,xn )c−b .

(2) Generalize the result to the weighted order (cf. 3.1.3): let w1 , . . . ,wn ∈ N be weights for x1 , . . . ,xn . Let f,g ∈ C {x1 , . . . ,xn } be regular of order b in xn , assume that w-ord(f ) = bwn and w-ord(g) = c, let g = qf + r as in 3.2.3. Prove that w-ord(q) = c − bwn if q 6= 0.

3.2.21. (1) Suppose I ⊂ C {x1 , . . . ,xn } contains an element which is regular of order one in xn . Use the Weierstraß Division Theorem to show that C {x1 , . . . ,xn }/I is isomorphic to a quotient of C {x1 , . . . ,xn−1 }. So we can “eliminate” the variable xn . (2) Give a different proof of the second part of Corollary 3.2.13.

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Applications

The first thing to note is that the Weierstraß Preparation Theorem implies the Implicit Function Theorem! Theorem 3.3.1 (Implicit Function Theorem). Let f ∈ C {x1 , . . . ,xn ,y} with f (0) = 0 and ∂f ∂y (0) 6= 0. Then there exists a unique ϕ ∈ (x1 , . . . ,xn )C {x1 , . . . ,xn } with f (x1 , . . . ,xn ,y) = 0 ⇐⇒ y = ϕ(x1 , . . . ,xn ). Proof. Note that the condition f (0) = 0 and ∂f ∂y (0) 6= 0 exactly means that f is regular of order one in y. By the Weierstraß Preparation Theorem f (x1 , . . . ,xn ,y) = u · (y − ϕ(x1 , . . . ,xn )) for some unit u ∈ C {x1 , . . . ,xn ,y} and some ϕ ∈ C {x1 , . . . ,xn }. The Implicit Function Theorem follows from this equation. Example 3.3.2. Consider f (x,y) = (y + 1)2 − x − 1 = y 2 + 2y −√x. Then f satisfies the conditions of the Implicit Function Theorem. Hence y = −1 + x + 1 exists as an analytic function. From the uniqueness statements in the Weierstraß Preparation Theorem the following follows immediately. Corollary 3.3.3. Let f ∈ C {x1 , . . . ,xn } be regular of order b in xn , and suppose that f = u · p, where u,p ∈ C [[x1 , . . . ,xn ]], u a unit and p a Weierstraß polynomial. Then u and p are in C {x1 , . . . ,xn }, that is, they are convergent. A similar statement holds for the Weierstraß Division Theorem. Example 3.3.4. Consider f (x,y) = 2y + y 2 − x. Then f is regular of order one in y. Hence, by the Weierstraß Preparation Theorem f (x,y) = u(y − ϕ(x)) for some unit u. We can generate ϕ and u as indicated by a proof of the formal Weierstraß Division Theorem (Exercise 3.2.18). ϕ(x) = 12 (2y + y 2 − x) + 12 x − 21 y 2

= ( 12 − 41 y)(2y + y 2 − x) + 12 x − 14 xy + 14 y 3

= ( 21 − 41 y − 18 x + 81 y 2 )(2y + y 2 − x) + 12 x − 81 x2 − 81 y 4 + 41 xy 2

= ....

So we see the beginning of the power series ϕ(x) = 12 x − 18 x2 + . . . which is the √ beginning of the power series of ϕ(x) = −1 + u = y + 1 + x + 1.

√ x + 1. Similarly, we obtain

The Implicit Function Theorem holds just as well for the formal case. From the uniqueness statement in the Implicit Function Theorem the following fact follows.

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Corollary 3.3.5. Let f ∈ C {x1 , . . . ,xn ,y} be regular of order 1 in y. Suppose ϕ(x) ∈ C [[x1 , . . . ,xn ]] is such that f (x1 , . . . ,xn ,ϕ(x)) = 0. Then ϕ(x) ∈ C {x1 , . . . ,xn }. We might phrase this by saying that a formal solution to f (x1 , . . . ,xn ,y) = 0 (solving for y), already is analytic. By induction we get the Implicit Mapping Theorem from the Implicit Function Theorem. Theorem 3.3.6 (Implicit Mapping Theorem).   ∂fi (0) 6= Let f1 , . . . ,fm ∈ C {x1 , . . . ,xn , y1 , . . . ,ym } such that fi (0) = 0 and det ∂y j

0. Then there exist uniquely defined ϕ1 , . . . ,ϕm ∈ (x1 , . . . ,xn )C {x1 , . . . ,xn } such that fi (x1 , . . . ,xn ,ϕ1 , . . . ,ϕm } = 0 for i = 1, . . . ,m.   ∂fi Proof. First suppose that the m×m matrix ∂y (0) is the identity matrix. We will prove j the statement by induction on m. The case m = 1 is the Implicit Function Theorem. For the induction step, first apply the Implicit Function Theorem to the element fm . We conclude that there exists a unique holomorphic function ψ = ψ(x1 , . . . ,xn ,y1 , . . . ,ym−1 ) with fm (x1 , . . . ,xn ,y1 , . . . ,ym−1 ,ψ) = 0. It is left as an exercise to show that the functions fi (x1 , . . . ,xn ,y1 , . . . ,ym−1 ,ψ), satisfy the induction hypothesis. For the general case, let A =



−1

∂fi ∂yj (0)

A·



i = 1, . . . m − 1,

. Then

 ∂fi (0) ∂yj

is the identity matrix. Replacing f1 , . . . ,fm by linear combinations of those given by A, we come back to the first case. The proof of the Inverse Function Theorem, which follows from the Implicit Mapping Theorem is standard, and left as Exercise 3.3.29. Corollary 3.3.7 (Inverse Function Theorem). Let f1 , . . . ,fn ∈ C {x1 , . . . ,xn } such that  ∂fi (0) = 6 0 if and only if the C –algebra homomor∂xj

f1 (0) = · · · = fn (0) = 0. Then det phism

C {x1 , . . . ,xn } −→ C {x1 , . . . ,xn } xi

7−→ fi

is an isomorphism. This again holds if and only if there exist open neighborhoods U,W of 0 such that F = (f1 , . . . ,fn ) defines a map F : U −→ W and this map has a holomorphic inverse. The Inverse Function Theorem is the basis to introduce coordinate functions and complex submanifolds of C n . Definition 3.3.8.

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(1) Let f1 , . . . ,fn be holomorphic on an open subset U . Let p ∈ U , and suppose f1 (p) = ··· =  fn (p)= 0. The set {f1 , . . . ,fn } is called a set of coordinate functions at p if det

∂f ∂xi (p)

6= 0.

(2) A subset X ⊂ C n is called a complex submanifold of C n if for every x ∈ X there exists an open subset U in C n and coordinate functions w1 , . . . ,wn of x such that X ∩ U = {y ∈ U : w1 (y) = · · · = wm (y) = 0} for some m ≤ n.

Remark 3.3.9. A complex submanifold X ⊂ C n locally looks like an open subset of C k for some k. Example 3.3.10. Let f1 , . . . ,fk be holomorphic functions on the open set U ⊂ C n .
Then X = a1 , . . . ,an , f1 (a1 , . . . ,an ), . . . ,fk (a1 , . . . ,an ) : (a1 , . . . ,an ) ∈ U is a complex submanifold of C n+k . To see this, let x1 , . . . ,xn ,y1 , . . . yk be coordinates on C n+k . Then X can be given by y1 − f1 (x1 , . . . ,xn ) = · · · = yk − fk (x1 , . . . ,xn ) = 0, and one checks that these are part of a set of coordinate functions at every point of X. We will see later in Chapter 4 that for an analytic subset of C n almost all points have a neighborhood which is a zero set of coordinate functions. More precisely, an analytic subset contains as an open set a complex submanifold. This gives rise to the following definition. Definition 3.3.11. Let U ⊂ C n be an open set and X ⊂ U an analytic subset. A point x ∈ X is called regular, or X is called smooth at x, if there exists an open subset V in C n , x ∈ V , such that X ∩ V is a complex submanifold of C n . If x ∈ X is not regular, then x is called singular, or a singularity. The set of singular points of X is denoted by Sing(X). Examples 3.3.12. (1) Every point in a complex submanifold of C n is a regular point. (2) Let X = {(x,y) ∈ C 2 : x2 − y 3 = 0}. Then X is an analytic subset of C 2 , 0 = (0,0) ∈ X is a singular point, see Exercise 3.3.37. In chapter 4 we will obtain a characterization of the singular points in terms of the Jacobian matrix (cf. Theorems 4.3.6 and 4.3.15). We now turn to Newton’s Lemma. Newton’s Lemma, as you learned it in a calculus course, is a method for approximating zeros of differentiable functions in one variable. Consider a differentiable function y = f (x) in one real variable. The question is to find a zero of f (x) = 0. The idea is to take an approximate solution ak , and to obtain a better approximation ak+1 by intersecting the tangent line of the graph of f at (ak ,f (ak )) with the x–axis.

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97

ak+1 ak

y =0

A simple calculation shows

ak+1 = ak −

(3.14)

f (ak ) . f ′ (ak )

Starting off with an initial approximation a0 we get a sequence a1 ,a2 , . . ., which, under appropriate assumptions, converges to a number a with f (a) = 0. We apply similar ideas to the equation f (x,y) = y 2 − 1 − x. We want to find √ a power series, y = ϕ(x) with ϕ(x)2 = x + 1, that is, we want to find a power series for x + 1. We start of with the initial approximation ϕ0 (x) = 1, and iterate with the obvious analog of (3.14):
f x,ϕk (x) (3.15) ϕk+1 (x) = ϕk (x) − ∂f
. ∂y x,ϕk (x) One calculates that

2

x x x x2 x 4 , ϕ2 (x) = 1 + − =1+ − + ..., 2 2 2(1 + x/2) 2 8 √ so one sees the beginning of the power series of 1 + x. We now generalize.

ϕ1 (x) = 1 +

Corollary 3.3.13 (Newton’s Lemma: simple case). Let f ∈ C {x1 , . . . ,xn ,y}, y(x) ∈ C {x1 , . . . ,xn } and c ≥ 1 with f (x,y(x)) ≡ 0

mod

Then there exists a y ∈ C {x1 , . . . ,xn } with




∂f x,y(x) ∂y

2

(x)c .

(1) f (x,y(x)) = 0. 
 c (x) . (2) y ≡ y mod ∂f ∂y x,y(x)

Proof. Write

∂f ∂y


x,y(x) = α ∈ C {x1 , . . . ,xn }. We choose a new variable t and substitute y = y(x) + t · α

in f (x,y). We therefore want to find a t ∈ C {x1 , . . . ,xn } with

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3 Basics of Analytic Geometry
g(x,t) := f x,y(x) + t · α = 0.

The existence will be guaranteed by applying the Implicit Function Theorem. Indeed Taylor’s formula gives



∂f g(x,t) := f x,y(x) + tα = f x,y(x) + tα x,y(x) + α2 R = f x,y(x) + tα2 + α2 R ∂y

with R ∈ (t)2 . By assumption f (x,y(x)) = α2 · r with r ∈ (x)c . Therefore g(x,t) = α2 (t + R + r). Hence it suffices to solve t + R + r = 0, with R ∈ (t)2 and r ∈ (x)c . Hence this equation satisfies the assumption of the Implicit Function Theorem. Even more, by Exercise 3.3.33 there exists a t ∈ (x)c with g(x,t) = 0. A more general version of Newton’s Lemma, whose proof is only notationally more difficult, will be proved in Exercise 3.3.34. This generalization will, for example, be used in the proof of Artin’s Approximation Theorem, see Chapter 8. Corollary 3.3.14. C {x1 , . . . ,xn } and C [[x1 , . . . ,xn ]] are Noetherian rings. Proof. As the proofs are similar, we only give the proof for the analytic case. The proof is by induction on n. For the case n = 0 we have the field C , which is obviously Noetherian. Now take an ideal I ⊂ C {x1 , . . . ,xn }. We have to prove that I is finitely generated. If I = (0), this is obviously the case. If I 6= (0), take 0 6= f ∈ I. After a general linear change of coordinates we may assume that f is regular in xn , say of order b, see 3.2.2. By 3.2.6, which is a corollary of the Weierstraß Division Theorem, it follows that C {x1 , . . . ,xn }/(f ) is a finitely generated (even free) C {x1 , . . . ,xn−1 }–module. Now C {x1 , . . . ,xn−1 } is Noetherian by induction, so that C {x1 , . . . ,xn }/(f ) is a Noetherian C {x1 , . . . ,xn−1 }–module by 1.2.15. In particular I modulo (f ) is finitely generated as C {x1 , . . . ,xn−1 }–module, and therefore certainly finitely generated as C {x1 , . . . ,xn }– module, say by the classes of f1 , . . . ,ft . It follows that f,f1 , . . . ,ft generate I. Remarks 3.3.15. (1) Analogous ideas also give a new proof for the Hilbert Basis Theorem for the polynomial ring C [x1 , . . . ,xn ]. (2) An alternative proof for the induction in the proof of Corollary 3.3.14 uses the Hilbert Basis Theorem. It runs as follows. Let I 6= (0) be an ideal, and 0 6= f ∈ I. Again, we may assume that C {x1 , . . . ,xn }/(f ) is a finitely generated C {x1 , . . . ,xn−1 }–module. We may also assume that f is a Weierstraß polynomial in xn . Hence C {x1 , . . . ,xn }/(f ) = C {x1 , . . . ,xn−1 }[xn ]/(f ). It follows from the Weierstraß Division Theorem that (I ∩ C {x1 , . . . ,xn−1 }[xn ]) C {x1 , . . . ,xn } = I. By induction and the Hilbert Basis Theorem, the ring C {x1 , . . . ,xn−1 }[xn ] is Noetherian. Hence I is finitely generated as C {x1 , . . . ,xn−1 }[xn ]–ideal, and in particular finitely generated as C {x1 , . . . ,xn−1 ,xn }–ideal. The next thing we want to show is that the power series ring is a unique factorization domain. For the proof we need the following fact, which is interesting in its own right.

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99

Lemma 3.3.16. Let h ∈ C {x1 , . . . ,xn−1 }[xn ] be a Weierstraß polynomial. Then h is reducible in C {x1 , . . . ,xn−1 }[xn ] ⇐⇒ h is reducible in C {x1 , . . . ,xn }. In particular, every irreducible factor of h in C {x1 , . . . ,xn } is xn –regular and, therefore, every decomposition of h in C {x1 , . . . ,xn } gives a decomposition of h into Weierstraß polynomials by multiplying with suitable units. Similar statements hold for the formal power series ring. Proof. We again only consider the analytic case, the formal case being similar. ⇐=: Let h = g1 · g2 be a factorization in C {x1 , . . . ,xn } with g1 and g2 nonunits. As h is regular in xn it follows that g1 and g2 are regular in xn . As they are both nonunits, they are regular of order at least one. By the Weierstraß Preparation Theorem we can write g1 = u1 h1 , and g2 = u2 h2 , where u1 and u2 are units and h1 and h2 are Weierstraß polynomials in xn . Hence h = u1 u2 h1 h2 . Now h1 h2 is a Weierstraß polynomial. It follows from the uniqueness statement in the Weierstraß Preparation Theorem that u1 u2 = 1. Therefore, h = h1 h2 , and h is reducible in C {x1 , . . . ,xn−1 }[xn ].

=⇒: Suppose h = g1 · g2 , with g1 and g2 nonunits in C {x1 , . . . ,xn−1 }[xn ]. We have to show that g1 and g2 are also nonunits in C {x1 , . . . ,xn }. But suppose, for example, that g1 is a unit. As g1 is not a unit in C {x1 , . . . ,xn−1 }[xn ] it follows that degxn (g1 ) ≥ 1. So degxn (g2 ) < degxn (h). But h = g1 · g2 , so g2 is regular in xn . This implies that h is xn –regular of order at most degxn (g2 ) which is a contradiction to the fact that h is a Weierstraß polynomial. Corollary 3.3.17. C {x1 , . . . ,xn } and C [[x1 , . . . ,xn ]] are factorial rings.

Proof. Again we only treat the convergent case. The proof is by induction on n. The case n = 0 is trivial. Suppose therefore that C {x1 , . . . ,xn−1 } is factorial. Then by the Lemma of Gauß (1.4.4) C {x1 , . . . ,xn−1 }[xn ] is factorial. Let f ∈ C {x1 , . . . ,xn }. After a general linear change of coordinates we may assume that f is regular in xn . By the Weierstraß Preparation Theorem f = uh, where u is a unit and h is a Weierstraß polynomial. As by the previous lemma, the polynomial factors of h in both C {x1 , . . . ,xn−1 }[xn ] and C {x1 , . . . ,xn } are equal, it follows that there is a unique factorization of h. For the case of two variables we obtain the following result which will be very useful in Chapter 5. Corollary 3.3.18. Let f ∈ C {x,y} and y ∈ (x)C [[x]] such that f (x,y) = 0. Then y ∈ C {x}.8 Proof. Suppose the converse, that is, assume that y 6∈ C {x}. First of all, we may assume that f = y n + an−1 (x)y n−1 + · · · + a0 (x) is an irreducible Weierstraß poly/ C {x} it follows that n > 1. From Lemma 1.4.5 it folnomial by (3.3.16). As y ∈ lows that f is irreducible in Q(C {x})[y]. The ring Q(C {x})[y] is a principal ideal domain, hence (f, ∂f ∂y ) = (1). By clearing denominators, we can find p,q ∈ C {x}[y] and

∂f 0 6= h ∈ C {x} with qf + p ∂f ∂y = h. Plugging in y = y(x) it follows that ∂y (x,y) 6= 0.
Let ord ∂f ∂y (x,y) = m and choose y1 ∈ C [x] such that ord(y − y1 ) ≥ 2m + 1. Then
2 2m+1 ). f (x,y1 ) ≡ 0 mod ∂f ∂y (x,y1 ) (x) = (x 8

This corollary is also true for n variables x1 , . . . ,xn and can be proved, for instance, using Artin’s Approximation Theorem from Chapter 8. In fact, it follows from Exercise 8.1.7.

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Using Newton’s Lemma 3.3.13 we obtain an y2 ∈ C {x} such that f (x,y2 ) = 0 and
y − y2 ∈ (x) ∂f ∂y (x,y1 ) . Now f (x,y) = g(x,y)(y − y2 ), and g(x,y) = 0. Hence g is not a unit, an we obtain a contradiction to the fact that f is irreducible. We now come to the Noether Normalization Theorem. Its proof is analogous to the proof in the affine case, see 2.2.9, but because it is such an important result, we spell it out. Corollary 3.3.19 (Noether Normalization). Let I ⊂ C {x1 , . . . ,xn } be an ideal. Then after a general linear coordinate change9 we have that there exists a k such that C {x1 , . . . ,xk } ⊂ C {x1 , . . . ,xn }/I and such that C {x1 , . . . ,xn }/I is a finitely generated C {x1 , . . . ,xk }–module. If we have such a situation we call C {x1 , . . . ,xk } ⊂ C {x1 , . . . ,xn }/I a Noether normalization. If moreover I 6= (0), then k < n. Proof. If I = (0), we can take k = n. Otherwise, the theorem is proved by induction on n. The case n = 0 is trivial. Take 0 6= g ∈ I. By 3.2.2 we have, after a general linear change of coordinates, that g is regular of order b in xn . By Corollary 3.2.6, we know that C {x1 , . . . ,xn }/g is a finitely generated C {x1 , . . . ,xn−1 }–module. Because g ∈ I, then certainly C {x1 , . . . ,xn }/I is a finitely generated C {x1 , . . . ,xn−1 }/(I ∩ C {x1 , . . . ,xn−1 })– module. We apply the induction hypothesis to the ideal I ∩ C {x1 , . . . ,xn−1 }. After a general linear coordinate change in x1 , . . . ,xn−1 we have that C {x1 , . . . ,xk } ⊂ C {x1 , . . . ,xn−1 }/(I ∩ C {x1 , . . . ,xn−1 }) is finitely generated. We compose this with the inclusion C {x1 , . . . ,xn−1 }/(I ∩ C {x1 , . . . ,xn−1 }) ⊂ C {x1 , . . . ,xn }/I. As a composition of finite rings extension is finite, see 1.5.2, we deduce that the ring extension C {x1 , . . . ,xk } ⊂ C {x1 , . . . ,xn }/I is finitely generated. This proves the corollary. Due to the fact that for general coordinate changes we get a Noether normalization, we can get the following situation, which will play an important role in the proof of the Nullstellensatz. Corollary 3.3.20. Suppose that p ⊂ C {x1 , . . . ,xn } is a prime ideal. Then we can get a Noether normalization C {x1 , . . . ,xk } ⊂ C {x1 , . . . ,xn }/p, with the property that the class of xk+1 in Q(C {x1 , . . . ,xn }/p) is a primitive element of the field extension Q(C {x1 , . . . ,xk }) ⊂ Q(C {x1 , . . . ,xn }/p). We call a Noether normalization with the above property a primitive Noether normalization. 9

More precisely, there exists a Zariski open subset U of Gln (C ) such that the statement holds for all coordinate transformations in U .

3.3 Applications

101

Proof. We know from the proof of the Theorem of the Primitive Element, see for example [Van der Waerden 1971], that the class of ck+1 xk+1 + · · · + cn xn in Q(C {x1 , . . . ,xn }/p) is a primitive element for the field extension Q(C {x1 , . . . ,xn }/p) ⊃ Q(C {x1 , . . . ,xk }) for general ck+1 , . . . ,cn ∈ C . Doing the coordinate change x′k+1 = ck+1 xk+1 + · · · + cn xn the corollary follows. Corollary 3.3.21 (Hensel’s Lemma). Let F ∈ C {x1 , . . . ,xn }[T ] and suppose that F ≡ (T − c1 )s1 · · · (T − ce )se mod (x1 , . . . ,xn ), where the ci ∈ C are pairwise different. Then F = F1 · · · Fe , Fi ∈ C {x1 , . . . ,xn }[T ], Fi ≡ (T − ci )si mod (x1 , . . . ,xn ), and the Fi have degree si in T . Proof. We use induction on e. The case e = 1 is trivial. We may assume that c1 = 0 (otherwise we perform the transformation T 7→ T + c1 ). Then F (0,T ) = T s1 · H(T ) and H(0) 6= 0. Now we use the Weierstraß Preparation Theorem (3.2.4) and obtain F = E·F1 , where F1 is a Weierstraß polynomial of degree s1 with respect to T , and E is a unit in C {x1 , . . . ,xn ,T }. On the other hand, E also has to be a polynomial in T : divide F by F1 with remainder in C {x1 , . . . ,xn }[T ], F = E1 F1 + R1 with E1 ,R1 ∈ C {x1 , . . . ,xn }[T ] and degT R1 < s1 . By the uniqueness statements in the Weierstraß Division Theorem, we obtain R1 = 0 and E = E1 . Now F1 (0,T ) = T s1 and, consequently, E(0,T ) = (T − c2 )s2 · · · (T − ce )se . The corollary is proved by induction hypothesis.

Zero set of F (0,T )

Zero set of F

Corollary 3.3.21 is equivalent to the following seemingly more general lemma: Remark 3.3.22 (General Hensel’s Lemma). Let F,G,H ∈ C {x1 , . . . ,xn }[T ] such that (1) F ≡ G · H mod (x1 , . . . ,xn ); (2) G is a monic polynomial; (3) (G,H,x1 , . . . ,xn ) = C {x1 , . . . ,xn }[T ]. Then there exist G1 ,H1 ∈ C {x1 , . . . ,xn }[T ] such that F = G1 · H1 , Moreover G1 ≡ G mod (x1 , . . . ,xn ), and H1 ≡ H mod (x1 , . . . ,xn ). The proof of the equivalence is left as part of Exercise 3.3.36. Corollary 3.3.23. Let f be a polynomial, f = xkn + a1 xnk−1 + . . . + ak ,

ai ∈ C {x1 , . . . ,xn−1 }.

Consider an open set U ⊂ C n−1 on which the ai converge, and a point p ∈ U such that the polynomial

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3 Basics of Analytic Geometry xkn + a1 (p)xnk−1 + . . . + ak (p)

has k different roots in C . Then there exists an open neighborhood V ⊂ U of p, and k different holomorphic functions α1 , . . . ,αk defined on V such that for all (x1 , . . . ,xn ) ∈ V ×C k Y
xn − αj (x1 , . . . ,xn−1 ) . f (x1 , . . . ,xn ) = j=1

Proof. This is in fact Hensel’s Lemma for the case that all si are equal to one. The following corollary we need in the proof of the Finite Mapping Theorem 6.3.5, Corollary 3.3.24. Consider a monic polynomial F ∈ C {x1 , . . . ,xn }[T ].PSuppose that F (0,T ) = (T − c1 )s1 · · · (T − ce )se for pairwise different ci ∈ C . Let s = ei=1 si be the degree of F . Let pi := (0, . . . ,0,ci ) ∈ C n+1 . Then for all g1 , . . . ,ge with gi ∈ On+1,pi there exist uniquely determined qi ∈ On+1,pi and a polynomial r ∈ C {x1 , . . . ,xn }[T ] of degree less than s such that gi = qi F + r holds in On+1,pi = C {x1 , . . . ,xn ,T − ci } for i = 1, . . . ,e. Proof. We first show existence. By Hensel’s Lemma, we may write F = F1 · · · Fe , with Fi (0,T ) = (T −Q ci )si , that is, the Fi are Weierstraß polynomials with respect to T − ci . We write Ei = j6=i Fj , so that F = Ei · Fi . Note that Ei is a unit in On+1,pi . We apply the Weierstraß Division Theorem in On+1,pi for i = 1, . . . ,e: Ei−1 gi = qi′ Fi + ri where ri ∈ C {x1 , . . . ,xn }[T ] has degree smaller than si . So gi = qi′ F + ri Ei . We set P ′ r = r1 E1 + . . . + re Ee . Thus gi = qi F + r − j6=i rj Ej . In the ring On+1,pi the elements Fj for j 6= i are units, so we get X gi = qi′ F + r − rj Fj−1 F. j6=i

P

Hence we may take qi = qi′ − j6=i rj Fj−1 . To show uniqueness, it suffices to show that from qi F − r = 0 for i = 1, . . . ,e with degree of r in T smaller than s, it follows that r = 0. Thus we may write r = F1 · · · Fe qi with qi ∈ On+1,pi . We define elements P1 :=

r r , P2 := , F1 F1 F2

. . . , Pe :=

r F1 F2 · · · Fe

Note that as r = F1 · · · Fe qi the Pi are elements of On+1,pi . Note that r = P1 F1 ;

Pj−1 = Pj Fj

for j = 2, . . . ,e.

As Fi and r are polynomials in T − ci it follows from the uniqueness in the Weierstraß Division Theorem in On+1,p1 and the fact we can do polynomial division that P1 is a polynomial in T . Inductively, we get that P2 , . . . ,Pe are polynomials in T . So we get r = Pe F1 F2 · · · Fe = Pe F . As the degree of F is equal to s, and r has smaller degree, it follows that r = Pe = 0.

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103

Corollary 3.3.25. Let R = On /I be an analytic algebra, and S be an R–algebra, which is finitely generated as R–module. Suppose that S is a local ring. Then S is also an analytic algebra. Proof. First of all, take a Noether normalization C {x1 , . . . ,xk } = Ok ֒→ R. Therefore, we may assume that R = Ok from the beginning. Let mR be the maximal ideal of R, and mS be the maximal ideal of S. We have to prove that S ∼ = C {y1 , . . . ,ym }/J for some m and J. Now S/mR S is a finitely generated R/mR = C –module, hence a finitely generated vector space. As mR S ⊂ mS , it follows that S/mS is a finite extension of the field C . As C is algebraically closed it follows that S/mS = C . Therefore for all u ∈ S, there exists a c ∈ C such that u − c ∈ mS . It follows that we can take generators 1 = u0 ,u1 , . . . ,ut of S as R–module with ui ∈ mS for i = 1, . . . ,t. We look at the surjective C –algebra homomorphism Ψ : Ok [y1 , . . . ,yt ] −→ S yi

7→

ui

and conclude that S ∼ = Ok [y1 , . . . ,yt ]/J for J = Ker(Ψ). But we want to show that S is a quotient of a power series ring. Therefore, we look at the canonical inclusion Ok [y1 , . . . ,yt ] ⊂ Ok+t , and put Je := J · Ok+t . We obtain a canonical map: e θ : Ok [y1 , . . . ,yt ]/J → Ok+t /J.

Therefore, it suffices to show that θ is an isomorphism. So we have to prove two things. Step 1: θ is surjective. As S is an integral extension of Ok , there exists for all i a monic polynomial pi ∈ Ok [X] such that pi (ui ) = 0. This shows, by definition of the ideal J, that pi (yi ) ∈ J. By the Weierstraß Division Theorem, any element in Ok+t is modulo Je equivalent to a polynomial in y1 , . . . ,yt . This shows that the map θ is surjective. Step 2: θ is injective. Take generators p1 , . . . ,ps for the ideal J. Take any element p ∈ e We have to show that p ∈ J. For this it suffices to show by Ker(θ) = Ok [y1 , . . . ,yt ] ∩ J. definition of J that Ψ(p) = 0. Now p is a polynomial in y1 , . . . ,yt , and we can write: p = g1 p1 + . . . + gs ps , gi ∈ Ok+t .

The gi are power series in general, so it does not make sense to talk about ψ(gi ). We therefore calculate modulo high powers of the yi , that is, for every natural number k we write gi = gik + hik , where hik ∈ (y1 , . . . ,yt )k+1 and the gik are polynomials. Because Ψ(yi ) = ui ∈ mS it follows that modulo mk+1 S : Ψ(p) = Ψ(gik p1 + . . . + gsk ps ). But Ψ(pi ) = 0, because the pi were supposed to be generators of J. It follows Ψ(p) ∈ mk+1 S for all k. It follows from Krull’s Intersection Theorem 1.3.5 that Ψ(p) = 0, or, what is the same, p ∈ J. At the end of the section we look at generalization of this result, namely at the case S is not necessarily local. Corollary 3.3.26. Let R be an analytic algebra, and S be an R–algebra, which is finitely generated as R–module. Then

104

3 Basics of Analytic Geometry S∼ =

r M Si i=1

and the Si are analytic algebras. So S is a finite direct sum of analytic algebras. The proof uses the case that S is a local ring, see 3.3.25 and the following lemma. Lemma 3.3.27. Let R = On , and S ⊃ R be an R–algebra, finitely generated as R– module. Let m be the maximal ideal of R. Let y ∈ S be an idempotent modulo mS, that is, y 2 − y ∈ mS. Then y can be lifted to an idempotent in S, that is, there exists a z ∈ S with z 2 = z and z ≡ y mod mS. √ Proof. As S is a finitely generated R–module, and y 2 − y ∈ mS ⊂ mS it follows from Lemma 1.5.11 that y 2 − y is integral over m. From this integral equation it follows that (y 2 − y)r =

m X

k=0

ck y k , for some ck ∈ m.

Define F (T ) := (T 2 − T )r − r

r

m X

k=0

ck T k ∈ R[T ].

So F (T ) ≡ T (T − 1) mod m. We can now apply Hensel’s Lemma 3.3.21 and get F = G · H, G(T ) ≡ T r mod m and H(T ) ≡ (T − 1)r mod m. In particular, from y 2 ≡ y mod mS, we obtain G(y) ≡ y r ≡ y mod mS and H(y) ≡ (y − 1)r ≡ (−1)r (1 − y) mod mS. Hence 1 = y + (1 − y) ≡ G(y) + (−1)r H(y) mod mS. Let M be a maximal ideal of S. Then M ∩ R = m is the maximal ideal of R. This can be proved directly, or follows from the Going-Up Theorem. It follows that mS is contained in all maximal ideals. Hence all elements of 1 + mS are units in S,
and the
inverse is also congruent to 1 modulo mS, see 1.3.25. Thus G(y) + (−1)r H(y) 1 + α = 1 for some α ∈ mS. We get 1 = c · G(y) + dH(y) for suitable c,d ∈ S, and c ≡ 1 mod mS. We now define z := c · G(y). Then z ≡ y mod mS, and
z = c · G(y) + dH(y) z = z 2 + dH(y)cG(y) = z 2 because 0 = F (y) = G(y)H(y). This is what we had to show.

Proof of Corollary 3.3.26. Let m be the maximal ideal of R. The proof is by induction on dimC (S/mS). The case that S is local has already been considered. So assume that S is not a local ring. As S is a finitely generated R–module we get that S/mS is an Artinian C –algebra (1.1.6). Because of Theorem 1.4.26 we have10 S/mS ∼ = 10

r M

Ci ,

Ci nonzero local Artinian C –algebras.

i=1

Recall that in the direct sum of algebras the multiplication is component wise.

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105

The number r is bigger than one, as S/mS is not a local ring. In particular S/mS contains an idempotent different from 1. (For example (1,0, . . . ,0) in ⊕ri=1 Ci ). This idempotent can be lifted by Lemma 3.3.27 to an idempotent y ∈ S, y 6= 1. We apply the Chinese Remainder Theorem 1.1.12 and get S∼ = S/(y) × S/(1 − y). To apply the Chinese Remainder Theorem, we need that (y) + (1 − y) = R (this is obvious), and (y) ∩ (1 − y) = (0). To show this, let a ∈ (y) ∩ (1 − y). Then we can write a = αy = β(1−y). Multiply with y and get αy 2 = 0. As y 2 = y it follows a = αy = 0. Note that both y and (1−y) are not in mS. Indeed, they were chosen such that modulo mS the element y is equivalent to (1,0, 1 − y is equivalent to (0,1, . . . ,1). So it follows  . . . ,0), and
 that dimC (S/mS) > dimC S/ mS + (y) . Similarly dimC (S/mS) > dimC S/ mS +
 (1 − y) . Thus we can apply the induction hypothesis, and the corollary follows.

Exercises

3.3.28. Compute the power series of y = −1 +

√

x + 1 using the ideas of Example 3.3.4.

3.3.29. Prove the Inverse Function Theorem 3.3.7. 3.3.30 (General Noether Normalization). Let I ⊂ C {x1 , . . . ,xn } be an ideal. Then, after a general linear coordinate change, we find that there exists a k such that C {x1 , . . . ,xk } ⊂ C {x1 , . . . ,xn }/I and for all j > k there are monic polynomials fj ∈ I ∩ C {x1 , . . . ,xj−1 }[xj ] (with respect to xj ) with degxj (fj ) = ord(fj ) (considered as a power series in C {x1 , . . . ,xj }). If, moreover, I is a prime ideal then Q(C {x1 , . . . ,xn }/I) = Q(C {x1 , . . . ,xk })[xk+1 ]/(fk+1 ). (Hint: Assume in the proof of 3.3.19 that the order of g ∈ I is b and apply 3.2.2 and the Weierstraß Preparation Theorem 3.2.4. In case of I being a prime ideal, combine this idea with the proof of 3.3.20.) 3.3.31. Consider a general Noether normalization Ok ⊂ OX,x = On /p in the sense of 3.3.30. Let f be a power series of order t. Show that modulo p the element f is equivalent to a polynomial in xk+1 , . . . ,xn which has order at least t. (Hint: Reduce to the case that p = (Pk+1 , . . . ,Pn ), where the Pj are Weierstraß polynomials in xj with coefficients in C {x1 , . . . ,xk }. For the case p = (Pn ), use Exercise 3.2.19, and apply induction to the coefficients.) 3.3.32. Consider a prime ideal p ∈ C {x1 , . . . ,xn }, and a Noether normalization C {x1 , . . . ,xk } ⊂ C {x1 , . . . ,xn }/p. Let Pj (X) be an integral equation of xj for j = k + 1, . . . ,n of minimal degree. Prove that the Pj are Weierstraß polynomials. (Hint: Use the Weierstraß Preparation Theorem.) 3.3.33. Let f ∈ C {x1 , . . . ,xn ,y} with f (0) = 0 and ∂f (0) 6= 0. Suppose moreover that f ∈ ∂y (y) + (x1 , . . . ,xn )c . Then there exists a ϕ ∈ (x1 , . . . ,xn )c with f (x1 , . . . ,xn ,ϕ) = 0. Generalize this to the case of m variables y1 , . . . ,ym . (Hint: Follow the steps in a proof of the formal Implicit Function Theorem.)

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3 Basics of Analytic Geometry

3.3.34 (Newton’s Lemma). Let f1 , . . . ,fm ∈ C {x1 , . . . ,xn , y1 , . . . ,yN }, m “≤ N” and J = ∂fi . Suppose J(f1 , . . . ,fm ) the ideal generated by the m–minors of the Jacobian matrix ∂y j ` ´ ` ´ 2 c N fi x,y(x) ≡ 0 mod J y(x) (x1 , . . . ,xn ) for i = 1, . . . ,m, y(x) ∈ C {x1 , . . . ,xn } and c ≥ 1. Then there exists a y ∈ C {x1 , . . . ,xn }N such that ` ´ fi x,y(x) = 0,` i =´ 1, . . . ,m y ≡ y mod J y(x) (x1 , . . . ,xn )c . Prove this lemma along the following lines. “ ” ` ´ ∂fi (1) Let M1 , . . . ,Ms , with s = N be the m–minors of ∂y . Take new variables tij (y(x)) m j ! f1 s P .. . Show that tij Mj . Let f = for 1 ≤ i ≤ N and 1 ≤ j ≤ s. Substitute yi = y i + . j=1 fm

it suffices to solve for tij an equation of type

(3.16)

0 = f (y) =

X

Mi Mj rij +

X

Mi Mj hij +

„

0 P

t1j Mj «B j ∂fi B .. (y) B ∂yj @ P . tNj Mj j

c

1

C C C. A

where the rij is a column vector of size m with entries in (x1 , . . . ,xn ) and hij is a column vector of size m with entries in (t11 , . . . ,tNs )2 . (2) Consider an m × N matrix A with entries in a ring R. Let M be an m–minor of A. Show that there exists a N × m matrix B such that AB = M Idm , where Idm is the m × m identity matrix. (Hint: If m = N , one can take for B the adjoint of A, see 1.2.9.) „ « ∂fi (3) Use this to replace Mi Mj in (3.16) by Mj (y) Bi for certain matrices Bi . ∂yj (4) Show that for all j with 1 ≤ j ≤ s it suffices to solve for tij the equation 0 1 t1j s X B C Bi (rij + hij ) + @ ... A = 0. i=1 tNj Apply the Implicit Mapping Theorem to solve this system of equations.

3.3.35. Newton’s Lemma cannot be strengthened in the following sense, that is, the following statement is not true in general. Let f ∈ C {x1 , . . . ,xn ,y}, and y(x) ∈ C {x1 , . . . ,xn ,y} with « „ ` ´ ´ ∂f ` f x,y(x) ≡ 0 mod x,y(x) (x)c . ∂y Then there exists a y ∈ C {x1 , . . . ,xn ,y} with ` ´ (1) f x,y(x) = 0. (2) y ≡ y mod (x)c .

Indeed, show that f (x,y) = y 2 − x3 is a counterexample.

3.3.36. A local C –algebra R with maximal ideal m is called Henselian if the statement of Hensel’s Lemma (see 3.3.21), where C {x1 , . . . ,xn } is replaced by R, and (x1 , . . . ,xn ) is replaced by m, holds. Prove the following statements. (1) Let R be an analytic algebra. Then R is Henselian.

3.4 Germs of Analytic Spaces

107

(2) Show that K[x](x) is not Henselian, by looking at the polynomial F = T 2 − x − 1. (3) Let p be a prime ideal. Prove that more generally K[x1 , . . . ,xn ]p is not Henselian. (4) Prove Remark 3.3.22. 3.3.37. Prove that the point 0 ∈ X in Example 3.3.12 (2) is singular.

3.4

Germs of Analytic Spaces

In this section we study germs of analytic spaces. We already know what an analytic space is, see 3.1.12, but we still have to define the notion of a germ. Definition 3.4.1. (1) Let X be a topological space and p ∈ X. Subsets A and B of X are called equivalent at p if there exists a neighborhood U of p such that A∩U = B∩U . It is easy to check that this is indeed an equivalence relation, see Exercise 3.4.33. The equivalence class of A at p we call the germ of A at p. A is called a representative of the germ. We write (A,p) for the germ of A at p. (2) Let (X,x) and (Y,x) be germs. We define (X,x) ⊂ (Y,x) if there are representatives X of (X,x) and Y of (Y,x) such that X ⊂ Y . It is not so difficult to see that (X,x) = (Y,x) if and only if (X,x) ⊂ (Y,x) and (Y,x) ⊂ (X,x). (3) Let (X,x) ⊂ (Y,x) and (Z,x) ⊂ (Y,x) be germs. Then we define (X,x) ∩ (Z,x) to be the germ of X ∩ Z at x for any representative X of (X,x), and Z of (Z,x) at x.

We define (X,x) ∪ (Z,x) to be the germ of X ∪ Z at x for any representative X of (X,x), and Z of (Z,x) at x.

We now define germs of analytic spaces. Definition 3.4.2. (1) A germ of an analytic space (X,x) is a germ at x of a locally analytic subset of C n .
(2) Let f ∈ On,x . Then we define a germ of an analytic hypersurface V (f ),x as follows. Consider an open neighborhood U of x on which f converges.
Consider the analytic subset V (f ) := {p ∈ U : f (p) = 0} of U . Then V (f ),x is the germ of V (f ) at zero, and is called the zero set of f . (3) Let I = (f1 , .
. . ,fs ) ⊂ On,x be an ideal. Then we define the germ of the analytic space V (I),x by s

\ V (fi ),x . V (I),x = i=1

This definition is independent of the choice of generators fi of
I, as is easily checked. Any germ of an analytic space is a germ of type V (I), 0 for some ideal I ⊂ C {x1 , . . . ,xn }.

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3 Basics of Analytic Geometry

(4) Let (X, 0) be a germ of an analytic space. Then define
I (X,x) = {f ∈ On,x : (X,x) ⊂ V (f ), 0 }.

The inclusion here is an inclusion of germs. So f ∈ I (X,x) if there exists a representative X of (X, 0) and an open neighborhood U of x such that X is an analytic subset of U , f converges on U , and its restriction to X is the zero function.
Remark 3.4.3.
We had to be a little careful with the definition of V (I),x . The defi
nition of V (I),x as ∩f ∈I V (f ),x would not make sense. In fact, in general it does not make sense to talk about infinite intersections of germs. Namely, if we have germs (Xi ,x) for i ∈ I, and if representatives are defined on open sets Ui , then it is very well possible that ∩i∈I Ui = {0}, if the set I has infinitely many elements. If I is a finite set however
then ∩i∈I Ui is still an open neighborhood of x. Note that in the definition of V (I),x we already used the basic fact that I is finitely generated, which follows from the fact that C {x1 , . . . ,xn } is Noetherian. We have properties similar to the statements in 2.2.2, except that, keeping in mind the remark above, we are only allowed to take finite unions and intersections! The proofs are all similar, and left to the reader. We also have:
√ Theorem 3.4.4 (Nullstellensatz). I V (I),x = I.

It obviously suffices to prove the Nullstellensatz for x = 0. For germs of analytic spaces the Nullstellensatz cannot be interpreted as a statement on maximal ideals. This is because the power series ring is a local ring, that is, it just has one maximal ideal! Moreover, it seems that also the idea of the proof of the Projection Theorem 2.2.8 does not work in the local case. The point is (in the notation of the Projection Theorem) that we first have to define representatives, and in particular generators h1 , . . . ,ht for I ′ = I ∩ C {x1 , . . . ,xn−1 }, which converge on some open neighborhood U of 0. The proof of the Projection Theorem provides us with a holomorphic function g on U , which vanishes on X and such that g(a) 6= 0. So it follows that the germ of g is in I ′ , but what is not clear is that g is in the ideal generated by the h1 , . . . ,ht all over U . It is therefore no surprise that proofs of the Nullstellensatz for the power series ring tend to be somewhat more difficult. As in the proof of the Nullstellensatz in the algebraic case, it suffices to prove the Nullstellensatz for prime ideals p. As the proof is completely similar, see 2.2.12, we leave the proof to the reader, and only formulate the result. Lemma 3.4.5. Suppose the Nullstellensatz holds for prime ideals. Then the Nullstellensatz holds for all ideals. So from now on we will look at prime ideals. As in the proof of the Nullstellensatz in the affine case, we will look at a Noether normalization,

C {x1 , . . . ,xk } ⊂ C {x1 , . . . ,xn }/p.
Put (X, 0) = V (p), 0 . Then we would like to show that the linear projection (X, 0) −→ (C k , 0) is surjective. This should be analogous to 2.2.10, where we showed that for π : X −→ K r a Noether normalization, the map π is surjective. This was the essential step in the proof of the Nullstellensatz. Note that we did, until now, not even define what a map between germs of spaces is!

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109

Definition 3.4.6. Let (X,p) and (Y,q) be two germs of topological spaces. Agerm of a continuous map f : (X,p) −→ (Y,q) is defined as an equivalence class of maps f : U −→ W , with f (p) = q, and where U and W are representatives of (X,p) and (Y,p) respectively. Two such maps f1 : U1 −→ W and f2 : U1 −→ W are called equivalent if they agree on an open neighborhood of p contained in U1 ∩ U2 . The problem with germs of continuous maps is, that it makes in general no sense to talk about the image. By this we mean that in general it might happen that for every subgerm (Z,q) ⊂ (Y,q), there does not exist an open neighborhood U of p in X such that (Z,q) is represented by f (U ). For an example, see Exercise 3.4.34. In particular, it does not make sense to talk about surjectivity. For special kind of maps, so-called finite maps, this problem does not occur. Let us define this notion. Definition 3.4.7. Let f : X → Y be a continuous map between topological spaces. (1) f is called closed if the image f (A) ⊂ Y is closed for all closed subspaces A ⊂ X. (2) f is called quasi-finite if for all p ∈ Y the fiber f −1 (p) consists of a finite number of points. (3) f is called finite if it is both closed and quasi-finite. The simplest example of a mapping which is quasi-finite but not finite probably is the inclusion mapping C \ {0} → C . (Here C has the standard Euclidean topology.) Lemma 3.4.8. Let X and Y be topological
spaces, and f : X −→ Y be closed. Let p ∈ X and let q = f (p). Assume that f −1 f (p) = {p}. Let A,B ⊂ X with p ∈ A, p ∈ B, defining the same germ at p. Then (f (A),q) = (f (B),q), that is, f (A) and f (B) define the same germ in q. Proof. By the definition of a germ, we have to find an open subset V of q in Y with V ∩ f (A) = V ∩ f (B). Step 1. We first show that there exists a neighborhood basis of p in X such that for each element U of this basis we have the equality U = f −1 (f (U )). So let W be an open neighborhood of p. Then X \ W is closed, so that f (X \ W ) is closed by assumption. Then T := Y \ f (X \ W ) is open. Define U = f −1 (T ). Then U is open and we have the following properties:
(1) As U = f −1 (T ), it follows f (U ) = f ◦ f −1 (T ) ⊂ T . Hence f −1 f
(U ) ⊂ f −1 (T ) = U . As the converse inclusion is trivial, it follows that f −1 f (U ) = U . (2) U ⊂ W , as, otherwise, there exists an x ∈ U , x 6∈ W . By definition of T , it immediately follows from x 6∈ W that f (x) ∈ f (X \ W ), so that f (x) 6∈ T . But we have f (U ) ⊂ T , so this is a contradiction.

(3) We have p ∈ U . To show this, note that as p ∈ W and f −1 (f (p)) = {p} it follows f (p) 6∈ f (X \ W ) and then f (p) ∈ T = Y \ f (X \ W ). Thus p ∈ f −1 (T ) = U .

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3 Basics of Analytic Geometry

Step 2. Let W be an open neighborhood of p such that A ∩ W = B ∩ W . Take U ⊂ W and T ⊂ Y as in Step 1. In particular U ∩ A = U ∩ B. We claim that f (U ∩ A) = T ∩ f (A). As f (U ) ⊂ T , the inclusion ⊂ is a tautology. To show the converse inclusion, take a point a ∈ T ∩ f (A). Then a = f (b) for some b ∈ A. Moreover, a ∈ T , so that b ∈ f −1 (T ) = U . This shows that b ∈ U ∩ A, showing the converse inclusion. As similarly f (U ∩ B) = T ∩ f (B), and U ∩ A = U ∩ B, it follows that T ∩ f (A) = T ∩ f (B). This is what we had to show. The lemma is the basis for the following definition: Definition 3.4.9. Let (X,p) and (Y,q) be two germs of topological spaces. Let f : (X,p) −→ (Y,q) be a germ of a continuous map. (1) f is called finite, if there exists a representative f : X −→ Y which is finite.
(2) Assume that f is finite. Then the image of f is given by Im(f ) := f (X),q , where
f : X −→ Y is a representative such that f −1 f (p) = {p} and f is finite. Note that because of Lemma 3.4.8 Im(f ) is well-defined.

(3) Assume that f is finite. Then f is called surjective if Im(f ) = (Y,q). Note that f is surjective, if and only if f : X
−→ Y is surjective for suitable representatives such that f is finite and f −1 f (p) = {p}. The first example we give of a finite map is the following.

Lemma 3.4.10. Let P = y r + a1 y r−1 + . . . + ar be a polynomial, with coefficients ai ∈ Ok . Consider an open neighborhood U ⊂ C k of 0 such that the power series a1 , . . . ,ar converge. Define: X := {P = 0} ∩ (U × C ).

Then the canonical projection π : X −→ U is finite.

Proof. As the map obviously is quasi-finite, we only have to show that π is closed. Therefore, let A ⊂ X be a closed subset, and take a point p ∈ π(A). We have to show that p ∈ π(A). As p ∈ π(A) we have a sequence (pi ,yi )

i = 1,2, . . .

with pi ∈ π(A), yi ∈ C , (pi ,yi ) ∈ A, and limi→∞ pi = p. Because (pi ,yi ) ∈ A ⊂ X, we have yir + a1 (pi )yir−1 + . . . + ar (pi ) = 0 for i ≥ 1.

As the aj are holomorphic, in particular continuous, we have that limi→∞ aj (pi ) = aj (p). Therefore, the sequence aj (pi ), i = 1,2, . . . is bounded. We claim that the sequence yi too. Indeed, |yi | < 1 or |yi | ≥ 1. But in the second case, we obviously have is bounded, aj (pi ) yj−1 ≤ |aj (pi )|, so that: i ar (pi ) |yi | = r−1 + . . . + a1 (pi ) ≤ |ar (pi )| + . . . + |a1 (pi )| , yi hence the sequence yi is bounded. There exists therefore a convergent subsequence yi1 ,yi2 , . . ., converging to say y. As A is closed, the point (p,y) = limj→∞ (pj ,yij ) belongs to A. Therefore, p ∈ π(A), which is what we had to prove.

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U

Typical picture of the zero set of a monic polynomial. Lemma 3.4.11 (Continuity of Roots). Consider a Weierstraß polynomial: P (x,y) = y r + a1 y r−1 + . . . + ar ;

ai ∈ m ⊂ C {x1 , . . . ,xk }.

(1) Then for every arbitrary small neighborhood V of 0 in C there exists a small neighborhood U of 0 in C k such that the power series a1 , . . . ,ar converge on U and such that for every p in U the r zeros (counted with multiplicity) of y r + a1 (p)y r−1 + . . . + ar (p) = 0 lie in V . (2) Let π :
C k+1 −→ C k be the projection on the first k coordinates. Then π : V (P ), 0 −→ (C k , 0) is surjective.

(3) Suppose that P is irreducible. Let ∆ be the discriminant of P and let U ⊂ C k be an open set such that a1 , . . . ,ar converge on U . Let X = {(x,y) ∈ U × C : P (x,y) = 0}, D = {x ∈ U : ∆(x) = 0} and let π : X −→ U be the projection on the first k coordinates. Then X \ π −1 (D) = X.

V

U

Typical picture of the zero set of of a Weierstraß polynomial.

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3 Basics of Analytic Geometry

Proof. (1) The proof runs as the proof of the previous lemma. Suppose the converse. Then there exists a small neighborhood V of 0 in C such that for all neighborhoods U of 0 in C k on which the power series a1 , . . . ,ar converge, there exists a point p ∈ U and a root of the polynomial P (p,y) = 0, which is not in V . Take a sequence Ui of open neighborhoods of 0 whose intersection is {0}, and points pi ∈ Ui such that there is a root yi of P (pi ,yi ) = 0 which does not lie in V . So there exist a C > 0, infinitely many yi with |yi | ≥ C and infinitely many pi ∈ C k such that {pi } converges to 0 and P (pi ,yi ) = 0. But then it follows as in the proof of 3.4.10 that |yi | ≤

1 C r−1

|ar (pi )| +

1 C r−2

|ar−1 (pi )| + . . . + |a1 (pi )| .

By assumption limi→∞ aj (pi ) = 0 for all j. So we can make yi as small as we want, in contradiction to |yi | ≥ C.

(2) Let U ⊂ C k be an open neighborhood of 0 such that the power series a1 , . . . ,ar converge. Define: X := {(x,y) ∈ U × C : P (x,y) = 0}.

The canonical map π : X −→ U is finite (Lemma 3.4.10) and, obviously, surjective. Moreover, because P is a Weierstraß polynomial, we have π −1 (0) = 0. This implies that V (P ), 0) −→ (C k , 0) is surjective, see 3.4.9(3). (3) We have to prove that for any point (a,b) ∈ π −1 (D) there exists a sequence (ai ,bi ) ∈ X \ π −1 (D) with limi→∞ (ai ,bi ) = (a,b). By Theorem 3.1.14 there exists a sequence ai in U \ D with limi→∞ ai = a. We consider P at the point (a,b). Here we can write P = U ·Q, U ∈ C {x,y} a unit, Q a Weierstraß polynomial with respect to x = x − a and y = y − b. Now we use (1) and obtain for all i that there exists an open neighborhood Vi of a such that for ai ∈ Vi all roots of Q(ai ,y) = 0 satisfy |y| < 1i . Therefore, we may choose a sequence bi such that (ai ,bi ) ∈ X converging to (a,b) and ai 6∈ D. Remarks 3.4.12. (1) Alternatively, Lemma 3.4.11 could have been proved by using Rouch´e’s Theorem R ′ of function theory. Even simpler, one could use the integral ff , which counts the number of zeroes of a holomorphic function in one variable and a continuity argument. The lemma in fact is the starting point for an alternative proof of the Weierstraß Division and Preparation Theorem, see the book of Gunning and Rossi, [Gunning-Rossi 1965]. (2) We will use the Nullstellensatz to generalize 3.4.11 (3). In fact we will prove that,
if X is a small representative of V (p),0 for a prime ideal p ⊂ C {x1 , . . . ,xn }, and h ∈ C {x1 , . . . ,xn }, h 6∈ I (X, 0), then X \ {h = 0} = X. Lemma 3.4.13. Let I ⊂ C {x1 , . . . ,xn } be an ideal. (1) Suppose that there exists an f ∈ I which is regular in xn . Let p : C n −→ C n−1 ;

(x1 , . . . ,xn ) 7→ (x1 , . . . ,xn−1 ).
Then there exists a representative X of V (I), 0 and an open neighborhood U of 0 ∈ C n−1 such that the map p : X −→ U is a finite map.

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(2) Suppose more generally, that C {x1 , . . . xn }/I is a finitely
generated C {x1 , . . . ,xk }– module. Then there exists a representative X of V (I), 0 and an open neighborhood U of C k such that the projection p : X −→ U is a finite map. Proof. By the Weierstraß Preparation Theorem, we may assume that f = up, where p is a Weierstraß polynomial, and u is a unit. So we may assume that f is a Weierstraß polyn−1 nomial. By 3.4.10 and a representative
there exists an open neighborhood U of 0 in C W of V
(f ), 0 such that p : W −→ U is finite. Consider a representative X ⊂ W of V (I), 0 . As X is a closed subset of W (being the zero set of finitely many holomorphic functions), it follows that p : X −→ U is finite too. For the second statement, it follows that each of the classes of xj for j = k + 1, . . . ,n satisfy an integral equation with coefficients in C {x1 , . . . ,xk }. Hence there exist Weierstraß polynomials fj (xj ) for j ≥ k + 1, with coefficients in C {x1 , . . . ,xk } which are contained in I. By Exercise 3.4.35 there exists an open neighborhood U ⊂ C k such that the projection
map p : X −→ U is finite for a suitable representative X of V (fk+1 , . . . ,fn ), 0 . So we can argue as above to deduce the lemma.

We will generalize the second part of Lemma 3.4.11 in the Local Parametrization Theorem 3.4.14. In particular we will prove
Let p ⊂ C {x1 , . . . ,xn } be a prime ideal and π : V (p), 0 −→ (C k , 0) be a primitive Noether normalization. Then π is finite and surjective. This is the basis to prove the Nullstellensatz.

Proof of the Nullstellensatz 3.4.4. (Using the Local Parametrization Theorem 3.4.14.) It suffices to show the Nullstellensatz for prime ideals p. We choose a primitive Noether
normalization π : V (p), 0 −→ (C k , 0) and use the fact that π is surjective. Then the proof is completely analogous to the proof of the Nullstellensatz in the affine case, see 2.2.11.
Recall that we want to prove that for some Noether normalization V (p), 0 = (X, 0) −→ (C k , 0) is surjective. This will be proved in two steps, namely we will first
show that (X, 0) projects surjectively onto a hypersurface V (P ), 0 in (C k+1 , 0), and then we project further onto (C k , 0). The first projection is “generically one to one”. This projection is an important tool in several proofs in this book, and is called the Local Parametrization Theorem.
Theorem 3.4.14 (Local Parametrization Theorem). Let (X, 0) = V (p), 0 for p ⊂ C {x1 , . . . ,xn } a prime ideal and C {x1 , . . . ,xk } ⊂ C {x1 , . . . ,xn }/p be a primitive Noether normalization (cf. 3.3.20). Let π : C n −→ C k+1 be the projection on the first k + 1 coordinates, p : C k+1 −→ C k be the projection on the first k coordinates. Then we have the following statements. (1) p ∩ C {x1 , . . . ,xk+1 } = (P ), for P ∈ C {x1 , . . . ,xk }[xk+1 ] a Weierstraß polynomial. (2) Let ∆ be the discriminant of P with respect to xk+1 . Then (a) there exist qk+2 , . . . ,qn ∈ C {x1 , . . . ,xk }[xk+1 ] such that Qj := ∆xj − qj ∈ p for j = k + 2, . . . ,n; (b) there exist f1 , . . . ,fs ∈ C {x1 , . . . ,xk }[xk+1 , . . . ,xn ] such that

114

3 Basics of Analytic Geometry i. (f1 , . . . ,fs ) = p in the ring C {x1 , . . . ,xn }. ii. fj ∈ C {x1 , . . . ,xk }[xk+j ] are Weierstraß polynomials for j = 1, . . . ,n − k and P,Qk+2 , . . . ,Qn ∈ {f1 , . . . ,fs }; iii. (P,Qk+2 , . . . ,Qn )C {x1 , . . . ,xk }∆ [xk+1 , . . . ,xn ] = (f1 , . . . ,fs )C {x1 , . . . ,xk }∆ [xk+1 , . . . ,xn ].

(3) There is an open neighborhood U of 0 in C k such that P,∆,Qk+2 , . . . ,Qn ,f1 , . . . ,fs converge on U × C n−k and for X = {x ∈ U × C n−k : f1 (x) = · · · = fs (x) = 0}

Y = {y ∈ U × C n−k : P (y) = Qk+2 (y) = · · · = Qn (y) = 0} X ′ = {x′ ∈ U × C : P (x′ ) = 0} D = {x ∈ U : ∆(x) = 0}

the following statements hold: (a) Y \ (p ◦ π)−1 (D) = X \ (p ◦ π)−1 (D);11

(b) π : X \ (p ◦ π)−1 (D) −→ X ′ \ p−1 (D) is biholomorphic;

(c) X \ (p ◦ π)−1 (D) is a complex submanifold of U × C n−k ;

(d) the map p ◦ π : X −→ U is surjective and finite and (p ◦ π)−1 (0) = {0}. (4) The map p ◦ π : (X, 0) −→ (C k , 0) is finite and surjective.
(5) The map π : (X, 0) −→ V (P ), 0 is finite and surjective.

Before giving the proof of the Local Parametrization Theorem we give an example. In this example all the power series are in fact polynomials. The neighborhood U can be taken to be the whole of the complex vector space. Example 3.4.15. Consider the map C −→ C 3 given by t 7→ (t3 ,t4 ,t5 ) = (x,y,z). Its image can be calculated, see 2.3.18, to be the zero set of the ideal I = (y 2 − xz,x3 − yz,z 2 − x2 y). A Noether normalization map is given by the projection on the x–axis. Indeed modulo I one has y 3 = xyz = x4 , and z 3 = x2 yz = x5 . This gives the integral equation z 3 − x5 = 0, and y 3 − x4 = 0.12 To show that this is a Noether normalization one has to show that I ∩ C {x} is zero. For this, see Exercise 3.1.28. We in fact have a primitive Noether normalization. Here y plays the role of xk+1 . This follows because 2 in the quotient field z = yx . The polynomial P is equal to y 3 − x4 . The discriminant is ∆ = 27x8 . So the Local Parametrization Theorem says that x8 z is a polynomial in y with coefficients in C {x}. Indeed, even xz = y 2 . So in this case the polynomial q is equal to 27x7 y 2 . The space X ′ is given by y 3 − x4 = 0. For each a 6= 0, and each y = b 2 with b3 − a4 , there is a unique point on X, with z–coordinate z = ba . Thus the map π in the Local Parametrization Theorem is indeed surjective. In fact, it is easily seen that for x 6= 0, the analytic set is a complex submanifold. 11

12

The equation implies that X \ (p ◦ π)−1 (D) =

n`

p,

´ qk+2 (p) n (p) , . . . , q∆(p) ∆(p)

o : p ∈ U × C , P (p) = 0 . This

(together with 3.3.23) is the reason to call 3.4.14 Local Parametrization Theorem. A different way to see this is by applying the general Weierstraß Division Theorem 3.2.10. Consider the C {x}–module C {x,y,z}/I. Divide out the maximal ideal (x) of C {x}. We get the C –vector space C {y,z}/(y 2 ,yz,z 2 ), which is finite-dimensional. A basis is given by the classes of 1,y,z.

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Proof of the Local Parametrization Theorem 3.4.14. Step 1. We first show (1) and (2a). We apply the Finiteness of Normalization Theorem 1.5.19. We obtain that p ∩ C {x1 , . . . ,xk }[xk+1 ] = (P ), where P is the minimal polynomial of the element xk+1 mod p in the field extension Q(C {x1 , . . . ,xn }/p) ⊃ Q(C {x1 , . . . ,xk }). By Exercise 3.3.32, P is a Weierstraß polynomial. Using the Weierstraß Division Theorem 3.2.3, we obtain p ∩ C {x1 , . . . ,xk+1 } = (P ), as, otherwise, we would find a nonzero element in p which is a polynomial in xk+1 of degree smaller than the degree of P . This is in contradiction to the fact that P is the minimal polynomial of the field extension. This proves (1). To prove (2a) we apply again 1.5.19. So for all f ∈ On there exists qf ∈ C {x1 , . . . ,xk }[xk+1 ] with ∆ · f − qf ∈ p. In particular, this holds for xk+2 , . . . ,xn . Therefore, there exist polynomials qk+2 , . . . ,qn ∈ C {x1 , . . . ,xk }[xk+1 ] such that Qj := ∆xj − qj ∈ p for j = k + 2, . . . ,n. This shows (2a).

Step 2. We now prove (2b). Consider the minimal polynomials Pj ∈ C {x1 , . . . ,xk }[xj ] of xj for j = k + 2, . . . ,n. Thus Pk+1 := P .

Step 2a. We first show that p has a system of generators in C {x1 , . . . ,xk }[xk+1 , . . . ,xn ]. Indeed, it follows that the extension C {x1 , . . . ,xk } ⊂ C {x1 , . . . ,xn }/(Pk+1 , . . . ,Pn ) is finite. We put P i := Pi (0, . . . ,0,xk+1 , . . . ,xn ). It then easily follows that the vector space C {xk+1 , . . . ,xn }/(P k+1 , . . . ,P n ) is generated by monomials in xk+1 , . . . ,xn . It follows from Corollary 3.2.12 that every element in C {x1 , . . . ,xn } is modulo (Pk+1 , . . . ,Pn ) equivalent to an element in C {x1 , . . . ,xk }[xk+1 , . . . ,xn ]. This then certainly holds modulo p. Hence we can choose generators f1 , . . . ,fs of p such that • f1 = Pk+1 = P, . . . ,fn−k = Pn ; • fn−k+1 , . . . ,fs ∈ C {x1 , . . . ,xk }[xk+1 , . . . ,xn ]; • Qk+2 , . . . ,Qn ∈ {f1 , . . . ,fs }. Step 2b. Put J∆ := (P,Qk+2 , . . . ,Qn )C {x1 , . . . ,xk }∆ [xk+1 , . . . ,xn ] I∆ := (f1 , . . . ,fs )C {x1 , . . . ,xk }∆ [xk+1 , . . . ,xn ]. To prove iii of 2b, We have to show J∆ = I∆ . The inclusion J∆ ⊂ I∆ is obvious and, therefore, we have a surjection (3.17)

C {x1 , . . . ,xk }∆ [xk+1 , . . . ,xn ]/J∆ ։ C {x1 , . . . ,xk }∆ [xk+1 , . . . ,xn ]/I∆ . q (x ,...,x

)

The equation Qj = 0 says that xj = j 1 ∆ k+1 , so we can eliminate the variables xk+2 , . . . ,xn in C {x1 , . . . ,xk }∆ [xk+1 , . . . ,xn ]/J∆ . After having done that, formula (3.17) gives a surjection C {x1 , . . . ,xk }∆ [xk+1 ]/(P ) ։ C {x1 , . . . ,xk }∆ [xk+1 , . . . ,xn ]/I∆ . We claim that is in fact a bijection, that is, I∆ ∩ C {x1 , . . . ,xk }∆ [xk+1 ] = (P ). As I generates p, it follows from the first part of the Theorem that I ∩ C {x1 , . . . ,xk }[xk+1 ] = (P ). As (P ) is a prime ideal, and ∆ ∈ / (P ), the claim follows.

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Step 3. The proof of (3a) now is easy. Let f1 , . . . ,fs ∈ C {x1 , . . . ,xk }[xk+1 , . . . ,xn ] be the generators of p we constructed in Step 2. By (2b) we have for some b = b(i): (3.18)

∆b fi =

n X

αj (fi )Qj + α(fi )P

j=k+2

for certain αj (fi ),α(fi ) ∈ C {x1 , . . . ,xk }[xk+1 , . . . ,xn ]. Choose U ⊂ C k an open neighborhood of 0 such that P,∆,Qk+2 , . . . ,Qn , α(fi ),αj (fi ) converge on U × C n−k for all i,j. Let p ∈ Y with ∆(p) 6= 0 be given. Then from (3.18) and P (p) = Qj (p) = 0 for j = k + 2, . . . ,n, it follows that fi (p) = 0 for i = 1, . . . ,s. This exactly means that Y \ (p ◦ π)−1 (D) = X \ (p ◦ π)−1 (D).

Step 4. We now prove (3b) and (3c). Consider a point a ∈ X ′ \ p−1 (D). From 3.3.23 X ′ is in a small neighborhood U ′ of a a complex submanifold of U ′ × W , given by the graph of a function xk+1 = α(x1 , . . . ,xn ). By definition of Qj = ∆xj − qj . Therefore, there is a unique point of X \ (p ◦ π)−1 (D) mapping to a. If the coordinates of a are (a1 , . . . ,ak+1 ) the unique point in π −1 (p) has coordinates   qn (a1 , . . . ,ak+1 ) qk+2 (a1 , . . . ,ak+1 ) . ,..., a1 , . . . ,ak ,ak+1 , ∆(a1 , . . . ,ak ) ∆(a1 , . . . ,ak )

So we have an inverse to the map π : X \ (p ◦ π)−1 (D) −→ X ′ \ p−1 (D), which we see to be biholomorphic. Indeed, locally in U ′ × W × C n−k−1 , the set Y \ (p ◦ π)−1 (D) is given by the set of equations xk+1 − α(x1 , . . . ,xk ) = xk+2 −

qn (x1 , . . . ,xk+1 ) qk+2 (x1 , . . . ,xk+1 ) = · · · = xn − = 0, ∆(x1 , . . . ,xk ) ∆(x1 , . . . ,xk )

and is therefore a complex submanifold. Step 5. We prove (3d), (4) and (5). By Lemma 3.4.13 the map p ◦ π : X −→ U is finite. The map p : X ′ \ p−1 (D) −→ U \ D is, obviously, surjective and because of (3b) the map p ◦ π : X \ (p ◦ π)−1 (D) −→ U \ D is surjective. As p ◦ π is closed, the closure of U \ D is in p ◦ π(X). Using Theorem 3.1.14 we obtain that U \ D = U and, therefore, π(X) = U . Now (4) and (5) are immediate consequences (cf. 3.4.9 and 3.4.13).
Remarks 3.4.16. Let p be a prime ideal, and (X, 0) = V (p), 0 . (1) Note that in the proof of the Local Parametrization Theorem, we showed that the number of points in the general fiber of a Noether normalization π : X −→ U is equal to the degree of P , that is, the degree of the field extension Q(Ok ) ⊂ Q(OX, 0 ). In fact, this holds generally for “finite maps between irreducible germs of analytic spaces”, as will be shown in the exercises. (2) We showed that locally, at a point a ∈ X \ (p ◦ π)−1 (D), the ideal I (X,a) ⊂ On,a is generated by Qk+2 , . . . ,Qn ,P . Moreover, On,a /I (X,a) is isomorphic to Ok . It follows that if (f1 , . . . ,fs ) = I (X, 0), the ideal of (X, 0), then there exists a neighborhood U of 0 in X such that the fi converge on U and for all a ∈ U ∩(X \D) the ideal I (X,a) is generated by f1 , . . . ,fs .13 13

In fact, this holds for all a ∈ U . This is the Coherence Theorem of Oka-Cartan to be proved in Chapter 6.

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(3) In the notation of the Local Parametrization Theorem, write OX ′ ,0 = Ok+1 /(P ). The cokernel K := OX,0 /OX ′ ,0 is an Ok –module. It was shown in the proof of the Local Parametrization Theorem, that there exists an a ∈ N such that ∆a · K = (0). In particular, we can view K as an Ok /(∆a )–module. Definition 3.4.17. Let (X,x) be a germ of an analytic space. Then (X,x) is called irreducible if from (X,x) = (X1 ,x) ∪ (X2 ,x), with (X1 ,x) and (X2 ,x) germs of analytic spaces it follows that either (X,x) = (X1 ,x) or (X,x) = (X2 ,x). Corollary 3.4.18. Let (X,x) be a germ of an analytic space. (1) (X,x) is irreducible if and only if I (X,x) is a prime ideal. (2) Let (X,x) ⊂ (C n ,x) be irreducible, and h ∈ On , h 6∈ I (X,x). Then for small enough representatives X \ {h = 0} = X. (3) There is a, up to permutation, unique decomposition (X,x) = (X1 ,x) ∪ · · · ∪ (Xr ,x), with (Xi ,x) irreducible and (Xj ,x) 6⊂ (Xi ,x) for i 6= j. This is called the irreducible decomposition of (X,x). The (Xi ,x) are called irreducible components of (X,x). Proof. The proof runs as in the algebraic case, see 2.2.15, 2.2.16 and 2.2.17. But note that in the proof the Nullstellensatz is used! Definition 3.4.19. Let (X,x) ⊂ (C n ,x) be a germ of an analytic space. A germ of an analytic function f : (X,x) −→ (C ,y) is a germ of a map f : (X,x) −→ (C ,y) such that some representative is the restriction to X of an analytic function on an open neighborhood of x in C n . As is easily shown, the germs of analytic functions on (X,x) form a C –algebra, which we denote by OX,x . This is called the ring of analytic functions on (X,x). Sometimes OC n , 0 , which is equal to C {x1 , . . . ,xn } is denoted by On . A germ of an analytic map or simply map, ϕ = (f1 , . . . ,fm ) : (X,x) −→ (Y,y ⊂ (C m ,y) is similarly defined. The map ϕ : (X,x) −→ (Y,y) is called an isomorphism if ϕ has a twosided inverse ψ : (Y,y) −→ (X,x) which is an analytic map. (X,x) and (Y,y) are called isomorphic if there exists an isomorphism ϕ : (X,x) −→ (Y,y). The following lemma is easy, and left to the reader. Lemma 3.4.20. (1) Let (X,x) ⊂ (C n ,x) be a germ of an analytic space, and I (X,x) be the ideal of X. Then OX,x = On,x /I (X,x). (2) Let (X,x) ⊂ (C n ,x) be a germ of a submanifold. Then OX,x ∼ = C {x1 , . . . ,xk } for some k. (3) Let ϕ : (X,x) −→ (Y,y) be a germ of an analytic mapping. Then by composition ϕ induces a map of C –algebras ϕ∗ : OY,y −→ OX,x : f 7→ f ◦ ϕ. Note that (ϕ ◦ ψ)∗ = ψ ∗ ◦ ϕ∗ for two germs of analytic mappings ϕ and ψ. The converse of the third part of the lemma also holds:

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Theorem 3.4.21. Let (X,x) ⊂ (C n ,x) and (Y,y) ⊂ (C m ,y) be germs of analytic spaces. Let α : OY,y −→ OX,x be a C –algebra homomorphism. Then there exists a unique germ of an analytic mapping ϕ : (X,x) −→ (Y,y) with ϕ∗ = α. Proof. . Without loss of generality, we may assume that x = 0 and y = 0. In 3.2.9 it was proved that α(mY,0 ) ⊂ mX,0 . As α is a ring homomorphism, we get α(mkY,0 ) ⊂ mkX,0 for all k > 0. Take generators w1 , . . . ,wm of the maximal ideal of Om . We have a surjective map Om −→ OY,0 = Om /I (Y, 0), sending wi to w i . Let f1 = α(w 1 ), . . . ,fm = α(w m ). So the fi are elements of OX,0 . Define ϕ = (f1 , . . . ,fm ) : (X, 0) −→ (C m , 0). Hence we get a map ϕ∗ : Om −→ OX,0 . Moreover consider the composition α e : Om −→ α OY,0 −→ OX,0 . We want to prove the following: (1) ϕ∗ = α e.


(2) ϕ (X, 0) ⊂ (Y, 0).

(1) By construction ϕ∗ (wi ) = α e(wi ). As they are both C –algebra homomorphisms, it follows that ϕ∗ (g) = α e(g) for all polynomials g ∈ C [w1 , . . . ,wm ]. But, we want to show that ϕ∗ (g) = α e(g) for all g ∈ Om . Given such a g, we can, for all k ∈ N write g = gk + gk′ ,

where gk is a polynomial of degree smaller than k and gk′ ∈ mk . (Here m is the maximal ideal of Om .) Now ϕ∗ (gk ) = α e(gk ). As gk is a polynomial, and both ϕ∗ and α e map mk into mkX,0 , it follows that ϕ∗ (g) − α e(g) = ϕ∗ (gk′ ) − α e(gk′ ) ∈ mkX,o .

As this holds for all k it follows that ϕ∗ (g) − α e(g) ∈ ∩k mkX,0 . But as OX,0 is a Noetherian k local ring, it follows that ∩k mX,0 = 0, by Krull’s Intersection Theorem 1.3.5. Therefore, ϕ∗ (g) = α e(g).

(2) Because (Y, 0) = V I (Y, 0) , 0 , it suffices to show that for all g ∈ I (Y, 0), the map g ◦ ϕ : (X, 0) −→ (C , 0), is the zero map. As α e is induced by α : OY,0 −→ OX,0 , the ideal I (Y, 0) is in the kernel of ϕ∗ : On −→ OX,0 . So the claim follows. The uniqueness of ϕ is clear.

Corollary 3.4.22. Two germs of analytic spaces (X,x) and (Y,y) are isomorphic if and only if OX,x and OY,y are isomorphic. So the study of germs of analytic spaces is equivalent to the study of reduced analytic algebras. In particular, a germ of a complex submanifold is isomorphic to (C k , 0) for some k.

3.4 Germs of Analytic Spaces

119

Proof. The Nullstellensatz says that there is a 1 − 1 correspondence between reduced analytic algebras and germs of analytic spaces. The previous theorem implies that the isomorphism classes of reduced analytic algebras correspond to isomorphism classes of germs of analytic spaces. Remark 3.4.23. More generally, we will later on consider germs of complex spaces (X,x). By this, we mean nothing else than analytic algebras OX,x which are, in general, nonreduced. They pop up naturally, when one considers intersections, as we will see later. Associated to such a germ of a complex space (X,x) is its reduction (Xred ,x), with local ring OXred ,x the reduction of OX,x , obtained by dividing out the nilpotent elements. We now characterize finite maps algebraically. Theorem 3.4.24. Let f : (X,x) −→ (Y,y) be a map between germs of analytic spaces. Then the following conditions are equivalent: (1) f is finite. (2) OX,x a finitely generated OY,y –module. (3) OX,x /mY,y OX,x is a finite-dimensional C –vector space. (4) f −1 (y) = {x}. Proof. (1) =⇒ (4) is easy. As f −1 (y) consist of a finite number of points, we can make X so small that it just contains x. (4) =⇒ (3). As f −1 (y) = {x}, the zero set of mY,y OX,x is {x}, and it follows from the Nullstellensatz, see Exercise 3.4.32, that OX,x /mY,y OX,x is a finite-dimensional vector space. (3) ⇐⇒ (2). This is the general Weierstraß Division Theorem, see 3.2.10.

(2) =⇒ (1). As (Y,y) ⊂ (C n , 0) for some n, that is, we have a surjection OC n , 0 ։ OY,y , we may reduce to the case (Y,y) = (C n , 0). This case was treated in 3.4.13(2). Example
3.4.25. A hypersurface singularity by definition is a germ of an analytic space V (f ), 0 defined by a single square free f ∈ m ⊂ C {x1 , . . . ,xn }. The isomorphism class of special such f have special names. • Ak : f (x1 , . . . ,xn ) = xk+1 + x22 + . . . + x2n , k ≥ 1, 1 • Dk : f (x1 , . . . ,xn ) = x1 x22 + x1k−1 + x23 + . . . + x2n , k ≥ 4, • E6 : f (x1 , . . . ,xn ) = x31 + x42 + x23 + . . . + x2n , • E7 : f (x1 , . . . ,xn ) = x31 + x1 x32 + x23 + . . . + x2n , • E8 : f (x1 , . . . ,xn ) = x31 + x52 + x23 + . . . + x2n .

Also for k = 0 the A0 is a germ of an analytic space, for which the local ring is isomorphic to C {x2 , . . . ,xn }. So the A0 “singularity” defines a germ of a smooth space. The A-D-E– singularities are the “simplest” singularities, and will be studied further in chapter nine. We draw pictures of the Ak and Dk for n = 2 and small k.

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A1

A2

A3

D4

D5

D6

Definition 3.4.26. Let f ∈ C {x1 , . . . ,xn }. ∂f ∂f (1) We define the Jacobian ideal by J(f ) := ( ∂x , . . . , ∂x ). 1 n

(2) The Milnor algebra of f by definition is the C –algebra C {x1 , . . . ,xn }/J(f ). The Milnor number of f , µ(f ), by definition is the C –vector space dimension of the Milnor algebra:
µ(f ) = dimC C {x1 , . . . ,xn }/J(f ) .
(3) The Tjurina algebra of f is by definition the C –algebra C {x1 , . . . ,xn }/ f,J(f ) . The Tjurina number of f , τ (f ) by definition is the C –vector space dimension of the Tjurina algebra. The Milnor number and Tjurina number might be infinite.

Theorem 3.4.27. Let f and g be in C {x1 , . . . ,xn }. Suppose that V (f ), 0 and V (g), 0 are isomorphic. Then the Tjurina algebras of f and of
g are isomorphic.14 In fact,
if ϕ is an automorphism of C {x1 , . . . ,xn } such that ϕ (f ) = (g), then ϕ (f ) + J(f ) = (g + J(g)). In particular the Tjurina numbers τ (f ) and τ (g) are equal. Proof. Let ϕ = (ϕ1 , . . . ,ϕn ) : C {x1 , . . . ,xn } −→ C {x1 , . .
. ,xn } be an automorphism
such that (ϕ(f )) = (g). It is not difficult to see that g,J(g) = ug,J(ug) for a unit u. Therefore, we may assume that ϕ(f ) = g. The chain rule says that for all i  ∂ϕ X ∂f X  ∂f  ∂ϕj ∂g j ϕ · = ◦ϕ · = . ∂xi ∂xj ∂xi ∂xj ∂xi j j

So we see that J(g) ⊂ ϕ(J(f )). As ϕ−1 is also an isomorphism, it follows by symmetry
J(f ) ⊂ ϕ−1 (J(g)). Compose this
with ϕ. We get ϕ(J(f )) ⊂ J(g). As moreover ϕ (f ) = (g), it follows that ϕ (f ) + J(f ) ⊂ (g + J(g)). The converse inclusion follows because of symmetry. 14

It is a famous theorem of Mather and Yau that the converse holds in case the Tjurina number is finite, see 9.1.8.

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121

Example 3.4.28. The Tjurina numbers of the Ak ,Dk and Ek singularities are equal to k.
The singularities Ak and Al are isomorphic if and only if k = l. Now let (X, 0) = V (f ), 0 for f 6= 0 be a germ of a regular point. Then one
calculates directly that τ (f ) = 0. In particular, all hypersurface singularities V (f ), 0 with 0 < τ (f ) < ∞ are genuine singularities. So now we proved that the Ak for k ≥ 1 are indeed singularities. Similarly Dk and Dl are isomorphic if and only if k = l, and these are indeed singularities. It is also true that the Milnor number is an invariant of the hypersurface singularity (X, 0). A proof however, is surprisingly much more difficult, and will be proved as Corollary 6.4.10. A1 –singularities can conveniently be described by the Tjurina, or Milnor number.
Theorem 3.4.29. Let (X, 0) = V (f ), 0 be a hypersurface singularity. Then µ(f ) = τ (f ) = 1 if and only if (X, 0) is isomorphic to an A1 –singularity. Proof. Suppose f ∈ m \ m2 . Then µ(f ) = τ (f ) = 0, as one easily checks. Hence we may assume that f ∈ m2 , so that τ (f ) ≥ 1. A direct calculation shows that for an A1 –singularity µ(f ) = τ (f ) = 1. Conversely, as τ (f ) ≤ µ(f ) it follows that either µ(f ) = 0 or 1. But µ(f ) = 0 cannot occur,Pas f ∈ / m \ m2 . So we may assume that µ(f ) = 1. We may write (nonuniquely) f = i,j xi xj Hij , and we may assume that Hij is symmetric. We put hij = Hij (0). Then X ∂f xj hij modulo m2 . =2 ∂xi j As µ(f ) = 1, it follows that J(f ) = m = (x1 , . . . ,xn ). So there exist αki with xk =

X

αki

X

aki hij xj .

i

∂f . ∂xi

Putting aij = αij (0) it follows that xk = 2

i,j

 2  P f So it follows that i aki hij = 12 δkj so that hij = ∂x∂i ∂x (0) is an invertible matrix. j The theorem therefore follows from the famous Morse Lemma, to be proved next. Lemma 3.4.30 (Morse Lemma). Let f ∈ C {x1 , . . . ,xn }, f ∈ m2 and suppose that   2 ∂ f ∂xi ∂xj (0) has maximal rank n. Then there exists an automorphism ϕ of C {x1 , . . . ,xn }

such that ϕ(f ) = x21 + . . . + x2n . Proof. We write

f=

n X

xi xj Hij (x1 , . . . ,xn ).

i,j=1

 2  f We may suppose that Hij = Hji for all i,j. The rank of ∂x∂i ∂x (0) is equal to the rank j of Hij (0). It follows from linear algebra that this rank is independent of the coordinates. By induction on s we suppose we found coordinates (y1 , . . . ,yn ) such that

122

3 Basics of Analytic Geometry 2 f (y1 , . . . ,yn ) = y12 + . . . + ys−1 +

X

yi yj Hij (y1 , . . . ,yn ).

i,j≥s

The case s = 0 is trivial. From the assumption that we have maximal rank, it follows that at least one of the Hij (0) is not equal to zero. After a linear coordinate change in the ys , . . . ,yn we may suppose that Hss (0) 6= 0. √ One can, therefore, by the Implicit Function Theorem, consider the square root g := Hss which is in fact a unit. We take new coordinates X His (w1 , . . . ,wn ) 1 yi = wi for i 6= s; ys (w1 , . . . ,wn ) = ws − 12 wi . g(w1 , . . . ,wn ) Hss (w1 , . . . ,wn ) i>s It follows from the Inverse Function Theorem that this is an invertible map. Now one calculates that f (w1 , . . . ,wn ) is of the desired form.

Exercises 3.4.31. Suppose X ⊂ C n , and Y ⊂ C m are affine varieties. Let ϕ : X −→ Y be a morphism and suppose that via ϕ∗ the coordinate ring K[X] is a finitely generated K[Y ]–module. Let x ∈ X, and y = f (x) ∈ Y . Show that for the germs of the analytic spaces (X,x) and (Y,y) the ring OX,x is a finitely generated OY,y module. (Hint: Use the general Weierstraß Division Theorem 3.2.10.) ` ´ 3.4.32. Let I ⊂ C {x1 , . . . ,xn } be an ideal. Use the Nullstellensatz to prove that V (I), 0 = {0} if and only if dimC C {x1 , . . . ,xn }/I < ∞. Prove that this is the case if and only if I is an m– primary ideal. Here m = (x1 , . . . ,xn ). 3.4.33. Prove that the relation in definition 3.4.1 is an equivalence relation. 3.4.34. Let f : (X,p) −→ (Y,q) be a germ of a continuous map. In general, it does not make sense to talk about the image of the map f . Indeed, look at the map f : R2 −→ R2 ,

(x,y) 7→ (z,w) = (x,xy).

Show that there exist arbitrary small representatives X of (X,p) and Y of (Y,q) such that for all U open in R2 U ∩ f (X) 6= U ∩ f (Y ).

So it is impossible to define the image as the class of the image of a small enough representative. Note that in this example, f is not finite. 3.4.35. Formulate and prove Lemma 3.4.10 and Lemma 3.4.11 for the case of finitely many Weierstraß polynomials. (For every Weierstraß polynomial one takes a new variable). 3.4.36. Let X,Y be topological spaces, f : X −→ Y be a closed map. Then for each open neighborhood U in X of a fiber f −1 (y) there exists an open neighborhood V of y in Y such that f −1 (V ) ⊂ U . Prove this. 3.4.37. Let X be an irreducible affine variety. Formulate and prove a Local Parametrization Theorem for X. 3.4.38. Let (X,x) be a germ of space, OX,x be its local ring. Define for I ⊂ OX,x ` an analytic ´ the germ of the analytic space V (I),x ⊂ (X,x). For a subgerm (A,x) ⊂ (X,x), define I (A,x). Prove the standard properties of these operations, including the Nullstellensatz. ` ´ 3.4.39. Let (X, 0) = V (f ), 0 be a hypersurface singularity. One says that (X, 0) has an isolated singularity, if there exists a representative of X such that X \ {0} is a complex submanifold. Using the Nullstellensatz, show that (X, 0) has an isolated singularity if and only if τ (f ) < ∞.

3.4 Germs of Analytic Spaces

123

3.4.40. Let ϕ : (X,x) −→ (Y,y) be finite map between germs of analytic spaces. The map ϕ is called generically s to 1 if there exists a hypersurface (D,y) in (Y,y) such that for all small enough representatives Y of (Y,y) and D of (D,y), the set π −1 (p) has cardinality s for all p ∈ X \ D. (1) Show that if ϕ is generically s to 1 for some s ≥ 1, then ϕ is surjective.

(2) Show that ϕ∗ induces an isomorphism of total quotient rings ϕ∗ : Q(OY,y ) ∼ = Q(OX,x ) if and only ϕ is generically 1 − 1. (Hint: First reduce to the case (X,x) and (Y,y) irreducible. Furthermore, if (Y,y) −→ (C k , 0) is a primitive Noether normalization of (Y,y), then the induced map (X,x) −→ (C k ,0) is a primitive Noether normalization of (X,x). Apply the Local Parametrization Theorem.)

(3) Suppose that OY,y is a normal ring. Show that ϕ is generically s to 1, where s is equal to the degree of the field extension Q(OX,x ) ⊃ Q(OY,y ). (Hint: First recall from 1.5.7 that OY,y is an integral domain. Then check that in the proof of the Local Parametrization Theorem we can replace C {x1 , . . . ,xk } by OY,y .)

(4) Show that this result can be generalized to general (Y,y) as soon as we know that a “normalization” exists, see Section 4.4. This remark also holds for the next exercise. This final item also holds for the next exercise

3.4.41. Let ϕ : (X,x) −→ (Y,y) be a finite analytic map.

(1) Suppose that ϕ∗ : OY,y −→ OX,x is injective. Show that ϕ is surjective, in the following two cases. (a) (X,x) and (Y,y) are irreducible, and ϕ∗ induces an isomorphism on quotient fields. (b) OY,y is a normal ring.

(2) Suppose ϕ is surjective. Show that ϕ∗ is injective. 3.4.42. (1) Let (X,x) and (Y,y) be isomorphic germs of analytic spaces. Show that for all k ∈ N the C –algebras OX,x /mkX,x and OY,y /mkY,y are isomorphic. In particular they are isomorphic as C –vector spaces. (2) Reprove that Ak is isomorphic to Al if and only if k = l. In fact, show that all of the A-D-E–singularities give different isomorphism classes. 3.4.43. Let (X, 0) ⊂ (C n , 0) and (Y, 0) ⊂ (C m , 0) be germs of analytic spaces, and let ϕ : (X, 0) −→ (Y, 0) be a germ of an analytic map. Define the graph Graph(ϕ) ⊂ (C n+m , 0), which is isomorphic to (X, 0). Show that we can view ϕ as the projection on the second factor.

