Monomial Algebras Villarreal

28

Chapter 1

1.4.21 If A is a unique factorization domain, show that A is normal.

1.4.22 Show an example of an integral extension of rings A C B, such that

A is a field and B is not an integral domain (cf. Lemma 1.4.5). 1.4.23 Let A C B be a ring extension and p a prime ideal of A. If S = A\p and Bp = S~~1B, prove that the map P i—> PB$ gives a bijection between

the set of prime ideals of P of B such that P n A = p and the set of prime ideals of Bp containing p5p. 1.4.24 Let B = A[x] be a polynomial ring over a ring A and let / be an ideal of A. Prove (A/J)[x] ~ B/IB, where the left hand side is a polynomial ring with coefficients in A/1. 1.4.25 If B = A[x] is a polynomial ring over a ring A and q is a p-primary ideal of A, then qB is a p5-primary ideal of B.

Hint Use Theorem 1.4.17.

1.4.26 Let B = A[x] be a polynomial ring over a ring A and let J be an ideal of A. If / = n[ =1 qj is a primary decomposition of 7, then /B = r\Ti=lc\iB is a primary decomposition of IB.

1.5

Koszul homology

Let a be an element of the ring R and Kit(a) be the complex defined as

R "* 0

*

for « = 0,1, otherwise,

with di:Ki(a) —> -K"o(a) being multiplication by a. Let / be an ideal of R generated by the sequence z = {x\ ,... ,xn}. The ordinary Koszul complex associated to z is defined as #*(z; R) = A"*(zi) ® • • • /£T*(z n ).

For an R- module M we shall write AT*(z; M) for K*(x; R)®M. The Koszul complex K*(x; R) is then the exterior algebra complex associated to E = Rn and the map 9:E— >R, defined as • +ZnXn.

That is, 6 defines a differential d = dO on the exterior algebra /\(E) of E given in degree r by r

d(ei / \ - - - f \ e r ) = '(-l)i~l9(ei)ei A • • • A &i A • • • A er.

Commutative Algebra______________________________ 29

A consequence of the definition of the differential of K+ (x_; R) is that if w and w1 are homogeneous elements of f \ ( E ) , of degrees p and q respectively, then d(w A w') = (-l)pw A d(w') + d(w) A w1.

This implies that the cycles Z(fiT*)'form a subalgebra of /\(E), and that the boundaries B(K+) form a two-sided ideal of Z(K^). As a consequence the homology of the Koszul complex, H*(x), inherits a skew commutative R- algebra structure.

One can also see that H*(x) is annihilated by I = (x). Indeed, if e £ E and uj £ Zr(K+~), we have from the last formula <9(e A w) = $(e)u>. The ordinary Koszul complex K*(x) = K*(x_;R) is simply the complex of free modules

#*(z) :

o -> A" -R" -» A71"1 #n - > • • • - » A1 -R" -> A° Rn -> °>

where f\kRn is the fct/i exterior power of f?"; thus f\kRn is a free .R-module of rank (™) with basis {ejj A • • • A ejjl < ij < • • • < i* < n}.

Proposition 1.5.1 // x_ = X i , . . . ,xn is a regular sequence in R, then the Koszul complex is acyclic, that is, the complex K*(x) is exact.

Proof. See [290, Theorem 2.3].

n

Sliding depth Let (R, m) be a Cohen-Macaulay local ring and let / be an ideal of R generated by xi, . . . ,xn, denote by H*(x) the homology of the ordinary Koszul complex built on the sequence x_ = {xi, . . . , xn}.

Definition 1.5.2 (i) (SD) I satisfies sliding depth if

depth Hi(x) > dim(R) -n + i, Vi > 0. (ii) (SCM) I is strongly Cohen-Macaulay if Hi(x) are C-M, Vi > 0. (Depths are computed with respect to maximal ideals. It is usual to set depth(O) equal to oo.)

Remark 1.5.3 (a) The (SD) condition localizes [156], (b) If / satisfies (SD) with respect to some generating set, then it will satisfy (SD) with respect to any other generating set of /. This follows from the isomorphisms:

..,xn,0}) ~ f l j ( z ) ® #i_i(z), and Hi({xi,...,xn,y}) ~ Hi({xi,...,xn,Q}), where y € (x).

Chapter 1

30

Linkage Let / and J be two ideals in a Cohen-Macaulay local ring R. The ideals J and J are said to be (algebraically) linked if there is an /^-sequence x = {xi,..., xn} in / n J such that I=((x):J]

and J=((x):I],

if in addition / and J are unmixed ideals of the same height n without common components and such that / fl J = (x), then / and J are said to be geometrically linked. When / and J are linked we shall write I ~ J. We say that J is in the linkage class of / if there are ideals l\,..., Im such that

J.

Ir,

The ideal J is said to be in the even linkage class of / if m is odd. Let _R be a Gorenstein ring and let / be a Cohen-Macaulay ideal of R. If J is linked to I, then Peskine and Szpiro [233] showed that J is CohenMacaulay. A dramatic result of Huneke [179] proves that (SCM) is preserved under even linkage, his method can be adapted to prove the following. Proposition 1.5.4 Let I and J be two ideals in a Gorenstein local ring R of dimension d, and let x_ = {x\,..., xn} be a generating set for I. Assume that J is evenly linked to I. If I satisfies the condition

(SDk)

depth Hi(x;R) > d-n + i, 0
then J satisfies the (SD^) condition as well.

Exercises 1.5.5 Let R = k [ x i : . . . , x e ] be a polynomial ring over a field k and let / = /2 (X) be the ideal of 2 x 2-minors of the symmetric matrix: v" _

Xi

X2

£3

X-2

£4

X5

Prove that / is linked to I2(X'), where X1 is the symmetric matrix Xi

3?9

—X**

X5

XQ

X1 = -£3

In particular if char(A;) = 2 the ideal / is self-linked. For other characteristics this is an open question.

Chapter 2

Affine and Graded Algebras A few topics connected with affine and graded algebras are studied in this chapter, e.g., Grobner bases, Hilbert Nullstellensatz, minimal resolutions and Betti numbers. We present the affine and graded versions of the Noether normalization lemma and some of their applications to affine and CohenMacaulay graded algebras. As before all base rings considered here are Noetherian and modules are finitely generated.

2.1

Noether normalizations

Affine algebras occur naturally in algebraic combinatorics and geometry. One of the key results to understand their behavior is the famous Noether normalization lemma, first we present its affine version.

Definition 2.1.1 Let k be a field and let S be a /c-algebra. We say that 5 is an affine k-algebra if S = k [ y i , . . . , yr] for some y\,..., yr € 5. Definition 2.1.2 Let a and /? be in N™. The lexicographical order in N" is obtained by declaring

/3>a if the last nonzero entry of /? — a is positive.

Notation The set of positive integers will be denoted by N+. If a,/3 G E n , here a • f3 will denote the usual inner product of a and /?.

31

32_______________________________________ Chapter 2

Lemma 2.1.3 Let ai > • • • > am be a sequence of m distinct points in N™ ordered lexicographically. Then there is w = (w\ ,..., wn) 6 N™ such that Wi = 1 and w • cti > w • on for i > 2. Proof. Let a; = (am, . . . ,ain) and fa = (an, . . . ,«*(„_!)). We proceed by induction on n > 2. There is k so that ccin = • • • = a^n and a.kn > Q-in for i > k. One may assume k < m; for otherwise fa > • • • > fan and one can use induction. Since fa > • • • > fa, by induction there is w' = (l,u>2, . . . , w