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Further Development of Analytic Geometry

In this chapter we will continue with the basics of the theory of germs of analytic spaces (and of affine varieties). Probably the most important invariant of (irreducible) germs of analytic spaces is the dimension. There are various ways to introduce it. We decided to give, in fact, four definitions of dimension. We introduce this many of them, so that we have a flexible development of the theory. Three of these definitions are given in Section 4.1, namely the Weierstraß dimension, the Krull dimension and the Chevalley dimension. They all have a geometric motivation. The Weierstraß dimension of a germ of an analytic space (X,x) is given by considering a Noether normalization (X,x) −→ (C k , 0). It turns out that k is independent of the Noether normalization of (X,x). This map is surjective and has finite fibers, so it should be geometrically clear to the reader that (X,x) has the same dimension as (C k , 0), which should be k. The Krull dimension is defined as the maximal length k of a chain of irreducible germs (X0 ,x) $ (X1 ,x) $ · · · $ (Xk ,x) ⊂ (X,x). The Chevalley dimension is defined by the minimal number k of hypersurfaces (H1 ,x), . . . ,(Hk ,x) one has to cut (X,x) with in order to get a point. The main part of the first section is devoted to the fact that these three definitions coincide. As suggested by the definition of Chevalley dimension, we will have to take intersections. Now it might happen that, even if one starts with a reduced space (X,x), the intersection with every hypersurface is nonreduced. Algebraically, this means it might happen that for a reduced local ring (R,m) and every f ∈ m the ring R/(f ) might be nonreduced. This is why we have to consider germs of complex spaces, that is, nonreduced spaces. Algebraically, this just means that we study analytic algebras. Our discussion on dimension applies just as well to algebraic sets. One has to keep in mind however, that dimension is a local property. For example, take X = V (zx,zy) and the point p on the line as indicated in the following figure.

p

Then one would like to say that the dimension of X at the point p is one. If, however, p is as in one of the following two pictures,

125

p

p

then the dimension of X at p is should be two. Therefore, when discussing dimension of algebraic sets, we will consider localizations K[V ]m . A Noether normalization for this case is discussed in Exercise 2.2.35. In Section 4.2 we study a fourth characterization of dimension, by means of the Hilbert-Samuel polynomial. This definition is less geometrically motivated, but it has the advantage that in many interesting examples it can be calculated by means of a computer. The idea of the Hilbert-Samuel polynomial is to “count” the number of functions defined on the germ of the analytic space
(X,x). More precisely, we will study the behavior of the function d 7→ dimC OX,x /md , the so-called Hilbert-Samuel function. The main result is that for d ≫ 0, the Hilbert-Samuel function behaves like a polynomial, whose degree is equal to the dimension of (X,x). As it is not more difficult, we will study more generally the Hilbert-Samuel function of an arbitrary local ring, or even finitely generated modules over a local ring. In order to prove that the degree of the Hilbert-Samuel polynomial is smaller than the (Chevalley) dimension, we have to study even more generally HilbertSamuel functions with respect
to an m–primary ideal q, that is, we have to study the function k 7→ dimC OX,x /qd . It turns out that also the leading coefficient of the Hilbert-Samuel polynomial has a nice geometrical interpretation. More precisely, if k is the dimension of (X,x), then k! times the leading coefficient of the Hilbert-Samuel polynomial is called the multiplicity of (X,x). If our space (X,x) has a representative X in a (small) open subset U in C n , then the multiplicity is the number of intersections points of X with a general linear space of dimension n − k. We have already studied this situation for hypersurfaces, see 3.4.11. There we showed that the intersection of a general line near zero with the zero set of a reduced power series of order d has d intersection points. It is, in fact, not difficult to show that the multiplicity in this case is equal to d. The general case will be reduced to the hypersurface case by using the Local Parametrization Theorem. Like the dimension, the multiplicity can, in many cases, be calculated by computer. In Section 4.3 we will consider tangent spaces, regular and singular points. We already defined in Chapter 3 what a smooth point of an analytic space is, see Definition 3.3.11. The first result is that the set of singular points is contained in a proper analytic subset. Essential use is made of the Local Parametrization Theorem. We have to postpone the proof that the set of singular points in fact is an analytic subset to Chapter 6. However, as a preliminary version, we will prove the Jacobian criterion. This says that the set of singular points is contained in the analytic subset given by the maximal minors of the Jacobian matrix. The fact that it is equal, uses the Coherence Theorem of Oka-Cartan, which is a deep theorem, to be proved in Chapter 6. We will introduce

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4 Further Development of Analytic Geometry

the Zariski tangent space and prove that singular points can be characterized by those points for which the embedding dimension, which is equal to the vector space dimension of the Zariski tangent space, is not equal to the dimension of the space. Finally we will prove the general Jacobian criterion, which gives a statement on when a general point of an analytic subset is smooth. In the final section of this chapter we study the normalization of a germ of an analytic e x) together with a space. In case (X,x) is irreducible, this is a normal analytic space (X,e e finite degree one map n : (X,e x) −→ (X,x). For reducible (X,x), we need the notion of a e x) of a space (X,x) always multi-germ. First of all, we will show that a normalization (X,e exists. The proof uses the Finiteness of Normalization Theorem 1.5.19, Corollary 3.3.25, and Corollary 3.3.26. These are used to show that the normalization of OX,x is a finite direct sum of analytic algebras. We continue to study properties of normal analytic spaces. It will be shown that the singular points of a normal analytic space are contained in a subset of codimension at least two. In particular, normal curve singularities are smooth. As the normalization always exists, this means that for irreducible curve singularities (X,x) there exists a germ of a finite, generically 1 − 1 map n : (C , 0) −→ (X,x). Hence, irreducible curve singularities can be parameterized! This property of curve singularities will be studied further in the next chapter. The final topic in this chapter is a function theoretic interpretation of normal analytic spaces, which can be phrased by saying that an analytic space (X,x) is normal, if and only if the “Riemann Extension Theorem” holds for (X,x). This means that if f is a holomorphic function on X \ Y , where Y contains the singular locus of X, and such that f is locally bounded, then there exists an extension of f to a holomorphic function on the whole of X. Again in the proof, Noether normalization is used to reduce the statement to the case of C k , a case we already know, see 3.1.15.

4.1

Dimension Theory

We start with a generalization of the notion of the germ of an analytic space. We need this, because in our development of dimension theory nonreduced analytic algebras occur naturally. Definition 4.1.1. A germ of a complex space (X,x) consists of (1) a germ of an analytic space (Xred ,x), called the reduction (X,x) and, (2) an analytic algebra OX,x such that (OX,x )red = OXred ,x .
Example 4.1.2. Let I = (x2 ) ⊂ C {x}. Then V (I), 0 = (Xred , 0) is the point 0, and OXred ,x = C . Now I defines a germ of a complex space (X, 0) using (Xred , 0) and OX,x = C {x}/I. Next we give three definitions of dimension.

Definition 4.1.3. Let (X,x) be a germ of a complex space and (OX,x ,m) be its local ring. • The Weierstraß dimension of (X,x) is the least number k, such that there exists a Noether normalization Ok ⊂ OX,x of (X,x). (It will turn out that the number k in a Noether normalization Ok ⊂ OX,x is independent of the Noether normalization.)

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127

• The Chevalley dimension of (X,x) is the least number of generators for an m– primary ideal of OX,x . If k is the Chevalley dimension of (X,x), and f1 , . . . ,fk generate an m–primary ideal of OX,x , then f1 , . . . ,fk is called a system of parameters for (X,x). Note that the definition of Chevalley dimension works for all (Noetherian) local rings. • The Krull dimension of (X,x) is the maximal length k of chains of prime ideals p0 $ p1 $ . . . $ pk in OX,x . Note that this definition also works for all (Noetherian) rings.

Although we will prove in this section that for germs of analytic spaces the three definitions of dimension coincide, we call the Krull dimension simply the dimension of (X,x), notation dim(X,x). To motivate the definition of Weierstraß dimension, consider a Noether normalization of (X,x). We have seen that we obtain a map (X,x) → (C k ,o) with finite fibers. It would be natural to call k the dimension of (X,x). However, one has to show that the number k is independent of the Noether normalization. This needs proof, and the proof will follow from the fact that the Weierstraß dimension is equal to the Krull dimension. In order to be careful, we defined the Weierstraß dimension to be the minimal k such that there exists a Noether normalization (X,x) −→ (C k , 0). If k is the Chevalley dimension of (X,x), then there exists a system of parameters f1 , . . . ,fk and an s ∈ Z such that ms ⊂ (f1 , . . . ,fk ), as (f1 , . . . ,fk ) is m–primary. In fact, the conditions are equivalent in this case, as obviously every element of OX,x /(f1 , . . . ,fk ) which is not a unit is nilpotent. It follows from the Nullstellensatz that the intersection of (X,x) with the k hypersurfaces {f1 = 0} = . . . = {fk = 0} consists of just one point, which should have dimension zero. The idea is that if (X,x) has dimension k, then a general hypersurface section should have dimension k − 1, so that one has to intersect (X,x) with at least k hypersurfaces to cut down to a single point. By taking zero sets, it follows that the Krull dimension of (X,x) is the supremum of the length n of a chain of irreducible subvarieties of (X,x): {0} $ (X1 ,x) $ . . . $ (Xn ,x) j (X,x). For the case OX,x = On , consider the following chain of prime ideals: (0) $ (x1 ) $ (x1 ,x2 ) $ . . . $ (x1 , . . . ,xn ) from which it follows that the Krull dimension of (C n , 0) is at least n. It needs proof, and is in fact nontrivial that the dimension of (C n , 0) is indeed equal to n. Of course, it follows from the fact that the Krull dimension is equal to the Weierstraß dimension, or alternatively, it is equal to the Chevalley dimension. We now start off proving that the three definitions coincide. Lemma 4.1.4. Let R ⊂ S be Noetherian local rings, mR ⊂ mS . Suppose that S is a finitely generated R–module. Then the Krull dimension of S is equal to the Krull dimension of R. In particular, if ϕ : (X,x) −→ (Y,y) is a finite surjective map of germs of complex spaces, then dim(X,x) = dim(Y,y).

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Proof. Step 1. First let p0 $ · · · $ pd be a chain of prime ideals in R. By the Going-Up Theorem 1.5.25 there exists a chain of prime ideals P0 $ · · · $ Pd with Pi ∩ R = pi . This shows that the Krull dimension of S is at least the Krull dimension of R. Step 2. Now let P0 $ · · · $ Pd be a chain of prime ideals in S. Let pi := Pi ∩ R. Then of course, the pi are prime ideals in R. It remains to show that for all i = 0, . . . ,d − 1 we have a proper inclusion pi $ pi+1 . By considering the quotient rings R/pi and S/Pi , which are integral domains, it obviously suffices to prove the following statement. Let R ⊂ S be an integral extension, and (0) 6= J ⊂ S be an ideal containing a nonzerodivisor. Then J ∩ R 6= (0). To prove this, let x ∈ J be a nonzerodivisor. It satisfies an integral equation, which we take of minimal degree xn + . . . + an−1 x + an = 0, ai ∈ R. Since x is a nonzerodivisor, it follows that an 6= 0, as otherwise we would get the integral equation xn−1 + . . . + an−1 = 0. Thus an ∈ J ∩ R is a nonzero element. Corollary 4.1.5. Let (X,x) be a germ of a complex space, and Ok ⊂ OX,x be a Noether normalization. Then the Krull dimension of (X,x) is equal to k. In particular, the Weierstraß dimension is equal to the Krull dimension. Proof. From the previous lemma, it follows that the Krull dimension of (X,x) is equal to the Krull dimension of (C k , 0). It remains to show that the Krull dimension of (C k , 0) is equal to k. This will be shown by induction on k. If k = 0, this is obvious. We already showed that the Krull dimension of (C k , 0) is at least k. Consider a chain of prime ideals p0 $ p1 $ · · · $ pd in Ok of maximal length. We have to show d ≤ k. Consider the ring Ok /p1 , which has Krull dimension d − 1. Since p1 6= (0), there exists a Noether normalization Os ⊂ Ok /p1 with s < k. By induction, the Krull dimension of Os is equal to s which, by the previous lemma, is equal to the Krull dimension of Ok /p1 . Therefore, d − 1 = s < k, so that d ≤ k. Lemma 4.1.6. Let (R,m) be a Noetherian local ring. (1) The Krull dimension of R is equal to the Krull dimension of Rred . A similar statement holds for the Chevalley dimension. (2) The Krull dimension of R is zero, if and only if the Chevalley dimension of R is zero. Proof. Step 1. A chain of prime ideals in Rred lifts to a chain of prime ideals in R. On the other hand, the nilpotent elements of a ring are contained in all prime ideals of R. Therefore, a chain of prime ideals in R gives also rise to a chain of prime ideals in Rred . Thus we showed that the Krull dimensions of R and Rred are equal. Step 2. Next we show that the Chevalley dimensions of R and Rred are equal. Let (f1 , . . . ,fd ) be an m–primary ideal of R. Then, obviously, the classes of f1 , . . . ,fd also generate an m–primary ideal in Rred .

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On the other hand, suppose that we have elements f1 , . . . ,fd in R that generate an m–primary ideal q = (f1 , . . . ,fd ) in Rred . This means that there exists a n ∈ N with mn + N ⊂ q + N , where N is the set of nilpotent elements of R. However, as R is Noetherian, there exists an s ∈ N with (q+N )s ⊂ q. It follows that mns ⊂ (mn +N )s ⊂ q. Thus q is an m–primary ideal. Step 3. First suppose that the Chevalley dimension is zero. We may now suppose that the ring is reduced. We will show that (0) is a maximal ideal. Because the Chevalley dimension is zero, it follows that there exists an m–primary ideal with 0 generators. Thus the ideal (0) is m–primary. From the definition of primary, it follows that every zero divisor is nilpotent, and hence zero, since R is reduced. Hence there are no zerodivisors different from 0, which means that (0) is a prime ideal. As, moreover, (0) is m–primary, it is equal to m. Hence the Krull dimension is zero. Step 4. Suppose the Krull dimension is zero. Then m is the only prime ideal of R, as otherwise we can construct a chain of length at least one. As the nilpotent elements of R is the intersection of all minimal prime ideals, it follows that all elements of m are nilpotent. Thus (0) is an m–primary ideal, which exactly says that the Chevalley dimension of R is zero. We now prove that the Chevalley dimension is smaller than or equal to the Krull dimension. The proof works for a general Noetherian local ring. Lemma 4.1.7. Let (R,m) be a Noetherian local ring. Then the Chevalley dimension is smaller than or equal to the Krull dimension. The main step in the proof is given by the following proposition. Proposition 4.1.8. Let R be a Noetherian ring of Krull dimension d. Suppose that f is an active element of R, that is, f is not contained in any of the minimal prime ideals of R 1 . Then the Krull dimension of R/(f ) is at most d − 1. Proof. Consider the ring R/(f ). Any chain of prime ideals p1 $ · · · $ ps in R/(f ) gives a chain of prime ideals in R, with (f ) ⊂ p1 $ · · · $ ps . The ideal p1 is not minimal, because otherwise f would be contained in a minimal prime ideal, contrary to the assumption. Therefore, we can find a minimal prime ideal p0 contained in p1 . Hence s ≤ d, and it follows that the Krull dimension of R/(f ) is at most d − 1. Proof of Lemma 4.1.7. Let d be the Krull dimension of R. The proof goes by induction on d. The case d = 0 was proved in Lemma 4.1.6. Therefore, we may suppose d > 0. Then m is not a minimal prime ideal. Take an element f ∈ m, which is not contained in ∪pi , where the pi are the minimal prime ideals of R. Such an element exists because of Prime Avoidance, see 1.1.13. Applying the previous lemma, we get that the Krull dimension of R/(f ) is at most d − 1. By induction we have elements f1 , . . . ,fs , with s ≤ d − 1 generating an m–primary ideal in R/(f ). Then f,f1 , . . . ,fs generate an m–primary ideal in R. Thus the Chevalley dimension is smaller than or equal to s + 1 ≤ d. Theorem 4.1.9. Let (X,x) be a germ of a complex space. Then the Weierstraß dimension of (X,x) is equal to the Chevalley dimension of (X,x), and thus all three definitions of dimension coincide for germs of complex spaces. 1

This exactly means that f is not a zerodivisor of Rred , see 1.4.24.

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Proof. By Corollary 4.1.5 and Lemma 4.1.7 it remains to show that the Weierstraß dimension is less than or equal to the Chevalley dimension of (X,x). This will be proved in two steps. Step 1. Let ϕ : Ok −→ OX,x be a C –algebra homomorphism, not necessarily injective. Suppose that OX,x is, via ϕ, a finitely generated Ok –module. Then we claim that the Weierstraß dimension of (X,x) is less than or equal to k. Indeed, consider the injective map Ok / Ker(ϕ) ⊂ OX,x . This is an integral extension. Now Ok / Ker(ϕ) is an analytic algebra. So there exists an s ≤ k and a Noether normalization Os ⊂ Ok / Ker(ϕ). It follows that Os ⊂ OX,x is a finitely generated extension, hence a Noether normalization. The claim follows. Step 2. Now let f1 , . . . fs be generators of an m–primary ideal of OX,x . We consider the map ϕ : C {y1 , . . . ,ys } −→ OX,x yi 7→ fi . By assumption OX,x /(f1 , . . . ,fs ) is a finitely generated C –vector space, so that, by the general Weierstraß Division Theorem 3.2.10, OX,x is a finitely generated C {y1 , . . . ,ys }– module. By step 1 it follows that the Weierstraß dimension is less than or equal to s. Corollary
4.1.10 (Active Lemma). Let f ∈ m ⊂ OX,x be an active element, (Y,x) := V (f ),x ⊂ (X,x). Then dim(Y,x) = dim(X,x) − 1.

Proof. We have shown in 4.1.8, dim(Y,x) ≤ dim(X,x) − 1. It cannot be smaller, as otherwise we would be able to find s generators of a primary ideal with s < dim(X,x).

Example 4.1.11. The condition that f is an active element is essential. For example,
the germ consisting coordinate axes (X, 0) = V (xy), 0 has dimension one. Take f = x.
Then (Y,x) = V (x), 0 also has dimension one.

The following corollary is typical for the germ (C n , 0) and certainly does not hold for general germs of analytic spaces.

Corollary 4.1.12 (Characterization of Hypersurfaces). Let (X, 0) ⊂ (C n , 0) be irreducible, and suppose dim(X, 0) = n − 1. Then there exists an f with I (X, 0) = (f ), that is (X, 0) is a hypersurface. More generally, if every irreducible component of (X, 0) has dimension n − 1, it follows that (X, 0) is a hypersurface. Proof. It follows that I (X, 0) is a prime ideal, so that we can take an irreducible f ∈ I (X, 0). We claim that (f ) = I (X, 0). Indeed, otherwise there would be a g ∈ I (X, 0) \ (f ). In particular g would
be a nonzerodivisor modulo (f ), and by the Active Lemma dim(X, 0) ≤ dim V (f,g), 0 = n − 2, a contradiction. To prove the second statement, let (X, 0) = (X1 , 0) ∪ · · · ∪
(Xr , 0) be an irreducible decomposition. Then

we just proved
that (Xi , 0) = V (fi ), 0 for some fi . Thus V (f1 · · · fr ), 0 = V (f1 ), 0 ∪ · · · ∪ V (fr ), 0 = (X, 0).

Theorem 4.1.13. Suppose (X,x) is a germ of a complex space. Let (X,x) = (X1 ,x) ∪ · · · ∪ (Xr ,x) be an irreducible decomposition of (X,x). Then dim(X,x) = max{dim(Xi ,x), i = 1, . . . ,r}.

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131

Proof. We may assume that OX,x is reduced. We use the Krull dimension. The inequality ≥ is obvious, since a chain of irreducible germs in (Xi ,x) gives one in (X,x). Let I (Xi ,x) = pi . Then pi is a minimal prime ideal of OX,x . Let (Y,x) be an irreducible subgerm of (X,x). Then there exists an i with (Y,x) ⊂ (Xi ,x). This is because I (Y,x) is a prime ideal, and therefore there exists a minimal prime ideal pi ⊂ I (Y,y). Taking zero sets gives (Y,x) ⊂ (Xi ,x). So a maximal chain of irreducible subgerms is contained in some irreducible component of (X,x). This shows the theorem. Now consider an irreducible germ of an analytic space (X,x), and let 0 6= f ∈ m ⊂
OX,x . Put (Y,x) := V (f ),x ⊂ (X,x). It may very well happen that (Y,x) is reducible. Note that we proved (Active Lemma) that the dimension of (Y,x) is equal to dim(X,x)−1, but we did not prove that each irreducible component of (Y,x) has dimension dim(X,x)− 1. We only proved that there is at least one component of this dimension. However, the statement that for an active element f each irreducible component of V (f ) has codimension one is in fact true, and holds for general Noetherian rings. This is Krull’s Principal Ideal Theorem, to be proved in the next section, see 4.2.16. In this section, we give a proof for germs of analytic spaces which uses the Going-Down Theorem 1.5.26. First we need a definition. Definition 4.1.14. Let R be a ring. (1) Let p be prime ideal. We define the height ht(p) of p, to be the maximal m such that there exists a chain of prime ideals p0 $ p1 $ · · · $ pm = p. (2) For a general ideal I in R we define the height of I by ht(I) := min{ht(p) : p ∈ Ass(I)}. Remark 4.1.15. Note that from the interpretation of prime ideals in a localization, see 1.3.15, it follows that the height of a prime ideal is equal to the Krull dimension of Rp . Furthermore, we have the inequality (4.1)

ht(I) + dim(R/I) ≤ dim(R),

as follows directly from the definition. Consider a germ of a complex space (X,x), and let I ⊂ OX,x be an ideal. Equality in (4.1) would allow us to interpret the height of I as the codimension of the zero set of I in (X, 0). The codimension of a germ of a complex (Y,x) in (X,x) is simply defined as dim(X,x) − dim(Y,x). Of course, it suffices to show this for irreducible germs . This will be shown now. There exist examples of Noetherian rings, for which we have strict inequality in (4.1), see Exercise 4.1.30. Theorem 4.1.16. Let (X,x) ⊂ (C n , 0) be an irreducible germ of a complex space. All maximal chains of prime ideals in OX,x have the same length, namely dim(X,x). In particular, formula (4.1) holds for R = OX,x . Proof. Let Ok ⊂ OX,x be a Noether Normalization, and consider a chain of prime ideals P0 $ P1 $ · · · $ Ps in OX,x which cannot be extended to a larger chain. We will show that s = k = dim(X,x). Obviously s ≤ k.

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Step 1. Let p be a prime ideal of height one in Ok . We will first show that the dimension of Ok /p is k − 1. Now p 6= (0), and there exists an element 0 6= f ∈ p. As p is prime, there exists an irreducible factor of f which lies in p. So we may assume that we have an irreducible f which lies in p. As Ok is a unique factorization domain, (f ) is a prime ideal. Since p has height one, it follows that p = (f ). By the Active Lemma 4.1.10, the dimension of Ok /p is k − 1. Step 2. Let pi = Pi ∩ Ok . Then we have seen in the proof of Lemma 4.1.4 that (4.2)

p0 $ p1 $ · · · $ ps

is a chain of prime ideals in Ok . We claim that this chain cannot be extended to a larger chain of prime ideals in Ok . Suppose the converse, and suppose that there exists a prime ideal q with say pi $ q $ pi+1 . (Here we take the convention ps+1 = Ok .) Take a Noether normalization Ot ⊂ Ok /pi . We thus have Ot ⊂ Ok /pi ⊂ OX,x /Pi . As Ok /pi ⊂ OX,x /Pi is also a finite extension, we have a Noether normalization Ot ⊂ OX,x /Pi . The prime ideals q and pi+1 induce prime ideals in Ok /pi which we denote by the same name. Moreover, (0) $ q $ pi+1 . Intersecting this chain with Ot , we get a chain of prime ideals in Ot (0) $ q′ $ p′i+1 . Moreover, by construction, Ot ∩ Pi+1 = p′i+1 . Furthermore, Ot is a normal ring (it is even factorial), so that we can apply the Going-Down Theorem 1.5.26: there exists a prime ideal Q in OX,x /Pi with (0) $ Q $ Pi+1 . We lift this prime ideal to a prime ideal in OX,x , which we also call Q. So Pi $ Q $ Pi+1 which is in contradiction to our assumption that the chain Po $ P1 $ · · · $ Ps is maximal. Hence the chain (4.2) is maximal after all. Step 3. We now show s = k = dim(X,x) by induction on k. The case k = 0 is obvious. If k > 0, then the maximal ideal of OX,x is nonzero. So in a maximal chain Ps is the maximal ideal and s ≥ 1. By step 1, Ok /p1 has dimension k − 1. Now p1 $ · · · $ ps induces a maximal chain in Ok /p1 by step 2, and by induction, s − 1 = k − 1. Corollary 4.1.17 (Krull’s Principal Ideal Theorem for OX,x ). Let (X,x) be an irreducible germ of an analytic space, and f ∈ OX,x . Then all irreducible components of
V (f ),x ⊂ (X,x) have dimension dim(X,x) − 1.

Definition 4.1.18.

(1) Let (X, 0) ⊂ (C n , 0) be a germ of an analytic space, and let I (X, 0) be the ideal of (X, 0), and k be the minimal number of generators of I (X, 0). Then (X, 0) is called a complete intersection if the dimension of (X, 0) is n − k. (2) A germ of an analytic space (X, 0) is called a set-theoretic complete intersection if
(X, 0) = V (I), 0 , and I can be generated by n − dim(X, 0) elements.

(3) Let R be a local ring. A sequence of nonunits {f1 , . . . ,fk } is called a regular sequence if for all i = 1, . . . ,k (the class of) fi is a nonzero divisor of R/(f1 , . . . ,fi−1 ).

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133

Example 4.1.19. We consider (X, 0), the coordinate axes in C 3 , given by the zero set of the ideal I = (xy,xz,yz). This is a radical ideal as it is the intersection of the prime ideals (x,y), (x,z) and (y,z). So I = I (X, 0), and by Nakayama’s Lemma it follows that the minimal number of generators of I is equal to three. It is easy to see (using for example the Chevalley dimension), that the dimension of (X, 0) is one. Therefore, (X, 0) is not a complete intersection. It is however a set-theoretic complete intersection. Indeed as one
easily checks, (X, 0) = V (J), 0 , where J = (xy,zy + zx). Corollary 4.1.20. Let (X, 0) be a germ of a complex space, and suppose I (X, 0) is minimally generated by f1 , . . . ,fk ∈ On . Then (X, 0) is a complete intersection, if and only if {f1 , . . . ,fk } form a regular sequence in On . In this case, dim(X, 0) = n − k. The proof is left as Exercise 4.1.25. Definition 4.1.21. Let I ⊂ R be an ideal. (1) I is called unmixed if all associated primes of I have the same height. (2) I (or R/I) is called equidimensional, if for all minimal associated primes p we have dim(Rp /IRp ) = dim(R/I). In case I ⊂ On is unmixed, I does not have embedded components and is equidimensional. By abuse of language, we also say that (X, 0) is unmixed, if the ideal I defining (X, 0) is unmixed. Note that it very well might happen that unmixed ideals are not radical. As a corollary of dimension theory, we can give a nice characterization of unmixed ideals in On in terms of a Noether normalization. Theorem 4.1.22. Let I ⊂ On be an ideal and OX,0 = On /I. Suppose we have a Noether normalization Ok ⊂ OX,0 of (X, 0). Then: I is unmixed ⇐⇒ Every f ∈ Ok is a nonzero divisor of OX,0 . Proof. Consider a primary decomposition of I: I = q1 ∩ . . . ∩ qs .

√ Let pi := qi for i = 1, . . . ,s. By the Weierstraß characterization of dimension the ideal I is unmixed, if and only if for all i = 1, . . . ,s the map Ok −→ On /pi , defined by the composition Ok ⊂ On /I −→ On /qi −→ On /pi , is injective. Therefore, I is unmixed if and only if for all f ∈ Ok we have f ∈ / pi for all i, that is, f is a nonzerodivisor according to the characterization of nonzerodivisors, see Theorem 1.4.24.
Example 4.1.23. Consider I = (x2 + xy,xz). Then (X, 0) = V (I), 0 consists of two components. The first one is given by x = 0, and the second by z = 0, x = −y. So (X, 0) is not equidimensional, and therefore not unmixed. A Noether normalization is given by C {y,z} ⊂ C {x,y,z}/(x2 + xy,xz). So geometrically we are projecting on the y, z plane. It is indeed an inclusion. Any function which vanishes on the union of the plane and the line must also vanish on the plane x = 0. The extension is integral. Indeed, x satisfies the integral equation x2 + xy = 0. So we see from the Noether normalization that (X, 0) is not unmixed, since the element z is a zerodivisor.

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Theorem 4.1.24 (Second Riemann Extension Theorem). Let U ⊂ C n be open, and consider an analytic set X in U of codimension at least 2, that is, for all points p ∈ X, we have dim(X,p) ≤ n − 2. Let f : U \ X −→ C

be a holomorphic function. Then there exists a unique holomorphic extension of f to U .

Proof. The basic idea of the proof is the same as in the proof of the case that X consists of a single point, see Theorem 3.1.16. We will show that f is locally bounded on U \ X, so that we can apply the First Riemann Extension Theorem 3.1.15. Consider a Noether normalization at p ∈ X: (X,p) −→ (C k , 0). The map is induced by the natural projection π : U −→ C k . By assumption k ≤ n − 2.

Step 1. We claim that for all small open balls V ⊂ C n−k there exists a small closed neighborhood B of 0 ∈ C k , such that B × V ⊂ U and π −1 (B) ∩ X ⊂ B × V . In particular (π −1 (B) ∩ X) ∩ (B × ∂V ) = ∅. To show this, note that, by the Noether Normalization Theorem, we may assume that OX,p ∼ = C {x1 , . . . ,xn }/J ⊃ C {x1 , . . . ,xk } is finite. We consider Weierstraß polynomials Pk+1 , . . . ,Pn ∈ J, with Pi ∈ C {x1 , . . . ,xk }[xi ]. We may assume that the Pi converge on U , and that X ⊂ {q ∈ U : Pk+1 (q) = · · · = Pn (q) = 0}. We now apply 3.4.11 to each of the Pi and obtain an open ball B0 ⊂ C k , 0 ∈ B0 , such that π −1 (B0 ) ∩ X ⊂ B0 × V . Let B ⊂ B0 be a closed ball with 0 ∈ B. Then π −1 (B) ∩ X ⊂ B × V .

Step 2. Now we choose a point q ∈ (B × V ) \ X. Then {π(q)} × V is also an open ball. As the restriction of π to X is finite, the intersection of {π(q)} × V with X consists of a finite number of points. We therefore can find a line Lq through q in {π(q)} × V which misses ({π(q)} × V ) ∩ X, because of the fact that V has at least dimension 2.

Lq q

V X

B The line Lq intersects the boundary {π(q)}×∂V . The restriction of f to Lq ∩({π(q)}×V ) is a function of one complex variable, so attains maximum absolute value on the boundary {π(q)} × ∂V . Hence

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135

|f (q)| ≤ max | f|{π(q)}×∂V |≤ max | f|B×∂V | . As f is continuous, and B × ∂V is compact, it follows that f is locally bounded. So we can apply the First Riemann Extension Theorem 3.1.15.

Exercises 4.1.25. Prove Corollary 4.1.20. 4.1.26. Let (X,x) and (Y,y) be germs of analytic spaces. Show that ` ´ dim X × Y,(x,y) = dim(X,x) + dim(Y,y).

4.1.27. Let X ⊂ C n be an affine algebraic set, m = (x1 , . . . ,xn ). Let mk = (x1 , . . . ,xk ) ⊂ C [x1 , . . . ,xk ] and suppose C [x1 , . . . ,xk ]m k ⊂ C [X]m

is a Noether normalization. Consider the germ of the analytic space (X, 0). Show that Ok ⊂ OX,0

is a Noether normalization, cf. 2.2.35. Deduce that the dimension of X at 0, considered as an algebraic set, is equal to the dimension of (X, 0), ` ´ 4.1.28. Let (X, 0) = V (f1 , . . . ,fr ), 0 ,fi ∈ On be a germ of an analytic space. Consider the map: ϕ : Onr −→ On ,

given by the matrix (f1 , . . . ,fr ). Show that for all i < j the element rij = fj ei − fi ej (here e1 , . . . ,er is the canonical basis of Onr ) is in the kernel of ϕ. Show that (X,0) has codimension r in (C n , 0) if and only if the kernel of ϕ is generated by the rij .

4.1.29. Let p be a prime ideal in C {x1 , . . . ,xn }, with ht(p) = k. Show that there exists a regular sequence f1 , . . . ,fk in p. Geometrically, this means that an irreducible space of codimension k is contained as component in a complete intersection of codimension k. 4.1.30. Consider the ring R = C [[x]][y]. Show that the dimension of R is two. Consider the principal ideal I := (xy − 1). Show that ht(I) = 1, and that I is a maximal ideal. Hence in (4.1) we have a strict inequality. 4.1.31. Prove that for an irreducible affine space X, dim(X,p) is independent of the chosen point p ∈ X. 4.1.32. Let X ⊂ C n be an irreducible affine variety and let C (X) be the quotient field of the coordinate ring C [X]. Prove that the dimension of X is equal to the transcendence degree of the field extension C ⊂ C (X). (In some books, this is used as the definition of dimension.)

4.2

Hilbert-Samuel Function and Multiplicity

Definition 4.2.1. Let (R,m) be a Noetherian local ring. The function HSR : N d

−→ N 7→

dimR/m R/md




is called the Hilbert-Samuel function of R. In case R = OX,x is the local ring of a germ of a complex space (X,x) we also write HSX,x and call it the Hilbert-Samuel function of (X,x).

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Example 4.2.2. We take (X,x) = (C n , 0). Then R/md is equal to the vector space of all polynomials of degree less than d in n variables. This is a vector space of dimension
n+d−1 . This can be proved by inductionP for example, but the easiest way to see this is as n αn 1 follows. To a monomial xα with · · · x i < d we assign a monomial of degree d − 1 n 1 i αP d−1− i αi α1 αn . So it suffices to count all monomials in n + 1 variables, namely x1 · · · xn · x0 αn 0 of degree d − 1 in n + 1 variables. A monomial xα 0 · · · xn we represent by circles and crosses. We first put α0 circles, then a cross, then α1 circles, then a cross, etc.

x

x

α0

αn

On the other hand, one can assign a polynomial to such a diagram. So we have to
choose n crosses out of n + d − 1 possibilities, and we get n+d−1 monomials. It follows n
n+d−1 n n that the Hilbert-Samuel function HSC ,0 of (C , 0) is given by d 7→ . n The aim of this section is to prove that for d ≫ 0, the Hilbert-Samuel function behaves like a polynomial, and interpret its degree as the dimension of the space. Finally, in case R is the local ring of a singularity we will interpret the leading coefficient of the polynomial as its “multiplicity”. Theorem 4.2.3. Let (R,m) be a Noetherian local ring. (1) There exists a polynomial HSPR ∈ Q[t], such that HSPR (d) = HSR (d) for d sufficiently large. We call HSPR the Hilbert-Samuel polynomial of R. In case R is the local ring of (X,x), we also write HSPX,x for the Hilbert-Samuel polynomial of R, and call it the Hilbert-Samuel polynomial of (X,x). (2) deg(HSPR ) = dim(R). The proof of 4.2.3 will be given on page 141. For the proof, we need to introduce a much more general notion of Hilbert-Samuel function and polynomial. Definition 4.2.4. Let (R,m) be a Noetherian local ring, q be an m–primary ideal and M be a finitely generated R–module. A q–filtration {Mi } = {Mi : i ∈ N} is a sequence of submodules M = M0 ⊃ M1 ⊃ . . . such that qMd ⊂ Md+1 for all d. It is called a stable q–filtration (or q–stable filtration) if moreover there exists an n0 such that qMd = Md+1 for all d ≥ n0 . Let {Mi } be a q–filtration. Then HS{Mi } : N −→ N with HS{Mi } (d) := dimR/m M/Md is called the Hilbert-Samuel function of the filtration {Mi }. (We will show in Theorem 4.2.6 that indeed dimR/m M/Md < ∞, so that the definition makes sense.) If we take the filtration {Mi } with Mi = mi · M , we simply write HSM for HS{Mi } , and this is called the Hilbert-Samuel function of M . The Hilbert-Samuel function of Definition 4.2.1 is the case M = R, q = m, and Mi = mi . Remarks 4.2.5.

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137

(1) We need the general concept of the Hilbert-Samuel function with respect to a filtration to obtain additivity. Namely, let N ⊂ M be a submodule, then the sequence 0 −→ N/N ∩ mn M −→ M/mn M −→ (M/N )/mn (M/N ) −→ 0 is exact and, therefore, HSM = HSM/N + HS{N ∩mnM} . But usually HSN 6= HS{N ∩mn M} . (2) We need the Hilbert-Samuel function with respect to an m–primary ideal q in order to be able to use the Noether Normalization Theorem. Namely, let Ok ⊂ OX,x be a Noether normalization, and m be the maximal ideal of OX,x . Then q := (x1 , . . . ,xk )OX,x is m–primary. {qi } is obviously a (x1 , . . . ,xk )–stable filtration of OX,x . Thus HSP{qi } is the Hilbert-Samuel polynomial of the Ok –module OX,x with respect to (x1 , . . . ,xk ), and at the same time it is the Hilbert-Samuel polynomial of OX,x with respect to the m–primary ideal q. Moreover, we will prove that {mi } is a stable (x1 , . . . ,xk )–filtration of OX,x if we have a general Noether normalization in the sense of 3.3.30. Therefore, the HilbertSamuel polynomial HSPX,x can also be calculated as the Hilbert-Samuel polynomial HSP{mi } of the stable (x1 , . . . ,xk )–filtration {mi } of the Ok –module OX,x .

The general notion of Hilbert-Samuel function with respect to an m– primary ideal q is also used in order to prove second part of 4.2.3, using the Chevalley dimension. In order to prove the first part of 4.2.3, we need to prove a more general version, as written down in the next theorem. Theorem 4.2.6. Let (R,m) be a local ring q be a primary ideal, and M be a finitely generated R–module. Let r be the minimal number of generators of q, and let {Mi } be a stable q-filtration of M . Then (1) HS{Mi } (d) < ∞ for all d ≥ 0, (2) there exists a polynomial HSP{Mi } of degree ≤ r, such that for all d sufficiently large HS{Mi } (d) = HSP{Mi } (d), (3) the degree and the leading coefficient of HSP{Mi } only depends on M and q, and not on the chosen stable q–filtration. The proof needs some more preparations, and will
be given on page 140. First note Pd−1 that, obviously HS{Mi } (d) = k=0 dimR/m Mk /Mk+1 . This leads to the following definition. Definition 4.2.7. Let (R,m) be a local ring, q an m–primary ideal and {Mi } be a stable q–filtration. The associated graded ring Grq = Grq (R) by definition is Grq (R) := ⊕d≥0 qd /qd+1. The associated graded module Gr{Mi } = Gr{Mi } (M ) by definition is Gr{Mi } := ⊕d≥0 Md /Md+1 . Note that Grq is a graded ring, and that Gr{Mi } is a graded Grq –module. This follows directly from the definition of a q–filtration. The following lemma motivates the condition in the definition of a stable q–filtration.

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4 Further Development of Analytic Geometry

Lemma 4.2.8. Let M be a finitely generated module over the Noetherian local ring (R,m), q be an m–primary ideal and {Mi } be a stable q–filtration of M . Then Grq is a Noetherian ring, and Gr{Mi } is a finitely generated Grq –module. Proof. Step 1. We first prove that Grq is Noetherian. By assumption R is Noetherian and, there
fore, q is finitely generated, say by a1 , . . . ,at . The algebra map R/q [x1 , . . . ,xt ] −→ Grq which sends xi to the class of ai is surjective. The ring (R/q)[x1 , . . . ,xt ] is Noetherian, and from the surjectivity it follows that Grq is Noetherian. Step 2. Now we show that Gr{Mi } is a finitely generated Grq –module. Because {Mi } is a stable q–filtration, there exists an n0 ∈ N such that M
n0 +ℓ = qℓ Mn0 for all ℓ ≥ 0. This says that Gr{Mi } is generated by ⊕k≤n0 Mk /Mk+1 . Furthermore, because R is Noetherian, the module Mk /Mk+1 is a finitely generated R–module. It
follows that Mk /Mk+1 is a finitely generated R/q–module. Therefore, ⊕k≤n0 Mk /Mk+1 is a finitely generated R/q–module. So indeed Gr{Mi } is a finitely generated Grq –module. This motivates us to study the graded situation. Definition 4.2.9. Let K be a field, A = ⊕i≥0 Ai be a graded K–algebra. Suppose that dimK A0 < ∞ and that A1 = (x1 , . . . ,xr )A0 generates A as an A0 –algebra. This means that A is the ring A0 [x1 , . . . ,xr ].2 Let M = ⊕i≥0 Mi be a finitely generated graded A–module. We define the Hilbert function HM : N −→ N of M by HM (n) = dimK Mn . The formal power series with integer coefficients X HPM (t) = HM (k)tk ∈ Z[[t]] k

is called the Hilbert-Poincar´e series of M . Proposition 4.2.10. With the notations of 4.2.9, there exists a polynomial QM ∈ Z[t] such that QM (t) HPM (t) = , (1 − t)r that is, the Hilbert-Poincar´e series is a rational function in t.

Proof. The proof is by induction on r. Step 1. We first consider the case r = 0. In this case A1 = 0, so that A = A0 . Therefore, M is a finitely generated module over A = A0 . Hence, from dimK (A0 ) < ∞, it follows that dimK M < ∞. Thus Mk = 0 for large k. Hence, by definition of the Hilbert-Poincar´e series it follows that HPM (t) ∈ Z[t]. Step 2. The induction step. Let r > 0. Consider the following exact sequence of graded modules: ϕ 0 −→ Ker(ϕ) −→ M (−1) −→ M −→ Coker(ϕ) −→ 0. Here M (−1) is isomorphic to M , but has different grading: M (−1)k = Mk−1 , see 1.2.17. The map ϕ is multiplication with xr . If we have an exact sequence 2

More precisely, via the map yi 7→ xi the algebra A is a quotient of the polynomial ring A0 [y1 , . . . ,yr ].

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0 −→ M1 −→ M2 −→ M3 −→ M4 −→ 0 of finite-dimensional K–vector spaces, then dimK (M1 ) + dimK (M3 ) = dimK (M2 ) + dimK (M4 ), see Exercise 1.2.26. It follows from the definition of the Hilbert-Poincar´e series that HPKer(ϕ) (t) + HPM (t) = HPM(−1) (t) + HPCoker(ϕ) (t). From the definition of ϕ, it follows that Ker(ϕ) and Coker(ϕ) are finitely generated A/xr –modules. By induction, we therefore have HPKer(ϕ) (t) =

QKer(ϕ) (t) , and (1 − t)r−1

HPCoker(ϕ) (t) =

QCoker(ϕ) (t) . (1 − t)r−1

From the definitions of M (−1) and HP it follows immediately that HPM(−1) (t) = t HPM (t), and we obtain (1 − t) HPM (t) = HPM (t) − HPM(−1) (t) = HPCoker(ϕ) (t) − HPKer(ϕ) (t) =

QCoker(ϕ) (t) − QKer(ϕ) (t) . (1 − t)r−1

This proves the proposition. Corollary 4.2.11. With the conditions as in Definition 4.2.9 we have the following statements. (1) There exists a polynomial PM ∈ Q[t] of degree ≤ r − 1, the so-called Hilbert polynomial of M , such that PM (d) = HM (d) for d sufficiently large. (2) Let N = ⊕i≥0 Ni be a finitely generated graded A–module and ϕ : M −→ N be a surjective homomorphism of graded A–modules. Then deg PM ≥ deg PN . (3) deg PM ≤ deg PA . (4) If Ann(m) = 0 for some homogeneous m ∈ M , then deg PM = deg PA . Proof. (1) By definition HM (d) is the coefficient of td in the Hilbert-Poincar´e series QM (t) HPM (t) = (1−t) r . We cancel powers of 1 − t in this fraction and get HPM (t) =

G(t) , (1 − t)s

G(1) 6= 0,

Pk with G(t) = i=0 gi ti ∈ Z[t] and s ≤ r. We expand (1 − t)−s in a power series. We leave it as an exercise to show that the power series expansion of (1 − t)−s is  ∞  X s−1+i i −s t. (1 − t) = s−1 i=0 By multiplying G with (1 − t)−s and taking the coefficient of td we obtain for d ≫ 0 HM (d) =

  k X s−1+d−i . gi s−1 i=0

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4 Further Development of Analytic Geometry

This is a polynomial in d of degree s− 1, which proves the first statement of the corollary. (2) The surjectivity implies that HM (d) = dimK Md ≥ dimK Nd = HN (d). Hence PM (d) ≥ PN (d) ≥ 0 for d sufficiently large. This is only possible if deg PM ≥ deg PN . (3) Let M be generated by m1 , . . . ,ms . Let di be the degree of mi . Then we have a surjective map of graded modules ⊕m . By the second part, deg P⊕A(di ) ≥ i=1 A(di ) ։ MP deg PM . But PA(di ) (d) = PA (d+di ) and P⊕A(di ) = si=1 PA(di ) . Therefore deg P⊕A(di ) = deg PA .

(4) If Ann(m) = 0, for some homogeneous m ∈ Mk , then there exists an injection of graded A–modules A(k) −→ M : 1→ 7 m ∈ Mk .

This implies HA (d + k) = HA(k) (d) ≤ HM (d). This proves the corollary. Example 4.2.12. Let A = C [x
1 , . . . ,xn ], Ai = {f ∈ A : f homogeneous of degree i}. Then HA (d) = PA (d) = n+d−1 n−1 . Hence, deg(PA ) = n − 1 and  ∞  X 1 i+n−1 i t = . HPA (t) = (1 − t)n n−1 i=0

We are now able to prove Theorem 4.2.6. Proof of Theorem 4.2.6. (1) We have seen that for
s all k the R/q module Mk /Mk+1 is finitely generated. So we have a surjection R/q
։ Mk /Mk+1 . As q is m–primary,
dimR/m R/q < ∞. Hence, dimR/m Mk /Mk+1 < ∞. It follows that HS{Mi } (d) =
Pd−1 k=0 dimR/m Mk /Mk+1 < ∞.

(2) Define Ai = qi /qi+1 for all i ≥ 0. Then Grq = ⊕i≥0 Ai , and satisfies the condition of Definition 4.2.9, and Gr{Mi } is a finitely generated Grq –module by 4.2.8. Let r be the minimal number of generators of q. In the proof of 4.2.8, we saw that then the minimal number of generators of A1 is at most r. It follows that the the Hilbert polynomial HPGr{Mi } has degree at most r − 1 by Corollary 4.2.11. Note that HPGr{Mi } (d) = HS{Mi } (d + 1) − HS{Mi } (d) for large d. It is a nice exercise, see 4.2.25, that this implies (2). fi } be another stable q–filtration of M . It is not difficult to show, see Exercise (3) Let {M fn and M fn+n0 ⊂ Mn for all n. This 4.2.31, that there exists an n0 such that Mn+n0 ⊂ M implies HS{Mi } (d) ≤ HS{M fi } (d + n0 ) and HS{M fi } (d) ≤ HS{Mi } (d + n0 ). As the degree fi }, of d 7→ HSP{Mi } (d + n0 ) in d is equal to the degree of HSP{Mi } , and similar for {M this shows that the degrees of HSP{Mi } and HSP{M fi } are equal. On the other hand the inequalities give HSP{Mi } (d) ≤ 1, and HSP{M fi } (d + n0 )

HSP{M fi } (d)

HSP{Mi } (d + n0 )

≤ 1,

for d large. This implies that the leading coefficients of HSP{Mi } and HSP{M fi } are equal and proves the theorem. Before giving the proof of Theorem 4.2.3 we need two more results.

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141

Lemma 4.2.13 (Artin-Rees Lemma). Let (R,m) be a Noetherian local ring, q be an m–primary ideal, and {Mi } be a q–stable filtration of a finitely generated R–module M . Let N ⊂ M be a submodule. Then the filtration {Ni } defined by Ni := Mi ∩ N is a stable q–filtration of N . Proof. Step 1. Consider the ring R∗ := ⊕k≥0 qk . This is a graded ring in the obvious way. If q is generated by a1 , . . . ,ar , then we get a surjective algebra map R[x1 , . . . ,xr ] −→ R∗ by sending xi to ai . Hence R∗ is Noetherian. Step 2. Now consider a filtration {Mi } of M , with qMk ⊂ Mk+1 , but which is not necessarily q–stable. Then we can form the graded R∗ –module M ∗ := ⊕k≥0 Mk . We claim: {Mi } is a stable q–filtration ⇐⇒ M ∗ is a finitely generated R∗ –module.

To see this, consider for each n the submodule Mn′ of M ∗ generated by ⊕nk=0 Mk . It is equal to Mn′ = M0 ⊕ · · · ⊕ Mn ⊕ qMn ⊕ q2 Mn ⊕ · · · .

The Mn′ form an ascending chain of submodules of M ∗ . As R∗ is Noetherian by Step 1, this chain becomes stable, if and only if M ∗ is a finitely generated R∗ –module. This is the case, if and only if Mn′ 0 = M ∗ for some n0 . This again is the case if and only if qr Mn0 = Mn0 +r for some n0 and all r ≥ 0, which is equivalent to the stability of the filtration.
Step 3. We now can prove the Artin-Rees Lemma. We have qNk = q N ∩ Mk ⊂ qN ∩ qMk ⊂ N ∩ Mk+1 = Nk+1 , for all k ≥ 0 showing that {Nk } is a q–filtration. Therefore N ∗ := ⊕k≥0 Nk is an R∗ –module, and in fact a submodule of M ∗ . As {Mi } is q–stable, the module M ∗ is finitely generated, and therefore Noetherian by the previous step. Therefore, N ∗ is finitely generated, and the filtration {Ni } is q–stable, again by step 2. Proposition 4.2.14. Let (R,m) be a local ring, q be an m–primary ideal, and {Mi } be the stable q–filtration defined by Mi = qi M . Let x ∈ R be a nonzerodivisor of M , M ′ := M/(x · M ), and {Mi′ } be the filtration on M ′ defined by Mi′ = qi M ′ . Then

deg HSP{Mi′ } ≤ deg HSP{Mi } − 1.

Proof. Consider N = x · M ⊂ M . Because x is not a zerodivisor, N is as R–module isomorphic to M . Consider the filtration {Ni } given by Ni = N ∩ qi M . By the Artin Rees Lemma 4.2.13, {Ni } is a q–stable filtration. So by 4.2.6 the degree and the leading coefficient of HSP{Ni } and of HSP{Mi } are the same. We have for all k the short exact sequence 0 −→ N/Nk −→ M/Mk −→ M ′ /Mk′ −→ 0. Therefore, for d ≫ 0 we have that HSP{Mi′ } (d) = HSP{Mi } (d) − HSP{Ni } (d), so we see that HSP{Mi′ } has smaller degree than HSP{Mi } . Proof of Theorem 4.2.3. We apply Theorem 4.2.6 to the case M = R, and get the first part. Now we show that the degree of the Hilbert-Samuel polynomial is equal to the dimension of R. This is done in several steps.

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4 Further Development of Analytic Geometry

Step 1. We first prove that the degree of HSPR is smaller than or equal to the Chevalley dimension of R. Let q be any m–primary ideal, and consider the filtration qi R. We claim that deg(HSP{qi } ) = deg(HSPR ) which was by definition equal to deg(HSP{mi R} ). To see this, first note that for some s ∈ N we have m ⊃ q ⊃ ms . Thus for all d ≥ 0, we have md ⊃ qd ⊃ msd . Thus for large d we get HSPR (d) ≤ HSP{qi } (d) ≤ HSPR (sd). Now HSPR (sd) is also a polynomial in d, and has the same degree as HSPR (d). Thus the degrees of the polynomials are equal, proving the claim. The second part of Theorem 4.2.6, says that this degree is at most the minimal number of generators of q, which is the Chevalley dimension of R. Step 2. We now prove that the Krull dimension of R is less than or equal to deg(HSPR ). As we already showed that the Chevalley dimension is less than or equal to the Krull dimension, see 4.1.7, this suffices to prove the theorem. The inequality is proved by induction on deg(HSPR ). Step 2a. Suppose that deg(HSPR ) = 0. It follows that dim(R/md ) is independent of d for d ≫ 0. Hence md = md+1 for d ≫ 0. It follows from Nakayama’s Lemma that md = 0.Thus the zero ideal is m–primary. So the Chevalley dimension, and hence the Krull dimension is zero. Step 2b. We now do the induction step. Let s = dim(R), and p0 $ · · · $ ps be a maximal chain of prime ideals in R. As the case s = 0 is trivial, we may suppose s > 0. In particular p1 6= 0. Take an element x ∈ p1 \ p0 , and consider the ring S := R/p0 .3 We claim that deg(HSPS/(x) ) ≤ deg(HSPR ) − 1. Indeed, x is a nonzerodivisor so that by 4.2.14 deg(HSPS/(x) ) ≤ deg(HSPS ) − 1. Moreover, we have a surjective map R/md ։ S/md . Therefore, deg(HSPS ) ≤ deg(HSPR ). Thus deg(HSPS/(x) ) ≤ deg(HSPR ) − 1. We apply the induction hypothesis. Any chain in S/(x) has length at most deg(HSPR ) − 1. But the image of the chain p1 $ · · · $ ps in S/(x) gives a chain of length s − 1. So dim(R) − 1 = s − 1 ≤ deg(HSPR ) − 1. This proves the theorem. Remark 4.2.15. Note that in this proof we showed that: (1) for any Noetherian local ring the Krull dimension is equal to the Chevalley dimension; (2) the degree of the Hilbert-Samuel polynomial of an R–module M does not depend on the choice of the filtration on M , and is smaller than or equal to the dimension of R. (Simply replace in the proof R by M .) Theorem 4.2.16 (Krull’s Principal Ideal Theorem). Let R be a Noetherian ring, f ∈ R, and p ⊃ (f ) be a minimal associated prime ideal of (f ). Then ht(p) ≤ 1. If moreover f is an active element of R, for example a nonzerodivisor, then ht(p) = 1. In particular, dim(R/(f ) = dim(R) − 1. Proof. Consider an irredundant primary decomposition (f ) = q1 ∩ · · · ∩ qr . 3

Note that if R is an integral domain, then S = R, because p0 = (0) is a prime ideal.

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143

√ √ So qi = p for some i. Since p is minimal it follows that for all j 6= i we have qj 6⊂ p. Consider the local ring Rp , with maximal ideal pRp . So we get the primary decomposition (f )Rp = qi Rp . Therefore (f )Rp is a primary ideal in Rp , and is even pRp –primary. It follows from the definition that the (Chevalley) dimension of Rp is at most one. Thus ht(p) ≤ 1, by using the Krull dimension and the first part of Remark 4.2.15. Suppose moreover that f is an active element. If ht(p) = 0, then p is a minimal prime ideal. As f ∈ p, this is in contradiction to the assumption that f is an active element. For a different proof of Krull’s Principal Ideal Theorem, which does not use the Hilbert-Samuel function, see Exercise 4.2.32. Definition P 4.2.17. Let (R,m) be a Noetherian local ring of dimension e. We write HSPR (t) = ek=0 ak tk . Then m(R) := e! · ae is called the multiplicity of R . If R = OX,x is the local ring of a singularity (X,x), then we also write m(X,x) for m(OX,x ) and call this the multiplicity of (X,x), or the multiplicity of X in x. is the coefficient Remark 4.2.18. The multiplicity m(X,x) is positive, because m(X,x) e! of the monomial of highest degree in HSPX,x and HSPX,x (d) > 0 for d sufficiently large. We leave it is as Exercise 4.2.25, to show that m(X,x) is indeed an integer. Examples 4.2.19.
(n+d−1)···d (1) Let (X,x) = (C n , 0). Then HSPX,x (d) = n+d−1 = = n n! is a polynomial in d of degree n. Therefore, m(X,x) = 1.

1 n n! d + . . . ,

which

(2) Consider the A2 –singularity with local ring R = C {x,y}/(y 2 − x3 ). Consider an element f ∈ C {x,y}. Then modulo (y 2 − x3 ) f has a representative Pf which is a polynomial in y of degree at most one. It is even true that ord(Pf ) ≥ ord(f ), see Exercise 3.2.19. So we get that R/md = C {x,y}/(y 2 − x3 ,xd−1 y,xd ), and it follows that HSPR (d) = 2d − 1. So the cusp singularity has dimension one, and multiplicity 2. We now generalize Example 4.2.19(2). Lemma 4.2.20. Let f ∈ C {x1 , . . . ,xn }, ord(f ) = m and OX,x = On /(f ). Then HSPX,x (d) =

 m  X n+d−j−1 . n−1 j=1

In particular the dimension of (X,x) is n−1, and the multiplicity is m. Note that we have a Noether normalization On−1 ⊂ On /(f ) = OX,x , so that In particular, if f irreducible, the multiplicity of (X,x) is equal to the degree of the field extension Q(OX,x ) ⊃ Q(On−1 ). Proof. Without loss of generality we may suppose that f is a Weierstraß polynomial in xn of degree equal to m = ord(f ). As in the example, any element g ∈ OX,x has a representative Pg which is a polynomial in xn of degree at most m − 1. Furthermore, ord(g) ≤ ord(Pg ). Now HSR (d) = dimC C {x1 , . . . ,xn }/((f ) + md ) is equal to

144

4 Further Development of Analytic Geometry dimC C {x1 , . . . ,xn }/(f, {xα : α1 + . . . + αn = d, αn < m}).

For d > m a basis of this vector space is B = ∪m Bj = {xα : αn = j − 1, α1 + j=1 Bj with
n+d−j−1 . . . + αn−1 ≤ d − j}. By Example 4.2.2 Bj has elements. n−1

We now turn our attention to a geometric interpretation of the multiplicity. First we need a lemma.

Lemma 4.2.21. Let (X,x) be an irreducible germ of an analytic space defined by a prime ideal p ⊂ C {x1 , . . . ,xn }. Let Ok ⊂ OX,x := C {x1 , . . . ,xn }/p be a general Noether normalization, that is, there exist monic polynomials Pj ∈ C {x1 , . . . ,xj−1 }[xj ] ∩ p for j = k + 1, . . . ,n, such that • degxj (Pj ) = ord(Pj ) =: nj , • Q(OX,x ) = Q(Ok )[xk+1 ]/(Pk+1 ). Such a general Noether normalization exists according to Exercise 3.3.30. Let m be the maximal ideal of OX,x , and mk be the maximal ideal of Ok . Consider the filtration {mi } of OX,x Then {mi } is a stable mk –filtration. Proof. Obviously, {mi } is an mk –filtration. Let f ∈ C {x1 , . . . ,xn } be a power series of order t. It follows from Exercise 3.3.31 that modulo p f is equivalent to a polynomial X νk+1 · · · xνnn , fνk+1 ,...,νn ∈ Ok . fνk+1 ,...,νn xk+1 Pf = νj
P Moreover, Pf has order at least t. So we see that, if t ≥ j≥k+1 nj , we can write every element of mt+1 as an element of mk · mt . This is what we needed to show. ∼ C {y} is not a general Noether Example 4.2.22. C {x} ⊂ C {x,y}/(y 2 − x) = OX,x = normalization (in the sense of 3.3.30) because degy (y 2 − x) = 2 > ord(y 2 − x) = 1. Now, mk = (x), m = (x,y) and m2i = (y 2i ) = (xi ) implies mk · m2i = (xi+1 ) = m2i+2 6= m2i+1 for all i. This shows that {mi } is not a stable mk –filtration of OX,x . Proposition 4.2.23. Let (X,x) be an irreducible germ of an analytic space, and Ok ⊂ OX,x be a general Noether normalization. Then m(X,x) = [Q(OX,x ) : Q(Ok )], that is, the multiplicity is equal to the degree of the field extension Q(OX,x ) ⊃ Q(Ok ). Proof. We apply the Local Parametrization Theorem 3.4.14, in order to reduce the statement to the case of a hypersurface singularity. This case was already considered in Lemma 4.2.20. Let ∆ be the discriminant of P := Pk+1 . Let OX,x := On /p, OX ′ ,x′ := Ok [xk+1 ]/(P ), and K = OX,x /OX ′ ,x′ . It was shown in the first part of the Local Parametrization Theorem that p ∩ Ok [xk+1 ] = (P ) so that we have an exact sequence of Ok –modules. (4.3)

0 −→ Ok [xk+1 ]/(P ) −→ On /p −→ K −→ 0.

Let mk be the maximal ideal of Ok . We consider the following filtrations on these three Ok –modules.

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145

• {mik OX,x } on OX,x , • {mik OX,x ∩ OX ′ ,x′ } on OX ′ ,x′ , • {mik K} on K. These are all stable mk –filtrations, where the second one follows from the Artin-Rees Lemma 4.2.13. It follows from exact sequence (4.3) that HSP{mik OX,x } = HSP{mik OX,x ∩OX ′ ,x } + HSP{mik K} . Let m be the maximal ideal of OX,x , and m′ be the maximal ideal of OX ′ ,x′ . Then the i filtrations {mi } and {m′ } are mk –stable by Lemma 4.2.21. It was noted in 3.4.16, that as a consequence of the Local Parametrization Theorem, the module K is in fact an Ok /∆a –module for some a ∈ N. The ring Ok /∆a has dimension k − 1. It follows from the second part of Theorem 4.2.6, and the second part of Remark 4.2.15 that the Hilbert-Samuel polynomial of K has degree at most k − 1. By definition m(X,x) is determined by the leading term of HSP{mi } , and m(X ′ ,x′ ) is determined by the leading term of HSP{m′i } , which both have degree k, and those are independent of the stable mk –filtration by Theorem 4.2.6. From these two facts it follows that m(X,x) = m(X ′ ,x′ ). So we reduced the problem to the case of an irreducible hypersurface, as promised. We finally come to the geometric interpretation of multiplicity. Theorem 4.2.24. Let (X,x) ⊂ (C n , 0) be an irreducible a germ of an analytic space, and let dim(X,x) = k. There exists a Zariski open subset U ⊂ Mn−k,n (C )4 with the following property: given M ∈ U , there exists e of (X,x) in the open ball Bε with center 0 and • an ε > 0 and a representative X radius ε,

• an open and dense subset V of Bε , such that for p ∈ V the plane defined by

x1

M

.. .

xn

!

= pt

e in exactly m(X,x) points. intersects X

To put it more down to earth, if k = dim(X,x), a general n − k–dimensional plane e of (X,x) in exactly m(X,x) points. near the origin cuts a representative X

e ⊂ Gln (C ) be the Zariski open subset such that for M f∈U e the correspondProof. Let U ing linear coordinate change defines, via the projection π : C n −→ C k onto the first k coordinates, a general Noether normalization (X,x) −→ (C k , 0). f consisting of the first n−k rows. This part of M f corresponds Let M be the part of M f e to the direction of the projection π. Let U = {M | M ∈ U }. 4

Mn−k,n (C ) is the vector space of (n − k) × n–matrices with entries in C .

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4 Further Development of Analytic Geometry

Choosing M ∈ U restricts the situation to the case that a general primitive Noether normalization OX,x ⊃ Ok is given and ! we may assume OX,x = C {x1 , . . . ,xn }/p ⊃ 1

C {x1 , . . . ,xk } and M =

0 0 ... 0

..

0

.

.. .

.. .

.

1 0 ... 0

x1

M

.. .

xn

!

=0 p

e X

D Ck Let V = {p | π(p) 6∈ D}, where D is the discriminant, given by V (∆), with ∆ the discriminant of the minimal polynomial of the field extension Q(OX,x ) ⊃ Q(Ok ). We ! now x1

.. apply 3.4.16 (1) to obtain that the (n−k)–dimensional plane E defined by M = p, . xn ( x1 ! )
.. that is E = : x1 = p1 , . . . ,xk = pk = π −1 π(p) , intersects a representative of . xn

(X,x) in [Q(OX,x ) : Q(Ok )] points. Now [Q(OX,x ) : Q(Ok )] is equal to m(X,x) by Proposition 4.2.23.

Exercises ` ´ 1 4.2.25. Define for all k ∈ Z the polynomial kt := k! t(t − 1) · · · (t − k + 1) ∈ Q[t]. For f ∈ Q[t] define ∆f by ∆f (t) := f (t + 1) − f (t) ∈ Q[t]. ` ´ ` t ´ (1) Prove that ∆ kt = k−1 . ` t ´ ` ´ + . . . + ck for suitable ci ∈ Q. Show (2) For any f ∈ Q[t] we can write f (t) = c0 kt + c1 k−1 this. (3) Suppose moreover that f (n) ∈ Z for n ≫ 0, and k = deg(f ). Prove that ci ∈ Z for i = 0, . . . ,k. (Hint: Use induction on k, and consider ∆f .)

(4) Let f : N −→ N be any function. Suppose that d 7→ ∆f (d) = f (d+1)−f (d) is a polynomial Q(d) for d ≫ 0 of degree k − 1. Prove that d 7→ f (d) is, for d ≫ 0, a polynomial of degree k. ` t ´ ` t ´ ` ´ ` t ´ (Hint: Write ∆f (t) = c0 k−1 + c1 k−2 + . . . + ck−1 , and consider P = c0 kt + c1 k−1 + `t´ . . . + ck−1 1 . Prove that ∆(f − P )(d) is a constant for d ≫ 0.)

4.2.26. Compute the Hilbert-Samuel function and polynomial of R = C {x,y,z}/(y 2 − xz,z 3 − x5 ,z 2 − x2 y). Check the result by using the computer algebra system Singular.

4.2.27. Let I = (xz − y 2 , xw − yz, yw − z 2 ) ⊂ C {x,y,z,w}. Let (X,x) be the germ at 0 defined by I. Compute the Hilbert-Samuel polynomial HSPX,x .

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4.2.28. Let I ⊂ C [x1 , . . . ,xr ] = C [x] be an ideal, define C [x]≤s = {f | deg(f ) ≤ s} and I≤s = I ∩ C [x]≤s . Define the affine Hilbert function by AHI (n) = dimC C [x]≤n /I≤n . d `´ P bi si , bi ∈ Z, bd > 0. Prove that for sufficiently large n AHI (n) is a polynomial API (n) = i=0

(Hint: Let I h ⊂ C [t,x1 , . . . ,xr ] be the homogenization of I. Prove that (with the canonical grading of C [t,x1 , . . . ,xr ]) HC [t,x1 ,...,xr ]/I h (n) = AHI (n).) 4.2.29. With the notations of Exercise 4.2.28 prove that deg API = dim C [x1 , . . . ,xr ]/I. (Hint: Let I h be as in 4.2.28. Prove that dim(C [t,x1 , . . . ,xr ]/I h ) = dim(C [x1 , . . . ,xr ]/I) + 1, by using Noether normalization.) 4.2.30. Prove the following version of the Artin-Rees Lemma: Let (R,m) be a local ring, q be an m–primary ideal, and M be a finitely generated module. If N ⊂ M a submodule then there exists an n0 such that ` ´ (qn M ) ∩ N = qn−n0 (qn0 M ) ∩ N .

fn } are stable q–filtrations of M then there exists an n0 such that 4.2.31. Prove: if {Mn }, {M fn+n0 ⊂ Mn for all n ∈ N. fn+n0 ⊂ Mn and M M fn = qn M for all n.) (Hint: It suffices to prove this for the special case M

4.2.32. Give a different proof of Krull’s Principal Ideal Theorem 4.2.16 along the following steps. (1) Reduce the second statement to the first, by using the characterization of zerodivisors.

(2) By looking at the localization Rp , reduce to the case that R is local, with maximal ideal m = p. From now one we suppose this. (3) From now suppose q 6= m is a prime ideal. Show that it suffices to show that height(q) = dim Rq = 0. (4) Show that it suffices to show that qRq is nilpotent. (5) Show that in R/(f ) every descending chain of ideals becomes stationary using the fact that mρ ⊂ (f ) for some ρ.

(6) Consider the symbolic powers q(n) := qn Rq ∩ R, see Exercise 1.3.23. Show that the chain of ideals (f ) + q(1) ⊃ (f ) + q(2) ⊃ (f ) + q(3) ⊃ . . . becomes stationary, that is, there exists an n ∈ N with (f ) + q(n) = (f ) + q(n+1) .

(7) Prove that for this n we have q(n) = (f )q(n) + q(n+1) . (Use the second part of Exercise 1.3.23.) (8) Apply Nakayama to see that q(n) = q(n+1) . (9) Use the definition of symbolic power and Nakayama to see that qn Rq = 0. (10) Deduce the theorem. S 4.2.33. Let (X,x) be a germ of an analytic space and (X,x) = n i=1 (Xi ,x) be Pan irreducible decomposition. We set I = {i : dim(Xi ,x) = dim(X,x)}. Prove that m(X,x) = i∈I m(Xi ,x). (Hint: Study the situation (X,x) = (X1 ,x) ∪ (X2 ,x), with (Xi ,x) reduced and defined by the ideals Ii ⊂ On . Compare the Hilbert-Samuel functions by looking at the exact sequence 0 → On /I1 ∩ I2 → On /I1 ⊕ On /I2 → On /I1 + I2 → 0. One may assume that dim(On /I1 + I2 ) < max{dim(On /I1 ), dim(On /I2 )}.)

148

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Regular Local Rings and the Jacobian Criterion

We now study regular and singular points of analytic spaces. According to Definition 3.3.11 a point x of an analytic space X is called regular if a neighborhood of this point is a complex submanifold of C n . In particular, the local ring OX,x of a smooth point x of X is isomorphic to C {x1 , . . . ,xk } for some k, which is an integral domain. Since for a reducible germ (X,x) we have that OX,x is not an integral domain the following proposition follows. Proposition 4.3.1. Let X be an analytic space, x ∈ X, and (X,x) be a reducible germ. Then x ∈ Sing(X). In particular, if X =X1 ∪ · · · ∪ Xr with Xi 6⊂ Xj for i 6= j, then
S S ∪i6=j Xi ∩ Xj . Sing(X) = i Sing(Xi )

It might not be a big surprise for the reader that “almost all” points of an analytic space are smooth. The proof however is a little bit tricky. For an irreducible space this is proved by means of the Local Parametrization Theorem 3.4.14 (cf. Remark 3.4.16 (2)), which essentially allows us to reduce the statement to the hypersurface case. Proposition 4.3.2. Let U be an open subset of C n , and X be an analytic subset of U . Then the set of singular points Sing(X) is contained in a proper analytic subset of X.

Proof. This is a local statement, that is, for all x ∈ X we have to prove it for a neighborhood of x ∈ X. If (X,x) is an irreducible germ, it was proved in the Local Parametrization Theorem that there exists a hypersurface (D,x) in (X,x), and representatives X and D such that X \ D is a complex submanifold of an open set in C n . Now if (X,x) is reducible, we can take an irreducible decomposition (X1 ,x) ∪ · · · ∪ (Xr ,x) of (X,x). For suitable representatives, we have X = X1 ∪ . . . ∪ Xr with Xi 6⊂ Xj for i 6= j. We therefore have that Xi ∩ Xj for i 6= j is a proper analytic subset of X. We may assume that Sing(Xi ) ⊂ Vi , Vi a proper analytic
subset of Xi . On the other hand, we have Sing(X) = ∪i Sing(Xi ) ∪ ∪i6=j (Xi ∩ Xj ) by the previous proposition. Now ∪i6=j (Xi ∩ Xj ) is a proper analytic subset of
X. Therefore, if x is not contained in the proper analytic subset ∪i Vi ∪ ∪i6=j (Xi ∩ Xj ) then x is a smooth point of X.

Our next goal would be to show that the set of singular points in fact forms an analytic subset of X. However, as we will explain later, we are not able to prove this in this section, since we will need the Oka-Cartan Theorem, see 6.3.2, concerning the coherence of the ideal sheaf. So the fact that the singular locus is an analytic subset is surprisingly deep. What we will show is, that we characterize the regular points by means of the Jacobian Criterion. Definition 4.3.3. Let (R,m) be a Noetherian local ring. The embedding dimension edim(R) of R is defined by edim(R) := dimR/m (m/m2 ). If (X,x) is a germ of an analytic space, then edim(X,x) := edim(OX,x ). The ring R is called a regular local ring, if the embedding dimension is equal to the dimension, that is, if edim(R) = dim(R).

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149

By Nakayama’s Lemma, the embedding dimension is the minimal number of generators of the maximal ideal. As the dimension of R is equal to the Chevalley dimension of R, see Remark 4.2.15, it follows that the number of generators of the maximal ideal of a Noetherian local ring cannot be less than dim(R). Example 4.3.4. Consider the power series ring C {x1 , . . . ,xn }. This is a ring of dimension n. The maximal ideal (x1 , . . . ,xn ) is generated by n = dim(C {x1 , . . . ,xn }) generators, so the power series ring is a regular local ring. Similarly, the formal power series ring C [[x1 , . . . ,xn ]], and the localization C [x1 , . . . ,xn ]m , m = (x1 , . . . ,xn ) are regular local rings. The following lemma motivates the name embedding dimension. Lemma 4.3.5. Let (X,x) be a germ of an analytic space, and let n ∈ N be the minimal number such that there exists an injective analytic map (X,x) ֒→ (C n , 0). Then edim(X,x) = n. In particular, (X,x) is smooth at x if and only if OX,x is a regular local ring. Proof. Suppose ϕ : (X,x) ֒→ (C n , 0). So we have a surjective C –algebra map ϕ∗ : OC n , 0 ։ OX,x . In particular ϕ induces a surjective map of vector spaces mC n ,0 /m2C n ,0 ։ mX,x /m2X,x . This shows that the minimal number of generators of mX,x is at most n, so that edim(X,x) ≤ n. On the other hand, let s = edim(X,x). Then s is the number of generators of mX,x , and it was proved in Corollary 3.2.13 that OX,x is a quotient of C {x1 , . . . ,xs }. By dualizing we get an injective analytic map ϕ : (X,x) ֒→ (C s , 0), so that s ≥ n. This proves the lemma. We now come to the Jacobian criterion. Theorem 4.3.6. Let (X, 0) ⊂ (C n , 0) be a germ of an analytic space, and let the ideal of (X, 0) be generated by f1 , . . . ,fs ∈ C {x1 , . . . ,xn }. We denote by rank0 (f1 , . . . ,fs ) the rank of the Jacobian matrix     ∂fi ∂fi (0) := (0) . ∂xj ∂xj 1≤i≤s,1≤j≤n Then edim(X, 0) + rank0 (f1 , . . . ,fs ) = n. Proof. Put e := edim(X, 0) and r = rank0 (f1 , . . . ,fs ). Step 1. Suppose we have functions f1 , . . . ,fs that generate the maximal ideal m of On . Then we claim that the Jacobian matrix   ∂fi (0) ∂xj 1≤i≤s,1≤j≤n has rank n. It follows from Nakayama’s Lemma that s ≥ n, and in fact already n of the fi generate the P maximal ideal m. Without loss of generality, f1 , . . . ,fn are generators. Now write fi = j Aij xj modulo m2 , for Aij ∈ C . As the fi generate m, we conversely P have xi = j Bij fj mod m2 , for some Bij ∈ C . Therefore (Aij ) is an invertible matrix   ∂fi with inverse (Bij ). Now ( ∂x (0) = (Aij ), so the Jacobian matrix has maximal rank n. j

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4 Further Development of Analytic Geometry

Step 2. We now prove r ≥ n − e. Let g1 , . . . ,ge ∈ OC n ,0 project to generators of the maximal ideal mX,0 . Now C ∼ = OX,0 /mX,0 = OC n ,0 /(f1 , . . . ,fs ,g1 , . . . ,ge ). So f1 , . . . ,fs ,g1 , . . . ,ge generate the maximal ideal of OC n , 0 . Hence the Jacobian matrix of (f1 , . . . ,fs ,g1 , . . . ,ge ) has rank n by step 1. Deleting a column reduces the rank by at most one. So deleting the columns corresponding to g1 , . . . ,ge shows that the rank r of  ∂fi the Jacobian matrix ( ∂xj (0) is at least n − e. Step 3. It remains to show that e ≤ n − r. This will be proved as an application of the Implicit Mapping Theorem. After renumbering the fi and xj we may suppose that  ∂f  i det 6= 0. (0) ∂xj n−r+1≤i,j≤n We can therefore apply the Implicit Mapping Theorem 3.3.6 and deduce that there exist functions ϕn−r+1 (x1 , . . . ,xn−r ), . . . ,ϕn (x1 , . . . ,xn−r ) such that
fi x1 , . . . ,xn−r ,ϕn−r+1 (x1 , . . . ,xn−r ), . . . ,ϕn (x1 , . . . ,xn−r ) = 0,

for i = n − r + 1, . . . ,n. This shows that (X, 0) is a subgerm of the graph of the map ϕ : C n−r −→ C r , that is, a subgerm of 
 V xn−r+1 − ϕn−r+1 , . . . ,xn − ϕn , 0 .

This is a smooth germ of (embedding) dimension n − r. Thus e ≤ n − r, as was to be proved.

Remark 4.3.7. Let X be an analytic subset of an open subset U of C n , say locally defined by holomorphic functions f1 , . . . ,fs on U . Suppose that for all x ∈ X, the germ (X,x) has dimension n − c. It follows directly from the Jacobian Criterion and Lemma 4.3.5 that the singular locus of X is contained in the zero set of the c–minors of the Jacobian matrix, which is an analytic set. However, a priori these two sets need not be equal. Indeed, to complete the proof, one needs that for all x ∈ X, the classes of f1 , . . . ,fs generate the ideal of (X,x), so it has to be a radical ideal. Now even if for some x ∈ X, the ideal generated by f1 , . . . ,fs is radical in OC n ,x , it is not clear that for all y in a small neighborhood of x, the ideal generated by the classes of f1 , . . . ,fs is radical in OC n ,y . That this is in fact the case, is surprisingly difficult to prove. It is equivalent to the Coherence Theorem of Oka-Cartan, see 6.3.2. From this Coherence Theorem it therefore follows that the singular locus of an analytic set itself is analytic. Definition 4.3.8. Let (X,p) be a germ of an analytic subset in C n , and let f1 , . . . ,fs generate the ideal of (X,p). The (Zariski) tangent space TX,p of X at the point p = (p1 , . . . ,pn ) is the affine space in C n defined by n o n X ∂fi (p) · qj = 0, i = 1, . . . ,s . TX,p = p + q = (q1 , . . . ,qn ) ∈ C n : ∂xj j=1

Lemma 4.3.9. TX,p is independent of the choice of f1 , . . . ,fs .

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151

Proof. Suppose that g1 , . . . ,gm also generate I (X,p). Then we have an equality gi =

s X

k=1

ξik · fk

for some ξik ∈ C {x1 , . . . ,xn }. We differentiate. By the product rule, and because fi (p) = 0 for all i, it follows s X ∂fk ∂gi (p) = ξik (p) (p). ∂xj ∂xj k=1

Now let (q1 , . . . ,qn ) ∈ C n be such that n X ∂gi j=1

∂xj

(p)qj =

n P

j=1

s X

k=1

∂fi ∂xj (p)qj

= 0 for i = 1, . . . ,s. Then

n X ∂fk ξik (p) (p)qj = 0. ∂xj j=1

This shows one inclusion. The other inclusion follows by symmetry.   ∂fi Note that the rank of ∂x (p) is n − edim(X,p) by 4.3.6, so that the tangent space j

has dimension edim(X,x). In particular, by 4.3.3, 4.3.5 and 3.4.20 we have the following. Theorem 4.3.10. Let (X,p) be a germ of an analytic space. Then the following conditions are equivalent. (1) p is a regular point of X. (2) OX,p is a regular local ring. (3) OX,p ∼ = C {x1 , . . . ,xk } for some k. (4) dim(OX,p ) = dimC (TX,p ). Examples 4.3.11. (1) Let X = {(x,y) ∈ C 2 : x2 − y = 0}. Then X is an analytic subset of C 2 , 0 = (0,0) ∈ X and TX, 0 = {(q,0) : q ∈ C } ∼ = C.

X

TX,0 0 (2) Let X = {(x,y) ∈ C 2 : x3 − y 2 = 0}. Then again X is an analytic subset of C 2 and 0 = (0,0) ∈ X, but TX,0 = C 2 . On the other hand, (1,1) ∈ X and TX,(1,1) = {(q, 23 q − 21 ), q ∈ C }.

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TX,(1,1)

0

Theorem 4.3.12. Let (X,p) ⊂ (C n ,p) and (Y,q) ⊂ (C m ,q) be germs of analytic subsets and f = (f1 , . . . ,fm ) : (X,p) −→ (Y,q) be a germ of a holomorphic map. Then there exists a canonical linear map Tf,p : TX,p −→TY,q   ∂fi p + a 7−→q + (p) · at . ∂xj If g : (Y,q) −→ (Z,r) is a further germ of a holomorphic map, then Tg◦f,p = Tg,q ◦ Tf,p . As TId,p = IdTX,p , it follows that if f is an isomorphism, then Tf,p is an isomorphism. Proof. Without loss of generality, p = 0 and q = 0. Let I (X, 0) = (g1 , . . . ,gs ), and I (Y, 0) = (h1 , . . . ,ht ). Then n X ∂gi (0)aj = 0, i = 1, . . . ,s} ∂xj j=1

TX,0 = {a ∈ C n : TY,0 = {b ∈ C m :

m X ∂hi j=1

∂yj

(0)bj = 0, i = 1, . . . ,t}.

Now h1 ◦ f, . . . ,ht ◦ f vanish on X, so they are in the ideal generated by g1 , . . . ,gs . Hence there exist sik ∈ OC n ,0 such that hi ◦ f =

s X

sik gk .

k=1

We differentiate and get, using the chain rule, m X ∂ ∂hi
∂fk
(hi ◦ f )(0) = 0 0 = ∂yk ∂xj ∂xj

k=1

s X ∂sik

k=1

∂xj

(0)gk (0)+

s X

k=1

s

X ∂gk ∂gk sik (0) (0) = sik (0) (0). ∂xj ∂xj

Now consider a point a = (a1 , . . . ,an ) ∈ TX,0 . So

k=1

Pn

j=1

∂gk (0)aj = 0. It follows that ∂xj

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153

m X ∂hi

k=1

n X ∂fk (0) (0)aj = 0. ∂yk j=1 ∂xj

 ∂fk (0) · at ∈ TY,0 . But this says exactly that ∂xj The second statement follows from the chain rule, and is left as an exercise. 

Examples 4.3.13. (1) Let π : C n −→ C k be the projection given by π(x1 , . . . ,xn ) = (x1 , . . . ,xk ), and p = (p1 , . . . ,pn ) ∈ C n . Then Tπ,p : TC n ,p = p + C n −→ TC k ,π(p) = π(p) + C k is defined by Tπ,p (p + a) = π(p) + π(a) for a ∈ C n .


(2) Let I = {f1 , . . . ,fm } ⊂ C {x1 , . . . ,xn } be an ideal and i : V (I), 0 −→ (C n , 0) be Pn ∂fi the canonical inclusion. Then TV (I),0 = {a = (a1 , . . . ,an ) ∈ C n : j=1 (0)aj = ∂xj 0, i = 1, . . . ,m}, and Ti,0 : TV (I),0 −→ C n = TC n ,0 is the canonical inclusion. (3) Let f : C −→ C 2 be defined by f (t) = (t2 ,t3 ),

2 0

Then Tf,0 : TC ,0 = C −→ TC 2 ,(0,0) = C 2 is the zero map. On the other hand Tf,2 : TC ,2 = 2 + C −→ TC 2 ,(4,8) = (4,8) + C 2 is defined by Tf,2 (2 + a) = (4,8) + (4a,12a). The following theorem we only need in Chapter nine. Theorem 4.3.14. Let X,Y ⊂ C n be irreducible affine algebraic sets, and ϕ : X −→ Y be a dominant map, see 2.3.11. Then there exists a point p ∈ X such that with q = f (p) Tϕ,p : TX,p −→ TY,q is surjective.

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Proof. In the proof of 2.3.12, it was shown that there exists an f ∈ C [Y ] ⊂ C [X], and an r ∈ N such that ϕ factorizes on X \ V (f ) as follows
ϕ e π X \ V (f ) −→ Y \ V (f ) × C r −→ Y \ V (f ).

Moreover C [X]f is a finitely generated C [Y ]f [x1 , . . . ,xr ]–module. Here π is the projection on the second factor, and thus for all q ′ ∈ Y \ V (f ) and a ∈ C r , the tangent map Tπ,(q′ ,a) is surjective. So it suffices to show that there exists a point p ∈ X \ V
(f ) with Tϕ,p is e surjective, as an application of 4.3.12. As both X \ V (f ) and Y \ V (f ) × C r are affine, see 2.3.10, we reduced the problem to the case of a finite map ϕ : X −→ Y of affine algebraic sets, such that C [X] is a finitely generated C [Y ]–module. So take a smooth point q ′ ∈ Y . By Theorem 2.3.9 the map ϕ is surjective, so there exists a point p′ in ϕ−1 (q ′ ). It follows from 3.4.24 that the germ of the analytic map ϕ : (X,p′ ) −→ (Y,q ′ ) is finite. Thus this is a Noether normalization, as by assumption q ′ is a smooth point of Y . It obviously suffices to prove the statement for an irreducible germ (X,p′ ). It was proved in the Local Parametrization Theorem, see 3.4.14 that for “almost all” points p in an open neighborhood of p′ the germ of the map ϕ : (X,p) −→ (Y,q) is an isomorphism. In particular the tangent map Tϕ,p is an isomorphism. This concludes the proof of the theorem. For later applications, for example in the interpretation of normal rings, and in the proof of Artin’s Approximation Theorem, we need a more general version of the Jacobian Criterion. Theorem 4.3.15 (Jacobian Criterion for Smoothness). Let I = (f1 , . . . ,fm ) ⊂ C {x1 , . . . ,xn }, p be a prime ideal associated to I, and q ⊃ p be a prime ideal containing p. Then5 
 ∂fi = ht(p). mod q C {x1 , . . . ,xn }q /Iq is a regular local ring ⇐⇒ rank ∂x j  
∂fi = k then f1 , . . . ,fk generate mod q Furthermore, if ht(p) = k and rank ∂x 1≤i≤k j the ideal Iq . We first discuss how to interpret this theorem.

  ∂fi mod m Examples 4.3.16. (1) For example, take q = m the maximal ideal. Then ∂x j   ∂fi is just the Jacobian matrix ∂x (0) . As all elements in C {x1 , . . . ,xn } \ m are already j invertible in C {x1 , . . . ,xn } we have the equality C {x1 , . . . ,xn }m /Im = C {x1 , . . . ,xn }/I. It follows that C {x1 , . . . ,xn }/I is a regular local ring if and only the Jacobian matrix has maximal rank. This we have already seen in 4.3.6. (2) Suppose that C {x1 , . . . ,xn }q /Iq is a regular local ring. Put k = ht(p). Then k is the codimension of V (p). Suppose p is a minimal prime of I. Consider the k–minors    associated 

∂fi ∂fi ∆1 , . . . ,∆s of the Jacobian matrix ∂x mod q = ht(p) exactly . Then rank ∂x j j means that there exists a minor ∆i which is not in q. It follows that for all a in an open
subset of V (I) is smooth at the point a. Namely V (∆1 , . . . ,∆s ), 0 ∩
V (q) the variety
V (q), 0 $ V (q), 0 is a proper analytic subset. Now take representatives of all spaces involved. It follows that for all a ∈ V (q)\V (∆1 , . . . ,∆s ) there is also a ∆i with ∆i (a) 6= 0. 5

“` ´” ∂fi Here rank ∂x mod q is the rank of the matrix over the quotient field Q(C {x1 , . . . ,xn }/q). j

4.3 Regular Local Rings and the Jacobian Criterion

155

(3) Note that in the special case I = q = p, that C {x1 , . . . ,xn }p /pp is a field. As a field obviously is a regular local ring, it follows that almost all points of V (p) are nonsingular. This we have already seen in the Local Parametrization Theorem, see 3.4.14. In the proof of the Jacobian criterion, this special case will be treated first and, in fact, in the proof the Local Parametrization Theorem is used. (4) We consider the explicit example of the Whitney umbrella defined by x2 − y 2 z = 0. So p = I = (x2 − y 2 z). We take q = (x,y). Then q is the ideal of the singular locus. So we expect that C {x,y,z}q /(x2 − y 2 z)q is not a regular local ring. Indeed, the minimal number of generators of ideal (x,y)q is two, as one checks by using Nakayama’s Lemma. However the dimension of C {x,y,z}q /(x2 − y 2 z)q is one. On the other hand, looking at the y–axis, a “general” point of the y–axis is smooth. We take the ideal q = (x,z), in which we localize. This is the same as localizing in the multiplicative set S = {1,y,y 2 , . . .}. So y becomes a unit, and we can eliminate z. Hence C {x,y,z}q /(x2 − y 2 z) ∼ = C {x,y}y , which one directly checks to be a regular local ring of dimension one. Before we can give the proof of the general Jacobian criterion, we need some basic facts on regular local rings. Theorem 4.3.17. Let (R,m) be a regular local ring. Then R is an integral domain. Proof. The proof is by induction on the Krull-dimension of R. Step 1. If dim(R) = 0, then edim(R) = 0, hence the maximal ideal has zero generators. Thus m = 0, hence R = R/m is a field, which obviously has no zerodivisors. Step 2. Now suppose dim(R) > 0. Let p1 , . . . ,ps be the minimal prime ideals of R. If we prove that pi = (0), for some i, then i = s = 1 (because of minimality), and therefore R = R/p1 = R/(0) is an integral domain. As dim(R) > 0, it follows by Nakayama that m 6= m2 . By prime avoidance 1.1.13, we can find an x ∈ m\m2 with x ∈ / ∪i pi . The last condition means that x is an active element of R. By Krull’s Principal Ideal Theorem 4.2.16, dim(R/(x)) = dim(R) − 1. Moreover,
m/ (x) + m2 is a vector space of dimension dim(R) − 1. Hence R/(x) is a regular local ring, and by induction hypothesis it is an integral domain. Therefore, the ideal (x) is a prime ideal. It cannot be a minimal prime ideal, since we have chosen x ∈ / ∪i pi . Therefore, there must be a minimal prime ideal, say pi , which is contained in (x). Let y ∈ pi be arbitrary. It follows that y = ax for some a ∈ R. As pi is prime, it follows a ∈ pi . Hence we proved the equality xpi = pi . From Nakayama’s Lemma it follows that pi = 0, which is what we had to prove. Corollary 4.3.18. Let (R,m) be an r–dimensional regular local ring and x1 , . . . ,xi ∈ m. The following conditions are equivalent. (1) There exist xi+1 , . . . ,xr ∈ R such that m = (x1 , . . . ,xr ). (2) The images of x1 , . . . ,xi in m/m2 are linearly independent over R/m. (3) R/(x1 , . . . ,xi ) is an (r − i)–dimensional regular local ring. Proof. Step 1. We first show (1) =⇒ (2). The images of x1 , . . . ,xr generate m/m2 as R/m–vector space. Since dimR/m (m/m2 ) = r they are linearly independent.

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4 Further Development of Analytic Geometry

Step 2. (2) =⇒ (3). We choose xi+1 , . . . ,xr ∈ m such that the images of x1 , . . . ,xr form a basis of m/m2 . Nakayama’s Lemma implies that m = (x1 , . . . ,xr ).
Now x1 is not a zero divisor of R by Theorem 4.3.17 . Thus dim R/(x1 ) = dim(R)− 1 by Krull’s Principal Ideal Theorem. The images of x2 , . . . ,xr generate the maximal ideal in R/(x1 ). This implies that R/(x1 ) is regular. By induction we get that R/(x1 , . . . ,xi ) is regular. Step 3. We finally show (3) =⇒ (1). Let m/(x1 , . . . ,xi ) be generated by the images of xi+1 , . . . ,xr ∈ m. Then m = (x1 , . . . ,xi , xi+1 , . . . ,xr ). Theorem 4.3.19. Let p be a prime ideal in C {x1 , . . . ,xn }. Then C {x1 , . . . ,xn }p is a regular local ring. Proof. The ring C {x1 , . . . ,xn }p has dimension ht(p). Consider a primitive Noether normalization C {x1 , . . . ,xk } −→ C {x1 , . . . ,xn }/p. Then ht(p), which is the codimension of V (p), is equal to n − k. Hence we have to find n − k generators of the maximal ideal pC {x1 , . . . ,xn }p . Let P ∈ C {x1 , . . . ,xk }][xk+1 ] be the minimal polynomial of xk+1 ∈ C {x1 , . . . ,xn }/p, and ∆ ∈ C {x1 , . . . ,xk } be the discriminant. In particular ∆ ∈ / p. Moreover, let Qj = xj ∆ − qj (xk+1 ) ∈ p for j = k + 2, . . . ,n. Here the qj are polynomials with coefficients in C {x1 , . . . ,xk } which exist according to the Finiteness of Normalization Theorem 1.5.19. Then Qn , . . . ,Qk+2 ,P ∈ p, and we proved already in 3.4.14 that they generate the ideal pC {x1 , . . . ,xn }∆ . As ∆ ∈ / p, they in particular generate the ideal pC {x1 , . . . ,xn }p . Remark 4.3.20. It is quite generally true that localizations of regular local rings in prime ideals are regular local rings. A proof of this more general result uses homological methods, see for example [Eisenbud 1995], Corollary 19.14. The proof which we gave works for the polynomial case too, and is a simple consequence of the Local Parametrization Theorem. Proof of Theorem 4.3.15. Step 1. We first consider the case (f1 , . . . ,fm ) = I = p = q. The ring C {x1 , . . . ,xn }p /pp is a field and hence a regular local ring. Therefore, in
∂fi = ht(p). Take a primitive Noether mod p) this case we have to prove rank ( ∂x i≤m,j≤n j normalization C {x1 , . . . ,xk } −→ C {x1 , . . . ,xn }/p. So we have to prove that the Jacobian matrix has rank n − k. We use the Local Parametrization Theorem, and keep the notations of the proof of the previous Theorem 4.3.19. We noted in the previous proof that the ideal pC {x1 , . . . ,xn }p is generated by (P,Qk+2 , . . . ,Qn ). By Exercise 4.3.24 the rank of the Jacobian matrix is independent of the generators of pC {x1 , . . . ,xn }p . As this ideal is generated by n − k elements, it follows that the rank of the Jacobian matrix modulo p is at most n − k. Now the final n − k columns of the Jacobian matrix for the generators (P,Qk+2 , . . . ,Qn ) look like

4.3 Regular Local Rings and the Jacobian Criterion  ∂P  ∂xk+1   ∂Q  k+2  J :=  ∂xk+1  .  .  .  ∂Q n ∂xk+1 So det(J) = ∆n−k−1 ·

... ...

...

∂P ∂xk+1 .

157



∂P ∂P  ∂xk+1  ∂xn     ∂qk+2  ∂Qk+2   − ∂xk+1   ∂xn  =  ..  . ..    .   .. . ∂Qn     ∂qn ∂xn − ∂xk+1 

0

 0    0 ... 0   .    ..  .   ∆ ...

∆ ..

.

Because P is the minimal polynomial of the field extension

C {x1 , . . . ,xn }p /pp ⊃ Q(C {x1 , . . . ,xk }) we have ∂x∂P 6∈ p. So det(J) 6∈ p. This implies k+1 that
∂fi rank ( mod p)i≤m, j≤n = rank(J mod p) = n − k = ht(p). ∂xj

This shows the theorem in the special case I = p = q.

Step 2. Now suppose that C {x1 , . . . ,xn }q /Iq is a regular local ring. Then we claim that (4.4)

Iq = pq .

Indeed, p ⊃ I. By assumption, C {x1 , . . . ,xn }q /Iq is a regular local ring, so in particular an integral domain by 4.3.17. So in the localization, C {x1 , . . . ,xn }q the ideal Iq is a prime ideal. Formula (4.4) now follows from 1.4.50 (5), which says how a primary decomposition behaves under localization. So if we suppose that C {x1 , . . . ,xn }q /Iq is a regular local ring, we may suppose that I = p is a prime ideal. Put a := ht(p), and b := ht(q). The dimension of C {x1 , . . . ,xn }q /Iq is equal to b − a. It is not difficult to see, see 4.3.23, that the ideal pq in the ring C {x1 , . . . ,xn }q can be generated by elements u1 , . . . ,ua ∈ q \ q2 . By 4.3.18 this sequence can be extended to a regular sequence u1 , . . . ,ub ∈ q \ q2 which generates the ideal qq in C {x1 , . . . ,xn }q . We apply step 1 and get  ∂u i rank ∂xj

mod q




i≤b, j≤n



= ht(q) = b,

which is the maximal possible rank. Deleting the columns corresponding to ua+1 , . . . ,ub we obtain  ∂u 
i rank mod q i≤a, j≤n = a = ht(p), ∂xj

which is again the maximal possible rank. We deduce that the Jacobian matrix for the f ’s also has maximal rank by applying again Exercise 4.3.24.   ∂fi Step 3. We now prove the converse. Therefore, assume rank ( ∂x = mod q) i≤m, j≤n j ht(p) = a. By renumbering, we may assume  ∂f i rank ( ∂xj

 mod q)i≤a, j≤a = a.

We first show that f1 , . . . ,fa is part of a generating set for qq . Assume

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4 Further Development of Analytic Geometry a X i=1

for some c1 , . . . ,ca ∈ C . Then

ci f i ∈ q 2

a a  X ∂fi ∂ X ci ci f i = ∈ q. ∂xj i=1 ∂x j i=1  mod q)i≤a, j≤a = a it follows that ci = 0 for i = 1, . . . ,a. By Theo-

 ∂f i As rank ( ∂xj rem 4.3.19 and Corollary 4.3.18, C {x1 , . . . ,xn }q /(f1 , . . . ,fa )q is a regular local ring. In particular (f1 , . . . ,fa ) is a prime ideal in C {x1 , . . . ,xn }q . As f1 , . . . fa ∈ I ⊂ p we have inclusions (f1 , . . . ,fa )q ⊂ Iq ⊂ pq .

But as the outer ideals have the same height, and are both prime in C {x1 , . . . ,xn }q all the ideals must be equal. In particular (f1 , . . . ,fa )q = Iq . Therefore, C {x1 , . . . ,xn }q /Iq is a regular local ring. This proves the theorem.

Exercises 4.3.21. Prove that Tg◦f,p = Tg,q ◦ Tf,p , see 4.3.12.

4.3.22. Let X ⊂ C n , and Y ⊂ C m be algebraic set. Suppose f : X −→ Y is defined by (f1 , . . . ,fm ). Consider the tangent map Tf,p : TX,p −→ TY,f (p) . Let ε be a new variable with ε2 = 0. Suppose p + a ∈ TX,p , and Tf,p (p + a) = f (p) + b. Prove that f (p + εa) = f (p) + εb. Make this to a precise statement! 4.3.23. Let (R,m) be a regular local ring, p ⊂ R a nonzero prime ideal. (1) Suppose p is contained in m2 . Show that R/p is not a regular local ring. (Hint: Use Krull’s Principal Ideal Theorem to prove that dim(R/p) < dim(R), and look at the generators of the maximal ideal in R/p.) (2) Suppose R/p is a regular local ring. Show that p can be generated by a regular sequence u1 , . . . us , with ui ∈ m \ m2 . 4.3.24. Let q ⊂ C {x1 , . . . ,xn } be a prime ideal, and I = (f1 , . . . ,fm ) be a further ideal. Let g1 , . . . ,gs ∈ I such that Iq = (g1 , . . . ,gs )q . Prove that ` ∂f ´ ` ∂g ´ i i rank ( ∂x mod q) = rank ( ∂x mod q) . j j 4.3.25. Let X ⊂ C n be an affine set. Prove that Sing(X) is an affine algebraic subset of X. (Hint: Use the Primary Decomposition Theorem.)

4.3.26. Let X be an analytic space. Prove the Identity Theorem, cf. 3.1.9, for X. (Hint: Use the fact that the singular points of X are contained in a proper analytic subset.) 4.3.27. Prove the following statement. Let (X, 0) ⊂ (C n , 0) be a germ of an analytic subset. Let m ⊂ OX,0 be the maximal ideal. There is a canonical isomorphism of affine spaces TX,0 ∼ = (m/m2 )∗ . (Hint: Choose a representative Y ⊂ U of (X, 0), U ⊂ C n open, and f1 , . . . ,fs holomorphic functions on U such that {y ∈ U : f1 (y) = · · · = fs (y) = 0} and I (X, 0) = (f1 , . . . ,fs ). “ Y =” Then TX,0 = {a ∈ C n :

∂fi ∂xj

(0) · at = 0}.

On the other hand, m is the maximal ideal in C {x1 , . . . ,xn }/(f1 , . . . ,fs ) generated by x1 , . . . ,xn , with xi = xi mod (f1 , . . . ,fs ). Define φ : TX,0 −→ (m/m2 )∗ by φ(a)(xi ) = ai , for a = (a1 , . . . ,an ). Pn Pn well-defined, P that is i=1 ci xi = 0 implies i=1 ci ai = 0. Use that Pn Prove that φ is P n s 2 c x = 0 implies c x + ξ f ∈ (x , . . . ,x ) for suitable ξi ∈ C {x1 , . . . ,xn }. i i i i i i 1 n i=1 i=1 i=1 P ∂fi = 0 in m/m2 .) (0)x Prove that φ is an isomorphism using the fact that n j j=1 ∂xj

4.4 Normalization

4.4

159

Normalization of Germs of Analytic Spaces

The first aim of this section is to show that every germ of an analytic space has a normalization. We start with the definition of a normal germ. Definition 4.4.1. A germ of an analytic space (X,x) is called normal if the local ring OX,x is normal. The first property of normal germs is that they are irreducible. Proposition 4.4.2. Let (X,x) a normal germ of an analytic space. Then (X,x) is irreducible. Proof. This is an immediate consequence of Theorem 1.5.7. So we see that normality is a nontrivial condition. The first goal in this section is to prove that for all germs there exists a normalization. But in order to reach this goal, we have to allow more general spaces then germs of analytic spaces: we have to consider multi-germs. Definition 4.4.3. A multi-germ of analytic spaces (X,x) is a finite disjoint union (X,x) = (X1 ,x1 ) ∪ . . . ∪ (Xr ,xr ) of germs of analytic spaces. The ring OX,x by definition is equal to ⊕ri=1 OXi ,xi . Let (Y,y) = (Y1 ,y1 )∪. . .∪(Ys ,ys ) be a further multi-germ. A map ϕ : (X,x) −→ (Y,y) is given by a system of maps of ϕi : (Xi ,xi ) −→ (Yα(i) ,yα(i) ) for i = 1, . . . ,r, and some α(i) ∈ {1, . . . ,s}. Such a map ϕ induces, and is induced by a C –algebra map ϕ∗ : OY,y −→ OX,x . In the obvious way, one defines such a map to be finite, generic s to 1, etc. A multi-germ (X,x) is called normal if OX,x is a normal ring. Remarks 4.4.4. (1) It is not so difficult to see that OX,x is a semi-local ring, see Exercise 4.4.16. This, by definition, means that OX,x has only finitely many maximal ideals. (2) It is an exercise, see 4.4.16, to show that a multi-germ (X,x) = (X1 ,x1 )∪. . .∪(Xr ,xr ) is normal if and only if (Xi ,xi ) is normal for i = 1, . . . ,r. Now we can define what we mean by a normalization of a germ of an analytic space. Definition 4.4.5. Let (X,x) be a germ of an analytic space. A normalization of (X,x) e x) together together with a finite, generically 1 − 1 map is a normal multi-germ (X,e e x) −→ (X,x). n : (X,e

Remark 4.4.6. It was proved in Exercise 3.4.40 that a generically 1 − 1 map ϕ between irreducible germs of analytic spaces induces an isomorphism of quotient fields. This was a simple consequence of the Local Parametrization Theorem. This holds more generally for multi-germs, see Exercise 4.4.16. As a normalization should be generically 1−1, it follows that the ring OX,e e x is the normalization of the ring OX,x . So a normalization, assuming it e x) has the same dimension exists, is uniquely determined. Note that the normalization (X,e e x) as (X,x). Furthermore, note that if (X,x) is irreducible, then the normalization (X,e must be a germ of an analytic space. This is because the total quotient ring of OX,x is equal to the total quotient ring of OX,e e x , which in case (X,x) is irreducible, is a field.

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4 Further Development of Analytic Geometry

Examples 4.4.7. (1) A smooth germ (C n , 0) is normal. This is because C {x1 , . . . ,xn } is a unique factorization domain by Corollary 3.3.17, and unique factorization domains are normal, see Theorem 1.5.5,6 (2) Consider the A1 –singularity, defined by the equation x2 − y 2 = 0. So OX,x = e x) of (X,x) consists of two copies of C {x,y}/(x2 − y 2 ). The normalization (X,e smooth spaces (X1 ,x1 ) with parameter t1 and (X2 ,x2 ) with parameter t2 . The normalization map n is given by two maps n1 : (X1 ,x1 ) −→ (X,x) which sends t1 to (t1 ,t1 ), and n2 : (X2 ,x2 ) −→ (X,x) which sends t2 to (t2 , − t2 ). (X1 ,x1 ) (X2 ,x2 )

n

(X,x) (3) Let (X,x) be the A2 –singularity, with local ring OX,x = C {x,y}/(y 2 −x3 ). Then the normalization of OX is equal to C {t}, where t = xy . Compare example 1.5.6. The e x) is a smooth one-dimensional space with parameter t, and the normalization (X,e normalization map is given by t 7→ (x(t),y(t)) = (t2 ,t3 ). Note that in this example the normalization map is “bijective”, but not an isomorphism. (4) Let (X,x) be the A1 –singularity in three space, defined by the equation z 2 − xy. Then X is already normal, cf. Exercise 1.5.33. The normalization of (X,x) is (X,x) itself, and the normalization map is the identity. (5) Let (X,x) be the “Whitney umbrella” defined by x2 − zy 2 = 0.

6

More generally, regular local rings are normal because they are unique factorization domains. But we do not prove this.

4.4 Normalization

161 2

2

The element s := xy is integral. Indeed, s2 = xy2 = zy y 2 = z. We consider the ring OX,x [s] generated by OX,x and s in the quotient ring Q(OX,x ) and claim that it is equal to the normalization. Indeed, by applying the Weierstraß Division Theorem, it is not so difficult to see that OX,x [s] = C {x,y,z,s}/(x2 − zy 2 ,ys − x,s2 − z). We can eliminate x and z and see that OX,x [s] is isomorphic to C {y,s}, which is normal. So the normalization is smooth, with parameters y,s and the normalization map is given by (y,s) 7→ (ys,y,s2 ).

Note that in this example the normalization map is indeed generically 1–1: for all points (x,y,z) on the Whitney umbrella not on the z–axis there is only one point in the preimage, but for all points on the z–axis except the origin there are two points in the preimage of the normalization map.

In the third example we saw that the normalization of a cusp singularity is smooth. This turns out to be the case for all curve singularities. So normality is a very strong condition for curve singularities! It follows that for an irreducible curve singularity (X,x) there always exists a finite generically 1–1 map from (C , 0) to (X,x). Such a map can even be given in a particularly simple form, which is then called Puiseux expansion. This will be studied in chapter five. More generally the singular locus of a normal space must have at least codimension two, compare the fifth example. For hypersurface singularities, this condition is also sufficient as suggested by the fourth example, but we are not able to prove this at the moment, but only in the section on Cohen-Macaulay spaces, see in particular Corollary 6.5.9. So in higher dimensions the normality condition is not as strong as in the case of curve singularities. e x) of (X,x) exists, the ring O e must be As noted before, if a normalization (X,e X,e x equal to the normalization of the ring OX,x . The normalization of a ring always exists, and it is our aim to show that this ring is a finite direct sum of analytic C –algebras. Theorem 4.4.8 (Existence of the Normalization). Let (X,x) be a germ of an analytic e x) −→ (X,x) of (X,x). space. Then there exists a normalization n : (X,e e x) is also an irreducible germ. If moreover (X,x) is irreducible, then (X,e

Proof. Consider an irreducible decomposition (X,x) = (X1 ,x) ∪ . . . ∪ (Xr ,x) of (X,x). By eX,x = O eX1 ,x ⊕. . .⊕ O eXr ,x . Splitting of Normalization, see Theorem 1.5.20, we have that O Thus we reduce to the case that (X,x) is irreducible. By the Finiteness of Normalization eX,x is a finitely generated OX,x –module. It was proved Theorem 1.5.19 we deduce that O eX,x is isomorphic to a direct sum ⊕s Si of nonzero analytic in Corollary 3.3.26 that O i=1 eX,x is a field, we have that O eX,x is an integral algebras. As the total quotient ring of O e domain. Therefore s = 1, and it follows that OX,x is an analytic algebra. Thus it is equal e x). to OX,e e x for some germ of an irreducible analytic space (X,e This normalization map is finite by 3.4.24, and it is generically 1-1 because of Exercise 3.4.40. To prove that curve singularities are not normal, it suffices to prove the following purely algebraic theorem.

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4 Further Development of Analytic Geometry

Theorem 4.4.9. Let R be a Noetherian local ring of dimension one. Then R is regular ⇐⇒ R is normal. Proof. Step 1. We first prove “=⇒”. As principal ideal domains are unique factorization domains, see Example 1.4.3, and those are normal by 1.5.5, it suffices to show that R is a principal ideal domain. The definition of regularity gives that dimC m/m2 = 1, so m is a principal ideal. To show that an arbitrary ideal I ⊂ R is principal, first write m = (t) for some t. Take an element x ∈ I, x 6= 0. We claim that there exists a unit u ∈ R and a natural number k(x) such that x = utk(x) . This is obvious if x is a unit. Otherwise put x = x1 t for some x1 ∈ R, and do the same with x1 . Either x1 is a unit, and we are done, or x1 = x2 t, etc. This process must stop because, otherwise, x = xk tk ∈ mk for all k. But ∩mk = 0 by Krull’s Intersection Theorem, which is in contradiction to the choice of x. This shows the claim. Now take k to be the minimum of all k(x) where x runs over the nonzero elements of I. It follows then I = (tk ), so, in particular, principal. Thus R is a principal ideal domain, as was to be proved. Step 2. “⇐=”. As a normal local ring, R is an integral domain by Theorem 1.5.7. Assume that R is not regular. We consider the R–module m∗ = (R : m) = {x ∈ Q(R) : xm ⊂ R}. Obviously, R ⊂ m∗ . We claim that this inclusion is strict. p To prove the claim, take an arbitrary pa ∈ m, a 6= 0. Then (a) = m. Indeed, by the Primary Decomposition Theorem, (a) is an intersection of prime ideals. But, by assumption on dim(R) = 1, and because R is an integral domain, there is just one nontrivial prime ideal, namely m. Because R is Noetherian, it follows that there exists an s ∈ N with ms ⊂ (a). Let s be minimal with this property. If s = 1, then m = (a), hence / R. R would be a regular local ring. Take a nonzero element b ∈ ms−1 \ (a). Then ab ∈ b b s ∗ However, a m ⊂ R, because bm ⊂ m ⊂ (a). Thus a ∈ m \ R. This proves the claim. From the definition of m∗ it follows that m ⊂ mm∗ ⊂ R. So either mm∗ = m or mm∗ = R. We claim that the first equality cannot occur. Indeed, it would follow that ab m ⊂ m. Therefore, m would be an R[ ab ]–module, finitely generated as an R– module. By the Cayley-Hamilton Theorem 1.5.8, ab would be an integral element, which, by assumption, would have to be in R. But ab was constructed in such a way that it does not belong to R. Therefore mm∗ = R. Now take any element t ∈ m \ m2 , which exists according to Nakayama’s Lemma. Then tm∗ ⊂ R. It is impossible to have tm∗ ⊂ m, for, otherwise, tR = tm∗ m ⊂ m2 , in contradiction to the choice of t. Therefore, tm∗ = R, and (t) = tR = tm∗ m = m. This shows that m is principal. Corollary 4.4.10. Let R = C {x1 , . . . ,xn }/I be a one-dimensional integral domain. e is isomorphic to C {t}. Then its normalization R

e is a one-dimensional analytic algebra, which is norProof. Because of Theorem 4.4.8 R e mal. By Theorem 4.4.9, R is regular of dimension one, hence isomorphic to C {t}. With a little bit more effort, one can give the following beautiful characterization of a normal ring, due to Serre. Theorem 4.4.11 (Serre’s R1 and S2 Criterion). Let R be a reduced ring. Then R is normal, if and only if the following two conditions are satisfied:

4.4 Normalization

163

(R1) for each prime ideal p of height one, Rp is a regular local ring; (S2) let f ∈ R be a nonzerodivisor, then the ideal (f ) in R has no embedded primes. Proof. Step 1. Suppose first that R is normal. Then Rp , for a prime ideal p, is normal too, by Exercise 1.5.32. Moreover, Rp has dimension one, so is regular by 4.4.9. To check the second condition, suppose, on the contrary, that we have an embedded prime p for some (f ). In particular, p is not a minimal prime ideal with p ⊃ (f ). By Lemma 1.4.17 we can find an element b ∈ R such that p = (f ) : (b). Let m = pRp . Considering m∗ := (Rp : m) = {x ∈ Q(R) : x · m ⊂ Rp }, one sees that fb ∈ m∗ . It follows, as in the proof of 4.4.9, that if mm∗ = Rp , then m is principal. But then it follows that the dimension of Rp is less than or equal to one. This is in contradiction to the fact that p is not minimal over (f ), as the dimension of Rp is equal to the height of p. So we may assume that mm∗ = m. We can now apply the Cayley-Hamilton Theorem 1.5.8: all elements of m∗ are integral over Rp . As we assumed R, and hence Rp , to be normal, we deduce that m∗ = Rp . In particular, as fb ∈ m∗ . we have fb ∈ Rp . Therefore, we can write fb = ac for some a ∈ / p. Now p = (f ) : (b) = (f a) : (ba) = (f a) : (cf ) = (a) : (c). In particular, a ∈ p, which is a contradiction. So our assumption that we had an embedded prime p for (f ) is wrong. This proves one direction. Step 2. Suppose on the other hand, that (R1) and (S2) hold, and suppose over R. Consider an irredundant primary decomposition of (f ):

b f

is integral

(f ) = q1 ∩ . . . ∩ qt .

√ As (f ) does not have embedded primes by (S2), all the pi := qi are minimal associated primes. By Krull’s Principal Ideal Theorem, see 4.2.16, all the pi are prime ideals of height one. By (R1) it follows that the Rpi are regular. Moreover they have dimension one, so the Rpi are normal by Theorem 4.4.9. It follows that fb ∈ Rpi for all i. To put it in another way: b ∈ f Rpi for all i. By the Second Uniqueness Theorem of Primary Decomposition 1.4.23 we have qi = R ∩ f Rpi for all i. Therefore, b ∈ qi for all i. Hence, b ∈ ∩qi = (f ), and thus fb ∈ R, which had to be proved. We will now consider a function-theoretic interpretation of normal spaces. For this, we will show that a space is normal if and only if “Riemann’s Extension Theorem” holds for X. First we need a definition. Definition 4.4.12. Let X be an analytic space, and Sing(X) be the singular locus of X. A function f : X \ Sing(X) −→ C is called weakly holomorphic on X if (1) f is holomorphic on X \ Sing(X); (2) f is locally bounded. Let (X,x) be a germ of an analytic space. A weakly holomorphic function germ f on (X,x) is an equivalence class of weakly holomorphic functions f : X \ Sing(X) −→ C . Here X is a representative of (X,x). Two such functions are called equivalent if they agree on the intersection of the corresponding representatives.

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Remarks 4.4.13. (1) The weakly holomorphic functions on X form a ring. In fact, as Sing(X) is contained in a proper analytic subset of X, it follows by continuity that the ring of holomorphic functions on X embeds in the ring of weakly holomorphic functions on X (an analytic function on X is already determined by its values on X \ Sing(X)). ′ (2) The germs of weakly holomorphic functions form a ring which we denote by OX,x . ′ As above one sees that OX,x ⊂ OX,x .

Examples 4.4.14. (1) Consider the coordinate axes, defined by xy = 0. Look at the function f which is identically 1 on the the x–axis, except at the origin, and identically 0 on the y–axis except at the origin. Then f cannot be extended to the origin (not even x . as a continuous function), but is bounded. Note that we can write f (x,y) = x+y 2 2 3 (2) Consider the analytic subset X = {y − x = 0} ⊂ C . The function f = xy is holomorphic, whenever √ x 6= 0. So f is a holomorphic function on X \ {(0,0)}. We have the estimate yx ≤ | x|, hence f defines a germ of a weakly holomorphic function on (X,x). In this case, f extends to a continuous function on X, by f (0,0) = 0. However f is not holomorphic. Theorem 4.4.15. (1) Let (X,x) be a germ of an analytic space. Then there is a canonical isomorphism ′ ∼ eX,x . To put it in another way, f is a weakly holomorphic function germ OX,x =O if and only if f is in the integral closure of OX,x .

(2) A germ of an analytic space (X,x) is normal if and only if every weakly holomorphic function germ on (X,x) can be extended to a holomorphic function on (X,x). To put it in another way, a space (X,x) is normal, if and only if the First Riemann Extension Theorem holds for (X,x). Proof. The second statement follows immediately from the first one. The proof of the first statement is quite long, and will be divided into several steps. Suppose the germ (X,x) is in (C n , 0). eX,x ⊂ O′ . Take f ∈ O eX,x . Then f satisfies an equation Step 1. First we prove O X,x (4.5)

f s + a1 f s−1 + . . . + as = 0, ai ∈ OX,x .

Consider a representative X of (X,x). Since in particular f ∈ Q(OX,x ) we can write f = hg , for some g ∈ OX,x and h ∈ OX,x a nonzerodivisor. As h is a nonzerodivisor it does not vanish identically on any of the components of (X,x). By definition, X \ Sing(X) is a locally complex submanifold of an open subset in C n , so that Riemann’s Extension Theorem holds for X \ Sing(X) (cf. Theorem 3.1.15). Therefore, it suffices to prove that f is bounded on X \ Sing(X). This will be proved as an application of the Continuity of Roots. We take X so small that the ai of Equation (4.5) are holomorphic on X, and that X is an analytic subset of U . So we may lift the ai to holomorphic functions on U . Now look at the equation (4.6)

F := T s + a1 T s−1 + . . . + as = 0.

We plug in 0 in the a1 , . . . ,as and factorize

4.4 Normalization

165 F (0,T ) = (T − c1 )s1 · · · (T − ce )se .

By Hensel’s Lemma 3.3.21 F factorizes as F = F1 · · · Fe , with Fi (0,T ) = (T − ci )si , and Fi a polynomial in T of degree si . In particular Fi (x1 , . . . ,xn ,T + ci ) is a Weierstraß polynomial in T . Take a small open neighborhood Vi of ci in C , for i = 1, . . . ,e. By the Continuity of Roots, there exists an open neighborhood Ui of 0, such for all p ∈ Ui , all si zeros (counted with multiplicity) of Fi (p,T ) lie in Vi . So if P we take V = V1 ∪ . . . ∪ Ve , and U = U1 ∩ . . . ∩ Ue it follows that for all p ∈ U the s = si zeros of F (p,T ) lie in V . Now formula (4.5) says that f (p) satisfies the equation F (p,T ). Therefore, if p ∈ U , and f (p) is defined then f (p) ∈ V , in particular it is bounded. So we can extend f to a holomorphic function on X \ Sing(X). By continuity, this extension is also bounded.

′ eX,x . We claim that it Step 2. We now turn our attention to the inclusion OX,x ⊂ O suffices to show this for irreducible (X,x). To see this let (X,x) = (X1 ,x) ∪ . . . ∪ (Xr ,x) be the irreducible decomposition of (X,x). Let f be a weakly holomorphic function on a representative X = X1 ∪ . . . ∪ Xr of (X,x). As f is a genuine function outside the singular locus
of X, it follows that f is a holomorphic function on each Xi \ Sing(Xi ) ∪ ∪j6=i (Xi ∩ Xj ) . By the Riemann Extension Theorem in C n , the function f can be extended to a weakly holomorphic function on Xi . So, once we have shown the inclusion for an eXi ,x for each each i. But by the Splitting of irreducible (X,x), f defines an element of O e e eX,x = ⊕r O Normalization, see 1.5.20, we get O i=1 Xi,x . This implies that f ∈ OX,x .

Step 3. We now reduce to the case that (X,x) is an irreducible hypersurface. By the Local Parametrization Theorem we have a surjective map π from (X,x) onto a hypersurface (X ′ ,x) which has the following properties: • OX ′ ,x ⊂ OX,x . • Q(OX,x ) = Q(OX ′ ,x ). • There exists a representative π : X −→ X ′ and a proper analytic subset D of X ′ such that π : X \ π −1 (D) −→ X ′ \ D is biholomorphic. • Sing(X ′ ) ⊂ D. Let f be a weakly holomorphic function on X. Then f gives a bounded holomorphic function on X ′ \ D. This is a smooth space, so by Riemann’s Extension Theorem 3.1.15 f can extend to a bounded holomorphic function on X ′ \ Sing(X ′ ). We obtain a weakly holomorphic function on X ′ . So if the theorem is proved for a hypersurface we get f ∈ Q(OX ′ ,x′ ) and an integral equation for f with coefficients in OX ′ ,x . As OX ′ ,x ⊂ OX,x eX,x , since f ∈ Q(OX ′ ,x ) = Q(OX,x ). this shows that f is in O Step 4. Now let (X,x) be a germ of an analytic hypersurface, defined by a Weierstraß polynomial P = xsn + a1 xns−1 + . . . + as = 0.

′ So ai is in the maximal ideal of On−1 . We will first show that OX,x ⊂ Q(OX,x ). Let ∆
be the discriminant of P , (D,x) = V (∆),x and f be a weakly holomorphic function germ on (X,x). We will show that on some representatives X and D we can write f as

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g ∆,

for a holomorphic function g on X.7 So we have to show that ∆ · f has a holomorphic extension to X. Consider the projection π : X −→ U ⊂ C n−1 . We choose U so small, so that there exists a constant C > 0 such that the absolute value of the xn –th coordinate of each point is smaller than C. This can be done by Continuity of Roots, see 3.4.11. Furthermore, we may assume that |f (a)| < C for all a ∈ X \Sing(X), because by assumption f is locally bounded. Take a point p ∈ U \ D. Locally around p, there exists an open neighborhood V of p, and s holomorphic functions α1 , . . . ,αs : V −→ C , such that π −1 (V ) is the disjoint union of s complex submanifolds, given by the equations xn = αi (x1 , . . . ,xn−1 ), see 3.3.23. U was chosen to be so small that the αi are bounded. Let q ∈ V , and consider the Lagrange interpolation polynomial in xn s X i=1

f (q,αi (q))

Y xn − αj (q) . αi (q) − αj (q) j6=i

For q ∈ V , this is indeed a well-defined polynomial in xn . By construction,
the Lagrange
interpolation polynomial has the same values as f on the s points q,α1 (q) , . . . , q,αs (q) . We now vary q, and we thus may assume that on π −1 (V ) f is equal to the above polynomial in xn whose coefficients are holomorphic functions on V . Now on V the discriminant is equal to Y ∆= (αi − αj ). j6=i

Thus we see that ∆·f has as coefficients holomorphic functions on V . As one easily checks, these functions are bounded by a number A(C), which only depends on C. Doing this for all p ∈ U \ D, we see that ∆ · f is a polynomial in xn whose coefficients are holomorphic functions on U \ D which are bounded. By the Riemann Extension Theorem for C n−1 , see 3.1.15, it follows that we can extend these to holomorphic functions on U . Hence ∆ · f can be extended to a holomorphic function on all of X. Step 5. We keep the notations of Step 4, and we finally show that any weakly holomorphic function is integral over OX,x . For p ∈ U \ D and V an open neighborhood of p, look at the holomorphic functions α1 , . . . ,αs : V −→ C , as above. The function q 7→ f (q,αi (q)) defines a holomorphic function on V . Moreover on π −1 (V ) the holomorphic function     f − f (x1 , . . . ,xn−1 ,α1 ) · · · f − f (x1 , . . . ,xn−1 ,αs )

is the zero function. Let bi be the i–th symmetric polynomial in the f (x1 , . . . ,xn−1 ,αj ). Then the bi define holomorphic functions on V , and we have an equation (4.7) 7

f s + b1 f s−1 + . . . + bs = 0,

In view of the fact that we try to prove that f is in the integral closure of OX,x , and the fact that any element in the integral closure has as universal denominator ∆, see 1.5.19, this is the obvious thing to do.

4.4 Normalization

167

as a holomorphic function on π −1 (V ). Moreover, the bi are bounded on V , because the f (x1 , . . . ,xn−1 ,αi ) are. Doing this for all p, we get bounded functions b1 , . . . ,bs on U \ D. By the Riemann Extension Theorem 3.1.15, they extend to a holomorphic function on g all of U . Now write f = ∆ . This we can do by Step 4. As ∆ does not vanish on U \ D, we can multiply equation (4.7) with ∆s and get (4.8)

g s + b1 ∆g s−1 + . . . + bs ∆s = 0,

as holomorphic functions on X \ π −1 (D). By continuity, it follows that (4.8) holds on X. We can now read (4.8) as an equation in OX,x . Upon dividing (4.8) by ∆s we get an g . integral equation for f = ∆

Exercises 4.4.16. Let (X,x) = (X1 ,x1 ) ∪ · · · ∪ (Xr ,xr ) be a multi-germ of analytic spaces.

(1) Show that OX,x has exactly r maximal ideals, so in particular OX,x is a semi-local ring. Describe those maximal ideals. (2) Show that (X,x) is normal if and only if (Xi ,xi ) is normal for i = 1, . . . ,r.

(3) Let ϕ : (X,x) −→ (Y,y) be a map between multi-germs of analytic spaces. Show that ϕ is generically 1 − 1 if and only if ϕ induces an isomorphism of total quotient rings ϕ∗ : Q(OY,y ) −→ Q(OX,x ). (4) Let ϕ : (X,x) −→ (Y,y) be a finite map between irreducible germs of analytic spaces. Suppose that ϕ∗ : OY,y −→ OX,x is injective and let s be the degree of the field extension Q(OX,x) ⊃ Q(OY,y ). Show that ϕ is generically s to 1 and that ϕ is surjective. (Hint: Use 3.4.41 and 3.4.40)

4.4.17. Let ϕ : (X,x) −→ (Y,y) be a finite map of germs of analytic spaces. Prove that the image of ϕ is a germ of an analytic space. (Hint: Let I = Ker(ϕ∗ ). Show that the image of ϕ consists of all points a ∈ Y with f (a) = 0 for all f ∈ I.) 4.4.18. The normalization of a local domain is not necessarily local. The local ring R, obtained from C [x,y]/(y 2 − x3 − x2 ) by localizing in the maximal ideal (x,y) is an integral domain. Show that the normalization of R is not local. e x) −→ (X,x) be a normalization. 4.4.19. Let (X,x) be a germ of an analytic space, and n : (X,e Let (Y,y) be a normal multi-germ, and ϕ : (Y,y) −→ (X,x) be a map of multi-germs. Show that e x) making the following there exists a uniquely determined map of multi-germs ϕ e : (Y,y) −→ (X,e diagram commutative. ϕ e / (X,e e x) FF FF FF n ϕ FF F# 

(Y,y)

(X,x)

(Hint: Prove a corresponding statement for rings.) 4.4.20. Let (X,x) be a germ of an analytic space, and Sing(X,x) be the singular locus of (X,x). Let I ⊂ OX,x be the ideal of germs of analytic functions vanishing on the singular locus. Show that (X,x) is normal if and only if HomOX,x (I,I) = OX,x . (Hint: Use the fact that regular analytic rings √ are normal. Follow the proof of 1.5.13, but now use the Nullstellensatz 3.4.4 to prove I ⊂ J (notations of 1.5.13).)

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4 Further Development of Analytic Geometry

“ ” ∂f ∂f 4.4.21. Let f ∈ C {x1 , . . . ,xn }. Prove that f k ∈ ∂x , . . . , for some k ≥ 0. ∂x n 1 ´ ` ∂f ∂f ∂f ∂f ), 0 and (Y, 0) = (V ( ), 0). Show that (X, 0) = , . . . , , . . . , ∂x (Hint: Let (X, 0) = V (f, ∂x ∂xn ∂x1 n 1 (Y, 0) in the following way. Suppose (X, 0) $ (Y, 0). Find a map n : (C , 0) −→ (Y, 0) with n((C , 0)) 6⊂ (X, 0) by using the normalization of a curve (C, 0) ⊂ (Y, 0) which is not contained d in (X, 0). Compute dt (f ◦ n).) e 0) −→ (C, 0) be given such 4.4.22. Let (C, 0) be an irreducible curve singularity. Let n : (C,e e that (C,e0) is smooth. Suppose that dimC (OC, /OC,0 ) < ∞. Show that n is the normalization ee 0 mapping. Generalize this result to reducible germs. (Hint: Show that the degree of the corresponding field extension is one.)

169

5

Plane Curve Singularities

This chapter is devoted to the study of plane curve singularities. First of all, from the existence of the normalization, and the fact that one-dimensional normal singularities are smooth, which are results proved in the previous chapter, it follows that one can parameterize curve singularities. Thus, if R is the local ring of an irreducible plane curve singularity, we have an injection R ⊂ C {t}. From the Finiteness of Normalization Theorem 1.5.19 it follows without much difficulty that the codimension of R in C {t} (as C –vector space) is finite. This codimension is called the δ–invariant. Here we prove the converse of this statement, that is, given a subring R of C {t} of finite codimension, generated by two elements, say x(t) and y(t), then R the local ring of an irreducible plane curve singularity, say given by f (x,y) = 0. In order to obtain information on the order of f , we introduce the intersection multiplicity of two (irreducible) plane curve singularities (C, 0) and (D, 0). This number can be calculated as follows. Consider the normalization C {t} of OC,0 , and let (D, 0) be given by g = 0. Via the inclusion OC,0 ⊂ C {t}, we can consider at g as an element of C {t}. Then the intersection multiplicity is given by the vanishing order of g. From this it easily follows that if R = C {x(t),y(t)}, and f ∈ C {x,y} is irreducible with f (x(t),y(t)) = 0, then the order of f in y is equal to ordt (x(t)), and similar for x. As an application we prove that if we have a convergent power series f ∈ C {x,y}, which is irreducible when considered as a formal power series, then it is already irreducible as a convergent power series. This in fact holds for power series in any number of variables, and is an easy consequence of the Artin Approximation Theorem, which will be proved in Chapter 8. Next we discuss Newton’s method. Here we start with an f ∈ C {x,y}, and try to find x(t), y(t) with f (x(t),y(t)) = 0. We give two proofs of the convergence. The first one is a consequence of Corollary 3.3.18, and is, to our knowledge, new. The second one is standard. Finally, we discuss the irreducibility of f in terms of its Newton polygon. It is shown that for an irreducible f the Newton polygon has only one face. Our proof is algorithmic, that is, if the Newton polygon has more than one face, we will give an algorithm to find a factorization of f . In Section 5.2 we will study several invariants of an irreducible plane curve singularity R. As above, we may assume that R ⊂ C {t}, which is the normalization. We already mentioned the δ–invariant. Moreover, we introduce the semigroup Γ(R) ⊂ N, which consists of the vanishing orders of the elements in R. Furthermore, the conductor c(R) is introduced. It is the smallest natural number α with α + N ⊂ Γ(R). Our first aim is to show the Theorem of Gorenstein, which says that c(R) = 2 · δ(R). The proof uses duality theory for fractional ideals. Fractional ideals are R–modules in the quotient field Q(R), which have a universal denominator. Given a fractional ideal I, we can look at e and the conductor ideal its dual I ∗ := HomR (I,R). For example, the normalization R e AnnR (R/R) are fractional ideals, which are dual to each other. We have a canonical map I −→ I ∗∗ , which one easily sees to be injective. One may ask oneself whether this map is an isomorphism for all fractional ideals. In general, this turns out not to be the case. Those curve singularities for which it is true are called Gorenstein, and we will prove that irreducible plane curve singularities are indeed Gorenstein. We then discuss how to compute the minimal generators of the semigroup Γ(R) from the parametrization. In

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5 Plane Curve Singularities

general, this would be much too so we first suppose that the parametrization P difficult, i is of type x(t) = tn , y(t) = i≥m ai t with m ≥ n. After a coordinate change in t, and interchanging x and y we may suppose that our parametrization is of this type. Characteristic exponents k0 , . . . ,kg are introduced, in terms of n, and the i ∈ N with ai 6= 0. A priori these are only invariants of the parametrization, but we will show that they determine, and are determined by, the minimal generators β0 , . . . ,βg of the semigroup Γ(R). We then discuss the Inversion Theorem, which is of importance in the next section. This gives a method for computing the characteristic exponents in case n > m. In Section 5.3 we consider resolutions of irreducible plane curve singularities. First of all, the blowing-up π : Bl0 C 2 −→ C 2 of C 2 in a point, say 0, is introduced. The space Bl0 C 2 is a complex manifold of dimension one, in which the point 0 has been replaced by all directions through 0. The set π −1 (0) is a projective line, called the exceptional curve or exceptional divisor. If one now has a plane curve singularity, which has representative C in some small open neighborhood U of 0, we can look at the strict transform C (1) := π −1 (C \ {0}), and iterate this process. That is, we blow up in the singular point of C (1) , if present. The Theorem of Resolutions of Plane Curve Singularities says that this process stops after finitely many steps. The main ingredient of the proof is the existence of a parametrization of the irreducible plane curve singularity, as studied in the first section. After some more blowing-ups, one can arrange that the strict transform intersects just one exceptional curve, and intersects this curve transversally. This is then called the standard resolution. Having such a resolution process, one can introduce two invariants. First of all, one has the multiplicity sequence. If the standard resolution is obtained after k blowing-ups, one has a sequence of numbers (m0 , . . . ,mk−1 ), where m0 is the multiplicity of C, and mi is the multiplicity of the strict transform C (i) after i blowingups. Furthermore, the resolution graph is introduced. It is a weighted graph with one vertex for each exceptional divisor. The weight of a vertex is i if the curve is created under the i–th blowing-up. Furthermore, for the strict transform we add a star ∗. Two vertices are connected by an edge, if the corresponding divisors intersect. Furthermore, the star is connected to the vertex corresponding to the unique exceptional divisor which it intersects. The main result of the section is that the multiplicity sequence, the resolution graph, the characteristic exponents, and the semigroup determine each other. In the proof, essential use of the Inversion Theorem is made. In Section 5.4 we study reducible plane curve singularities. Again we can study a standard resolution. Its existence easily follows from the irreducible case. As a new invariant the contact number between two irreducible components, also called branches of the plane curve singularity, is introduced. The contact number of two irreducible plane curve singularities is the number of blowing-ups one needs to separate the strict transforms. It turns out that this number is finite. As in the irreducible case, we introduce a resolution graph, but of course we need a star ∗ for each branch. We then show that the resolution graph determines the resolution graphs of the irreducible components and the contact numbers between the branches, and conversely. Finally, as an application of the standard resolution we give two results. First of all, given irreducible plane curve singularities, we show how to compute the intersection number from the multiplicity sequences and the contact numbers. Secondly, for irreducible plane curve singularities, we will prove a theorem of Max Noether, which gives a formula for the δ–invariant of an irreducible plane curve singularity in terms of its multiplicity sequence. This result will be generalized to the reducible case in the exercises.

5.1 Puiseux Expansion

5.1

171

Puiseux Expansion

From the results of the previous chapter it is easy to see that one can parameterize irreducible (plane) curve singularities: Theorem 5.1.1 (Puiseux Expansion). Let 0 6= f ∈ C {x,y} be irreducible. Then there exist x(t), y(t) ∈ C {t} such that
(1) f x(t),y(t) = 0,
(2) dimC C {t}/C {x(t),y(t)} < ∞.

Proof. Let R be the normalization of C {x,y}/(f ). Because of Corollary 4.4.10 R is isomorphic to C {t}. Therefore, we have an injective map C {x,y}/(f ) −→ C {t}, and C {t} is a finitely generated C {x,y}/(f )–module via this map. Let x(t) respectively y(t) be the images of x mod (f ) respectively y mod (f ) in C {t}. Then obviously (1) holds. To prove (2) we may assume that f is a Weierstraß polynomial in y. Let ∆ be its discriminant. As f is irreducible, it does not have multiple factors. In particular ∆ ∈ / (f ). It follows from the Theorem of Finiteness of Normalization, see 1.5.19, that ∆C {t} ⊂ C {x(t),y(t)}. We can view ∆ as an element of C {t}. Let k be the vanishing order of ∆, that is, ∆ = tk u for some unit u. Then it follows that tk C {t} ⊂ C {x(t),y(t)}, so that {1, . . . ,tk−1 } generate C {t}/C {x(t),y(t)} as C –vector space. Example 5.1.2. For the A2 –singularity y 2 −x3 = 0 we can take x(t) = t2 , and y(t) = t3 . We have the following converse to 5.1.1. Theorem 5.1.3. Let two power series x(t),
y(t) ∈ tC {t} be given. Define R = C {x(t),y(t)} and suppose that dimC C {t}/R < ∞. Then there exists an irreducible f ∈ C {x,y} with R ∼ = C {x,y}/(f ). Moreover, R ⊂ C {t} is the normalization of R.

Proof. Consider the map ϕ : C {x,y} −→ C
{t} defined by ϕ(x) = x(t), ϕ(y) = y(t). Then C {x,y}/ Ker(ϕ) ∼ = R. From dimC C {t}/R < ∞ it follows that R ⊂ C {t} is a finite ring extension, and as C {t} has dimension one, it follows that R has dimension one, see 4.1.4. Because R ⊂ C {t}, it is an integral domain. Thus R is the local ring of an irreducible plane curve singularity. Therefore Ker(ϕ) is a principal ideal by the Characterization of Hypersurfaces 4.1.12. So Ker(ϕ) = (f ) for some f ∈ C {x,y}. As R is an integral domain, we get that Ker(ϕ) is a prime ideal, and it follows that f is irreducible.
It remains to prove that C {t} is the normalization of R. As dimC C {t}/R < ∞, it follows that I := AnnR (C {t}/R) 6= 0 (Exercise 5.1.30). Now I is an ideal in R = C {x(t),y(t)}, but from the definition it follows immediately that I is also an ideal in C {t}. As C {t} is a principal ideal domain it follows that I = tc ·C {t} for some c > 0. This implies that tc C {t} ⊂ C {x(t),y(t)}. Let ψ ∈ tC {t}, then for all k ≥ c we have ψ k ∈ R. k+1 In particular ψ = ψψk is in the quotient field of C {x(t),y(t)}. As certainly the constants are in the quotient field of C {x(t),y(t)} too, it follows that the quotient fields of C {t} and C {x(t),y(t)} are equal. As C {t} is a finitely generated C {x(t),y(t)}–module, every element in C {t} is integral over C {x(t),y(t)}. Therefore C {t} is in the normalization of C {x(t),y(t)}. As C {t} is normal itself, it follows that C {t} is the normalization of C {x(t),y(t)}.

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5 Plane Curve Singularities

In this theorem we showed that we can find such an f , but we still have no information on its order. In order to answer this question, we introduce the intersection multiplicity. Definition 5.1.4. Let (X, 0) and (Y, 0) be plane curve singularities, not necessarily reduced, not having a component in common. Let (X, 0) be defined by f = 0, and let (Y, 0) be defined by g = 0. The intersection multiplicity or intersection number of (X, 0) and (Y, 0) at 0 is defined by
(X, 0) · (Y, 0) := dimC C {x,y}/(f,g) . In case (X, 0) and (Y, 0) are reduced, this intersection multiplicity has a geometric interpretation as the number of intersection points in a small neighborhood of 0 after a small perturbation of f and g. We are not able to prove this at the moment, but it will be done in Chapter 6. The following important lemma tells us how to compute the intersection number with the help of the normalization in case one of the curves, say (X, 0), is irreducible.

Lemma 5.1.5. Suppose that (X, 0) is an irreducible plane curve singularity. Let OX,0 ⊂ ∼ OX,e e 0 = C {t} be its normalization. Let (Y, 0) be a plane curve singularity, defined by g = 0. We can consider the class of g in OX,0 and, therefore, we can view g as an element of C {t}. Then (X, 0) · (Y, 0) = ordt (g). Proof. Consider the diagram

OX,e e 0 O

·g

/ OX,e e 0 O

? OX,0

·g

? / OX,0 .


·g ·g Then ordt (g) = dimC Coker(OX,e e 0 −→ OX,e e 0 ) , and (X, 0)·(Y, 0) = dimC Coker(OX,0 −→
OX,0 ) . The problem is now reduced to a problem in linear algebra, and its solution is left as Exercise 5.1.31. Corollary 5.1.6. Under the conditions of Theorem 5.1.3, suppose that ord(x(t)) = n, and that ord(y(t)) = m. Let f ∈ C {x,y} be irreducible with C {x(t),y(t)} ∼ = C {x,y}/(f ). Then (1) f is regular in y of order n, and regular in x of order m. (2) The multiplicity ord(f ) of f is equal to min{n,m}. Proof. Let (X, 0) is the singularity defined by f = 0, and (L, 0) the germ of the line defined by x = 0. The first statement follows because, by Lemma 5.1.5,
ordy (f ) = dimC C {x,y}/(f,x) = (X, 0) · (L, 0) = ordt (x(t)).

Similarly for x. To prove the second statement we may without loss of generality assume that n ≤ m, and that f is regular in y of order ord(f ). This is always possible after a linear coordinate change. It follows from the first part that ord(f ) = ordy (f ) = (X, 0) · (L, 0) = ordt (x(t)) = n = min{n,m}, as claimed.

5.1 Puiseux Expansion

173

Theorem 5.1.7. Let ε be a primitive n–th root of unity. Consider the ring extension C {x,y} ⊂ C {t,y}, where x = tn . (1) Let f = y n + an−1 (x)y n−1 + · · · + a0 (x) be an irreducible Weierstraß polynomial. Then there exists a y(t) ∈ tC {t} such that f=

n Y

i=1

One also writes this as f =

Qn

i=1


y − y(εi t) .


y − y(εi x1/n ) .


(2) Let y(t) ∈ tC {t} such that dimC C {t}/C {tn ,y(t)} < ∞ be given. Then f =
Qn i i=1 y − y(ε t) is an irreducible Weierstraß polynomial in C {x,y}.

Similar statements hold in the formal case.

Proof. (1) We first prove the
first part. By Theorem 5.1.1 there exist x(t) and y(t) in C {t} such that f x(t),y(t) = 0. By Corollary 5.1.5 it follows that ord(x(t)) = n, and we may after a coordinate change in t, assume that x(t) = tn . Consider f (tn ,y) ∈ C {t,y}. As f (tn ,y(t)) = 0, it follows from the Weierstraß Division Theorem that y − y(t) is a factor of f (tn ,y). For all i with 1 ≤ i ≤ n it follows that f ((εi t)n ,y(εi t)) = f (tn ,y(εi t)) = 0. So y − y(εi t) is also a factor of f (tn ,y). We claim that all these factors are different. That is, we claim y(εi t) = y(εj t) ⇐⇒ n | (i − j). P The direction “⇐=” is trivial. For the converse, write y(t) = i≥m ai ti . It is an elementary fact that for h ∈ C {tn ,y(t)} the order ordt (h) is in the semigroup generated by n and all i with ai 6= 0. As C {t}/C {tn ,y(t)} is a finite-dimensional vector space we can find k1 . . . ,ks ∈ N with gcd(n,k1 , . . . ,ks ) = 1, and akℓ 6= 0 for ℓ = 1, . . . ,s. Now y(εi t) = y(εj t) implies that εkℓ i = εkℓ j for ℓ = 1, . . . ,s. Thus n | kℓ (i − j) for ℓ = 1, . . . ,s. It follows that n | (i − j). (This is elementary number theory.) So the first part follows, because C {t,y} is a unique factorization domain. (2) Consider the kernel of the map ϕ : C {x,y} −→ C {t} which sends x to x(t) and y to y(t). Now in Theorem 5.1.3 it was shown that Ker(ϕ) is principal, and in fact can be generated by an irreducible Weierstraß polynomial in y of degree n by Corollary 5.1.6. This polynomial we call f . From the first part it follows that f has in C {t,y} the desired factorization. The following corollary is most easily proved as an application of Artin’s Approximation Theorem. The proof we give now is more elementary, however. Corollary 5.1.8. Consider an f ∈ C {x,y}. Then we have the following statements. (1) Suppose f is irreducible in C {x,y}. Then f is also irreducible when considered as element of C [[x,y]]. (2) Suppose f = f1 ·f2 with f1 , f2 ∈ C [[x,y]], and suppose f1 is a Weierstraß polynomial in y. Then f1 and f2 are convergent. (3) Suppose y = y(x) is a formal solution of f (x,y) = 0. Then y(x) is convergent.1 1

This we already proved as 3.3.18.

174

5 Plane Curve Singularities

Proof. (1) We may assume that f is an irreducible Weierstraß polynomial. As f is irreQ ducible we can, by Corollary 5.1.7, write f = ni=1 (y − y(εi t)), where x = tn . Moreover
dimC C {t}/C {tn ,y(t)} is finite. So for all k ≫ 0 the element tk can be
written as a n n power series in t and y(t). Thus it follows that dimC C [[t]]/C [[t ,y(t)]] is finite. Applying the second part of Theorem 5.1.7 in the formal case, shows that f is irreducible in C [[x,y]]. (2) Take a factorization f = u · g1 · · · gn of f , where the gi are Weierstraß polynomials and u is a unit. Then u and g1 . . . ,gn are uniquely determined. By the first part, the gi are also irreducible in C [[x,y]]. Taking similar factorizations of f1 and f2 . It follows that f1 is a product of the g ′ s, hence convergent. But then f2 = ff1 is also convergent. (3) If y = y(x) is a solution of f (x,y) = 0, we have that y − y(x) is a factor of f . Thus (3) is a special case of the second part. Now we wish to explain an explicit construction of the parametrization obtained in Theorem 5.1.1. As a tool we need to introduce the Newton polygon. P Definition 5.1.9. Let f = aij xi y j ∈ C {x,y}. We call supp(f ) = {(i,j) : aij 6= 0}

the support of f . Sometimes we will identify the pair (i,j) with the monomial xi y j . Let
S (1) Γ+ (f ) = (i,j) + R2+ . (i,j)∈ supp(f )

(2) Γ(f ) the boundary of the convex hull of Γ+ (f ) in R2 .

Then (1) Γ(f ) is called the Newton polygon of f . (2) The compact segments of Γ(f ) we call faces. The slope of the line through a face is called the slope of the face. (3) Let P∆ be ia jface of Γ(f ). Then the restriction of f to ∆ by definition is f∆ = aij x y . (i,j)∈∆

(4) The face ∆ induces a grading on C {x,y} in the following way: if ∆ has slope w1 −µ = − w , in lowest terms, and w1 , w2 > 0. Then we set w-deg(x) = w1 , and 2 w-deg(y) = w2 . Note that f∆ is quasi-homogeneous with respect to these weights.

Example 5.1.10. Let f = x5 + x3 y 3 + y 3 + xy. Then Supp(f ) = {(5,0),(3,3),(0,3),(1,1)}.

Two faces

Γ+ (f )

Γ(f )

5.1 Puiseux Expansion

175

Let ∆ be the face defined by (1,1) and (5,0). Then f∆ = x5 + xy. The weights corresponding to the grading defined by ∆ are w-deg(x) = 1 and w-deg(y) = 4. Obviously, different f can have the same Newton polygon. For example, x5 + x3 y 3 + y 3 + xy and x5 + y 3 + xy have the same Newton polygon. The element (3,3) does not enlarge the Newton polygon. Example 5.1.11. f = x3 y 3 + x2 y 4 .

One face Γ(f )

Definition 5.1.12. f is called convenient if there exist positive integers n,m such that (n,0),(0,m) ∈ supp(f ), that is, the noncompact parts of Γ(f ) are on the coordinate axis. Remark 5.1.13. f in Example 5.1.10 is convenient, f in Example 5.1.11 is not convenient. For any f ∈ C {x,y}, we can always write f = xa · y b · f1 , where f1 is convenient. Therefore, from the point of view of the zeros of f the convenient case is the important one. We now come to Newton’s method to find a parametrization of a branch of a plane curve singularity. Theorem 5.1.14. Let f ∈ C {x,y}, and suppose that f is convenient. We have the following algorithm for finding a parametrization of an irreducible component (also called branch) of f = 0. w1 , which is in lowest terms, and with (1) Let ∆ be the face of Γ(f ) with smallest slope − w 2 w1 , w2 > 0. Then f∆ is quasi-homogeneous of say, degree d0 . Consider a zero of f∆ (1,y) which we call a0 . Take new variables x1 and y1 and substitute 1 x = xw 1 ,

Then f (1) (x1 ,y1 ) :=

1 xd10

2 y = xw 1 (a0 + y1 ).


w2 1 f (xw 1 ,x1 (a0 + y1 )) ∈ C {x1 ,y1 }.

(2) Inductively find a parametrization x1 = ta , y1 = ϕ(t) of a branch of f (1) (x1 ,y1 ) in case f (1) 6= 0. Then x = taw1 , y = a0 taw2 + taw2 ϕ(t), gives a parametrization of a branch of f = 0. Before giving the proof we give an example.

176

5 Plane Curve Singularities

Example 5.1.15. Let f (x,y) = y 4 − 2x3 y 2 + x6 − x7 . The Newton polygon consists of one face.

Γ(f )

We have w1 = 2, w2 = 3. Then d0 = 12 and f∆ (x,y) = (y 2 − x3 )2 . Thus f∆ (1,y) = (y 2 − 1)2 . We have the solution a0 = 1. We set x = x21 , y = x31 (1 + y1 ). (1) We plug this into f (x,y) to get x12 (x1 ,y1 ). We obtain as result 1 f

f (1) (x1 ,y1 ) = y14 + 4y13 + 4y12 − x21 . Of course, we can go on as in the first step, but we can also use some common sense. Indeed, f (1) (x1 ,y1 ) = (2y1 + y12 )2 − x21 ,

so that x1 = ±(2y1 + y12 ). We, therefore, see two solutions for y1 : √ √ y1 = −1 + 1 − x1 ; y1 = −1 + 1 + x1 . Hence, we obtain the solutions (setting x1 = t) √ (1) x = t2 , y = t3 1 − t, √ (2) x = t2 , y = t3 1 + t.

Hence, √ we can deduce easily, that f is reducible, as we can find g1 ∈ C {x,y} with g1 (t2 ,t3 1 −√t) = 0, and g1 is a Weierstraß polynomial in y. Similarly we
can find a g
2 with g2 (t2 ,t3 1 + t) = 0. Note that the plane curve singularities V (g1 ), 0 and V (g2 ), 0 √ √ are such that their intersection is zero. Indeed (t2 ,t3 1 − t) = (t2 ,t3 1 − t) for t small if and only if t = 0, as a simple calculation shows. Thus (g1 ) 6= (g2 ), and we get the factorization f = g1 · g2 , since degy (gi ) = 2 for i = 1,2. Proof of Theorem 5.1.14. As f is convenient, it is in particular regular in y, of order say b. Step 1. First of all, we show that the definition of f (1) makes sense. We can write X fd , f= d≥d0

where fd is quasi-homogeneous of degree d with weights w1 and w2 . From the definition of quasi-homogeneity it follows that

5.1 Puiseux Expansion

177

X
w2 1 xd1 fd (1,a0 + y1 ), f xw 1 ,x1 (a0 + y1 ) = d≥d0

(1)

so that indeed the definition of f makes sense. Obviously, if we can find a parametrization of a branch of f (1) = 0, we get a parametrization of a branch of f = 0. Step 2. We now look at f (1) . Its definition was X 0 xd−d fd (1,a0 + y1 ). (5.1) f (1) (x1 ,y1 ) = 1 d≥d0

We claim that f (1) is again regular in y1 . In view of formula (5.1) we have to show that fd0 (1,a0 + y1 ) is regular in y1 . We can factorize fd0 (1,y): (Recall that f is regular of order b.) fd0 (1,y) = c · (y − a0 ) · · · (y − ab−1 ),

for some constants c,a0 , . . . ,ab−1 ∈ C . From this we see that

fd0 (1,a0 + y1 ) = c · y1 (y1 + a0 − a1 ) · · · (y1 + a0 − ab−1 ),

so fd0 (1,a0 + y1 ) is regular in y1 . Thus f (1) is also regular in y1 , of order at most b. There are two cases to consider. (1) f (1) is also regular of order b. Then it follows that a0 = a1 = · · · = ab−1 , so that fd0 (1,y) = c · (y − a0 )b . As fd0 is quasi-homogeneous and convenient it follows that fd0 = c · (y − a0 xm )b . So w1 = 1, and m · b = d0 . Furthermore w2 = m, and m x1 = x, y = xm 1 (a0 + y1 ) = x (a0 + y1 ). (2) f (1) is regular of order b1 < b. So if we iterate this procedure, that is, we define xi+1 := (xi )1 , yi+1 := (yi )1 , f (i+1) := (f (i) )(1) , then the second case above can occur only finitely many times in the iteration. For the rest of the proof, we may therefore assume that we are always in the first case, that is, we may assume that x = x1 = x2 = · · · . So in this case one tries to find a parametrization of type x = t, y = ϕ(t), that is a smooth branch y − y(x) = 0 of f = 0. Step 3. So assuming we are always in the first case, we want to show that the algorithm gives a solution y = y(x) of f (x,y) = 0. By Corollary 3.3.18, or alternatively, the third part of Corollary 5.1.8 it suffices to show that y = y(x) is a formal solution of f (x,y) = 0. So if ai is a solution of f (i) (0,yi ) = 0, the iteration gives y = xm0 (a0 + y1 ),

y1 = xm1 (a1 + y2 ), etc,

for certain m0 ,m1 , . . .. Plugging in we get y = xm0 (a0 + xm1 (a1 + y2 )) = a0 xm0 + a1 xm0 +m1 + y2 xm0 +m1 = · · · . We claim that the formal power series y(x) =

∞ X

ak xm0 +...+mk

k=0

is a formal solution of f (x,y) = 0. Indeed, f (x,y) = xd0 f (1) (x,y1 ) = · · · = xkd0 f (k) (x,yk ). Plugging in y = y(x) we get f (x,y(x)) = xkd0 f (k) (x,yk (x)) for some power series yk (x). Thus f (x,y(x)) is divisible by all powers of x, and hence is zero.

178

5 Plane Curve Singularities

The next thing we study in this section is the irreducibility of a power series f , in terms of the Newton polygon. Let us look at the simplest case. Suppose h = f · g, f and g convenient, and suppose that the Newton polygon of both f and g has just one face. Let the slope of the face of Γ(f ) be given by the endpoints (p1 ,0) and (0,q1 ), and similarly the slope of the face of Γ(g) by the endpoints (p2 ,0) and (0,q2 ). We assume that the slope of the face of Γ(f ) is less than or equal to the slope of the face of Γ(g). Then h = f · g is convenient. The Newton polygon of h has two faces, resp. one face if the slopes are equal. The first face has endpoints (0,q1 + q2 ), and (p1 ,q2 ), the second has endpoints (p1 ,q2 ) and (p1 + p2 ,0). q1

q1 + q2 Γ(f ) q2

p1

q2

Γ(g)

p2

p1

p1 + p2

This is easy to prove. One has to show that each monomial xα y β of h lies right to both faces. The condition that a monomial xα y β lies right to a line whose intersections with the axes are (p,0) and (0,q) say, is simply αq + βp ≥ pq. Let xα1 y β1 be a monomial of f and xα2 y β2 be a monomial of g. Then, by assumption on Γ(f ) and Γ(g) we have (1)

q1 p1

≥

q2 p2 ;

(2) β1 ≥ − pq11 α1 + q1 ; (3) β2 ≥ − pq22 α2 + q2 . This implies that β1 +β2 ≥ − pq11 (α1 +α2 )+q1 +q2 and β1 +β2 ≥ − pq22 (α1 +α2 )+ pq22 (p1 +p2 ). The first inequality follows by adding (2) and (3) and using (1). The second inequality follows by adding pp21 q2 to (3) and using the inequality α1 q2 + β1 p2 ≥ p1 q2 . Therefore, xα1 +α2 y β1 +β2 lies right to both faces. Moreover if xα1 +α2 y β1 +β2 lies on both faces, it follows that β1 = 0, from which α1 = p1 , β2 = q2 and α2 = 0 easily follow. So the monomial xp1 y q2 indeed occurs in the product f · g. More generally, we need the following lemma. The proof is just more technical, and therefore left to the reader. Lemma 5.1.16. Let f1 ,f2 ∈ C {x,y} be convenient. Assume Γ(f2 ) has one face ∆1 defined by (p2 ,0) and (0,q1 − q2 ). Denote the faces of f1 by ∆2 , . . . ,∆r and let ∆i be defined by (pi ,q i ) and (pi+1 ,q i+1 ) with p2 = 0, q 2 = q2 and q r+1 = 0. Assume furthermore that the slope of ∆1 is smaller than the slopes of the ∆i , i ≥ 2. Then the Newton polygon Γ(f ) of f = f1 · f2 has faces ∆1 , . . . ,∆r , which are given by: • ∆1 defined by (0,q1 ) and (p2 ,q2 ); • ∆i defined by (pi + p2 ,q i ) and (pi+1 + p2 ,q i+1 ) for i = 2, . . . ,r.

5.1 Puiseux Expansion

179

q1 − q2

q1

Γ(f1 · f2 )

∆1 ∆1 Γ(f2 ) q2 p2

q2

∆2

∆2 Γ(f1 ) p2

We now come to a converse of this lemma. It follows without much difficulty from the explicit construction of the Puiseux expansion, that the converse must hold. We want an explicit construction of the factorization, however. Theorem 5.1.17. Let f ∈ C {x,y} be convenient, ∆1 , . . . ,∆r be the faces of Γ(f ) and di the slope of ∆i . Then f = f1 · · · fr , where fi is convenient such that Γ(fi ) has only one face of slope di , i = 1, . . . ,r. In the proof of this theorem, we use the following lemma. Lemma 5.1.18. Consider the polynomial ring C [x,y], with grading defined by w-deg(x) = w1 > 0 and w-deg(y) = w2 > 0. Let g = xb + . . . be quasi-homogeneous of weighted degree bw1 . Let h ∈ C [x,y] be quasi-homogeneous with weighted degree at least bw1 + aw2 for some a ∈ N. Then h ∈ (y a ,g). Proof. We apply Division with Remainder, see Theorem 2.1.5, and get h = q · g + r, where r is a polynomial in x of degree at most b − 1. From the algorithm of division with remainder, it follows that also r is quasi-homogeneous, in fact w-deg(r) = w-deg(h) if r 6= 0. Take a monomial xi y j in r. Then i < b. Hence bw1 + aw2 ≤ w-deg(r) = w-deg(xi y j ) = iw1 + jw2 < bw1 + jw2 . It follows that j > a, and every monomial of r is divisible by y a . Thus indeed h ∈ (y a ,g). P Proof of Theorem 5.1.17. Let f = aij xi y j , assume r > 1 and let ∆1 be the face with minimal slope. Because f is convenient, ∆1 is the line segment from (0,q1 ) to (p2 ,q2 ):

180

5 Plane Curve Singularities (0,q1 )

∆1

(p2 ,q2 )

We define g :=

1 · f∆1 , h := f − y q2 g. y q2

We obviously have f = y q2 g + h. As in Definition 5.1.9 we introduce the weighted degree on the monomials of C {x,y} corresponding to ∆1 , that is, we put w-deg(x) = w1 = (q1 − q2 )/c, and w-deg(y) = w2 = p2 /c, where c = gcd(q1 − q2 ,p2 ). With these weights, g is weighted homogeneous of degree (q1 − q2 )p2 /c. We consider the elements of C {x,y} to be ordered by the corresponding weighted order function, w-ord(f ) = min{w-deg(xi y j ) : (i,j) ∈ supp(f )}. (n) (n) We will now inductively construct f1 , f2 and h(n) such that (n)

• f = f1

(n)

· f2

+ h(n) ,

(n)

= y q2 + terms of higher w-order not divisible by y q2 ,

(n)

= g + terms of higher w-order,

• f1 • f2

• p2 q1 /c < w-ord(h) < w-ord(h(2) ) < · · · < w-ord(h(n) ). (1)

(1)

The initialization of course is f1 = y q2 , f2 = g and h(1) = h. Note that the weighted order of h is indeed bigger than p2 q1 /c, as w-ord(h) > w-ord(y q2 g) = p2 q1 /c. This is because all monomials of h lie right to the line through the face ∆1 . Now let k be the quasi-homogeneous part of h(n) of w–degree equal to w-ord(h(n) ). Because of w-deg(k) > w-ord(y q2 g) we can find by Lemma 5.1.18 weighted homogeneous ξ,η such that k = ξy q2 + ηg. Moreover one can arrange that no monomial of η is divisible by y q2 . Now set (n+1)

= f1

(n+1)

= f2

• f1

• f2

(n)

+ η,

(n)

+ ξ, (n+1)

• h(n+1) = f − f1

(n+1)

· f2

. (n)

(n)

It is not difficult to check, using w-ord(f1 − y q2 ) > p2 q2 /c and w-ord(f2 p2 (q1 − q2 )/c, that w-ord(h(n+1) ) > w-ord(h(n) ).

− g) >

Continuing this way, we get formal power series f1 and f2 with f = f1 ·f2 . In particular, by construction, f1 is a Weierstraß polynomial with respect to y. Therefore, the convergence of f1 and f2 follows from Corollary 5.1.8. The Newton polygon Γ(f2 ) = Γ(g) looks like,

5.1 Puiseux Expansion

181 q1 − q2

p2

Using Lemma 5.1.16, the lemma follows by induction on r. Example 5.1.19. We consider f = y 7 + x3 y 5 + x2 y 2 + x5 .

∆1

∆2

(1)

h

(1)

(1)

We have w1 = 5, w2 = 2, g = f2 = y12 · f∆1 = y 5 + x2 . So with f1 = x5 + x3 y 5 we get (1) (1) f = f1 f2 + h(1) .

= y 2 and

Now h(1) is weighted homogeneous of degree 25. The monomials of h(1) all are divisible by x3 . In fact h(1) = (x2 + y 5 ) · x3 = g · x3 , so that f = (y 2 + x3 )(y 5 + x2 ). We have the following generalization. Theorem 5.1.20. Let f ∈ C {x,y} P and suppose Γ(f ) has just one face P be convenient, ∆. Suppose that there exist g = ν≥d gν , h = ν≥d hν and R such that (1) f = g · h + R;

(2) the gν and hν are homogeneous of degree ν with respect to the grading introduced by ∆; (3) gν = hν for ν = d, . . . ,p − 1, and gp 6= hp ;

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5 Plane Curve Singularities

(4) (gd ,hp ) is (x,y)–primary; (5) w-ord(R) > d + p. Then f is reducible. The proof runs as in the previous theorem, and is therefore left to the reader as an exercise. One needs the following lemma, the proof of which will be given in Chapter seven, see Lemma 7.3. (0)

(0)

(0)

(0)

Lemma 5.1.21. Let h1 and h2 be homogeneous and assume that (h1 ,h2 ) is (x,y)– (0) (0) (1) (0) (1) (1) primary. Let h1 = h1 + h1 , h2 = h2 + h2 , with w-ord(hi ) > w-ord(hi ). Suppose i j that iw1 + jw2 > w-ord(h1 · h2 ) for some i,j ∈ N. Then x y ∈ (h1 ,h2 ). Examples 5.1.22. (1) In the previous example we were “lucky” to get factors of f which are polynomials. In general of course, one cannot expect this. We now look at the standard example f = y 2 − x2 (x + 1). We get f = (y − x)(y + x) + h(1) . Then k = h(1) = −x3 ∈ (y − x,y + x). In fact −x3 = 21 x2 (y − x) − 12 x2 (y + x). Hence, f = (y − x − 12 x2 )(y + x + 12 x2 ) + 14 x4 . Now 41 x4 = − 81 x3 (y − x) + 81 x3 (y + x). Hence f = (y − x − 12 x2 + 18 x3 ) · (y + x + 21 x2 − 81 x3 ) − 18 x5 +

1 6 64 x ,

etc. √ will not stop, because we get as the factorization f (x,y) = (y − √ This process x x + 1)(y + x x + 1). (2) We consider f = y 4 − 2x3 y 2 + x6 − x7 . We have already seen in Example 5.1.15 that f is reducible. To apply the theorem, we note that f = (y 2 − x3 − x2 y + 12 x4 )(y 2 − x3 + x2 y + 12 x4 ) − 14 x8 . so that the conditions of Theorem 5.1.20 are satisfied, and we get a second proof that f is reducible.

Exercises 5.1.23. Use the algorithm for the Puiseux expansion to find a solution of y 2 + 2y − x = 0, that √ is, find the power series expansion of y = −1 + 1 + x. 5.1.24. Find a factorization of f (x,y) = y 4 − 2x3 y 2 + x6 − x7 in C {x,y}, see Example 5.1.22. 5.1.25. Find the Puiseux expansions of the following (irreducible) power series. (1) f = y 4 − 2x3 y 2 − 4x5 y + x6 − x7 .

(2) f = x33 − 3x22 y 2 + 6x21 y − x20 + 3x11 y 4 + 2x10 y 3 − y 6 . 5.1.26. Prove Lemma 5.1.16. 5.1.27. Let C ((x)) be the quotient field of C {x} and C ((x)) its algebraic closure. Prove that C ((x)) = ∪ C ((x1/n )). n∈N

5.2 Invariants

183

5.1.28. This is an exercise for the reader who knows about resultants. Let (X, 0) and (Y, 0) be plane curve singularities, which are given by Weierstraß polynomials f and g in y. Thus in particular f, g ∈ C {x}[y]. Consider the resultant Rf,g ∈ C {x}. Prove that (X, 0) · (Y, 0) = ordx (Rf,g ) (Hint: Reduce to the case that f and g are irreducible. Let x = tn , y = ϕ(t) be a Puiseux expansion of (X, 0), and factorize f in C {t}[y]. Compute the resultant in C {t}.) 5.1.29. This exercise uses the previous exercises. Let (X, 0) be a plane curve singularity given by a Weierstraß polynomial f in y. Let ∆ ∈ C {x} be the discriminant of f . Prove ordx (∆) = µ(f ) + (X, 0) · (L, 0), where (L, 0) is the line given by y = 0, and µ(f ) is the Milnor number. ∂f (Hint: Consider (Y, 0) given by ∂x = 0. Compute the intersection number (X, 0) · (Y, 0) in two ways. One way is by using the previous exercise, the other way is by using Puiseux expansion for the branches of (Y, 0): if (ϕ1 (t),ϕ2 (t)) is a parametrization of a branch then look at the vanishing d order of t 7→ dt (f (ϕ1 (t),ϕ2 (t))), which differs by one from the intersection multiplicity.) 5.1.30. Let (R,m) be an analytic C –algebra and M a finitely generated R–module. Assume that dimC M < ∞. Prove that AnnR (M ) 6= {0}. (Hint: Choose x ∈ m, x 6= 0 and consider M ⊃ xM ⊃ . . ..) 5.1.31. Consider a vector space V over k and a subspace W ⊂ V . Let T : V −→ V be a linear map, such that T (W ) ⊂ W and Ker(T ) ⊂ W . Suppose that both V /W and V /T (V ) are finite-dimensional. Prove that dimk (V /T (V )) = dimk (W/T (W )). 5.1.32. Prove Theorem 5.1.20.

5.2

Invariants

In this chapter we wish to study invariants of irreducible plane curve singularities. An invariant is here a map from the set of isomorphy classes of singularities to Zk for some k; for example, the dimension of some vector space associated to the singularity, as we shall soon see. Invariants play an important role in studying singularities. Let R = OX,x be the local ring of an irreducible plane curve singularity and m the maximal ideal. By the e the normalization of result of the previous section, we may assume that R ⊂ C {t} =: R, R. Definition 5.2.1.

e (1) We call δ(R) := dimC (R/R) the δ–invariant of R.

(2) We call Γ(R) := {ordt (a) | a ∈ R, a 6= 0} the semi–group of values of R.

(3) We define the conductor by c(R) := min{α ∈ Γ(R) : α + N ⊂ Γ(R)}. (4) Γ(R) is called symmetric, if there exists a k ∈ N with α ∈ Γ(R) ⇐⇒ k − α ∈ / Γ(R). Lemma 5.2.2.
(1) δ(R) = # N \ Γ(R) .

184

5 Plane Curve Singularities

e e (2) AnnR (R/R) = tc(R) R.

Proof.

e (1) Suppose that a1 , . . . ,aδ is a basis of R/R. Suppose that ord(ai ) = ord(aj ) = k, c so that ai = ci tk + . . . ,aj = cj tk + . . .. Then we may replace aj by aj − cji ai , which has higher vanishing order. Using this operation several times we can get a basis a1 . . . . ,aδ with ord(ai ) 6= ord(aj ) for i 6= j. Similarly, we can assure that ord(ai ) 6∈ Γ(R), then the ord(a1 ), . . . , ord(aδ ) are different elements of N \ Γ(R). On the other hand, if we have different elements c1 < . . . < cγ of N \ Γ(R), then b1 tc1 + . . . + bγ tcγ with bi ∈ C has vanishing order min{ci : bi 6= 0}. So b1 tc1 + . . . + bγ tcγ ∈ R only when b1 = . . . = bγ = 0, so that δ(R) ≥ γ.

e e (2) In the proof of 5.1.3 we showed already that AnnR (R/R) is an ideal in R and R ce e and, therefore AnnR (R/R) = t R for some c. We have to prove that c − 1 6∈ Γ(R). e ⊂ R implies that tc−1 R e ⊂ R and, consequently, tc−1 ∈ But c − 1 ∈ Γ(R) and tc R e e e AnnR (R/R), which contradicts AnnR (R/R) = tc R. This shows the Lemma.

Example 5.2.3. The δ invariant of A2 defined by y 2 − x3 is equal to one, the conductor is equal to two. For the E8 , defined by y 3 − x5 = 0, the δ invariant is four, and the conductor is eight. Theorem 5.2.4 (Gorenstein). c(R) = 2δ(R). For the proof of this Theorem, we need a definition. Definition 5.2.5. (1) Let R be as above, K the quotient field of R. A set I ⊂ K is called a fractional ideal if (a) I is via multiplication in K an R–module, that is r · a ∈ I for all r ∈ R and a ∈ I and, of course, a + a′ ∈ I for all a,a′ ∈ I.

(b) There exists a “universal denominator”, that is, there exists a d ∈ R,d 6= 0, with d · I ⊂ R. (2) Let I ⊂ K be a fractional ideal, then I ∗ := HomR (I,R).

(3) The ring R, or the curve singularity is called Gorenstein if for all fractional ideals I the canonical map I −→ I ∗∗ = HomR (HomR (I,R),R) a 7→ a ˆ,ˆ a(ϕ) := ϕ(a) is an isomorphism. Remarks 5.2.6. e of R is a fractional ideal, (1) Any ideal in R is a fractional ideal. The normalization R which, in general, is not an ideal.

5.2 Invariants

185

(2) For fractional ideals I,J different from (0) one has the equality HomR (I,J) = (J : I)K := {x ∈ K : x · I ⊂ J}. In the remaining part of this Chapter we shall use (J : I) for (J : I)K to simplify the notation. The map (J : I) −→ HomR (I,J) sends an element a ∈ (J : I) to the multiplication with this element in K. On the other hand, a homomorphism α ∈ HomR (I,J) can be extended to an R–linear map α from K to K by putting α( rs ) := 1s α(r) for an element r ∈ I and s ∈ R. Then one shows that the element α(1) ∈ (J : I), giving the inverse map. In particular, we have for (0) 6= J ⊂ I always I ∗ ⊂ J ∗ . e ∗ = (R : R) e R = {c ∈ R : cR e ⊂ R} = AnnR (R/R) e e (3) (R) = tc(R) R.

(4) For finite dimensional vector spaces V over a field k, one knows that the canonical map V −→ V ∗∗ is an isomorphism. The corresponding statement for rings is false, in general, even if one takes only fractional ideals. An example is given in the Exercises. The canonical map is injective, but need not be surjective.

(5) As a fractional ideal is as R–module isomorphic to an ideal in R, it follows that one has to prove the duality only for ideals. Lemma 5.2.7. Let R be as above. Then R is Gorenstein if and only if dimC (m∗ /R) = 1. Proof. First suppose that R is Gorenstein. Suppose the statement is wrong. Then we can find a C –vectorspace I with R $ I $ m∗ . We claim that I is a fractional ideal. The fact that I has a universal denominator follows immediately from the fact that m∗ has a universal denominator. So we only have to show that I is an R–module. Obviously a + a′ ∈ I if a,a′ ∈ I. Let f ∈ R, and a ∈ I. Then we can write f = c + r, with c ∈ C and r ∈ m. Then r · a ∈ R, as a ∈ m∗ . Furthermore ca ∈ I as I is a vector space. Hence f a ∈ I. It follows from the assumption that m $ I ∗ $ R. This is a contradiction, as dimC (R/m) = 1. The converse is more difficult, and is proved in two steps. Step 1. Let (0) 6= J ⊂ I be nonzero fractional ideals. Then we claim that dimC (I/J) ≥ dimC (J ∗ /I ∗ )2 . If J ⊂ L ⊂ I then dimC (I/J) = dimC (I/L)+dimC (L/J) and dimC (J ∗ /I ∗ ) = dimC (J ∗ /L∗ ) + dimC (L∗ /I ∗ ). We can always find a vector space L with J $ L $ I, and mL ⊂ J. Indeed, take any vector space L such that J has codimension one in L. By Nakayama’s Lemma, it follows that mL ⊂ J. It follows as above that L is a fractional ideal. Therefore, we can use induction and it suffices to prove the claim for the case dimC (I/J) = 1. We will construct an injective map ϕ : J ∗ /I ∗ −→ m∗ /R, 2

Because of R being one–dimensional we have dimC (I/J) < ∞.

186

5 Plane Curve Singularities

which, using the assumption, proves the claim. Take an element a ∈ I with (J,a) = I. Then m(I/J) $ (I/J) by Nakayama’s Lemma. This implies m(I/J) = 0 because dimC (I/J) = 1. In particular a · m ⊂ J. We now construct the map ϕ. Let b ∈ J ∗ . Then ba · m = b · am ⊂ b · J ⊂ R. Thus ba ∈ m∗ . Thus the map ϕ sends b to ba. This gives a well-defined map, as b ∈ I ∗ sens a ∈ I to an element in R. We now prove ϕ is injective. So suppose ba ∈ R, for some b ∈ J ∗ . But then it sends all elements of I to R. Thus b ∈ I ∗ . Step 2. Let I be a nonzero fractional ideal. We may assume that I is an ideal in R. Take an element x ∈ I. We claim that we have the following equalities: 
 (∗) dimC (I/(x)) = dimC R/ (x) : I , dimC (R/I) = dimC (I ∗ /R).

(∗∗) Indeed, applying Step 1 gives


dimC (R/(x)) = dimC (R/I) + dimC I/(x)


≥ dimC (I ∗ /R) + dimC (x)∗ /I ∗ 


= dimC (x) : I/(x) + dimC (x) : (x) / (x) : I   

 = dimC (x) : I /(x) + dimC R/ (x) : I
= dimC R/(x) .


Here we used xI ∗ = x(R : I) = (x) : I and x(x)∗ = x R : (x) = (x) : (x)
. So all inequalities have to be equalities and the claim follows because dimC I/(x) =


∗ ∗ ∗ ∗ R/(x) : R . We apply (∗) to the ideal x(x) /xI = dim /I = dim dimC (x) C C
(x) : I and obtain   

dimC (R/I) = dimC (I ∗ /R) = dimC (x) : I /(x) = dimC R/ (x) : ((x) : I) .




Now (x) : I = xI ∗ and (x) : ((x) : I) = (x) : (xI ∗ ) = x(xI ∗ )∗ . As furthermore (xI ∗ )∗ = x1 I ∗∗ , it follows that dimC (R/I) = dimC (R/I ∗∗ ). As I ⊂ I ∗∗ is trivial, the equality I = I ∗∗ follows. This is what we had to prove.

Corollary 5.2.8. Let R be the local ring of a plane curve singularity. Then R is Gorenstein. Remark 5.2.9. It follows from the proof that for any two fractional ideals I and J, we have dimC (I/J) = dimC (J ∗ /I ∗ ) if both are finite. In particular, they are isomorphic as C –vector spaces. Proof. We will show that dimC (m∗ /R) = 1. Let 0 = 6 ϕ ∈ m ⊂ R. We consider the so-called socle
Soc R/(ϕ) := {q ∈ R/(ϕ); m · q = 0 ∈ R/(ϕ)}.

We claim

5.2 Invariants

187
Soc R/(ϕ) ∼ = m∗ /R.


Indeed, given g ∈ Soc R/(ϕ) we can assign to this the element ϕg . This is an element of m∗ , because for all f ∈ m it follows that g · f ∈ (ϕ) and, hence, ϕg · f ∈ R. On the other
hand, suppose we have an element ab ∈ m∗ /R. Then we claim that ab ϕ ∈ Soc R/(ϕ) . Let f ∈ m be arbitrary. Then ab ϕ · f = ab · f · ϕ ∈ (ϕ). As the maps are obviously inverse to each other, the claim follows. Now assume that R is defined by a Weierstraß polynomial f in y of degree m. It suffices to show that the socle of C {x,y}/(f,x) has vector-space dimension 1. But as C {x,y}/(f,x) ≃ C {y}/(y m ), we can easily calculate the socle. It is equal to C · y m−1 , hence has dimension 1. Theorem 5.2.10. Let R be a Gorenstein curve singularity. Then c(R) = 2δ(R) Proof. By duality R/I ≃ I ∗ /R as C –vector spaces for every nonzero ideal I. We apply e ∗ . Then I ∗ = R, e and we get this to the conductor ideal, which by definition is I = (R) e e R/I ≃ R/R. As the conductor is equal to dimC (R/I), (Lemma 5.2.2 (2)), and δ is equal e to dimC (R/R), the Theorem follows. Definition 5.2.11. Let

β0 = min{a ∈ Γ(R); a 6= 0}, β1 = min{a ∈ Γ(R); gcd (a,β0 ) < β0 }, βν = min{a ∈ Γ(R); gcd (a,β0 , . . . ,βν−1 ) < gcd (β0 , . . . ,βν−1 )}, ν ≥ 2. By this construction we obviously obtain only finitely many βν , say β0 , . . . ,βg′ . Let G(R) := {β0 , . . . ,βg′ }. Lemma 5.2.12. (1) The set G(R) is the minimal set of generators of Γ(R). (2) Let ϕ : R −→ R′ be an isomorphism, then G(R) = G(R′ ). We leave the proof of the Lemma as an Exercise. Examples 5.2.13. (1) Let the cusp singularity be given by the parametrization x = t2 ,y = t3 . So R = C {t2 ,t3 }, and the semi-group Γ(R) is generated by β0 = 2 and β1 = 3. (2) We consider the parametrization x = t4 ,y = t6 + t7 . Then obviously β0 = 4 and β1 = 6 are in Γ(R). But also β2 = 13 ∈ Γ(R), as it is the vanishing order of y 2 − x3 = 2t13 + t14 . We shall soon see that Γ(R) is generated by 4,6,13. The question is how to calculate generators of the semi-group Γ(R) from the parametrization. To answer this question, we introduce other invariants, the characteristic exponents.

188

5 Plane Curve Singularities P

Definition 5.2.14. Let R = C {tn ,y(t)}, y(t) = k0 = min{n,m},   min{i|ai 6= 0, k1 = min{i|ai 6= 0,   n ( min{i|ai 6= 0, kν = min{i|ai 6= 0,

i≥m

gcd (i,k0 ) < k0 } gcd (i,k0 ) < k0 } + n − m

ai ti and am 6= 0. We define

if n ≤ m, if n > m and m | n, , if n > m and m ∤ n,

gcd (i,k0 , . . . ,kν−1 ) < gcd (k0 , . . . ,kν−1 )} gcd (i,k0 , . . . ,kν−1 ) < gcd (k0 , . . . ,kν−1 )} + n − m

if n ≤ m, if n > m,

ν ≥ 2. By this construction again, we obviously obtain only finitely many kν , say k0 , . . . ,kg . Note that in all cases k0 < k1 < . . . < kg and gcd(k0 , . . . ,kg ) = 1 because dimC (C {t}/R) < ∞. The set E(R) := {k0 ,k1 , . . . ,kg } is called the set of characteristic exponents of R. Examples 5.2.15. (1) A curve with just one characteristic exponent k0 = 1 is smooth. (2) Consider the curve with parametrization x = t4 ,y = t6 + t7 . The characteristic exponents are k0 = 4,k1 = 6,k2 = 7. We have seen in Example 5.2.13 that β0 = 4,β1 = 6 and β2 = 13. A priori, the k0 , . . . ,kg depend on the parametrization, but in fact they do not! More precisely, we have the following Theorem, implying this. Theorem 5.2.16. The minimal set of generators of Γ(R), G(R) = {β0 , . . . ,βg′ }, and the set of characteristic exponents, E(R) = {k0 , . . . ,kg }, satisfy the relations: (1) (2)

(3)

g = g′, ki = βi − βi = ki +

i−1  X gcd(β0 , . . . ,βj−1 ) j=1

gcd(β0 , . . . ,βj )

i−1  X gcd(k0 , . . . ,kj−1 ) j=1

gcd(k0 , . . . ,ki−1 )



− 1 βj , −

gcd(k0 , . . . ,kj ) gcd(k0 , . . . ,ki−1 )



kj .

It is left as Exercise 5.2.24 to show that the formulas in (2) and (3) are equivalent. Corollary 5.2.17. Let ϕ : R −→ R′ be an isomorphism, then E(R) = E(R′ ). Definition 5.2.18. Let E(R) = {k0 , . . . ,kg } be the characteristic exponents. Define n1 P (R) = {(m1 ,n1 ), . . . ,(mg ,ng )}, the set of Puiseux pairs as follows: let kk01 = m such 1 nν kν that gcd(m1 ,n1 ) = 1. k0 = m1 ·...·mν such that gcd(mν ,nν ) = 1, ν = 2, . . . ,g. To digest this definition a little bit, let us write the Puiseux expansion as X ak xk . y= k∈Q

5.2 Invariants

189

If the curve is not smooth, there is a smallest k 1 with ak1 6= 0 and k 1 not an integer. We n1 write k1 = m in lowest terms, and then (m1 ,n1 ) is the first Puiseux pair of the curve. To 1 find the second Puiseux pair, we look for the smallest k 2 > k 1 which cannot be written 2 with gcd(m2 ,n2 ) = 1 and call (m2 ,n2 ) the second in the form ma1 . We write k 2 = m1n·m 2 Puiseux pair. To find the third, we look for the smallest k 3 > k 2 which cannot be written n3 with gcd(m3 ,n3 ) = 1. Then (m3 ,n3 ) the third in the form m1a·m2 . Write k 3 = m1 ·m 2 ·m3 Puiseux pair, etc. Remark 5.2.19. k0 = m1 · . . . · mg and ki = k0 k i , i = 1, . . . ,g. Therefore, P (R) determines also E(R). To prove Theorem 5.2.16 we need the following Lemma: Lemma 5.2.20. Let R = C {tn ,y(t)} and f = of unity, then ord Proof. From f =

∂f ∂y n Q

i=1

n Q

i=1


y − y(εi x1/n ) , ε a primitive n–th root

g

P gcd(k0 , . . . ,kj−1 )−gcd(k0 , . . . ,kj ) kj +max{n−m,0}. t ,y(t) = n

j=1


y − y(εi x1/n ) it follows that

n Y X
∂f = y − y(εj x1/n ) , ∂y i=1 j6=i

and, hence, Y

n−1 ∂f n y(t) − y(εi t) . t ,y(t) = ∂y i=1

It thus suffices to calculate the vanishing order for the different y(t) − y(εi t), their sum n being the vanishing order of ∂f ∂y (t ,y(t)). We shall do the calculation for n ≤ m and leave the other cases to the reader. In our situation k0 = n, and we may expand y(t) − y(εi t) as a power series: X y(t) − y(εi t) = aν (1 − εiν )tν . ν≥m

n

k0 0 | i, gcd(kk0 ,...,k ∤i Setting Nj = i; 1 ≤ i ≤ n − 1, gcd(k0 ,...,k j−1 ) j)

o

and noting that Sg gcd(k0 , . . . ,kg ) = 1, we consider the following partition {1, . . . ,n − 1} = j=1 Nj . We
claim that ord y(t) − y(εi t) = kj for any i ∈ Nj . If εiν 6= 1, then k0 = n ∤ iν, and hence gcd(k0 , . . . ,kj−1 ) ∤ ν, by assumption on i. Therefore, if in addition aν 6= 0 then ν ≥ kj , by definition of kj , and thus the order of y(t) − y(εi t) is at least kj . Conversely, we know that akj 6= 0, and thus it suffices to show gcd(k ,...,k ) that εikj 6= 1, i. e. k0 = nmidikj . Writing q = gcd(k0 0 ,...,kj−1 we get j) (∗)

 gcd q,

kj gcd(k0 , . . . ,kj )



= gcd



k0 kj ,..., gcd(k0 , . . . ,kj ) gcd(k0 , . . . ,kj )



= 1.

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5 Plane Curve Singularities

By assumption on i we know that

qk0 gcd(k0 ,...,kj−1 )

=

k0 gcd(k0 ,...,kj )

∤ i. And hence,

i gcd(k0 ,...,kj−1 ) q ∤ , which is an integer, again by assumption on i. From equation k0 i gcd(k0 ,...,kj−1 )kj deduce that q ∤ k0 gcd(k0 ,...,kj ) , and finally, k0 ∤ ikj . This proves the claim. n o Writing Mj = i; 1 ≤ i ≤ n − 1, gcd(kk0 0,...,kj ) |i we easily see that #Mj =

∗ we

gcd(k0 , . . . ,kj )−1 and Nj = Mj−1 \Mj . Hence, #Nj = gcd(k0 , . . . ,kj−1 )−gcd(k0 , . . . ,kj ), which finishes the proof. P ai ti ,am 6= 0. We define β0 , . . . ,βg Proof of Theorem 5.2.16. Let R = C {tn ,y(t)},y(t) = i≥m

by the formulas in 5.2.16(3). So we have to show that β0 , . . . ,βg generate Γ(R). This will be done by induction on g. The case g = 0 is easy, as we then have a smooth curve. g=1 In this case it follows from the definition of the ki that gcd(k0 ,k1 ) = 1. We have to prove that β0 = k0 and β1 = k1 generate Γ(R). We first claim that we can arrange that R = C {tk0 + . . . , tk1 + . . .}, gcd(k0 ,k1 ) = 1 and k0 < k1 . For this we consider the different cases.

(1) If n ≤ m, thenP n = k0 and we may subtract suitable powers of tk0 from y(t) and obtain y(t) = ai ti with ak1 6= 0 and R = C {tk0 ,tk1 + . . .}. i≥k1

(2) If n > m, m ∤ n then m = k0 and n = k1 and R = C {tk0 + . . . , tk1 }.

(3) If n > m and m | n then m = k0 and there exists a polynomial q ∈ C [t] such that
tn − q y(t) = tn−m+k1 + terms of higher order, k 1 = min{i|ai 6= 0, gcd(i,k0 ) = 1}. We leave it as Exercise 5.2.25 to find q. Then, by definition, we have n−m+k1 = k1 and R = C {tk1 + . . . , tk0 + . . .}. This proves the claim. By Corollary 5.1.6 there is a Weierstraß polynomial f ∈ C {x}[y], degy (f ) = k0 such that f (tk0 + . . . , tk1 + . . .) = 0. For an arbitrary h ∈ C {x,y}, we can, by the Weierstraß division Theorem, write h = a · f + r for a suitable polynomial r ∈ C {x}[y], whose degree in y is less than k0 . Hence, to calculate the vanishing order of h(tk0 + . . . ,tk1 + . . . ), we may suppose that h is a polynomial in y of degree less than k0 = n. So write kX 0 −1 ri (x)y i . h= i=0

k0 +...

k1 +... i

The vanishing order of ri (t )(t ) obviously is in k0 N + k1 N. To see that the order of h(th0 + . . . , + th1 + . . .) is in k0 N + k1 N it is enough to show that all orders of ri (tk0 +. . . )(tk1 +. . . )i are different, so that no canceling can occur in the sum. Therefore, suppose

ord ri (tk0 + . . .)(tk1 + . . .)i = ord rj (tk0 + . . .)(tk1 + . . .)j Then it follows that ik1 ≡ jk1 mod k0 . Since gcd(k0 ,k1 ) = 1, it follows k0 | i − j. As both i and j are smaller than k0 , this can happen only if i = j. Induction Step. Now assume g > 1.

5.2 Invariants

191

P i Let x = tn ,y = ai t be thePparametrization of the curve and f an irreducible Weierstraß polynomial with f (tn , ai ti ) = 0. For simplicity, we use for short Kj for gcd(k0 , . . . ,kj ). We shall show by induction on d, that for all h ∈ C {x}[y] with degy (h) < n Kd we have d 
 X ord h(tn ,y(t) ∈ βν N. ν=0

This is sufficient because any h ∈ C {x,y} can be written as h = p · f + r, r ∈ C {x}[y] with degy (r) < n as before. d=1 We define a new curve by the parametrization X ai ti/K1 . x = tn/K1 , y = m≤i
Here k 2 = k2 − max{n − m,0}. This defines a curve with two characteristic exponents k0 k1 n K1 , K1 . Now let h ∈ C {x}[y], degy (h) < K1 . Applying the case g = 1 it follows that 
 β0 β1 ord h x(t), y(t) ∈ N+ N. K1 K1

Hence


 ord h tn , y(tK1 ) ∈ β0 N + β1 N,

as we replaced t by tK1 . By Taylor’s Theorem

h tn , y(t) = h tn , y(tK1 ) + R.

Therefore, the vanishing order of the first term is in β0 N + β1 N. We claim that R has higher vanishing order. Because h is a polynomial in y of degree less than Kn1 , say P i h= qi y . As in case g = 1 we can show that the vanishing orders of qi (tn )tK1 i are pairwise different. Hence, there exists an i such that the vanishing order of qi (tn )tK1 i is
K1 n equal to the vanishing order of h t , y(t ) . Now in R terms of type

occur, and

∂h ∂y

=

P


∂h n t , y(tK1 ) ak2 tk2 + . . . ∂y

iqi y i−1 . The difference in the vanishing order between qi (tn )tK1 (i−1) (tk2 + . . .)

and qi (tn )tK1 i is k 2 − K1 which is a positive number. Similar for the higher order derivatives of h. Therefore, the vanishing order of h(tn ,y(t)) is in β0 N + β1 N. d > 1 We shall first construct a function with vanishing order βd . To define this function, we define a new curve R by the following parametrization: X ai ti/Kd−1 . x = tn/Kd−1 ; y = m≤i
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5 Plane Curve Singularities

Here k d = kd −max{n−m,0}. We consider a Weierstraß polynomial f ∈ C {x}[y] of degree kd−1 k0 n Kd−1 such that f (x,y) = 0. The curve R has characteristic exponents Kd−1 , . . . , Kd−1 . It follows from 5.2.20 ord



 d−1  X  Kj−1 ∂f  n/Kd−1  Kj = t ,y kj + max{n − m,0}. − 2 2 ∂y Kd−1 Kd−1 j=1

Now consider

∂f n
t ,y(tKd−1 ) · (akd tkd + . . .) + R. f tn ,y(t) = f tn ,y(tKd−1 ) + ∂y

The first term is zero, the second term has vanishing order kd +

d−1 X j=1

Kj−1 Kj − Kd−1 Kd−1



kj + max{n − m,0} = kd +

d−1 X j=1

Kj−1 Kj − Kd−1 Kd−1



kj = βd .

The fact that R has higher vanishing order is proved as in the case d = 1. Hence, we have constructed an f with vanishing order βd . Let h ∈ C {x}[y], degy (h) < Knd . By successive application of the Weierstraß division Ps i n Theorem we can write h = i=0 hi (x,y)f with hi ∈ C {x}[y], degy (hi ) < Kd−1 and Kd−1 . s< Kd 
 d−1 P By induction we have ord hi tn ,y(t) ∈ βj N. To conclude the Theorem, again j=0

we have to show that no canceling occurs. Therefore, suppose  

i 
j 
= ord hj tn ,y(t) f tn ,y(t) . ord hi tn ,y(t) f tn ,y(t)

 

 Pd−1 Then (i − j)βd = ord hj tn ,y(t) − ord hi tn ,y(t) so that (i − j)βd ∈ j=0 βj N. By Exercise 5.2.24, gcd(β0 , . . . ,βd−1 ) = Kd−1 | (i − j)βd . From the definition of βd it Kd−1 then follows that Kd−1 | (i − j)kd , and, hence, | i − j. Kd Kd−1 But as i − j ≤ s < Kd , this implies i = j. This proves that {β0 , . . . ,βg } defined by (3) in the Theorem generate Γ(R). It remains to prove that {β0 , . . . ,βg } is minimal, which is left as Exercise 5.2.26.

Remark 5.2.21. Usually the k0 , . . . ,kg are introduced as invariants of the parametrization and not of R as follows: k0 = n, kν = min{i|ai 6= 0, gcd (i,k0 , . . . ,kν−1 ) < gcd (k0 , . . . ,kν−1 )}, for 1 ≤ ν ≤ g. P bν τ ν , y = τ m , is the parametrization obtained by τ m = y(τ ), and If, then, x = ν≥n

′ ′ k0′ = m, kν′ = min{i|bi 6= 0, gcd (i,k0′ , . . . ,kν−1 ) < gcd (k0′ , . . . ,kν−1 )}, 1 ≤ ν ≤ g ′ , then ′ the so–called inversion Theorem gives a relation between {k0 , . . . ,kg′ ′ } and {k0 , . . . ,kg } as follows:

5.3 Resolutions of Irreducible Plane Curve Singularities

193

′ If n < m < k1 then g ′ = g + 1, k1′ = n and kν+1 = kν + n − m. If n = m or k1′ = m then g ′ = g and kν′ = kν + n − m. Using our way of defining k0 , . . . ,kg the proof of Theorem 5.2.16 includes also a proof of the inversion Theorem.

Exercises 5.2.22. Let R be the local ring of a curve singularity, H its semi-group. (1) Let α ∈ H. Prove that c(R) − 1 − α ∈ / H.

(2) Prove that 2δ(R) ≤ c(R).

(3) Prove that 2δ(R) = c(R) if and only if H is symmetric, that is, α ∈ H ⇐⇒ c(R) − 1 − α ∈ / H.

5.2.23. Let ϕ : R −→ R′ be an isomorphism, then Γ(R) = Γ(R′ ). /Hint: use the fact that ϕ extends to an isomorphism ϕ : C {t} −→ C {t} of the normalizations of R and R′ , ϕ(t) = t· unit.) 5.2.24. (1) Show that (2) ⇐⇒ (3) in 5.2.16.

(2) Show that for d = 1, . . . ,g we have gcd(k0 , . . . ,kd ) = gcd(β0 , . . . ,βd ). P 5.2.25. Let x = tn ,y = i≥m ai ti , with am = 1. Suppose n > m and m | n. Then there exists a ` ´ polynomial q ∈ C [t] such that tn −q y(t) = tn−m+k1 + terms of higher order, k1 = min{i|ai 6= 0, gcd(i,k0 ) = 1}. P i ci t , and if ci 6= 0 for Hint: let n = r · m and assume am = 1, then tn − y(t)r = i>n `r´ m(r−1) i < n − m + k1 = (r − 1)m + k1 , the exponent of 1 t ak1 tk1 in y(t)r ), then m | i. If ci 6= 0 for i < n − m + k1 , subtract again a suitable power of y(t) and continue up to the order n − m + k1 . 5.2.26. Prove that G(R) = {β0 , . . . ,βg } for β0 , . . . ,βg defined by (3) of Theorem 5.2.16. Hint: we know that β0 , . . . ,βg generate Γ(R). Use (3) to prove that gcd(β0 , . . . ,βi ) < gcd(β0 , . . . ,βi−1 ) for i = 1, . . . ,g, and gcd(β0 , . . . ,βg ) = 1.

5.3

Resolutions of Irreducible Plane Curve Singularities

In this picture we blow up a point to a sphere to remove a singularity. This section is devoted to the study of embedded resolutions of plane curve singularities, which will be done by successive “blowing-ups” of the curve singularity. In order to define the notion of blowing–up we need some other definitions first.

194

5 Plane Curve Singularities

Definition 5.3.1. We define P1 = P1C to be the set of lines in C 2 which pass through the origin. Similarly one can define PnC for n ≥ 2. Alternatively, but very similarly, one could define P1 as P1 = (C 2 \ {0,0})/ ∼

Here (x,y) ∼ (x′ ,y ′ ) if and only if there exists a λ ∈ C \ {0} with (x′ ,y ′ ) = λ(x,y). The equivalence class of (x,y) 6= (0,0) is denoted by (x : y). An equivalence class is called a point of P1 . The (x : y) are called homogeneous coordinates of P1 . As any line through (0,0) has a nonempty intersection with the unit sphere in C 2 = R4 we also could have defined P1 as {(x,y) ∈ C 2 :| x |2 + | y |2 = 1}/ ∼= §3 / ∼ ,

where (x,y) ∼ (x′ ,y ′ ) if ∃ λ ∈ C with |λ| = 1 so that (x,y) = λ(x′ ,y ′ ). We give P1 the quotient topology. As a quotient of a compact topological space, P1 is compact. Moreover, it is left to the reader to show that P1 is Hausdorff. In fact, P1 is homeomorphic to §2 . This can be seen as follows. Consider a line with homogeneous coordinates (x : y) and suppose that y 6= 0. Then there is a representative with y = 1, and x can be an arbitrary element of C . So this subset of P1 is homeomorphic to C . There is only one point of P1 with y = 0, as all (x,0) are equivalent to (1,0). Thus, it can be seen that P1C is homeomorphic to the one point compactification of C = R2 , hence to §2 . Another way to see this, is by considering an open covering P1 = U0 ∪ U1 , where U0 = {(x : y) ∈ P1 : x 6= 0}, and U1 = {(x : y) ∈ P1 : y 6= 0}. Now U0 is homeomorphic to C via the map f0 : U0 −→ C : (x : y) 7→ xy , and similarly U1 is homeomorphic to C via the map f1 : U1 −→ C : (x : y) 7→ xy . Note that f0 ◦ f1−1 : C \ {0} −→ C \ {0} is the map which sends z to z1 , so is holomorphic. Therefore, P1 carries the structure of a complex manifold. A complex manifold in general is a topological Hausdorff space X which has an open covering X = ∪i∈I Ui . These Ui are all homeomorphic to an open subset Vi in C n via fi , that is, fi : Ui −→ Vi is a homeomorphism. Moreover, for all i,j ∈ I, the map fi ◦ fj−1 : fj (Ui ∩ Uj ) −→ fi (Ui ∩ Uj ) should be holomorphic. This makes sense, as these sets are subsets of C n . We shall study complex manifolds in Chapter 6. To study the blowing up here we need only to know the P1 . We now can define the blow-up of C 2 in the origin.

Definition 5.3.2. u : v) | xv = yu} ⊂ C 2 × P1C and π : X −→
Consider X = {(x,y, 2 2 C ,π (x,y, u : v) = (x,y). π : X −→ C is called the blowing–up of (0,0) in C 2 . This map has the following properties:
(1) π −1 (0,0) = {(0,0)} × P1C ,

(2) π|X\({(0,0)}×P1C ) : X \ ({(0,0)} × P1C ) −→ C 2 \ {(0,0)} is an isomorphism.

5.3 Resolutions of Irreducible Plane Curve Singularities

195

Remark 5.3.3. Let P1C = U1 ∪ U2 , U1 = {(1 : v) ∈ P1C }, U2 = {(u : 1) ∈ P1C }. Then X ∩ (C 2 × U1 ) = {(x,xv,1 : v)} ≃ C 2 2

X ∩ (C × U2 ) = {(yu,y,u : 1)} ≃ C

with coordinates (x,v),

2

with coordinates (y,u).

We shall now study the behavior of curves under blowing–up. Let, as before, π : X −→ C 2 be the blowing–up of (0,0) ∈ C 2 . Let C ⊂ C 2 be a curve through (0,0).
Definition 5.3.4. E = π −1 (0,0) = {(0,0)} × P1C is called the exceptional line or
exceptional divisor, C1 = π −1 C \ {(0,0)} , the closure of the preimage of the curve without the origin is called the strict transform of C. First case: C is a line. If the equation of C is y = ax then on C 2 × U1 = {(x,xv,1 : v) | x,v ∈ C } we have C1 ∩ (C 2 × U1 ) = {(x,ax, 1 : a) | x ∈ C }, E ∩ (C 2 × U1 ) = {(0,0, 1 : v) | v ∈ C }. v

E C1 a y

y C

x

x

On C 2 × U2 = {(yu,y,u : 1) | y,u ∈ C } we have  o (n y ,y,1 : a | y ∈ C 2 a C1 ∩ (C × U2 ) = ∅

if a 6= 0,

if a = 0,

2

E ∩ (C × U2 ) = {(0,0,u : 1) | u ∈ C }.

Altogether we see that C1 is a line intersecting E in (0,0,1 : a). If C is defined by x = 0, then we obtain similarly C1 ∩ (C 2 × U1 ) = ∅

C1 ∩ (C 2 × U2 ) = {(0,y,0 : 1) | y ∈ C },

and again C1 is a line intersecting E in (0,0,0 : 1). Corollary 5.3.5. Let C be the union of the two lines defined by x = 0, respectively y = 0, then C1 is the union of two lines which have no intersection point.

196

5 Plane Curve Singularities

(1 : a) π

E (1 : b)

Second case: C is an P irreducible curve. Because of Theorem 5.1.17 we may assume that f = (y p + αxq )c + aij xi y j defines C with q ≥ p, α 6= 0 and gcd(p,q) = 1. ip+jq>pqc

Let us assume first that q > p.

y 2 − x3 = 0 C

C1 ∩ (C 2 × U1 ) = {(x,xv,1 : v) | f1 (x,v) = 0} with f1 (x,v) = P aij xi+j−pc v j . αxq−p )c +

f (x,xv) xord(f )

= (v p +

ip+jq>pqc

E ∩ (C 2 × U1 ) = {(0,0,1 : v) | v ∈ C },

C1 ∩ E(∩C 2 × U1 ) = {(0,0,1 : 0)}. C (1)

x x − v2 = 0

E

v

f (uy,y) And C1 ∩ (C 2 × U2 ) = {(yu,y,u : 1) | f2 (u,y) = 0} with f2 (u,y) = ord(f ) = y P (1 + αy q−p uq )c + ip+jq>pqc aij ui y i+j−pc . E ∩ (C 2 × U2 ) = {(0,0,u : 1) | u ∈ C },

C1 ∩ E ∩ (C 2 × U2 ) = ∅.

5.3 Resolutions of Irreducible Plane Curve Singularities

197

y f2 = 1 − yu3 u

E

The case p = q, that is p = q = 1 can be reduced by the transformation y 7→ y − αx to the first case. Altogether we obtain that the strict transform C1 intersects E at one point. Outside this intersection point, C is smooth. If [ pq ] > 1, then the multiplicity m(C1 ) = ord(f1 ) = pc = ord(f ) = m(C). If [ pq ] = 1 then the multiplicity drops: m(C1 ) = ord(f1 ) = (q − p)c. Now we can iterate this procedure. ∼ We identify X ∩ (C 2 × U1 ) − → C 2 : (x,xv,1 : v) 7→ (x,v) and blow up (0,0) in C 2 again. We shall denote the ambient space after the i–th blowing–up by Xi (we always take the affine chart U1 or U2 , in which the curve meets the origin), the corresponding exceptional line by Ei and the strict transform of Ci−1 by Ci : π

k Xk −→ Xk−1 −→ . . . ∪ ∪ Ck Ck−1 mk = m(Ck ) mk−1 = m(Ck−1 ) ...

−→

π

1 −→ C2 X1 ∪ ∪ C1 C m1 = m(C1 ) m0 = m(C).

If we denote by mi = m(Ci ) the multiplicity of Ci at (0,0), then mi ≤ mi−1 , as we have already seen, and after a while it has to drop strictly. This implies that there is a k such that Ck is smooth, that is, we have resolved the singularity after k times blowing–up. If k

∪ Ev intersects Ck not transversally, we blow up again. Let us denote by π = π1 ◦ · · ·◦ πk

v=1

k

the composition of the blowing–ups and E = π −1 (0,0) = ∪ Ei the exceptional divisor. i=1

π

π

1 k C 2 is called standard resolution of Definition 5.3.6. Xk −→ Xk−1 −→ . . . −→ X1 −→ the singularity of the irreducible curve C at (0,0) if either Ci−1 ⊂ Xi still has a singular point and πi is the blowing–up in Xi−1 of this point or Ci−1 is smooth but the intersection with Ei−1 in Xi−1 is not transversal and πi is the blowing–up of the intersection point

k

of Ei−1 and Ci−1 in Xi−1 and if, furthermore, Ck is smooth and intersects E = ∪ Ei transversally. (m0 , . . . ,mk−1 ) is called the multiplicity sequence of the singularity of C.

i=1

We have proved that every irreducible plane curve singularity has a standard resolution. Furthermore, we leave it as an Exercise to prove that Ei ∩ Ej ∩ Ek = ∅ if i,j,k are pairwise different. Let us look for an example. Example 5.3.7. Let C ⊂ C 2 be defined by f = y 4 − 2x3 y 2 − 4x5 y + x6 − x7 = (y 2 − x3 )2 − 4x5 y − x7 .

198

5 Plane Curve Singularities

m(C) = 4

The first blowing–up defined by y = xv gives C1 defined by f1 = (v 2 − x)2 − 4x2 v − x = 0. E1 is defined by x = 0. 3

m(C1 ) = 2 E1 The second blowing–up defined by x = vu gives C2 defined by f2 = (v − u)2 − 4vu2 − vu = 0. E2 is defined by v = 0. 3

E1

E2 m(C2 )) = 2

Now we change coordinates v v + u and obtain fe2 = f2 (v + u,u) = v 2 − 4u3 − 4 3 4u v − u − u v. The third blowing–up v = xu gives C3 defined by f3 = x2 − 4u − 4ux − u2 − u2 x. E3 is defined by u = 0. 2

E3

E2 m(C3 ) = 1 E1

5.3 Resolutions of Irreducible Plane Curve Singularities

199

The fourth blowing–up u = xy gives C4 defined by f4 = x − 4y − 4xy − xy 2 − x2 y 2 . E4 is defined by x = 0.

E4

E3 E2

m(C4 ) = 1 E1 We change again coordinates x x + 4y and obtain fe4 (x,y) = f4 (x + 4y,y) = 2 2 3 2 2 x − 4xy − 16y − xy − 4y − (x + 4y) y . The fifth blowing up x = yu gives C5 defined by f5 = u − 4uy − 4y − uy 2 − 4y 2 − (u + 4)2 y 3 . E5 is defined by y = 0.

E5

E4

E3 m(C5) = 1 E2

E1

Now C5 is smooth and intersects the exceptional locus E = ∪Ei transversally. The multiplicity sequence is (4,2,2,1,1). Definition 5.3.8. Let π : Xk −→ C 2 be a standard resolution of the singular point k

0 ∈ C ⊂ C 2 . Let π −1 (0) = E = ∪ Ei be the exceptional divisor and Ck the strict i=1

transform of C. To this resolution we can associate a weighted graph, the the resolution graph. The vertices of the graph are the exceptional lines Ei and the curve Ck . We connect two vertices Ei and Ej if Ei ∩ Ej 6= ∅ and, similarly, we connect a vertex Ei to Ck if Ei ∩ Ck 6= ∅, that is, we connect Ck to Ek . The weight of Ei in the graph is i. We shall draw “•” for an Ei and “∗” for Ck . For our example 5.3.7 we obtain:

200

5 Plane Curve Singularities

∗ 5

1

4

3

2

Now we are ready to prove the following Theorem: Theorem 5.3.9. For a given irreducible curve singularity the following data determine each other: (1) the Puiseux pairs P (R); (2) the characteristic exponents E(R); (3) the minimal set of generators of the semi-group G(R); (4) the multiplicity sequence (m0 , . . . ,mk−1 ); (5) the resolution graph. We proved this already for (1), (2), and (3). To prove that (2), (4) and (5) determine each other we need to P study the resolution more carefully. Let x = tk0 , y = ai ti define a singularity with characteristic exponents k0 , . . . ,kg . i≥k1

Let

X1 = k1 ,

r1,1 = k0 . We perform the Euclidean algorithm and obtain: X1 = µ1,1 r1,1 + r1,2 ,

r1,1 = µ1,2 r1,2 + r1,3 , .. . r1,w(1)−1 = µ1,w(1) r1,w(1) , For i = 2, . . . ,g, let Xi = ki − ki−1 and define inductively ri,1 = ri−1,w(i−1) . We perform the Euclidian algorithm and obtain: Xi = µi,1 ri,1 + ri,2 ,

ri,1 = µi,2 ri,2 + ri,3 , .. . ri,w(i)−1 = µi,w(i) ri,w(i) ,

5.3 Resolutions of Irreducible Plane Curve Singularities

201

with 0 ≤ ri,j+1 < ri,j .

Notice that by the definition of the ki , always w(i) ≥ 2. Using this data we construct a weighted graph as follows: The building blocks are the graphs Sij : • π(i,j) + 1

• π(i,j) + 2

... • π(i,j) + 3

•

• π(i,j) + µij − 1

• π(i,j) + µij

with π(1,1) = 0,


π(i,1) = π i − 1,w(i − 1) + µi−1,w(i−1) , for i = 2, . . . ,g, π(i,j) = π(i,j − 1) + µi,j−1 , for 2 ≤ j ≤ w(i) and i = 1, . . . ,g.

The point with the smallest weight (respectively largest weight) is called the start point (respectively end point) of the graph Sij . Now we connect the end point of Sik with the start point of Sik+2 if k ≤ w(i) − 2, w(i) w(i)−1 for all i. Then we connect the end with the end point of Si the end point of Si w(i) 1 with the start point of Si+1 if µi+1,1 6= 0, otherwise with the start point point of Si 3 2 of Si+1 if w(i + 1) ≥ 3 or with the end point of Si+1 if w(i + 1) = 2. Because w(i) ≥ 2, w(i) is, therefore, connected with three other points and usually called the end point of Si w(g) contact point of the graph. The end point of Sg we connect with the star. The result is a graph of the following type:

∗

Contact points

... ... .. . ... ...

...

...

We have g − 1 contact points and draw the graph in such a way that each contact point creates a new level. On the other hand it is obvious how to obtain the characteristic exponents from the resolution graph: g − 1 is the number of contact points, points which are connected to three other points. One can identify the Sij as the maximal connected subsets of the graph such that the weights are strictly increasing (respectively decreasing). Starting w(g) with multiplicity 1 which occurs µg,w(g) –times in Sg we obtain rg,w(g)−1 = µg,w(g) · 1. This way, going from bottom to top in the Euclidean algorithm we can reconstruct the ki .

202

5 Plane Curve Singularities Let us now look at an example.

Example 5.3.10.

∗

10 11 12 9

1

2

6

13 14 16 15

8

7

5

4

3

From the diagram we can extract the number g of levels and the number ω(i) of blocks in each level i, as well the blocks Sij themselves and, hence, the µi,j = #Sij . The remaining data in the following table have to be calculated: g=3 ω(1) = 3 S11 = {1,2} µ1,1 = 2 r1,1 = 28 ω(2) = 2 S21 = ∅ µ2,1 = 0 r2,1 = 4 ω(3) = 2 S31 = {10,11,12,13,14} µ3,1 = 5 r3,1 = 2

S12 = {3,4,5} S13 = {6,7} µ1,2 = 3 µ1,3 = 2 r1,2 = 8 r1,3 = 4 χ1 = 64 S22 = {8,9} µ2,2 = 2 r2,2 = 2 χ2 = 2 S32 = {15,16} µ3,2 = 2 r3,2 = 1 χ3 = 11

In order to do the calculations we proceed as follows: the initialization step in level three is done by setting r3,2 = rg,ω(g) = 1. We then go on with the following formulas: k3 − k2

=

r3,1 χ3

= µ3,2 · r3,2 = 2 · 1 = 2, = µ3,1 · r3,1 + r3,2 = 5 · 2 + 1 = 11.

This finishes level three, and we may continue with level two:

k2 − k1

r2,2 r2,1 = χ2

= r3,1 = 2, = µ2,2 · r2,2 = 2 · 2 = 4, = µ2,1 · r2,1 + r2,2 = 0 · 2 + 2 = 2.

Finally, in level one we obtain:

k0 k1

r1,3 r1,2 = r1,1 = χ1

= = = =

r2,1 = 4, µ1,3 · r1,3 = 2 · 4 = 8, µ1,2 · r1,2 + r1,3 = 3 · 8 + 4 = 28, µ1,1 · r1,1 + r1,2 = 2 · 28 + 8 = 64.

5.3 Resolutions of Irreducible Plane Curve Singularities

203

We, therefore, find as characteristic exponents: k0 = 28, k1 = 64, k2 = 66, k3 = 77. Now Theorem 5.3.9 is a consequence of the following Theorem: Theorem 5.3.11. (1) The graph described above is the resolution graph of the singularity. (2) The multiplicity sequence contains exactly µi,j times the multiplicity ri,j , i = 1, . . . ,q,j = 1, . . . ,w(i). In our Example 5.3.7 we have (4,2,2,1,1) as the multiplicity sequence and

∗

S22

S11

S12

as the resolution graph. C was defined by f = y 4 − 2x3 y 2 − 4x5 y + x6 − x7 . As we have already seen in Exercise 5.1.25 and Example 5.2.15, C has the parametrization x = t4 , y = t6 + t7 with the characteristic exponents k0 = 4, k1 = 6, k2 = 7. We want now to calculate the ri,j and the µi,j using their definition. The initial values in the first level are given by χ1 = k1 = 6 and r1,1 = k0 = 4. Proceeding along the lines of the algorithm, i. e. doing division with remainder, we find: χ1 r1,1

= 6 = 1 · 4 + 2 = µ1,1 · r1,1 + r1,2 = 4 = 2 · 2 = µ1,2 · r1,2

⇒ µ1,1 = 1, r1,2 = 2, ⇒ µ1,2 = 2, ω(1) = 2.

The second level thus starts with χ2 = k2 − k1 = 1 and r2,1 = r1,ω(1) = r1,2 = 2. Doing division with remainder we then get: χ2 r2,1

= 1 = 0 · 2 + 1 = µ2,1 · r2,1 + r2,2 = 2 = 2 · 1 = µ2,2 · r2,2

⇒ µ2,1 = 0, r2,2 = 1, ⇒ µ2,2 = 2, ω(2) = 2.

And this finishes the algorithm. One easily sees that starting from the diagram we would get the same values for the µi,j and for the ri,j and, moreover, the reader immediately verifies the connection between the (ri,j ,µi,j ) and the multiplicity sequence.

204

5 Plane Curve Singularities

Proof of the Theorem. We prove the Theorem using induction on the length k of the multiplicity sequence. If the length k is 1, that is, the curve is smooth, defined by x = t, y = t and k0 = k1 = 1. Consequently X1 = 1, r1,1 = 1 µ1,1 = 1. Thus, from now on we may assume k > 1. P ai ti , ak1 = 1 and k0 , . . . ,kg be the characteristic exponents and Let x = tk0 , y = i≥k1

(m0 , . . . ,mk−1 ) be the multiplicity sequence.

gcd(k0 ,k1 )  k0 k1 gcd(k ,k ) gcd(k ,k ) 0 1 0 1 + −y The singularity has an equation f of the following type: f (x,y) = x P bij xi y j .

k1 k0 ik1 +jk0 > gcd(k 0 ,k1 )

blowing up given by y = vx leads to the parametrization x = tk0 , y = P The i−k0 . ai t

i≥k1

There are three possible cases:

(1) If k0 ≤ k1 − k0 , then k0 < k1 − k0 and we obtain as characteristic exponents k0 ,k1 − k0 , . . . ,kg − k0 . (2) If k0 > k1 − k0 and k1 − k0 ∤ k0 , then we have as characteristic exponents k1 − k0 ,k0 ,k2 + k0 − k1 , . . . ,kg + k0 − k1 . (3) If k0 > k1 − k0 and k1 − k0 | k0 , then k1 − k0 = gcd(k0 ,k1 ) and we obtain as characteristic exponents k1 − k0 , k2 + k0 − k1 , . . . ,kg + k0 − k1 . Now recall the following definitions: X1 = k1 , r1,1 = k0 , X1

r1,1

= µ1,1 r1,1 + r1,2 , = µ1,2 r1,2 + r1,3 , .. .

r1,w(1)−1

=

µ1,w(1) r1,w(1) ,

X2

=

k2 − k1 ,

r2,1

=

X2

r1,w(1) ,

= µ2,1 r2,1 + r2,2 , .. . If we denote by X i ,r i,j ,µi,j ,k i the corresponding data after blowing up. In case (1) we have: X1 = X1 − k0 = (µ1,1 − 1)r1,1 + r1,2 , r1,1 = µ1,2 r1,2 + r1,3 , .. . This implies:

5.3 Resolutions of Irreducible Plane Curve Singularities

205

µ1,1 = µ1,1 − 1, µi,j = µi,j if (i,j) 6= (1,1), r i,j = ri,j , X i = Xi i > 1. In case (2) we have: k1 = X1 = k0 + r1,2 , that is, we start with X 1 = r1,1 = µ1,2 r1,2 + r1,3 , .. . This implies: X 1 = r1,1 , µ1,j = µ1,j+1 , r 1,i = r1,j+1 , X i = Xi ,i > 1, r i,j = ri,j ,i > 1, µi,j = µi,j ,i > 1. In case (3) we have: X1 = k1 , µ1,1 = 1, r1,1 = r1,w(1)−1 = k0 , r1,2 = r1,w(1) = k1 − k0 , r1,w(1)−1 = k0 = µ1,w(1) (k1 − k0 ), X2 = k2 − k1 , r2,1 = k1 − k0 , X2 = µ2,1 r2,1 + r2,2 , .. . We start with: X 1 = X2 + k0 = X2 + µ1,w(1) (k1 − k0 ), r 1,1 = k1 − k0 . This implies: X 1 = µ1,1 r 1,1 + r1,2 , µ1,1 = µ2,1 + µ1,w(1) , and we have: X i = Xi+1 , i ≥ 2, r i,j = ri+1,j , (i,j) 6= (1,1), µi,j = µi+1,j , (i,j) 6= (1,1). Now we use the induction hypothesis. In the first case, after blowing up the multiplicity, r1,1 = k0 occurs µ1,1 − 1 times and the multiplicities ri,j occur µi,j times if (i,j) 6= (1,1). This proves the Theorem in this case. In the second case, the multiplicity changes after blowing up, that is, m0 occurs once. Then X1 = k1 = 1 · k0 + (k1 − k0 ),k0 = X 1 , and k1 − k0 = r 1,1 determine the rest, as before. In the third case, the multiplicity also changes after blowing up, that is, m0 occurs once. m1 = r1,2 = r1,1 occurs (µ1,w(1) + µ2,1 ) times. This is covered by X 1 = X2 + µ1,w(1) (k1 − k0 ). The rest remains unchanged. This proves the second part of the Theorem.

206

5 Plane Curve Singularities

Now, using induction hypothesis, assume that the singularity after one blowing up has the resolution graph as described before and denote the corresponding building blocks j by S i .

∗

2

S11

S1

We have only to study the behavior at the first building block:

S12

S11 If we assume, as before, f (x,y) = (xq − y p )c +

P

bij xi y j to be the equation

ip+jq>pq

for our singularity with gcd(p,q) = 1 and q > 1, then there are two possibilities for the intersection of the exceptional divisor E1 with the curve after one blowing up:

E1 E1 : x = 0,

(1) q−p > p gives rise to a transversal intersection: (2) q−p < p means that the intersection has maximal contact:

E1 : x = 0. The second case occurs if and only if the multiplicity changes.

E1

In the first case the next blowing–up gives: E2

j

or

E1

1

This implies S i = Sij (with the weights shifted by one) for (i,j) 6= (1,1) and if S is 1

2

3

ℓ−2 ℓ−1

E2

.

5.3 Resolutions of Irreducible Plane Curve Singularities

207

then S11 1

2

ℓ−1

3

ℓ

In the second case we have two possibilities, the number of characteristic exponents remains constant (if q −p > 1) or drops (if q −p = 1) after the blowing–up (corresponding to case (2) and (3) in the proof of the second part of the Theorem). The next blowing–up gives:

E1 : x = 0

E2 : y = 0 The sequence of blowing up continues like

E1

E1

E3

Eℓ

E2

E2 Eℓ−1

up to the next multiplicity change in case q − p > 1, or after p − 1 steps in case q − p = 1. In the latter case we have after p blowing–ups the equation fp = (x − y)c + . . ., so that we have to change coordinates. The next blowing–up in the case q − p > 1 gives:

Eℓ+1 Eℓ

E1

Eℓ−1

E2

208

5 Plane Curve Singularities

that is, in the graph the point corresponding to E1 is connected with the end of 1 j−1 S 1 = S12 (weights shifted by 1) and S11 = {•}. Moreover, S1j = S 1 (weights shifted by j 1) for j > 1 and Sij = S i (weights shifted by 1) for i > 1.

E2

E3

E4

Eℓ+1

E1

1

S1 In the case q − p = 1 we obtain:

Eℓ+1 Eℓ

E1

Eℓ−1

E2

that is, in the graph the point corresponding to E1 is connected with the ℓ-th point E1 and the points corresponding to E2 , . . . ,Eℓ+1 give the first building block of in the resolution graph: 1 S1.

E2

E3

Eℓ

Eℓ+1 E1

We obtain S11 : the point defined by E1 . 1 S12 the first ℓ–points of S 1 (weights shifted by 1), connected with S11 at the ℓ–th point. 1 S21 the rest of S 1 (weights shifted by 1), which is not in S12 . j S2j = S 1 (weights shifted by 1) for j > 1. j j Si = Si−1 for i ≥ 3. This proves the Theorem.

5.4

Reducible Plane Curve Singularities

As in the irreducible case, we can define the standard resolution of a reducible plane curve singularity, as follows:

5.4 Reducible Plane Curve Singularities

209

Definition 5.4.1. Let C = ∪ri=1 Ci ⊂ U be a small representative of a reducible plane π1 π2 πi U X1 −→ . . . −→ curve singularity, with branches C1 , . . . ,Cr , r ≥ 2. Assume that Xi −→ is already defined. Denote by E (i) = (π1 ◦ · · · ◦ πi )−1 (0) the exceptional divisor and by C (i) = (π1 ◦ · · · ◦ πi )−1 (C \ {0}) the strict transform of C. πi+1 Let Xi+1 −−−→ Xi be the blowing-up of Xi in all points of C (i) ∩ E (i) which are still singular on C (i) or nontransversal intersection of C (i) with E (i) , that is, points with intersection multiplicity of C (i) and E (i) greater than one, or where two exceptional divisors and C (i) meet. πk π1 X1 −→ U is called a standard resolution of (C, 0) if all branches Then Xk −→ · · · −→ (k) of C are smooth, do not intersect each-other, do intersect just one component of E (k) , and do intersect this component transversally. Theorem 5.4.2. Every germ of a plane curve singularity has a standard resolution. πi+1

Proof. We have to prove that the process of constructing Xi+1 ←−−− Xi as described above stops after a while, that is, there exists an n such that C (n) has no singularities on C (n) ∩ E (n) and intersects E (n) transversally. From the irreducible case, we know that we can resolve the singularities of all branches such that after a while we have at most nontransversal intersections of C (n) and E (n) . Now it might still happen that two (smooth) branches of C (n) intersect each other. But we studied the intersections of smooth curves under blowing-up, and we know that the intersection multiplicity drops with each blowing-up, see ??. This proves the theorem. Definition 5.4.3. Consider the standard resolution of a reducible plane curve singularity (j) (0) (C, 0) = ∪(Ci , 0). Write Ci := Ci . Furthermore, for some i and k assume that Ci ∩ (j+1) (j) (j+1) ∩ Ck = ∅. Then γ(Ci ,Ck ) := j + 1 is called the contact number Ck 6= ∅, but Ci of the branches Ci and Ck . Definition 5.4.4. With the notations of Definition 5.4.1 we consider the following weighted graph, the resolution graph of C: (1) To each component of E (k) a point • is associated. (2) To each component of C (k) a point ∗ is associated. (3) Two points are connected by an edge if the corresponding components intersect. (4) The points of type (1) are weighted. Let E be a component of E (k) . We give to the corresponding point the weight i, if E is created in the i–th level of the blowing ups, that is, i is minimal such that πi+1 ◦ · · · ◦ πn (E) is not a point. Theorem 5.4.5. Let (C, 0) = ∪ri=1 (Ci , 0) be a plane curve singularity. The following data determine each other: (1) The resolution graph of C. (2) The resolution graph of the branches (Ci , 0), together with the contact numbers γ(Ci ,Cj ) for i 6= j.

210

5 Plane Curve Singularities

Proof. Step 1. First suppose we know the resolution graph of C. Take a branch (Ci , 0), and delete all stars from the resolution graph which belong to the branches (Cj , 0) for j 6= i. Call a point of the remaining graph contractible, if it satisfies one of the following conditions: (1) It is not connected to the star, and connected only to points of lower weight. (2) It is connected to a star, and to just one other point of lower weight. This exactly means that the corresponding curve was created in a blowing-up, which would not have been necessary for getting the standard resolution of the branch (Ci , 0). From the resolution process, it follows that such a point is connected to at most two points in the graph. If it is connected to just one other point, then delete the point of highest weight, and the edge connected to it. If it is connected to two points A and B say, then delete the point of highest weight, the connecting edges, and connect A and B by an edge. This corresponds to “blowing-down” the curve corresponding to this highest weight. Now iterate this process, until it is no longer possible. We then arrive at the resolution graph of (Ci , 0). For example, if we have the following resolution graph * * *

2

5

3

1

4

3

2

we can get the resolution graph of the curve corresponding to the star on the left by the following process * *

2

delete *

5

3

1

3

2

2

*

2

3

*

4

3

4

1

3

*

1

3

2

2

3

2 *

1

2

2

3

1

To get the contact number γ(Ci ,Cj ), we remove all the stars corresponding to the branches Ck with k 6= i,j, and then successively remove points with property (1) and (2). If this is no longer possible, then we are left with the resolution graph of the union of (Ci , 0) and (Cj , 0). Consider the following numbers. (1) The maximal natural number k such that the numbers 1, . . . ,k −1,k occur as weight just once in the resolution graph of the union of (Ci , 0) and (Cj , 0)}. (2) The weights of the vertices connected to the stars in the resolution graph of the union of (Ci , 0) and (Cj , 0).

5.4 Reducible Plane Curve Singularities

211

The contact number of (Ci , 0) and (Cj , 0) is the minimum of these numbers. So for example in the resolution graph above, the contact number between the curves belonging to the right two stars is equal to min{5,5,4} = 4, between the curve belonging to the star on the left and the other two is one. Step 2. Conversely, suppose we know the resolution graphs of the (Ci , 0) and the contact numbers. We give an algorithm for finding the resolution graph of (C, 0). The proof goes by induction on r. The case r = 1 is trivial. So suppose that the resolution graph Γ of ∪r−1 i=1 (Ci , 0) has been constructed out off the resolution graphs of the (Ci , 0) and the contact numbers. For i = 1, . . . ,r − 1, define ℓ(i) = γ(Cr ,Ci ), and let l(r) be the maximum of γ(Ci ,Cr ) for i 6= r. Let w(i) be the weight of the vertex in Γ belonging to the exceptional curve which intersects the strict transform of Ci , for i = 1, . . . r − 1. Let w(r) be the vertex of highest weight in the resolution graph Γr of (Cr , 0). It is clear that the point in the resolution graph connected to the star, which corresponds to (Ci , 0) has weight max{ℓ(i),w(i)}. So if ℓ(i) > w(i), we need, starting from the standard resolution, ℓ(i) − w(i) extra blowing-ups. Therefore, if ℓ(i) > w(i), we modify the graph Γ as suggested by the following picture. ∗ ℓ(i)

.. .

∗ ...

... w(i) Γ

...

... w(i) Γ′

If ℓ(i) ≤ w(i) we leave Γ unchanged at the star corresponding to (Ci 0). We get a modified graph Γ′ . Similarly, we get a modified graph Γ′r . Let k = max{ℓ(j) : j < r} be the maximum of the contact numbers of (Cr , 0) with (Cj , 0) for j < r. Without loss of generality, we may suppose that this maximum is achieved for j = 1. Let π1 ◦ · · · ◦ πk : Xk −→ U be the map which occurs in the standard resolution process of ∪ri=1 (Ci , 0) after (k) (k) k steps. So by definition of contact number, the strict transforms C1 and Cr intersect (k) a common component E1 of E (k) , but do not intersect each other. In fact, by choice of (k) (k) k, Cr does not intersect any of the Cj . Let p be the point of Γ′ which corresponds to (k)

E1 . If k > 1, it follows from the resolution process that there is a unique component (k) E ′ of E (k) which intersects E1 and was created under πk−1 . The point of Γ′ which ′ corresponds to E we call q. It has weight k − 1. There are two possible cases. (k)

(1) k > 1, and the curve Cr intersects E ′ . Note that the points p and q in Γ′ are connected by an edge. The following is a possible picture of a part of the resolution graph after k blowing-ups.

212

5 Plane Curve Singularities (k)

C1

E′

(k)

E1

(k)

Cr

(k)

(2) k = 1, or the curve Cr

does not intersect E ′ . (k)

C1

E′

(k)

E1

(k)

Cr

In the second case, consider the subgraph Γ′′r of Γ′r which one gets by deleting all points (k−1) with all weights shifted of weight less than k. So this is the resolution graph of Cr (k) (k) by k − 1. As we already noted that none of the Cj intersects Cr , it follows that the resolution graph of (C, 0) is obtained by identifying the unique point of Γ′′r of weight k with p, which also has weight k. In the first case, consider the subgraph Γ′′r of Γ′r which one gets by deleting all points of weight less than k − 1. The graph Γ′′r looks like

k−1

k

and we get the resolution graph (C, 0) by doing surgery: In Γ′ one replaces the subgraph

p k

q k−1

by Γ′′r . So we see that we can construct the resolution graph (C, 0) from the resolution graphs of (Ci 0) and the contact numbers. Example 5.4.6. We consider the reducible plane curve singularity given by f = 0, where f = f1 f2 f3 with f1 = (y 2 − x3 )2 − 4x5 y − x7 , f2 = y 2 − x3 , f3 = y 3 − x2

5.4 Reducible Plane Curve Singularities

213

The second and third branch are cusp singularities, and we know how the resolution for those looks like, see the fourth part of Example ??. The resolution of the first branch has been studied in Example 5.3.7. So we have to put the following three resolution graphs into one graph. *

5

1

*

4

3

2

2

3

*

1

2

3

1

For this we have to determine the contact numbers. (1) (1) The interesting parts of the strict transforms C1 and C2 of the blowing-up of the branches (C1 , 0) and (C2 , 0) are in the first chart: (1) C1 : defined by (y 2 − x)2 − 4x2 y − x3 = 0, (1) C2 : defined by y 2 − x = 0, E1 : defined by x = 0. In the second chart we obtain: (1) C3 : defined by y − x2 , E1 : defined by y = 0. So we have the following picture (1)

C3

(1)

C2

(1)

C1

E1

and deduce that the contact number of (C1 , 0) and (C3 , 0) is equal to one. Similarly, the contact number of (C2 , 0) and (C3 , 0) is equal to one. We still have to determine the contact number of (C1 , 0) and (C2 , 0). So we look at the resolution of the union of (C1 , 0) and (C2 , 0). One can calculate that the sequence of blowing-ups continues as suggested in the following picture. (3)

C2

(3)

C1

(2)

C2

(2)

E3

C1

E2

E1 E1 E2

214

5 Plane Curve Singularities (4)

(4)

We see that the on the fourth blowing-up the branches C1 and C2 are separated. Thus the contact number is four. It follows that the resolution graph of the curve (C, 0) is equal to * * *

2

3

5

1

4

3

2

by using the algorithm described in the proof. Of course, we could also have gotten this resolution graph by simply doing the whole resolution process. Finally, summarizing all results, we obtain the following theorem: Theorem 5.4.7. Let (C, 0) be a germ of a plane curve singularity. The following data determine each other: (1) the multiplicity sequences of the branches of C and the contact numbers between the branches; (2) the resolution graphs of the branches of C and the contact numbers between the branches; (3) the resolution graph of C. Next we want to characterize the contact numbers of the branches by their intersection multiplicity. Theorem 5.4.8. Let (C, 0) respectively (D, 0) be irreducible plane curve singularities with multiplicity sequences (m0 , . . . ,mg ) respectively (n0 , . . . ,nh ). For k ≥ g + 1 define mk = 1, and similarly, for k ≥ h + 1 define nk = 1. Then γ(C,D)

(C, 0) · (D, 0) =

X i=0

mi · n i .

Proof. Of course we use Lemma 5.1.5 and induction on γ(C,D). After a general linear coordinate change we may assume that (D, 0) is given by a Weierstraß polynomial f = y n + a1 y n−1 + . . . + an of degree n = n0 in y, and that (C, 0) has as parametrization x(t) = tm , y(t) = tq + . . . , where m = m0 . By Lemma 5.1.5 we have (C, 0)·(D, 0) = ordt (f ). We look at what happens after blowingup. Then x(t) = tm , v(t) = tq−m + . . . . is a parametrization of the strict transform C (1) . An equation for the strict transform D(1) of (D, 0) is given by f (x,v) = 0, where f (x,v) is defined by f (x,v) = x−n f (x,xv).

5.4 Reducible Plane Curve Singularities

215

Therefore,  

 ordt f x(t),y(t) = ordt f x(t),x(t)v(t)
= ordt x(t)n · ordt (f (x(t),v(t))) =m0 · n0 + C (1) · D(1)

by induction. Example 5.4.9. Consider the curve (C, 0) given by (y 2 − x3 )2 − 4x5 y − x7 = 0, and (D, 0) given by g = y 2 − x3 = 0. The multiplicity sequence of (C, 0) is (4,2,2,1,1), and the multiplicity sequence of (D, 0) is (2,1,1). We have seen that the contact number is four, see Example 5.4.6. It follows that the intersection multiplicity is equal to 4·2+2·1+2·1+1·1 = 13. The curve (C, 0) is given by the parametrization x = t4 , y = t6 + t7 . So g is in C {t} equal to (t6 + t7 )2 − (t4 )3 = 2t13 + t14 , so its order is thirteen. Therefore, conversely, one can now deduce that the contact number is four. As in the irreducible case, we define the δ–invariant. e 0) its normalization. Then Definition 5.4.10. Let (C, 0) be a curve singularity and (C,e we define the δ–invariant by δ(C, 0) = dimC (OC,e e 0 /OC,0 ).

The following easy lemma is useful if one wants to calculate the δ–invariant for reducible singularities. Lemma 5.4.11 (Lemma of Hironaka). Let (C, 0), and (D, 0) in (C n , 0) be reduced curve singularities which do not have a component in common. Then δ((C, 0) ∪ (D, 0)) = δ(C, 0) + δ(D, 0) + (C, 0) · (D, 0). Proof. Set I = I (C, 0) ⊂ On and J = I (D, 0) ⊂ On . Consider the following exact sequence 0 −→ On /(I ∩ J) −→ On /I ⊕ On /J −→ On /(I + J) −→ 0. We conclude
from this exact sequence that the vector space dimension of OC,0 ⊕ OD,0 /OC∪D,0 and the intersection multiplicity (C, 0) · (D, 0) are equal. By splitting of the normalization we know that OC,e e 0 ⊕ OD,e e 0 = OC∪D, ^0 . Therefore δ(C ∪ D, 0) is
equal to dimC OC,e e 0 ⊕ OD,e e 0 /OC∪D,0 , which is equal to

dimC (OC,e e 0 ⊕ OD,e e 0 )/(OC,0 ⊕ OD,0 ) + dimC (OC,0 ⊕ OD,0 )/OC∪D,0 = dimC (OC,e e 0 /OC,0 ) + dimC (OD,e e 0 /OD, 0 ) + (C, 0) · (D, 0) = δ(C, 0) + δ(D, 0) + (C, 0) · (D, 0).

We wish to finish this chapter with a proof of Max Noether’s Theorem. This theorem tells us how to compute the δ–invariant from the multiplicity sequence of an irreducible plane curve singularity. In the proof of Noether’s Theorem, we need the following lemma.
Lemma 5.4.12. Let (C, 0) = V (f ), 0 be an irreducible plane curve singularity. Let the multiplicity be m = mult(C, 0), and c = c(C, 0) be the conductor. Suppose g ∈ OC,0 ⊂ OC, e 0 = C {t} with ordt (g) ≥ c. Then the multiplicity of g is at least m − 1.

216

5 Plane Curve Singularities

Proof. Suppose the converse. Then after a general linear coordinate change, we may assume that g is regular in y of order k < m − 1 in y. We may also assume that f is regular of order m in y, so that (C, 0) has parametrization x = tm , y = ts + . . . We now y s−m look at h := xy · g ∈ OC, · unit ∈ C {t} = OC, e 0 . This is indeed in OC, e 0 , since x = t e 0. Obviously ordt (h) ≥ c(C, 0), hence h ∈ OC,0 . Thus in OC,0 we have the equality yg = xh. Therefore, qf = yg − xh for some q ∈ C {x,y}. Now on the right-hand side a term y k+1 occurs, as it occurs in yg but not in xh. But it cannot occur on the left-hand side, as the multiplicity of f is m, and k + 1 < m. This is a contradiction.

Theorem 5.4.13 (M. Noether). Let (C, 0) be an irreducible plane curve singularity, (m0 , . . . ,mr ) be the multiplicity sequence of (C, 0). Then δ(C, 0) =

r X mi (mi − 1) i=0

2

.

Pr Proof. As c(C, 0) = 2δ(C, 0) by 5.2.4, it suffices to show c(C, 0) = i=0 mi (mi − 1). The formula, obviously, holds for the case of a smooth curve, since c(C, 0) = 0 and m0 = 1. Let C (1) be the strict transform of (C, 0) under blowing-up U in the origin for some sufficiently small neighborhood U . We claim that if the formula holds for C (1) , then it holds for (C, 0). By induction on r, this suffices to prove the theorem. Suppose (C, 0) is given by the parametrization x = tm , y = ts + . . . , s > m.

Step 1. We first show α ∈ Γ(C (1) ) =⇒ α + m(m − 1) ∈ Γ(C, 0). The strict transform C (1) has parametrization (x = x,y = xv): x = tm , v = ts−m + . . . . Let g = g(x,v) ∈ OC (1) . As we are looking at the vanishing order of g in OC, e 0 = C {t}, and C (1) is the zero set of a function which is regular of order ≤ m, we may, by applying the Weierstraß Division Theorem, assume that degv (g) ≤ m − 1. We write g = am−1 v m−1 + am−2 v m−2 + . . . + a0 , ai ∈ C {x}. Multiplying g with xm−1 we get a function that we call g(x,y): g(x,y) := xm−1 g(x,v) = am−1 y m−1 + xam−2 y m−2 + . . . + a0 xm−1 . As (C, 0) and C (1) have the same normalization, it follows that g(x,y) ∈ OC,0 and has vanishing order ordt (g) + m(m − 1). Hence it follows that α ∈ Γ(C (1) ) =⇒ α + m(m − 1) ∈ Γ(C, 0), as claimed. Step 2. To finish the proof we show that

5.4 Reducible Plane Curve Singularities

217

6 ∃ g ∈ OC,o with ordt (g) = c(C (1) ) + m(m − 1) − 1. Suppose the converse, that is, let g ∈ OC,0 with ordt (g) = c(C (1) ) + m(m − 1) − 1. It is left as Exercise 5.4.15 to show that we may take such g which is moreover reduced. By the first step we know that c(C, 0) ≤ c(C (1) ) + m(m − 1) − 1. By the previous lemma, we know that the multiplicity of g,
say q, is at least
m − 1. Let g(x,v) = x−q g(x,xv). Then 5.4.8 gives that (C, 0) · V (g), 0 = C (1) · V (g), 0 + qm. Hence


C (1) · V (g), 0 = c(C (1) ) + m(m − 1) − 1 − qm,
because (C, 0) · V (g), 0 = ordt (g). Thus, we have found a function g ∈ OC (1) which has vanishing order ordt (g) = c(C (1) ) + m(m − 1) − qm − 1. Now q ≥ m − 1, so that xq−m+1 · g ∈ OC (1) . This function has vanishing order c(C (1) ) − 1. This is a contradiction to the definition of the conductor c(C (1) ). Example 5.4.14. Consider the plane curve singularity given by (y 2 − x3 )2 − 4x5 y − x7 = 0. The multiplicity sequence is (4,2,2,1,1), so the delta–invariant is equal to 6 + 1 + 1 = 8. This can also be calculated immediately, with the aid of the characteristic exponents, which are 4,6 and 7. Thus one can calculate the minimal set of generators of the semigroup to be 4,6 and 13. The fact that δ = 8 is then easily seen by counting.

Exercises 5.4.15. Prove that one may take g in the proof of 5.4.13 to be reduced. (Hint: Add elements of high order to some factors which are multiple.) 5.4.16. Let (C, 0) = ∪(Ci , 0) be a curve singularity. Consider the standard resolution of (C, 0), for some small U : Xk −→ . . . −→ X1 −→ U.

Let C (j) be the strict transform of C in Xj . An infinitely near point of C is, by definition, a singular point of C (j) for some j, C (0) = C. Prove that δ(C, 0) =

X

p infinitely

mp (mp − 1) 2

near point

where mp is the multiplicity of C (j) at p. (Hint: Use the Lemma of Hironaka 5.4.11, Lemma 5.4.8 and Lemma 5.4.13.) 5.4.17. Let (C, 0) be the plane curve singularity defined by f = (x3 − y 4 )(x3 − y 7 )(y 11 − x5 ) = 0. Compute the multiplicity sequences of the branches of (C, 0), the contact numbers, and the resolution graph of (C, 0). Also construct the resolution graph of (C, 0) out off the resolution graphs of the branches and their contact numbers. 5.4.18. Compute the δ–invariant of (1) the D4 –singularity: y(y 2 − x2 ) = 0,

(2) the E7 –singularity: y(y 2 − x3 ) = 0, and

(3) m different lines through the origin.

5.4.19. Let (C, 0) be a reduced plane curve singularity given by a Weierstraß polynomial f in y of degree b and multiplicity m. Let C (1) be the strict transform under blowing-up, E the exceptional divisor. Prove: P (1) (1) ,p) = b. p∈C (1) ∩E (E,p) · (C

218

5 Plane Curve Singularities

(2) µ(X, 0) − 1 =

P

p∈C (1) ∩E

µ(C (1) ,p) + m(m − 1).

(Hint: Use Exercise 5.1.29, and study the behavior of the terms in that formula under blowing-up; use part 1 of this exercise and use Lemma 5.4.8.)

(3) Deduce Milnor’s formula µ(C, 0) = 2 · δ(C, 0) − r(C, 0) + 1, where r(C, 0) is the number of branches of (C, 0).

219

6

The Principle of Conservation of Number

The aim of this chapter is to study the behavior of certain invariants of singularities in a family. The idea is that the invariants should be constant in the following sense: if in the family a singular point splits into several singular points by varying the parameter, then the sum of the invariants of those singular points should be equal to the invariant of the point we started with.

0

s

The picture describes a typical situation. In the fiber above the point 0, we have an intersection of a line with a cusp and the intersection multiplicity at that point is three. In a small neighborhood we have in the fiber three different intersection points and a transversal intersection at each point of multiplicity 1. Altogether we see a conservation of the intersection number. We will see that other invariants, such as the Milnor number and the δ–invariant have the same property. The basis to prove such this property is the theory of sheaves. Therefore, we start the chapter with an overview of the basic facts of sheaf theory. In short, a presheaf S of rings (or modules, or . . .) on a topological space gives for all open subsets U a ring S (U ) (or module, or . . .), with some natural conditions. An element of S (U ) is called a section over U . One can talk about the germ of a section at any point of U . The set of germs at x is called the stalk at x, notation Sx . A sheaf is a presheaf, which is, loosely speaking, determined by its stalks. The application we have in mind is that these stalks are modules whose vector space dimension gives an invariant of a germ of an analytic space. Thus, we want to compare the various stalks for a given sheaf. Now general sheaves are like “sand” but coherent sheaves are not. This is reflected in the statement that a map of sheaves which induces an isomorphism at the stalk at x, is an isomorphism in a small neighborhood of x. Loosely speaking, the stalk determines the sheaf in an open neighborhood. The problem with the theory of coherent sheaves is that it is difficult to give nontrivial examples. The only examples one gets from the definition are the zero sheaf, and finitely generated sheaves over a point. However, it is true that most sheaves we consider are coherent. For example OX , that is, the sheaf of holomorphic

220

6 The Principle of Conservation of Number

functions, the ideal sheaf of an analytic set and the normalization sheaf of an analytic space are coherent. Moreover, for finite maps coherence is preserved (Finite Mapping Theorem). These theorems are all proved in Section 6.3. In Section 6.2 we develop the technical tools. Here we prove the Meta-Theorem, which says that coherence of sheaves is preserved under all “reasonable” algebraic operations. In Section 6.2 we discuss these coherent sheaves, but give only two examples. On this basis, in Section 6.4, the principle of conversion of number is studied. In particular we study the behavior of the invariants mentioned above. Finally, in Section 6.5, we study Cohen-Macaulay spaces. This notion is closely related to the principle of conservation of number. Hypersurfaces, more generally, complete intersections are CohenMacaulay.

The space in the picture is not Cohen-Macaulay. There is an algebraic method of deciding whether spaces are Cohen-Macaulay. This method is by using resolutions and the Auslander-Buchsbaum formula. From this it follows that a space is Cohen-Macaulay if and only if the resolution has the right length. For lack of a better site, we will finally discuss the Hilbert-Burch Theorem. It says that the ideal of Cohen-Macaulay codimension two spaces can be described by the t minors of a certain t × (t + 1) matrix. It is used in Chapter 10 to give examples of singularities for which there exists a semi-universal deformation over a smooth space.

6.1

Sheaves

Definition 6.1.1. Let X be a topological space. A presheaf S of rings on X consists of the following data (1) for each open set U of X a ring S (U ); elements of S (U ) we call sections of S over U , elements of S (X) are called global sections of S ; (2) for each inclusion V ⊂ U of open sets in X a ring homomorphism (called the restriction map): ρUV : S (U ) −→ S (V ). These data have to satisfy the following conditions: 1. S (∅) = 0, 2. ρUU : S (U ) −→ S (U ) is the identity map, 3. if W ⊂ V ⊂ U are three open subsets of X then ρUW = ρV W ◦ ρUV .

6.1 Sheaves

221

For s ∈ S (U ) one usually writes s|V instead of ρUV (s). For all open subsets V we get in a natural way a presheaf S|V which we call the restriction of S to V . Similarly one defines presheaves of abelian groups, of vector spaces, C –algebras, etc. Examples 6.1.2. (1) Let X be the topological space consisting of just one point. Then a presheaf of rings on X can be identified with a ring. (2) Let X be a topological space. For all open U ⊂ X put C (U ) := {f : U −→ C , f is continuous}. For V ⊂ U , let ρUV be the obvious restriction of functions from U to V . Then one shows without difficulty that this defines a presheaf of rings, or even of C –algebras on X. (3) Consider an open set X ⊂ C n . Put for U ⊂ X open: O(U ) := {f : U −→ C , f is holomorphic}, and again for an open subset V of U , the map ρUV is given by restriction. This is a presheaf of C –algebras. (4) Consider an open set X ⊂ C n . For U ⊂ X open we put O′ (U ) := {f : U −→ C , f is holomorphic}, but we define different restriction maps. For V $ U we put ρUV = 0, and ρUU = id. Then O′ is a presheaf. (5) Let A be a ring, X a topological space. The constant presheaf C is given by C (U ) = A if U 6= ∅, and C (∅) = 0. The maps ρUV is the identity map if V 6= ∅. Moreover, ρU∅ is the zero map for all open U . Definition 6.1.3 (Sheaf Conditions). Let X be a topological space, S a presheaf on X. Then S is called a sheaf if the following two sheaf axioms are satisfied. (1) Let U be an open set in X, and U = ∪i∈I Vi be an open covering of U . Let s ∈ S (U ) be a section of S over U , such that s|Vi = 0 for all i. Then s = 0. (2) Let U be an open set in X, and U = ∪i∈I Vi be an open covering of U . Let si ∈ S (Vi ) be given such that si|Vi ∩Vj = sj|Vi ∩Vj for all i,j ∈ I. Then there exists an s ∈ S (U ) such that s|Vi = si . A presheaf which satisfies the sheaf axioms, so is in fact a sheaf, is also called a canonical presheaf. Remark 6.1.4. It follows from the first condition that such an s, existing according to the second condition, is uniquely determined. Example 6.1.5. In 6.1.2, the first three examples, in fact, are sheaves. The fourth example, the presheaf O′ is not a sheaf, as it does not satisfy the first condition. Moreover, the constant presheaf A is, in general, not a sheaf. In fact, suppose U,V are two open subsets of X with U ∩ V = ∅. Let a,b ∈ A, with a 6= b. Then the restrictions of a and b to U ∩ V = ∅ are equal, but there does not exist a section of U ∪ V , restricting to a and b, respectively.

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6 The Principle of Conservation of Number

Definition 6.1.6. Let S be a presheaf (of rings) on a topological space X and p ∈ X be a point. Consider the collection of rings: {S (U ) : U ∋ p an open neighborhood of p}. We define an equivalence relation on the disjoint union of this collection: s ∈ S (U ) ∼p t ∈ S (V ) ⇐⇒ there exists an open neighborhood W ⊂ U ∩ V of p such that s|W = t|W . The equivalence class of s in p we call the germ of s in p, notation sp . The set of equivalence classes we call the stalk of S at p, and will be denoted by Sp . It is not difficult to show that, in a natural way, Sp is a ring again, see Exercise 6.1.28. The support of S is the set of all points p ∈ X such that Sp 6= 0. The support of S is denoted by supp(S ). Definition 6.1.7. Let two presheaves S and T (of rings) on a topological space, with restriction mappings ρUV and ρ′UV respectively, be given. A morphism of presheaves f : S −→ T is a collection of ring homomorphisms fU : S (U ) −→ T (U ), for all open subsets U of X, such that for all V ⊂ U ⊂ X open, the diagram S (U )

fU

ρ′U V

ρU V

 S (V )

/ T (U )

fV

 / T (V )

commutes. A morphism of sheaves is given by a morphism of the underlying presheaves. A subsheaf of a sheaf F is a sheaf G such that for all open subsets U of X G (U ) is a subring of F (U ), and the restriction homomorphism are those induced by F , that is, the inclusion mapping is a morphism of presheaves. Examples 6.1.8. (1) Consider the presheaves C and O on C n defined in Example 6.1.2 and f : O −→ C defined by the canonical injection O(U ) ⊂ C (U ). Then f is a morphism of sheaves. (2) The map defined by the canonical injection O′ (U ) ⊂ C (U ) is not a morphism of presheaves. Lemma 6.1.9. Let f : S −→ T be a morphism of presheaves on X. Then there is, for all p ∈ X, a naturally defined ring homomorphism fp : Sp −→ Tp . This map is defined as follows. Take an element ap ∈ Sp , and a representative a ∈ S (U ) of a in some open neighborhood U of a. Then define fp (ap ) as the germ in p of f (a). The fact that this map is well defined is left as Exercise 6.1.29. The second part of the following theorem says, loosely speaking, that a sheaf is determined by its stalks.

6.1 Sheaves

223

Theorem 6.1.10. Let f : S −→ T be a morphism of sheaves on a topological space X. Let, for all p ∈ X, fp : Sp −→ Tp be the induced map on stalks. Then (1) fU is injective for all U open in X if and only if fp is injective for all p; (2) f is an isomorphism if and only if fp is an isomorphism for all p. Proof. In both cases, one direction is quite obvious. If fU is injective (resp. an isomorphism) for all U , then fp is injective (resp. an isomorphism) for all p. (1) We prove that fU is injective. Let s ∈ S (U ), with fU (s) = 0. In particular, for all p ∈ U , the germs (fU (s))p = 0. Thus fp (sp ) = 0. By assumption sp = 0 for all p ∈ U . This means that for all p there exists a neighborhood Up of p with s|Up = 0. By the first sheaf axiom, it follows that s = 0. Thus fU is injective. (2) We need to show that for all U open in X, the map fU : S (U ) −→ T (U ) is an isomorphism. Namely, if this is the case, we can define the inverse map f −1 by (f −1 )U := (fU )−1 . It is a direct check that this gives indeed an inverse map. The first part gives that fU is injective. So it remains to prove the surjectivity of fU . Take a section t ∈ T (U ). Let p ∈ U , and consider tp . As fp is surjective by assumption, there exists an element sp with fp (sp ) = tp . Take a representative sep ∈ S (Up ) of sp for some small open neighborhood sp ) = t|Up . So we Up of p. We may, by making Up smaller if necessary, assume that fUp (e have (1) an open covering {Up }p∈U of U , (2) sections sep ∈ S (Up ),

sq )|Up ∩Uq for all p,q ∈ U . (3) such that (e sp )|Up ∩Uq = (e sq )|Up ∩Uq are both sections of S (Up ∩ Uq ) The third condition follows as (e sp )|Up ∩Uq and (e which are mapped under f|Up ∩Uq to t|Up ∩Uq . They are equal, because of the injectivity proved in (1)! It follows from the second sheaf property, that there exists a section s ∈ S (U ) with s|Up = sep . By construction fU (s)|Up = t|Up for all p, so that we may conclude fU (s) = t by using the first sheaf axiom. This results motivates the following definition.

Definition 6.1.11. A sequence of morphisms of sheaves on X f

g

R −→ S −→ T is called exact, if for all p ∈ X, the induced map on stalks fp

gp

Rp −→ Sp −→ Tp is exact. In particular, a morphism S −→ T is called injective, if for all p ∈ X, the induced map on stalks fp : Sp −→ Tp is injective. Similarly surjectivity is defined. Remark 6.1.12. It is important to note that from the surjectivity of fp for all p it does not follow that fU is surjective for all U in general. To give an example, consider the following map of sheaves on X = C \ {0}:

224

6 The Principle of Conservation of Number ∗ f : OX −→ OX ,

fU : g 7→ e2πig .

∗ Let U be an open subset of X, and OX (U ) = {h ∈ O(U ): h(p) 6= 0 for all p ∈ U }. Then ∗ fp is surjective for all p. Indeed, take hp ∈ OX,p , and a small simply connected open neighborhood U of p on which h is a representative of hp . By continuity, we may assume that U is so small that h is nonzero throughout U . From function theory, we know that log(h) exists, and e2πi log(h) = h. Thus log(h)p maps to hp , showing that the map of sheaves is surjective. However fp is not injective. With log(h) also log(h) + k for all k ∈ Z maps to h. Let z be a coordinate on C . Then z ∈ O∗ (X) but z is not in the image of f , as log(z) is not a well defined holomorphic function on X \ {0}.

This example also shows that
given a map f : G −→ H of sheaves the presheaf defined by U 7→ H(U )/fU G (U ) is not necessarily a sheaf. The first sheaf axiom is not satisfied: locally defined logarithms do not necessarily glue to a globally defined logarithm. Thus we have problems in defining the cokernel of a map as sheaf. These kinds of problems occur for several other constructions with sheaves as, for example, the tensor product. This is the reason for the following construction, called sheafifying. Theorem 6.1.13. Let X be a topological space and F be a presheaf on X. There exists a sheaf F + and a morphism j : F −→ F + of presheaves with the following properties: (1) jp : Fp −→ Fp+ is an isomorphism for all p. (2) (F + ,j) is universal, that is, for any morphism ϕ : F −→ G of F to a sheaf G there exists a unique morphism ψ : F + −→ G such that ϕ = ψ ◦ j. Definition 6.1.14. The (uniquely) determined sheaf F + is called the sheaf associated to the presheaf F . One says that F + is obtained from F by sheafifying. Proof of Theorem 6.1.13. Let U be an open set and define F + (U ) to be the set of functions f : U −→ ∪ Fx having the following two properties: x∈U

(1) for each x ∈ U , f (x) ∈ Fx , (2) for each x ∈ U there is a neighborhood W ⊂ U of x and an element g ∈ F (W ) such that the germ gy of g in y ∈ W is equal to f (y) for all y ∈ W . Now it is not difficult to verify, see Exercise 6.1.36, that F + together with the canonical restriction maps is a sheaf. It is also not difficult to check that the canonical map j : F −→ F + defined by jU (g)(y) = gy has the universal property mentioned, see Exercise 6.1.36. Note that not all nonzero section of F (U ) give nonzero sections of F + (U ). Namely, it might happen that all stalks are zero, but F (X) 6= 0. On the other hand, in the ∗ example f : OX −→ OX , fU (g) = e2πig the function z is in Im(f )(U ) as sheaf, but not as presheaf. So F + (U ) might have more sections. This is because by definition sections of F + (U ) are not sections of F (U ), but functions which are locally equal to sections of F. We are now prepared to give a lot of examples for sheaves. Definition 6.1.15. Let X be a topological space, F , G be sheaves on X and f : F −→ G a morphism.

6.1 Sheaves

225

(1) The kernel of f , Ker(f ) is the sheaf defined by Ker(f )(U ) = Ker(fU ). (2) The cokernel of f , Coker(f ) is the sheaf associated to the presheaf defined by
U 7→ G (U )/fU F (U ) . A cokernel of a map of sheaves is also called a quotient sheaf. (3) The image
of f , Im(f ) is the sheaf associated to the presheaf defined by U 7→ fU F (U ) . (Note that Im(f ) is a subsheaf of G , Exercise 6.1.37.) (4) Let A be a ring, the constant sheaf CA is the sheaf associated to the constant presheaf defined by U 7→ A if U 6= ∅. In this case if U is the disjoint union of two connected subsets, we have C (U ) = A ⊕ A. Sections of CA over U are now allowed to differ on the two connected subsets. The proof that the presheaf defined in (1) is already a sheaf will be given in Exercise 6.1.37. We now relate (pre)sheaves which are defined on different topological spaces. Definition 6.1.16. Let a continuous map f : X −→ Y between topological spaces be given. (1) Let S be a sheaf on X. We define a presheaf f∗ S on Y by putting for an open set U ⊂Y
f∗ S (U ) := S f −1 (U ) ,

with the obvious restriction maps. (It can be proven without difficulty, see Exercise 6.1.30, that f∗ S is in fact a sheaf.)

(2) Let F be a sheaf on Y . f −1 F is the sheaf on X associated to the presheaf defined by U 7→ lim F (V ).1 V ⊃f (U)

(3) If f is an injection and F a sheaf on Y , we define the restriction F|X of F to X by f −1 F . As a simple example, if f is the map which sends X to a point, then f∗ S can be identified with a sheaf on a point, hence a ring. In this case we can identify f∗ S with the global sections S (X) of X. On the other hand, if f : p −→ X is the injection of the point p in the space X, then f −1 F is the sheaf on the point p, consisting of the stalk Fp . More generally, if f : X −→ Y is the injection of a closed subspace, then f −1 G is a sheaf whose stalks at p ∈ X are equal to Gp . It is a direct and easy consequence of the definitions, that if one has a morphism α : S −→ T of sheaves on X, and a continuous map f : X −→ Y , then one has an induced map α∗ : f∗ S −→ f∗ T . One cannot hope in general, that for a surjective morphism of sheaves α : S −→ T , the induced map α∗ : f∗ S −→ f∗ T is surjective (look at the constant map f , that is, look at global sections). But, on the other hand, we have the following theorem. 1

By definition

lim F (V ) :=

V ⊃f (U )

∪

V ⊃f (U )

F (V )/ ∼ where s ∼ s′ for s ∈ F (V ), s′ ∈ F (W ) if and only if

s|V ∩W = s′|V ∩W . Note that this is a generalization of the notion of germ, which corresponds to the case that f is the injection of a point.

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6 The Principle of Conservation of Number

Theorem 6.1.17. Let X and Y be Hausdorff topological spaces and f : X −→ Y be a finite continuous map. Consider an exact sequence of sheaves on X: 0 −→ S1 −→ S2 −→ S3 −→ 0. Then the induced sequence of direct image sheaves 0 −→ f∗ S1 −→ f∗ S2 −→ f∗ S3 −→ 0 is also exact. One phrases this by saying that for finite f , the functor f∗ is exact. This theorem will play an important role in the proof of the Finite Mapping Theorem, see 6.3.5. The theorem follows immediately from the fact that exactness means exactness of the stalks and the following theorem. Theorem 6.1.18. Let X,Y be Hausdorff spaces, and f : X −→ Y be a finite continuous map. Let S be a sheaf on X. Let y ∈ Y , and x1 , . . . ,xt be the points in the fiber f −1 (y). Then we have the equality of stalks: (f∗ S )y =

t M

Sxi .

i=1

Proof. As X is Hausdorff, we can choose pairwise disjoint open neighborhoods U1′ , . . . ,Ut′ of x1 , . . . ,xt in X. As f is closed, there exists an open neighborhood V of y in Y , such that f −1 (V ) ⊂ ∪ti=1 Ui′ , see Exercise 3.4.36. x1

U1′

xt

Ut′

f y

V

We define Ui := f −1 (V ) ∩ Ui′ . Then (1) The U1 , . . . ,Ut are pairwise disjoint. (2) f −1 (V ) = ∪ti=1 Ui . Therefore, f∗ S (V ) := S (f −1 (V )) = S (∪ti=1 Ui ) = ⊕ti=1 S (Ui ). The last equality is due to the fact that the Ui are pairwise disjoint, and because of the sheaf properties. Now use the definition of stalk to deduce the theorem. There is a method to produce a lot of examples of sheaves, if we know them locally. Proposition 6.1.19 (Gluing Sheaves). Let X be a topological space, X = ∪ Ui , Ui open i∈I ∼

subsets in X. For every i let Fi be a sheaf on Ui and ϕij : Fi|Ui ∩Uj −→ Fj|Ui ∩Uj be isomorphisms such that (1) ϕii is the identity, (2) ϕik = ϕjk ◦ ϕij on Ui ∩ Uj ∩ Uk .

6.1 Sheaves

227 ∼

Then there exists a unique sheaf F on X and isomorphisms λi : F|Ui −→ Fi such that for all i,j, ϕij ◦ λi = λj on Ui ∩ Uj . In fact, for an open subset V of X we define ( ) Y F (V ) := {si }i∈I ∈ Fi (Ui ∩ V ) : ϕij (si|Ui ∩Uj ∩V ) = sj|Ui ∩Uj ∩V . i∈I

The proof that this defines a sheaf, and the uniqueness is left as Exercise 6.1.40. Definition 6.1.20. A ringed space (X,OX ) is a topological space X together with a sheaf of rings OX . Usually, if it is clear from the context what the sheaf of rings is, we simply write X for the pair (X,OX ). A sheaf F on X is called an OX –module if F (U ) is an OX (U )–module for all U ⊂ X open, such that the restriction mapping F (U ) −→ F (V ) is OX (U )–linear for V ⊂ U . (Note that F (V ) for V ⊂ U is also an OX (U )–module.) An OX –module sheaf F which is a subsheaf of OX is called an ideal sheaf. Examples 6.1.21 (Ringed Spaces). (1) Let U be an open set of C n and OU be the sheaf of holomorphic functions on U .

(2) Let U be an open set in C n , f1 , . . . ,ft be holomorphic functions defined on U . Consider X := V (f1 , . . . ,ft ). Let V ⊂ U be open. Consider the presheaf defined by OX (V ∩ X) := O(V )/(f1 , . . . ,ft )O(V ) on X. This is easily checked to be a sheaf. The ringed space constructed in this way is called a complex model space. Consider X = {x ∈ U : f1 (x) = · · · = ft (x) = 0}. Consider the presheaf I (X) on U defined by I (X)(W ) = {f ∈ OC n (W ) : f (x) = 0 for all x ∈ W ∩ X}. This is easily checked to be a sheaf and is called the ideal sheaf of X. (3) Let U ⊂ Rn be an open subset, and OU the sheaf of differentiable functions on U . Similarly let CU be the sheaf of continuous functions on U . (4) An affine variety X (with the Zariski topology) can also be seen as a ringed space by taking as canonical presheaf the presheaf which assigns to each U ⊂ X the ring g OX (U ) = {f ∈ K(X) : ∀p ∈ U ∃g,h ∈ K[X] with f = and h(p) 6= 0}. h An element f in OX (U ) therefore can be viewed as a rational function f : U −→ K which have a well-defined value for all p ∈ U . (5) Spectrum of a ring: Let R be a ring (commutative with 1 as always). Then the spectrum Spec(R) is defined as follows: • As a set Spec(R) is the set of prime ideals p ⊂ R. • One defines a topology on Spec(R) by defining the closed sets to be the sets of type V (I) := {p ∈ Spec(R) : I ⊂ p}. This topology is called the Zariski topology on Spec(R).
• We define a sheaf O of rings on Spec(R). Let U = D(f ) := Spec(R) \ V (f ) be an open set of Spec(R). Then O(D(f )) := Rf . We have to show that this is independent of the chosen f . For this we refer to Exercise 6.1.33. As the D(f ) form a basis for the topology by Exercise 6.1.34 this induces a presheaf on Spec R. This presheaf is in fact a sheaf, see Exercise 6.1.34.

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6 The Principle of Conservation of Number

Definition 6.1.22. Let X be a ringed space and F , G be OX –module sheaves on X. (1) HomX (F ,G ) is the sheaf defined by HomX (F ,G )(U ) = HomOX|U (F|U ,G|U ). (2) F ⊗OX G is the sheaf associated to the presheaf defined by U 7→ F (U )⊗OX (U) G (U ). (3) F ⊕ G is the sheaf defined by (F ⊕ G )(U ) = F (U ) ⊕ G (U ). (4) Let F ,G ⊂ H, H an OX –module sheaf. Then F ∩ G is the sheaf defined by (F ∩ G )(U ) = F (U ) ∩ G (U ). F + G is the sheaf associated to the presheaf defined by U 7→ F (U ) + G (U ). F : G is the sheaf associated to the presheaf U 7→ F (U ) : G (U ). (5) Ann(F ), the
annihilator sheaf of F , is the sheaf associated to the presheaf U 7→ Ann F (U ) = {f ∈ OX (U ) : f · F (U ) = 0}.

The proof that the presheaves defined in (1), (3) and (4) are already sheaves is left as Exercise 6.1.38. Note that in all the examples considered above the stalk at each point is a local ring. We put this phenomenon in a definition. Definition 6.1.23. A locally ringed space (X,OX ) is a ringed space, such that for all points p ∈ X the stalk OX,p is a local ring. A morphism between locally ringed spaces is a pair: (f,f ∗ ) : (X,OX ) → (Y,OY ), where f : X → Y is a continuous map, and f ∗ is a morphism of sheaves of OY −modules: f ∗ : OY → f∗ OX such that the induced map on stalks: fp∗ : OY,f (p) → OX,p sends the maximal ideal of OY,f (p) into the maximal ideal of OX,p . Usually, if there is no danger of confusion, we just write f for the pair (f,f ∗ ). Examples 6.1.24. (1) The examples of Examples 6.1.21. (2) The projective space PnC , the set of lines through the origin in C n+1 . As a set we have PnC := {c ∈ C n+1 , c 6= 0}/ ∼ and c ∼ c′ if there exists a λ ∈ C such that λc = c′ . The equivalence class of c = (c0 , . . . ,cn ) we write as (c0 : c1 : · · · : cn ). We consider PnC with the quotient topology of the ordinary topology of C n+1 . Let Ui = {c = (c0 : · · · : cn ) ∈ PnC , ci 6= 0}. Then Ui is an open subset n

of PnC and ∪ Ui = PnC . We have homeomorphisms ϕi : Ui −→ C n defined by i=0 
 ci+1 cn ϕi (c0 : · · · : cn ) = cc0i , cc1i , . . . , ci−1 ci , ci , . . . , ci . The map ϕj ◦ ϕ−1 : ϕi (Ui ∩ Uj ) −→ ϕj (Ui ∩ Uj ) maps (z1 , . . . ,zn ) to i  zj−1 zj+1 z1 , . . . , , , . . . , z1j , . . . , zznj and is obviously holomorphic. zj zj zj OPn is the sheaf obtained by gluing the sheaves OC n on Ui .

6.1 Sheaves

229

The second example can be generalized as follows: Proposition 6.1.25. Let X be a Hausdorff space and X = ∪i∈I Ui for open sets Ui . Let ϕi : Ui −→ Wi be homeomorphisms to domains Wi in C n such that for all i,j and Ui ∩ Uj 6= ∅ ϕj ◦ ϕ−1 : Wi ∩ ϕi (Uj ) −→ Wj ∩ ϕj (Ui ) i is holomorphic. Then X is a locally ringed space. Proof. the structure sheaf OX is defined to be the gluing of the sheaves OUi = ϕ∗i OWi (Proposition 6.1.19). Definition 6.1.26. Let X be a Hausdorff space. (1) Let (X,OX ) be a locally ringed space. (X,OX ) is called a complex manifold if every x ∈ X has a neighborhood U such that (U,OX|U ) is isomorphic to (V,OV ), V ⊂ C n a domain. (2) (X,OX ) is called an analytic space if every x ∈ X has a neighborhood U such that (U,OX|U ) is isomorphic to (V,OV ), V an analytic subset of an open set W in some
C n , and OV = OW /I (V ) |V .

(3) (X,OX ) is called a complex space if if every x ∈ X has a neighborhood U such that (U,OX|U ) is isomorphic to (V,OV ), where (V,OV ) is a complex model space. This means that V is an analytic subset of an open set W in √ some C n and OV =
OW /J |V , J ⊂ OW an ideal sheaf such that for all x ∈ V , Jx = I (V )x . One says that (Y,OY ) is a closed complex subspace of (X,OX ), if there exists a map of locally ringed spaces (i,i∗ ) : (Y,OY ) −→ (X,OX ) such that i is injective, and i∗ is surjective. For the case (X,OX ) = (U,OU ) for U ⊂ C n open, we obtain that a closed complex subspace is a complex model space.

Exercises 6.1.27. Consider the following presheaf on R. Let U ⊂ X be open and B(U ) := {f : U −→ R : f bounded}, and ρU V the ordinary restrictions. Show that B is not a sheaf. 6.1.28. Let S be a sheaf of rings on a topological space X. Show that for each point x ∈ X the stalk Sx is in a natural way a ring again. 6.1.29. Prove Lemma 6.1.9. 6.1.30. Prove that for f : X −→ Y continuous, S a sheaf on X, the direct image f∗ S is also a sheaf. 6.1.31. Let R be a Noetherian ring. Prove that the topology on Spec(R) is indeed a topology by proving the following statements: ` ´ ` ´ (1) V (0) = Spec(R); V (1) = ∅, (2) V (I · J) = V (I) ∪ V (J), P (3) V ( j Ij ) = ∩j V (Ij ).

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6 The Principle of Conservation of Number

6.1.32. For a subset A of Spec(R) let I (A) = ∩ p. Prove the analog of Hilbert’s Nullstellenp∈A √ satz: I = I (V (I)). (Hint: Use primary decomposition of an ideal). 6.1.33. Let R be a ring and f,g ∈ R such that D(f ) = D(g) (equivalently V (f ) = V (g)). Prove that Rf = Rg . 6.1.34. Let R be a ring. Show that the open sets D(f ) form a basis for the topology of Spec(R). Show that the presheaf O on Spec(R), defined in 6.1.21, is in fact a sheaf. 6.1.35. Let f : X −→ Y be a continuous map, A ⊂ Y a closed subset, S a sheaf on X. Prove that ` ´ (f∗ S )|A = f|f −1 (A) ∗ S|f −1 (A) . 6.1.36. Prove that F + defined in the proof of Theorem 6.1.13 is a sheaf and the canonical map j : F −→ F + has the universal property. 6.1.37. Let f : F −→ G be a morphism of sheaves. (1) Prove that there is a canonical injection i : Im(f ) −→ G .

(2) Prove that the presheaf defined by U 7→ Ker(fU ) is already a sheaf. 6.1.38. Let X be a ringed space and F ,G be OX –module sheaves on X. (1) Suppose that both F and G are subsheaves of H . Prove that the presheaf defined by U 7→ F (U ) ∩ G (U ) is a sheaf. (2) Prove that the presheaf defined by U 7→ F (U ) ⊕ G (U ) is a sheaf.

(3) Prove that the presheaf defined by U 7→ HomOX|U (F|U ,G|U ) is a sheaf.

6.1.39. Let f : X −→ Y be a continuous map between the topological spaces X,Y , F be a sheaf on X and G be a sheaf on Y . (1) Prove that Gf (x) = (f −1 G )x . (2) Show that there are canonical maps λ : f −1 f∗ F −→ F and ρ : G −→ f∗ f −1 G . (3) Prove that Hom(f −1 G ,F ) = Hom(G ,f∗ F ).

6.1.40. Prove Proposition 6.1.19. 6.1.41. One can view X = C 2 \ {0} as a locally ringed space (X,OX ) just as one can see an affine variety as a locally ringed space. (1) Show that OX (X) = C [x,y].

(2) Show that X, as locally ringed space, is not isomorphic to an affine variety.

6.2

Fundamental Properties of Coherent Sheaves

In this section, unless otherwise stated explicitly, X will be a complex space, with structure sheaf OX , mX ⊂ OX the maximal ideal. S ,T , etc., will be sheaves of OX modules. If (X,OX ) is fixed we shall often use O and m instead of OX and mX . What is a coherent sheaf? Well, before defining it, let us mention some fundamental properties of coherent sheaves, give some examples, and deduce some properties which show that coherence is indeed a very useful property a sheaf can have. The properties we shall mention will also justify the name “coherence”.

6.2 Fundamental Properties of Coherent Sheaves

231

Theorem 6.2.1 (Meta-Theorem for coherent sheaves). Let S ,T be coherent sheaves of OX –modules then every reasonable operation with S ,T (as taking Hom,⊗, finitely generated subsheaves . . .) produces again a coherent sheaf. Theorem 6.2.2. Let S ,T be coherent sheaves of OX - modules. Let α : S −→ T be a map of OX -modules. Suppose that at some point x ∈ X the maps of stalks αx : Sx −→ Tx is an isomorphism. Then there exists an open neighborhood U of x, such that for all points p ∈ U the maps of stalks αp : Sp −→ Tp are isomorphisms, that is, the sheaves S and T are isomorphic, when restricted to U . The Meta Theorem will be proved in Lemma 6.2.11, Corollary 6.2.12, Lemma 6.2.13 and Lemma 6.2.14. Theorem 6.2.2 is an immediate consequence of 6.2.12 (2) and 6.2.9. So, to show that two coherent sheaves S and T are isomorphic on some open set containing a point x, one has to find a map of sheaves α : S −→ T , and then show that αx is an isomorphism. The last condition usually is pure algebra! As a sheaf is determined by its stalks, the above property gives an easy method to give examples of sheaves which are not coherent. Example 6.2.3. Consider the ringed space consisting of the line C and the sheaf OC of holomorphic functions. For S we take the zero-sheaf (this sheaf will be coherent). To define T , consider a sequence of points p1 ,p2 , . . . , converging to 0. Let T be the subsheaf of OC consisting of all holomorphic functions vanishing at each point pi . We let α be the zero map. By the identity Theorem for holomorphic functions in one variable, every holomorphic function defined on some open neighborhood of U of 0 and vanishing at all points pi which lie in U actually is zero. This shows that the stalk T0 is zero, hence α0 is an isomorphism. However, T is not the zero sheaf (there exist nonzero sections over any open set not containing zero). Therefore T is not a coherent sheaf. In fact, for all points p 6= pi and p 6= 0, we have Tp = Op . For the pi we have Tpi = mpi , the germs of holomorphic functions vanishing at the pi . Notice that T is an ideal sheaf but not finitely generated. To prove this we show that for every domain U ⊂ C containing 0, every OC –module homomorphism ϕ : OC |U −→ T|U has to be the zero map. Namely, because U contains 0 it contains almost all of the pi , then T (U ) = {0} because of the Identity Theorem (Remark 3.1.10). Now let W ⊂ U be an open subset and consider the diagram OC (U )

ϕU

ρ′U W

ρU W

 OC (W )

/ T (U ) = {0}

ϕW

 / T (W ).

As ϕU (1) = 0 we have ϕW (1) = ϕW ρUW (1) = ρ′UW ϕU (1) = 0. This implies that ϕW (g) = g · ϕW (1) = 0 for all g ∈ OC (W ). Hence ϕ = 0.

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6 The Principle of Conservation of Number

So for coherent sheaves such an ugly thing does not happen. The problem with the theory of coherent sheaves (in complex analysis, not in algebra, see Exercise 6.2.19.) is that one has to work hard to produce examples. It follows relatively easy from the definition that sheaves behave well under most algebraic operations on sheaves (the Meta– Theorem for coherent sheaves): kernels, cokernels of maps between coherent sheaves are coherent. Direct sums of coherent sheaves are coherent. Finitely generated subsheaves of coherent sheaves are coherent. Taking Hom, tensor product etc. preserves coherence. These things will be discussed in this section. It is much harder to prove the four basic coherence Theorems: (1) The structure sheaf On is coherent. More generally, the structure sheaf of any complex space is coherent. (2) Coherence is preserved under finite mappings of complex spaces. (3) The ideal sheaf of any analytic space is coherent. (4) The normalization sheaf of any analytic space is coherent. These Theorems will be proved in Section 6.3. Assuming these, we give some applications. Corollary 6.2.4. Let J ⊂ OX be a coherent ideal sheaf then V (J) := {x ∈ X : Jx ⊂ mX,x } is an analytic space. Proof. Let x ∈ X and assume that Jx ⊂ mX,x . Since OX,x is Noetherian, we have Jx = (f1 , . . . ,fm ). Let U ⊂ X be an open set such that f1 , . . . ,fm ∈ OX (U ). Let J ⊂ OX|U be the ideal sheaf generated by f1 , . . . ,fm . This is a coherent sheaf (Theorem 6.2.1) and we have a canonical map α : J −→ J such that αx : J x −→ Jx is the identity. This implies that J |W = JW for an open set W ⊂ U , x ∈ W (Theorem 6.2.2). Therefore, V (J) ∩ W = {x ∈ W : f1 (x) = · · · = fm (x) = 0}. Corollary 6.2.5. Let M be a coherent OX –module. Then the support of M : supp(M ) := {x ∈ X : Mx 6= 0} is an analytic space. Proof. Along with M , the annihilator is coherent (cf. Theorem 6.2.1 or Lemma 6.2.14), and the annihilator is an ideal sheaf. Now the support of M is given by the zero set of
the annihilator of M : Supp(M ) = V Ann(M ) . Indeed, let x ∈ / Supp(M ). Then Mx = 0, so that Ann(Mx ) = Ann(M )x = OX,x . Therefore, x does
not lie in the zero set of Ann(M ). Suppose on the other hand that x∈ / V Ann(M ) . Then there exists an f ∈ Ann(Mx ) with f (x) 6= 0 and fx m = 0 for all m ∈ Mx . As f (x) 6= 0 it follows that the germ fx is a unit, so that m = 0 for all m ∈ Mx . Hence Mx = 0, and thus x ∈ / Supp(M ). Hence the support of M is an analytic subset. Corollary 6.2.6. Let F be a coherent sheaf and supp(F ) = {y : Fy 6= 0} = {x} then (1) F (X) = Fx ; (2) dimC Fx < ∞.

6.2 Fundamental Properties of Coherent Sheaves

233

Proof. The first property holds without using the coherence. Consider the map F (X) −→ Fx . If for some s ∈ F (X) sx = 0 then there exists a neighborhood U of x such that s|U = 0 because sy = 0 for all y 6= x. Let W := X \ {x} then X = U ∪ W and s|W = 0 implies s = 0. To prove the surjectivity let f ∈ Fx and choose a neighborhood U of x such that f ∈ F (U ). On W := X \ {x} we have f|U∩W = 0. So the section 0 on W and f on U glue to a section on X. To prove (2) let I = Ann(F ) and assume that, in a neighborhood U of x, I is generated by f1 , . . . ,fr . In OX,x we consider the primary decomposition of Ix = q1 ∩· · ·∩qs and claim that s = 1 and q1 is mX,x –primary. If this is true, then mcX,x annihilates Fx for a suitable c and, therefore, Fx is a finitely generated OX,x /mcX,x –module. But dimC OX,x /mcX,x < ∞ implies dimC Fx < ∞. √ To prove the claim assume Ix ⊂ q primary but q 6= mX,x . In this case V (I) ⊃ V (q) % {x}. This is a contradiction to the fact that Iy = OX,y for all y 6= x. Definition 6.2.7. (1) The sheaf S is said to be of finite type if for every point x ∈ X there is an open neighborhood U of X and a surjective morphism of sheaves: q OX|U −→ S|U −→ 0.

(2) The sheaf S is said to be of relation finite type if for any open set U and any q morphism of sheaves: α : OX|U −→ S|U the kernel Ker(α) is a sheaf of finite type. (3) A sheaf S of OX modules is called coherent, if S is of finite type, and of relation finite type. Lemma 6.2.8. (1) A quotient sheaf of a sheaf of finite type, is a sheaf of finite type. (2) A subsheaf of a sheaf of relation finite type, is a sheaf of relation finite type. (3) A finitely generated subsheaf of a coherent sheaf is coherent. Proof. Exercise 6.2.18. Lemma 6.2.9. Let s1 , . . . ,sq be holomorphic functions on an open set U containing x. Let S be a sheaf of finite type, and suppose we have a surjection of stalks: (s1x ,...,sqx )

Oxq −−−−−−−→ Sx −→ 0. Then on some open neighborhood V ⊂ U of x, we have a surjection: (s1 ,...,sq )

q OX|V −−−−−−→ S|V −→ 0.

In particular, if a sheaf of finite type has zero stalk in x, then the sheaf is identically zero in some open neighborhood V of x.

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6 The Principle of Conservation of Number

Proof. Because S is of finite type there exist an open neighborhood W of x, and sections t1 , . . . ,ts generating S|V . But, because the Psix generate the stalk at x, we can find germs of holomorphic functions such that tix = j fijx sjx . Take an open neighborhood V ⊂ W of x such that all these functions converge on V . Because the t′ s generate S , it follows that the si also generate S on V . The following Lemma is the basic Lemma, from which almost everything in this section follows. Proposition 6.2.10. Consider an exact sequence of sheaves on X: ϕ2

ϕ1

ϕ3

ϕ4

S1 −→ S2 −→ S3 −→ S4 −→ S5 . If Si is coherent for i = 1,2,4,5, then S3 is coherent.2 Proof. As S1 is of finite type, we may assume S1 = Oq for some q. Consider a map β : Op −→ S3 . We have to prove that Ker(β) is finitely generated. As S4 is of relation finite type, the map ϕ3 β : Op −→ S4 has a finitely generated kernel. This means that we ψ

ϕ3 β

have an exact sequence Os −→ Op −→ S 4. Obviously Ker(β) ⊂ Ker(ϕ3 β) = Im(psi). By construction ϕ3 (β ◦ ψ) = 0, that is Im(β ◦ ψ) ⊂ Ker(ϕ3 ) which by exactness is equal to Im(ϕ2 ). Therefore, we can find for every generator ei of Os an element ω(ei ) ∈ S2 with ω(ei ) = β ◦ ψ(ei ). Thus we get a map ω making a commutative diagram

Oq

ϕ1

Os D DD ψ DD DD D! ω Op C CC CC β CC C   ϕ2 ϕ3 ! / S3 / S4 / S2

Consider the map ϕ1 ⊕ ωOq ⊕ Os −→ S2 . As S2 is of relation finite type, the kernel K is finitely generated. Let π : Oq ⊕ Os be the projection on the second summand. As images of finitely generated sheaves are finitely generated, it follows that ψ ◦ π(K) is finitely generated. So we are done if we show that Ker(β) = ψ ◦ π(K). (1) We first show ψ ◦ π(K) ⊂ Ker(β). Hence suppose (a1 ,a2 ) ∈ K, that is ϕ1 (a1 ) = −ω(a2 ). Then 0 = ϕ2 ϕ1 (a1 ) = −ϕ2 ω(a2 ) = −βψ(a2 ). The first equality follows from exactness, the third because the diagram is commutative. Hence ψ(a2 ) ∈ Ker(β). As a2 = π(a1 ,a2 ) the claim follows. (2) We now show the converse inclusion. Let a ∈ Ker(β). Then certainly a ∈ Ker(ϕ3 β) = Im(ψ). Hence there exists a b ∈ Os with ψ(b) = a. Then 0 = β(a) = βψ(b) = ϕ2 ω(b). Thus ω(b) ∈ Ker(ϕ2 ) = Im(ϕ1 ). Hence there exists a c ∈ Os with ϕ1 (c) = ω(b). But then (−c,b) ∈ K , and thus a = ψπ(−c,b) ∈ ψπ(K). 2

The proof shows that we only need that S2 , S3 are coherent, that S1 is of finite type, and that S5 is of relation finite typ.

6.2 Fundamental Properties of Coherent Sheaves

235

Lemma 6.2.11. Consider a short exact sequence of sheaves of OX -modules: φ

ψ

0 −→ S1 −→ S2 −→ S3 −→ 0. Then all sheaves S1 ,S2 ,S3 are coherent if two of them are coherent. Corollary 6.2.12. (1) The direct sum of finitely many coherent sheaves is coherent. (2) Let α : S −→ T be a morphism of coherent sheaves. Then the sheaves Im(α), Ker(α) and Coker(α) are coherent. (3) Let S and T be coherent subsheaves of a coherent sheaf U . Then the sum S + T and the intersection S ∩ T are coherent sheaves. (4) Suppose that the structure sheaf OX is coherent. Then S is coherent, if and only if for every x ∈ X there exists a small neighborhood U of x and an exact sequence: q p OX|U −→ OX|U −→ S|U −→ 0.

(5) If S and T are coherent, then S ⊗ T is coherent. Proof. Exercise 6.2.17 It is somewhat more tricky to show that the sheaf Hom of coherent sheaves is coherent: Lemma 6.2.13. Let S and T be coherent sheaves. (1) The natural map HomO (S ,T )x −→ HomOx (Sx ,Tx ) is bijective. (2) HomO (S ,T ) is coherent. Proof. (1) The injectivity is easy: let ϕ : S|U −→ T|U be a map and assume that Ker(ϕx ) = Sx then Theorem 6.2.2 implies that Ker(ϕ)|V = S|V for an open neighborhood of x. This implies that ϕ is the zero map in HomO (S ,T )x . To show that the map is surjective, we have to lift any map φx between germs to some open neighborhood p of x. This is easy for the case that S = OX , because we then just have to take a lift in T of the images of the basis elements. Because S is finitely generated, p we can find a map ψ : O|U −→ S|U . By the remark made above, we can also find p a lift of the map ψx φx to χ : O|U −→ T|U , after possibly shrinking U . We have Ker ψx ⊂ Ker χx . Because S is of relation finite type, Ker ψ is finitely generated, and hence Ker ψU ⊂ Ker χU after possibly shrinking U . Therefore, χ induces the p desired map χ : S|U = O|U / Ker ψU −→ T|U . Note that in this proof we only made use of the coherence of S .

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6 The Principle of Conservation of Number

(2) It follows directly from the definition of coherence, that if S is coherent we (locally) have an exact sequence Oq −→ Op −→ S −→ 0. We want to show that 0 −→ HomO (S ,T ) −→ HomO (Op ,T ) −→ HomO (Oq ,T )

(∗)

is exact. This suffices, because HomO (Op ,T ) is isomorphic to T p , which is coherent by assumption. As a kernel of a morphism between coherent sheaves is coherent again, it then follows that HomO (S ,T ) is coherent. To prove the exactness of (∗), look at the following commutative diagram: 0

// HomO (S ,T )x

// HomO (Op ,T )x

// HomO (Oq ,T )x

0

 // HomOx (Sx ,Tx )

 // HomOx (Oxp ,Tx )

 // HomOx (Oxq ,Tx ).

The sequence at the bottom is exact, because that is just a statement about rings and modules. The left vertical arrow is an isomorphism by the first part. Because, as already remarked, HomO (Op ,T ) is isomorphic to T p , the fact that the other vertical arrows are isomorphisms is trivial. Therefore, the upper row is exact. As this is true for all x ∈ X, the claim follows. The following now follows easily: Lemma 6.2.14. Suppose that the structure sheaf is coherent. Let S be coherent, and suppose that P and Q are coherent submodules of S . Then the sheaves Ann(S ) and the transporter sheaf (Q : P) are coherent. Proof. The annihilator sheaf is the kernel of the morphism O −→ H omO (S ,S ) by sending an element to multiplication with this element. Therefore, by the previous result, the Annihilator sheaf is coherent. The other result is Exercise 6.2.21. The final basic property of coherent sheaves is the following extension principle: Theorem 6.2.15 (Extension Principle). Let (Y,OY ) be a closed subspace of the complex space (X,OX ), i : Y −→ X the inclusion. Suppose that the sheaf OX is coherent. Let S be a sheaf of OY –modules. Then S is coherent, if and only if i∗ S is coherent. In particular,it follows that OY is coherent. Proof. Step 1. We first prove that S is of finite type if and only if i∗ S is of finite type. Let i∗ OY = OX /I , where I is a finitely generated (and hence coherent) OX –ideal sheaf. Suppose that S is a finitely generated sheaf of OY –modules. Then we have a surjective map OVp −→ S|V for some p ∈ N, and an open set V in Y . Then V = Y ∩ U , for some open set U in X. As OU ։ i∗ OV = OU /I|U is surjective, it follows that the induced p map OU −→ i∗ S is surjective. p On the other hand, if OU −→ i∗ S is surjective, the fact that i∗ S is an i∗ OV = OU /I|U –module implies that the map induces a surjective map OVp −→ S .

6.2 Fundamental Properties of Coherent Sheaves

237

Step 2. Suppose that i∗ S is OX –coherent. By Step 1, it remains to show that S is of relation finite type. Therefore, consider an open set V in Y , and a map of OV –modules ϕ : OVp −→ S|V . Now V = U ∩Y for some open set U in X. Let π : OU ։ i∗ OV be the canonical surjection. p The composed map ϕ ◦ π : OU −→ (i∗ S )|U of OU –modules has a finitely generated
kernel, because we assumed that i∗ S is OX –coherent. But as i∗ Ker(ϕ) = π Ker(ϕ ◦ π) it follows that Ker(ϕ) is finitely generated. So, in particular, as i∗ OY as a cokernel of a coherent sheaf by a finitely generated ideal sheaf is coherent, it follows that OY itself is coherent. Step 3. Now suppose that S is OY –coherent. It follows that we can write S locally as a cokernel. Thus for all points p ∈ Y there exists an open neighborhood V of p in Y and an exact sequence OVq −→ OVp −→ S −→ 0. Again we write V = U ∩Y for an open set U in X. Then i∗ OV is OU –coherent, and therefore i∗ OVq and i∗ OVp are OU –coherent. It follows that i∗ S , as cokernel of two coherent OU –modules, itself is OU –coherent. Lemma 6.2.16. Suppose the structure sheaf OX is coherent and let S be a coherent OX –module. Then the set A = {x ∈ X : Sx is not a free OX,x -module} is a proper analytic subset of X. Proof. Let x ∈ X and U be an open neighborhood such that S|U has a representation ϕ

q P −→ S|U −→ 0. −→ OX|U OX|U

This is possible because of 6.2.12 (4). For y ∈ U , Sy is a free OX,y –module of rank p − r if and only if for the Fitting ideals holds Ir (ϕy ) = OX,y and Ir+1 (ϕy ) = (0), (cf. Proposition 1.3.8). We may assume that
on U , Ir+1 (ϕ) = (0) and Ir (ϕy ) 6= (0) for some y ∈ U , then A ∩ U = V Ir (ϕ) $ U .

Exercises

6.2.17. Prove Corollary 6.2.12. 6.2.18. Prove Lemma 6.2.8. 6.2.19. Let X be an affine variety, and M be a finitely generated K[X]–module. Show that the ` ´ f, defined by M f D(f ) = Mf , is a coherent OX –module. associated sheaf M

6.2.20. Assume that OX is coherent. Prove that the support of a coherent sheaf S is an analytic set, by using the exact sequence M

q p O|U −→ O|U −→ S|U −→ 0

and looking at minors of M . 6.2.21. Prove the second assertion of 6.2.14.

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6.3

6 The Principle of Conservation of Number

The Four Basic Coherence Theorems

In this section we prove the four basic Coherence Theorems in complex analysis. Theorem 6.3.1 (Oka). The sheaf OC n is coherent. More generally, for any complex space X, the sheaf OX is coherent. Proof. Step 1. A sheaf S is coherent, if for every point there exists an open neighborhood of this point such that S|U is coherent. Thus to show that OX is coherent, it suffices to show this for a complex model space (D,OD ). Such a complex model space injects, by definition, into an open subset of C n . By the extension principle 6.2.15 it suffices therefore to prove the coherence of OC n . This will be done by induction on n. The case n = 0 is trivial, see the second example of ??. Step 2. For a short notation we write On instead of OC n and, since there is no possible confusion, OU instead of OC n |U for an open subset U ⊂ C n . So assume that On−1 is coherent. The statement is local, so we look at a neighborhood of a point p ∈ C n . Without loss of generality we may assume p = 0. As the sheaf On is obviously of finite type, we only have to show that it is of relation finite type. Therefore, we have to show that for any map of sheaves: F =(f1 ,...,fs )

Ons −−−−−−−−→ On the kernel Ker(F ) is ofP finite type. Finding elements in Ker(F ) means finding relations, s that is, r1 , . . . ,rs with i=1 ri fi = 0. After a linear change of coordinates (if necessary), we may assume that f1 , . . . ,fs are regular in xn . As it is easy to relate the relations between f1 , . . . ,fs and u1 f1 , . . . ,us fs for units ui , we may, by applying the Weierstraß Preparation Theorem, assume that the fi are Weierstraß polynomials in xn . Let di be the degree of fi in xn , and put d = max{di }. We may assume, without loss of generality, that the degree of fs is d. For any k we denote by On−1 [xn ]≤k the free On−1 –module of rank k +1 of polynomials in xn of degree less than or equal to k. By induction, On−1 is coherent and, therefore, the free On−1 –module On−1 [xn ]≤k is also coherent for all k by 6.2.12. By definition of coherence, the kernel of the following map of On−1 –modules: (On−1 [xn ]≤d )s −→ On−1 [xn ]≤2d

(a1 , . . . ,as ) 7−→ a1 f1 + . . . + as fs .

is finitely generated. This exactly means that Ker(F ) ∩ (On−1 [xn ]≤d )s is a finitely generated On−1 –module. Step 3. It remains to show that in some small open neighborhood of 0, the sheaf of s relations Ker(F ) is as an On –module generated by Ker(F ) ∩ (On−1 [xn ]≤d ) . This means s the following: Take generators h1 , . . . ,hl of Ker(F ) ∩ On−1 [xn ]≤d . Take a small open neighborhood U of 0 on which the h1 , . . . ,hl and the f1 , . . . ,fs converge. Then the germs h1,p , . . . hl,p of the hi in p generate the kernel of the map Fp for all p ∈ U . To prove this, first note that elements of the following type (so-called Koszul relations): (−fs ,0, . . . ,0,f1 ), . . . , (0, . . . ,0, − fs ,fs−1 )

6.3 The Four Basic Coherence Theorems

239

are in Ker(F ) ∩ On−1 [xn ]s≤d . Indeed, they are in Ker(F ), and also of degree less than or equal to d in xn . We take an arbitrary point p = (p1 , . . . ,pn ) ∈ U . At the point p the function fs,p is regular of order d′ ≤ d in xn − pn . Consider any relation between the fi at the point p: a1 f1,p + . . . + as fs,p = 0. For i = 1, . . . ,s − 1 we perform Weierstraß division with respect to fs at p: ai = qi fs,p + ci , Moreover, if we define cs := as +

Ps−1 i=1

deg(ci ) < d′ ≤ d.

fi,p qi we see that

(a1 , . . . ,as ) = (c1 , . . . ,cs ) − (−fs,p ,0, . . . ,0,f1,p )q1 − . . . − (0, . . . ,0, − fs,p ,fs−1,p )qs−1 . From this it follows that c1 f1,p + . . . + cs fs,p = 0 and, therefore, we expressed our relation as an On –linear combination of elements of Ker(F ) ∩ On−1 [xn ]s≤d , as soon as we have proved that cs is a polynomial of P degree less than or equal to d. This is easy, because we s−1 have just proved that cs fs,p = − i=1 ci fi,p . The right-hand side is a polynomial in xn ′ of degree less than d + d . As we have assumed that the degree of fs is d, it follows that cs has degree at most d′ ≤ d. This concludes the proof of Oka’s Theorem. Theorem 6.3.2 (Oka-Cartan). Let X ⊂ U ⊂ C n be an analytic set. Then the ideal sheaf I := I (X) of X is coherent. Remark 6.3.3. Let
the ideal I = (f1 , . . . ,fs ) ⊂ C {x1 , . . . ,xn } be a radical ideal , and let (X, 0) = V (I), 0 . Take small enough representatives. Then it is a tautology that for all a ∈ X V (f1,a , . . . fs,a ) = (X,a) but it is not clear that the ideal of (X,a) is generated by the fi,a . To put it in another way, it is not clear a priori that the ideal (f1,a , . . . fs,a ) is a radical ideal in OX,a . That this is indeed the case is the content of the Coherence Theorem of Cartan and Oka. Proof of Theorem 6.3.2. Step 1. Consider a point x ∈ X. Without loss of generality, we may assume that x = 0. Consider the ideal I0 ⊂ On,0 = C {x1 , . . . ,xn } of the germ (X, 0). As this ring is Noetherian, this ideal is finitely generated. We take finitely many generators: I0 = (f1 , . . . ,fs ) and choose a small open neighborhood U of 0 such that all fi converge on U . As a finitely generated subsheaf of the coherent sheaf On the sheaf generated by (f1 , . . . ,fs ) is coherent by Lemma 6.2.8. We have a map of sheaves (f1 , . . . ,fs ) −→ I . This is an isomorphism, if it induces an isomorphism on all stalks. So we only have to show (6.1)

Ip = (f1,p , . . . ,fs,p )

for all points p in a small open neighborhood V of 0. Step 2. We reduce to the case that the germ (X, 0) is irreducible. Indeed, suppose (X, 0) = (X1 , 0) ∪ · · · ∪ (Xr , 0). Take representatives of (Xi , 0), so that X = X1 ∪ . . . ∪ Xr in a small open set U . For the ideal sheaves on U we have

240

6 The Principle of Conservation of Number I (X) = I (X1 ) ∩ . . . ∩ I (Xr ).

Therefore, if we show that the I (Xj ) are coherent it follows that I (X) is coherent by by 6.2.12. Step 3. Thus we may suppose that (X, 0) is irreducible. By Remark 3.4.16, as a consequence of the Local Parametrization Theorem, there exists a representative X of (X, 0) and a hypersurface D = V (∆) ⊂ X such that for p ∈ X \ D, Ip = (f1 , . . . ,fs )p . Hence (6.1) holds for all points p outside a proper analytic subset. We now consider the following ideal sheaf on U : J := (f1 , . . . ,fs ) : (∆) ⊂ On .

This is a coherent sheaf. Indeed, since (f1 , . . . ,fs ) is a finitely generated subsheaf of the coherent sheaf On , it is coherent by 6.2.8. Hence J is coherent by 6.2.14. Therefore, we can take finitely many generators g1 , . . . ,gt for J . Looking at the stalk at zero, these elements have the property that ∆0 gi,0 ∈ (f1 , . . . ,fs ) = I0 . But this ideal was supposed to be prime, hence gi,0 ∈ (f1 , . . . ,fs ) ⊂ On,0 . By taking U smaller, we may assume that gi,p ∈ (f1,p , . . . ,fs,p ) for all p ∈ U . Hence we reduced to the case that we have the following equality of ideal sheaves: (6.2)

(f1 , . . . ,fs ) : (∆) = (f1 , . . . ,fs ).

Step 4. Take a point p in U and an element g ∈ Ip , the stalk of the ideal sheaf of X at p. Take a representative g vanishing on V ∩ X, for a (very small) neighborhood V ⊂ U of p. We have to show g ∈ (f1,p , . . . ,fs,p ). U

V

p

On V , the sheaf S := (f1 , . . . ,fs ) : (g) is coherent by 6.3.1, 6.2.8 and 6.2.14. Hence, S has finitely many generators, say h1 , . . . ,hu , for holomorphic functions hi defined on V . Since (6.1) was true for all points p ∈ X \ D, it follows that gp ∈ (f1,p , . . . ,fs,p ) for all p ∈ V \ D, so that we deduce Sp = On,p for all p ∈ V \ D. Thus, the zero set V (h1 , . . . ,hu ) ⊂ D ∩ V . As ∆ also vanishes on D, it follows from the Nullstellensatz that there exists an r ≥ 0 such that ∆r ∈ (h1,p , . . . ,hu,p ) for all p ∈ D ∩ V . From the defining properties of the hi it follows that ∆r g ∈ (f1,p , . . . ,fs,p ). Let r ≥ 0 be minimal with this property. To prove the theorem, we have to show that r = 0. Suppose the converse, that is, r > 0. But then we have ∆r−1 g ∈ (f1,p , . . . ,fs,p ) : (∆) = (f1,p , . . . ,fs,p ), by (6.2), giving a contradiction. Corollary 6.3.4. Let X ⊂ C n be an analytic space. The set of singular points Sing(X) of X is an analytic subspace of X.

6.3 The Four Basic Coherence Theorems

241

Proof. Let x ∈ X. By applying Proposition 4.3.1, we may assume that (X,x) is irreducible. Now we may apply the Jacobian Criterion, 4.3.6, and the coherence of ideal sheaves, 6.3.2. Theorem 6.3.5 (Finite Mapping Theorem). Consider a finite mapping π : X −→ Y of complex spaces, and let S be a coherent OX –sheaf. Then π∗ S is a coherent OY –sheaf. Proof. Using the Extension principle (6.2.15) it is enough to prove the theorem for analytic spaces, that is, both X and Y are reduced. Let y ∈ Y be a point. Since coherence is a local property it is enough to prove that (π∗ S )|U is coherent for a sufficiently small neighborhood U of y. We may assume that y = 0. The theorem is proved in several steps.
Step 1. We first consider the case that X is a small representative of (X, 0) = V (f ), 0 , where f ∈ C {x1 , . . . xn } is a Weierstraß polynomial of degree b in xn , Y an open subset of C n−1 , π : X −→ Y the projection on the first n − 1 coordinates, and S = OX . Let V ⊂ C be a small open neighborhood of 0. We choose an open neighborhood U ⊂ C n−1 of 0 such that for all p ∈ U the b zeros (counted with multiplicity) of f (p,xn ) lie in V (continuity of roots, see 3.4.11). We may assume that X = {(p,x) : p ∈ U, f (p,x) = 0} ⊂ U × V and π : X −→ U is induced by the canonical projection. For an open subset W ⊂ U we have a canonical map: b OU (W ) −→ OU (W )[xn ] ⊂ OU×V (W × V ) −→ OX (W × V ) = π∗ OX (W )

(s0 , . . . sb−1 ) 7→

b−1 X

sj xjn .

j=0

We will prove that for all p ∈ U the induced map of the stalks (6.3)

b OU,p −→ (π∗ OX )p

b is coherent by Oka’s is an isomorphism. This will prove this step. Namely, as On−1|U Theorem 6.3.1 , it follows that π∗ OX|(U×V ) is a coherent On−1|U –module. Now let p ∈ U , x1 , . . . ,xt be the zeros of f (p,x Ptn ) and b1 , . . . ,bt their multiplicities. Put pi := (p,xi ) for i = 1, . . . ,t. In particular, i=1 bi = b. Using 6.1.18, we obtain (π∗ OX )p = ⊕ti=1 OX,pi . To prove that (6.3) is surjective, consider an element (g1 , . . . ,gt ) ∈ Pb−1 (π∗ OX )p = ⊕ti=1 OX,pi . By Corollary 3.3.24 there exists an r = j=0 ri xjn , such that r = gi in OX,pi for i = 1, . . . ,t. This shows the surjectivity. The injectivity of (6.3) follows from the uniqueness of the decomposition of 3.3.24.

Step 2. We consider now the situation as in Step 1 but now S is an arbitrary coherent sheaf. Because S is coherent, we have locally an exact sequence: q p OX −→ OX −→ S → 0.

We use Theorem 6.1.17 and obtain an exact sequence: q p π∗ OX −→ π∗ OX −→ π∗ S → 0.

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6 The Principle of Conservation of Number

q p We proved in Step 1 that π∗ OX is coherent. Hence π∗ OX = (π∗ OX )q and π∗ OX are coherent. Therefore, π∗ S as a cokernel is coherent.

Step 3. We now consider the case that i : X ⊂ V (f ) is a closed subspace, f regular in xn , Y an open subset of C n−1 , π ′ : V (f ) −→ Y the projection on the first n − 1 coordinates and π = π ′ ◦ i. Then i∗ S is coherent by the extension principle, and π∗ S = (π ′ ◦ i)∗ S = π∗′ (i∗ S ) is coherent by the previous step. Step 4. We now consider the case that X is a small representative of (V (I), 0) ⊂ (C n , 0), where C {x1 , . . . ,xn }/I is a finitely generated C {x1 , . . . ,xk }–module, Y is a small open neighborhood of 0 in C k , and π : X −→ C k the obvious projection on the first k coordinates. This case follows from Step 3 and induction on n − k. Step 5. We now consider the general case. We can (locally) view any finite mapping as a projection:  i / n X C π′

π

  Y

j



/ C k.

This is the graph construction, see 2.3.14 and Exercise 3.4.43. Here π ′ is the projection on the first k coordinates. By the extension principle π∗ S is coherent if and only if j∗ (π∗ S ) = π∗′ (i∗ S ) is coherent. But the last sheaf is coherent by Step 4. For the proof of the final coherence theorem in this chapter, the following theorem is of crucial importance. Theorem 6.3.6 (Oka). Let X be an analytic space. Then the set of nonnormal points of X forms an analytic subspace of X. In particular, the set of normal points is open in X. Proof. Consider a point 0 ∈ X, and let I ⊂ R := OX,0 be the ideal of the singular locus of X at the point 0. Then Exercise 4.4.20, which is the local version of Theorem 1.5.13, gives: = R ֒→ HomR (I,I) ⇐⇒ R is normal. We now sheafify. We have proved in 6.3.2 that the ideal sheaf I of the singular locus (it is an analytic space) is coherent. It follows from 6.2.13 that the sheaf HomOX (I ,I ) is coherent, and that HomOX (I ,I )p = HomOX,p (Ip ,Ip ) for all p ∈ X. It follows that the set of nonnormal points is given by the support of the sheaf HomOX (I ,I )/OX . As this is the quotient of two coherent sheaves, it is a coherent sheaf and, therefore, its support is an analytic set, see 6.2.5. This proves the theorem. eX Theorem 6.3.7 (Oka). Let (X,OX ) be an analytic space. The normalization sheaf O eX,x at x by definition is the normalization of OX,x . Then the is a sheaf whose stalk O eX is OX –coherent. normalization sheaf O

6.4 The Principle of Conservation of Number

243

Proof. For simplicity of notation, we will assume that X is irreducible. We showed in e x) and a finite map n : (X,e e x) −→ (X,x). 4.4.8 that there is a germ of a normal space (X,e e In particular, there exist functions f1 , . . . ,fs in Q(OX,x ) which generate OX,e e x = OX,x as Ps OX,x –module, that is, OX,e e x = i=1 fi OX,x . e e x) and X be a representative of (X,x) such that X e Let X be a representative of (X,e e and the induced map n : X e −→ X is a normal analytic space, f1 , . . . ,fs are defined on X, is finite. This is possible because the set of normal points is open (6.3.6). Then n∗ OXe is coherent by the Finite Mapping Theorem 6.3.5. P As the fi have a universal denominator s ∼ ∆, multiplication with ∆ gives an isomorphism i=1 fi OX = S , where S is a finitely Ps generated ideal sheaf. Thus i=1 fi OX is coherent. To prove the theorem, it therefore Ps eX . suffices to show that n∗ OXe = Pi=1 fi OX and is equal to the normalization sheaf O s holds as both sheaves are coherent, and we Using ?? the equality n∗ OXe = i=1 fi OX P s can lift the isomorphism on stalks OX,e e x = i=1 fi OX,x to an isomorphism on a small open neighborhood. To show that it is equal to the normalization sheaf, note that the fi satisfy an integral equation (i)

(i)

fiki + a1 fiki −1 + . . . + ak = 0.

(6.4)

(i)

Consider an even smaller open neighborhood U of x, on which the aj converge. Then P eX,a . the equation (6.4) on U . In particular, for all a ∈ U , we have si=1 fi OX,a ⊂ O Pholds s −1 This shows that i=1 fi OX|U is contained in the normalization sheaf. Now let n (a) = e and Theorem 6.1.18 we {p1 , . . . ,pt }. By construction of the germ of the analytic space X have an equality s X i=1

fi OX,a = n∗ (OXe )a = OX,p e 1 ⊕ . . . ⊕ OX,p e t.

e Therefore, the direct All of the rings OX,p for i = 1, . . . ,t are normal by the choice of X. e Ps i eX,a . sum, which is i=1 fi OX,a , is normal and, therefore, equal to the normalization O This proves the theorem.

Exercises

6.3.8. Prove that any compact complex subspace of a domain in C n consists of a finite number of points. (Hint: Use the Finite Mapping Theorem, and induction on n.)

6.4

The Principle of Conservation of Number

To illustrate the principle of conservation of number, let us formulate three theorems, whose proofs will be given later in this Section. Theorem 6.4.1. Let (C, 0) and (D, 0) be two plane curve singularities, given by f (x,y) = 0 and g(x,y) = 0, respectively. Suppose that (C, 0) and (D, 0) do not have components in common. Consider deformations of f and g, that is, F (x,y,s) and G(x,y,s) in C {x,y,s} with F (x,y,0) = f and G(x,y,0) = g. We can choose for all sufficiently small open neighborhoods U of 0 in C 2 , an open neighborhood V of 0 in C 2 such that

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6 The Principle of Conservation of Number

(1) both F and G converge on U × V , (2) if C resp. D are representatives of (C, 0) respectively (D, 0), then C and D intersect only in 0 and C \ {0} and D \ {0} are smooth. For s fixed, denote by Cs , respectively Ds the curve {F (x,y,s) = 0} ⊂ U , respectively {G(x,y,s) = 0} ⊂ U . Then, for all U sufficiently small there exists a V such that for all s∈V X (C, 0) · (D, 0) = (Cs ,p) · (Ds ,p). p∈Cs

Here (Cs ,p) · (Ds ,p) denotes the intersection number of Cs and Ds in the point p, see 5.1.4. This theorem gives a geometric interpretation of the intersection number. Suppose that Cs and Ds intersect transversally, that is, the intersection number (Cs · Ds )p = 1 for all p ∈ Cs ∩ Ds . The theorem then says, that the intersection number of C and D is equal to the number of intersection points appearing after a small perturbation of C and D. Example 6.4.2. Let C be given by f (x,y) = y 2 − x3 = 0, and D be given by g(x,y) = y = 0. Then (C, 0) · (D, 0) = dimC C {x,y}/(y 2 − x3 ,y) = dimC C {x}/(x3 ) = 3.

We consider the small perturbations: F (x,y,s) = y 2 − x2 (x − s),

G(x,y,s) = y − s.

We see indeed three intersection points appearing. Theorem 6.4.3. Let (C, 0) = ∪ri=1 (Ci , 0) be a germ of a plane curve singularity, and let e 0) = ∪r (C ei , 0) −→ (C, 0) fi = 0 be the equation of Ci . Consider the normalization (C, i=1 of (C, 0). Let W ⊂ C be an open neighborhood of 0 and (x(ti ),y(ti )) : W −→ C 2 for i = 1, . . . ,r

6.4 The Principle of Conservation of Number

245

be a parametrization of the branch Ci , and consider a one-parameter deformation of the parametrization with base space S ⊂ C (x(ti ,s),y(ti ,s)) : W × S −→ C 2 ,

x(ti ,0) = x(ti ),

y(ti ,0) = y(ti )

for i = 1, . . . ,r.

Then we can take W and S so small that the image of (x(ti ,s),y(ti ,s)) is Q an analytic space r in W × S given by Fi (x,y,s) = 0 with Fi (x,y,0)Q = fi . Then F (x,y,s) = i=1 Fi (x,y,s) is r the induced one-parameter deformation of f = i=1 fi . Moreover for all sufficiently small neighborhoods U of 0 in W , there exists an open neighborhood V of 0 in S such that: (1) F converges on U × V . (2) The only singular point of f on U is 0. Denote for fixed s by Cs the curve {F (x,y,s) = 0} ∩ U , then for all s ∈ V we have X δ(C) = δ(Cs ,p), p∈Cs

where δ(Cs ,p) denotes the δ–invariant of Cs in the point p. Example 6.4.4. We consider the previous example. We consider the curve singularity (C ∪ D, 0) given by the equation y(y 2 − x3 ) = 0. The deformation in fact is given by the following deformation of the parametrization: x(t1 ,s) = t21 + s, y(t1 ,s) = t31 + st1 ;

x(t2 ) = t2 , y(t2 ) = s,

(check this). As the δ–invariant of an A1 –singularity is one, we see from the picture

that δ(C ∪ D, 0) = 4. Theorem 6.4.5. Let f = f (x1 , . . . ,xn ) : (C n , 0) −→ (C , 0) be a germ of a holomorphic function with isolated singularity, µ(f ) its Milnor number. Consider a deformation of f : F (x1 , . . . ,xn ,s) : (C n × S, 0) −→ (C × S, 0),

F (x1 , . . . ,xn ,0) = f (x1 , . . . ,xn ).

Then for all sufficiently small open neighborhoods U of 0 there exists an open neighborhood V of 0 such that for all s ∈ V : X µ(f ) = µ(fs ,p) p∈U

where µ(fs ,p) is the Milnor number of the germ of the holomorphic function: F (x1 , . . . ,xn ,s) = fs : (C n ,p) −→ (C ,fs (p))

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6 The Principle of Conservation of Number

Example 6.4.6. We consider the E7 –singularity given by f (x,y) = y(y 2 − x3 ), and the deformation F (x,y,s) = (y − s)(y 2 − x2 (x − s)). The zero sets of f and fs are given in the previous figures. In the picture of the zero set we see four A1 –singularities. But since a function on a compact set must have a maximum and a minimum, we have that in each compact region in the real plane bounded by the zero set of fs there must be a singularity, too. One can in fact calculate that fs has seven A1 –singularities, so that µ(f ) = 7.

How to prove that “an invariant is constant under deformation”, or, to put it in another way, when do we have conservation of number? Usually, a numerical invariant is defined by the vector space dimension of a certain module M over the local ring of a singularity OX,0 . In the above theorems, we have respectively as local ring and as module: (1) The local ring is OX,0 = C {x,y}/(f,g), the module M is OX,0 . (2) The local ring is OC,0 , the module M is OC, e 0 /OC,0 .

∂f ∂f , . . . , ∂x ) (3) The local ring is On , and the module M is On /J(f ), where J(f ) = ( ∂x 1 n is the Jacobian ideal.

When deforming, the local ring OX,0 will be a quotient of the local ring OXS ,0 , and similarly M is a factor module of MS . Both the local ring and the module are in fact also C {s}–modules. In the above examples: (1) OXS ,0 = C {x,y,s}/(F,G), MS = OXS ,0 . eCS,0 /OCS ,0 . (2) The local ring is OCS ,0 = C {x,y,s}/F , the module MS is O

(3) The local ring is C {x1 , . . . ,xn ,s}, and the module is MS = C {x1 , . . . ,xn ,s}/J(F ), ∂F ∂F where J(F ) = ( ∂x , . . . , ∂x ) is the “relative” Jacobian ideal. 1 n In order to be able to compare the invariants at different points in the parameter space S, we need to sheafify. Let the singularity (X, 0) be embedded in (C n , 0), and let U be a sufficiently small open neighborhood of 0. Then we consider a representative X of (X, 0) in U . Furthermore, we choose a small open neighborhood V of 0 in the parameter space S = C . Let π : U × V −→ V be the canonical projection and XS ⊂ U × V be a complex space such that XS ∩ π −1 (0) = X. For s ∈ V we put Xs := XS ∩ π −1 (s).

6.4 The Principle of Conservation of Number

247

U X0

V

Xs

0

s

Theorem 6.4.7 (Principle of Conservation of Number). Let M be a sheaf on XS with the following properties: (1) M is a coherent OXS –module.3 (2) M0 , the stalk of M at 0, is a free C {s}–module of finite rank. Then for all sufficiently small neighborhoods U ′ ⊂ U of 0 in C n there exists an open neighborhood V ′ ⊂ V of 0 in C such that for all s ∈ V ′ : X

dimC (M|X0 )0 = dimC (M|Xs )p , p∈Xs′

where Xs′ = Xs ∩ (U ′ × {s}). Proof. As support of a coherent analytic sheaf, Z := Supp(M ) is an analytic space, see 6.2.5. We claim that the natural projection map f : (Z, 0) −→ (S, 0) is finite. This we prove in three steps. Step 1. We give Z the structure of a complex space by means of the ideal sheaf Ann(M ). This means that OZ = OXS /Ann(M ). By Theorem 6.2.15 M is a coherent OZ –module. From the second assumption it follows that M0 /mS,0M0 is a finite-dimensional vector space. Step 2. By construction AnnOZ,0 (M 0 ) = (0). We claim I := AnnOZ,0 /mS,0 OZ,0 (M0 /mS,0 M0 ) ⊂

p (0),

that is, I is a nilpotent ideal. To show this, let M0 as C {s}–module be generated by m1 , . . . ,mr . Let P f ∈ OZ,0 be a representative of an element in I, so that f M0 ⊂ mS,0 M0 , r that is, f mi = j=1 ξij mj , i = 1, . . . ,r for suitable ξij ∈ mS,0 . By Cramer’s rule 1.2.8,
we get that det (ξij )
− f Idr mk = 0 for k = 1, . . . ,r. As AnnOZ,0 (M0 ) = (0), it follows that det (ξij ) − f Idr = 0. This says that f r ∈ mS,0 OZ,0 . 3

The invariants we are interested in are the dimensions of the stalks of the restriction of M to the fibers.

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6 The Principle of Conservation of Number

Step 3. Put R = OZ,0 /mS,0 OZ,0 and M = M0 /mS,0 M0 . Thus M is a finite-dimensional C –vector space, and an R/I algebra. As R/I–algebra, the annihilator of M is zero. It follows from Exercise 1.2.31 that R/I is a finite-dimensional vector space, hence R/I p has dimension zero. On the other hand, because I ⊂ (0) the dimensions dim(R) and dim(R/I) are equal, hence dim(R) is zero. This implies that dimC (R) < ∞ because R is a factor ring of a convergent power series ring. By Theorem 3.4.24 it follows that f : (Z, 0) −→ (S, 0) is a finite map, as claimed.

We now finish the proof. We take a representative f : Z −→ S such that f −1 (0) = {0}, and such that f is finite. From the Finite Mapping Theorem 6.3.5 it follows that f∗ M is coherent, whose stalk at 0 by assumption is C {s}t for a suitable t. From the coherence of f∗ M it follows that there exists an open neighborhood V ′ of 0 in S such that f∗ M|V ′ ∼ = C {s}t = OCt |V ′ (consider the map OCt |V ′ −→ f∗ M|V ′ induced by (f∗ M ) 0 ∼ ′ on a neighborhood V ⊂ S and use Lemma ??). Take a point s0 ∈ V ′ , and let q1 , . . . ,qℓ ∈ Z be the points in the fiber of f . Because f is finite, one has by Theorem 6.1.18: C {s}t ∼ = (f∗ M )s0 = ⊕ℓi=1 Mqi . Restricting this to the point s0 ∈ V ′ and applying Exercise 6.1.35 implies that Ct ∼ = ⊕ℓi=1 (M|Xs0 )qi . This proves the theorem. Proofs of Theorems 6.4.1, 6.4.3 and 6.4.5. Proof of 6.4.1. We have to show that we can apply the previous theorem to the sheaf M := OU×V /(F,G). By the results of the previous section we know that M is coherent and, therefore, the first assumption is satisfied. So we have to check the second condition. Therefore, consider the stalk of M at 0 which is the ring C {x,y,s}/(F,G). This is a C {s}–module. By assumption (C, 0) and (D, 0) have no component in common, so that by the Nullstellensatz 3.4.4 the vector space C {x,y}/(f,g) is finite-dimensional. By the general Weierstraß Division Theorem 3.2.10 it follows that C {x,y,s}/(F,G) is a finitely generated C {s}–module. To prove that it is free, we have to show, by Theorem 1.3.9, that it is torsion free, hence it remains to show that s is a nonzerodivisor. As (C, 0) ∩ (D, 0) consists of one point by assumption, it follows from 4.1.20 that f, g is a regular sequence in C {x,y}, that is, g is a nonzerodivisor of C {x,y}/(f ). From Exercise 4.1.28 it follows that if a · f + b · g = 0, then a = Lg and b = −Lf for some element L in C {x,y}. Now suppose that s·Q ∈ (F,G). We have to show that Q ∈ (F,G). We can write s·Q = A·F +B·G for some A,B ∈ C {x,y,s}. We put a = A(x,y,0) and b = B(x,y,0). By plugging in it follows that af + bg = 0, so that a = Lg, b = −Lf . We now add 0 = (−LG)F + (LF )G to the expression for sQ and get s · Q = (A − LG)F + (B + LF )G.

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The power series A − LG and B + LG are divisible by s so that Q=

A − LG B + LF F+ G. s s

Hence Q ∈ (F,G) as we wanted to show. Proof of 6.4.3. As above, we have to show that the conditions of Theorem 6.4.7 are satisfied. For notational convenience we will only consider the case that the curve C is irreducible. So we have a map W × S −→ C 2 × S

(t,s) 7→ (x(t,s),y(t,s),s).

Let CS = {F = 0} be the image. F is a generator of the kernel of the map C {x,y,s} −→ C {t,s}, x 7→ x(t,s), y 7→ y(t,s), s 7→ s. In particular F is irreducible. Let m = (x,y,s)
be the maximal ideal in C {x,y,s}/(F ). Then C {t,s}/mC {t,s} = C {t}/ x(t,0),y(t,0) C {t} is a finite-dimensional C –vector space. This implies, using the General Weierstraß Division Theorem 3.2.10, that C {t,s} is a finitely generated C {x,y,s}/(F )–module. Using Theorem 3.4.24 we may assume (after perhaps shrinking W and S) that the map n : W × S −→ CS is finite. Using Theorem 6.3.5 we obtain that n∗ OW ×S is a coherent OCS – module and, therefore, M := n∗ OW ×S /OCS is a coherent OCS –module, too. The first condition of Theorem 6.4.7 is therefore fulfilled. The stalk at 0 is M 0 = OW ×S,0 /OCS ,0 = C {t,s}/C {x(t,s),y(t,s),s}. Because M 0 /(s)M 0 = C {t}/C {x(t,0),y(t,0)} is a finitedimensional C –vector space, we can apply Corollary 3.2.14, and conclude that M0 is a finitely generated C {s}–module. We now show that s is a nonzerodivisor of OW ×S,0 /OCS,0 . To show this, we claim that F (x,y,0) is a defining equation for the curve C, that is, up to multiplication with a unit it is equal to f . Now we have the following isomorphisms:

C {x,y}/ F (x,y,0),x ∼ = C {x,y,s}/(F,x,s) ∼ = C {x(t,s),y(t,s),s}/ x(t,s),s
∼ = C {x(t),y(t)}/ x(t) ∼ = C {x,y}/(f,x). This implies that both f and F (x,y,0) are y–regular of the same order. So the fact that f is up to a unit equal to F (x,y,0) now follows from 3.3.16. Now suppose s · Q ∈ OCS,0 for some Q ∈ OC ×S,0 . We have to show that Q ∈ OCS,0 . We can write s · Q = G(x(t,s),y(t,s),s) for some G ∈ C {x,y,s}. Plugging in s = 0 yields that G(x(t),y(t),0) = 0, so that G(x,y,0) is divisible by f , say G(x,y,0) = Lf . We may now, without loss of generality, assume that F (x,y,0) is equal to f . Now add 0 = −LF to the expression for s · Q. Then s · Q = (G − LF )(x(t,s),y(t,s),s). In the ring C {x,y,s} we can divide G − LF by s, so that Q = G−LF (x(t,s),y(t,s),s). s Hence Q is already an element of OCS,0 , so that s is a nonzerodivisor, as was to be shown. eC ,e /OCs ,p for s ∈ S and p ∈ Cs . Looking at the It remains to show that (M|Cs )p = O s p integral equations for x(t,s) and y(t,s) for fixed s it follows that the map n : W × {s} −→ Cs is finite. As we just proved that in particular OW ×{s} /OCs is a finite-dimensional vector space it follows from Exercise 4.4.22 that n : W × {s} −→ Cs is the normalization for all points in Cs . This is what we needed to show. Proof of 6.4.5. This is analogous to the proof of 6.4.1, and left as Exercise 6.4.11.

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We may ask ourselves whether, for example, in the case of the Milnor number, a deformation with only A1 –singularities always exists. This question is answered in the following theorem. Theorem 6.4.8. Consider a function f ∈ C {x1 , . . . ,xn } with an isolated singularity. Let F : (C n × C n , 0) −→ (C , 0) be defined by F (x1 , . . . ,xn ,t1 , . . . ,tn ) := f + t1 x1 + . . . + tn xn . Then there exists a hypersurface (D, 0) ⊂ (C n , 0), such that for all small enough representatives D and T of (D, 0) and (C n , 0), there exists a small neighborhood U of 0 in C n such that for all t ∈ T \ D Ft (x1 , . . . ,xn ,t1 , . . . ,tn ) : U −→ C has only A1 –singularities. The number of these A1 –singularities is equal to the Milnor number µ(f ) of f . Proof. Let U be a neighborhood of 0 in C n such that f converges on U and let t ∈ C n . A point p ∈ U is a singular point of Ft exactly if all partial
derivatives with respect to ∂Ft t the xi vanish. In short, if p is in the zero set of ∂F ∂x1 , . . . , ∂xn . We therefore consider the n n space (Σ, 0) in (C × C , 0) defined by the ideal     ∂F ∂f ∂f ∂F = ,..., + t1 , . . . , + tn . ∂x1 ∂xn ∂x1 ∂xn Let Σ be a representative of (Σ, 0), We want a condition for a point (p,t) in Σ to have an A1 –singularity. For this, we look at the projection to the second factor π : Σ −→ C n . Suppose that given (p,t) ∈ Σ, there exists a neighborhood of (p,t) which is mapped biholomorphically under π onto an open subset T of C n . Then we claim that Ft has an A1 –singularity at p. Indeed, by assumption we have that the inclusion of stalks OT,t ⊂ OΣ,(p,t) is a isomorphism. Dividing out the local parameters at t, we have   ∂Ft ∂Ft ∼ C = On,p / (p), . . . , (p) . ∂x1 ∂xn Hence Ft has Milnor number one at p. This implies that it is an A1 –singularity at p. We therefore have to investigate the map π : Σ −→ C n . We first look at the space (Σ, 0). In the local ring OΣ,0 of this space, we can eliminate the functions t1 , . . . ,tn . Hence OΣ,0 ∼ = C {x1 , . . . ,xn }, so we see that (Σ, 0) is smooth. The local ring of the fiber of π above 0 ∈ C n is given by OC n ×C n ,0 /

  ∂f  ∂f ∂f ∂f  , + t1 , . . . , + tn ,t1 , . . . ,tn = OC n ,0 / ,..., ∂x1 ∂xn ∂x1 ∂xn

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which is a finite-dimensional vector space by assumption. Therefore, the map π : (Σ, 0) −→ (C n , 0) is finite by 3.2.10 and 3.4.24. As dim(Σ, 0) = dim(C n , 0) and π is closed, it follows that π is surjective. Therefore, π : (Σ, 0) −→ (C n , 0) is a Noether normalization. We proved in 3.4.14 that there exists a hypersurface D ⊂ T , such that for all t ∈ T \ D, there exists a small neighborhood V of t, such that π −1 (V ) is the union of s open subsets in Σ, each of which is mapped biholomorphically on V . Therefore, by the discussion in the beginning of the proof, for all t ∈ T \ D, the function Ft has only (finitely) many A1 –singularities. The fact that this number is equal to µ(f ) is proved by means of conservation of number, see 6.4.5. Example 6.4.9. Let f = x3 . Consider F = x3 + tx. Then Σ is given by ∂F = 3x2 + t = 0. ∂x The map π : Σ −→ T is given by the following picture.

Σ

T So we see that for t 6= 0 we get two A1 –singularities. Corollary 6.4.10. Consider f ∈ On with µ(f ) < ∞, and let u ∈ On be a unit. Then µ(uf ) = µ(f ). In particular, for a germ of an analytic hypersurface (X, 0) with an isolated singularity the Milnor number µ(X, 0) is well defined. Proof. By symmetry, it suffices to show µ(uf ) ≥ µ(f ). Take a small open neighborhood V of 0 in C n . Consider a family Ft ∈ C {x1 , . . . ,xn ,t} with F (x1 , . . . xn ,0) = f . Let U be a small open neighborhood of 0, and suppose that the function Ft for t fixed has exactly µ(f ) A1 –singularities in V . Then uFt is a family of functions with uF (x1 , . . . xn ,0) = uf . If Ft has an A1 –singularity at p, then uFt also has an A1 –singularity at p. From the principle of conservation of number, it follows that µ(uf ) ≥ µ(f ). This is what we wanted to show.

Exercises 6.4.11. Prove 6.4.5. 6.4.12. For plane curve singularities (X, 0) and (Y, 0), prove Hironaka’s formula, see 5.4.11,using the principle of conservation of number, in particular Theorems 6.4.1 and 6.4.3.

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6.4.13. In the Theorem of the Principle of Conservation of Number, replace the second condition by the condition that M0 is a finitely generated C {s}–module, or, what is the same by Nakayama, that (M|X0 )0 is a finitely generated C –vector space. Prove that X ´ ` ´ ` dimC (M|Xs )p . dimC (M|X0 )0 ≥ p∈Xs′

6.4.14. An ordinary m–tuple point is by definition a plane curve singularity (X, 0) = ∪m i=1 (Xi , 0) such that all (Xi , 0) are smooth, and the intersection number (X , 0) · (Xj , 0) = 1 for all i 6= j. i ` ´ Prove that the δ–invariant of an ordinary m–tuple point is m using 2 (1) Hironaka’s formula 5.4.11 or

(2) the principle of conservation of number. Find a C –basis of OX, e 0 /OX,0 .

` ´ 6.4.15. Consider an irreducible plane curve singularity (X, 0) = V (f ), 0 in (C 2 , 0) of multiplicity m. Without loss of generality we may suppose that (X, 0) is given by the parametrization x(t) = tm ,

y(t) = tm ϕ(t).

Consider a deformation of the parametrization as follows x(t,s) = tm + s, y(t,s) = (tm + s)ϕ(t). ` ´ Let F ∈ C {x,y.x} be irreducible with F x(t,s),y(t,s) = 0 and F (x,y,0) = f . (1) Show that one can take f as follows:

f (x,y) = y m + a1 (x)y m−1 + . . . + am (x) = 0, where the order of ai is at least i. (2) Show that F looks like F = ym +

a1 (x − s) m−1 am (x − s) m y x +... + x = 0. x−s (x − s)m

(3) Show that for small s 6= 0 the zero set of fs (x,y) := F (x,y,s) has two singular points: 1. an ordinary m–tuple point.

2. the strict transform of the blowing-up of the origin in C 2 . For which values of t do these occur? (4) Use the principle of conservation of number, in particular 6.4.3, to reprove Noether’s formula 5.4.13. (5) Generalize these results to reducible curves.

6.5

Cohen-Macaulay Spaces

The Cohen-Macaulay property (CM for short) of a space is a local property, related to the principle of conservation of number. In this section we deal with the first properties of Cohen-Macaulay spaces, hopefully showing that it is a good (and, therefore, quite strong) property a space can have. Nevertheless, we will prove that complete intersections (in particular hypersurface singularities) are Cohen-Macaulay. First we introduce the notion of depth, needed in order to define Cohen-Macaulay. We introduce these matters directly for modules over a local ring, in order to get a more flexible development of the theory.

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Definition 6.5.1. Let (R,m) be a local ring and M be an R–module. (1) A sequence f1 , . . . ,fr of elements in m is called a regular sequence of M if f1 is not a zerodivisor of M , and fi is not a zerodivisor of M/(f1 , . . . ,fi−1 )M for i = 2, . . . ,r. (2) Let I ⊂ R be an ideal with IM 6= M . Then the I–depth of M , depth(I,M ) is the maximal length of a regular sequence of M in I. If IM = M we define depth(I,M ) = ∞. (3) The depth of M , depth(M ) is the maximal length of a regular sequence of M , that is, depth(M ) = depth(m,M ). If we want to emphasize the ring R, we will write depthR (M ). (4) The module M is called Cohen-Macaulay, if the depth of M is equal to the dimension of M . The dimension of the module M is defined to be the dimension of R/ Ann(M ). (5) R is called Cohen-Macaulay, if R is a Cohen-Macaulay R–module. (6) A germ (X,x) is called Cohen-Macaulay, if OX,x is Cohen-Macaulay. As usual, the depth of a space at a point is defined as the depth of the corresponding local ring. Note that because of the Active Lemma 4.1.10 we always have the inequality depth(M ) ≤ dim(M ). If IM 6= M then depth(I,M ) ≤ depth(M ). Notice that depth(I,M ) > depth(M ) implies IM = M and, especially for M = R, it implies I = R. Furthermore, it is immediate from the definition that the depth of a module is zero exactly if every element of m is a zerodivisor for M . From 1.4.52 it follows that m is an associated prime ideal of M , so that m = M : f for some f ∈ m. Hence the following lemma follows. Lemma 6.5.2. Suppose that M is an R–module, and that depth(M ) = 0. Then m is an associated prime of M , that is, there exists an f ∈ M \ {0} such that m · f = 0. In case that M = R is the local ring of a germ of a complex space, the depth is zero exactly if R is a zero-dimensional ring, or the special point is an embedded point. More generally, the depth of a germ of a complex space at a point is the number of hypersurface sections one has to take before one gets an embedded point. Note that if a nonzero-dimensional germ of a complex space is reduced, it must have positive depth. The Cohen-Macaulay property, therefore, states that an embedded point will not occur when taking dim(X) successive hypersurface sections. It will turn out that the length of a maximal regular sequence is independent of the maximal regular sequence. Let us, before giving examples, prove this statement, for which we need the following lemma. Lemma 6.5.3. Let R be a Noetherian local ring, M a finitely generated R–module. Let {f,g} be a regular sequence of M . Then {g,f } is also a regular sequence. Proof. The proof will be given in two steps. Step 1. We first show that g is a nonzerodivisor. Suppose ga0 = 0, a0 ∈ M . In particular, ga0 ∈ f · M . The assumption that {f,g} is a regular sequence, allows us to deduce that a0 ∈ f · M . Write a0 = f a1 , a1 ∈ M . Then ga0 = f ga1 = 0, and f is not a zero divisor, so ga1 = 0, etc. Therefore, for all k there exists an ak ∈ M such that a0 = ak f k . Therefore, a0 ∈ ∩mk M which is zero according to Krull’s Intersection Theorem (Exercise 1.3.24).

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Step 2. We have to show that f is a nonzerodivisor of M/g · M . Suppose f a ∈ g · M , that is, f a = gb for some a,b ∈ M . Since by assumption, g is a nonzerodivisor of M/f · M , it follows that b ∈ f · M , that is, b = f c for some c ∈ M . Therefore, f (a − gc) = 0. Because f is not a zero divisor, a − cg = 0, that is, a ∈ g · M . This is what we had to show. Lemma 6.5.4. Let R be a Noetherian local ring, I ⊂ R an ideal and M an R–module. Any two maximal regular sequences in I for M have the same number of elements. Proof. Let f1 , . . . ,fn and g1 , . . . ,gp be regular sequences in I, which cannot be extended. We may assume that n ≥ p. We have to show that p = n. The lemma will be proved by induction on n. The case n = 1 is trivial. We now claim that there exists an f ∈ I such that the sequences f1 , . . . ,fn−1 and g1 , . . . ,gp−1 can be extended to regular sequences f1 , . . . ,fn−1 ,f and g1 , . . . ,gp−1 ,f . Let S be the union of the associated primes of (f1 , . . . ,fn−1 )M and of (g1 , . . . ,gp−1 )M , then, by the characterization of nonzerodivisors, see 1.4.52, the ideal (fn ,gp ) 6⊂ S, because (fn ,gp ) ⊂ S would imply (fn ,gp ) ⊂ p, p one of the associated primes (cf. 1.1.13, Prime Avoidance). But by assumption fn or gp are nonzerodivisors in M/(f1 , . . . ,fn−1 )M , resp. M/(g1 , . . . ,gp−1 )M . This implies that afn + bgp =: f 6∈ S for suitable a,b ∈ R and has, therefore, the required property. Applying the case n = 1 to the ring R/(f1 , . . . ,fn−1 ) gives that the sequence f1 , . . . ,fn−1 ,f is maximal. Similarly g1 , . . . ,gp−1 ,f is a maximal regular sequence. By repeatedly applying the previous lemma, we find that f,f1 , . . . ,fn−1 and f,g1 , . . . ,gp−1 are maximal regular sequences. Applying the induction hypothesis to the ring R/(f ) proves the lemma. Corollary 6.5.5. Let R be a Noetherian local ring, M be an R–module, I ⊂ R an ideal and f ∈ I a nonzerodivisor of M . Then: (1) depth(I,M/f M ) = depth(I,M ) − 1, (2) M is Cohen-Macaulay ⇐⇒ M/f M is Cohen-Macaulay. Proof. The first part follows from Lemma 6.5.4. For the second part, the implication “⇐=” is trivial. For the other implication, we use the first part. It remains to show dim(R/(Ann(M ),f ) = dim(R/ Ann(M ))−1. This holds, as from the assumption it follows easily that f is a nonzerodivisor of Ann(M ), so that we can apply Krull’s Principal Ideal Theorem 4.2.16. Examples 6.5.6. (1) Smooth spaces are Cohen-Macaulay. (2) By applying the second part of Corollary 6.5.5 we obtain that complete intersection singularities (in particular hypersurface singularities) are Cohen-Macaulay. (3) Reduced curve singularities are Cohen-Macaulay. This is also obvious, as the depth is positive and, therefore, only can be one, because the depth cannot be bigger than the dimension. (4) The singularity in three–space given by the ideal (xz,yz)

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is reduced, but not Cohen-Macaulay. An example of a nonzerodivisor is z − x, which cuts down the space to the following space with embedded point.

The local ring of the intersection is C {x,y,z}/(xz,yz,x−z) ∼ = C {x,y}/(x2 ,xy), hence
we see the embedded point appear. The depth of the germ the analytic space V (xz,yz), 0 is one. This space is therefore not Cohen-Macaulay.

(5) Consider C 4 with coordinates x,y,z,w, and consider the union X of the two planes x = y = 0 and z = w = 0, intersecting in one point. The ideal of this union is (x,y) ∩ (z,w) = (xz,xw,yz,yw). It is geometrically clear that if we intersect with a generic threedimensional linear space, we have the union of two lines. But we obtain an embedded point. If we intersect with y − w = 0, the local ring becomes: C {x,y,z,w}/(xz,xw,yz,yw,y − w) ∼ = C {x,y,z}/(xz,xy,yz,y 2). A primary decomposition of (xz,xy,yz,y 2 ) is given by: (xz,xy,yz,y 2 ) = (y,x) ∩ (y,z) ∩ (y 2 ,x,z). So we see an embedded point appearing on the A1 –singularity. The depth is one, and, therefore, (X, 0) is not Cohen-Macaulay. (6) To give an example of an irreducible space which is not Cohen-Macaulay, we consider the map: (C 2 , 0) −→ (C 4 , 0); (s,t) 7→ (x,y,z,u) = (s,st,t2 ,t3 ). By Exercise 2.3.18 the ideal of the image is: (y 2 − zx2 ,yz − ux,z 2 x − uy,u2 − z 3 ). A general hyperplane section is given by x − z = 0. This section gives a space given by the ideal (y 2 − x3 ,x(y − u),x3 − uy,u2 − x3 ). From x(y − u) = 0 it follows that either x = 0 or y = u. From the first possibility, it follows that y = u = 0. The second possibility gives y = u, y 2 = x3 , a cusp singularity. In total we get a cusp singularity with an embedded point. In fact, a primary decomposition of the ideal is: (y − u,y 2 − x3 ) ∩ (x,y 2 ,uy,u2 ). Indeed, we need embedding dimension four to find an example of a reduced irreducible space which is not Cohen-Macaulay, due to the Principal Ideal Theorem and the second example.

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The examples four and five are quite typical. In fact, a Cohen-Macaulay space is equidimensional, even unmixed. For the definition of unmixed, see 4.1.22. In the fifth example, the intersection of the two components has dimension zero. A necessary condition for the union to be Cohen-Macaulay is the condition that the intersection has codimension one, as we will see in 6.5.19. At least from these examples we see that the Cohen-Macaulay property is indeed a nontrivial one. Proposition 6.5.7. Let (R,mR ) and (S,mS ) be local rings, and R ⊂ S be a finite ring extension of local rings. Let M be an S–module. Suppose depthR (M ) = 0. Then depthS (M ) = 0. Proof. As depthR (M ) = 0, we have that m is an associated prime of M . Thus there exists an x ∈ M with x · m = 0. Take an element g ∈ mS . We have to show that g is a zerodivisor of M . By 1.5.11 g is integral over mR , so we have an integral equation: g n + a1 g n−1 + . . . + an = 0, ai ∈ mR . As depthR (M ) = 0, we have ai x = 0 for i = 1, . . . ,n. Hence g n · x = 0. Thus g is a zerodivisor of M . This proposition will be generalized in Exercise 6.5.29. Corollary 6.5.8. (1) Let OX,0 be a singularity, Ok ֒→ OX,0 be a Noether normalization. Then (X, 0) is Cohen-Macaulay if and only if OX,0 is a free Ok –module. (2) A Cohen-Macaulay space is unmixed. Proof. (1) Recall from 1.3.9 that if M is a finitely generated R–module, x ∈ m a nonzerodivisor of M , then M is a free R–module if and only if M/xM is a free R/(x)-module. By 6.5.7 we can find an x ∈ Ok which is a nonzerodivisor of OX,0 . The first statement follows by induction, since Corollary 6.5.5 gives that OX.0 /(x) is also Cohen-Macaulay. (2) We know by the first part, that OX,0 is a free Ok –module, so, in particular, every element of Ok is a nonzerodivisor of OX,0 . Therefore, (X, 0) is unmixed, by Theorem 4.1.22. The second part of this corollary, applied to the case of a complete intersection is usually called Macaulay’s Unmixedness Theorem. Corollary 6.5.9. Let (X, 0) be an irreducible complete intersection singularity, and suppose the singular locus of (X, 0) has codimension at least two. Then (X, 0) is normal. Proof. By assumption, OX,0 satisfies (R1). We only need to show that it satisfies (S2) by 4.4.11. The condition (S2) means that every hypersurface in (X, 0) has no embedded primes. Because of Corollary 6.5.5 this hypersurface is also Cohen-Macaulay, and therefore unmixed. Thus (S2) holds for Cohen-Macaulay singularities. Theorem 6.5.10. Let (X, 0) be Cohen-Macaulay germ of an analytic space and π : (X, 0) −→ (C k , 0) be a Noether normalization. Then there exist representatives X of (X, 0) and U of (C k , 0) and a hypersurface D ⊂ U such that for all t ∈ U \ D the fiber π −1 (t) consists of dimC OX,0 /mC k ,0 OX,0 many points.

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Proof. As a Cohen-Macaulay space is unmixed, in particular equidimensional, we can reduce to the case that (X, 0) is irreducible. As in the proof of 6.4.7 we choose X and U s . This is possible because π∗ OX is coherent and (π∗ OX )0 = OX,0 is such that π∗ OX ∼ = OU a free OC k ,0 –module of finite rank s (Corollary 6.5.8). Let p ∈ U and π −1 (p) = {q1 , . . . ,qr } then s ∼ OU,p = (π∗ OX )p = ⊕ri=1 OX,qi and, therefore, s=

r X i=1

dimC OX,qi /mC k ,p OX,qi .

As in the proof of Theorem 6.4.8, we use the Local Parametrization Theorem 3.4.14 to find a hypersurface D ⊂ U with the following property: every point p ∈ U \ D has a neighborhood V such that π −1 (V ) is a union of open subsets of X each of which is mapped biholomorphically to V . This implies OX,qi ∼ = OU,p for all i and, therefore, r = s. Remark 6.5.11. The previous theorem does not hold if (X, 0) is not Cohen-Macaulay, see the following examples. We can apply the theorem to the theory of intersection number. We consider f,g such that dimC C {x,y}/(f,g)
< ∞, and deformations FS ,GS of f and g, with OS,0 = Ok . Let (X, 0) = V (FS ,GS ), 0 . Then (X, 0) −→ (C k , 0) is a Noether normalization. (X, 0) is a complete intersection, hence Cohen-Macaulay. The number of points in Xs = V (Fs ,Gs ) for general s, therefore, is equal to dimC C {x,y}/(f,g). Examples 6.5.12.
(1) Let I = (xy,x(x − z)). Then (X, 0) = V (I), 0 is the union of a plane and a line. The projection (X, 0) −→ (C 2 , 0) which sends (a,b,c) to (b,c) is a Noether normalization. On the level of rings we have a map C {y,z} ⊂ C {x,y,z}/(xy,x2 − xz). The number of points in a general fiber obviously is one. But

dimC C {x,y,z}/(I + (y,z)) = dimC C {x}/(x2 ) = 2.

Therefore, (X, 0) is not Cohen-Macaulay.

(2) We consider example 6.5.6 (6). So let I
= (y 2 − zx2 ,yz − ux,z 2 x − uy,u2 − z 3 ). We consider the projection π : (X, 0) = V (I), 0 −→ (C 2 , 0), which sends (a,b,c,d) to (a,c). Recall that (X, 0) was given by the image of the map (C 2 , 0) −→ (C 4 , 0) which sends (s,t) to (s,st,t2 ,t3 ). The composition is the map sending (s,t) to (s,t2 ). So we see that (X, 0) −→ (C 2 , 0) is surjective, and in fact in a general fiber of π we have two elements. Since π is surjective we have an inclusion of rings C {x,z} ֒→ C {x,y,z,u}/I. Now C {x,y,z,u}/(I + (x,z)) = C {y,u}/(y 2,uy,u2 ) which is a three-dimensional vector space. By the General Division Theorem, it follows that C {x,y,z,u}/I is a finitely generated C {x,z}–module. Hence π : (X, 0) −→ (C 2 , 0) is a Noether normalization, has two points in a general fiber, but the vector space C {x,y,z,u}/(I + (x,z)) has dimension three. Thus we see that (X, 0) is not Cohen-Macaulay.

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6 The Principle of Conservation of Number

There is also a completely algebraic method by which one can decide whether a ring (or module) is Cohen-Macaulay. It is based on resolutions of modules. This method is often used in the literature for proving that for a certain invariant the principle of conservation of number holds. Definition 6.5.13. Let (R,m) be a Noetherian local ring and M be a finitely generated R–module. A free resolution of M is an exact sequence (infinite of finite) (6.5)

αk+1

α

α

0 1 M −→ 0, F0 −→ · · · −→ Fk+1 −−−→ Fk −→ · · · −→

with finitely generated free R–modules Fi for i ≥ 0. One says that M has finite projective dimension if there exists an exact sequence (6.6)

α

α

α

0 1 n M −→ 0, F0 −→ Fn−1 −→ · · · −→ 0 −→ Fn −→

with finitely generated free R–modules Fi . The minimal n with Fn 6= 0 is called the projective dimension of M , denoted by pd(M ) = pdR (M ). If M does not have finite projective dimension, we write pdR (M ) = ∞. The resolution (6.5) is called minimal, if for all k ≥ 1, the map αk maps Fk into mFk−1 . Theorem 6.5.14. (Minimal) free resolutions of finitely generated modules M over a Noetherian local ring (R,m) exist. If a module M has a finite minimal resolution α

α

α

0 1 n M −→ 0, F0 −→ Fn−1 −→ · · · −→ 0 −→ Fn −→

with Fn 6= (0), then n = pdR (M ). To put it in other words, two minimal resolutions have the same length. Moreover, the rank of Fk is independent of the minimal resolution, and is called the k–th Betti number of M . Proof. Consider a minimal set of generators b1 , . . . ,bs0 of M . We get a surjective map α

0 F0 := Rs0 −→ M

with α0 (a1 , . . . ,as0 ) = a1 b1 + . . . + as0 bs0 . The map α0 is surjective, because b1 , . . . ,bs0 generate M . By Nakayama, α0 induces an isomorphism α0 : F0 /mF0 ∼ = M/mM. Let K1 be the kernel of α0 . Then K1 ⊂ mF0 , as α0 is an isomorphism. As K1 is a submodule of the Noetherian module F0 , it is finitely generated. Hence, there exists a surjective map F1 ∼ = Rs1 −→ K1 −→ 0, where s1 is the minimal number of generators of K1 . We denote the composition F1 −→ K1 −→ F0 by α1 . As K1 ⊂ mF0 , it follows that α1 (F1 ) ⊂ mF0 . We get an exact sequence α

α

0 1 M −→ 0. F0 −→ F1 −→

Now put K2 = Ker(α1 ), take minimal number of generators of K2 etc. Continuing this way, we get a minimal free resolution of M . To show the invariance of the Betti-numbers, consider two minimal resolutions of M:

6.5 Cohen-Macaulay Spaces

259 αk+1

α

α

α′k+1

α′

α′

0 1 M −→ 0, F0 −→ · · · −→ Fk+1 −−−→ Fk −→ · · · −→ 1 0 ′ −−−→ Fk′ −→ · · · −→ F0′ −→ M −→ 0. · · · −→ Fk+1

As both α1 and α′1 both have entries in m, we get isomorphisms F0 /mF0 ∼ = M/mM ∼ = F0′ /mF0′ . Therefore rank(F0 ) = rank(F0′ ). ′ ′ Let e1 , . . . ,es0 be a basis of F0 , and Pf1 , . . .′ ,fs0 be a basis of F0 . As the α0 (fj ) generate M , there exist βij ∈ R with α0 (ei ) = βij α0 (fj ). Consider the map h1 = (βij ) : F0 −→ F0′ . So we get a commutative diagram F0

α0

/M

h1

 F0′

α′0

/ M.

Modulo m the map h1 : F0 /mF0 −→ F0′ /mF0′ is an isomorphism, that is, det(βij ) mod m 6= 0. Then det(βij ) is a unit in R which implies that h1 is an isomorphism. Hence h1 induces an isomorphism h1 : Ker(α0 ) = K1 −→ K1′ = Ker(α′0 ).

As α2 and α′2 have entries in m, and we have surjections F1 −→ K1 and F1′ −→ K1′ , it follows that rank(F1 ) = rank(F1′ ), etc. Examples 6.5.15. (1) An R–module M has projective dimension zero if and only if M is a free R–module. (2) Let R = C {x,y}, I = (x,y), and M = R/I. We will show that (−y (x y) res x ) 0 −→ R −−−→ R2 −−−→ R −−→ R/I −→ 0 is a free resolution of R/I as R–module. Here res is just taking residue classes, and is obviously surjective. The kernel of res is generated by x and y. We have to compute the (x y)

kernel of R2 −−−→ R. So let (α,β) be in the kernel. Then x · α + y · β = 0. As (x,y) is a regular sequence in C {x,y} it follows that α = y · m, for a certain m. Similarly β = x · ℓ for a certain ℓ. Then xy · m + xy · ℓ = 0, hence ℓ = −m. Therefore, (α,β) = ℓ(−y,x). So (−y (x y) x ) R −−−→ R2 −−−→ R

(−y x ) is exact. Moreover, it is easy to see that R −−−→ R2 is injective.

(3) We consider the ring R = C {x,y}/(xy), and M = C {y} ⊕ C {x} = R/(x) ⊕ R/(y). e the normalization of R. M can be described as the cokernel of the map Then M = R, 0 α1 =(x 0 y) R2 −−−−−−→ R2 −→ M −→ 0.

The kernel of α1 is easily seen to be generated by (y,0) and (0,x). Hence we get an exact sequence 0 α1 =(x α2 =(0y x0) 0 y) R2 −−−−−−→ R2 −−−−−−→ R2 −→ M −→ 0. Going on like this, we see that M has a minimal free resolution which is 2-periodic. In particular we see that M does not have finite projective dimension.

260

6 The Principle of Conservation of Number

So one sees that a finitely generated module over a local ring does not, in general, have finite projective dimension. But it is true in the case that R is a regular local ring. The analogous statement for graded modules over a polynomial ring is a famous theorem of Hilbert. Theorem 6.5.16 (Hilbert’s Syzygy Theorem). Let (R,m) be a regular local ring of dimension n, and M be a finitely generated R–module. Then pdR (M ) ≤ n. In particular, the projective dimension of M is finite. For the proof we need the following lemma. Lemma 6.5.17. Let (R,m) be a Noetherian local ring, let M be a finitely generated R– module, and x ∈ m be a nonzerodivisor of R and M . Consider a minimal free resolution of M 0 −→ Fn −→Fn−1 −→ · · · −→F0 −→M −→ 0. Then the induced sequence (6.7)

0 −→ Fn /xFn −→Fn−1 /xFn−1 −→ · · · −→F0 /xF0 −→M/xM −→ 0

is a minimal free resolution of M/xM as R/(x)–module. Moreover, pdR (M ) = pdR/(x) (M/xM ). Proof. By 1.2.35 the sequence (6.7) is indeed exact. Moreover, from Fn /xFn = 0 it would follow that Fn = xFn , hence Fn = 0 by Nakayama. Hence pdR (M ) = pdR/(x) (M/xM ). Proof of Hilbert’s Syzygy Theorem 6.5.16. The proof is by induction on n. If n = 0, then R is a field and M is a finitely generated vector space over R. Hence M is free, and the projective dimension pdR (M ) = 0. Otherwise, consider a minimal free resolution · · · −→ Fk −→ Fk−1 −→ · · · −→ F1 −→ F0 −→ M −→ 0 of M . We split this exact sequence into two exact sequences · · · −→ Fk −→ Fk−1 −→ · · · −→ F1 −→ K −→ 0, 0 −→ K −→ F0 −→ M −→ 0,

with K = Ker(F0 → M ). Take any element x ∈ m \ m2 . As R is an integral domain, see 4.3.17, x is a nonzerodivisor of F0 . As K is a submodule of F0 , x is a nonzerodivisor of K. Therefore x is a nonzerodivisor of Fi for all i ≥ 1 and of K, so that we get an exact sequence · · · −→ Fk /xFk −→ Fk−1 /xFk−1 −→ · · · −→ F1 /xF1 −→ K/xK −→ 0 The ring R is Noetherian, so that K/xK is a finitely generated R/(x)–module. As R/(x) is a regular local ring of dimension n−1, it follows by induction that pdR/(x) (K/xK) ≤ n−1. Hence Fk /xFk = 0 for k ≥ n + 1, because the resolution is minimal. By Nakayama, it follows that Fk = 0 for k ≥ n + 1. This proves the theorem. Lemma 6.5.18 (Depth Lemma). Let R be a Noetherian local ring, and consider a short exact sequence of R–modules 0 −→ M1 −→ M2 −→ M3 −→ 0.

6.5 Cohen-Macaulay Spaces Then

261


depth(M2 ) ≥ min depth(M1 ), depth(M3 ) .

In case this inequality is strict we have, furthermore,

depth(M1 ) = depth(M3 ) + 1. Proof. Suppose that all three modules have positive depth. From Prime Avoidance, see Exercise 6.5.30, we can find an f ∈ R which is a nonzerodivisor of M1 ,M2 and M3 . By the Snake Lemma 1.2.12, it follows that 0 −→ M1 /f M1 −→ M2 /f M2 −→ M3 /f M3 −→ 0 is also exact. Since the depth drops by one if one divides out a nonzerodivisor (Corollary 6.5.5), we can immediately reduce to the case that the depth of one of the three modules M1 ,M2 or M3 is zero. So we distinguish three cases. Case 1. Suppose depth(M1 ) = 0. We will prove that depth(M2 ) = 0. But this is obvious, as any nonzerodivisor of M2 can be viewed as a nonzerodivisor of M1 , as M1 may be viewed as a submodule of M2 . Case 2. Suppose depth(M2 ) = 0. Then we have to show that either depth(M1 ) = 0 or depth(M3 ) = 0. Suppose not, and let (by prime avoidance) f ∈ R be a nonzerodivisor of M1 and M3 . Therefore, the multiplication by f on M1 and M3 has a trivial kernel. From the Snake Lemma it follows that multiplication with f on M2 has trivial kernel, so f is a nonzerodivisor of M2 . But as depth(M2 ) = 0 this cannot be the case, contradiction. Case 3. Suppose depth(M3 ) = 0. We have to show that if, moreover, depth(M2 ) > 0, then depth(M1 ) = 1. Take a nonzerodivisor f of M2 . As M1 is a submodule of M2 it is also a nonzerodivisor of M1 . Hence, depth(M1 ) ≥ 1. By the Snake Lemma we obtain an inclusion 0 −→ M ′ −→ M1 /f M1 ,

(6.8) ·f

where M ′ := Ker(M3 −→ M3 ). As depth(M3 ) = 0, f is a zerodivisor. Thus M ′ 6= 0. One even has that there exists an element g ∈ M3 \ {0} with mg = 0. Hence, in particular, g ∈ M ′ . Any nonzerodivisor of M1 /f M1 would give a nonzerodivisor of M ′ by the inclusion (6.8). But M ′ cannot have one by definition of g ∈ M3 . Therefore, depth(M1 /f M1) = 0, and depth(M1 ) = 1.
Corollary 6.5.19.
Let I,J ⊂ On be radical ideals. Suppose that (X, 0) = V (I), 0 and (Y, 0) = V (J), 0
both have dimension k. Suppose (X, 0) intersects (Y, 0) properly, that is, dim (X ∩ Y ), 0 ≤ k − 1. Suppose (X, 0) and (Y, 0) are Cohen-Macaulay. Then (X ∪ Y, 0) is Cohen-Macaulay ⇐⇒ (X ∩ Y, 0) is Cohen-Macaulay of dimension k − 1.

Proof. Look at the exact sequence: 0 −→ On /I ∩ J −→ On /I ⊕ On /J −→ On /(I + J) −→ 0. From the assumption dim(X ∩ Y, 0) ≤ k − 1 it follows that depth(On /(I + J)) ≤ k − 1. Because (X, 0) and (Y, 0) are Cohen-Macaulay of dimension k, we know that depth(On /I ⊕ On /J) = k. The Depth Lemma gives that depth(On /(I ∩ J)) = depth(On /(I + J)) + 1, which can be interpreted as the statement of the theorem.

262

6 The Principle of Conservation of Number

Theorem 6.5.20 (Auslander-Buchsbaum Formula). Let (R,m) be a Noetherian local ring and M be a finitely generated R–module of finite projective dimension. Then: depth(M ) + pdR (M ) = depth(R). Proof. The proof is by induction on depth(R) and depth(R) − depth(M ). Step 1. Suppose first that depth(R) = 0. Then we have to show that M is free, as pdR (M ) = depth(M ) = 0 follows from it. As depth(R) = 0, there exists a nonzero x ∈ R with x · m = 0. Now let α

αn−1

α

1 n F0 −→ M −→ 0 Fn−1 −−−→ · · · −→ 0 −→ Fn −−→

be a minimal free resolution of M as an R–module. Suppose n > 0. We can view Fn as a submodule of Fn−1 via αn . Because of the minimality, Fn ⊂ mFn−1 . Then xFn = xmFn−1 = 0, hence Fn = 0 because Fn is free, a contradiction. It follows that pdR (M ) = 0, hence M is free. Step 2. Suppose that depth(M ) = depth(R). Then we can find a maximal regular sequence f1 , . . . ,fs for both M and R. As M is free if and only M/xM is free for x an nonzerodivisor, it follows that M is free. Hence pdR (M ) = 0. Step 3. For the induction step we consider two cases. Case 1. Suppose depth(M ) > 0 and depth(R) > 0. This case is done by induction on depth(R). Take an element f ∈ m which is a nonzerodivisor of M and of R. Take a presentation: 0 −→ K −→ F −→ M −→ 0. From the Snake Lemma: 0 −→ K/f · K −→ F/f · F −→ M/f · M −→ 0 is exact. Because the projective dimension stays the same if one divides out a nonzerodivisor, see 6.5.17, and the depth drops by one if one divides out a nonzerodivisor, the theorem follows by induction on depth(R). Case 2. Finally, suppose that depth(M ) = 0 and depth(R) > 0. The case pdR (M ) = 0 is trivial, as M is then a free R–module. Otherwise, consider a minimal a presentation 0 −→ K −→ F −→ M −→ 0 with F free and K 6= 0. Then pdR (K) = pdR (M )−1. Since depth(M ) = 0, the inequality in the Depth Lemma is strict, so that depth(K) = depth(M ) + 1 = 1. As depth(R) − depth(K) = depth ∗R) − depth(M ) − 1, the Auslander Buchsbaum Formula holds for K by induction, so depth(K) + pdR (K) = depth(R). Hence depth(M ) + pdR (M ) = depth(R).
Corollary 6.5.21. Let I ⊂ C {x1 , . . . ,xn } = R be an ideal, (X, 0) = V (I), 0 . Consider a minimal free resolution of R/I as R–module 0 −→ Fk −→Fk−1 −→ · · · −→F0 = R−→R/I −→ 0.
Then depth (X, 0)
= n − k. In particular, (X, 0) is Cohen-Macaulay if and only if k = n − dim (X, 0) .

6.5 Cohen-Macaulay Spaces

263

This gives a possibility of checking on a computer whether a singularity, defined by polynomials, is Cohen-Macaulay. Example 6.5.22. Take the union of two planes in C 4 intersecting in a point again. The local ring is R := C {x,y,z,w}/(xz,xw,yz,yw). A free resolution of R as O := C {x,y,z,w}– module is given by 0

y1 @ −w A z −x

w y 0 01 0 y 0A @ −z 0 −x 0 w 0 0 −x −z 4 0

(xz xw yz yw)

0 −→ O −−−−−→ O −−−−−−−−−−−−→ O 4 −−−−−−−−−−→ O −→ R −→ 0,

which shows that R as a O–module has projective dimension three. It follows from the Auslander-Buchsbaum Formula that, as O–module, R has depth one. It follows that R (as R–module) has depth one. We see again that R is not Cohen–Macaulay. The following corollary of the Auslander-Buchsbaum formula will be used in the proof of the Hilbert-Burch Theorem 6.5.26. Corollary 6.5.23. Let (R,m) be a local ring and M be an R–module. Then
pdR (M ) ≥ depth Ann(M ),R .

Proof. We may assume that pdR (M ) < ∞. The proof is by induction on depth(M ). If depth(M ) = 0, then, by the Auslander Buchsbaum Formula, pdR (M ) = depth(R) ≥
depth Ann(M ),R . We now assume that depth(M ) > 0. In particular, from the Auslander-Buchsbaum Formula, it follows that d := depth(R) > 0. Let 0 −→ Fn −→ · · · −→ F0 −→ M −→ 0 be a minimal free resolution of M . Choose a regular sequence x1 , . . . ,xd in R such that xr+1 , . . . ,xd is a maximal regular sequence in Ann(M ). Since depth(M ) > 0, there exists a nonzerodivisor of M . Thus in particular x1 is a nonzerodivisor of R/ Ann(M ). Hence we have that r ≥ 1 and, by renumbering, we may suppose x1 is a nonzerodivisor of M . We obtain, using 1.2.35, the exact sequence 0 −→ Fn /x1 Fn −→ · · · −→ F0 /x1 F0 −→ M/x1 M −→ 0. By induction on depth(M ) we obtain
pdR/(x1 ) (M/x1 M ) ≥ depth Ann(M/x1 M ),R/(x1 ) .


Now pdR (M ) = pdR/(x1 ) (M/x1 M ) because of 6.5.17 and depth Ann(M/x1 M ),R/(x1 ) ≥
depth Ann(M ),R because of the special choice of x1 , . . . ,xd . Finally, in this chapter we will prove the Hilbert-Burch Theorem. The proof needs some preparations. First of all we need a characterization for a complex of type ϕn

ϕn−1

ϕ1

0 −→ Fn −→ Fn−1 −→ · · · −→ F0 being exact. We start with a lemma. ϕn

ϕ1

· · · −→ F0 be Lemma 6.5.24. Let R be a Noetherian local ring and 0 −→ Pn Fn −→ j−ν a complex4 of finitely generated free R–modules. Let rν = (−1) rank(Fj ) for j=ν ν = 0, . . . ,n. Then the induced sequence 4

Note that there is no “−→ 0” at the right hand side of the complex.

264

6 The Principle of Conservation of Number

0 −→ Fn ⊗R Q(R) −→ · · · −→ F0 ⊗R Q(R)
is exact if and only if depth Irν (ϕν ),R ≥ 1 for all ν. Moreover, if this is the case, then Im(ϕν ) has rank rν for all ν. (6.9)

Proof.
Step 1. First suppose that (6.9) is exact. To prove that depth Irν (ϕν ),R ≥ 1 means to prove that Irν (ϕν ) contains a nonzerodivisor, which means that Irν (ϕν ) ⊗ Q(R) = Q(R). ϕn ϕ1 Therefore, we may suppose that R = Q(R), and that 0 −→ Fn −→ · · · −→ F0 is exact, and we have to prove that Irν (ϕν ) = R. For j = n, . . . ,1 we have exact sequences 0 −→ Im(ϕj ) −→ Fj−1 −→ Im(ϕj−1 ). We proceed by descending induction. First of all Fn has rank rn , and thus, as ϕn is injective, Im(ϕn ) has rank rn . From 1.4.32 it follows that Irn (ϕn ) = R, and Irn +1 (ϕn ) = (0). From 1.3.8 it follows that Im(ϕn−1 ) has rank rn−1 . Again 1.4.32 implies that Irn−1 (ϕn−1 ) = R and Irn−1 +1 (ϕn−1 ) = (0). The induction step is proved in a completely similar manner. Step 2. To prove the other direction, assume that depth(Irν (ϕν ),R) ≥ 1 for all ν. This condition gives that Irν (ϕν ) contains a nonzerodivisor. As a nonzerodivisor is a unit in Q(R), we get Irν (ϕν ) ⊗ Q(R) = Q(R). Therefore, we may assume that R = Q(R) and thus that Irν (ϕν ) = R. Under these conditions we have to show exactness, that is, Im(ϕν+1 ) = Ker(ϕν ) for all ν, and that Im(ϕν ) is free of rank rν . By applying 1.3.8 the freeness of Im(ϕν ) follows if we moreover show that Irν +1 (ϕν ) = (0). We will proceed by descending induction on ν. So, assume that we have already proved that Irj +1 (ϕj ) = (0) for j ≥ ν + 1, and that the complex 0 −→ Fn −→ · · · −→ Fν+1 −→ Fν is exact. Now consider Fν+1 −→ Fν −→ Coker(ϕν+1 ) −→ 0. By assumption we have Irν+1 (ϕν+1 ) = R, and by 1.3.8 it follows that Im(ϕν+1 ) is free, necessarily of rank rν+1 . Moreover Proposition 1.3.8 says that Fν = Im(ϕν+1 ) ⊕ Fν′ for a free module Fν′ . It easily follows from the definition of the rj that the rank of Fν′ is equal to rν . Since Im(ϕν+1 ) ⊂ Ker(ϕν ) it follows from the definition of Fitting ideals that (R =)Irν (ϕν ) = Irν (ϕν|Fν′ ). Because of size reasons (rank(Fν′ ) = rν ), Irν +1 (ϕν|Fν′ ) = 0 and therefore Irν +1 (ϕν ) = (0). Again from Proposition 1.3.8 it follows that Im(ϕν ) has rank rν , and that ϕν|Fν′ is injective. This exactly says that Im(ϕν+1 ) = Ker(ϕν ), which is what we had to prove. Theorem 6.5.25 (Buchsbaum-Eisenbud). Let R be a Noetherian local ring, and let (6.10)

ϕn

ϕ1

0 −→ Fn −→ · · · −→ F0

Pn be a complex of free R–modules of finite rank. Let rν = j=ν (−1)j−ν rank(Fj ) for ν =
0, . . . ,n. Then the complex (6.10) is exact if and only if depth Irν (ϕν ),R ≥ ν for ν = 1, . . . ,n.

6.5 Cohen-Macaulay Spaces

265

Proof.
Step 1. Assume the complex is exact. We have to prove that depth Irν (ϕν ),R ≥ ν. We use induction on d = depth(R). The case d = 0 is not difficult. The exactness of
the
complex gives that pdR Im(ϕν ) < ∞ for ν = 1, . . . ,n. This implies that pdR Im(ϕν ) = 0 (Auslander-Buchsbaum 6.5.20), that is, Im(ϕν ) is free and of rank rν . By 1.4.32 Irν (ϕν ) contains a nonzerodivisor. As depth(R) = 0, the only nonzerodivisors are the units.
Therefore Irν (ϕν ) = R and, consequently, depth Irν (ϕν ),R = ∞.

Now assume that d > 0. By Exercise 1.3.22 the sequence 0
−→ Fn ⊗Q(R) −→ · · · −→ F0 ⊗ Q(R) is exact. By the previous lemma depth Irν (ϕν ),R ≥ 1 for all ν. Therefore we can find a nonzerodivisor x ∈ ∩nν=1
Irν (ϕν ). If x is a unit then Irν (ϕν ) = R for all ν and thus by definition depth Irν (ϕν ),R = ∞. Otherwise, x ∈ m is a nonzerodivisor, so that by 1.2.35 we get that 0 −→ Fn /xFn −→ · · · −→ F1 /xF1
is exact. The induction hypothesis implies that
depth Irν (ϕν ⊗ idR/(x) ),R/(x) ≥ ν − 1 for ν = 2, . . . ,n. This implies depth Irν (ϕν ),R ≥ ν because of 6.5.5 and the definition of the Fitting ideals, see 1.2.18. Step 2. For the converse we consider three cases. Case 1. This is the case that d = depth(R) = 0. Then the only nonzerodivisors in R are units. Thus R = Q(R) and the statement follows from the previous lemma. Case 2. The case that d = depth(R) ≥ 2. This is proved by induction on the dimension of R and on n. The case n = 0 is trivial, as obviously the sequence 0 −→ F0 is exact. The case dim(R) ≤ 1 follows because then depth(R) ≤ 1, hence we are in the first case or in the third case. Now assume that dim(R) > 0 and n > 0. By induction hypothesis we may suppose that 0 −→ Fn −→ · · · −→ F1 is exact. Let M = Im(ϕ2 ) and K = Ker(ϕ1 ). It remains to prove that K/M = 0. Let p 6= m be a prime ideal. By Exercise 1.3.22 the localization of the sequence 0 −→ Fnp −→ · · · −→ F1p −→ F0p is exact. As dim(Rp ) ≤ dim(R) − 1, it follows from the induction hypothesis with respect to the dimension of the ring that (K/M )p = 0 for all prime ideals p ⊂ R different from the maximal ideal m. By Exercise 1.4.54 this implies that a certain power of m annihilates K/M . Thus K/M is a module of dimension zero. Hence depth(K/M ) = 0. We consider the exact sequences (6.11)

0 −→ Im(ϕν ) −→ Fν−1 −→ Im(ϕν−1 ) −→ 0

for ν = 2, . . . , min{d,n}. Recall that d = depth(R). Suppose that n ≤ d. Then Fn = Im(ϕn ) is free, and so has depth d, see Exercise 6.5.27, and thus certainly depth ≥ n. Suppose on the other hand that j > d. By assumption depth(Irj (ϕj ),R) ≥ j. As R has only depth d, it follows that Irj (ϕj ) = R. Following the first part of the proof of Lemma 6.5.24 we obtain that Im(ϕj ) is free of rank rj for j ≥ d, In particular, Im(ϕd ) has depth at least d. Applying the Depth Lemma 6.5.18 successively to the exact sequences (6.11) gives that depth(Im(ϕj )) ≥ j for j ≥ 2. As M = Im(ϕ2 ) we get depth(M ) ≥ 2. Using that F1 has positive depth, and applying the Depth Lemma to the exact sequence

266

6 The Principle of Conservation of Number 0 −→ M −→ F1 −→ F1 /M −→ 0,

gives that depth(F1 /M ) ≥ 1. Note that also depth(F1 /K) ≥ 1. This holds because we have an embedding F1 /K ⊂ F0 . Now assume that K/M 6= 0. We apply the Depth Lemma to the exact sequence 0 −→ K/M −→ F1 /M −→ F1 /K −→ 0. As both F1 /M and F1 /K have positive depth, it follows that depth(K/M ) ≥ 1. This is a contradiction, as we have already proved that depth(K/M ) = 0. Thus K/M = 0, and the sequence is exact. Case 3. This is the case d = depth(R) = 1. By assumption depth(Ir2 (ϕ2 ),R) ≥ 2. It follows that Ir2 (ϕ2 ) = R. Following the proof of case 2, it follows that Im(ϕ2 ) is free, and F1 ∼ = Im(ϕ2 ) ⊕ F1′ for a suitable free module F1′ (1.3.8). It remains to prove that the restriction of ϕ1 to F1′ is injective. This is trivial if F1′ = 0. By 6.5.24 we have that rank(F1′ ) = r1 . As in the second case we have that Ker(ϕ1 |F1′ ) is annihilated by a power of the maximal ideal m and thus, if Ker(ϕ1 |F1′ ) 6= 0, it follows that depth(Ker(ϕ1 |F1′ )) = 0, and we have an exact sequence 0 −→ Ker(ϕ1 |F1′ ) −→ F1′ −→ Im(ϕ1 ) −→ 0 As F1′ has depth one, it follows from the Depth Lemma that depth(Ker(ϕ1 |F1′ )) > 0. This is a contradiction. Therefore, ϕ1 |F1′ is injective and 0 −→ Fn −→ · · · −→ F0 is exact. Theorem 6.5.26 (Hilbert-Burch). Let R be a Noetherian local ring and I ⊂ R be an ideal. α

α

1 2 R −→ R/I −→ 0 is a minimal free resolution. F1 −→ (1) Assume that 0 −→ F2 −→ Then

(1.1) rank(F1 ) = rank(F2 ) + 1, (1.2) depth(Irank(F2 ) (α2 ),R) = 2, (1.3) I = aIrank(F2 ) (α2 ) for a suitable nonzerodivisor a ∈ R. More precisely the i–th entry of the matrix for α1 is (−1)i a times the minor obtained from α2 by leaving out the i–th row.
(2) Let α2 be a (n − 1) × n–matrix and suppose that depth In−1 (α2 ),R ≥ 2. Let a ∈ R be a nonzerodivisor and define α1 as in the description of (1.3). Let I := aIn−1 (α2 ). α1 α2 R −→ R/I −→ 0 is a free resolution of R/I. Rn −→ Then 0 −→ Rn−1 −→ Proof. The statement (2) is an immediate consequence of 6.5.25, as r2 = n−1, and r1 = 1. We now prove (1.3). Let α1 be the map defined by the 1 × n–matrix having as i–th entry (−1)i times the minor obtained from α2 by deleting the i–th row. Then Laplace expansion gives α1 ◦ α2 = 0. On the other hand, by definition, I1 (α1 ) = Im(α1 ) = Irank(F2 ) (α2 ). By
6.5.25 we obtain that depth Irank(F2 ) (α2 ),R ≥ 2. Therefore, we can apply again 6.5.25 to the dual complex

6.5 Cohen-Macaulay Spaces

267 α∗

α∗

1 2 F1∗ −→ F2∗ , 0 −→ R −→

using that I1 (α∗1 ) = I1 (α1 ) = Irank(F2 ) (α2 ), to obtain that this complex is exact. Now consider the commutative diagram R  σ   /R

0

α∗ 1

α∗ 1

/ F1∗

/ F1∗

α∗ 2

α∗ 2

/ F2∗

/ F2∗

The map σ with α∗1 ◦σ = α∗1 exists because of the exactness of the second row. Because σ ∈ HomR (R,R) there exists an a ∈ R such that σ(x) = ax for all x ∈ R. It remains to prove that a is a nonzerodivisor. So suppose x · a = 0 for some x ∈ R. This implies xI1 (α1 ) = 0. Therefore, I1 (α1 ) consists of zerodivisors. This is not possible, as the sequence 0 −→
α1 α2 F0 is exact and thus depth I1 (α1 ),R ≥ 1 by Theorem 6.5.25. F1 −→ F2 −→ To prove (1.1) note that we have now proved that I contains a nonzerodivisor. Hence I ⊗ Q(R) = Q(R) and R/I ⊗ Q(R) = 0. We tensorize the exact sequence with Q(R) and obtain by Exercise 1.3.22 an exact sequence 0 −→ F2 ⊗R Q(R) −→ F1 ⊗R Q(R) −→ Q(R) −→ 0. This implies rank(F1 ) = rank(F2 ) + 1. It remains to prove (1.2). Put I ′ = Irank(F2 ) (α2 ). By dividing by a we get an exact sequence α

α1

2 a F1 −→ 0 −→ F2 −→ R −→ R/I ′ −→ 0.

Obviously I ′ = Ann(R/I ′ ), so that from 6.5.23 we obtain 2 ≥ pdR (R/I ′ ) ≥ depth(I ′ ,R). On the other hand, by 6.5.25, we have depth(I ′ ,R) ≥ 2. This implies depth(I ′ ,R) = 2.

Exercises 6.5.27. Suppose M is a free R–module. Then depth(M ) = depth(R). 6.5.28. Let R ⊂ S be a finite ring extension of local rings, and M be an S–module. (1) Let I ⊂ R be be an ideal. Show that (I · R) · M = (I · S) · M .

(2) Let f1 , . . . ,fs be a regular sequence of M as R–module. Show that f1 , . . . ,fs is a regular sequence of M as S–module.

6.5.29. Let R ⊂ S be a finite ring extension of local rings, and M be an S–module. Show that depthR (M ) = depthS (M ). 6.5.30. Let M,N be R–modules of positive depth. Show that there exists an f ∈ R which is a nonzerodivisor for both M and N . Generalize this statement to finitely many modules. (Hint: Use Exercise 1.4.52 and Prime Avoidance 1.1.13.) 6.5.31. The assumption on R to be local in Lemma 6.5.3 is an essential assumption! Show that in the ring C [x,y,z]/(xz − z) the sequence x,xy − y is a regular sequence, but xy − y,x is not a regular sequence. 6.5.32. Consider the map f (s,t) = (s,st,t2 ,t3 ) = (x,y,z,u) of Exercise 2.3.18. The image is given by the ideal I = (yz − ux,y 2 − zx2 ,z 2 x − uy,z 3 − u2 ), as shown in that exercise. Let R = C {x,y,z,u}. Show that

268

6 The Principle of Conservation of Number 0

u1 @ −y A −z x

0

1 y u xz z 2 B −z 0 −u C 0 @ A −x z −y u 0 −x 0 −y 4

(yz−ux,y 2 −zx2 ,z 2 x−uy,z 3 −u2 )

0 −→ R −−−−−→ R − −−−−−−−−−−−−−−→ R4 −−−−−−−−−−−−−−−−−−−−−−−→ R −→ R/I −→ 0

is a minimal free resolution of´ I. Conclude by the Auslander-Buchsbaum formula that the germ ` of the complex space V (I), 0 is not Cohen-Macaulay.

6.5.33. Let R = C {x,y,z},I =´ (xy,xz,yz). Compute a free resolution of R/I as R–module. ` Conclude that (X, 0) = V (I), 0 is Cohen-Macaulay.

6.5.34. Let R = C `{x,y,z},I ´ = (xy,xz). Compute a free resolution of R/I as R–module. Conclude that (X, 0) = V (I), 0 is not Cohen-Macaulay.

e be the normalization of R. Show that M is not a 6.5.35. Let R = C {x,y}/(xy), and M = R free R–module. Show by using the Auslander Buchsbaum formula, that M does not have finite projective dimension.

269

7

Standard Bases

Let an ideal I ⊂ C {x} be given. In this chapter we will define and study standard bases of the ideal I. The main idea here is that one wants to have a “good” representative for any coset of I. In fact, the Weierstraß Division Theorem provides us with the simplest example of a standard basis. Consider f ∈ C {x} which is regular of order b in xn . For any g ∈ C {x} we have a unique representation g = qf + r with r a polynomial of degree less than b. From the uniqueness statement in the Weierstraß Division Theorem, it follows without difficulty that the remainder r only depends on the coset g + (f ) of (f ). Of course, the remainder r depends on the choice of xn . In order to generalize, we would like to have division with remainder through an ideal. The “choice” of xn is now replaced by the choice of a so-called monomial ordering on the monomials of C {x}. Having such a monomial ordering, one can talk about the leading term L(f ) of f , and the ideal L(I) = (L(f ) : f ∈ I \ {0}) of leading terms of I. A standard basis of I is a set of elements f1 , . . . ,fm of I such that L(f1 ), . . . ,L(fm ) generate L(I). It is quite trivially true that a standard basis exists. Here one uses only the fact that C {x} is Noetherian. It also turns out that a standard basis of I generates the ideal I. Standard bases of ideals in C {x} generated by polynomials can be computed effectively. The algorithms are explained in the exercises. For example, the computer algebra system SINGULAR can be used for computation (cf. [Singular 2000]). The first thing one would have is a kind of division with remainder through the fi , that is, one would like for any f ∈ C {x} a representation (7.1)

f=

m X

qi fi + r

i=1

where no term of r is divisible by the leading term of any fi . This replaces the condition that r is polynomial of degree less than b in the Weierstraß Division Theorem. In fact, we will give alternative proofs of the Weierstraß Division and Preparation Theorems. In order to control the convergence of the qi and r, we will define seminorms1 on C {x} P α n or even C [[x]] depending on ρ ∈ R+ defined by kf kρ := α |fα |ρ . This is also the norm which is used in the books of Grauert and Remmert, see [Grauert-Remmert 1971], and [Grauert-Remmert 1984] to give proofs of the Weierstraß Division and Preparation Theorem. It turns out that if f1 , . . . ,fm is a standard basis, then the r is uniquely determined, and is called the normal form NF(f ) = NF(f |I) of f . It depends, of course, on the monomial ordering. In particular, it follows that f ∈ I if and only if NF(f ) = 0. This will be used in an essential way in the proof of Grauert’s Approximation Theorem in Chapter 8. We will give some applications in Section 7.3. More generally, there is a theory of standard bases of modules. As it is not more difficult, we will treat the general case from the beginning. 1

Here semi means that the norm might take infinite value.

270

7.1

7 Standard Bases

The Division Theorem

Definition 7.1.1. Let M = {xα |α ∈ Zn+ } be the set of monomials in C [x]. A monomial ordering is a total ordering on M (respectively Zn+ ) compatible with the semi–group structure, that is, xα > xβ implies xα+γ > xβ+γ . We assume furthermore that the orderings considered here are well–orderings, that is, every set of monomials has a minimal element with respect to the ordering. This is equivalent to the property that 1 is the smallest monomial (cf. Exercise 7.1.9). Examples 7.1.2. (1) The lexicographical ordering >

lex

α

β

x > x if and only if α 6= β and there is an i ≥ 1 such that αj = βj for i > j and lex

αi > βi .

(2) The degree lexicographical ordering

>

deglex

Let w = (w1 , . . . ,wn ) ∈ Rn and wi > 0, i = 1, . . . ,n. For α ∈ Zn+ define |α|w = P n i=1 αi wi (we shall also simply use |α| = |α|w if w is fixed).

xα

> xβ if and only if |α| > |β| or |α| = |β| and xα > xβ

deglex

lex

Definition 7.1.3. Now let > be a monomial ordering. Let f ∈ C {x} be a power series. Then we may write X f= ai xαi i≥1

such that ai 6= 0, ai ∈ C and xαi < xαi+1 for all i. We define (1) L(f ) = xα1 the leading monomial of f ;

(2) C(f ) = a1 the coefficient of the leading monomial; (3) tail(f ) = f − C(f ) · L(f ). Let I ⊂ C {x} be an ideal, then L(I) = h{L(f )|f ∈ I, f 6= 0}i, the leading ideal, is the ideal generated by all leading monomials of non–zero elements of I. Example:. Let f = 2x51 + x62 + x42 + x1 and I = (f ). With respect to the orderings introduced before (with w = (1,1)), we obtain for L(f ), C(f ), tail(f ) and L(I) the following (the monomials of tail(f ) always ordered with respect to the monomial ordering): c(f ) L(f ) L(I)

tail(f ) x62 + x1 + 2x51

lex

1

x42

hx42 i

deglex

1

x1

hx1 i

x42 + 2x51 + x62

We wish to define orderings also for the monomials of the free module C {x}N . For a compatible notation we choose, as a basis, the vectors (0, . . . ,1,0 . . . 0) and identify them, as before, with xi−1 n+1 , xn+1 a new variable.

i

7.1 The Division Theorem

271

PN −1 Definition 7.1.4. Let < be a monomial ordering on C {x}. Let C {x}N = i=0 C {x}xin+1 N and M = {xα |α ∈ Zn+1 + ,0 ≤ αn+1 ≤ N −1} be the set of monomials of C {x} . A module ordering <m is a total ordering of M compatible with the ordering < on C {x}, that is (1) f,f ′ ∈ M and f <m f ′ implies xα f <m xα f ′ .

(2) f ∈ M and xα < xβ implies xα f <m xβ f .

Exercise 7.1.11 gives several examples for module orderings. From now on in this chapter we use the following ordering: Let w = (w1 , . . . ,wn+1 ) ∈ Rn+1 be fixed and < be the degree lexicographical ordering. + For xα and xβ ∈ M we define xα < xβ if and only if |α| < |β| or if |α| = |β| then α < β with respect to the lexicographical ordering. As in Definition 7.1.3 we define for f ∈ C {x}N the leading monomial L(f ), the leading coefficient C(f ), the tail, tail(f ) and the leading module L(I) for a submodule I ⊂ C {x}N . Let m,m′ ∈ M be two monomials, m = xα , m′ = xβ . We define m | m′ if αn+1 = βn+1 and xα | xβ . The aim of this Chapter is to obtain information about ideals by studying their leading ideals (respectively modules) which are generated by monomials and are, therefore, simpler. Definition 7.1.5. Let I ⊂ C {x}N be
a submodule, S = {f1 , . . . ,fm } ⊂ I is a standard basis of I if L(I) = L(f1 ), . . . ,L(fm ) . A standard basis is reduced if C(fi ) = 1 for all i and L(fi ) does not divide any monomial occurring in fj for i 6= j and occurring in tail(fi ).

Example:. Let I = (x3 + y 2 ,y), w = (1,2), then {x3 + y 2 ,y,y 5 } is a standard basis of I in C {x} but not a reduced standard basis of I, {x3 ,y} is a reduced standard basis.

Remark:. Standard bases were introduced by Hironaka (cf. [Hironaka 1964]), Grauert (cf. [Grauert 1972]) for the power series case and Buchberger (cf. [Buchberger 1965]) for polynomial rings. In case of polynomial rings, they are called Gr¨obner bases. Proposition 7.1.6. Let I ⊂ C {x}N be a submodule, then I has a standard basis.

Proof. L(I) is, by definition, the submodule in C {x}N generated by the leading monomials of the elements of I. C {x} is Noetherian and, therefore L(I) is generated by finitely many such leading monomials, say xα1 , . . . ,xαm . Now choose f1 , . . . ,fm ∈ I such that L(fi ) = xαi then {f1 , . . . ,fm } is a standard basis of I. Later we shall see that a standard basis of I generates the module I. We shall also prove that reduced standard bases exist and are uniquely determined. Therefore, we need the notion of a normal form of an element f ∈ C {x}N with respect to a set S = {f1 , . . . ,fm } of elements of C {x}N , that is, a representation in C {x}N f=

m X

qi fi + r

i=1

such that no monomial of r is divisible by L(fi ) for all i. The following result about the normal form is also a generalization of the Weierstraß division Theorem and, therefore, also often called division Theorem. It is the essential Theorem of this section with many consequences. We will need a quite general version. The case s = 0 in the Theorem is the most important case, however.

272

7 Standard Bases

Theorem 7.1.7 (Grauert’s Divison Theorem). Let s ≥ 0 and f1 , . . . ,fm ∈ C {xs+1 , . . . ,xn }N , then for every f ∈ C {x}N there exists a representation m X f= qν fν + r ν=1

with the following properties.

(1) No monomial of r is divisible by L(fr ),ν = 1, . . . ,m; (2) L(qν fν ) ≥ L(f ),ν = 1, . . . ,m2 n+1 (3) Let f ∈ Bρn for some ρ ∈ Rn+1 such that + . For every ε > 0 there exist ρ ∈ R+

• ρ ≤ ρ and ρi = ρi for i = 1, . . . ,s,

• for 0 < λ ≤ 1 let λ ◦ ρ = (ρ1 , . . . ,ρs ,λws+1 ρs+1 , . . . ,λwn+1 ρn+1 ), then 1 kf kλ◦ρ 1−ε 1 kqν kλ◦ρ kL(fν )kλ◦ρ ≤ kf kλ◦ρ . 1−ε krkλ◦ρ ≤

Proof. be the Zn+ –semi– Special case: fi = xαi , i ∈ {1, . . . ,m}, let ∆ = hα1 , . . . ,αm i ⊂ Zn+1 + module generated by α1 , . . . ,αm . We define inductively a partition of ∆ by

Let f =

P

∆1 = hα1 i ∆i = hαi i ∩ (∆ r hα1 , . . . ,αi−1 i). fα xα , we define r :=

X

fα xα

α∈∆ /

qi =

1 X fα xα xαi α∈∆i

P

then f = qi fi + r. Let kf kρ < ∞, we choose ρ = ρ, then krkρ ≤ kf kρ and kqi kρ kfi kρ ≤ kf kρ . We shall use the construction of this step in the next one. Therefore, we define for S = {xα1 , . . . ,xαm } being fixed maps qi : C {x}N −→ C {x}N by qi (f ) =

1 x αi

P

r : C {x}N −→ C {x}N

α∈∆i

fα xα resp. r(f ) =

P

α∈∆ /

fα xα and obtain f =

P

qi (f )xαi + r(f ).

General case: Let L(fi ) = xαi and assume C(fi ) = 1. Let S = {xα1 , . . . ,xαm }. We define a sequence νi ∈ C {x}N by 2

This condition is a kind of minimality condition. Note that in a sum one can always add two terms which cancel. The condition written here, avoids this phenomenon.

7.1 The Division Theorem

273 ν0 = f m X

νj+1 = νj −

ν=1

The idea of the proof is now to define ν =

qν (νj )fν − r(νj ).

∞ P

νj and prove that

j=0

• f=

P

qj (ν)fj + r(ν).

• ν ∈ C {x}N . Lemma 7.1.8. Let g1 , . . . ,gm ∈ C {xs+1 , . . . ,xn }N for some s ≥ 0. Then, for every ε > 0 such that ρ ≤ ρ and ρi = ρi for i = 1, . . . ,s and for and ρ ∈ Rn+1 , there exist ρ ∈ Rn+1 + all λ, 0 < λ < 1, k tail(gi )kλ◦ρ < εkL(gi )kλ◦ρ , i = 1, . . . ,r.

We shall prove the Lemma later.

ε Now let 1 > ε > 0 be given. We apply the Lemma to f1 , . . . ,fm and m and choose ε αi and ρ ≤ ρ such that ktail (fP )k < (λ ◦ ρ) ρ and ρ = ρ for i = 1, . . . ,s. i λ◦ρ i i m Pm m Now, kνj+1 kλ◦ρ ≤ i=1 k tail(fi )kλ◦ρ kqi (νj )kλ◦ρ because νj+1 = − i=0 qi (νj ) tail(fi ). 1 By definition of the qi we have kqi (νj )kλ◦ρ ≤ (λ◦ρ)αi kνj kλ◦ρ and, therefore, kνj+1 kλ◦ρ ≤ εkνj kλ◦ρ . Pi Let si := j=0 νj then

ksi kλ◦ρ = k

i X j=0

νj kλ◦ρ ≤ ≤

i X j=0

i X j=0

kνj kλ◦ρ

εj kν0 kλ◦ρ

≤

1 kf kλ◦ρ . 1−ε

kνkλ◦ρ ≤

1 kf kλ◦ρ . 1−ε

This implies

Especially, kxαi kλ◦ρ · kqi (ν)kλ◦ρ ≤ kνkλ◦ρ and kr(ν)kλ◦ρ ≤ kνkλ◦ρ implies 1 kf kλ◦ρ 1−ε 1 ≤ kf kλ◦ρ . 1−ε

kL(fi )kλ◦ρ kqi (ν)kλ◦ρ ≤ kr(ν)kλ◦ρ

274

7 Standard Bases

It remains to prove that f=

m X


qi (ν)fi + r(ν) and L qi (ν)fi ≥ L(f ).

∞ X

(νj − νj+1 )

i=1

Now f= = =

j=0 ∞ X

νj − νj +

j=0 m X ∞ X

m X ℓ=1

qℓ (νj )fℓ +


qℓ (νj )fℓ + r(νj ) ∞ X

r(νj ).

j=0

ℓ=1 j=0

P∞ P∞ But j=0 qℓ (νj ) = qℓ (ν) and j=0 r(νj ) = r(ν) and, therefore, the first part is proved.



P∞ Now L qi (ν)fi = L j=0 qi (νj )fi ≥ L qi (ν0 )fi = L qi (f )fi because L(νj+1 ) > L(νj ). Therefore,

L qi (ν)fi ≥ L qi (f )fi ≥ L(f ). This proves the Theorem.

Proof of the Lemma. Because the g1 , . . . ,gm do not depend on x1 , . . . ,xs we have a free choice for ρ1 , . . . ,ρs . So we may assume that s = 0. P (i) Let ε > 0 be given and assume that C(gi ) = 1 for all i. Let gi = α gα xα and (i) L(gi ) = xα . We have to prove that there exist ρ ∈ Rn+1 such that for all 0 < λ < 1 + (i)

sup{(2|α| + 2)n+2 | gα(i) | (λ ◦ ρ)α } < ε(2|α(i) | + 2)n+2 (λ ◦ ρ)α . Now (λ ◦ ρ)α = λ|α| · ρα and, therefore, it is enough to find a ρ ∈ Rn+1 such that + (i)

|gα(i) | ρα < ερα

for |α| = |α(i) | and i = 1, . . . ,m.

(i)

(i)

Now |α| = |α(i) | and xα > xα implies xα > xα with respect to the lexicographical (i) (i) ordering, that is, for α = (α1 , . . . ,αn+1 ) and α(i) = (α1 , . . . ,αn+1 ) exist k such that (i)

(i)

(i)

α1 = α1 , . . . ,αk−1 = αk−1 and αk > αk .

(∗)

(i)

We choose k ∈ {1, . . . ,n + 1} maximal such that for at least one i and one α with gα 6= 0 (*) holds. We choose any ρk+1 , . . . ,ρn+1 ∈ R+ smaller than 1 and choose ρk such that (i)

α

(i)

α

α

(i)

(i)

α

α

n+1 n+1 k+1 n+1 (i) k k |gα | ρα k ρk+1 · . . . · ρn+1 < ερk ρk+1 · . . . · ρn+1 for all i and α such that |α| = |α |,

(i)

gα 6= 0 and (*) holds, that is, (∗∗)

(i)

α −αk

ρk k

<

ε (i) |gα |

(i)

α

k+1 · ρk+1

−αk+1

(i)

α

n+1 · . . . · ρn+1

−αn+1

.

7.2 Characterizations and Properties of Standard Bases

275

Assume ρℓ , . . . ,ρn+1 are already defined. We choose k ∈ {1, . . . ,ℓ − 1} maximal such (i) that for at least one i and one α with gα 6= 0 (∗) holds. We choose any ρk+1 , . . . ,ρℓ−1 ∈ R+ smaller than 1 and choose ρk such that (∗∗) holds. If such k does not exist we choose any ρ1 , . . . ,ρℓ−1 ∈ R+ smaller than 1. Let ρ = (ρ1 , . . . ,ρn+1 ) then, by construction, we (i) (i) have |gα | ρα < ερα for |α| = |α(i) |. This proves the Lemma.

Exercises 7.1.9. Prove that a monomial ordering is a well–ordering if and only if 1 is the smallest monomial. 7.1.10. Give an algorithm which constructs in C {x} the functions q and r up to a given order.

7.1.11 (Examples for module orderings). Let < be a monomial ordering on C {x} and C {x}N = PN−1 n+1 k α i=0 C {x}xn+1 be the identification as in Definition 7.1.4 and M = {x |α ∈ Z+ , 0 ≤ αn+1 ≤ N − 1}. Let w = (w1 , . . . ,wn+1 ) ∈ Rn+1 , wn+1 6= 0. For xα and xβ ∈ M we define xα <m xβ if and only if P Pn+1 • |α|w = n+1 i=1 αi wi < i=1 βi wi = |β|w or β1 βn αn 1 • |α|w = |β|w and xα 1 · . . . · xn < x1 · . . . · xn .

Prove that <m is a module ordering. Notice that the following module orderings are often used: (1) degree orderings w1 > 0, . . . ,wn+1 > 0; (2) elimination orderings w1 = · · · = wn = 0.

7.1.12. For xα and xβ ∈ M we define xα <m xβ if and only if β1 βn n • x1α1 · . . . · xα n < x1 · . . . · xn or • if (α1 , . . . ,αn ) = (β1 , . . . ,βn ) then αn+1 < βn+1 .

Prove that <m is a module ordering.

7.1.13. Let <m be an elimination ordering (7.1.11 (2)) with w1 · · · = wn = 0, wn+1 = 1. Let I ⊂ C {x}N be aPsubmodule and S a standard basis of I. P Prove that S ∩ ti=0 C {x}xin+1 is a standard basis of I ∩ ti=0 C {x}xin+1 for all t ≤ N − 1.

7.2

Characterizations and Properties of Standard Bases

N Definition 7.2.1. Let S = {f1 , . . . ,fm } ⊂ C {x}N be Pan ordered set. For f ∈ C {x} we can, by Grauert’s Division Theorem ??, write f = qi fi + r, such that no monomial of r is divisible by any of the L(fi ). The r constructed in the proof of Grauert’s Division Theorem is uniquely determined with respect to the ordered set S, see 7.1.7. This r is called the normal form of f with respect to the (ordered) set S. We will write r = NF(f |S), and if S is fixed, just NF(f ).

Remark 7.2.2. In general, the normal form depends on the ordering of S. For example, f = x2 y + xy 2 + y 2 , f1 = y 2 − 1, f2 = xy, S1 = {f1 ,f2 } and S2 = {f2 ,f1 }. Then for the degree lexicographic ordering with w = (1,1) we get v q1 (v) q2 (v) r(v) = NF(f |S1 )

= f +x = x+1 = x = x+1

v q1 (v) q2 (v) r(v) = NF(f |S2 )

= f +1 = x+y = 1 = 1.

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7 Standard Bases

We will show that if S is a standard basis however, then the normal form does not depend on the ordering. First the definition. Definition 7.2.3. Let I ⊂ C {x}N be a submodule.
Then S = {f1 , . . . ,fm } ⊂ I is called a standard basis of I if L(I) = L(f1 ), . . . ,L(fm ) . A standard basis is called reduced if C(fi ) = 1 for all i and L(fi ) does not divide any monomial occurring in fj for i 6= j and occurring in Tail(fi ). Example 7.2.4. Let I = (x3 + y 2 ,y), w = (1,2). Then {x3 + y 2 ,y,y 5 } is a standard basis of I in C {x} but not a reduced standard basis of I. The standard basis {x3 ,y} is reduced. Remark 7.2.5. Standard bases were introduced by Hironaka (cf. [Hironaka 1964]), Grauert (cf. [Grauert 1972]) for the power series case and, for polynomial rings, by Buchberger (cf. [Buchberger 1965]). In case of polynomial rings, they are also called Gr¨obner bases. The idea of Gr¨obner bases is already contained in the work of Gordan, see [Gordan 1899]. We now give some basic results. Proposition 7.2.6. Let I ⊂ C {x}N be a submodule. Then I has a standard basis. Proof. The module L(I) is, by definition, the submodule in C {x}N generated by the leading monomials of the elements of I. The ring C {x} is Noetherian and, therefore L(I) is generated by finitely many such leading monomials, say xα1 , . . . ,xαm . Choose f1 , . . . ,fm ∈ I such that L(fi ) = xαi . Then {f1 , . . . ,fm } is a standard basis of I. Corollary 7.2.7. Let S = {f1 , . . . ,fm } be a standard basis of I. Then I = (f1 , . . . ,fm ). P Proof. For f ∈ I we have f = qi fi + NF(f |S). This implies that NF(f |S) ∈ I. If

NF(f |S) 6= 0, then L NF(f |S) ∈ L(I) = L(f1 ), . . . ,L(fm ) , which is a contradiction to the definition of NF(f |S). Corollary 7.2.8. Let S = {f1 , . . . ,fm } and T = {g1 , . . . ,gs } be standard bases of I. Then for all f ∈ C {x}N we have NF(f |S) = NF(f |T ). P P Proof. f = qν fν + NF(f |S) = qν gν + NF(f |T
) implies NF(f |S) − NF(f |T ) ∈ I. If NF(f |S) 6= NF(f |T ), then L NF(f |S) − NF(f |T ) ∈ L(I) is a monomial which appears in NF(f |S) or in NF(f |T ) which is impossible by definition of NF. This result makes the following definition possible. Definition 7.2.9. Let I ⊂ C {x}N be a submodule and S a standard basis of I. Then for any f ∈ C {x}N we define NF(f |I) = NF(f |S) to be the normal form of f with respect to I. Corollary 7.2.10. Let I be a submodule of C {x}N , and S = {f1 , . . . ,fm } be a standard basis of I. Let f ∈ C {x}N . Then (1) f ∈ I if and only if NF(f |S) = 0, (2) f ∈ I if and only if NF(f |I) = 0, (3) f − N F (f |I) ∈ I.

7.2 Characterizations and Properties of Standard Bases

277


P Proof. We know that f = qν fν +NF(f |S) and I = f1 , . . . ,fm . Therefore, NF(f |S) = 0 implies f
∈ I. On the other hand, f ∈ I implies NF(f |S) ∈ I. If NF(f |S) 6= 0 then L NF(f |S) ∈ L(I) which is not possible by the definition of NF.

Corollary 7.2.11. Let I ⊂ C {x}N be a submodule. Then I has a reduced standard basis S = {f1 , . . . ,fm }, with the property L(f1 ) < · · · < L(fm ). Such a standard basis is uniquely determined.

Proof. Step 1. We first prove the existence. Let T = {g1 , . . . ,gm } be a standard basis of I, and NF the normal form with respect to T . After reordering, and leaving out some of the gi , we may assume L(g1 ) < · · · < L(gm ). Suppose that L(gi )|L(gj ) for some i 6= j. Then {g1 , . . . ,gj−1 , gj+1 , . . . ,gm } is also a standard basis for I. Thus we may assume that L(gi ) ∤ L(gj ) for i 6= j. Furthermore, dividing gi by C(gi ) we may assume that C(gi ) = 1 for all i. We define
fi := L(gi ) + NF Tail(gi )|I , i = 1, . . . ,m. We claim standard basis of I. Because gi = L(gi )+ Tail(gi ) =
Pthat f1 , . . . ,fm is a reduced L(gi ) + qiν gν + NF Tail(gi )|T , the fi are in I. Because L(gi ) = L(fi ) we obtain that S := {f1 , . . . ,fm } is a standard basis of I. Furthermore, L(fi ) ∤ L(fj ) for i 6= j, and by definition of normal form, no term of Tail(fi ) is divisible by any of the L(g1 ) = L(f1 ), . . . ,L(gm ) = L(fm ). Thus f1 , . . . ,fm is a reduced standard basis.

Step 2. We now show uniqueness. Let S = {f1 , . . . ,fm } and T = {g1 , . . . ,gs } be two reduced standard bases and L(f1 ) < · · · < L(fm ), L(g1 ) < · · · < L(gs ). We first prove
that s = m and L(fi ) = L(gi ). Suppose the converse. Now L(I) = L(f1 ), . . . ,L(fm ) and L(f
i ) ∤ L(fj ) for i 6= j (S is a reduced standard basis). Similarly L(I) = L(g1 ), . . . ,L(gs ) , L(gi ) ∤ L(gj ) for i 6= j. Assume that for some i ≥ 1 L(gj ) = L(fj ) for j < i and L(gi ) 6= L(fi ). By symmetry we may
assume L(gi ) < L(fi ).
As they are both standard basis, it follows that L(g1 ), . . . ,L(gs ) = L(f1 ), . . . ,L(fm ) . Hence L(fν ) | L(gi ) for some ν. There are two cases. (1) If ν ≥ i, then L(fν ) ≥ L(fi ). Furthermore, because of L(gi ) < L(fi ) we get L(gi ) < L(fν ). Thus it is not possible that L(fν ) | L(gi ), contradiction. (2) If ν < i, we have, by choice of i, L(fν ) = L(gν ). But as we have a reduced standard basis L(gν ) ∤ L(gi ) for ν 6= i. This gives a contradiction because we had L(fν ) | L(gi ). So we proved that s = m and L(gi ) = L(fi ) for all i. Now consider fi − gi =
Tail(fi ) − Tail(gi ) ∈ I. If Tail(fi ) 6= Tail(gi ) then L Tail(fi ) − Tail(gi ) is a monomial in L(I) and a monomial appearing in Tail(fi ) or Tail(gi ), which is a contradiction to the property of both standard bases being reduced. We will now give characterization of standard bases which is the theoretical background for Buchberger’s algorithm to compute standard bases. Definition 7.2.12. Let f, g ∈ C {x}N , with L(f ) = xβ and L(g) = xγ .
α (1) We define the least common multiple lcm
L(f ),L(g) to be the monomial x with α = max(β1 ,γ1 ), . . . , max(βn ,γn ),βn+1 if βn+1 = γn+1 and to be 0 if βn+1 6= γn+1 .

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7 Standard Bases

(2) Let m = lcm(L(f ),L(g). Then we define the syzygy polynomial of f and g by spoly(f,g) =

m C(f ) m ·f − · · g. L(f ) C(g) L(g)

(Note that in case m 6= 0 we have L(spoly(f,g)) > m.) Examples 7.2.13. (1) Let f = 2xy 2 + x4 , g = 5x2 y + y 4 ∈ C {x,y} and w = (1,1), then spoly(f,g) = x5 − 25 y 5 . (2) Consider the free C {x1 ,x2 }–module C {x1 ,x2 }2 = C {x1 ,x2 } ⊕ C {x1 ,x2 }x3 , and w = (1,1,1) Let f = (x21 + 1,x2 ) = 1 + x21 + x2 x3 and g = (x2 ,x31 ) = x2 + x31 x3 . Then L(f ) = 1, and L(g) = x2 , so that spoly(f,g) = x2 f − g = (x21 x2 ,x22 − x31 ) = x21 x2 + x22 x3 − x31 x3 . Definition 7.2.14. Let f1 , . . . ,fm ∈ C {x}N . Write C [x]m = (0, . . . ,0,1,0, . . . ,0).

Pm

i=1

C [x] · ei , ei =

(1) P A syzygy on the leading terms C(f1 )L(f1 ), . . . ,C(fm )L(fm ) is an element s = m m such that i=1 si ei = (s1 , . . . ,sm ) ∈ C [x] s(f1 , . . . ,fm ) :=

m X

si C(fi )L(fi ) = 0.

i=1

(2) Let syz(f1 , . . . ,fm ) be the submodule of C [x]m of all syzygies of the leading terms of f1 , . . . ,fm . (3) Write xα(i,j) = lcm(L(fi ),L(fj )). Then s(i,j) :=

C(fi ) xα(i,j) xα(i,j) ei − ej ∈ syz(f1 , . . . ,fm ). L(fi ) C(fj ) L(fj )

(Note that s(i,j) · (f1 , . . . ,fm ) = spoly(fi ,fj ), where the dot denotes the dot product.) Lemma 7.2.15. Consider f1 , . . . ,fm ∈ C {x}N , and keep the notations of the previous definition. (1) The C [x]–module syz(f1 , . . . ,fm ) admits a Zn+1 –grading, that is, we can write syz(f1 , . . . ,fm ) = ⊕ν∈Nn+1 Sν with Sν :=

m X i=1

(2) s(i,j) ∈ Sα(i,j) .

ci xα(i) ei ∈ syz(f1 , . . . ,fm ) : ci ∈ C , xα(i) L(fi ) = xν .

7.2 Characterizations and Properties of Standard Bases

279

(3) The set {s(i,j) } generates syz(f1 , . . . ,fm ). Proof. P Pm (1) Let s = i=1 si ei ∈ syz(f1 , . . . ,fm ) and write si = α∈Nn ciα xα ∈ C [x]. For ν ∈ Nn+1 put ( ciβ xβ if xβ L(fi ) = xν siν = 0 if L(fi ) ∤ xv . P Pm Define sν := i=1 siν ei . Then obviously sν ∈ Sν and s = ν sν . (2) This is clear.

(3) To prove the third statement, we may assume that C(fi ) = 1 for all i. Let L(fi ) = xαi . Because ofP (1) it is enough to prove the statement for homogeneous s ∈ Sγ . We thus m write s = i=1 ci xβi , ci ∈ C and βi + αi = γ for all i with ci 6= 0. Let {i1 , . . . ,ik } be the indices such that ciν 6= 0. We will prove by induction on k that s is in the module generated by the s(i,j) . This is obvious for k = 0, as then s = 0. If k 6= 0, then k ≥ 2. Note that as αij + βij = γ, it follows in particular that xγ ≥ lcm(L(fi1 ),L(fi2 )) =: xα . Thus we can consider xγ s′ := s − c1 α s(i1 ,i2 ) . x We have s′ ∈ syz(f1 , . . . ,fm ), and s′ ∈ ⊕kν=2 C [x]eiν . By induction, this proves the lemma. Theorem 7.2.16 (Buchberger’s Criterion). Let I = (f1 , . . . ,fm ) ⊂ C {x}N be a submodule. The following conditions are equivalent: (1) S = {f1 , . . . ,fm } is a standard basis for I.
(2) NF spoly(fi ,fj )|S = 0 for all i,j.

(3) Let G be a subset of {s(i,j) } generating syz(f1 , . . . ,fm ). Then NF spoly(fi ,fj )|S) = 0 for all i,j with s(i,j) ∈ G.

Proof. The implication (1) =⇒ (2) follows from Corollary 7.2.10, and (2) =⇒ (3) is obvious. It remains to show (3) =⇒ (1). Without loss of generality, we may assume that C(fi ) = 1 for all i. Let L(fi ) = xαi for i = 1, . . . ,m. Consider f ∈ I. We have Pmto prove that L(f ) is divisible by L(fi ) for some i. As f ∈ I we can write f = i=1 gi fi . To such a representation we associate the monomial xδ = min{L(gi fi )}. Assume that the representation is chosen with maximal xδ between all possible representations. We will prove that L(f ) = xδ . This would imply L(f ) = L(gi fi ) = L(gi ) · L(fi ) for some i, and thus the theorem. Suppose the converse, that is L(f ) > xδ . We define I1 := {i : L(gi fi ) = xδ }, I2 := {i : L(gi fi ) 6= xδ }. Then we can write f=

X

i∈I1

C(gi )L(gi )fi +

X

i∈I1

Tail(gi )fi +

X

i∈I2

gi fi .

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7 Standard Bases

P The assumption L(f ) > xδ implies i∈I1 C(gi )L(gi )L(fi ) = 0. Thus we have a homogeneous syzygy on the leading terms of the fi , in fact in Sδ ⊂ syz(f1 , . . . ,fm ). By assumption on G and Lemma 7.2.15 this syzygy on the leading terms is of type P αij (i,j) s ∈ Sδ for suitable cij ∈ C , and αij ∈ Nn . Taking the dot product i,j∈I1 cij x s(i,j) ∈G

with (f1 , . . . ,fm ) we obtain X C(gi )L(gi )fi = i∈I1

X

cij xαij spoly(fi ,fj ).

i,j∈I1 s(i,j) ∈G

Now consider a nonzero term of the right hand side. Then there exist i,j ∈ I1 , with s(i,j) ∈ G and cij spoly(fi ,fj ) 6= 0. As the syzygy on the leading terms is in Sδ we have xαij lcm(L(fi ),L(fj )) = xδ . It was noted in the second part of Definition 7.2.12 that L(xαij spoly(fi ,fj )) > xδ .
Now the third assumption implies NF spoly(fi ,fj )|S = 0
for s(i,j) ∈ G and, therePm ij fore, spoly(fi ,fj ) = ν=1 qν fν and L(qνij fν ) ≥ L spoly(fi ,fj ) . Thus in particular we get xαij L(qνij fν ) > xδ . We get X

C(gi )L(gi )fi =

i∈I1

=

X

cij xαij

i,j∈I1 s(i,j) ∈G m  X X ν=1

i,j∈I1 s(i,j) ∈G

m X

qνij fν

ν=1

cij xαij qνij

 fν .

Now we define g ν by gν =

X

cij xαij qνij +

i,j∈I1 s(i,j) ∈G

(

gν , for ν ∈ I2 Tail(gν ), for ν ∈ I1 .

Pm Then f = ν=1 g ν fν and L(gν fν ) > xδ for all ν. This is a contradiction to the choice of δ and proves the theorem. Example 7.2.17. Let f1 = x4 + x3 y 2 , f2 = y 4 + x2 y 3 , w = (1,1). Then {f1 ,f2 } is a standard basis: spoly(f1 ,f2 ) = x3 y 6 − x6 y 3 = −x2 y 3 f1 + x3 y 2 f2 . The following lemma gives the possibility to find proper subsets of {s(i,j) } which generate syz(f1 , . . . ,fm ). This is very useful for the application of 7.2.16. Lemma 7.2.18 (Chain Criterion). Let f1 , . . . ,fm ∈ C {x}N and assume that a subset G of {s(i,j) } generates syz(f1 , . . . ,fm ). Suppose s(i,j) , s(i,k) and s(k,j) are in G and
L fi )| lcm(L(fj ),L(fk ) . Then G r {s(j,k) } generates syz(f1 , . . . ,fm ). Proof. We may assume that C(fi ) = 1 for all i, then one computes that

lcm L(fj ),L(fk ) (i,j) lcm L(fj ),L(fk ) (i,k)
s
s − . s(k,j) = lcm L(fi ),L(fj ) lcm L(fi ),L(fk )

Thus the lemma follows from 7.2.16.

7.2 Characterizations and Properties of Standard Bases

281

Exercises 7.2.19 (Product Criterion). Let f1 ,f2 ∈ C {x}, and assume that lcm(L(f1 ),L(f2 )) = L(f1 ) · L(f2 ). Prove that NF(spoly(f1 ,f2 )|{f1 ,f2 }) = 0. 7.2.20. Let <m be an elimination ordering (see 7.1.11 (2)) with w1 = · · · = wn = 0, wn+1 = 1. Let I ⊂ C {x}N P be a submodule and S a standard basis ofP I. Prove that S ∩ ti=0 C {x}xin+1 is a standard basis of I ∩ ti=0 C {x}xin+1 for all t ≤ N − 1.

7.2.21 (Gr¨ obner bases). In this exercise we will develop the theory of standard bases for polynomial rings rewriting some passages of the Sections 7.1 and 7.2. • Replace C {x} by C [x].

• Replace in Definition 7.1.3 xαi < xαi+1 by xαi > xαi+1 .

• Rewrite the examples.

• Replace (2) by L(qν fν ) ≤ L(f ), for ν = 1, . . . ,m in Theorem ??, and cancel Remark 7.1.7. • Replace the proof of Theorem ?? by the following NF–algorithm (normal form algorithm): r = NF(f |{f1 , . . . ,fm }) r=f S = {f1 , . . . ,fm } while r 6= 0 and L(fi ) | L(r) for some i choose i minimal with L(fi ) | L(r) C(r) L(r) r = r − C(f fi i ) L(fi )

if r = 6 0 r = C(r)L(r) + NF(Tail(r)|S) return(r). Prove that this algorithm P terminates. Use the fact that < is a well-ordering and Definition 7.1.3. Prove that f = qν fν + r, r = NF(f |{f1 , . . . ,fm }) for suitable qν ∈ C [x] and (1) and (2) of Theorem ?? are true.

• Check the proofs of 7.2.7, 7.2.8, 7.2.11 and 7.2.10, 7.2.15, and 7.2.16 (here we have to change the definition of xδ = max{L(gi fi )} to min{L(gi fi )} and replace several times > by <), and 7.2.18. 7.2.22 (Buchberger’s algorithm). S = standard({f1 , . . . ,fm }) S = {f1 , . . . ,fm } P = {(fi ,fj )}i<j while P 6= ∅ choose (f,g) ∈ P P = P r {(f,g)} h = NF(spoly(f,g)|S) if h 6= 0 P = P ∪ {(f,h)|h ∈ S} S = S ∪ {h} return(S). • Prove that the algorithm terminates.

• Use 7.2.16 to prove the S is a standard basis of (f1 , . . . ,fm ). 7.2.23 (Standard bases for formal power series). Standard bases for formal power series can be defined as in 7.2.3. • Prove that the theory developed in this section holds in the formal case.

• Prove that {f1 , . . . ,fm } ⊂ C {x}N is a standard basis if and only if {f1 , . . . ,fm } ⊂ C [[x]]N is a standard basis.

282

7 Standard Bases • Let {f1 , . . . ,fm } ⊂ C {x}N be a standard basis and f ∈ C {x}N . Prove that NF(f |{f1 , . . . ,fm }) is the limit of the following sequence: (1) v0 = f . (2) Suppose vi is defined. Then vi+1 := vi − hfπ(i) . Here h = 0 and π(i) = 1 if no monomial in vi is divisible by L(fj ) for all j. Otherwise, let cxα be the smallest term in vi such that L(fj ) | xα for some j. Let π(i) be minimal with fπ(i) | xα . Then we α define h := c L(fx ) . π(i)

7.2.24 (Computation of standard bases in the power series ring). Let < be a well ordering on the set of monomials of C [x]. We choose a new variable t and extend this ordering in the following way to C [x,t]: x α ta > x β tb if deg(xα ) + a > deg(xβ ) + b or in case of equality xα < xβ . This extension has the following property: P Let f ∈ C [x], f = i≥1 ai xαi , ai 6= 0 and xαi < xαi+1 , that is, the monomials in f are ordered ` ´ in the sense of a power series ring (cf. Definition 7.1.3). Let F = f h = tdeg(f ) f xt1 , . . . , xtn be P αi αi the homogenization of f , then F = i≥1 ai xαi tdeg(f )−deg(x ) and we have xαi tdeg(f )−deg(x ) > αi+1

) xαi+1 tdeg(f )−deg(x , that is, the monomials of F are ordered in the sense of a polynomial ring (cf. Exercise 7.2.21). Prove the following: Let f1 , . . . ,fm ∈ C [x] and I = (f1 , . . . ,fm )C {x}, respectively I = (f1 , . . . ,fm )C [[x]]. Let < be a well-ordering on C [x] extended as before to C [x,t]. Let J = h (f1h , . . . ,fm )C [x,t] and {G1 , . . . ,Gr } be a standard basis of J. Then {G1 (t = 1), . . . ,Gr (t = 1)} is standard basis of I.3

7.3

Applications

Proposition 7.3.1. Let I ⊂ C {x} be a 0–dimensional ideal, then dimC C {x}/I = dimC C {x}/L(I). Proof. {xα |xα 6∈ L(I)} induces a C –basis of C {x}/L(I). We shall prove thatP this is also a C –basis of C {x}/I. Let SP= {f1 , . . . ,fm } be a standard basis of I. Assume cα xα ∈ I for some cα ∈ C , then NF( Pcα xα |S) = 0 but this is impossible, because S is a standard basis and all monomials in cα xα are not divisible by L(fi ) for all i. This implies that α α the {x |x 6∈ L(I)} induces a linearly independent set in C {x}/I. t Let xβ be a monomial in L(I) then NF(xβ |S) is a linear combination of the {xα |xα 6∈ L(I)} and, therefore, this set is also a basis. Now we want to prove that for w = (1, . . . ,1) the Hilbert–Samuel function of an ideal I ⊂ C {x} coincides with the corresponding function of the leading ideals (cf. Chapter 4.3). Corollary 7.3.2. Let I ⊂ C {x} be an ideal, then HSC {x}/I = HSC {x}/L(I) . 3

In practice this way to compute a standard basis is not very efficient. In the computer algebra system Singular for this purpose Mora’s algorithm, a modified version of Buchberger’s algorithm, see [Mora 1982] is implemented.

7.3 Applications

283


Proof. HSC {x}/I (c) = dimC C {x}/I + (x)c
= dimC C {x}/L I + (x)c by 7.3.1
= dimC C {x}/L(I) + (x)c = HSC {x}/L(I) (c). Finally, we shall prove Lemma 5.1.21 from Chapter 5.
Let f ,g ∈ C [x,y] quasi-homogeneous deg(x) = q, deg(y) = p and (x,y)–primary, let f = f + terms of higher degree, g = g+ terms of higher degree. Then xi y j ∈ (f,g) if iq + jp > w-ord(f · g). Proof. We choose w = (p,q) and compute a standard basis for (f,g). Because of the fact that f ,g are quasi-homogeneous and (f ,g) is (x,y)–primary, we may assume that L(f ) = xc0 and define g0 = f and g1 = NF(g|{g0 }), L(g1 ) = xc1 y b1 with c0 > c1 and b1 > 0. Assume we have already constructed {g0 , . . . ,gi−1 } with the following properties: (1) gj ∈ (f,g). (2) L(gj ) = xcj y bj , cj > cj+1 ,bj+1 > bj .
(3) gj = NF spoly(gj−2 ,gj−1 )|{g0 , . . . ,gj−1 } .


We define gi := NF spoly(gi−2 ,gi−1 )|{g0 , . . . ,gi−1 } .
Let L(gi ) = xci y bi then bi > bi−1 because L spoly(gi−2 ,gi−1 ) < xci−2 y bi−1 . Similarly, ci < ci−1 . Because the ci are strictly decreasing and (f ,g) is (x,y)–primary there is an m such
,g ) | {g , . . . ,g } . If
that L(gm ) = y bm and gm+1 = 0. Namely, gm+1 = NF spoly(gm−1 m 0 m
cm−1 bm y and, therefore, deg L(gm+1 ) > deg spoly(gm−1 ,gm gm+1 6= 0 then L(gm+1 ) > x
) = pbm + qcm−1 . We shall prove now that in this situation L(gm+1 ) ∈ L(g0 ), . . . ,L(gm ) , which is a contradiction for gm+1 being in normal form. At the same time we prove that {g0 , . . . ,gm } is a standard basis of (f,g). We use Theorem 7.2.16 and Lemma 7.2.18. The property cj > cj+1 and bj < bj+1 implies that L(gj−1 ) = xcj−1 y bj−1 |
ck bj lcm L(gj ),L(gk ) = x y for k ≤ j − 2. This implies that already G = {s(i,i−1) }i=1,...,m generates syz(g
0 , . . . ,gm ). (Lemma 7.2.18.) But by definition gi =
NF spoly(gi−2 ,gi−1 ) | {g0 , . . . ,gi−1 } and, therefore, NF spoly(gi−2 ,gi−1 )|{g0 , . . . ,gm } = 0. (4) of Theorem 7.2.16 implies that {g0 , . . . ,gm } is a standard basis of (f,g). Because of (f ,g) being (x,y)–primary, L(gm ) = y bm . Now deg(g0 ) = c0 p (the weighted degree) and deg(gi ) = deg(gi−1 ) + (ci−2 − ci−1 )p = deg(g1 ) + (c0 − ci−1 )p.

This implies that deg(gm ) = deg(g1 ) + (c0 − cm−1 )p = deg(g0 · g1 ) − cm−1 p = deg(f · g) − cm−1 p

284

7 Standard Bases y

bm

xcm−1 y bm ·xi y j

b1

cm−1

c0

x

The monomial xcm−1 y bm has, therefore, degree deg(f · g). It is the biggest monomial which is not in (f,g). This implies that xi y j ∈ (f,g) if iq + jp > deg(f · g) and proves the Lemma. As a totally different application, we can study the ideal of the product of germs of analytic spaces. Definition 7.3.3. Let (X, 0) ⊂ (C n , 0) and (Y, 0) ⊂ (C m , 0) be germs of analytic spaces. Let X and Y be representatives We define the product (X × Y, 0) to be the germ of (X × Y ) in 0. One checks without difficulty that this definition is independent on the choice of the representatives. Let I ⊂ C {x} be the ideal of (X, 0), and J ⊂ C {y} be the ideal of (Y, 0). Then it is a direct check that the product (X × Y, 0) is given as the zero set of (I + J) · C {x,y}. That this are all functions in the ideal of (X × Y, 0) is not at all trivial: Theorem 7.3.4. With the above notation one has I (X × Y, 0) = (I + J) · C {x,y}. Proof. The inclusion ⊃ is easy. For the converse, let f ∈ I (X × Y, 0). We have to show that f ∈ (I + J) · C {x,y}. Take a standard basis of J. By Grauert’s division Theorem 7.1.7, we may assume that no monomial of f is in L(J). We write X f= fα y α , fα ∈ C {x}. α

Therefore, for all α with fα 6= 0 we have that y α is not in the leading ideal of J. We claim that all fα are in I, which suffices to prove the Theorem (Exercise 7.3.7). We choose representatives X of (X, 0) and Y of (Y, 0) such that f is defined on X × Y and f (a,b) = 0 for all (a,b) ∈ X × Y . Suppose the converse. Then there exists an a ∈ X with fα (a) 6= 0 for some α. This is beause I = I (X, 0). For all b ∈ Y we have

7.3 Applications

285 f (a,b) =

X

fα (a)bα = 0,

α

that is, f (a,y) =

X α

fα (a)y α ∈ I (Y, 0) = J.

This is a contradiction, as the leading term of f (a,y) does not lie in L(J). This result motivates the following definition. Definition 7.3.5. Let A = C {x}/I and B = C {y}/J be analytic algebras. Then we b by define the analytic tensor product A⊗B b := C {x,y}/(I + J). A⊗B

Thus in case that A and B are reduced, and are the local rings of (X, 0) and (Y, 0) respectively, the analytic tensor product is the local ring of the product (X × Y, 0). b b Remark 7.3.6. We have obvious maps inclusions i : A −→ A⊗B, and j : B −→ A⊗B, which corresponds geometrically to the projection on the factors, if A and B are reduced. b −→ A. This corresponds On the other hand, we also have a natural surjective map A⊗B geometrically with the inclusion of one of the factors in the product. The analytic tensor product satisfies the following universal property. Let C be an analytic algebra, and γ : A −→ C and δ : B −→ C be analytic maps. Then there exist a b −→ C such that γ = θ ◦ i and δ = θ ◦ j. uniquely determined map θ : A⊗B In the case that A and B are reduced, this is easy to prove and follows from the fact that analytic maps (Z, 0) −→ (X, 0) and (Z, 0) −→ (Y, 0) induce obviously an analytic map (Z, 0) −→ (X × Y, 0). For the proof of the universal property and more facts on the analytic tensor product we refer to the book Analytische Stellenalgebren, see [Grauert-Remmert 1971].

Exercises: P 7.3.7. Let I ⊂ C {x1 , . . . ,xn } be an ideal and f ∈ C {x1 , . . . ,xn , y1 , . . . ,ym }, f = fα y α such that fα ∈ I for all α. Prove that f ∈ IC {x1 , . . . ,xn ,y1 , . . . ,ym }. (Hint: Use the fact that the normal form with respect to IC {x1 , . . . ,xn ,y1 , . . . ,ym } and the degree ordering with w = (1, . . . ,1) is additive and maps (x1 , . . . ,ym )c to (x1 , . . . ,ym )c .) 7.3.8. Let I ⊂ C {x0 , . . . ,x4 } be the ideal generated by the 2–minors of the matrix ( xx01 xx12 xx23 xx34 ). Prove that I is a prime ideal. (Hint: Consider the map ϕ : C {x0 , . . . ,x4 } −→ C {x0 ,x1 }x0 defined by ϕ(x2 ) = x10 x21 , ϕ(x3 ) = 1 3 x and ϕ(x4 ) = x13 x41 . Prove that Ker(ϕ) = I : (x0 )∞ . Then prove that I = I : (x0 ). To x2 1 0

0

see this, prove that (x1 ,x2 ,x3 )2 = L(I) the leading ideal of I with respect to the degree reverse lexicographical ordering. The six 2–minors of the matrix above are already a standard basis of I with respect to this ordering.)

286

8

Approximation Theorems

A canonical construction in local analytic geometry, which is possible without knowing the convergence of the solution, should have a convergent solution. This is always our wish and the underlying philosophy of the Approximation Theorems. We saw already in Chapters 3, 5, 6 and 7 that a lot of canonical formally constructed power series simply (and luckily) converge. Unfortunately, this is not always the case, as we will see in this chapter. This is the reason why we need different Approximation Theorems, adapted to the problem. We start with the famous Approximation Theorem of M. Artin: if a system of analytic equations has a formal solution (which is sometimes really not convergent) then we can find a convergent solution to this system. Even more, we can approximate the formal one by convergent ones up to a given order. Unfortunately, special properties of the formal solution (as not depending on some variables) do not survive during the approximation process in general. Therefore, for special situations, as needed later in deformation theory, we need special Approximation Theorems as Grauert’s Approximation Theorem proved in Section 8.2. In Section 8.3 we will give an overview about general Approximation Theorems and prove some applications.

8.1

Artin’s Approximation Theorem

Here it comes: Theorem 8.1.1 (Artin’s Approximation Theorem). Consider two sets of variables x = (x1 , . . . ,xn ) and y = (y1 , . . . ,yN ). Let f1 , . . . ,fm ∈ C {x,y} and y ∈ (x)C [[x]]N such that fi (y) = 0 for i = 1, . . . ,m. Let c be a positive integer. Then there exists a y ∈ C {x}N such that: • fi (y) = 0 for i = 1, . . . ,m. • y ≡ y modulo (x)c . The proof is by induction on n. The case n = 0 is trivial. The proof of the induction step is quite long, and will be divided in several lemmas. The first two lemmas will prepare us to prove the Artin Approximation Theorem for a special case. The general case will then be reduced to this special case. Lemma 8.1.2. Suppose that Artin’s Approximation Theorem holds for n − 1 variables x1 , . . . ,xn−1 . Consider the following data: • c, a positive integer. • f1 , . . . ,fm ∈ C {x,y}. • A formal Weierstraß polynomial P = xsn + a1 xns−1 + . . . + as , ai ∈ (x1 , . . . ,xn−1 )C [[x1 , . . . ,xn−1 ]].

8.1 Artin’s Approximation Theorem

287

• Elements: r1 , . . . ,r N ∈ C [[x1 , . . . ,xn−1 ]][xn ] of degree smaller than s in xn . Suppose that fi (r(x)) ≡ 0

modulo P for i = 1, . . . ,m.

Then there exists an analytic Weierstraß polynomial P of degree s in xn , and elements r1 , . . . ,rN ∈ C {x1 , . . . ,xn−1 }[xn ] of degree smaller than s with respect to xn such that fi (r(x)) ≡ 0

modulo P

and P ≡ P and ri ≡ ri modulo (x)c . Proof. We may write: ri =

s−1 X

(r ij + cij )xjn

j=0

r ij ∈ (x1 , . . . ,xn−1 ),

cij ∈ C .

We consider new variables: A1 , . . . ,As ;

Rij , 1 ≤ i ≤ N, 0 ≤ j ≤ s − 1

and define Pe := xsn + A1 xns−1 + . . . + As

ei := R

s−1 X (Rij + cij )xjn

1 ≤ i ≤ N.

j=0

We do m Weierstraß division with remainder for each fi and obtain e1 , . . . ,R eN ) = Qi · Pe + fi (R

s−1 X

Fij xjn

i = 1, . . . ,m,

j=0

for some Fij ∈ C {x1 , . . . ,xn−1 ,A1 , . . . ,As ,{Rij }}. Plugging in a1 , . . . ,as for A1 , . . . ,As and, similarly, r ij for Rij for 1 ≤ i ≤ N, 0 ≤ j ≤ s − 1, we see that we have formal solutions to the equations Fij = 0. We may apply the induction hypothesis to find the P convergent solutions ai and the rij . Then clearly P = Pe (a) and ri = (rij + cij )xjn satisfy the statement of the lemma. Lemma 8.1.3. Suppose that Artin’s Approximation Theorem holds for n − 1 variables x1 , . . . ,xn−1 . Let g,f1 , . . . ,fm ∈ C {x,y}, c ∈ N and formal power series y ∈ (x)C [[x]]N be given such that g(y) 6= 0 and fi (y) ≡ 0

modulo g(y)

i = 1, . . . ,m.

Then there exists a y ∈ C {x}N with y ≡ y modulo (x)c such that g(y) 6= 0 and fi (y) ≡ 0

modulo g(y)

i = 1, . . . ,m.

288

8 Approximation Theorems

Proof. Consider g(y) ∈ C [[x]]. After a linear coordinate change in the x1 , . . . ,xn , we may assume that g(y) is regular of order s in xn . We may enlarge c so that c > s. We apply the Weierstraß Preparation Theorem: g(y) = U · P , where P is a Weierstraß polynomial in xn of degree s, and U is a unit. In particular it follows that fi (y) ≡ 0 modulo P . For i = 1, . . . ,m we do Weierstraß divisions with remainder: y i = Qi P + r i , where r i ∈ C [[x1 , . . . ,xn−1 ]][xn ] are polynomials of degree smaller than s in xn . We will approximate the Qi ,P , and the r i . From yi ≡ ri modulo P it follows that g(r) ≡ g(y) ≡ 0

modulo P .

The first equivalence follows from the Taylor expansion, the second equivalence holds by construction of P . Similarly fi (r) ≡ fi (y) ≡ 0

modulo P .

By the previous lemma, applied to the functions g,f1 , . . . ,fm , we can find convergent ri ∈ C {x}, and a convergent Weierstraß polynomial P with ri ≡ r i and P ≡ P modulo (x)c such that g(r) ≡ 0 and fi (r) ≡ 0

modulo P

for i = 1, . . . ,m.

To define the yi , take any Qi ∈ C {x} such that Qi ≡ Qi modulo (x)c and put: yi := Qi P + ri . By construction yi ≡ y i modulo (x)c . Using Taylor’s formula again, as above, we see that g(y) ≡ 0 and fi (y) ≡ 0 modulo P

for i = 1, . . . ,m.

We are finished if we show that (8.1)

g(y) = U · P

for a unit U ∈ C {x}. It then follows immediately that fi (y) ≡ 0 modulo g(y) for i = 1, . . . ,m, and g(y) 6= 0 because P 6= 0. To show (8.1), observe that by the Weierstraß Division Theorem and the fact that g(y) ≡ 0 modulo P there exists a U ∈ C {x} with g(y) = U · P . We only have to show that U is a unit. By assumption, g(y) is regular of order s in xn . Because y ≡ y modulo xc and c > s it follows that g(y) is regular of order s in xn . As P is a Weierstraß polynomial of degree s in xn , it follows that U is a unit. This finishes the proof of the lemma. Lemma 8.1.4. Suppose that Artin’s Approximation Theorem holds for n − 1 variables x1 , . . . ,xn−1 . Then Artin’s Approximation Theorem for n variables x1 , . . . ,xn holds under the following assumptions: (1) m ≤ N .

8.1 Artin’s Approximation Theorem

289

(2) There exists an m–minor ∆ of the Jacobian matrix



∂fi ∂yj



such that ∆(y) 6= 0.

Proof. This we will prove as an application of Newton’s Lemma, which itself follows from the Implicit Mapping Theorem. Take any nonzero function h ∈ (x)c . Then because fi (y) = 0, we certainly have: fi (y) ≡ 0 modulo ∆2 (y)h. We can apply the previous lemma to find a y ′ ∈ C {x}N , y ′ ≡ y modulo (x)c , such that fi (y ′ ) ≡ 0 modulo ∆2 (y ′ )h. Putting J to be the ideal generated by the m–minors of the Jacobian matrix we certainly have fi (y ′ ) ≡ 0 modulo J 2 (y ′ )(x)c , so that Newton’s Lemma 3.3.34 applies: there exists a y ∈ C {x}N , y ≡ y ′ modulo J(y ′ )(x)c , such that fi (y) = 0 for all i = 1, . . . ,m. (Note that we needed m ≤ N in order to apply Newton’s Lemma.)

Proof of Artin’s Approximation Theorem 8.1.1. Define the ideal p to be the kernel of the map: α : C {x,y} −→ C [[x]] defined by x 7→ x, and y 7→ y(x). Note that as we have an injective map C {x,y}/p ֒→ C [[x]], it follows that p is a prime ideal. As f1 , . . . ,fm ∈ p it suffices to find a y(x) ∈ C {x}N such that g(x,y(x)) = 0 for all g ∈ p. So we may have assumed p = (f1 , . . . ,fm ) from the beginning. Let ht(p) = k. By the Jacobian Criterion 4.3.15 applied to the case I = p = q, there exist k variables z1 , . . . ,zk among the xi and the yi , such that (after renumbering the fi )   ∂fi ∈ / p. ∆′ := det ∂zj 1≤i,j≤k In particular ∆′ (x,y(x)) 6= 0. As 0 = fi (x,y(x)) ∈ C [[x]], it follows that for all j. From the chain rule it follows that

∂fi (x,y(x)) ∂xj

=0

X ∂fi ∂y (x) ∂fi (x,y(x)) = − (x,y(x)) · k . ∂xj ∂yk ∂xj k

Hence, ∆′ (x,y(x)) is in the ideal generated by the k–minors of the matrix   ∂fi . (x,y(x)) 1≤i≤m ∂yj 1≤j≤N

In particular, it follows that one of those minors is nonzero. We may, therefore, apply the previous lemma to find a convergent y ∈ C {x}N with y ≡ y modulo (x)c and f1 (x,y(x)) = . . . = fk (x,y(x)) = 0. Now C {x,y}p /p is a regular local ring. (It is even a field.) In the Jacobian Criterion 4.3.15 it was shown that f1 , . . . ,fk generate pC {x,y}p , that is,

290

8 Approximation Theorems (f1 , . . . ,fk )C {x,y}p = pC {x,y}p .

Because C {x,y} is Noetherian, we can find h ∈ / p such that (8.2)

((f1 , . . . ,fk ) : h) = p.

As h ∈ / p = Ker(α) it follows that h(x,y(x)) 6= 0. By enlarging c we may assume that h(x,y(x)) ∈ / (x)c . Hence y ≡ y modulo (x)c implies that h(x,y(x)) ∈ / (x)c . In particular h(x,y(x)) 6= 0. For any g ∈ p it follows from formula (8.2) that hg ∈ (f1 , . . . ,fk ). As h(x,y(x)) 6= 0 it follows that g(x,y(x)) = 0. This is what we had to show.

Exercises 8.1.5. Deduce from the Artin Approximation Theorem the following somewhat more general version. N ` Let ´f1 , . . . ,fm ∈ C {x,y}, and J ⊂ C {x} be an ideal. Let y(x) ∈ C [[x]] such that fi x,y(x) ∈ J · C [[x]] for i = 1, . . . ,m. Then there exists a y(x) ∈ C {x}N such that fi (x,y(x)) ∈ J for all i = 1, . . . ,m. 8.1.6. Let the conditions be as in the previous exercise. Let I ⊂ C {x} be an ideal. Suppose we have a formal solution y(x) whose components lie in I · C [[x]]. Then we can find a convergent solution in I. Prove this. 8.1.7. Let p ⊂ C {x} be a prime ideal. Show that p · C [[x]] is a prime ideal. (Hint: Consider f,g ∈ C [[x]] with f · g ∈ p · C [[x]]. Use Exercise 8.1.5 to find for all c ∈ N fc ,gc ∈ C {x}, fc ≡ f modulo (x)c and gc ≡ g modulo (x)c , and fc · gc ∈ p. Deduce that either f = limc→∞ fc ∈ p · C [[x]] or g = limc→∞ gc ∈ p · C [[x]].) 8.1.8. Let p ⊂ C {x} be a prime ideal. Suppose that C {x}/p is normal. Use Exercise 8.1.5 to see that C [[x]]/p · C [[x]] is normal.

8.2

Grauert’s Approximation Theorem

In this section we prove Grauert’s Approximation Theorem. This approximation theorem looks rather complicated, but is very useful. For example, we will use this theorem to prove the existence of a ‘semi-universal’ deformation of an isolated singularity in Chapter 10. Definition 8.2.1. We consider four sets of variables, x = (x1 , . . . ,xn ), s = (s1 , . . . ,sl ), Φ = (Φ1 , . . . ,Φp ) and Ψ = (Ψ1 , . . . ,Ψq ). Let I ⊂ C {s} be an ideal. We can extend I to an ideal I ⊂ C {x,s}. Consider an element F = (F1 , . . . ,Fk ) ∈ C {x,s,Φ,Ψ}k . (1) An analytic solution of the equation F ≡ 0 modulo I is a pair (φ,ψ) ∈ C {s}p × C {x,s}q , that is power series φ1 , . . . ,φp ∈ C {s}, and ψ1 , . . . ,ψq ∈ C {x,s} such that F (x,s,φ(s),ψ(x,s)) ≡ 0 modulo I · C {x,s}k , that is, the φ and the ψ have the property that F1 (x,s,φ(s),ψ(x,s)), . . . ,Fk (x,s,φ(s),ψ(x,s)) ∈ I.

8.2 Grauert’s Approximation Theorem

291

(2) A solution of order e of the equation F ≡ 0 modulo I is a pair (φ,ψ) ∈ C [s]p × C {x}[s]q such that F (x,s,φ(s),ψ(x,s)) ≡ 0

modulo (I + me+1 ) · C {x,s}k ,

where m = (s) = (s1 , . . . ,sl ). We can now formulate the theorem. Theorem 8.2.2 (Grauert’s Approximation Theorem). Let the notation be as in the previous definition. Let e0 ∈ N. Suppose that the system of equations F ≡ 0 modulo I has a solution (φ(e0 ) ,ψ (e0 ) ) of order e0 . Suppose, moreover, that for all e ≥ e0 every solution (φ(e) ,ψ (e) ) of order e with φ(e) ≡ φ(e0 ) modulo me0 +1 , and ψ (e) ≡ ψ (e0 ) modulo me0 +1 , extends to a solution (φ(e) + δ (e) ,ψ (e) + γ (e) ) of order e + 1, with δ (e) ∈ C [s]p and γ (e) ∈ C {x}[s]q homogeneous of degree e + 1 in s. Then the system of equations F ≡ 0 modulo I has an analytic solution (φ,ψ), with φ ≡ φ(e0 ) modulo me0 +1 , and ψ ≡ ψ (e0 ) modulo me0 +1 . Before giving the proof, we need some preparations. As in Chapter 7, we choose as a basis for the free module C {x,s,Φ,Ψ}k the vectors ei := (0, . . . ,0,1,0, . . . ,0) (the 1 is on the i–th spot) and identify ei with xi−1 n+1 by adding a new variable xn+1 and enlarging x to x = (x1 , . . . ,xn+1 ). We choose on C {x,s,Φ,Ψ}k the degree lexicographical ordering as in Chapter 7 with weights 1 for all variables. For an element F ∈ C {x,s,Φ,Ψ}k there exists a polyradius n+1+l+p+q λ ∈ R+ such that F ∈ Bλk . We will write, in order to distinguish between the p+q l variables x,s and φ,ψ, the polyradius as λ = (ρ,τ,σ), ρ ∈ Rn+1 + , τ ∈ R+ , σ ∈ R+ . Proposition 8.2.3. Let b1 , . . . bt ∈ C {x}k and ρ ∈ Rn+1 such that b1 , . . . bt ∈ Bρk be + given. Then for all ρ ≤ ρ there exists a K(ρ) ∈ R+ such that the following holds. For all C ∈ C {x}[s]k , C = C [s] such that

P

|β|=e Cβ s

β

with Cβ ∈ C {x} and for all y1 , . . . ,y t ∈

(1) yj is a homogeneous polynomial of degree e in s for j = 1, . . . ,t, (2)

t P

bj y j = C,

j=1

there exist yj ∈ C [s], j = 1, . . . ,t such that (1) yj is a homogeneous polynomial of degree e in s for j = 1, . . . ,t, (2)

t P

bj yj = C,

j=1

(3) kyj kτ ≤ K(ρ)kCk(ρ,τ ) for all j and all ρ ≤ ρ, τ ∈ Rl+ . The important thing to note here, is that K(ρ) does not depend on C, e and τ . However, it depends (of course) on the bi and on ρ.

292

8 Approximation Theorems

Proof. This is an application of Cramer’s rule. Step 1. We write bi =

X

bi,α xα , C =

α∈Nn+1 +

X

Cα,β xα sβ , y j =

α∈Nn+1 + |β|=e

X

y j,β sβ .

|β|=e

The condition then reads t X j=1

bj,α y j,β = Cα,β for all α ∈ Nn+1 and |β| = e. +

This we can view as system of linear equations. More precisely, consider the linear subspace in C t spanned by the vectors b(α) := (b1,α , . . . ,bt,α ), α ∈ Nn+1 + . Let r be the dimension of this subspace and choose a basis {b(αi ) = (b1,αi , . . . ,bt,αi ), i = 1, . . . ,r}. Pr So for any α ∈ Nn+1 , we have an expression b(α) = i=1 ki,α b(αi ) for some ki,α ∈ C . By linear algebra, there is an r–minor M of the matrix
bi,αj 1≤i≤t 1≤j≤r

which is nonzero. After renumbering the bi we may assume that M = det(bi,αj )1≤i,j≤r . We define P (i,β) (1) for i = 1, . . . ,r, yi := |β|=e yi,β sβ with yi,β = MM , (2) yr+1 := · · · := yt := 0.

Here M (i,β) is obtained from M by replacing the i–th column of M by the column vector Pt (Cα1 ,β , . . . ,Cαr ,β )t . We want to show j=1 bj,α yj,β = Cα,β for all α. By Cramer’s rule, we know this is true for α = α1 , . . . ,αr . Now X X X X X bj,α yj,β , ki,α bj,αi yj,β = ki,α bj,αi y j,β = ki,α Cαi ,β = bj,α y j,β = Cα,β = j

i,j

i,j

i

j

as was to be shown. Step 2. It remains to define the K = K(ρ) and to show the estimates. We take K = max i,j

|Mij | . ραj |M |

Here Mij is the determinant obtained from M by deleting the i–th row and the j–th column. As yr+1 = · · · = yt = 0, we only have to show the estimates for the y1 , . . . ,yr . By definition X |M (i,β) |

X M (i,β) β s τ = τβ. kyi kτ = M |M | |β|=e

|β|=e

8.2 Grauert’s Approximation Theorem

293

Moreover, by linear algebra we have M (i,β) = r

Pr

j=1

Cαj ,β Mij and, therefore,

|Mij | |M (i,β) | X |Cαj ,β | ≤ . |M | |M | j=1 This implies kyi kτ ≤

r 1 X X |Cαj ,β | |Mij |τ β . |M | j=1 |β|=e

On the other hand by the definition of the norm kCk(ρ,τ ) ≥

r X X

|β|=e j=1

|Cαj ,β |τ β ραj .

Thus with our choice of K the inequality kyi kτ ≤ K · kCk(ρ,τ ) holds. We need the following generalization of this proposition. Lemma 8.2.4 (Cartan’s Lemma). Let a1 , . . . ,ar , b1 , . . . ,bt ∈ C {x}k and ρ ∈ Rn+1 + , τ ∈ Rl+ . Suppose that a1 , . . . ,ar , b1 , . . . ,bt ∈ Bρk . There exists a ρ ∈ Rn+1 ρ, and , ρ ≤ + K = K(ρ) ∈ R+ with the following property.

P (α) α k s with C (α) ∈ C {x} For all τ ∈ Rl+ with τ ≤ τ and for all C ∈ B(ρ,τ |α|=e C ), C = and for all y1 , . . . ,y t ∈ C [s] and z 1 , . . . z r ∈ C {x}[s] such that (1) yj is a homogeneous polynomial of degree e in s for j = 1, . . . ,t, and z i is a homogeneous polynomial of degree e in s for i = 1, . . . ,r with coefficients in C {x}, (2)

r P

ai z i +

i=1

t P

j=1

bj yj = C,

there exist zi ∈ C {x}[s], i = 1, . . . ,r and yj ∈ C [s], j = 1, . . . ,t, such that (1) yj is a homogeneous polynomial of degree e in s for j = 1, . . . ,t, and zi is a homogeneous polynomial of degree e in s for i = 1, . . . ,r with coefficients in C {x}, (2)

r P

ai z i +

i=1

t P

bj yj = C,

j=1

(3) kzi k(ρ,τ ) ≤ KkCk(ρ,τ ) and kyj kτ ≤ KkCk(ρ,τ ) for all i and j. The important thing to note here, is that K does not depend on C, e, and τ . However, it depends (of course) on the ai , the bi and ρ.

294

8 Approximation Theorems

Proof. Step 1. The case r = 0 is dealt with in the previous proposition. We consider the normal form NF with respect to the submodule (a1 , . . . ,ar ). Note that as the y j do not Pr Pt depend on x, we have NF(yj ) = y j . We take the normal form of i=1 ai z i + j=1 bj yj = C and get t X

(8.3)

NF(bj )y j = NF(C).

j=1

We apply the previous proposition. There exists a K0 (ρ) depending on the NF(bj ) and P ρ, and y1 , . . . ,yt ∈ C [s] homogeneous of degree e such that tj=1 NF(bj )yj = NF(C) and (8.4)

kyj kτ ≤ K0 (ρ)k NF(C)k(ρ,τ ) for all ρ ≤ ρ.

To define the zi , note that from 7.2.10 it follows that we can write (8.5)

bj =

r X

αij ai + NF(bj ),

C=

r X

βi ai + NF(C).

i=1

i=1

The precise αij and βi we take will be determined in the second step. Plugging (8.5) in Pr Pt i=1 ai z i + j=1 bj y j − C = 0 and using (8.3) we get r X

zi +

i=1

t X j=1

Now we define zi := −


αij y j − βi ai = 0.

t X

αij yj + βi .

j=1

Obviously, r X

zi +

t X


αij yj − βi ai = 0.

j=1

NF(bj )yj = NF(C) we get

j=1

i=1

Working backwards, that is, using

Pt

r X i=1

ai z i +

t X

bj yj = C.

j=1

Step 2. We still have to tell which αij and βi we take that fulfill (8.5), and show the estimates. To do this, we take a standard basis f1 , . . . ,fm of the submodule (a1 , . . . ,ar ). We can write r X hνi ai fν = i=1

for some hνi ∈ C {x}. Choose ρ ≤ ρ so small that a1 , . . . ,ar ,b1 , . . . ,bt ,f1 , . . . ,fm , h11 , . . . ,hmr ∈ Bρk . We apply Grauert’s Division Theorem, see Remark 7.1.7 with ε = 21 . Therefore, by making ρ smaller we get equations

8.2 Grauert’s Approximation Theorem

295

X

bj =

bjν fν + NF(bj )

ν

X

C=

Cν fν + NF(C),

ν

with the following estimates. kb k

ρ ; (1) kbjν kρ ≤ 2 kL(fjν )k ρ

kCk

) (2) kCν k(ρ,τ ) ≤ 2 kL(f(ρ,τ ; ν )kρ

(3) k NF(C)k(ρ,τ ) ≤ 2kCk(ρ,τ ). In particular from formula (8.4) we get for K0 := K0 (ρ) (8.6)

kyj kτ ≤ K0 k NF(C)k(ρ,τ ) ≤ 2K0 kCk(ρ,τ ) ,

so that we have an estimate for the yi . We define X X αij := bjν hνi , βi := hνi Cν . ν

ν

With these αij and βi the equations (8.5) hold. Let L = 2m · maxν,i kαij kρ ≤

X ν

khνi kρ kbjν kρ ≤ Lkbj kρ ,

n

khνi kρ kL(fν )kρ

o

. Then

kβi k(ρ,τ ) ≤ LkCk(ρ,τ ) . From the definition of the zi and formula (8.6) we get kzi k(ρ,τ ) ≤ ≤

t X j=1

t X j=1

kαij kρ kyj kτ + kβi k(ρ,τ ) Lkbj kρ kyj kτ + LkCk(ρ,τ )

≤ L max{kbj kρ kyj kτ } + LkCk(ρ,τ ) j

≤ L max{2K0 kbj kρ + 1}kCk(ρ,τ ). j

Let K ′ = L maxj {2K0 kbj kρ + 1} and K = max{2K0,K ′ }. With this K the desired estimates hold. We need one more result before we can give the proof of Grauert’s Approximation Theorem. Lemma 8.2.5. Let F (x,s,Θ) ∈ C {x,s,Θ}k be a vector of power series in three sets of variables x = (x1 , . . . ,xn ), s = (s1 , . . . ,sl ) and Θ = (Θ1 , . . . ,Θt ). Let I ⊂ C {s} be an n+1+l+t k . Then for ideal. Assume that F ∈ B(ρ,τ ,σ) for a suitable polyradius (ρ,τ ,σ) ∈ R l all ε > 0, there exists a K = K(ε) > 0 and a τ ∈ R+ , τ ≤ τ , with the following property.
(e) (e) Let e ∈ N and θ(e) (x,s) = θ1 (x,s), . . . ,θt (x,s) such that

296

8 Approximation Theorems

(1) the θi (x,s) are polynomials of degree ≤ e in s with coefficients in C {x}, (2) the θi (x,s) are in normal form with respect to I, (e)

(3) kθν (x,s)k(ρ,λτ ) ≤ K for ν = 1, . . . ,t and some λ, 0 < λ ≤ 1. Let Fe ∈ C {x,s}k be the degree e + 1 part in s of the normal form of F (x,s,θ(e) (x,s)) with respect to I, then: kFe k(ρ,λτ ) ≤ Kε. Proof. Let 0 < α < 1 be given. Then there exists a polyradius τ ∈ Rl+ such that for all f ∈ B(ρ,τ ) k NF(f |I)k(ρ,τ ) ≤

(8.7)

1 kf k(ρ,τ ). 1−α

Given α and (ρ,τ ), this inequality still holds if we replace τ by λτ for any λ with 0 < λ ≤ 1, see Remark 7.1.7. We develop F : X F (x,s,Θ) = Fν (x,s)Θν . ν∈Nt

As this series is convergent, there exists an A > 0 such that kFν (x,s)k(ρ,τ ) ≤ A|ν| . (cf. Exercise 8.2.6). We choose K so small that AK < 1. It is an exercise to show that we can choose K = K(ε) so small that even X

(AK)|ν| =

|ν|≥2

1 ε(1 − α) − 1 − tAK < K , t (1 − AK) 3

see Exercise 8.2.7. We write: Fν (x,s) = Fν (x,0) + Fν′ (x,s)

with Fν′ ∈ (s)

for |ν| ≤ 1.

, and δ < ε(1−α) We choose δ so small that δ < ε(1−α)K 3 3t . By making τ smaller we may ′ assume that kFν (x,s)k(ρ,τ ) ≤ δ for all ν with |ν| ≤ 1. Now +

X

|ν|=1

F (x,s,θ(e) (x,s)) = F0 (x,0) + F0′ (x,s) X X Fν (x,s)θ(e) (x,s)ν . Fν′ (x,s)θ(e) (x,s)ν + Fν (x,0)θ(e) (x,s)ν + |ν|≥2

|ν|=1

By assumption the θ(e) (x,s) are in normal form with respect to P I. Moreover the Fν (x,0) for all ν with |ν| ≤ 1 do not depend on s. It follows that F0 (x,0)+ |ν|=1 Fν (x,0)θ(e) (x,s)ν P is in normal form with respect to I. Furthermore F0 (x,0) + |ν|=1 Fν (x,0)θ(e) (x,s)ν does not have terms of degree e + 1 in the variables s. It follows that Fe , which by definition is the degree e + 1 part in s of the normal form of F (x,s,θ(e) (x,s)), is equal to the degree e + 1 part of the normal form of X X Fν (x,s)θ(e) (x,s)ν . Fν′ (x,s)θ(e) (x,s)ν + F0′ (x,s) + |ν|=1

Thus, because of (8.7)

|ν|≥2

8.2 Grauert’s Approximation Theorem (1 − α)kFe k(ρ,λτ ) ≤ kF0′ (x,s) +

X

297

Fν′ (x,s)θ(e) (x,s)ν +

|ν|=1

≤ δ + tδK +

X

X

|ν|≥2

Fν (x,s)θ(e) (x,s)ν k(ρ,λτ )

|ν|

(AK) .

|ν|≥2

By choice of K and δ each term on the right hand side is smaller than or equal to (1 − α) εK 3 . Hence kFe k(ρ,λτ ) < εK, as was to be shown. Proof of Theorem 8.2.2. ∂F Step 1. We first reduce to the case that φ(e0 ) and ψ (e0 ) are in m := (s) and ∂Φ , s=0 i ∂F k ∂Ψi s=0 ∈ C {x} , that is, they do not depend on the Φ and Ψ. Indeed, we can write φ(e0 ) = c +

X

0 )′ sν (cν + φ(e ) ν

ψ (e0 ) = d +

(e )′

(e )′

X

sν (dν + ψν(e0 )′ ). (e0 )′

Here c,d,cν and dν are in C {x}. Moreover φν 0 = {φi,ν0 }i=1,...,p , ψν (e )′

(e )′

(e )′

= {ψi,ν0 }i=1,...,q

with φi,ν0 ∈ m, ψi,ν0 ∈ m · C {x,s}. Define new variables Φ′ = {Φ′iν },Ψ′ = {Ψ′iν }, and define X X F ′ (x,s,Φ′ ,Ψ′ ) := F (x,s,c + sν (cν + Φ′ν ),d + sν (dν + Ψ′ν )). ′ ′
′ ′ Now F ′ has a solution φ(e0 ) ,ψ (e0 ) of order e0 . Replacing F,φ(e0 ) , ψ (e0 ) by F ′ ,φ(e0 ) ,ψ (e0 ) proves the claim.

Step 2. We can decide whether F (φ,ψ) ∈ I by looking at the normal form NF(F (φ,ψ)|I). In fact, in 7.2.10 it is proved that: F (φ,ψ) ∈ I ⇐⇒ NF(F (φ,ψ)|I) = 0. The normal form NF(F (φ,ψ)|I) is a power series, all of whose terms are not in L(I), the ideal of leading terms of I. Moreover, NF(F (φ,ψ)|I) is congruent to F (φ,ψ) modulo I. If (φ,ψ) is a solution of F ≡ 0 modulo I, then it follows immediately that (NF(φ|I), NF(ψ|I)) is also a solution of F ≡ 0 modulo I. Similar remarks hold for solutions of order e, as P P obviously aα sα being in normal form implies |α|≤e aα sα is in normal form. So we will and may assume that all our solutions are in normal form. n+1+l+p+q k Step 3. We now choose the polyradius. Let (ρ,τ ,σ) ∈ R+ such that F ∈ B(ρ,τ ,σ) . We consider the elements

∂F ∂F ∈ C {x}k and ∈ C {x}k , for i = 1, . . . ,p and j = 1, . . . ,q. s=0 ∂Φ i ∂Ψ j s=0

n+1 = M (ρ) ≥ 1 depending By Cartan’s Lemma 8.2.4 we can choose a ρ ∈ R+ , ρ ≤ ρ and M ∂F ∂F k on ∂Ψ s=0 and ∂Φ s=0 with the following property. Let G ∈ B(ρ,τ ) , and δ 1 , . . . ,δ p ∈ C [s] and γ 1 , . . . ,γ q ∈ C {x}[s] such that

G+

p X i=0

δi

q X ∂F ∂F γi + = 0, s=0 ∂Φi ∂Ψi s=0 i=0

and G, δ 1 , . . . ,δ p , γ 1 , . . . ,γ q are homogeneous with respect to s of degree e + 1 for e ≥ e0 . Then there exist δ1 , . . . ,δp ∈ C [s] and γ1 , . . . ,γq ∈ C {x}[s] satisfying this equation being homogeneous as before, where, moreover,

298

8 Approximation Theorems kδi kτ , kγi k(ρ,τ ) ≤ M kGk(ρ,τ ) for all τ ≤ τ .

1 . Let K = K(ε) and τ ≤ τ as in the previous lemma, applied to F (x,s,Φ,Ψ) We take ε = M (e0 ) (e0 ) and let (φ ,ψ ), a solution of order e0 . Because φ(e0 ) ∈ m and ψ (e0 ) ∈ mC {x,s} we can choose, by scaling, the polyradius τ so small that, moreover, kφ(e0 ) kτ < K and kψ (e0 ) k(ρ,τ ) < K.

Step 4. We will for all e ≥ e0 +1 construct solutions (φ(e) ,ψ (e) ) of order e with (φ(e) ,ψ (e) ) ≡ (φ(e−1) ,ψ (e−1) ) modulo me and with the estimates: kφe kτ , kψe k(ρ,τ ) < K. Here φe and ψe are the degree e parts in s of φ(e) and ψ (e) . Let (φ(e) ,ψ (e) ) be a solution of order e, such that (φ(e) ,ψ (e) ) are polynomials in s of degree ≤ e, in normal form with respect to I and kφe kτ ≤ K, kψe k(ρ,τ ) ≤ K. Let Fe be the degree e + 1 part in s of K . By the normal form of F (x,s,φ(e) ,ψ (e) ). By the previous lemma, kFe k(ρ,τ ) < εK = M (e)

assumption, there exist δ and γ (e) homogeneous of degree e + 1 in s and in normal (e) form with respect to I such that (φ(e) + δ ,ψ (e) + γ (e) ) is a solution of order e + 1. By Taylor expansion and taking the degree e + 1 part of the normal form we obtain Fe +

p q X X ∂F ∂F (e) (e) |s=0 δ i + |s=0 γ i = 0. ∂Φ ∂Ψ i i i=1 i=1 (e)

By choice of M there exist such δi

(e)

and γi , having moreover the estimates:

(e)

(e)

kδi kτ , kγi k(ρ,τ ) < M kFe k(ρ,τ ) < M

K = K. M

Now φ(e+1) := φ(e) + δ (e) and ψ (e+1) := ψ (e) + γ (e) . Then φe+1 = δ (e) and ψe+1 = γ (e) , and the desired estimate holds. P P∞ Step 5. We now look at the formal solution φ = ∞ e=0 φe , ψ = e=0 ψe . Then, because φe and ψe are homogenous of degree e with respect to s, and kφe kτ < K, kψe k(ρ,τ ) < K we obtain kφkτ /2 <

∞ X e=0


1 e 2

· K = 2K, kψk(ρ,τ /2) <

∞ X e=0


1 e 2

· K = 2K,

showing the convergence of φ and ψ. This concludes the proof of Grauert’s Approximation Theorem.

Exercises P 8.2.6. Let f (x,Θ) = ν∈Nt fν Θν . Let (ρ,τ ) be a polyradius with kf k(ρ,τ ) < ∞. Show that there exists an A with kfν kρ ≤ A|ν| . 1 8.2.7. Let c > 0, prove that for all x with |x| sufficiently small (1−x) t < (c + t)x + 1. (Hint: Check the monotony behavior of the real function corresponding to the above equality.)

8.3 Some other Approximation Theorems

8.3

299

Some other Approximation Theorems

Now we would like to discuss, without proof, a generalization of Artin’s Approximation Theorem 8.1.1: Theorem 8.3.1. Let f1 , . . . ,fm ∈ C {x,y}. There is a function Θ : N → N with the following property. Let fi (y) ≡ 0 mod (x)Θ(c) , i = 1, . . . ,m, for y ∈ (x)C {x}N , then there exist y ∈ C {x}N ,

fi (y) = 0,

y≡y

mod (x)c and

i = 1, . . . ,m.

The point here is that one does not even have to find a formal solution, but only a solution up to high enough order. The weakness of this theorem however is, that it seems not to be possible in general to compute the function Θ. The theorem is a consequence of Artin’s Approximation Theorem and the following theorem, proved in [Pfister-Popescu 1975], [Kurke et al. 1978]. Theorem 8.3.2. Let K be a field and f1 , . . . ,fm ∈ K[[x,y]], then there exists a function Θ : N → N with the following property: Let fi (y) ≡ 0 mod (x)Θ(c) , i = 1, . . . ,m, for y ∈ (x)K[[x]]N . Then there exists y ∈ K[[x]]N such that y ≡ y mod (x)c and fi (y) = 0, i = 1, . . . ,m. Artin’s Approximation Theorem 8.1.1 has been generalized in many directions. The most general result is due to Dorin Popescu (cf. [Popescu 1985], [Popescu 1986, 1990]) and has been reproved by Marc Spivakovsky, see [Spivakovsky 1999] and Richard Swan [Swan 1998]. Theorem 8.3.3 (Popescu). Let A be an excellent Noetherian, and with respect to an bN ideal a, Henselian ring. Let f1 , . . . ,fm ∈ A[Y ]. Suppose fi (y) = 0, i = 1, . . . ,m, y ∈ A a (the so-called a–adic completion). Let c > 0 be an integer. Then there exists a y ∈ AN such that y ≡ y mod ac and fi (y) = 0, i = 1, . . . ,m. We do not want to explain the notion of excellence. This can be found in Matsumura’s book (cf. [Matsumura 1979]). Almost all rings occurring in this book, or more generally in algebraic geometry, are excellent, as, for instance, the rings K[[x]],C {x},K[x] their quotients and their polynomial extensions. The ring A is Henselian with respect to the ideal a if Hensel’s Lemma holds with respect to a (cf. Corollary 3.3.21, replace there (x) by a, cf. also [Kurke et al. 1975]). Another important example of an excellent Henselian local ring is the ring Khxi of algebraic power series over a field K, that is Khxi = {f ∈ K[[x]], F (f ) = 0 for a suitable F ∈ K[x,T ], F 6= 0}. We would like to finish this chapter with some important remarks concerning generalizations of the previous theorem. Remark 8.3.4. A generalization of Theorem 8.3.3 similar to Theorem 8.3.1 is not possible (cf. a counterexample of Spivakovsky [Spivakovsky 1994]). Remark 8.3.5. Assume that some of the y i in Theorem 8.1.1 do not depend on some of the xi . In general, it is not possible to find convergent solutions y = (y1 , . . . ,yN ) with the same property. Gabrielov, see [Gabrielov 1971] gave a counterexample of the following type:

300

8 Approximation Theorems

f = aY1 + bY2 + cY3 + d ∈ C {x1 ,x2 ,x3 , Y1 ,Y2 ,Y3 }, y = (y 1 ,y2 ,y3 ) ∈ C [[x1 ,x2 ,x3 ]] such that f (y) = 0, y2 ,y3 ∈ C [[x1 ,x2 ]] and there are no y = (y1 ,y2 ,y3 ) ∈ C {x1 ,x2 ,x3 } with f (y) = 0 and y2 ,y3 ∈ C {x1 ,x2 }. Remark 8.3.6. In the following special situation the problem of Remark 8.3.5 can be solved. Let K be a field and f1 , . . . ,fm ∈ Khx,Y i, assume fi (y) = 0 for y ∈ (x)K[[x]]. Assume yi ∈ K[[x1 , . . . ,xvi ]] 1 ≤ v1 ≤ · · · ≤ vN ≤ n. Let c > 0 be an integer, then there exists a y ∈ KhxiN such that y ≡ y mod (x)c yi ∈ Khx1 , . . . ,xvi i fi (y) = 0,

i = 1, . . . ,m.

As we will not explain here, this theorem is a consequence of Popescu’s general theorem (cf. [Popescu 1985], [Popescu 1986, 1990]). We obtain the following corollary: Corollary 8.3.7 (Grauert’s Theorem for Algebraic Power Series). We take the notations of and assumptions of 8.2.1 and 8.2.2. Assume furthermore that I ⊂ C hsi and F ∈ C hx,s,φ,ψik . Then the system of equations F ≡ 0 modulo I has an algebraic solution (φ,ψ) with φ ≡ φ(e0 ) modulo me0 +1 , and ψ ≡ ψ (e0 ) modulo me0 +1 . Artin’s Approximation Theorem and its generalizations have many applications. As an application we give Artin’s answer to a question of Grauert. Proposition 8.3.8. Let A,B,C be analytic C –algebras and w : C −→ B a homomorphism. (1) Assume that for a surjective homomorphism ϕ : A −→ B the following diagram is commutative: /B b B OO G OO  ϕ ϕ ˆ  w b 7/ A A  ooo o  o ooo  ooooo u o C Then, for a given positive integer c, there exists a u : C −→ A such that ϕ ◦ u = w and u ≡ u mod mcAb.

(2) Assume the following diagram is commutative:

b _? A O ? ?? u ?? ?? AO ?B ~~ ~ v ~~w ~~ C Then for a given positive integer c there exists a u : B −→ A such that v = u◦w and b −→ A b is an isomorphism, then u is also an isomorphism. ˆ:B u ≡ u mod mcAb. If u

8.3 Some other Approximation Theorems

301

Proof. Let A = C {x}/(f1 , . . . ,fa ), B = C {y}/(g1 , . . . ,gb ), C = C {z}/(h1, . . . ,hc ), x = (x1 , . . . ,xn ), y = (y1 , . . . ,ym ), z = (z1 , . . . ,zr ).
In case of (1), we may assume y = x and ϕ xi mod (f1 , . . . ,fa ) = xi mod (g1 , . . . ,gb ). Let zi (x) ∈ C {x} be representatives of w(zi ) and y i (x) ∈ C [[x]] be representatives b = C [[x]]/(f1 , . . . ,fa )). Then we have, because of the commutativity of u(zi ) (note that A of the diagram zi (x) ≡ yi (x) mod (g1 , . . . ,gb ), that is, zi (x) − yi (x) =

b X

αij gj , i = 1, . . . ,r

j=1


for suitable αij ∈ C [[x]]. Furthermore, we have hj y(x) ≡ 0 mod (f1 , . . . ,fa ) because
of the definition of y and the fact that C = C {z}/(h1, . . . ,hc ). This implies hi y(x) = a P β ij fj for suitable β ij ∈ C [[x]]. j=1

Now consider the following systems of equations zi (x) − yi − hi (y) −

b X

j=1 a X

Aij gj = 0 Bij fj = 0

j=1

in the variables y,{Aij },{Bij }. It has the formal solution (y,{αij },{β ij }). Using Artin’s
Approximation Theorem, we obtain for a given c an analytic solution y,{αij },{βij } such that (y,{αij },{βij }) ≡ (y,{αij },{β ij }) mod (x1 , . . . ,xn )c . We define u : C −→ A by u(zi ) = yi (x) mod (f1 , . . . ,fa ). Because of hi (y) ≡ 0 mod (f1 , . . . ,fa ) it is well-defined and because of zi (x) ≡ yi (x) mod (g1 , . . . ,gb ) we have ϕ ◦ u = w.

To prove (2), let zi′ (x) ∈ C {x} (respectively zi′′ (y) ∈ C {y}) be representatives of v(zi ) (respectively w(zi )). Furthermore, let yi (x) ∈ C [[x]] be representatives of u(yi ). Similar to case (1), we obtain a X
zi′′ y(x) − zi′ (x) = αij fj


gi y(x) =

j=1 a X

β ij fj

j=1

for suitable αij ,β ij ∈ C [[x]]. Similar to the first case, the approximation of this formal solution defines the morphism u : B −→ A with the required properties. b −→ A b was ˆ : B We leave it as an exercise to prove that u is an isomorphism if u already an isomorphism.

302

9

Classification of Simple Hypersurface Singularities

This chapter is devoted to the study of hypersurface singularities. There are various equivalence relations one can consider on the set of all functions f ∈ C {x1 , . . . ,xn }. The two most important ones are right equivalence and contact equivalence. Two functions f and g are called right equivalent if there exists an automorphism ϕ of C {x1 , . . . ,xn } such that ϕ(f ) = g. They are called contact equivalent if there exists an automorphism ϕ and a unit u in C {x1 , . . . ,xn } such that ϕ(f ) = u · g. Until now we considered contact equivalence in this book (without using this name), but in this chapter we will mainly consider right equivalence, because in this case the theorems are somewhat easier to formulate and to prove. (One does not have to worry about the unit.) The aim of this chapter is to give a beginning of the classification of functions up to right equivalence. The first thing is to consider Finite Determinacy Theorems. Let m be the maximal ideal of C {x1 , . . . ,xn }. A function f is called (right) k–determined if for all elements g with f − g ∈ mk+1 the functions f and g are right equivalent. In particular, a k–determined f is right equivalent to a polynomial of degree at most k. A function is called finitely determined, if there exists a k ∈ N such that f is k–determined. It is maybe a little surprising that all functions which have an isolated singularity are finitely determined. The proof, however, is not difficult at all: it is a direct application of Newton’s Lemma, see 3.3.34. This result is not strong enough for our purposes, however. We want to find a small k such that f is k–determined. So we prove that, if mk+1 ⊂ m2 J(f ), then f is k–determined. (Here J(f ) is the Jacobian ideal, that is, the ideal generated by the partial derivatives of f .) This is what is usually called the Finite Determinacy Theorem, and is more difficult to prove. The idea of the proof is to connect f and g via the path f + a(g − f ). So for a = 0 we get f , and for a = 1 we get g. For all a, we want to prove that there exists a small neighborhood of a in C such that for all b ∈ U the functions f + b(g − f ) and f + a(g − f ) are right equivalent. In order to reach this goal, we need a criterion of local triviality. So, consider an element F ∈ C {x1 , . . . ,xn ,t} which we view as a family of functions with parameter t. Then we prove that this family is trivial if   ∂F ∂F ∂F . In order to prove this, we construct an ∈ (x , . . . ,x ) , . . . , and only if ∂t 1 n ∂x1 ∂xn automorphism in the formal power series ring. The existence of a convergent one then follows from Artin’s Approximation Theorem. We will show that this condition holds in the family f + ag, so that compactness of [0,1] then completes the proof of the Finite Determinacy Theorem. As an application of the local triviality we will prove the Mather-Yau Theorem. It says that f and g are contact equivalent, if and only if their Tjurina algebras are isomorphic. We will also prove a corresponding statement for right equivalence. The first reduction in the proof of the Mather-Yau Theorem is to the case that (f ) + J(f ) = (g) + J(g). Again we connect the functions by a path f + a(g − f ). But now one has to be more careful, because it is certainly not true that for all a ∈ C the functions f + a(g − f ) and f are contact equivalent. For example, the functions f and −f are contact equivalent, but for a = 12 we get that f + 21 (−f − f ) = 0, which is certainly not contact equivalent to f . But at least we will show that for all a ∈ C , except for a discrete subset, the functions

9.1 Finite Determinacy of Hypersurface Singularities

303

f and f + a(g − f ) are contact equivalent. As the complement of a discrete subset is still path connected, this is sufficient for the proof of the Mather-Yau Theorem. In Section 9.2 we start with the classification of functions up to right equivalence. We first define what a simple function is. This is done by putting a topology on C {x1 , . . . ,xn }. A function f is then called simple, if there are only finitely many equivalence classes in a sufficiently small neighborhood of f . The goal of this chapter is to prove that the simple singularities are given by the A-D-E–singularities. We first prove the Splitting Lemma. This allows us to study functions which lie in m3 . Then, via a sequence of arguments, it is shown that the A-D-E–singularities are simple, and that for nonsimple singularities f we may assume that either n ≥ 3 and f ∈ m3 , or n = 2 and f ∈ m4 , or n = 2 and f ∈ (x,y 2 )3 or f is nonisolated. In Section 9.3, we then show that functions with one of these properties are not simple. The idea is to use the Finite Determinacy Theorem, to reduce the study of right equivalence classes to the study of orbits under a group action in C N for some N ≫ 0. In fact, we study the group action on the k–jets, which are the Taylor series of f up to order k + 1. These orbits are orbits under the action of an algebraic group. This means that all the actions are given by polynomial functions. We then apply some algebraic geometry to prove that orbits of algebraic group actions are locally closed in the Zariski topology. (That is, the intersection of an open and closed subset.) As all orbits contain a smooth point, we can use the group action to show that all points of the orbit are smooth. The next thing to do is to compute the tangent space. We will see that it is equal to mJ(f ) + mk+1 /mk+1 . In particular, we get the dimension of the orbit of f . We deduce that if g ∈ / mJ(f ), then in the family f + t · g there are only finitely many t such that f + tg is right equivalent to f . The knowledge of the dimension of the orbits is our main tool for showing that certain functions are not simple.

9.1

Finite Determinacy of Hypersurface Singularities

Definition 9.1.1. Let f, g ∈ m ⊂ C {x1 , . . . ,xn }, where m is, as usual, the maximal ideal. (1) One says that f is right equivalent to g, f ∼ g, if there exists an automorphism ϕ of C {x1 , . . . ,xn } such that ϕ(f ) = g.

R

(2) One says that f is contact equivalent to g, f ∼ g, if there exists an automorphism C

ϕ of C {x1 , . . . ,xn } such that (ϕ(f )) = (g), that is, there exists a unit u such that ϕ(f ) = ug. Thus f and g are contact equivalent exactly if the C –algebras C {x1 , . . . ,xn }/(f ) and C {x1 , . . . ,xn }/(g) are isomorphic. This is exactly the same as saying that the germs of the analytic hypersurfaces defined by f and g are isomorphic.

In this chapter, we will mainly work with right equivalence. Similar results hold for contact equivalence, but the formulation and proofs will be left to the reader. Definition 9.1.2. Let f ∈ m ⊂ C {x1 , . . . ,xn }. Then f is called k–determined if all g ∈ C {x1 , . . . ,xn } with f − g ∈ mk+1 are right equivalent to f . If f is k–determined for some k ∈ N, then f is called finitely determined. In particular, a k–determined f is right equivalent to a polynomial.

304

9 Simple Hypersurface Singularities

The fact that functions which have an isolated singularity are finitely determined follows quite easily from Newton’s Lemma 3.3.34, as we will show now. Theorem 9.1.3. Let f ∈ m2 ⊂ C {x1 , . . . ,xn }. Suppose mk+1 ⊂ mJ(f )2 . Then f is k–determined. In particular, functions with isolated singularities are finitely determined. Proof. Let g ∈ m such that f − g ∈ mk+1 . Consider the following system of equations, which we want to solve for y1 , . . . ,yn ∈ C {x1 , . . . ,xn }: F (x,y) := f (y) − g(x) = 0. We start with the initial approximation y i (x) = xi . Obviously ∂f ∂xi .

∂F ∂yi 2


x,y(x) =

∂f ∂yi


y(x) =

As by assumption F (x,y) = f (x) − g(x) ∈ mk+1 ⊂ mJ(f ) , we can apply Newton’s Lemma 3.3.34. There exist y1 , . . . ,yn ∈ C {x1 , . . . ,xn } with • F (x,y) = 0, • yi (x) ≡ xi mod mJ(f ). Note that mJ(f ) ⊂ m2 . This says that the y give a coordinate transformation transforming f into g. So the fact that germs of holomorphic functions with an isolated singularity are right equivalent to a polynomial is a relatively easy consequence of Newton’s Lemma. For our purposes we need, however, a stronger version of the Finite Determinacy Theorem. The statement of the above theorem can be strengthened to the following statement. Theorem 9.1.4 (Finite Determinacy Theorem). Let f ∈ m2 ⊂ C {x1 , . . . ,xn }. Suppose that mk+1 ⊂ m2 J(f ). Then f is k–determined. In particular, if k ≥ µ(f ) + 1, then f is k–determined. Note that from mk ⊂ mJ(f ) it follows that mk+1 ⊂ m2 J(f ). The last condition is somewhat weaker. See Exercise 9.1.13. The basic idea of the proof of the Finite Determinacy Theorem is to connect f and g via the path Ft = f + t · (g − f ), for 0 ≤ t ≤ 1. We need a criterion whether for given t,s the element Ft is right equivalent to Fs . The following theorem gives us such a criterion “locally”. By this we mean that we give a criterion that Ft is right equivalent to F0 , for small t. Theorem 9.1.5. Let F ∈ C {x1 , . . . ,xn ,t} and c ≥ 0 be an integer. The following conditions are equivalent:   ∂F ∂F c ∈ (x , . . . ,x ) , . . . , (1) ∂F 1 n ∂t ∂x1 ∂xn . (2) There exists a ϕ = (ϕ1 , . . . ,ϕn ) ∈ C {x1 , . . . ,xn ,t}n such that (a) ϕi (x1 , . . . ,xn ,0) = xi , (b) ϕi − xi ∈ (x1 , . . . ,xn )c ,

(c) F (ϕ1 , . . . ,ϕn ,0) = F (x1 , . . . ,xn ,t). Corollary

9.1.6. Suppose that

∂F ∂t

∈

(x1 , . . . ,xn )



∂F ∂F ∂x1 , . . . , ∂xn



for some

F ∈ C {x1 , . . . ,xn ,t}. Then there exists a small neighborhood U of 0 in C such that for all fixed a ∈ U the function Fa := F (x1 , . . . ,xn ,a) is right equivalent to F0 .

9.1 Finite Determinacy of Hypersurface Singularities

305

Proof. Consider the elements ϕ1 , . . . ,ϕn ∈ C {x1 , . . . ,xn ,t} which exist according to 9.1.5. There exists a smallopen neighborhood U of 0 in C , so that for all a ∈ U , the ϕi (x,a) are  convergent and det

∂ϕi (x,a) (0) ∂xj

6= 0. Thus for fixed a ∈ U , (ϕ1 , . . . ,ϕn ) is an automor-

phism of C {x1 , . . . ,xn }. Moreover, formula (c) in Theorem 9.1.5 says that it transforms F (x1 , . . . ,xn ,0) into F (x1 , . . . ,xn ,a).

Proof of Theorem 9.1.5. Step 1. We first prove the easy implication (2) =⇒ (1). As ϕi (x1 , . . . ,xn ,0) = xi it follows easily from the Inverse Function Theorem that (ϕ1 , . . . ,ϕn ,t) is an automorphism of C {x1 , . . . ,xn ,t}. Let ψ = (ψ1 , . . . ,ψn ,t) be its inverse. Then ψi − xi ∈ (x1 , . . . ,xn )c , as one easily checks. By assumption, we have F (ϕ,0) = F (x,t). Compose this with ψ. Then F (x,0) = F (ψ,t). Hence the derivative of F (ψ1 , . . . ,ψn ,t) with respect to t is zero. Thus (9.1)

n X ∂F

∂xi i=1

(ψ1 , . . . ,ψn ,t)

∂ψi ∂F + (ψ1 , . . . ,ψn ,t) = 0. ∂t ∂t

Compose (9.1) with the automorphism (ϕ1 , . . . ,ϕn ,t). Then (9.2)

n X ∂F ∂ψi ∂F · (ϕ1 , . . . ,ϕn ,t) + = 0. ∂x ∂t ∂t i i=1

c i We already noted that ψi −xi ∈ (x1 , . . . ,xn )c . But then also ∂ψ ∂t (ϕ1 , . . . ,ϕn ,t) ∈ (x1 , . . . ,xn ) follows. So formula (9.2) gives the implication (2) =⇒ (1).   ∂F ∂F c Step 2. We now assume that ∂F ∈ (x , . . . ,x ) , . . . , 1 n ∂t ∂x1 ∂xn . By Artin’s Approxima-

tion Theorem, or better its (slightly) more general version given in Exercise 8.1.6, it follows that it suffices to find a formal ϕ = (ϕ1 , . . . ,ϕn ) ∈ C [[x1 , . . . ,xn ,t]]n such that (1) ϕi (x1 , . . . ,xn ,0) = xi , (2) ϕi − xi ∈ (x1 , . . . ,xn )c , (3) F (ϕ1 , . . . ,ϕn ,0) = F (x1 , . . . ,xn ,t). We now construct those formal ϕ1 , . . . ,ϕn . By assumption, we can write n

X ∂F ∂F ξi , ξi ∈ (x1 , . . . ,xn )c . = ∂t ∂x i i=1 ∂ Consider the differential operator δ := − ∂t + consider the map

Pn

∂ i=1 ξi ∂xi .

Let s be a new variable and

Φ : C [[x1 , . . . ,xn ,t,s]] −→ C [[x1 , . . . ,xn ,t,s]] ∞ X k 1 k a 7−→ exp(sδ)(a) := k! δ (a)s s 7−→ s.

k=0

for a ∈ C [[x1 , . . . ,xn ,t]]

306

9 Simple Hypersurface Singularities

Put Φi := Φ(xi ) for i = 1, . . . ,n, and Φn+1 = Φ(t). Note that Φn+1 = Φ(t) = t − s. Now define ϕi := Φi (x1 , . . . ,xn ,t,t) for i = 1, . . . ,n. We have to show that these ϕi satisfy our conditions. (1) The first condition is easy. Indeed, ϕi (x1 , . . . ,xn ,0) by definition is equal to δ 0 (xi ) = xi . (2) The second condition is also not too difficult. Indeed from the first part we have X k 1 k Φ(xi ) = xi + k! δ (xi )s . k≥1

From the definition of δ, and the fact that the ξi ∈ (x1 , . . . ,xn )c it follows immediately that Φ(xi ) − xi ∈ (x1 , . . . ,xn )c . The second condition now follows. (3) The third condition is the most difficult one. The crucial thing to show is ∂F (Φ1 , . . . ,Φn+1 ) = 0, ∂s that is, F (Φ1 , . . . ,Φn ,Φn+1 ) does not depend on s. Indeed, assuming (9.3) it follows that (9.3)

F (ϕ1 , . . . ,ϕn ,0) = F (Φ1 , . . . ,Φn ,Φn+1 )s=t = F (Φ1 , . . . ,Φn ,Φn+1 )s=0 = F (x1 , . . . ,xn ,t). The first equality holds by definition of the ϕi , the second equality because of (9.3). The third equality is because Φi |s=0 = xi for i ≤ n, and Φn+1 |s=0 = t. It therefore remains to show (9.3). We have the following formulae: ∂Φi = ξi (Φ1 , . . . ,Φn ,Φn+1 ), ∂s ∂Φn+1 = −1, ∂s δ(F ) = 0.

(9.4) (9.5) (9.6)

The second and the third formula follow immediately from the fact that Φn+1 = t − s, and the definition of δ. To prove formula (9.4), consider an arbitrary a ∈ C [[x1 , . . . ,xn ,t]], and differentiate: ∞ ∞ X 1 1 ∂Φ(a) X = δ k (a)sk−1 = δ k−1 (δ(a))sk−1 = Φ(δ(a)). ∂s (k − 1)! (k − 1)! k=1

k=1

In particular for i = 1, . . . ,n we have
∂Φi = Φ δ(xi ) = Φ(ξi ) = ξi (Φ1 , . . . ,Φn ,Φn+1 ). ∂s The last equality follows as Φi = Φ(xi ) for 1 ≤ i ≤ n, and Φ(t) = Φn+1 , by definition. By using the chain rule and the formulae (9.4), (9.5) and (9.6) we get n ∂Φi ∂F ∂F (Φ1 , . . . ,Φn+1 ) X ∂F (Φ1 , . . . ,Φn+1 ) = − (Φ1 , . . . ,Φn+1 ) ∂s ∂xi ∂s ∂t i=1 =

n X ∂F ∂F (Φ1 , . . . ,Φn+1 )ξi (Φ1 . . . ,Φn+1 ) − (Φ1 , . . . ,Φn+1 ) ∂x ∂t i i=1

= δ(F )(Φ1 , . . . ,Φn+1 ) = 0.

9.1 Finite Determinacy of Hypersurface Singularities

307

Proof of Theorem 9.1.4. Step 1. Let g ∈ mk+1 be given. Let 0 ≤ a ≤ 1. Consider Fa (x1 , . . . ,xn ,t) := f + (a + t) · g. For all a we will show that there exists an open neighborhood U of a such that for all a+t∈ U f + (a + t) · g ∼ f + a · g.

(9.7)

R

Because the unit interval [0,1] is compact we can cover the unit interval with finitely many U . As right equivalence is an equivalence relation, it then follows that f = F0 is right equivalent to F1 = f + g. Step 2. To prove (9.7) we will apply Corollary 9.1.6, and it therefore suffices to prove   ∂Fa ∂Fa ∂Fa 2 (9.8) . ,..., ∈m ∂t ∂x1 ∂xn Here m = (x1 , . . . ,xn ). By applying Artin’s Approximation Theorem, it suffices to prove this in the formal power series ring C [[x1 , . . . ,xn ,t]]. Step 3. We will first show the following intermediate result. Let f ∈ C {x1 , . . . ,xn } and assume mk+1 ⊂ m2 J(f ). Then mk+1 ⊂ m2 J(f + g) for all g ∈ mk+1 . In order to prove this, put F := f + g. Let h ∈ mk+1 be given. We have to find ξ i ∈ m2 such that (9.9)

h=

n X i=1

ξi

∂F . ∂xi

2 By applying Artin’s Approximation Theorem, it suffices to find ξP i ∈ m C [[x1 , . . . ,xn ]] n ∂f k+1 , ξi ∈ m2 . satisfying (9.9). Given h ∈ m , we can by assumption write h = i=1 ξi ∂x i But then n n X X ∂F ∂g ξi ξi h= − . ∂x ∂x i i i=1 i=1

(9.10)

∂g As g ∈ mk+1 and ξi ∈ m2 it follows that ξi ∂x in mk+2 . By assumption mk+1 ⊂ m2 J(f ), i so that we can write n X

n

X (1) ∂f ∂g (1) ξi ξi = , ξi ∈ m3 . ∂x ∂x i i i=1 i=1

(9.11)

(This is the reason we need mk+1 ⊂ m2 J(f ), as the condition mk+1 ⊂ mJ(f ) does not suffice.) Plugging (9.11) in (9.10) we get h=

n X i=1

n

(1)

(ξi + ξi )

X (1) ∂g ∂F ξ − ∂xi i=1 i ∂xi

with

n X i=1

(1)

ξi

∂g ∈ mk+3 . ∂xi

308

9 Simple Hypersurface Singularities

Repeating the procedure, we get ξ i ∈ m2 C [[x1 , . . . ,xn ]] satisfying formula (9.9). Step 4. The rest of the proof runs similar to the third step. Applying the third step to ag we obtain that mk+1 ⊂ m2 J(f + ag). Hence g=

n X ∂(f + ag) ξi , for some ξi ∈ m2 . ∂x i i=1

This implies n

n

X ∂g X ∂Fa ∂Fa ξi ξi −t· . =g= ∂t ∂xi ∂xi i=1 i=1

(9.12)

∂g ∈ mk+2 . Thus we can write As g ∈ mk+1 , it follows that ξi ∂x i

(9.13)

n X

n

X (1) ∂(f + ag) ∂g (1) ξi ξi = , ξi ∈ m3 . ∂x ∂x i i i=1 i=1

Plugging in (9.13) in (9.12) we get n

g=

X ∂Fa (1) ∂Fa (ξi − tξi ) = ∂t ∂xi i=1

mod t2 mk+3 .

Iterating this procedure we obtain n

X ∂Fa ∂Fa ξi , ξ i ∈ m2 C [[x1 , . . . ,xn ,t]]. = ∂t ∂x i i=1 Step 5. For the final statement we apply Exercise 1.3.20. It follows that mk−1 ⊂ J(f ), from which mk+1 ⊂ m2 J(f ) follows. So we can apply the first part of the theorem. Theorem 9.1.5 can be generalized as follows: Theorem 9.1.7 (Characterization of Local Analytic Triviality). Let f ∈ C {x1 , . . . ,xn ,y1 , . . . ,ym } and c ≥ 0 be an integer. The following conditions are equivalent:   ∂f ∂f ∂f c (1) ∂y ∈ (x , . . . ,x ) , . . . , 1 n ∂x1 ∂xn + (f ) for i = 1, . . . ,m. i (2) There exist ϕ1 , . . . ,ϕn ,u ∈ C {x1 , . . . ,xn ,y1 , . . . ,ym } such that • u(x1 , . . . ,xn ,0, . . . ,0) = 1

• ϕi (x1 , . . . ,xn ,0, . . . ,0) = xi • ϕi − xi ∈ (x1 , . . . ,xn )c

• f (x1 , . . . ,xn ,y1 , . . . ,ym ) = u · f (ϕ1 , . . . ,ϕn ,0, . . . ,0).   ∂f ∂f ∂f for all i, then we can choose u = 1. If moreover ∂y ∈ (x1 , . . . ,xn )c ∂x , . . . , ∂x i 1 n

9.1 Finite Determinacy of Hypersurface Singularities

309

The proof is similar to the proof of 9.1.5 and will be left as Exercise 9.1.15. We now come to the Theorem of Mather and Yau. Theorem 9.1.8 (Mather-Yau). Let f and g ∈ m = (x1 , . . . ,xn ) ⊂ C {x1 , . . . ,xn }. Then the following conditions are equivalent: (1) C {x1 , . . . ,xn }/(f ) ∼ = C {x1 , . . . ,xn }/(g), that is, f and g are contact equivalent.

(2) C {x1 , . . . ,xn }/ (f ) + mJ(f ) ∼ = C {x1 , . . . ,xn }/ (g) + mJ(g) as C –algebras.

If f and g define isolated singularities these conditions are equivalent to

∼ C {x1 , . . . ,xn }/ (g) + J(g) as C –algebras. (3) C {x1 , . . . ,xn }/ (f ) + J(f ) =

To put it in another way, the function f and g are contact equivalent if and only if the Tjurina algebras of f and g are isomorphic.

Remark 9.1.9. It is not true that functions are right equivalent if and only if their Milnor algebras are isomorphic. Indeed, consider the family of functions ft = x4 + y 5 + t · x2 y 3 . Then one can show, see Exercise 9.1.16: (1) ft ∼ f1 ⇐⇒ t = a2 b2 , a4 = 1, b5 = 1. R

5 (2) Consider the automorphism ϕα of C
{x,y} defined by ϕα (x) = α x, and ϕα (y) = 4 2 −1 α y, with α = t . Then ϕα J(ft ) = J(f1 ).

To find the correct theorem for the case of right equivalence, we consider the Milnor algebra C {x1 , . . . ,xn }/J(f ) as C {t}–algebra with multiplication t · h := f h for h ∈ C {x1 , . . . ,xn }/J(f ). Theorem 9.1.10 (Mather-Yau for Right Equivalence). Let f and g ∈ m = (x1 , . . . ,xn ) ⊂ C {x1 , . . . ,xn }. Then the following conditions are equivalent: (1) f ∼ g. R

(2) C {x1 , . . . ,xn }/mJ(f ) ∼ = C {x1 , . . . ,xn }/mJ(g) as C {t}–algebras. If f and g define isolated singularities these conditions are equivalent to ∼ C {x1 , . . . ,xn }/J(g) as C {t}–algebras. (3) C {x1 , . . . ,xn }/J(f ) =

The proof of this theorem is similar to the proof of 9.1.8 and left as Exercise 9.1.20.

Proof of the Mather-Yau Theorem for Contact Equivalence 9.1.8. The equivalence (1) ⇐⇒ (2) is left as Exercise 9.1.17. The implication (1) =⇒ (3) is easy, and has been proved in 3.4.27. The implication (3) =⇒ (1) is the most difficult case, and will be proved in several steps. Step
1. We first reduce to the case (f
) + J(f ) = (g) + J(g). Let ϕ : C {x1 , . . . ,xn }/ (f ) + J(f ) ∼ = C {x1 , . . . ,xn }/ (g) + J(g) be a C –algebra isomorphism. We may assume that ord(f ϕ induces an isomorphism between (f ) + J(f ) +
), ord(g) ≥ 2. The isomorphism
m2 /m2 and (g) + J(g) + m2 /m2 . These are finite-dimensional C –vector spaces. Let k be its dimension, and let z1 , . . . ,zn ∈ C {x1 , . . . ,xn } induce a basis of m/m2 such that z1 , . . . ,zk ∈ (f ) + J(f ). Let w1 , . . . ,wn ∈ C {x1 , . . . ,xn } induce a basis of m/m2 such that w1 , . . . ,wk ∈ (g) + J(g) and ϕ([zk+i ]) = [wk+i ] for i ≥ 1. Define an automorphism ψ on C {x1 , . . . ,xn } by ψ(zi ) = wi for all i. This is an automorphism by the Inverse Function Theorem 3.3.7. Then the diagram

310

9 Simple Hypersurface Singularities ψ

C {x1 , . . . ,xn } 
C {x1 , . . . ,xn }/ (f ) + J(f )

ϕ

/ C {x1 , . . . ,xn } 
/ C {x1 , . . . ,xn }/ (g) + J(g)


commutes, that is, ψ (f ) + J(f )
= (g) + J(g). Now let h
:= ψ(f ). Then (h) + J(h) = (g) + J(g), because ψ (f ) + J(f ) = (ψ(f )) + J(ψ(f )) , see 3.4.27. This is what we wanted to show. Step 2. We may now assume that (f ) + J(f ) = (g) + J(g). Let h := f − g and F := g + th ∈ C {x1 , . . . ,xn ,t}. So we are connecting f and g by a (complex) path: for t = 0 we have g, and for t = 1 we have f . Unlike in the proof of the Finite Determinacy Theorem, it is not true however that for all a, the functions g and g + ah are contact equivalent. In this step we will show that at least for t small the functions g and g + th are contact equivalent. From (f ) + J(f ) = (g) + J(g), it follows that (h) + J(h) ⊂ (g) + J(g). Hence n X ∂g ∂F ξi = h = ug + ∂t ∂xi i=1 n X

n X ∂F ∂h ξi ξi = uF + − t uh − ∂x ∂x i i i=1 i=1 n X

!

n X ∂F (1) ∂g ξi ξi = uF + + t u(1) g + ∂xi ∂xi i=1 i=1

!

Iterating this process we obtain n

X ∂F ∂F = uF + ξi ∂t ∂xi i=1 for suitable u,ξ i ∈ C [[x1 , . . . ,xn ,t]]. Using Artin’s Approximation Theorem 8.1.1 we obtain   ∂F ∂F ∂F . ,..., ∈ F, ∂t ∂x1 ∂xn

Now we can apply Theorem 9.1.7 to obtain
ϕ F (x1 , . . . ,xn ,0) = wF (x1 , . . . ,xn ,t)

for a suitable map ϕ = (ϕ1 , . . . ,ϕn ,t) : C {x1 , . . . ,xn ,t} −→ C {x1 , . . . ,xn ,t} and w a unit in C {x1 , . . . ,xn ,t}. Now we want to prove that for small t fixed the map (ϕ1 , . . . ,ϕn ,t) is an automorphism of C {x1 , . . . ,xn }. This is the place where we need the assumption that f and g define isolated singularities.1 At least the map (ϕ1 , . . . ,ϕn ,t) is an automorphism of C {x1 , . . . ,xn ,t}, so has an inverse (ψ1 , . . . ,ψn ,t), so that g = F (x1 , . . . ,xn ,0) = w′ F (ψ1 , . . . ,ψn ,t) 1

Note that this is not necessary for the proof of (1) ⇐⇒ (2). Here the fact that for small t fixed the map (ϕ1 , .“. . ,ϕn ,t) is an”automorphism follows directly from Theorem 9.1.7 because we have ∂F ∂t

∈ (F ) + m

∂F ∂F , . . . , ∂x ∂x1 n

.

9.1 Finite Determinacy of Hypersurface Singularities

311



for some unit w′ . Thus V (F ), 0 is as germ of an analytic space isomorphic to V (g), 0 .
This is the product of an isolated singularity V (g), 0 in (C n , 0) with the germ of a

smooth one-dimensional space (C , 0). The
singular locus of V (g), 0 in C n+1 , 0 is a line, so that the singular locus of V (F ), 0 is a smooth curve. Now this curve must be given by x1 = · · · = xn = 0, as F = g + t ·  (f − g), and we assumed that the order of  ∂F ∂F ∂F both f and g is at least two, and therefore F, ∂x1 , . . . , ∂xn , ∂t ⊂ (x1 , . . . ,xn ). Define

ai (t) := ψi (0, . . . ,0,t). Then ψi − ai ∈ (x1 , . . . ,xn ), thus for fixed sufficiently small t (ψ1 − a1 , . . . ,ψn − an ) gives an automorphism of C {x1 , . . . ,xn }. We want to show that ai = 0. For some small representative, the singular point (0, . . . ,0,t) is mapped via the automorphism (ψ1 , . . . ,ψn ,t) to a singular point of V (g). As the only singular points of V (g) are given by x1 = · · · = xn = 0, it follows that ψi (0, . . . ,0,t) = 0. As moreover ψi (x1 , . . . ,xn ,0) = xi , it follows that ψi − xi ∈ t · (x1 , . . . ,xn ). Thus for fixed small t, the (ψ1 , . . . ,ψn ), and hence (ϕ1 , . . . ,ϕn ) gives an automorphism of C {x1 , . . . ,xn } by the Inverse Function Theorem. (Note that in this step we only used the inclusion (h) + J(h) ⊂ (g) + J(g).) Step 3. We want to redo the argument in step 2 for the function F := g + ah + th for almost all a, that is, for all a except for a discrete subset of C . In order for the argument to work, we need that (h) + J(h) ⊂ (g + ah) + J(g + ah). This would follow from the statement (9.14)

(g + ah) + J(g + ah) = (g) + J(g)

Indeed, if this were true, then (h) + J(h) ⊂ (g) + J(g) = (g + ah) + J(g + ah) and we can redo the argument of Step 2. Now (9.14) does not hold for all a ∈ C , but we will show that it holds for almost all a ∈ C . To prove (9.14) for almost all a ∈ C , we use the theory of coherent sheaves. Let U ⊂ C n be an open subset such
that f and g converge on U and ∂G ∂G 0 ∈ U . Define G := g + th, and J(G) = ∂x , . . . , ∂xn . We have (G) + J(G) ⊂ (g) + J(g). 1 The ideal sheaves (G) + J(G) and (g) + J(g) in OU×C are coherent sheaves. Therefore,
the quotient M := (g) + J(g) /((G) + J(G)) is a coherent sheaf. Supports of coherent analytic sheaves are analytic sets by 6.2.5, so that Z := supp(M ) is an analytic subset of U × C . Therefore, the intersection Ze := Z ∩ ({0} × C ) is an analytic subset of {0} × C . Now Ze is the set of all points (0,a) such that M(0,a) 6= (0), and therefore includes the set of all points for which the inclusion of stalks (G) + J(G) ⊂ (g) + J(g) is strict. As an analytic subset of {0} × C , it is either {0} × C or a discrete set (Identity Theorem e therefore, Z e is discrete. This proves (9.14) for all a 3.1.10). But obviously (0,0) 6∈ Z, except for a discrete subset of C . This discrete subset we call D. Step 4. Because of (f ) + J(f ) = (g) + J(g), we see that 1 ∈ / D. Choose a path γ : [0,1] −→ C \ D from 0 to 1. For all a ∈ γ ([0,1]), we have that there exists an open neighborhood U of a such that for all b ∈ U , the function f + bh is contact equivalent to f + ah by Step 3. As γ([0,1]) is compact, we can cover γ([0,1]) by a finite number of U ′ s. It follows that f and f + g − f = g are contact equivalent.

Remark 9.1.11. The Morse Lemma follows from the Mather-Yau Theorem. As for g = x21 + · · · + x2n we have J(g) = (x1 , . . . ,xn ) we have that for all f with J(f ) = m are right equivalent to x21 + · · · + x2n . The C {t} module structure can only be trivial, that is, multiplication with t is the zero map. Of course, this is a very difficult proof of the Morse Lemma.

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9 Simple Hypersurface Singularities

Exercises 9.1.12. Suppose that f and g are right equivalent. Let k ∈ N. Suppose that f is k–determined. Prove that g is k–determined. 9.1.13. Show by counterexample that the statement mk+2 ⊂ m3 J(f ) does not imply that f is k–determined in general. 9.1.14. Let R be a commutative ring and u ∈ R[[t]]. Prove that for given a ∈ R the differential equation dy =u·y dt has a unique solution Pwith y(0) = a. P y ∈ R[[t]] (Hint: Write u = uk tk , y = yk tk and compare both sides of the equation.) k

k

9.1.15. Prove Theorem 9.1.7. (Hint: As an application of Artin’s Approximation Theorem, it suffices to find a formal automorphism. Now use the same idea as in Theorem 9.1.5 and induction on the number of parameters. (k) (k) Let ϕ(0) = (x1 , . . . ,xn ,y1 , . . . ,ym ) and u(0) = 1. Assume ϕ(k) = (ϕ1 , . . . ,ϕn ,y1 , . . . ,ym ) and (k) u are defined such that “ ” ∂f ∂f ∂f c • u(k) = 1 in case of ∂y ∈ (x , . . . , 1 , . . . ,xn ) ∂x1 ∂xn i • u(k) (x1 , . . . ,xn ,y1 , . . . ,ym−k ,0, . . . ,0) = 1 (k)

• ϕi (x1 , . . . ,xn ,y1 , . . . ,ym−k ,0, . . . ,0) = xi (k)

• ϕi − xi ∈ (x1 , . . . ,xn )c . ´ ` (k) (k) • f ϕ1 , . . . ,ϕn ,y1 , . . . ,ym−k ,0, . . . ,0 = u(k) f (x1 , . . . ,xn ,y1 , . . . ,ym ).

Choose ξ1 , . . . ,ξn ∈ (x1 , . . . ,xn )c C {x1 , . . . ,xn ,y1 , . . . ,ym } and v ∈ C {x1 , . . . ,xn ,y1 , . . . ,ym } such n P ∂f ξj ∂x (the case v = 0 included). Let ϕ be the automorphism defined that ∂y∂f = v · f + j m−k

i=1

by the integral curve of the corresponding vector field and define ϕi := ϕ(xi ) i = 1, . . . ,n. Note i that ϕ(yi ) = yi − δi,m−k · t. Then ∂ϕ = ξi (ϕ1 , . . . ,ϕn ,y1 , . . . ,ym−k−1 ,ym−k − t,0, . . . ,0). Use this ∂t property to prove that ∂f (ϕ1 , . . . ,ϕn ,y1 , . . . ,ym−k−1 ,ym−k − t,0, . . . ,0) = ∂t v(ϕ1 , . . . ,ϕn ,y1 , . . . ,ym−k−1 ,ym−k − t,0, . . . ,0) · f (ϕ1 , . . . ,ϕn ,y1 , . . . ,ym−k−1 ,ym−k − t,0, . . . ,0). Use Exercise 9.1.14 to choose s ∈ C [[x1 , . . . ,xn ,y1 , . . . ,ym−k ,t]] such that ∂s = v(ϕ1 , . . . ,ϕn ,y1 , . . . ,ym−k−1 ,ym−k − t,0, . . . ,0) · s, ∂t and s(t = 0) = 1. Use Exercise 9.1.14 again to prove that f (ϕ1 , . . . ,ϕn ,y1 , . . . ,ym−k−1 ,ym−k − t,0, . . . ,0) = s · f (x1 , . . . ,xn ,y1 , . . . ,ym−k ,0, . . . ,0). Let ϕi := ϕi |t=−ym−k and s = s |t=−ym−k then f (ϕ1 , . . . ,ϕn ,y1 , . . . ,ym−k−1 ,0, . . . ,0) = s · f (x1 , . . . ,xn ,y1 , . . . ,ym−k ,0, . . . ,0). (k+1)

Use the ϕi , s to define ϕi

,u(k+1) .

9.1.16. Prove the two statements in Remark 9.1.9. 9.1.17. Prove the equivalence (1) ⇔ (2) of the Theorem 9.1.8.

9.2 The A-D-E–singularities are simple.

313

9.1.18. Let g and f in m ⊂ C {x1 , . . . ,xn } be given such that J(g) ⊂ m2 J(f ). Prove that f and g are right equivalent. (Hint: Follow the steps in the proof of the Finite Determinacy Theorem 9.1.4.) 9.1.19. Let f ∈ C {x,y} of type

f = x3 + xy 3 + ψ,

with ψ ∈ (y 6 ,xy 4 ,x2 y 3 ,x3 y 2 ,x4 ). Prove that f is right equivalent to x3 + xy 3 . 9.1.20. Prove Theorem 9.1.10. (Hint: Use as in the proof of 9.1.8 the reduction to the case that J(f ) = J(g), resp. mJ(f ) = mJ(g). The C {t}–algebra isomorphism implies that h := f − g ∈ J(g), resp. h := f − g ∈ mJ(g). Now just continue as in the proof of 9.1.8.)

9.2

The A-D-E–singularities are simple.

We consider the automorphism group G of On = C {x1 , . . . ,xn }. Let f ∈ C {x1 , . . . ,xn }. The right equivalence class of f is exactly the orbit of f under the following group action.2 G × On −→ On ;

(G,f ) 7→ G(f ).

We denote the orbit of f by G · f := {G(f ) : G ∈ G }3 . The problem with the study of such orbits is, that On as C –vector space is infinite-dimensional! In case of isolated singularities, this is not a severe problem, as we can use the Finite Determinacy Theorem, which allows us to reduce the study of this problem to the finite-dimensional case. Definition 9.2.1. Let coordinates x1 , . . . ,xn on (C n , 0) be given, and let m be the maximal ideal of On . Let k ≥ 1 be a natural number. (1) We define the P k–jet space by Jk := On /mk+1 . Each element in Jk has a representative of type |ν|≤k aν xν . Note that Jk is a finite-dimensional C –vector space. P P (2) Let f = ν aν xν ∈ On be given. The k − jet of f is defined by j k f = |ν|≤k aν xν . (3) We can view G as a subset of ⊕ni=1 m. Indeed if G ∈ G then put gi := G(xi ). Then G is uniquely determined by the element (g1 , . . . ,gn ), see Exercise 3.1.22. We now define Gk := {(g1

mod mk+1 , . . . ,gn

mod mk+1 ) : (g1 , . . . ,gn ) ∈ G }.

(4) We get group actions Gk × Jk −→ Jk .

induced by those of G . Here G · f := j k (G(f )). One immediately checks that this is a group action and is well defined, see Exercise 9.2.17. 2 3

Group action means that (G1 · G2 )(f ) = G1 (G2 (f )). ∗ of G with O ∗ . Here O ∗ In case of contact equivalence, we have to use the semi-direct product G ⋊On n n is the group of units in On = C {x1 , . . . ,xn }. As a set the semi-direct product is just the Cartesian product, but the multiplication is given by (ϕ1 ,w1 ) · (ϕ2 ,w2 ) := (ϕ1 ϕ2 , w1 · ϕ1 (w2 )). One needs this ∗ on C {x , . . . ,x } sending ((ϕ,u),f ) to u · ϕ(f ) strange looking product so that the action of G ⋊ On n 1 is a group action.

314

9 Simple Hypersurface Singularities

Example 9.2.2. Let n = 1 and k = 2. Then J2 = {a0 + a1 x + a2 x2 : ai ∈ C }, and G2 = {αx+ βx2 : α 6= 0}. So an element G = αx+ βx2 acts on a0 + a1 x+ a2 x2 by sending it to a0 +a1 (αx+βx2 )+a2 (αx+βx2 )2 which in Jk is equal to a0 +αa1 x+(a1 β +a2 α2 )x2 . Lemma 9.2.3. Let coordinates x1 , . . . ,xn on (C n , 0) be given. Consider f ∈ On , and k ∈ N such that f is k–determined. Consider the group operation of Gk on Jk , and let g ∈ On be given. Then j k g ∈ Gk (j k f ) ⇐⇒ g is right equivalent to f. Proof. The implication ⇐= is obvious. Suppose on the other hand that j k g ∈ Gk (j k f ). Then there exists a G ∈ G such that f ≡ G(g) mod mk+1 . Because f is k–determined, and f and G(g) have the same k–jet, we have that f and G(g) are right equivalent. By transitivity, f and g are right equivalent. Definition 9.2.4. On all jet spaces Jk , which are as vector spaces isomorphic to some C N , we take the usual (Euclidean) C –topology. Consider the projection maps πk : On = C {x1 , . . . ,xn } −→ Jk = C {x1 , . . . ,xn }/mk+1 . We define a topology on C {x1 , . . . ,xn } as the coarsest topology such that all πk are continuous. Thus a basis for the topology for On is given by πk−1 (V ) for V ⊂ Jk . Informally speaking, a power series is near zero when “some of its coefficients are small”, and the coefficients of monomials which are not small are in a high power of the maximal ideal m. On G we similarly define a topology. It is the topology induced by the product topology of mn . The action of G on On is continuous. We omit the boring proof, as we do not need the result. Definition 9.2.5. We consider the action of G on On . Let f ∈ m ⊂ On . Then f is called simple, if there exists an open neighborhood U of f , such that the number of orbits which intersect U is finite. Remark 9.2.6. There is a different way to phrase the definition of simple. Namely, consider families F (x,t) ∈ C {x1 , . . . ,xn ,t} such that F (x,t) ∈ (x1 , . . . ,xn ), and F (x,0) = f . Families with this property are also called deformations. Then f is called simple, if there exist finitely many right equivalence classes G1 , . . . ,Gs such that for all families F (x,t) as above and t small and fixed the germ Ft ∈ C {x1 , . . . ,xn } is right equivalent to one of the germs in G1 , . . . ,Gs . In case that f is finitely determined, the proof that these two definitions are equivalent is left as Exercise 9.2.16. We will see later on, that simple germs have an isolated singularity. In fact, we have the following Classification Theorem, due to Arnold, whose proof will occupy the remainder of this chapter. Theorem 9.2.7 (Classification of Simple Singularities). Suppose that f ∈ On is simple. Then f is right equivalent to one of the singularities in the following list.

9.2 The A-D-E–singularities are simple. • Ak : f (x1 , . . . ,xn ) = xk+1 + x22 + . . . + x2n , 1

315 k ≥ 0,4

• Dk : f (x1 , . . . ,xn ) = x1 x22 + x1k−1 + x23 + . . . + x2n ,

k ≥ 4,

• E6 : f (x1 , . . . ,xn ) = x31 + x42 + x23 + . . . + x2n , • E7 : f (x1 , . . . ,xn ) = x31 + x1 x32 + x23 + . . . + x2n , • E8 : f (x1 , . . . ,xn ) = x31 + x52 + x23 + . . . + x2n . We give real pictures in dimension two of A2 , A3 , D5 , D6 , E6 and E7 .

In this section we will show that the A-D-E–singularities of Arnold’s list are simple. In the next section we will show that there are no other simple singularities. We first define the corank of a function. It is the coarsest invariant of right equivalence classes. Definition 9.2.8. Let f ∈ m2 ⊂ C {x1 , . . . ,xn }. The Hesse matrix of f is defined by   2 ∂ f (0) . H(f ) = ∂xi ∂xj 1≤i,j≤n The corank of f is defined as Corank(f ) := n − rank(H(f )). The following easy lemma is left as Exercise 9.2.20. Lemma 9.2.9. Let f be right equivalent to g. Then Corank(f ) = Corank(g). Moreover, let Ft be a holomorphic family of functions with Ft ∈ m2 for all t. Then for all t small Corank(Ft ) ≤ Corank(F0 ). We now prove the Splitting Lemma, which is a generalization of the Morse Lemma. 4

Note that A0 is the germ of a smooth map.

316

9 Simple Hypersurface Singularities

Lemma 9.2.10 (Splitting Lemma). (1) Let f ∈ C {x1 , . . . ,xn }, with f ∈ m2 and Corank(f ) equal to s. Suppose that f is finitely determined. Then f is right equivalent to an element of type g(x1 , . . . ,xs ) + x2s+1 + . . . + x2n . where g ∈ C {x1 , . . . ,xs } and g ∈ m3 . (2) Let g1 and g2 ∈ m3 ⊂ C {x1 , . . . ,xs } both have isolated singularities. Then g1 (x1 , . . . ,xs ) + x2s+1 + . . . + x2n is right equivalent to g2 (x1 , . . . ,xs ) + x2s+1 + . . . + x2n if and only if g1 is right equivalent to g2 . (3) Let f ∈ C {x1 , . . . ,xs } have an isolated singularity. Then f (x1 , . . . ,xs ) + x2s+1 + . . . + x2n is simple if and only if f is simple. Proof. (1) By following the proof of the Morse Lemma 3.4.30, we may (after renaming coordinates) assume that f is equivalent to a germ of type X (9.15) x2s+1 + . . . + x2n + xi xj Hij . 1≤i,j≤s

As the Corank is s, it follows that none of the Hij for 1 ≤ i,j ≤ s is a unit, hence Hij ∈ m. We may rewrite (9.15) as x2s+1 + . . . + x2n + g(x1 , . . . ,xs ) +

n X

xi Gi (x1 , . . . ,xn ) =

i=s+1

g+

n  X

i=s+1

 (xi + 12 Gi (x1 , . . . ,xn ))2 − 41 Gi (x1 , . . . ,xn )2 .

As Hij ∈ m, it follows that Gi ∈ m2 for all i, and g(x1 , . . . ,xs ) ∈ m3 . We can therefore do the following coordinate change: x′i = x′i

= xi +

xi 1 2 Gi (x1 , . . . ,xn )

for i = 1, . . . ,s, for i = s + 1, . . . ,n.

After this coordinate change we have that f is right equivalent to a germ of type (9.16)

x2s+1 + . . . + x2n + g(x1 , . . . ,xs ) +

n X

xi Gi (x1 , . . . ,xn ).

i=s+1

with Gi ∈ m3 . (Both the g and the Gi may have changed in this process). So we improved the situation. Now we have Gi ∈ m3 , which is a stronger statement than Gi ∈ m2 . We can continue like this, and we obtain that f is right equivalent to a germ of type (9.16) with Gi ∈ mk . As f is k–determined it follows from Exercise 9.1.12 that (9.16) is also k–determined. Thus (9.16) is right equivalent to its k–jet, and the first part of the theorem follows.

9.2 The A-D-E–singularities are simple.

317

(2) If g1 and g2 are right equivalent, then obviously g1 + x2s+1 + . . . + x2n and g2 + x2s+1 + . . . + x2n are right equivalent. The converse follows without difficulty from the Mather-Yau Theorem for right equivalence, see 9.1.10, but let us give a direct argument. Suppose therefore that f1 = g1 (x1 , . . . ,xs ) + x2s+1 + . . . + x2n is right equivalent to f2 = g2 (x1 , . . . ,xs ) + x2s+1 + . . . + x2n . Hence there exist  ϕ1 , . . . ,ϕn ∈ C {x1 , . . . ,xn }, with ϕi = li + hi , hi ∈ m2 and li linear forms with det

(9.17)

g1 (ϕ1 , . . . ,ϕs ) +

n X

∂li ∂xj

6= 0 such that

ϕ2i = g2 (x1 , . . . ,xs ) +

Note that that

2 i=s+1 li

+

Pn

i=s+1

hi (2li + hi ) = n X

(9.18)

x2i .

i=s+1

i=s+1

Pn

n X

li2

i=s+1

=

Pn

i=s+1

n X

ϕ2i . By assumption, g1 ,g2 ∈ m3 , so

x2i .

i=s+1

Now suppose we can find ψs+1 , . . . ,ψn ∈ C {x1 , . . . ,xs } such that (9.19)

(2li + hi )(x1 , . . . ,xs ,ψs+1 , . . . ,ψn ) = 0

for i = s + 1, . . . ,n.

Then we can consider the endomorphism of C {x1 , . . . ,xs } given by ϕ′1 := ϕ1 (x1 , . . . ,xs ,ψs+1 , . . . ,ψn ), . . . , ϕ′s := ϕs (x1 , . . . ,xs ,ψs+1 , . . . ,ψn ). This is in fact an automorphism. By the Inverse Function Theorem it suffices to show that (ϕ′1 , . . . ,ϕ′s ) = (x1 , . . . ,xs ). Now (ϕ1 , . . . ,ϕs ,ls+1 + 21 hs+1 , . . . ,ln + 12 hn ) is an automorphism of C {x1 , . . . ,xn }, again by the Inverse Function Theorem. Hence (ϕ1 , . . . ,ϕs ,ls+1 + 1 1 2 hs+1 , . . . ,ln + 2 hn ) is the the maximal ideal of On . It follows that for all k with 1 ≤ k ≤ s we can write s n X X 1
xk = ξν ϕν + ξν lν + hν , k = 1, . . . ,s. 2 ν=1 ν=s+1 By plugging in ψi instead of xi for i ≥ s + 1 we get by using (9.19) xk =

s X

ξν (x1 , . . . ,xs ,ψs+1 , . . . ,ψn )ϕ′ν .

ν=1

Therefore

(ϕ′1 , . . . ,ϕ′s )

= (x1 , . . . ,xs ). Now g1 (ϕ′1 , . . . ,ϕ′s ) = g2 (x1 , . . . ,xs )

follows from formulae (9.17), (9.18) and (9.19) so that we have an automorphism of C {x1 , . . . ,xs } transforming g1 into g2 . In order to construct the ψi , we therefore have to solve the system of equations Fi = 0 with Fi := 2li + hi , for i = s + 1, . . . ,n. Now det



   ∂Fi ∂li . (0) = 2 det ∂xj ∂xj i,j≥s+1 i,j≥s+1

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9 Simple Hypersurface Singularities

Therefore, if we show that the right hand side is nonzero, we can apply  Implicit  the ∂li is Mapping Theorem to solve this system of equations. The fact that det ∂x j i,j≥s+1

nonzero, is linear algebra, and treated in Exercise 9.2.21.

(3) First suppose that f is simple. We have to prove that f P + x2s+1 + . . .+ x2n is simple too. n Consider Ft ∈ (x1 , . . . ,xn )C {x1 , . . . ,xn ,t} with F0 = f + i=s+1 x2i . Let k ≥ µ(f ) + 1. For t small we have µ(Ft ) ≤ µ(f ). Then for all t small, Ft is k–determined by the Finite Determinacy Theorem 9.1.4. We redo the argument of the first step simultaneously for the whole family Ft . For t = 0, we get a coordinate change (in fact the identity), so that n for all small t we have a coordinate small Pnchange2of (C , 0). As Ft is k–determined for Pall n t we get a family Gt (x1 , . . . ,xs )+ i=s+1 xi , with G0 = f and Gt (x1 , . . . ,xs )+ i=s+1 x2i is right equivalent to Ft for t small. As f is simple, there exist only finitely many right equivalence classes in which Gt (x1 , . . . ,xs ) can lie. Hence there exist only finitely many equivalence classes in which Ft can lie. The converse statement is left to the reader. Corollary 9.2.11. Let f ∈ m2 with µ(f ) < ∞, and Corank(f ) = 1. Then there exists a k ≥ 2 such that f is right equivalent to Ak , that is, xk+1 + x22 + . . . + x2n . Moreover, the 1 Ak singularities are simple. Proof. By the Splitting Lemma, we only have to consider a function in one variable f (x) ∈ C {x}. If f (x) = 0 then µ(f ) = ∞. As this is not the case, there exists a unique k with f (x)√= xk+1 u, where u is a unit in C {x}. By the Implicit Function Theorem, the function k+1 u exists, is holomorphic and a unit in C {x}. We do the coordinate change √ k+1 x′ = k+1 ux. In this coordinate f (x′ ) = x′ . This proves that f is right equivalent to Ak . If Ft is a deformation of f , then µ(Ft ) ≤ µ(f ) for all t small. It follows that an Ak singularity only deforms into an Al singularity with l ≤ k. In particular, the Ak singularities are simple. Now we turn our attention to functions of corank two. By the Splitting Lemma, we only have to consider f (x,y) ∈ (x,y)3 ⊂ C {x,y}. Proposition 9.2.12. Let f ∈ C{x,y} and f ∈ m3 . After a linear coordinate change we may assume that j 3 f is of one of the following types. 1. j 3 f = xy(x + y), the zero set of j 3 f consists of three different lines. 2. j 3 f = x2 y, two lines coincide. 3. j 3 f = x3 , the three lines coincide. 4. j 3 f = 0. Suppose Ft is a holomorphic family with F0 = f and Ft ∈ m2 for all small t. Then for all t 6= 0 and small we have for the types (1), (2), resp (3): (1) Either Corank(Ft ) = 0 or 1 or j 3 Ft = 0 consists of three different lines. (2) Either Corank(Ft ) = 0 or 1 or j 3 Ft = 0 consists of three different lines, or two lines coincide. (3) Either Corank(Ft ) = 0 or 1 or j 3 Ft = 0 consists of three different lines, or two lines coincide, or the three lines coincide.

9.2 The A-D-E–singularities are simple.

319

Proof. Write j 3 f = ax3 + bx2 y + cxy 2 + dy 3 . Case 1. a = d = 0. Then j 3 f = xy(bx + cy). (i) If b = c = 0 we get 4. (ii) Suppose b = 0 or c = 0, but not both. Without loss of generality, let c = 0. Then j 3 f = xy(bx). As b 6= 0, we can do the coordinate change y ′ = by. Then j 3 f = x2 y ′ which is 2. (iii) Both b and c are nonzero. After the√coordinate change x′ = bx, y ′ = cy we may √ 3 3 ′ ′ assume that b = c. By putting x = b · x, y = b · y we get j 3 f = x′ y ′ (x′ + y ′ ) which is 1. Case 2. Suppose a 6= 0 or d 6= 0. Without √ loss of generality, we may assume that a 6= 0 and, after the coordinate change x′ = 3 a · x, even a = 1. We factorize j 3 f : j 3 f = (x − αy)(x − βy)(x − γy). We do the coordinate change x′ = x − αy, so that we may assume α = 0. Hence j 3 f = x · (x − βy)(x − γy). (i) If β = γ = 0 we get 3. (ii) Suppose that β 6= 0 or γ 6= 0. Without loss of generality β 6= 0. Take y ′ = x − βy. Then j 3 f = x′ y ′ (x′ − δy ′ ) for some δ, and we are back in Case 1. This proves the first part of the proposition. The second part of the proposition follows from the fact that the corank can only drop, see 9.2.9, and the fact that the zeros of the polynomial j 3 (Ft (x,1)) depend continuously on t. Proposition 9.2.13. Let f ∈ C {x,y}, µ(f ) < ∞. (1) Suppose j 3 f = xy(x + y). Then f is right equivalent to D4 . (2) Suppose j 3 f = x2 y. Then there exists a k ≥ 5 such that f is right equivalent to Dk . (3) The germs Dk for k ≥ 4 are simple. Proof. (1) Suppose that j 3 f = xy(x + y). It is an exercise to show that from the Finite Determinacy Theorem it follows that xy(x + y) is 3–determined. It follows without difficulty that f is right equivalent to D4 . (2) Suppose j 3 f = x2 y. As we suppose µ(f ) < ∞, the element f is k–determined for some k. Suppose that we have an s ≥ 4, such that j s f = x2 y + ay s + bxy s−1 + x2 ϕ, for a suitable ϕ ∈ ms−2 , and a,b ∈ C . This we certainly have for s = 4. We do the coordinate transformation x′ = x + 21 by s−2 , y ′ = y + ϕ. Then a simple calculation shows 2 s that j s f = x′ y ′ + ay ′ . Renaming x′ to x and y ′ to y we may therefore suppose that s 2 s j f = x y + ay . There are two cases to consider.

320

9 Simple Hypersurface Singularities

√ 1 Case 1. We have a 6= 0. Do the coordinate change y ′ = s a · y, x′ = 2s√ x. Then a s 2 s 2 s j f = x y + y . As x y + y is s–determined (exercise), it follows that f is a Ds+1 singularity. Case 2. We have a = 0. In this case we write down j s+1 f . Iterate the above coordinate changes until we come in Case 1, or until s ≥ k. We will show that s ≥ k leads to a contradiction. As f is k–determined it follows that f is right equivalent to j s f = x2 y+ay s . If a = 0, then f does not have an isolated singularity, contrary to our assumption. (3) Consider D4 , and a holomorphic family Ft with F0 = f = xy(x + y). We have that µ(Ft ) < ∞. If for fixed t small, the corank of Ft is zero, then Ft has an A1 –singularity by the Morse Lemma. If the corank is one, then Ft is right equivalent to Ak for some k by 9.2.11. As µ(Ft ) ≤ µ(D4 ), it follows that k ≤ 4.5 Hence D4 can only deform in finitely many germs. If Corank(Ft ) = 2, it follows that j 3 Ft = 0 consists of three different lines by 9.2.12. Hence there exist coordinates such that j 3 Ft = xy(x + y) for fixed t. Hence, by what we just proved, j 3 Ft is right equivalent to D4 . This shows that D4 can deform only in finitely many germs. The proof that the Dk for k ≥ 5 are simple, is similar. So in view of Proposition 9.2.12, the case to consider now are functions f ∈ C {x,y} whose 3–jet is x3 . Proposition 9.2.14. Let f ∈ C {x,y} with µ(f ) < ∞. Suppose j 3 f = x3 . Then either (1) f is right equivalent to E6 , E7 or E8 or (2) f is right equivalent to a function which lies in the ideal (x,y 2 )3 = (x3 ,x2 y 2 ,xy 4 ,y 6 ). Moreover, E6 , E7 and E8 are simple. Proof. Suppose j 3 f = x3 . We write down the four jet. j 4 f = x3 + ay 4 + bxy 3 + x2 ϕ, with a,b ∈ C and ϕ ∈ m2 . We do the coordinate change x′ = x + 13 ϕ. Hence we may assume j 4 f = x3 + ay 4 + bxy 3 . √ Case 1. Suppose a 6= 0. Then by doing the coordinate change y ′ = 4 ay we may assume a = 1. After doing the coordinate change y ′ = y + 14 bx we get j 4 f = x3 + y 4 + x2 ψ, with ψ ∈ m2 . By the coordinate change x′ = x + 13 ψ, we can get rid of ψ to obtain j 4 f = x3 + y 4 . As x3 + y 4 is 4–determined it follows that f is right equivalent to E6 . Redoing this proof for a holomorphic family Ft with F0 = x3 + y 4 shows that E6 can only deform into singularities of type Ak , Dk or E6 . As µ(Ft ) ≤ µ(f ) it follows that E6 is simple. Case 2. Suppose a =√ 0, and b 6= 0. We may assume then that b = 1, by the coordinate change y ′ = 3 by. From the Finite Determinacy Theorem it follows that f is 5–determined, but it does not follow from this theorem that f is 4–determined. Indeed, one calculates that (xy 4 ,x2 y 3 ,x3 y 2 ,x4 ) ⊂ m2 J(f ), but y 5 ∈ / m2 J(f ). But in fact f is 4–determined. To prove this, we write down the 5–jet of f 5

It follows from Exercise 9.3.22 that even k ≤ 3.

9.2 The A-D-E–singularities are simple.

321

j 5 f = x3 + xy 3 + cy 5 + xϕ, with c ∈ C and ϕ ∈ m4 . We do the coordinate change x′ = x + cy 2 , and get 3

2

j 5 f = x′ + x′ y 3 − 3cx′ y 2 + x′ ϕ′ , with ϕ′ ∈ m4 . Now we do the coordinate change y ′ = y − cx′ . Then 3

3

3

j 5 f = x′ + x′ y ′ − 3cx′ y ′ + ψ,

√ with ψ ∈ (x′ y ′ 4 ,x′ 2 y ′ 3 ,x′ 3 y ′ 2 ,x′ 4 ) ⊂ m2 J(f ). Replacing x′ by x′ 3 1 − 3cy ′ , and renaming the coordinates we get j 5 f = x3 + xy 3 + ψ, with ψ ∈ (xy 4 ,x2 y 3 ,x3 y 2 ,x4 ) ⊂ m2 J(x3 + xy 3 ). By Exercise 9.1.19 it follows that f is right equivalent to x3 + xy 3 , that is, we have an E7 –singularity. Similarly, one shows that E7 is simple, and this proof is left to the reader. Case 3. The final case to consider is a = b = 0, that is, j 4 f = x3 . So we consider the 5–jet: j 5 f = x3 + cy 5 + dxy 4 + x2 ϕ, with c,d ∈ C and ϕ ∈ m3 . Again we can get rid of the x2 ϕ term. The two remaining cases are: √ Case 3a. Suppose c 6= 0. We can after the coordinate change y ′ = 5 cy get that c = 1. Now 1 5 3 ′5 2 3 ′ put y = y+ 5 dx. Then j f = x +y +x ψ, with ψ ∈ m . By doing the coordinate change 3 5 x′ = x + 31 ψ, we can attain ψ = 0. Hence j 5 f = x′ + y ′ . By the Finite Determinacy Theorem, it follows that f is right equivalent to an E8 –singularity. Similarly, it is proved that E8 is simple. Case 3b. Consider the case c = 0. Then f ∈ (x3 ,xy 4 ) + (x,y)6 ⊂ (x,y 2 )3 , as was to be proved. We now succeeded in our goal of proving that the A-D-E-singularities are simple. In the next section, we will show that these are the only ones. Up to now, we already observed the following Remark 9.2.15. Suppose f ∈ On is not simple. Then either (1) f has a nonisolated singularity, or (2) Corank(f ) ≥ 3, or (3) Corank(f ) = 2, and we may suppose f ∈ C {x,y} with f ∈ m4 , or (4) Corank(f ) = 2, and we may suppose f ∈ C {x,y} with f ∈ (x,y 2 )3 .

Exercises 9.2.16. For finitely determined germs prove that the two definitions of simple, see Remark 9.2.6, are equivalent. 9.2.17. Show that the group action Gk on Jk is well-defined and indeed a group action. 9.2.18. Prove the Morse Lemma by using only elementary linear algebra, and applying the Finite Determinacy Theorem.

322

9 Simple Hypersurface Singularities

9.2.19. (1) Let f be a germ of corank s. Prove that µ(f ) ≥ 2s .

(2) Let f ∈ m4 ⊂ C {x,y}. Show that µ(f ) ≥ 9.

(3) Let f ∈ (y,x2 )3 ⊂ C {x,y}. Prove that µ(f ) ≥ 10.

Conclude that germs with µ(f ) < 8 are simple. 9.2.20. Prove Lemma 9.2.9.

9.2.21. In this exercise we complete proof of the second part of the Splitting Lemma 9.2.10. Consider matrices B of size (n P − s) × s and C of size (n − s) × (n − s). Let A = (aij )s+1≤i≤n,1≤j≤n := (B,C). Let li = n j=s+1 aij xj . Suppose that n X

(9.20)

i=s+1

li2 =

n X

x2i .

i=s+1

To complete the proof of the Splitting Lemma it suffices to show det(C) = 1. Prove this. (Hint: Use (9.20) to show that C is orthogonal.) 9.2.22. Let f ∈ C {x,y} with j 3 f = xy(x + y). Prove, without using the Finite Determinacy Theorem, that f is right equivalent to D4 .

9.3

Orbits

In order to show that functions with one of the properties mentioned in Remark 9.2.15 are not simple, we need to study orbits of group actions more carefully. First a definition. Definition 9.3.1. An affine algebraic group G is an affine variety G, together with regular maps • µ : G × G −→ G, • ι : G −→ G, which satisfy the groups axioms. Examples 9.3.2. (1) Consider Ga (k), k a field. As an affine variety, G ∼ = k. Furthermore µ(x,y) := x + y and ι(x) := −x. This is called the additive group . (2) Gm (k), the multiplicative group. As an affine variety it is equal to k \ {0}. Furthermore, µ(x,y) := x · y, and ι(x) := x−1 . (3) Consider the group of GLn (k) of invertible n×n matrices. This is an algebraic group. 2 Indeed it is an affine algebraic set, because it is equal to k n \ {det(aij ) = 0}, so as a complement of an affine hypersurface it is itself affine. (4) The group Gk is also an algebraic group because, as in the previous example, it is the complement of an affine hypersurface.

9.3 Orbits

323

Definition 9.3.3. An affine algebraic group acts on an affine variety X, if there exists a regular map ϕ : G × X −→ X : (g,x) 7→ g · x such that (g1 g2 ) · x = g1 · (g2 · x). Example 9.3.4. The action of Gk on Jk . The main result we need concerning group actions is the following theorem. Theorem 9.3.5. Let G be an affine algebraic group, acting on X. Then (1) Every orbit G · x is locally closed in the Zariski topology.6 (2) Every orbit is smooth. (3) The boundary of each orbit is a union of orbits of smaller dimension. The proof will occupy the first part of this section. In fact we need to do some more algebraic geometry, in particular, we need to study so-called constructible sets. Definition 9.3.6. Let X be a topological space, Z ⊂ X a subset.
Z is called constructible if it is a finite union of locally closed subsets: Z = ∪ni=1 Ui ∩ Fi , with Ui open and Fi closed. Example 9.3.7. Consider C 2 with the Zariski topology, and let Z = U1 ∪ F2 , where U1 is defined by U1 := {(x,y) : x 6= 0} and F2 = {(x,y) : y = 0}.

The following set-theoretic lemma is left as Exercise 9.3.20. Lemma 9.3.8. Let Z and W be constructible subsets of X. Then Z ∪ W , Z ∩ W and Z \ W are constructible. In particular Z c (the complement) is constructible. Lemma 9.3.9. Let X be an affine variety, and U ⊂ X be a (Zariski) open subset. Then there exists a finite cover U = ∪ni=1 Ui with Ui affine. Example 9.3.10. We have seen in Exercise 6.1.41 that C 2 \ {0}, as a locally ringed space, is not isomorphic to an affine space. Let f ∈ C [x,y]. Then C 2 \ {f = 0} is affine with coordinate ring C[x,y,z]/(z · f − 1). More generally, the complement of f = 0 in an irreducible variety X is affine. Note that we can write

C 2 \ {0} = C 2 \ {y = 0} ∪ C 2 \ {x = 0} . 6

Recall that this means that it is the intersection of an open and a closed subset.

324

9 Simple Hypersurface Singularities

Proof of Lemma 9.3.9. Consider Y = X \ U . Then Y is Zariski closed. Consider the ideal I (Y ) ⊂ K[X]. Write I (Y ) = (f1 , . . . ,fs ) ⊂ K[X] and define Ui := X \ V (fi ). Then Ui is open and affine. We claim that U = ∪ni=1 Ui . As Ui ⊂ U , the inclusion ∪ni=1 Ui ⊂ U is obvious. For the other inclusion, suppose we have an element a ∈ U \ ∪ni=1 Ui . Then for all i, we have a ∈ / Ui . Therefore fi (a) = 0 for all i. Hence a ∈ V (f1 , . . . ,fs ) = V (I (Y )) = Y . Hence a ∈ / U , a contradiction. Theorem 9.3.11 (Chevalley). Let X,Y be affine, and ϕ : X −→ Y be a morphism. Then ϕ maps constructible sets in X to constructible sets in Y . Proof. It suffices to prove that a locally closed subset of X is mapped to a constructible set of Y . Let U ∩ F , with U open and F closed, be a locally closed subset. By Lemma 9.3.9, we can write U as a finite union of open affines. Therefore, it suffices to prove the theorem for the case X = U , that is we have to show that the image of a closed subset F of X is constructible. As F now is a finite union of irreducible varieties, it suffices to prove the theorem for the case X = F is irreducible. So we have to show that the image of X is constructible. As ϕ(X) is closed, it suffices to show that ϕ(X) is a finite union of locally closed subsets of ϕ(X). Thus we may suppose Y = ϕ(X), and Y is irreducible. The claim now follows by induction on dim(Y ). The case dim(Y ) = 0 is trivial. So assume that the theorem has been proved for all Y ′ with Y ′ $ Y . As ϕ(X) = Y , the image ϕ(X) contains a nonempty open subset U , see Theorem 2.3.12. Consider Y \ U = Z1 ∪ · · · ∪ Zs . The set ϕ−1 (Zi ) is a closed subset of X, and has finitely many irreducible components. Let Wij be such a component. By restriction we get a map ϕ : Wij −→ Zi . By induction, the image is constructible. Therefore ϕ(X) is the union of U and a constructible set, and therefore constructible itself. We can now prove the important Theorem 9.3.5. Proof of Theorem 9.3.5. (1) Let x ∈ X. Then G · x is constructible. Indeed, A := G · x is the image of the map G −→ X; g 7→ g · x. Thus the orbit A = G · x is constructible by the previous Theorem 9.3.11. Write A = ∪si=1 Ui ∩ Fi . We may suppose that U1 ∩ F1 cannot be removed from this  union. Then 

F1 \ (F2 ∪· · · ∪Fs ) is an open subset of F1 ∪· · · ∪Fs . Hence U := U1 ∩(F1 \ F2 ∪· · · ∪Fs )

is an open subset of F1 ∪ . . . ∪ Fs . As A is a closed subset of F1 ∪ . . . ∪ Fs , it follows that U ⊂ A ⊂ A is open in A. Furthermore, by continuity we have that for all g ∈ G the set g −1 (U ) := {g −1 · x : x ∈ U } is open in A. Applying this to g −1 ∈ G we see that g(U ) is open in A. As A is an orbit, it follows that A = ∪g∈G g(U ). Thus A is open in A, or what is the same, A is locally closed.

(2) Let g ∈ G be fixed. The morphism g : X −→ X which sends x to g · x is an isomorphism. Now let x ∈ X. Then G · x contains a smooth point. But if g · x = y, it follows that y is also a smooth point. Therefore, all points of the orbit are smooth. (3) Let A = G·x. Consider A, the closure of A. As multiplication with g ∈ G is continuous, it follows that g −1 (A) is closed. As g −1 (A) = A, it follows by definition of closure that g −1 (A) ⊃ A. Similarly g(A) ⊃ A. Now A = g ◦ g −1 (A) ⊃ g(A) ⊃ A.

9.3 Orbits

325

Hence we have equality everywhere. Therefore, A is G–stable, and ∂A = A \ A is a union of orbits. As A is open in A all the boundary components have smaller dimension. (Use the Krull-dimension.) Corollary 9.3.12. All orbits of the group action of Gk on Jk are complex submanifolds of Jk . The next step is the computation of the (co-)dimension of the orbits. It turns out that this is closely related to the Milnor number. Theorem 9.3.13. Let the irreducible algebraic group G act on the vector space C N . Let x ∈ C N . Consider the orbit map Ox : G −→ G · x; g 7→ g · x. The differential of the orbit map TOx ,id : TG,id −→ TG·x,x is surjective. Thus the tangent space to the orbit of x is equal to the image of the differential map at the identity. Proof. We proved in 4.3.14 that for a dominant map ϕ : X −→ Y there exists a point p ∈ X such that Tϕ,p : TX,p −→ TY,ϕ(p) is surjective. We apply this to X = G, and Y = G · x. So there exists a g ∈ G such that the differential of the orbit map at g is surjective. As we already proved that the orbit is smooth, we only have to show that the rank of the differential of the orbit map is independent of the point g ∈ G. For fixed h ∈ G, consider the automorphism mh : G −→ G which sends g to hg. Let nh : G · x −→ G · x be the automorphism which sends y to h · y. Consider the following commutative diagram of holomorphic maps Ox −−→ G·x G ↓ mh ↓ nh G

O

x −−→ G·x

If we take derivatives we get a commutative diagram of linear maps TG,id ↓ Tmh ,id TG,h

TO

,id

TO

,h

x −−− −→

−−−x−→

TG·x,x ↓ Tnh ,x

TG·x,h·x,

in which the vertical arrows are isomorphisms. It follows that the differential of the orbit map, the horizontal arrows, have constant rank. Theorem 9.3.14. The tangent space to the orbit of f ∈ Jk under the action of Gk is k+1 . equal to mJ(fm)+m k+1 Proof. As Gk can be viewed as an open subset of ⊕ni=1 m/mk+1 , it follows that the tangent space of Gk is equal to the tangent space of ⊕ni=1 m/mk+1 . The differential of the orbit map Gk −→ Jk , h 7→ f ◦ h = h(f ),

∂f ∂f at the identity is given by (h1 , . . . ,hn ) ∈ ⊕ni=1 m/mk+1 7→ h1 ∂x + . . . + hn ∂x , because 1 n
∂f ∂f f (x1 + εh1 , . . . ,xn εhn ) − f = ε h1 ∂x1 + . . . + hn ∂xn , as a simple calculation shows. (Use Exercise 4.3.22.)

326

9 Simple Hypersurface Singularities

Remark 9.3.15. Let f ∈ m2 ⊂ On have an isolated singularity, and suppose k satisfies mk+1 ⊂ mJ(f ). Then the codimension of the orbit of f under Gk in Jk is equal to n+µ(f ). Proof. Indeed, applying the previous theorem shows that this codimension is equal to the vector space dimension of On /mJ(f ). Now we have the exact sequence (9.21)

0 −→ J(f )/mJ(f ) −→ On /mJ(f ) −→ On /J(f ) −→ 0.

The vector space dimension of J(f )/mJ(f ) is by Nakayama’s Lemma the minimal number of generators of J(f ). As f has an isolated singularity, V (J(f )) = {0}, or better, J(f ) is an m–primary ideal. As the (Chevalley) dimension of (C n , 0) is n, it follows that number of generators of J(f ) is at least n. But then the number of generators of J(f ) is exactly ∂f ∂f n, because ∂x , . . . , ∂x are n generators of J(f ). The statement therefore follows from 1 n the exact sequence (9.21). Theorem 9.3.16. Let f ∈ m2 ⊂ C {x1 , . . . ,xn }, with finite Milnor number. Let g ∈ / m · J(f ). Then there exist only finitely many t ∈ C such that f + t · g is right equivalent to f . Proof. Choose k such that mk+1 ⊂ mJ(f ). Hence g does not lie in the tangent space to the orbit through f in Jk under the action of Gk . This implies that the line f +tg in the jet space Jk does not lie entirely in the orbit of f . Therefore, the line f + tg in Jk intersects the orbit in only finitely many points, as the intersection of the orbit with the line is a constructible subspace of the line. This is exactly the statement of the theorem. Proposition 9.3.17. Let f ∈ C {x1 , . . . ,xn } be simple. Then f has an isolated singularity. Proof. Suppose not. Then the Milnor number
µ(f ) is not finite. As mJ(f ) ⊂ J(f ), it follows that in particular dimC C {x}/mJ(f ) = ∞, that is, for all k > 0, there exists a gk ∈ mk \mJ(f )+mk+1 , as otherwise mk ⊂ mJ(f ) by Nakayama. Hence for all sufficiently small t 6= 0 the germ fk := f + tgk is not right equivalent to f . This is because otherwise, gk had to be in the tangent space of the orbit of f under Gk , but by assumption it is not. Thus even fk and f are not in the same orbit under Gk . Moreover for k < s also fk is not right equivalent to fs . Indeed otherwise the classes of fs and fk in the jet space Jk would be Gk equivalent. But they are not, as in Jk the class of fs is equal to the class of f . Hence there are infinitely many nonequivalent function germs in any small neighborhood of f , and therefore f is not simple. Examples 9.3.18. (1) Let a complex number a with a2 6= 4 be given. Then fa (x,y) = x4 + y 4 + ax2 y 2 has an isolated singularity, hence a finite Milnor number. One calculates that m5 ⊂ m2 J(fa ). Hence fa is 4–determined. Now J(fa ) has two generators, and mJ(fa ) has 4 generators. Therefore, not all of the degree four monomials in x,y can be in mJ(fa ). Indeed, mJ(fa ) looks like,

9.3 Orbits

327 y

x

Here dots denote nonzero monomials in O/mJ(fa ), that are however equal up to a nonzero constant modulo mJ(fa ). We see that x2 y 2 ∈ / mJ(fa ). As this holds for all a ∈ C with a2 6= 4, it follows that all germs fa are not simple. Indeed, suppose the converse. Then there exists a (small) t0 such that fa+t = x4 + y 4 + (a + t)x2 y 2 are mutually right equivalent for infinitely many t with 0 < t < t0 . This is impossible as for t small x2 y 2 is not in mJ(fa+t ), so there can only be finitely many t’s in the family fa+t that are right equivalent. (2) Consider the family fa := x3 + y 6 + axy 4 , such that a is a complex number with 4a3 + 27 6= 0. The zero set of fa consists of three (different) parabolas. One calculates that fa then has an isolated singularity. Moreover, one shows that the monomial xy 4 ∈ / mJ(fa ), see Exercise 9.3.21 so that all germs fa are not simple as in the first example. (3) Consider the family fa := x3 + y 3 + z 3 + axyz with a3 + 27 6= 0. Then fa has an isolated singularity, and xyz ∈ / mJ(fa ), see Exercise 9.3.21. Thus again fa is not simple. We now can reach our goal of proving that there are no other simple singularities except for the A-D-E–singularities. Theorem 9.3.19. Germs f ∈ On with one of the following properties are not simple. (1) Nonisolated singularities. (2) n = 2 and f ∈ m4 . (3) n = 2 and f ∈ (x,y 2 )3 . (4) f with Corank(f ) ≥ 3. In particular, Arnold’s Classification Theorem of Simple Singularities 9.2.7 holds. Proof. It was shown in the previous section, see 9.2.15 that either f is right equivalent to an A-D-E–singularity, or right equivalent to a germ with one of the above listed properties. So if we show that singularities with one of the above properties are not simple, Arnold’s Classification Theorem follows. (1) The fact that nonisolated singularities are not simple was proved in 9.3.17. (2) We calculate modulo m5 . If f is simple, then j 4 f ∈ J4 has the property that in a neighborhood U of j 4 f there are only finitely many orbits under the action of G4 . We look at the tangent space at the orbit. As f ∈ m4 , the tangent space mJ(j 4 f ) + m5 /m5 is a subset of m4 /m5 . This tangent space has at most dimension 4, as mJ(j 4 f ) has four generators which all lie in m4 . It follows that the orbit of j 4 f has dimension at most 4. This holds for all f ∈ m4 . Note that m4 /m5 is G4 –stable, and has dimension 5. As a finite

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9 Simple Hypersurface Singularities

union of subvarieties of dimension four never can fill up an open subset of a vector space of dimension 5, it follows that all neighborhoods of j 4 f intersect infinitely many orbits. In particular f is not simple. (4) The argument is similar, but now we calculate modulo
m4 . We know that mJ(f )/m4 has dimension at most n2 , but m3 /m4 has dimension n+2 3 , which is bigger.

(3) In this case the argument is somewhat more subtle. We calculate modulo m7 , so we consider the action of G6 on the jet space J6 . We take the weighted order by defining the weight of x to be equal to two, and the weight of y to be equal to one. So all terms of elements in (x,y 2 )3 have weighted order at least six. Let f ∈ (x,y 2 )3 . The ∂f ∂f ∂f four generators x ∂f ∂x ,y ∂x ,x ∂y ,y ∂y of mJ(f ) have weighted order at least six, five, seven and six respectively. So we have at most three linearly independent elements in mJ(f ) of ∂f 2 ∂f weighted order six, namely those given by x ∂f ∂x , y ∂x , and y ∂y . There are however four of them in (x,y 2 )3 . We draw those weighted order six elements for the example f = x3 + y 6 . The box with the cross means that that monomial is not in mJ(f ). y

x

A counting argument shows that (x,y 2 )3 + m7 /m7 is a vector space of dimension sixteen. It follows that the orbit G6 · f has dimension at most sixteen, as its tangent space has at most dimension 16. However, G6 · f does not contain the whole of (y,x2 )3 modulo m7 , as otherwise the tangent space of G6 ·f would contain (y,x2 )3 modulo m7 . So the intersection of G6 · f with (y,x2 )3 modulo m7 has dimension at most fifteen. It now follows as in the other cases that f is not simple.

Exercises 9.3.20. Prove Lemma 9.3.8. 9.3.21. (1) Consider the family fa := x3 +y 6 +axy 4 for a ∈ C . Show that fa has an isolated singularity exactly when 4a3 + 27 6= 0. Show that xy 4 ∈ / mJ(fa ).

(2) Do the same for the family fa = x3 + y 3 + z 3 + axyz.

9.3.22. Let Ft be a holomorphic family. Suppose that for all sufficiently small nonzero t,s the germs Ft and Fs are right equivalent. Suppose that F0 is not right equivalent to Ft for t sufficiently small and nonzero. Prove that µ(Ft ) < µ(F0 ). (Hint: Use the third part of Theorem 9.3.5 and Remark 9.3.15.)

329

10

Deformations of Singularities

In the final chapter of this book, we study deformations of germs of complex spaces. The ultimate goal is to prove the existence of a semi-universal deformation of (X, 0), in case it has an isolated singularity. As the proof of this theorem is quite involved, we will first treat some special cases. So, in Section 10.1, we will consider isolated hypersurface singularities (X, 0) given by f = 0. A deformation (XS , 0) −→ (S, 0) over some germ of a b S,0 , with complex space (S, 0) is simply given by FS = 0 for some element FS ∈ C {x}⊗O FS (x,0) = f . There is a natural notion of isomorphism of deformations. Moreover, there ϕ is the notion of induced family, also called pull-back. If (T, 0) −→ (S, 0) is a holomorphic b S,0 −→ C {x}⊗O b T,0 , which we denote by the same name. map, we get a map ϕ : C {x}⊗O b T,0 , which gives a deformation (XT , 0) −→ (T, 0) of Then ϕ(FS ) is an element of C {x}⊗O (X, 0). A very important special case is that (T, 0) ֒→ (S, 0), in which case one just restricts the family to the subspace (T, 0). A versal deformation (XS , 0) −→ (S, 0) is a deformation from which every other deformation (XT , 0) −→ (T, 0) can be induced. If, moreover, the map on tangent spaces (mT,0 /m2T,0 )∗ −→ (mS,0 /m2S,0)∗ is uniquely determined, it is called semi-universal. We will show in Remark ?? , that it is too much to ask for uniqueness of the map (T, 0) −→ (S, 0). The main tool in the proof of the existence of a semi-universal deformation of an isolated hypersurface singularity is Grauert’s Approximation Theorem. In Section 10.2 we will consider deformations of germs of complex spaces, defined by say f1 = · · · = fk = 0. In general, defining a deformation by simply deforming f1 , . . . ,fk to F1S , . . . ,FkS as in the hypersurface case, does not lead to a good notion. This is because the dimension might change, or components might disappear. In case (S, 0) is smooth and one-dimensional, so that OS,0 = C {s}, we will show in Example 10.2.1 that it is a b S,0 /(F1S , . . . ,FkS ). good idea to take the condition that s is a nonzerodivisor of C {x}⊗O We will show that this property is equivalent to the fact that one can lift all the relations (r1 , . . . ,rk ), that is f1 r1 + . . . + fk rk = 0, to a relation between the F1S , . . . ,FkS . This is called flatness, and can be defined for general OS,0 , for example, C [ε]/(ε2 ), the local ring of the “double point” (T, 0). As ε is always a zerodivisor, the flatness definition by the lifting property of the relations makes more sense. We then classify all deformations of (X, 0) over the double point (T, 0). These deformations are classified by the so-called 1 TX, , which is a finite-dimensional vector space in case (X, 0) has an isolated singularity. 0 If one has a deformation over a germ of a complex space (S, 0), one can
∗ define the Kodaira-Spencer map. To every element of the tangent space mS,0 /m2S,0 , one assigns 1 a deformation over (T, 0), hence an element of TX, . We then prove the existence of a 0 versal deformation, again by applying Grauert’s Approximation Theorem, in case there exists a deformation over a smooth space (S, 0) such that the Kodaira-Spencer map is surjective. This case applies, for example, to isolated complete intersection singularities, see Exercise 10.2.33, and to isolated Cohen-Macaulay codimension two singularities. An essential ingredient is the Hilbert-Burch Theorem. In the final section, we will prove the existence of a semi-universal deformation in 1 is a finite-dimensional a more general setting, namely under the assumption that TX, 0 vector space. The basic idea is to do this order by order. Let (X, 0) be defined by f ∈

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10 Deformations of Singularities

1 C {x}k . On first order, one does the following. Take a basis g1 , . . . ,gτ of TX, , and take 0 a new ring C [[s1 , . . . ,sτ ]] with maximal ideal m = (s1 , . . . ,sτ ). We then have a flat deformation

(10.1)

f + s1 g1 + . . . ,sτ gτ

over the space with local ring C [[s1 , . . . ,sτ ]]/m2 . In general, it is not possible to lift this family to a flat deformation over C [[s1 , . . . ,sτ ]]/m3 . We will show however, that there is a minimal I with m3 ⊂ I ⊂ m2 such that there exists a deformation over C [[s1 , . . . ,sτ ]]/I lifting (10.1). In general, suppose we have a deformation of (X, 0) over a space with local ring O/J. We will show that there exists a minimal I with mJ ⊂ I ⊂ J, such that there exists an extension of the deformation over the space with local ring O/I. The existence of this minimal I will be shown by obstruction theory. We will construct 2 2 an element in TX, ⊗ I for all I with mJ ⊂ I ⊂ J, TX, a vector space, the so called 0 0 obstruction space. It is zero if and only if the family can be lifted over the space with local ring O/I. Then we will successively extend the family in a maximal way order by order. This leads to a formal deformation. The fact that it is formally semi-universal is not very difficult. In fact, by applying the ideas of Section 10.2, one can show that it is semi-universal, by applying Grauert’s Approximation Theorem, in case the formal deformation is in fact convergent. The final subtle point is to prove that there exists a convergent deformation which is formally semi-universal. Also in this proof Grauert’s Approximation Theorem is used. We moreover prove the following criterion for (formal) semi-universality. Suppose a formal deformation of (X, 0) over a formal algebra C [[s]]/J 2 is given. Consider the obstruction element in TX, ⊗ J/mJ, which leads to the obstruction 0
∗ 2 map TX,0 −→ J/mJ. Then the formal deformation is semi-universal if and only if the Kodaira-Spencer map is an isomorphism, and the obstruction map is surjective.

10.1

Deformations of Functions

We make precise in the following definition the notion of deformations of functions. Definition 10.1.1. Let f ∈ C {x1 , . . . ,xn }, S a germ of a complex space. A deformation b S such that FS |s=0 = f .1 FS of f over S is a an element FS ∈ C {x1 , . . . ,xn }⊗O

Example 10.1.2. Let f,g ∈ C {x1 , . . . ,xn }. Then FS = f + s · g ∈ C {x1 , . . . ,xn ,s} = b {s} is a deformation of f . C {x1 , . . . ,xn }⊗C

Definition 10.1.3. Let FS be a deformation of f over S, and T −→ S be an analytic map between germs of complex spaces. Then we have an induced map OS −→ OT on the ring level, and hence a map: b S −→ C {x1 , . . . ,xn }⊗O b T. C {x1 , . . . ,xn }⊗O

The image of FS under this map we denote by FT . Then FT is a deformation of f over T , which we call the pullback of FS over T −→ S, or the deformation of f induced by T −→ S. 1

b S −→ C {x1 , . . . ,xn }⊗O b S /mS ∼ Recall that FS |s=0 is the image of the natural map C {x1 , . . . ,xn }⊗O = C {x1 , . . . ,xn }.

10.1 Deformations of Functions

331

Examples 10.1.4. Restrictions, etc... Definition 10.1.5. Let FS and GS be deformations of f over S. A morphism between FS and GS is given by an analytic isomorphism b S −→ C {x1 , . . . ,xn }⊗O b S Φ : C {x1 , . . . ,xn }⊗O

such that it transforms fibers into fibers, that is the diagram b S C {x1 , . . . ,xn }⊗O ↓ OS

Φ b S −→ C {x1 , . . . ,xn }⊗O ↓ = OS

is commutative, such that • Φ(FS ) = GS , • Φ|s=0 = Id. Definition 10.1.6. A deformation FS of F over S is called versal if for any deformation GT of f over T there exists a map T −→ S such that GT is isomorphic to FT , the deformation induced by T −→ S. ∂f ∂f Theorem 10.1.7. Let f ∈ C {x1 , . . . ,xn } with µ = µ(f ) = dimC C {x1 , . . . ,xn }/( ∂x , . . . ∂x )< 1 1 ∂f ∂f ∞. Let g1 , . . . ,gµ ∈ C {x1 , . . . ,xn } represent a basis of C {x1 , . . . ,xn }/( ∂x1 , . . . ∂x1 ). Then

FS (x1 , . . . ,xn ) := f +

µ X i=1

si · gi ∈ C {x1 , . . . ,xn ,s1 , . . . ,sµ }

is a versal deformation of f . Before giving the proof of this Theorem, we give some examples. Example 10.1.8. We take the function f = x3 ∈ C {x}. A basis for the Milnor algebra is given by {1,x}. Therefore, by the Theorem, a versal deformation is given by F (x,s1 ,s2 ) = x3 + s1 x + s2 . Look at the deformation G(x,t) = x3 + tx2 of f . Its zero set looks like

x t

Versality says that we can find a map T −→ S, that is, dually a map of rings C {s1 ,s2 } −→ C {t}, or to put it in another way, two power series s1 (t),s2 (t) such that F (x,s1 (t),s2 (t)) ≃ G(x,t). This means that we have to find an automorphism Φ : C × T −→ C × T with

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10 Deformations of Singularities

• Φ = (Φ(x,t),t), • Φ(0,t) = Id, • F (x,s1 (t),s2 (t)) = G(Φ(x,t),t). We rewrite:

1 1 1 x3 + tx2 = (x + t)3 − t2 x − t3 . 3 3 27

With Φ(x,t) = x − 31 t we get 1 2 G(Φ(x,t),t) = x3 − t2 x + t3 . 3 27 We can therefore take

2 3 1 t . s1 (t) = − t2 ; s2 (t) = 3 27 therefore the map T −→ S is given by “the picture”

t

s2 s1

Example 10.1.9. We take G = f + ε · h ∈ C {x1 , . . . ,xn ,ε}/(ε2 ). Hence we can take h ∈ ∂f ∂f C {x1 , . . . ,xn }. Then G is a deformation of f over the double point T. As C {x1 , . . . ,xn }/( ∂x , . . . , ∂x ) 1 n is a finite dimensional vectorspace we can write h=

µ X i=1

bi gi +

n X i=1

hi

∂f ∂xi

with bi ∈ C and hi ∈ C {x1 , . . . ,xn } and gi as in the statement of the Theorem. We define the automorphism of C n × T by Φ(xi ,ε) = xi − hi · ε. By using Taylor we get G(φ(x,ε),ε) = P f (x1 − h1 ε, . . . ,xn − hn ε) + εh = Pnexpansion µ ∂f f (x1 , . . . ,xn ) − i=1 εhi ∂x + εh = f (x , . . . ,x ) + 1 n i=1 bi gi . i So G(Φ(x,ε),ε) = F (x,εb1 , . . . ,εbn ), which shows that G is isomorphic to a deformation induced by F and the map: C {s1 . . . ,sµ } −→ C {ε}/(ε2 ) si 7→ εbi .

10.2 Deformations of Singularities

333

Proof of Theorem 10.1.7 The proof uses Grauert’s approximation Theorem and the ideas of the last example. Let GT be a deformation of f over T , where OT = C {t1 , . . . ,tq }/J for some ideal J. We have to find • A map S −→ T , that is µ power series s(t) = s1 (t1 , . . . ,tq ), . . . ,sµ (t1 , . . . ,tq ) with s ≡ 0 modulo t, and • an automorphism Φ(x,t) with Φ(x,0) = Id, such that (∗)

G(Φ(x,t),t) = F (x,s(t)).

This we can view as a system of equations, which we have to solve for Φ and s. Here Φ may depend on x and t, but the µ functions si are only allowed to depend on t1 , . . . ,tq . We show that this equation has a solution by applying Grauert’s approximation Theorem 8.2.2. To apply this Theorem, we have to show that every solution (Φ,s) of (∗) of order i in t can be extended to a solution of (∗) of order i + 1. Having this solution (Φ,s) of order i, we look at the homogeneous part of degree i + 1 of G(Φ(x,t),t) − F (x,s(t)) and write it as X tν h ν |ν|=i+1

with hν ∈ C {x1 , . . . ,xn }. Just as in the second example we can write for all ν; hν =

µ X j=1

bjν gj +

n X j=1

hνj

∂f ∂xj

with bjν ∈ C and hνj ∈ C {x1 , . . . ,xn }. We put: Φ′j =P Φj − tν hνj j = 1, . . . ,n s′j = sj + |ν|=i+1 tν bjν j = 1, . . . ,µ.

As G ≡ f modulo t, it follows from Taylor expansion that (Φ′ ,s′ ) = (Φ′1 , . . . ,Φ′n ,s′1 , . . . ,s′µ ) is a solution of (∗) of order i + 1. This proves the Theorem.

10.2

Deformations of Singularities

We now consider deformations of germs of complex spaces. So consider a singularity (X, 0) defined by the equations f1 (x) = . . . = fk (x) = 0, for fi ∈ C {x1 , . . . ,xn } = C {x}. Naively, one probably would like to say that a deformation of (X, 0) over a space (S, 0) is given by the zero set of functions, F1 (x,s) = . . . = Fk (x,s) = 0 b S,0 and Fi (x,0) = fi (x) for i = 1, . . . k. However, this kind of deforwith Fi ∈ C {x}⊗O mation is too general, because a “general fiber” of (XS , 0) −→ (S, 0) does not sufficiently reflect the properties of (X, 0). To see this, look at the following example.

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10 Deformations of Singularities

Example 10.2.1. Let (XS , 0) −→ (S, 0) be an analytic map. Here (XS , 0) is defined by F1 F2 F3

= xy − s = 0 = xz − s = 0 = yz − s = 0

Although this is a space defined in 4-space, we would like to imagine the situation by the following picture:

s=0

s 6= 0

s

The problem here is that s is zerodivisor of C {x,y,z,s}/(F1 ,F2 ,F3 ). The total space (XS , 0) has four components, of which three are in the fiber s = 0. In fact, a primary decomposition of (F1 ,F2 ,F3 ) is (cf. 1.4.12): (F1 ,F2 ,F3 ) = (x,y,s) ∩ (x,z,s) ∩ (y,z,s) ∩ (x − y,y − z,x2 − s). We therefore have a decomposition (XS , 0) = (X0 , 0) ∪ (X1 , 0), where (X0 , 0) is the union of three coordinate axes, and (X1 , 0) is the parabola. Take a function f , vanishing on (X1 , 0), but not on (X0 , 0). Then obviously s · f = 0 as elements of OXS ,0 , but both s and f are nonzero in OXS ,0 . This expresses the fact that s is a zerodivisor. In fact it can be proved by using the Active Lemma 4.1.10, see Exercise 10.2.28, that Lemma 10.2.2. s is a zerodivisor ⇐⇒ there exists a component of (XS , 0) in the fiber above zero. The above example hopefully makes clear to the reader that for a ”nice” one parameter family (with parameter s), one should impose the condition that s is a nonzerodivisor. This we call flatness: Definition 10.2.3. A C {s}-module M is called flat if and only if s is a nonzerodivisor of M . For example a finitely generated C {s}-module M is flat if and only if M is free, see 1.3.9. It turns out that it is not so easy at all to construct (nontrivial) one parameter flat deformations of singularities. Therefore, it is a natural idea to construct deformations by “power-series expansion”. That is, construct deformations over (T, 0), then try to lift over the germ of the complex space with local ring C [s]/(s3 ), etc. But then one has the problem of defining flatness of a C [s]/(s2 )-module, as s is always a zerodivisor of C [s]/(s2 ). So we need to give a different definition of flatness of a C {s}-module, which gives a good generalization to rings with nilpotent elements. The following lemma is of importance for solving this problem.

10.2 Deformations of Singularities

335

Lemma 10.2.4. Let (S, 0) = (C , 0), and consider a deformation of (X, 0) as above. Then the deformation is flat (that is, s is a nonzerodivisor) if and only if for every relation between the fi : f1 r1 + . . . + fk rk = 0, we can find a lift Ri (x,s), that is, Ri (x,0) = ri (x) and F1 R1 + . . . + Fk Rk = 0. P Proof. Suppose that s is a nonzerodivisor, and take a relation fi ri = 0. Take any lift Ri′ (x,s) of the ri , and look at F1 R1′ + . . . + Fk Rk′ . This might not be zero, but we know it is if we plug in s = 0. This shows that this expression is divisible by s: F1 R1′ + . . . + Fk Rk′ = sΦ. This says that sΦ = 0 as element of OXS ,0 . As Ps is a nonzerodivisor it follows that Φ = 0 as element of OXS ,0 . Hence we can write Φ = αi Fi for some αi . Now put Ri = Ri′ −sαi . It follows that F1 R1 + . . . + Fk Rk = 0, so we found a lift of the ri . On the other hand, suppose that we can lift every relation. We need to show that s is a nonzerodivisor. So suppose that sΦ = 0 ∈ OXS ,0 , that is sΦ = F1 R1 + . . . + Fk Rk . P Putting P s = 0 we get a relation fi ri = 0, which by assumption can be lifted to a relation Fi Ri′ = 0. Then X sΦ = Fi (Ri − Ri′ ).

As both Ri and Ri′ are lifts of the ri , it follows that Ri − Ri′ is divisible by s. Hence R −R′ the power-series i s i exists. Because in the power-series ring s is a nonzerodivisor it follows that X Ri − R′ i Φ= Fi , s expressing the fact that Φ = 0 as element of OXS ,0 . This is what we had to show. This result motivates the following definition. Definition 10.2.5. Let (X, 0) ⊂ (C n , 0) be a germ of a complex space. Let I (X, 0) = (f1 , . . . ,fk ). Let (S, 0) be a further germ of a complex space. Let (XS , 0) ⊂ (C n × S, 0) b S,0 . Then (XS , 0) −→ (S, 0) is called a (flat)2 be defined by F1 , . . . ,Fk , Fi ∈ C {x}⊗O deformation of (X, 0) over (S, 0) if (1) Fi (x,0) = fi (x) for i = 1, . . . k,

2 3

(2) for every relation r = (r1 , . . . ,rk ) ∈ C {x}k between the fi , that is f · r := Pk k b i=1 fi ri = 0, there exists a lift R = (R1 , . . . ,Rk ) ∈ (C {x}⊗OS,0 ) , that is Ri (x,0) = ri (x), which is a relation between the Fi . This means that F · R := Pk 3 i=1 Fi Ri = 0. Later on we will often write deformation instead of flat deformation. b S,0 /(F1 , . . . ,Fk ) is a flat OS,0 –module in the general This is, in fact, equivalent to saying that C {x}⊗O set-up of flatness, see [Matsumura 1979], Theorem 49.

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10 Deformations of Singularities

A priori the definition depends on the choice of the generators of (XS , 0) (and of (X, 0)), but it is not so difficult to show that it in fact does not. See Exercise 10.2.29. Proposition 10.2.6. Let (X, 0) be a germ of a complex space, defined by the ideal I = (f1 , . . . ,fk ) in C {x}. Let g1 , . . . ,gk be further elements in C {x} =: O. Consider (T, 0), the double point. Then f1 +εg1 , . . . ,fk +εgk defines a (flat) deformation (XT , 0) −→ (T, 0) if and only if f1 7→ g1 , . . . ,fk 7→ gk gives a well-defined element of HomO (I,OX,0 ) =: NX,0 .4 The module NX,0 is called the normal module of (X, 0).

Proof. Suppose first that f + εg := (f1 + εg1 , . . . ,fk + εgk ) defines a (flat) deformation (XT , 0) −→ (T, 0). We have to show that P the map fi −→ gP i induces a well defined map in HomO (I,OX,0 ). Thus an element h = rP ri gi . In order i fi is sent to P to show that this is well-defined, we have to show that if r f = 0, it follows that ri gi ∈ I. But i i P ri fi = 0 means that r = (r1 , . . . ,rk ) is a relation, which by assumption can be lifted to a relation r + εs between f + εg, hence 0 = (f + εg) · (r + εs) = f · r + ε(f · s + r · g) = ε(f · s + r · g) because ε2 = 0. Hence r · g ∈ (f1 , . . . ,fk ) as was to be shown. The converse is proved by reversing the arguments. Lemma 10.2.7. Let (X, 0) be a germ of a complex space. Then (flat) deformations (XT , 0) −→ (T, 0) are in one-one correspondence with the elements of the normal module NX,0 . Proof. We apply the previous proposition. Step 1. Suppose that g1 , . . . ,gk and g1′ , . . . ,gk′ give the same element in NX,0 . We have to show that the ideals (f1 + εg1, . . . ,fk + εgk ) and (f1 + εg1′ , . . . ,fk + εgk′ ) are equal. For this ′ it suffices, by symmetry, to show one inclusion. P Now, by assumption, gi − gi ∈ 2I, that is, ′ there exist aij ∈ C {x} such that gi − gi = j aij fj . It then follows that, as ε = 0 fi + εgi = fi + εgi′ + ε

X j

aij fj = (fi + εgi′ ) + ε

X

aij (fj + εgj′ ),

j

This gives the inclusion (f1 + εg1 , . . . ,fk + εgk ) ⊂ (f1 + εg1′ , . . . ,fk + εgk′ ). Step 2. The fact that equality of the ideals (f1 +εg1 , . . . ,fk +εgk ) and (f1 +εg1′ , . . . ,fk +εgk′ ) implies that g1 , . . . ,gk and g1′ , . . . ,gk′ give the same element in NX,0 = HomO (I,OX ) is proved similarly. Example 10.2.8. We consider the germ of the complex space (X, 0) defined by the 2–minors of the matrix5 4 5

Note that NX, 0 is not an invariant of the isomorphism class of (X, 0), as it might depend on the embedding. Readers knowing some algebraic geometry will recognize that this singularity is the cone over the rational normal curve of degree four.

10.2 Deformations of Singularities 

337 x0 x1

x1 x2

x2 x3

x3 x4



.

We will denote by fij the minor obtained by taking the i–th and j–th column. So for example f13 = x0 x3 −x1 x2 . It has been shown in Exercise 7.3.8 that the ideal I defined by the minors is a prime ideal. Thus (X, 0) is irreducible. We determine the normal module NX,0 . Step 1. We have to determine the relations between the fij . We get eight relations in the following way: take three columns, double one of the rows, and compute the determinant. As two rows are the same, the determinant is zero, so that we indeed get a relation. We get relations rijk if we double the first row, and sijk if we double the second row. As an example, we compute r123 . So we take the first three columns, and put the first row on top. We get the following matrix   x0 x1 x2  x0 x1 x2  , x1 x2 x3

and hence the relation x0 f23 −x1 f13 +x2 f12 = 0. We claim that all relations are generated by the relations constructed in this way. We will only indicate how to prove this, leaving the reader to fill in the details. First of all, we have that C {x0 ,x4 } ⊂ OX,0 is a Noether normalization. Because OX,0 is a free C {x0 ,x4 }–module, see Exercise 7.3.8, it follows that x0 ,x4 is a regular sequence in OX,0 . Thus all relations between the elements −x21 =f12 (0,x1 ,x2 ,x3 ,0),

−x1 x2 =f13 (0,x1 ,x2 ,x3 ,0), −x1 x3 =f14 (0,x1 ,x2 ,x3 ,0), −x22 =f23 (0,x1 ,x2 ,x3 ,0)

−x2 x3 =f24 (0,x1 ,x2 ,x3 ,0), −x23 =f34 (0,x1 ,x2 ,x3 ,0)

can be lifted to relations between the fij . The relations between these monomials are easily determined, and there are indeed eight of them. Step 2. We can easily generate 8 elements n1 , . . . ,n8 of the normal module NX,0 . These we get by putting an extra ε at one of the eight spots of the matrix. The fact that the relations can be lifted, follows because we can lift the relations rijk and sijk by again taking three columns and double a row! The n1 , . . . ,n4 correspond to the four entries in the top row, and the n5 , . . . ,n8 correspond to the four entries in the bottom row. To give an example we determine n1 . We get the following matrix   x0 + ε x1 x2 x3 . x1 x2 x3 x4 We take the first minor, and get x0 x2 + εx2 − x21 . Thus f12 → x2 etc. It turns out that the normal module is generated by nine elements, namely by n1 , . . . ,n8 and one further element n9 . We put these in the following table. The rows give the elements of NX,0 by giving the value on fij in the columns.

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10 Deformations of Singularities

n1 n2 n3 n4 n5 n6 n7 n8 n9

f12 x2 −x1 0 0 −x1 x0 0 0 x0

f23 0 x3 −x2 0 0 −x2 x1 0 −x2

f34 0 0 x4 −x3 0 0 −x3 x2 x4

f13 x3 0 −x1 0 −x2 0 x0 0 0

f24 0 x4 0 −x2 0 −x3 0 x1 0

f14 x4 0 0 −x1 −x3 0 0 x0 x2

The fact that n9 is also an element of the normal module is a direct check. We want to show that n1 , . . . ,n9 generate the normal module. To show this, we first note that an element in NX,0 is determined by the values on f12 ,f23 and f34 , that is, by the first three columns. To see this, look for example at the relation r123 . It says that x0 f23 − x1 f13 + x2 f12 = 0. So if ϕ ∈ NX,0 , it follows that x1 ϕ(f13 ) = x0 ϕ(f23 ) + x2 ϕ(f12 ), so we can compute x1 ϕ(f13 ). As x1 is a nonzerodivisor of OX,0 we can compute ϕ(f13 ). Similarly one sees that one can compute ϕ(f24 ) and ϕ(f14 ). By doing column operations, we see that we have normal module elements ϕ of the following type: (1) ϕ(f12 ) = xi , ϕ(f23 ) = 0, ϕ(f34 ) = 0, for i = 0,1,2; (2) ϕ(f12 ) = 0, ϕ(f23 ) = xi , ϕ(f34 ) = 0, for i = 1,2,3; (3) ϕ(f12 ) = 0, ϕ(f23 ) = 0, ϕ(f34 ) = xi , for i = 2,3,4. Take an element ϕ of NX,0 . We claim that ϕ(f23 ) ∈ (x1 ,x2 ,x3 ). So calculate modulo (x1 ,x2 ,x3 ). The ideal I modulo (x1 ,x2 ,x3 ) is generated by one element, namely x0 x4 . We look at 0 = ϕ(r123 ) = x0 ϕ(f23 ) − x1 ϕ(f13 ) + x2 ϕ(f12 ).

Modulo (x1 ,x2 ,x3 ) we get x0 ϕ(f23 ) = 0. Hence a representative (also called ϕ(f23 )) of ϕ(f23 ) in C {x} modulo (x1 ,x2 ,x3 ) is divisible by x4 . Similarly, using the relation s234 , we get that ϕ(f23 ) is divisible by x0 . Thus ϕ(f23 ) is divisible by x0 x4 , which is modulo I equal to x1 x3 . It is somewhat more difficult, but similar, to show that ϕ(f12 ) = 0 mod (x0 ,x1 ,x2 ), and ϕ(f34 ) = 0 mod (x2 ,x3 ,x4 ). Altogether this shows that NX, 0 has the nine generators written down above. Definition 10.2.9. Let two deformations (XS , 0) −→ (S, 0) and (XS′ , 0) −→ (S, 0) of (X, 0) over (S, 0) be given. A morphism of (XS , 0) −→ (S, 0) to (XS′ , 0) −→ (S, 0) is given by a map Φ : (XS , 0) −→ (XS′ , 0) such that: (1) Φ restricts to the identity map on (X, 0). (2) Φ preserves fibers, that is, the diagram Φ

(XS , 0) −→ (XS′ , 0) ↓ ↓ (S, 0)

Id

−→

(S, 0)

10.2 Deformations of Singularities

339

is commutative. Theorem 10.2.10. Let Φ : (XS , 0) −→ (XS′ , 0) be a morphism of deformations of (X, 0) over (S, 0). Then Φ is an isomorphism. Proof. This is a consequence of the Inverse Function Theorem. A morphism of deformations of (X, 0) induces the identity on (X, 0), and preserve fibers. Therefore, Φ∗ is induced by a morphism which, by abuse of notation, we also call Φ∗ b S,0 C {x1 , . . . ,xn }⊗O xi

s ∈ OS,0

b S,0 , −→ C {x1 , . . . ,xn }⊗O 7→

7→

xi + hi , i = 1, . . . n,

s.

b S,0 , mS,0 the maximal ideal of OS,0 . To prove that Φ∗ is an where hi ∈ mS,0 · C {x}⊗O isomorphism, we may suppose that OS,0 = C {s}. Therefore, the determinant of the Jacobian matrix of Φ∗ is equal to one. Hence the inverse of Φ∗ , and hence of Φ exists, by the Inverse Function Theorem. Remark 10.2.11. Let (X, 0) ⊂ (C n , 0) be a singularity, given by f = (f1 , . . . ,fk ). Consider deformations (XS , 0) −→ (S, 0) and (XS′ , 0) −→ (S, 0) over (S, 0). Suppose (XS , 0) is defined by FS = (F1 , . . . ,Fk ), and (XS′ , 0) is defined by FS′ = (F1′ , . . . ,Fk′ ), both lifting (f1 , . . . ,fk ). To give an isomorphism Φ : (XS , 0) −→ (XS′ , 0) means that one has to find: b S,0 with Φi (x,0) = xi . • Elements Φ1 , . . . ,Φn ∈ C {x1 , . . . ,xn }⊗O

b S,0 such that Λ|s=0 = Id. • A k × k–matrix Λ with entries in C {x1 , . . . ,xn }⊗O

satisfying

b S,0 )k . FS′ = FS (Φ1 , . . . ,Φn ) ◦ Λ ∈ (C {x}⊗O

(Here ◦ denotes matrix multiplication.) The elements Φi give the relative isomorphism, inducing the identity on (X, 0), whereas the matrix Λ changes generators for the ideal of (XS , 0). Now we want to classify deformations of (X, 0) over the double point (T, 0). Suppose that the ideal of (X, 0) is equal to I ⊂ On . Consider Θn , the free On –module of all ∂ + . . . + hn ∂x∂ n for hi ∈ On . Note derivations of On . Elements θ of Θn look like θ = h1 ∂x 1 that we have a map

So we can define:

α : Θn −→NX,0 = HomOn (I,OX,0 )
θ 7→ f 7→ θ(f ) .

1 Definition 10.2.12. TX, := Coker(α).6 0

Theorem 10.2.13. Let (X, 0) ⊂ (C n , 0) be a germ of a complex space. Isomorphism classes of deformations of (X, 0) over (T, 0) are in one-one correspondence with elements 1 of TX, . 0 6

1 It is, in fact true that TX, is an invariant of the isomorphism class of (X, 0), see Exercise 10.2.26. 0

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10 Deformations of Singularities

Proof. Suppose (X, 0) is defined by f1 , . . . ,fk . Consider two deformations (XT , 0) −→ (T, 0) and (XT′ , 0) −→ (T, 0) given by f1 + εg1 , . . . ,fk + εgk and f1 + εg1′ , . . . ,fk + εgk′ respectively. A morphism of deformations between (XT , 0) and (XT′ , 0) is induced by a map C {x} ⊗ C [ε]/(ε2 ) −→ C {x} ⊗ C [ε]/(ε2 ) with (x1 , . . . ,xn ,ε) 7→ (x1 + εh1 , . . . ,xn + εhn ,ε) for some h1 , . . . hn ∈ C {x}. Then (XT′ , 0) is given by the zero set of   n X ∂f i + gi  for i = 1, . . . k. fi (x1 + εh1 , . . . ,xn + εhn ) + εgi = fi + ε  hj ∂x j j=1

This holds because ε2 = 0. Hence we see that the map I −→ OX,0 given by fi 7→ gi − gi′ is in the image of Θn . Similarly, one shows that elements of Θn induce isomorphisms. This is what we needed to show. Examples 10.2.14. 1 (1) Let (X, 0) be a germ of an isolated hypersurface, defined by f ∈ C {x}. Then TX, = 0 C {x}/(f,J(f )). This follows from Example ??. (2) We take the singularity of Example 10.2.8. Recall that the normal module elements are determined by their values on f12 , f23 and f34 . A direct calculation shows that ∂ ∂ ∂ ∂ ∂ 1 ∂x0 = n1 , ∂x1 = n2 + n5 , ∂x2 = n3 + n6 , ∂x3 = n7 + n4 , ∂x4 = n8 . Thus TX,0 is generated 1 by four element n2 , n3 , n4 , and n9 . We claim that (x0 ,x1 ,x2 ,x3 ,x4 )TX, = 0. Thus we 0 have to do twenty checks!! The reader hopefully forgives us for just doing one of those. 1 We will show that the class of x0 n4 is the zero element in TX, .We look at the values on 0 f12 , f13 and f34 . They are 0, 0 and −x0 x3 which modulo I is equal to 0, 0 and −x2 x1 . ∂ 1 Thus it is equal to −x1 n8 = −x1 ∂x , hence the trivial element in TX, . Altogether we see 0 4 1 that dimC (TX,0 ) = 4, with a basis given by the classes of the normal module elements n2 ,n3 ,n4 and n9 . Theorem 10.2.15. Let (X, 0) be a germ of a complex space with an isolated singularity. 1 Then TX, is a finite-dimensional C –vector space. 0 Proof. Let X ⊂ U be a small representative of (X, 0) such that 0 is the only singular point of X. Consider the sheaf TX1 := HomOU (I ,OX )/ΘU . Here ΘU is the sheaf of derivations on U , I is the sheaf of ideals defining X. The map ΘU −→ HomOU (I ,OX ) 1 is defined as in the definition of TX, . By the basic coherence theorems TX1 is a coherent 0 1 sheaf on X, whose stalks in p ∈ X is equal to TX,p . As X has only one singularity, these are zero for p 6= 0, by Exercise 10.2.27. Hence TX1 is a coherent sheaf concentrated in the 1 point 0, and therefore the set of global sections Γ(TX1 ,U ) = TX, is a finite-dimensional 0 C –vector space by Theorem 6.2.6. The converse of this theorem does not hold, see Exercise 10.2.30. We now come to the definition of induced family and of versal deformations. Definition 10.2.16. Let (X, 0) be a germ of a complex space. (1) Suppose (XS , 0) −→ (S, 0) is a deformation of (X, 0), defined by F1 , . . . ,Fk ∈ b S,0 . Let p : (T, 0) −→ (S, 0) be an analytic map. We define the pull-back, C {x}⊗O or the induced deformation (X ×S T, 0) −→ (T, 0)

10.2 Deformations of Singularities

341

as follows. The map p induces a map p∗ : OS,0 −→ OT,0 , which in turns induces a b S,0 −→ C {x}⊗O b T,0 . Then (XT , 0) is given by p∗ (F1 ), . . . ,p∗ (Fk ). map p∗ : C {x}⊗O It is proved in Exercise 10.2.32 that the map (X ×S T, 0) −→ (T, 0) is indeed flat.

(2) A deformation (XS , 0) −→ (S, 0) is called versal if for every deformation (XT , 0) −→ (T, 0) there exists a map p : (T, 0) −→ (S, 0) such that the deformations (XT , 0) −→ (T, 0) and (X ×S T, 0) −→ (T, 0) are isomorphic. The versal deformation is called semi-universal if moreover the map on tangent spaces is uniquely determined. Definition 10.2.17. Let (XS , 0) −→ (S, 0) be a deformation of (X, 0). Suppose (X, 0)
k b is defined by f ∈ C {x}k , and that the deformation is given by F ∈ C {x} P⊗OS,0 . Suppose mS,0 has a minimal system of generators s1 , . . . ,sp . Write F = f + si gi . We 1 denote by [gi ] the class of gi mod mS,0 in TX, . Then the map 0
∗ TS,0 = mS,0 /m2S,0 X ai s∗i i

1 −→ TX, 0 X ai [gi ] 7→ i

is called the Kodaira-Spencer map. It is easily shown that the definition of the KodairaSpencer map is independent of the choices made. The main goal in this chapter is to prove that for (X, 0) with an isolated singularity, a semi-universal deformation exists. In general, the proof of this theorem needs many additional preparations, but we proved it already in the first section for hypersurface singularities. In the remainder of this chapter, we will prove a special case of this theorem, which includes the case of hypersurface singularities and complete intersection singularities (see Exercise 10.2.33). Moreover we treat the case of Cohen-Macaulay singularities of codimension two by applying the Hilbert-Burch Theorem 6.5.26. Theorem 10.2.18. Let (X, 0) be a germ of a complex space, defined by f ∈ C {x}k . 1 Suppose dimC (TX, ) < ∞. Let (S, 0) = (C τ , 0). Suppose that 0 (1) there exists a (flat) deformation (XS , 0) −→ (S, 0) defined by7 FS (x,s) := f + s1 g1 + . . . + sτ gτ ∈ (C {x} ⊗ C {s1 , . . . ,sτ })k , (2) the Kodaira-Spencer map of (XS , 0) −→ (S, 0) is surjective. Then (XS , 0) −→ (S, 0) is a versal deformation of (X, 0). If moreover the Kodaira-Spencer map is an isomorphism, then (XS , 0) −→ (S, 0) is a semi-universal deformation. For the proof of this theorem, we need the following preparations, which we will also need in the next section. Definition 10.2.19. For a local Artinian C –algebra A we denote by (A, 0) the germ of the complex space consisting of one point 0, but with local ring A. Definition 10.2.20. Let A and B be local C –algebras, m the maximal ideal of B. Suppose ϕ : B −→ A is a surjective morphism of C –algebras. Then (1) ϕ is called a small surjection if m · Ker(ϕ) = 0, 7

Note that not necessarily gi ∈ C {x}.

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10 Deformations of Singularities

(2) ϕ is called a simple surjection if Ker(ϕ) is a one-dimensional C –vector space. Example 10.2.21. The surjection C [ε]/(ε3 ) −→ C [ε]/(ε2 ) is a simple surjection. The surjection C [ε,η]/(ε,η)2 −→ C [ε,η]/(ε,η) is small, but not simple. The surjection ϕ : C [ε,η]/(ε2 ,η 2 ) −→ C [ε,η]/(ε,η) is not small, as εη is not zero in C [ε,η]/(ε2 ,η 2 ). Lemma 10.2.22. Let (X, 0) be a germ of a complex space, with f ∈ C {x}k generators of the ideal of (X, 0). Let ϕ : B −→ A be a small surjection of local Artinian C –algebras.
b k . Let (XA , 0) −→ Consider a deformation (XB , 0) −→ (B, 0), defined by FB ∈ C {x}⊗B (A, 0) be the restriction of (XB , 0) −→ (B, 0) to (A, 0), defined by the image FA of FB in
b k . Let rA be a relation between the FA , that is, FA · rA = 0. Then there exists C {x}⊗A a lift rB of rA with FB · rB = 0.

Proof. The proof is by induction on the vector space dimension dimC (A). The case dimC (A) = 1 is easy. In this case A = C , and the deformation (XA , 0) is equal to (X, 0). By assumption on flatness, every relation r = rA can be lifted to a relation rB between the FB . Now assume dimC (A) > 1. We can always find 0 6= α ∈ A with mA · α = 0. Consider ′ ′ ′ A′ := A/(α). By induction we can find a lift rB of rA with rB ·FB = 0. Moreover, take any ′ ′ ′ lift reB of rA . As both ϕ(e rB ) and ϕ(rB ) project to rA , it follows that ϕ(e rB )−ϕ(rB ) = h·α k ′ for some h ∈ C {x} . Since both ϕ(e rB ) and ϕ(rB ) are relations between the FA we get h · FA · α = 0. Moreover FA · α = f · α, because mA · α = 0. Thus h is a relation between the f , and can by flatness be lifted to a relation hB between the FB . Now define ′ rB := rB + α′ hB for any α′ with ϕ(α′ ) = α. Then rB is a relation, and lifts rA . This proves the lemma. Lemma 10.2.23. Consider: (1) A singularity (X, 0) defined by f ∈ C {x}k . (2) ϕ : B −→ A a small surjection of local Artinian C –algebras. Let α1 , . . . ,αp be a C –basis of Ker(ϕ), (3) A deformation (XA , 0) −→ (A, 0), defined by FA and two lifts (XB , 0) −→ (B, 0), and (XB′ , 0) −→ (B, 0) of (XA , 0) −→ (A, 0), defined by FB and FB′ respectively, both being lifts of FA . Then there exist elements n1 , . . . ,np ∈ NX,0 with FB − FB′ = α1 n1 + . . . + αp np . Conversely, let FB be given. Then FB + α1 n1 + . . . + αp np also defines a deformation of (X, 0) over (B, 0) for all n1 , . . . ,np ∈ NX,0 . Proof. We can write: FB = FA +

X

αi gi ;

FB′ = FA +

X

αi gi′ .

Let r be a relation, that is, r · f = 0. We will see that r · (gi − gi′ ) ∈ (f1 , . . . ,fk ), that is, ′ gi − gi′ defines an element of NX,0 . Consider lifts RB and RB of r, which are relations ′ between the FB and the FB respectively. By Lemma 10.2.22, we may assume that the ′ restrictions of RB and RB to C {x}k ⊗ A are equal. Therefore we can write: X X ′ RB = RA + ci · αi ; RB = RA + c′i · αi .

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343

for complex numbers ci and c′i . As B −→ A is a small extension, it follows that mαi = 0 for all i. Hence X X 0 = RB · FB = RA · FA + (r · gi )αi + (ci · f )αi . Similarly

′ 0 = RB · FB′ = RA · FA +

X

(r · gi′ )αi +

X

(c′i · f )αi .

By subtracting and looking at the coefficients of αi we see that for all i = 1, . . . ,p: r · (gi − gi′ ) ∈ (f1 , . . . ,fk ). Therefore, ni := gi − gi′ is an element of NX,0 . The converse statement is much simpler, and left as Exercise 10.2.34. Proof of Theorem 10.2.18. Let OT,0 = C {t}/J, and (XT , 0) −→ (T, 0) be a deformation of (X, 0), given by equations FT (x,t) = 0. Versality means that we have to find a map (T, 0) −→ (S, 0) such that the deformation (X ×S T, 0) −→ (T, 0) is isomorphic to (XT , 0) −→ (T, 0). Dually the map (T, 0) −→ (S, 0) is given by an analytic map OS,0 −→ OT,0 . Thus we have to find τ power series s1 (t), . . . ,sτ (t) ∈ C {t} to define this map. The space (X ×S T, 0) is given by the equations FS (x,s(t)) = 0. To say that this deformation is isomorphic to FT (x,t) = 0 means that we can find an automorphism Φ of C {x,t} with Φ(t) = t, Φ(x)|t=0 = x, and a matrix Λ ∈ C {x,t} with Λ|t=0 the identity matrix such that (10.2)

FS (x,s(t)) ≡ (FT ◦ Φ)(x,t) ◦ Λ mod J · C {x,t}k ,

with OT, 0 = C {t}/J. Let Φ(xi ) = φi and φ = (φ1 , . . . ,φn ). We want to apply Grauert’s Approximation Theorem 8.2.2 in order to find a solution to the equation (10.2). We have to solve for φ, Λ and s, but the s is only allowed to depend on t, and not on x. The equation (10.2) has a solution of order zero, as it then reads f = f . Thus we have to show that every solution of (10.2) of order e with respect to t extends to a solution of (10.2) of order e + 1. Therefore, suppose (φ,Λ,s) is a solution of (10.2) of order e. Thus (10.3)

FS (x,s(t)) ≡ FT (φ,t) ◦ Λ

mod (J + (t)e+1 ) · C {x,t}k .

Here (t) is the maximal ideal of C {t}. We will change the order e + 1 terms in t of s,φ and Λ in such a way that (10.3) holds modulo (J + (t)e+2 ). Consider the small surjection σ : B := C {t}/(J + (t)e+2 ) −→ C {t}/(J + (t)e+1 ) =: A. Take a basis α1 , . . . ,αp of Ker(σ). We can factorize σ into p simple surjections σi , for i = 1, . . . ,p. Thus, using induction, we may suppose that σ : B −→ A is a small surjection, the kernel of σ is of dimension one, and Ker(σ) is generated by a monomial α of degree e + 1. Step 1. Change of s(t): We apply 10.2.23 to σ : B −→ A, FB = FS (x,s(t)) and FB′ = FT (φ,t) ◦ Λ. The s(t), φ(x,t) and Λ(x,t) are given by induction. It follows that (10.4)

FS (x,s(t)) − FT (φ,t) ◦ Λ = n · α

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for an element n ∈ C {x}k such that f 7→ n is an element of the normal module NX,0 . By assumption the Kodaira-Spencer map is surjective, that is, the elements ai : f −→ gi 1 1 mod (s) for i = 1, . . . ,τ project to generators of TX, . Thus as elements of TX, , we can 0 0 write n = b 1 a1 + . . . + b τ aτ for suitable bi ∈ C . Now we can redefine the s(t): s1 (t) := s1 (t) − b1 α, . . . , sτ (t) := sτ (t) − bτ α. Step 2. Change of φ(x,t). After the change of s(t) we have just done, we have reached 1 the stage that the n = (n1 , . . . ,nk ) ∈ NX,0 in (10.4) defines a trivial element in TX, . 0 −1 Then, obviously, nΛ|t=0 is a trivial element too. Therefore, there exist c1 , . . . ,cn ∈ C {x} ∂ ∂ such that θ(f ) = nΛ−1 |t=0 , with θ := c1 ∂x1 + . . . + cn ∂xn . We redefine the automorphism (φ1 , . . . ,φn ,t) in the following way. φ1 := φ1 − c1 α, . . . , φn := φn − cn α. It follows from Taylor’s expansion that after this redefinition, the n defines the trivial element of NX,0 , that is, ni is the zero element in OX,0 for i = 1, . . . ,k. Step 3. Change of Λ. After these redefinitions the elements n in (10.4) have entries ni which all give the zero element P in OX,0 = C {x}/(f1 , . . . ,fk ). Thus we can find elements cji ∈ C {x} such that ni = j cji fj . Redefine the matrix Λ by putting Λ := Λ − (cji )α. After this redefinition, we reached the stage that n = 0 ∈ C {x}k . This is what we had to show. To finish this section, we will prove the existence of a versal deformation for (X, 0) ⊂ (C n , 0) that have an isolated singularity, are Cohen-Macaulay, and have dimension n − 2. The Auslander-Buchsbaum formula 6.5.20 implies that the projective dimension of OX,0 is equal to two. The main ingredient is now the Hilbert-Burch Theorem 6.5.26 and Theorem 10.2.18. Indeed, by the Hilbert-Burch Theorem, one has the following free R := On – resolution of OX,0 = R/I, where I = I (X, 0): α

α

1 2 R −→ R/I −→ 0. F1 −→ 0 −→ F2 −→

Here α2 is an (t − 1) × t matrix for some t. Generators f1 , . . . ,ft for the ideal I are obtained by taking the (t−1)–minors of the matrix α2 . Equivalently, one has the following resolution of the ideal I: α1 α2 I −→ 0. F1 −→ 0 −→ F2 −→ Proposition 10.2.24. With the above notation, we get deformations of (X, 0) over any germ of a complex space (S, 0) in the following way. Let E be a matrix with entries in b S,0 . Consider the matrix α2 + E, and consider It−1 (α2 + E). This is an ideal in On ⊗m b S,0 , and defines a space (XS , 0) such that (XS , 0) −→ (S, 0) is a flat deformation On ⊗O of (X, 0). Moreover, all deformations of (X, 0) are obtained this way. Proof. Step 1. Consider α1 + E ′ , the 1 × t matrix where the i–th entry equals (−1)i times the minor obtained from α2 + E by leaving out the i–th row. Then the sequence α +E

α +E ′

b S,0 )t−1 −−2−−→ (On ⊗O b S,0 )t −−1−−→ It−1 (α2 + E) −→ 0 (On ⊗O

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345

is certainly a complex, that is, (α1 + E ′ ) ◦ (α2 + E) = 0. This implies that we can lift all the relations between the minors of α2 , which are generators of the ideal I. So we get indeed a flat deformation. Step 2. Conversely, suppose we have a deformation (XS , 0) of (X, 0) over (S, 0). The flatness says that we can lift the relations. Hence we get a complex (10.5)

α′

α′

2 1 b S,0 )t−1 −→ b S,0 )t −→ b S, 0 −→ OXS ,0 −→ 0. (On ⊗O (On ⊗O On ⊗O

where α′1 = α1 modulo mS,0 and α′2 = α2 modulo mS,0 . We leave it as Exercise 10.2.37 to prove that the sequence (10.5) is exact, that is, the kernel of α′1 is generated by the lifts of the set of t − 1 generators of the kernel of α. We claim that the map α′2 is in fact injective. To show this suppose α′2 (a1 , . . . ,at−1 ) = 0. Calculating modulo mS,0 , we see that (a1 , . . . at−1 ) is zero modulo mS,0 . This is because α2 is injective. Take a basis P b1 , . . . ,bp of mS,0 /m2S,0 , and write (a1 , . . . ,at−1 ) = (ai1 , . . . ,ait−1 )bi modulo m2S,0 for aij ∈ On . From α′2 (a1 , . . . ,at−1 ) = 0 it follows by calculating modulo m2S,0 that aij = 0. Thus (a1 , . . . at−1 ) is zero modulo m2S,0 . Going on like this, we see that (a1 , . . . at−1 ) is zero modulo mkS,0 for all k, hence 0 by Krull’s Intersection Theorem. So we get an exact sequence α′

α′

2 1 b S,0 )t−1 −→ b S,0 )t −→ b S, 0 −→ OXS ,0 −→ 0. 0 −→ (On ⊗O (On ⊗O On ⊗O

b S,0 . It follows from the Hilbert-Burch Theorem that I = Let I = I (XS , 0) ⊂ On ⊗O aIt−1 (α′2 ) for some nonzerodivisor a. Calculating modulo mS,0 we see that a ≡ 1 mod mS,0 . b S,0 . Then we change the minors of α′2 to a−1 times the minors, Thus a is a unit in On ⊗O to get the desired generators of the ideal I.

Theorem 10.2.25. Let (X, 0) be a Cohen-Macaulay codimension two isolated singularity, whose ideal is given by the (t − 1)–minors of a matrix A with entries in On . Let 1 τ = dimC (TX, ). Consider matrices αi such that the deformations given by the (t − 1)– 0 minors of a matrix A + εαi form a basis of the vector space of isomorphism classes of deformations over (T, 0). Then the ideal given by the (t − 1)–minors of the matrix Pτ b {s1 , . . . ,sτ } is the ideal of a space (XS , 0) A + i=1 si αi , which has entries in On ⊗C which gives a semi-universal deformation of (X, 0) over (S, 0) ∼ = (C τ , 0). Proof. The theorem follows from Proposition 10.2.24 and Theorem 10.2.18.

Exercises 10.2.26. (1) Let φ : C {x} −→ C {x} =: O be an automorphism, and I ⊂ O be an ideal. Prove that λ 7→ ϕλϕ−1 induces an isomorphism HomO (I,O/I) −→ HomO (ϕ(I),O/ϕ(I)). 1 ∼ 1 (2) Let (X, 0) ∼ = (Y, 0) be germs of complex spaces. Prove that TX, 0 = TY,0 . 1 10.2.27. Let (X, 0) be a germ of a smooth space. Show that TX, 0 = 0.

10.2.28. Prove Lemma 10.2.2. 10.2.29. Prove that the definition of flat deformation is independent of the choice of generators of (XS , 0).

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10.2.30. (1) Consider the germ of the complex space (X, 0) defined by the ideal I = (xz,xw,yz,yw) ⊂ C {x,y,z,w}. Thus V (I) is the union of two planes in C 4 intersecting in a point. Show 1 that TX, 0 = 0. Singularities with this property are called rigid. 1 (2) Give an example of a nonisolated singularity (X, 0) with finite-dimensional TX, 0. ` ´ 10.2.31. Consider the curve singularity V (xy,xz,yz), 0 . Compute a semi-universal deformation of this curve singularity.

10.2.32. Prove that for a (flat) deformation (XS , 0) −→ (S, 0), and an analytic map p : (T, 0) −→ (S, 0) the map (X ×S T, 0) −→ (T, 0) is flat. 10.2.33. Let (X, 0) be a complete intersection with isolated singularity. Describe a versal deformation of (X, 0). 10.2.34. Finish the proof of Lemma 10.2.23. 10.2.35. Consider t(t − 1) variables xij for 1 ≤ i ≤ t − 1 and 1 ≤ j ≤ t. Consider the generic determinantal variety (X, 0) ⊂ (C t(t−1) , 0) defined by the ((t − 1)–minors of the matrix (xij ). Prove that (X, 0) is rigid. 10.2.36. Use a computer algebra system (for example SINGULAR) to compute Example 10.2.8 (Hint: Try in SINGULAR the following: ring r =0,(x(0..4)),(c,dp); matrix m[2][4] = x(0),x(1),x(2),x(3),x(1),x(2),x(3),x(4); ideal i = minor(m,2); module s = syz(i); qring q = std(i); module s = imap(r,s); syz(transpose(s));) 10.2.37. Prove that the sequence (10.5) is exact. (Hint: Use the same method as in the proof of the injectivity of α′2 .)

10.3

Existence of a Semi-Universal Deformation

In this section we will finally give the proof that for an isolated singularity a versal deformation exists. We need many preparations. Definition 10.3.1. Let (X, 0) be a germ of a complex space, A a local Artinian C – algebra with maximal ideal mA . Suppose me+1 = 0. Then a deformation (XA , 0) −→ (A, 0) A is called a deformation of order e. Definition 10.3.2. Let (X, 0) be a germ of a complex space with local ring OX,0 = C {x}/(f1 , . . . ,fk ). Let A = C [[s]]/J be a local formal C –algebra in the variables s1 , . . . ,sp , with maximal ideal mA . By abuse of notation, we denote by J also the ideal J · C {x}[[s]]. (1) A formal deformation (XA , 0) of (X, 0) over A is given by a quotient of C {x}[[s]]/J by an ideal (F1 , . . . Fk ) such that (a) Fi (x,0) = fi for i = 1, . . . ,k. (b) For every relation r ∈ C {x}k , that is r · f = 0, there exists R ∈ C {x}[[s]]k such that R · F = 0 as elements of C {x}[[s]]/J and R(x,0) = r.

10.3 Existence of a Semi-Universal Deformation

347

(2) For all e ∈ N, consider the local Artinian C –algebra Ae defined by Ae = C [[s]]/(J + me+1 A ). Note that a formal deformation induces a deformation (XAe , 0) −→ (Ae , 0) of order e for all e. (3) A formal deformation (XA , 0) is called versal if for all local Artinian C –algebras B, and all deformations (XB , 0) −→ (B, 0) there exists a homomorphism Ae −→ B for some e ≫ 0 and an isomorphism of deformations (XAe ×Ae B, 0) ∼ = (XB , 0). It is called semi-universal, if the induced homomorphism on the tangent spaces (mA /m2A )∗ −→ (mB /m2B )∗ is uniquely determined. Remark 10.3.3. Let (XA , 0) and (XB , 0) be two formal semi-universal deformations of a germ of a complex space (X, 0). Then the formal analytic algebras A and B are isomorphic. To see this, note that, because (XA , 0) is semi-universal, we have a unique map α : A1 −→ B1 , inducing (XB1 , 0) −→ (B1 , 0). Similarly, we get a map β : B1 −→ A1 . By semiuniversality, the composition β ◦ α is the identity on A1 , and α ◦ β is the identity on B1 . Hence A1 and B1 are isomorphic. Now let e ∈ N be given and consider (XBe , 0) −→ (Be , 0). By semi-universality, for some q there exists a map Aq −→ Be , so that (XBe , 0) −→ (Be , 0) is induced by (XAq , 0) −→ (Aq , 0). As me+1 Be = 0 by definition, we can take q = e. Thus we get a map φ : Ae −→ Be . Conversely, we get a map ψ : Be −→ Ae . The composition ψ ◦ φ : Ae −→ Ae is such that the pull-back of the deformation (XAe , 0) −→ (Ae , 0) is isomorphic to (XAe , 0) −→ (Ae , 0). By semi-universality, this map is uniquely defined up to first order, and is the identity modulo the square of the maximal ideal. By the Inverse Function Theorem8 , the map ψ ◦ φ is an isomorphism, and similarly φ ◦ ψ is an isomorphism. Note that in particular dimC (Ae ) = dimC (Be ) for all e ∈ N. Remark 10.3.4. Suppose that (XA , 0) is a formal semi-universal deformation of (X, 0). 1 Then the Kodaira-Spencer map (mA /m2A )∗ −→ TX, is an isomorphism. 0 1 Namely, to construct the inverse mapping, consider an element of TX, , that is, 0 2 a deformation over the space defined by the local ring C [[ε]]/ε . Because the formal deformation is semi-universal, we get a unique map A1 −→ C [[ε]]/ε2 . By looking at the coefficients of ε of the values of all elements in mA /m2A we get an element of (mA /mA )∗ . One checks that this gives the inverse of the Kodaira-Spencer map. Thus, as already said, we first want to show the existence of a formal semi-universal deformation. The idea is to do this order by order, that is, define deformations (XAe , 0) → (Ae , 0) of order e for all e ∈ N. For e = 1, we define a deformation over (A1 , 0), with 1 A1 = C [[s1 , . . . ,sτ ]]/m2 , and τ is the dimension of TX, . Given Ae , it might be quite 0 obvious that we would like to have a deformation of order e + 1 over an as big as possible space Ae+1 . In order to see that such a space exists, we will develop obstruction theory. 2 To formulate this, we need to define a further OX,0 –module, called TX, . 0 Definition 10.3.5. Let a singularity (X, 0), with local ring OX,0 = On /I be given. For simplicity, we write On = O. Take generators f1 , . . . ,fk of I. We consider the map F := Ok −→ O, which sends the ei = (0, . . . ,0,1,0, . . . ,0) to fi . We let R be the kernel of this map. Thus, we have an exact sequence 0 −→ R −→ F −→ O −→ OX,0 −→ 0. 8

This theorem holds in the formal case too, just ignore the convergence.

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10 Deformations of Singularities

Let R0 ⊂ R be the submodule generated by the so-called Koszul relations fi ej −fj ei ∈ R. Note that fi ej − fj ei is indeed in R, as obviously fi · fj − fj · fi = 0. Now we can define 2 TX, := HomO (R/R0 ,OX,0 )/ HomO (F ,OX,0 ). 0 2 Lemma 10.3.6. TX, is a well-defined OX,0 –module. 0

Proof. The fact that it is an OX,0 –module is easy, as we are looking at homomorphisms to OX,0 . Thus if we multiply a homomorphism with something in the ideal of (X, 0), it will certainly become the zero homomorphism. As R ⊂ F we get dually a map HomO (F ,OX,0 ) −→ HomO (R,OX,0 ) Now R0 is generated by elements of type fi ej − fj ei . If ϕ ∈ Hom(F ,OX,0 ), then this element is sent to fi ϕ(ej ) − fj ϕ(ei ) = 0 ∈ OX,0 . This shows that we get an induced map HomO (F ,OX,0 ) −→ HomO (R/R0 ,OX,0 ). It remains to prove that the definition is independent of the choice of the generators f1 , . . . ,fk of the ideal of OX,0 . This is left as Exercise 10.3.25. Theorem 10.3.7. Let (X, 0) be a germ of a complex space with an isolated singularity. 2 Then TX, is a finite-dimensional C –vector space. 0 The proof is left as Exercise 10.3.26. The main theorem on obstruction theory is the following. Theorem 10.3.8. Let (X, 0) be a germ of a complex space. Consider a local Artinian (resp. formal, resp. analytic) C –algebra A, and a small surjection B −→ A, with kernel J. This means that we have an exact sequence 0 −→ J −→ B −→ A −→ 0. and m · J = 0, m the maximal ideal of B. Therefore, J is a vector space over C = B/m. Suppose we have a deformation (XA , 0) −→ (A, 0) of (X, 0). For a short notation we write ξ for this deformation. Then we have the following statements. (1) There exists a well-defined obstruction element 2 ob(ξ)B→A ∈ TX, ⊗ J. 0

This obstruction element ob(ξ)B→A is zero, if and only if there exists a flat deformation of (XB , 0) −→ (B, 0) extending the given deformation ξ, (2) Suppose we have a commutative diagram 0

// J

// B

// A

// 0

0

 // J ′

 // B ′

 // A′

// 0

such that the rows are small extensions. Let ξ ′ be the deformation over A′ induced 2 2 by ξ. Let β : TX, ⊗ J −→ TX, ⊗ J ′ be the induced map. Then 0 0
ob(ξ ′ )B ′ →A′ = β ob(ξ)B→A .

10.3 Existence of a Semi-Universal Deformation

349

Proof. We give the proof for local Artinian C –algebras, the remaining cases we leave as Exercise 10.3.22. We first construct the obstruction element ob(ξ)B→A . Suppose the
k deformation ξ, that is (XA , 0) −→ (A, 0), is given by the elements FA ∈ C {x} ⊗ A . Let r ∈ R be a relation. By flatness we get a relation rA between the FA . Now take arbitrary lifts FB ∈ C {x} ⊗ B of FA and a lift rB of rA . Then we consider rB · FB ∈ OX,0 ⊗ B. As rB is a lift of rA , and FB is a lift of FA it follows that the image of rB ·FB in C {x} ⊗ A is zero. Thus we have in fact rB · FB ∈ OX,0 ⊗ J. We claim that the map r 7→ rB · FB defines an element in HomO (R/R0 ,OX,0 ⊗ J) = 2 HomO (R/R0 ,OX,0 ) ⊗ J,9 which induces the obstruction element ob(ξ)B→A ∈ TX, ⊗ J. 0 Step 1. We first show that the element rB · FB depends on r and on FB , but not on the ′ particular lift rB . In particular it is independent of the lift rA . So, take another lift rA ′ ′ ′ of r, with rA · FA = 0, and a lift rB of rA . First we show that we may assume that the ′ ′ restrictions rA of rB and rA of rB to A are equal. The idea of the proof is the same as in the proof of Lemma 10.2.22, so it is done by induction on dimC (A). This part of the proof does not use the fact that B −→ A is a small extension. The case dimC (A) = 1 is our assumption. To do the induction step, take an element 0 6= α ∈ A with mA · α = 0. ′ We may suppose by induction that the image of rB and rB in A/(α) are equal. So we ′ k ′ can write rA − rA = h · α for some h ∈ C {x} . By assumption rA · FA = rA · FA = 0. Hence h · FA α = 0. Now mA α = 0, and FA ≡ f modulo mA , so that h · f = 0. Thus h is a relation, which can be lifted to a relation hA between the FA . We lift hA to hB , and ′ ′′ ′ ′′ change rB to rB := rB − hB α′ , for a suitable α′ which maps to α. Then rB is also a lift ′ ′′ ′′ of r. Moreover rB · FB is equal to rB · FB , and the restrictions of rB and rB to A are equal. ′ Thus we now reached the stage that rB − rB has entries in C {x} ⊗ J, J being the kernel of the map B −→ A. As mB J = 0 by assumption, and FB ≡ f modulo mB it ′ ′ follows that (rB − rB ) · FB = (rB − rB ) · f , which, as the fi are zero in OX,0 , is zero in OX,0 ⊗ J. This shows that rB · FB is independent of the lift rB of r. Therefore the map r 7→ rB · FB gives a well-defined element in HomO (R,OX,0 ) ⊗ J. This element might depend on the choice of the lift FB . Step 2. The restriction of the homomorphism to the submodule R0 generated by the Koszul relations is zero. Indeed, we can take as lift of r = fi ej − fj ei simply the element rB := FBi ej − FBj ei . Then rB · FB = 0.

2 Step 3. We now show that the class of rB ·FB gives a well-defined element in TX, ⊗J. Take 0 ′ ′ k ′ another lift FB of FA . Then FB −FB ∈ C{x} ⊗J. We look at rB ·(FB −FB ). As mB J = 0, and rB ≡ r modulo mB , it follows that this is equal to r · (FB − FB′ ). It follows that the map r 7→ rB · (FB − FB′ ) is equal to the map r 7→ r · (FB − FB′ ) ∈ HomO (F ,OX,0 ) ⊗ J. This is what we had to show.

Step 4. Suppose that an extension of ξ to a deformation (XB , 0) −→ (B, 0) exists. Then we can find FB and for every r ∈ R an element rB such that rB · FB = 0 ∈ C {x} ⊗ J. Thus, certainly, the obstruction element is zero. 9

This equality holds because the tensor product is taken over C and, therefore, all modules are free.

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10 Deformations of Singularities

Step 5. On the other hand, suppose that ob(ξ)B→A = 0. Take an arbitrary lift FB′ = ′ ′ (FB1 , . . . ,FBk ) of FA . Because ob(ξ)B→A = 0, the element in HomO (R/R0 ,OX,0 ) ⊗ J given by r 7→ rB · FB′ , rB any lift of r, is, in fact, in HomO (F ,OX,0 ) ⊗ J. Hence, there exist h1 , . . . ,hk ∈ C {x} ⊗ J such that the map is given by k X ri hi . r = (r1 , . . . ,rk ) 7→ i=1

Now we define

′ ′ ) − (h1 , . . . ,hk ). FB = (FB1 , . . . ,FBk ) := (FB1 , . . . ,FBk

By construction, the element in HomO (R,OX,0 ) ⊗ J = HomO (R,OX,0 ⊗ J), given by r 7→ rB · FB , rB any lift of r, is the zero map. We claim that the FB defines a deformation which extends FA . To show ′ ′ this, for every relation r ∈ R a lift rB such that rB · FB = 0 has to be found. Now by construction rB · FB = 0 as element of OX,0 ⊗ J. Therefore, there exist a1 , . . . ,ak in J with rB · FB = a1 f1 + . . . + ak fk . Put a = (a1 , . . . ,ak ). As mB J = 0, and FB ≡ f modulo ′ ′ mB , it follows that a · FB = a · f . Now define rB := rB − a. Then rB is a lift of r with ′ rB · FB = 0. As we can do this for all r, it follows that FB gives a (flat) deformation of (X, 0). Step 6. The second statement of the theorem is immediate from the construction of the obstruction element. Corollary 10.3.9. Let (X, 0) be a germ of a complex space, and A = O/J a local Artinian (resp. formal, resp. analytic) C –algebra, and A′ := O/mJ. Let ξ be a deformation (XA , 0) −→ (A, 0), and consider the small extension 0 −→ J/mJ −→ O/mJ −→ O/J −→ 0, 2 and let ob(ξ)A′ →A ∈ TX, ⊗ (J/mJ) be the obstruction element. By Exercise 1.2.46 this 0 obstruction element gives a map which we, by abuse of notation, give the same name:
∗ 2 −→ J/mJ. ob(ξ)A′ →A : TX, 0

Let I mod mJ be the image of the map ob(ξ)A′ →A . Then I is the unique minimal ideal (with respect to inclusion) with the properties: (1) mJ ⊂ I ⊂ J

(2) Let B = O/I. Then there exists a deformation (XB , 0) −→ (B, 0) of (X, 0) inducing the deformation ξ. The small surjection B −→ A we call a maximal extension with respect to the deformation (XA , 0) −→ (A, 0).

10.3 Existence of a Semi-Universal Deformation

351

Proof. We can write ob(ξ)A′ −→A = α1 ⊗ h1 + . . . + αp ⊗ hp , 2 for some linear independent elements α1 , . . . ,αp of TX, and h1 , . . . hp of J. Let I = 0 (h1 , . . . ,hp ) + mJ. To show that a deformation (XB , 0) −→ (B, 0) exists, we consider the local Artinian C –algebra O/I. Then ob(ξ)O/I→A is equal to α1 ⊗ h1 + . . . + αp ⊗ hp , 2 but this time considered as element of TX, ⊗ (J/I), thus ob(ξ)O/I→A is zero. (See the 0 second part of Theorem 10.3.8.) Hence we forced the obstruction element to be zero, and therefore there exists an extension of ξ over the Artinian C –algebra O/I. To show that I is minimal and unique with respect to the two properties, let K be another ideal satisfying mJ ⊂ K ⊂ J, such that there exists an extension of ξ over O/K. The obstruction element is α1 ⊗ h1 + . . . + αp ⊗ hp , and therefore is zero considered as element of J/K. Hence I = (h1 , . . . ,hp ) + mJ ⊂ K, which is what we had to prove.

Lemma 10.3.10. Let A = C [[s]]/J be a formal algebra, (XA , 0) be a formal deformation of (X, 0) over A, and (XAe , 0) be the induced deformation of order e for all e. Consider the small surjection (10.6)

C [[s]]/J + me+1 −→ C [[s]]/J + me .

Assume that (1) C [[s]]/J + me −→ C [[s]]/J + me−1 is a maximal extension for the deformation (XAe−1 , 0) −→ (Ae−1 , 0),     J+me J+me+1 = dim (2) dimC mJ+m C e+1 mJ+me .

Then (10.6) is a maximal extension for the deformation (XAe , 0).

Proof. If C [[s]]/J + me+1 is not maximal, then there exists an ideal I $ J + me+1 with mJ + me+1 ⊂ I ⊂ J + me , and such that the deformation extends to a deformation over C [[s]]/I. Then the inclusion I + me ⊂ J + me is in fact an equality because of the first e condition. Therefore, the minimal number of generators of  I modulo  to ethe  m is equal  e I+m J+m minimal number of generators of J modulo me , that is, dimC mI+me = dimC mJ+m . e As I $ J + me+1 , it follows that there is a minimal set of generators of J modulo me+1 , which has an element of degree e. In formula’s       J+me I+me I dimC mJ+m = dimC mI+m ≤ dimC mI+m = e e e+1     J+me+1 I < dimC mJ+m . dimC mJ+m e+1 e+1 But this is in contradiction to the second condition. It is time for an example. Example 10.3.11. We consider the germ of the complex space (X, 0) defined by the 2–minors of the matrix   x0 x1 x2 x3 . x1 x2 x3 x4

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10 Deformations of Singularities

1 The modules NX,0 and TX, have been described in 10.2.8 and 10.2.14. Thus we have the 0 following deformation ξ over C [[s1 ,s2 ,s3 ,t]]/m2 =: A, with m = (s1 ,s2 ,s3 ,t):

F12 = f12 − s1 x1 + tx0 F23 = f23 + s1 x3 − s2 x2 − tx2

F34 = f34 + s2 x4 − s3 x3 + tx4 F13 = F24 =

F14 =

f13 − s2 x1 f24 + s1 x4 − s3 x2 f14 − s3 x1 + tx2

The fij are the 2–minors of the above matrix as in Example 10.2.8. The Fij give a (flat) deformation over C [[s1 ,s2 ,s3 ,t]]/(s1 ,s2 ,s3 ,t)2 , so we must be able to lift the relations. Now we give the lift Rijk and Sijk of the relations rijk and sijk . R123 = R124 =

x0 F23 − (x1 + s1 )F13 + (x2 + s2 )F12

x0 F24 − (x1 + s1 )F14 + (x3 + s3 )F12 − tF13

R134 = x0 F34 − (x2 + s2 )F14 + (x3 + s3 )F13 − tF14 − tF23 R234 = (x1 + s1 )F34 − (x2 + s2 + t)F24 + (x3 + s3 )F23 S123 = S234 =

S124 = S134 =

x1 F23 − (x2 + t)F13 + x3 F12 x2 F34 − x3 F24 + x4 F23

x1 F24 − (x2 + t)F14 + x4 F12 − tF23 x1 F34 − x3 F14 + x4 F13 − tF24

These Rijk and Sijk are all zero modulo (s1 ,s2 ,s3 ,t)2 expressing the fact that we have a (flat) deformation. We obtain the obstruction element by looking at the quadratic terms in s and t, which one gets by simply expanding the terms: r123 → 7 R123 = s2 tx0 r124 → 7 R124 = s3 tx0 − s1 tx2 + s2 tx1

r134 → 7 R134 = r234 → 7 R234 =

s3 tx1 − s1 tx3 0

s134 7→ S134 = s234 7→ S234 =

−s1 tx4 + s3 tx2 0

s123 7→ S123 = s2 tx1 s124 7→ S124 = s3 tx1 − s1 tx3 + s2 tx2

2 This map of R −→ OX,0 ⊗ J gives the element ob(ξ)A′ →A in TX, ⊗ J, where A′ = 0 3 2 3 C [[s1 ,s2 ,s3 ,t]]/m and J = m /m . Now, according to Corollary 10.3.9, we need a basis
∗ 2 2 ′ of TX, to describe the map ob(ξ) : T −→ m2 /m3 , and compute I, the image A →A 0 X,0 of this map. We will see that we already computed part of a basis, which is enough for our purpose. We have in particular that the elements α1 , α2 and α3 given in the first three columns of the following table, (obtained by looking at the coefficients of s1 t, s2 t and s3 t) give three elements in HomO (R/R0 ,OX,0 ). The last six columns correspond to the images of the relations with respect to the canonical basis of HomO (F ,OX,0 ). We will see that α1 , α2 and α3 are linearly independent modulo HomO (F ,OX,0 ), and therefore 2 . induce linear independent elements of TX, 0

10.3 Existence of a Semi-Universal Deformation

(10.7)

r123 r124 r134 r234 s123 s124 s134 s234

7→ 7 → 7→ 7 → 7→ 7 → 7 → 7 →

0 −x2 −x3 0 0 −x3 −x4 0

x0 x1 0 0 x1 x2 0 0

0 x0 x1 0 0 x1 x2 0

x2 x3 0 0 x3 x4 0 0

x0 0 0 x3 x1 0 0 x4

353 0 0 x0 x1 0 0 x1 x2

−x1 0 x3 0 −x2 0 x4 0

0 x0 0 −x2 0 x1 0 −x3

0 −x1 −x2 0 0 −x2 −x3 0

2 To see whether these elements are linearly independent in TX, we have to look at the 0 2 “trivial” elements in TX,0 , that is, elements of Hom(F ,OX,0 ) which are given by the final six columns of the above table (10.7). These six columns are the duals of f12 , f23 , f34 , f13 , f24 , and f14 respectively. Thus, for example, the first row and the final six columns says that r123 is the relation x2 f12 +x0 f23 +0f34 −x1 f13 +0f24 +0f14 = 0. An elementary but boring check now shows that α1 , α2 and α3 represent linearly independent elements 2 2 in TX, . They can therefore be considered as part of a basis of TX, and ob(ξ)A′ →A = 0 0 α1 ⊗ s1 t + α2 ⊗ s2 t + α3 ⊗ s3 t. Hence the minimal ideal I of the corollary is given by the ideal (s1 t,s2 t,s3 t) + (s1 ,s2 ,s3 ,t)3 . Note, however, that the family considered above, in fact, defines a deformation over C {s1 ,s2 ,s3 ,t}/(s1 t,s2 t,s3 t), as all the Rijk · F and Sijk · F vanish modulo (s1 t,s2 t,s3 t) and not only modulo (s1 t,s2 t,s3 t) + (s1 ,s2 ,s3 ,t)3 .

Theorem 10.3.12. Let A = C [[s]]/J be a formal local C –algebra, and (X, 0) be a germ of a complex space. Let (XA , 0) be a formal deformation of (X, 0). For all k put Ak := A/mk+1 and Jk = J + mk+1 . Suppose that (1) the Kodaira-Spencer map of (XA , 0) is surjective (resp. an isomorphism), (2) for all e ≥ 2, the small extension Ae+1 −→ Ae is a maximal extension with respect to the deformation (XAe , 0) −→ (Ae , 0). Then (XA , 0) is a formal versal (resp. semi-universal) deformation. Example 10.3.13. The family written down in the previous example is formally versal. It is a deformation over A = C [[s1 ,s2 ,s3 ,t]]/(s1 t,s2 t,s3 t). Details are left as an exercise. The basic idea of the proof is as the proof of Theorem 10.2.18, but there is a subtle difficulty popping up, which is taken care of by Proposition 10.3.15. In its proof we need the following lemma. Lemma 10.3.14. Suppose that ϕ : B −→ A is a simple surjection,and let ψ : C −→ B be a map of local Artinian C –algebras such that (1) ϕ ◦ ψ : C −→ A is surjective, (2) ϕ does not have a section. Then ψ is surjective. Proof. Suppose the converse. Let J = Ker(ψ). Then we have a injective map ψ : C/J −→ B. Moreover ϕ ◦ ψ : C/J −→ A is surjective. As ψ : C/J −→ B is not surjective, and dimC (B) = dimC (A) + 1, it follows that dimC (C/J) = dimC (A). In Exercise 10.3.23 you will prove that a surjective algebra map of Artinian C –algebras of the same (vector space) dimension is an isomorphism. Thus ϕ ◦ ψ is an isomorphism. Denote the inverse by β : A −→ C/J. Then the map ψ ◦ β is a section of ϕ. This is a contradiction.

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10 Deformations of Singularities

Proposition 10.3.15. Let local Artinian C –algebras A = C [[s1 , . . . ,sp ]]/I, and A′ = C [[s1 , . . . ,sp ]]/J be given, for some ideals I, J with mJ ⊂ I ⊂ J. Let two further Artinian C –algebras B and B ′ be given. Let ψ : A −→ A′ be the canonical surjection. Let ξ : (XA′ , 0) −→ (A′ , 0) be a deformation of (X, 0). Suppose that (1) the small extension A −→ A′ is a maximal extension with respect to the deformation ξ : (XA′ , 0) −→ (A′ , 0);

(2) we have a simple surjection π : B −→ B ′ ; (3) we have a map φ′ : A′ −→ B ′ and a deformation (XB , 0) −→ (B, 0) inducing Cartesian diagrams (XB′ , 0) −→ (XB , 0) ↓ ↓

(XB′ , 0) −→ (XA′ , 0) ↓ ↓ (B′ , 0)

φ′

(A′ , 0)

−→

(B′ , 0)

−→

(B, 0)

Then there exists a map α : A −→ B making the following diagram commutative: (10.8)

A ↓ψ

A′

α

−→ φ′

−→

B ↓π

B′.

Proof. We consider the so-called fibered product A′ ×B ′ B := {(a,b) ∈ A′ × B : φ′ (a) = π(b)}. Defining addition and multiplication component wise A′ ×B ′ B is a local C –algebra. We have a natural projection pr1 : A′ ×B ′ B −→ A′ which sends (a,b) to a. Note that pr1 is also a simple surjection. Indeed, pr1 (a,b) = 0 implies that a = 0, and thus, by definition π(b) = 0. Thus b is in the kernel of π, which by assumption is one-dimensional. It is immediate from the definition that making the diagram (10.8) commutative is the same as making the diagram A′:: ×B ′ B v φ vv pr1 vv v v  vv ψ // A′ A commutative. Case 1. Suppose that pr1 has a section, that is, there exists a map s : A′ −→ A′ ×B ′ B with pr1 ◦ s the identity. This is the easy case, as we then can define φ := s ◦ ψ. Case 2. Suppose that pr1 does not have a section. Step 1. We will show that A′ ×B ′ B is a quotient of C [[s1 , . . . .sp ]].10 We first construct a map ω making the diagram C [[s1 , . . . .sp ]]

ω

// A′ ×B ′ B

ψ

 // A′

pr1

 A 10

Note that both A and A′ are quotients of the same C [[s1 , . . . .sp ]].

10.3 Existence of a Semi-Universal Deformation

355

commutative. Indeed, by assumption A′ is a quotient of C [[s1 , . . . .sp ]]. Consider the composition φ′

β : C [|s1 , . . . ,sp ]] −→ A′ −→ B ′ .

As π : B −→ B ′ is surjective, we can find bi ∈ B with π(bi ) = β(si ) for all i = 1, . . . ,p. Now define ω(si ) = (si ,bi ) ∈ A′ ×B ′ B. As C [[s1 , . . . ,sp ]] is a formal free algebra, we can extend ω to an algebra map. Applying Lemma 10.3.14 to the case C = C [[s1 , . . . ,sp ]]/ Ker(ω), ψ = ω,B = A′ ×B ′ B, A = A′ and ϕ = pr1 , we obtain A′ ×B ′ B ∼ = C [[s1 , . . . ,sp ]]/ Ker(ω). Step 2. Write O = C [[s1 , . . . ,sp ]]. Put K := Ker(ω). Thus we get A′ ×B ′ B ∼ = O/K ։ O/J = A′ . Recall that A = O/I. We will show that ω induces a map φ : A −→ A′ ×B ′ B. To show this we have to show that I ⊂ K. Now O/K ։ O/J is a simple surjection, thus in particular small by Exercise 10.3.24. Hence mJ ⊂ K ⊂ J. As by assumption A −→ A′ is a maximal extension with respect to (XA′ , 0) −→ (A′ , 0), we have I ⊂ K, if there exists a deformation over O/K inducing the deformation ξ. This we show in Step 3. Step 3. We now write D = A′ ×B ′ B for short. To complete the proof, it therefore suffices to show that there exists a deformation (XD , 0) −→ (D, 0)


b ′ k . Moreover, by assumption inducing ξ. Suppose ξ is given by functions FA′ ∈ C {x}⊗A
b k , and (XB′ , 0) given by we have deformations (XB , 0) given by say FB ∈ C {x}⊗B
b ′ k . As the deformation over (A′ , 0) and (B, 0) induces the same say FB ′ ∈ C {x}⊗B deformation over B ′ , it can be easily seen that we may assume that φ′ (FA′ ) = FB ′ = π(FB ).


b k . To show that it defines a (flat) Thus the pair (FA′ ,FB ) defines an element in C {x}⊗D deformation, we have to lift relations. So let r ∈ C {x}k with r · f = 0 be an arbitrary relation. By flatness we can find RA′ ∈ C {x}k ⊗ A′ with RA′ · FA′ = 0. Let RB ′ be the image of RA′ under the natural map C {x}k ⊗ A′ −→ C {x}k ⊗ B ′ . By Lemma 10.2.22 we can find an RB ∈ C {x}k ⊗ B lifting RB ′ with RB · FB = 0. The pair (RA′ ,RB ) therefore b = C {x}k ⊗(A b ′ ×B ′ B), and (RA′ ,RB ) · (FA′ ,FB ) = 0. Hence is an element of C {x}k ⊗D the relation r lifts. This finishes the proof of the proposition.

Proof of Theorem 10.3.12. Consider a local Artinian C –algebra B, and suppose that we have a deformation (XB , 0) −→ (B, 0) of (X, 0). Let Ae := A/me+1 = C [[s]]/(J + me+1 ). We have an induced deformation (XAe , 0) −→ (Ae , 0). Consider the following small surjection (10.9)

0 −→ J + me+1 /(mJ + me+2 ) −→ C [[s]]/(mJ + me+2 ) −→ Ae −→ 0

By assumption, for all e the ideal J mod me+2 is minimal among those ideals for which there exists an extension of (XAe , 0) −→ (Ae , 0) over some small extension of Ae .

356

10 Deformations of Singularities

To show the versality, we have to show that there exists an e ∈ N and a morphism of C –algebras Ae −→ B such that the deformation (XB , 0) −→ (B, 0) is isomorphic to (XAe ×Ae B, 0) −→ (B, 0). This we will prove by induction on the vector space dimension dimC (B). We suppose that XAe is defined by elements F (e) (x,s) =
(e) (e) F1 (x,s), . . . ,Fk (x,s) ∈ C {x}k ⊗ Ae = C {x}k ⊗ C [[s]]/Je . (Recall that Je = J + me+1 .) Step 1. Consider the induced deformation over B/m2B . As the Kodaira Spencer map is surjective, we can find a map A1 −→ B inducing the deformation over B/m2B . This map is uniquely determined in case the Kodaira-Spencer map is an isomorphism.

Step 2. Now take a nonzero element α ∈ B, such that mB α = 0. Put B ′ = B/(α), and denote by π : B −→ B ′ the canonical projection. Then dimC (B ′ ) = dimC (B) − 1, and therefore, by induction, we have for suitable e ∈ N a morphism of C –algebras σ′

Ae −→ B ′ that induces a deformation which is isomorphic to the given deformation (XB′ , 0) −→ (B′ , 0), induced by (XB , 0) −→ (B, 0) via the projection π. By Proposition 10.3.15, there exists a map σ : Ae+1 −→ B, making the diagram Ae+1 ↓ Ae

σ

−→ σ′

−→

B ↓π B′

commutative. Step 3. Now that we have our map σ we get two different deformations over (B, 0): The given deformation (XB , 0) −→ (B, 0), and the deformation which is induced by σ. By assumption the restriction of the two deformations to (B′ , 0) are isomorphic. Hence there exist (1) a k × k matrix Λ′ with entries in C {x} ⊗ B ′ , lifting the identity matrix and (2) a B ′ – automorphism Φ′ defined by φ′1 = Φ′ (x1 ), . . . ,φ′n = Φ′ (xn ) ∈ C {x} ⊗ B ′ lifting the identity, such that F (e) (x,σ ′ (s1 ), . . . ,σ ′ (sp )) = FB′ (φ′1 , . . . ,φ′n ,t) ◦ Λ′ .

We now take any lift Λ of Λ′ , and φ1 , . . . ,φn of φ′1 , . . . ,φ′n . We will change σ, Λ, and φ1 , . . . ,φn in such a way that: F (e+1) (x,σ(s1 ), . . . ,σ(sp )) = FB (φ1 , . . . ,φn ,t) ◦ Λ. This change of σ, φi , and Λ can be done exactly as in the proof of 10.2.18. Theorem 10.3.16 (Schlessinger). Let (X, 0) be a germ of a complex space. Suppose 1 that TX, is a finite-dimensional vector space, of dimension p, and a1 , . . . ap are elements 0 1 of NX,0 projecting to a basis of TX, . The following construction gives a formal semi0 universal deformation of (X, 0).

10.3 Existence of a Semi-Universal Deformation

357

(1) Consider new variables s1 , . . . ,sp , and consider A := C [[s1 , . . . ,sp ]], with maximal ideal m = (s1 , . . . ,sp ). We set A1 := A/m2 and define a first order deformation:

by taking F (1) (x,s) = f +

Pp

(XA1 , 0) −→ (A1 , 0),

i=1 si ai

∈ C {x}k ⊗ A1 .

(2) Now suppose the deformation (XAe , 0) −→ (Ae , 0), for Ae = O/Je has already been constructed. We apply 10.3.9 to find an ideal Je+1 , minimal with respect to the properties mentioned in that corollary. Then define Ae+1 := O/Je+1 . So by 10.3.9, we can find (XAe+1 , 0) −→ (Ae+1 , 0) defined by F (e) (x,s). Continuing like this we get formal power series F ∈ C {x}[[s]]/J, which defines a (flat) deformation over J = ∩e Je . Moreover, all formal semi-universal deformations of (X, 0) are obtained by this construction. Lemma 10.3.17. Consider the ideals J and Je constructed in the statement of Theorem 10.3.16. Then Je = J + me+1 . Proof. As Je+k ⊂ Je for all k by construction, and J = ∩e Je , it suffices to show Je+1 + me+1 = Je and iterate this equation. This is proved by induction on e. The case e = 1 is trivial, as J1 = m2 . So assume we proved Je +me = Je−1 . As me+1 ⊂ mJe−1 ⊂ Je ⊂ Je−1 , and Je+1 ⊂ Je by construction, the inclusion Je+1 +me+1 ⊂ Je follows. To show the other inclusion, observe that there is a deformation of order e over the space with local ring O/Je+1 + me+1 (lifting the deformation (XAe−1 , 0) −→ (Ae−1 , 0)) by simply restricting. Hence, by minimality of Je we only have to show mJe−1 ⊂ Je+1 + me+1 ⊂ Je−1 . The inclusion Je+1 + me+1 ⊂ Je−1 is easy, as Je+1 ⊂ Je ⊂ Je−1 , and me+1 ⊂ me ⊂ Je−1 by construction. For the other inclusion, we use mJe−1 = mJe +me+1 ⊂ Je+1 +me+1 . The equality follows from the induction hypothesis, the inclusion holds because mJe ⊂ Je+1 by definition. Proof of Theorem 10.3.16. The fact that the constructed deformation is semi-universal is an immediate consequence of Theorem 10.3.12, Lemma 10.3.17 and Corollary 10.3.9. It remains to show that all formal versal deformations come from a construction as in Theorem 10.3.16. This we do by contradiction. Thus let (XA , 0) be a formal semiuniversal deformation. Now certainly, because of the versality, the Kodaira-Spencer map 1 is surjective. Furthermore dimC (mA /m2A ) is equal to the dimension of TX, , because the 0 deformation is semi-universal. This implies that the Kodaira-Spencer map is an isomorphism. Now suppose that for some e the small extension Ae+1 −→ Ae is not maximal with respect to the deformation (XAe , 0) −→ (Ae , 0). We can follow the construction of Theorem 10.3.16 to find a formal semi-universal deformation over A′ with A′e = Ae , and A′e+1 ։ Ae+1 . But we have formal semi-universal deformations over both A and A′ it follows that dimC (A′e+1 ) = dimC (Ae+1 ), see Remark 10.3.3. Thus Ae+1 was maximal after all, proving the claim.

358

10 Deformations of Singularities

Remark 10.3.18. Note that in fact we proved something stronger. Indeed, under the assumptions of Theorem 10.3.16, suppose we have a deformation (XB , 0) −→ (B, 0) of order q. We have an induced deformation of order q − 1 over the space defined by the Artinian ring Bq−1 := B + mq /mq . Suppose we have a map ϕ : Ae −→ Bq−1 inducing this deformation. Then there exists a lift φe : Af −→ B for some f ≥ e, that is, we have a commutative diagram Af

e φ

// B

 Ae

φ

 // Bq−1 ,

such that φe induces the given deformation (XB , 0) −→ (B, 0). This property of the constructed versal deformation is of great use in the proof of the existence of a convergent semi-universal deformation. Theorem 10.3.19. Let (XS , 0) and (S, 0) be germs of complex spaces, and suppose (XS , 0) −→ (S, 0) is formally semi-universal. Then (XS , 0) is a semi-universal deformation. Proof. Let a deformation (XT , 0) −→ (T, 0) be given. Suppose that this deformation is
b T,0 k . We have to find a map (T, 0) −→ (S, 0) such given by functions FT (x,t) ∈ C {x}⊗O that the deformation (XT , 0) −→ (T, 0) is isomorphic to (XS ×S T, 0) −→ (T, 0). Write OS,0 = C {s}/J, and OT,0 = C {t}/J ′ . To give a map (T, 0) −→ (S, 0) means that dually we have to find a map φ : C {s} −→ C {t} with φ(J) ⊂ J ′ . The induced deformation (XS ×S T, 0) −→ (T, 0) is given by FS (x,φ(s)) ∈ C {x,t}k . This has to be isomorphic to the given deformation defined by FT . Therefore, we have to find an automorphism Φ : C {x,t}n −→ C {x,t}n , and a matrix Λ ∈ C {x,t}k such that (10.10)

FS (x,φ(s)) = (FT ◦ Φ)(x,t) ◦ Λ

mod J ′ · C {x,t}.

This is a system of system of equations we want to solve for φ, Φ and Λ. We want to apply Grauert’s Approximation Theorem, and we therefore have to show that every solution of (10.10) of order e can be lifted to a solution of order e + 1. By assumption, (XS , 0) is formally semi-universal, hence is given by a sequence of maximal extensions by Theorem 10.3.16. Note that Remark 10.3.18 says that every solution of order e can be lifted to a solution of order e + 1. Thus we may indeed apply Grauert’s Approximation Theorem. Corollary 10.3.20. Let (X, 0) be a germ of a complex space, and let (XA , 0) be a (formal) deformation over A = C [[s]]/J. Then (XA , 0) is (formally) semi-universal if and only if (1) The Kodaira-Spencer map is an isomorphism,
∗ 2 (2) The obstruction map TX, −→ J/mJ is surjective. 0

A similar statement holds for convergent deformations.

Proof. From Theorem 10.3.19, it suffices to consider the formal case. Step 1. First suppose that the Kodaira-Spencer map is an isomorphism, and that the obstruction map is surjective. Consider the small extensions

10.3 Existence of a Semi-Universal Deformation

359

J/mJ

// C [[s]]/mJ

// C [[s]]/J

 (J + me+1 )/(mJ + me+2 )

 // C [[s]]/mJ + me+2

 // C [[s]]/J + me+1 .

We obtain,
using the second part of Theorem 10.3.8, for the corresponding obstruction ∗ 2 map TX, −→ J + me+1 /mJ + me+2 a commutative diagram 0 2 TX, 0


∗

// J/mJ QQQ QQQ QQQ QQQ Q((  J + me+1 /mJ + me+2 .


∗ 2 As by the second assumption, the obstruction map TX, −→ J/mJ is surjective, the 0 image of the obstruction map of the second extension is thus (J + me+2 )/(mJ + me+2 ). By Corollary 10.3.9 we see that for all e the extension C [[s]]/J + me+2 = Ae+1 −→ Ae = C [[s]]/J + me+1 is maximal with respect to the deformation (XAe , 0) −→ (Ae , 0). Thus we have a formal versal deformation by Theorem 10.3.12. Step 2. Now suppose (XA , 0) is formally semi-universal. Consider the formal deformation constructed in Theorem 10.3.16. By construction, for all e, the obstruction element associated to the extension 0 −→ (J + me+1 )/(mJ + me+2 ) −→ C [[s]]/mJ + me+2 −→ C [[s]]/J + me+1 −→ 0, maps surjectively onto Je+1 /mJe+1 = J + me+2 /mJ + me+2 . Now for some e ≫ 0 the minimal number of generators of Je+1 mod me+2 is equal to the minimal number of generators of J by Exercise 1.3.19. Thus the obstruction map is surjective. The KodairaSpencer map is an isomorphism by definition of the first order deformation. Now we come to the proof of the main theorem. Theorem 10.3.21 (Grauert). Let (X, 0) be a germ of a complex space. Suppose that 1 dimC (TX, ) < ∞. Then there exists an analytic semi-universal deformation of (X, 0). 0 Proof. The previous theorem says that it remains to give a convergent deformation which is formally semi-universal. Let (X, 0) be defined by f = (f1 , . . . ,fk ) ∈ C {x}k , and let r1 , . . . ,rt ∈ C {x}k generate the module of relations between the f1 , . . . ,fk . Take a formal semi-universal deformation of (X, 0) over A = C [[s]]/J, and let m = (s). Let e be a natural number, such that the minimal number of generators of J is equal to the minimal number of generators of J modulo me+1 . As C [[s]] is Noetherian local ring, such an e exists according to Exercise 1.3.19. Let g1 , . . . ,gp be the generators of J +me+1 /me+1 . We may assume that the gi are polynomials in s. In order to apply Grauert’s Approximation Theorem, we will have to interpret the flatness condition in terms of equations. Now flatness means that we can find lifts F,R1 , . . . ,Rt ∈ C {x,s} of f,r1 , . . . ,rt and vectors M1 , . . . ,Mt ∈ C {x,s}p with (10.11)

Ri · F − g · Mi = 0 for i = 1, . . . t.

360

10 Deformations of Singularities

We want to apply Grauert’s Approximation Theorem 8.2.2 to this set of equations. Our formal semi-universal deformation of (X, 0) over A = C [[s]]/J gives solutions modulo (q) (q) mq+1 of (10.11), F (q) , Ri , g (q) , and Mi for all q ≥ e, lifting those for q = e. Now ′ ′ ′ ′ (q) (q) is a solution modulo mq+1 of (10.11) lifting F (e) , suppose F (q) , Ri , g (q) and Mi (e) (e) (e) Ri , g , and Mi . In order to apply Grauert’s Approximation Theorem, we have to ′ ′ (q)′ (q)′ prove that the F (q) , Ri , g (q) and Mi can be lifted to solutions modulo mq+2 . The ′ ′ (q) (q) ideal Jq′ = (g1 , . . . ,gp ) has p generators, just as many as Je . From Lemma 10.3.10 it follows that the surjection C [[s]]/Jq′ −→ C [[s]]/Je is induced by a sequence of maximal small extensions. Therefore Jq′ can by 10.3.16 be extended to an ideal J ′ , so that there exists a formal semi-universal deformation over C [[s]]/J ′ . Hence from Remark 10.3.3 it follows that the formal algebras C [[s]]/J ′ and C [[s]]/J are isomorphic, and thus that C [[s]]/Jq′ and C [[s]]/Jq are isomorphic for all q. Thus for dimension reasons the minimal ideal I with ′ (1) mJq+1 ⊂ I ⊂ Jq′ and,

(2) there exists a lift of the deformation over (A′q , 0) to one over (A′q+1 , 0) with A′q+1 = C [[s]]/I, ′ is the ideal Jq+1 , as this is also the case for Jq+1 . Thus there does indeed exists a lift ′

(q+1)′

′

(q+1)′

′

(q)′

′

(q)′

of F (q) , Ri , g (q) and Mi , g (q+1) and Mi F (q+1) , Ri Approximation Theorem 8.2.2 applies, and the theorem is proved.

. Thus Grauert’s

Exercises 10.3.22. Prove Theorem 10.3.8 for the formal and the analytic case. (Hint: Use Exercise 1.3.19 to show that there exists an n ∈ N with OX,0 ⊗J ∼ = OX,0 ⊗(J +mn /mn ), and use the theorem already proved for the Artinian case.) 10.3.23. Let ϕ : B −→ A be a surjective map of Artinian C –algebras. Suppose that dimC (A) = dimC (B). Show that ϕ is an isomorphism. 10.3.24. Prove that a simple surjection of Artinian C –algebras is small. 2 10.3.25. Show that the definition of of TX, 0 is independent of the choice of generators of the ideal of (X, 0).

10.3.26. Prove Theorem 10.3.7. 1 (Hint: Use similar arguments as in the corresponding proof for TX, 0 .) 10.3.27. Prove Theorem 10.3.21 for the algebraic case: Let (X, 0) be an isolated singularity, 1 defined by algebraic power series. Suppose dimC (TX, 0 ) < ∞. Then there exists a semi-universal deformation of (X, 0) which is defined by algebraic power series. (Hint: Use Corollary 8.3.7.)

361

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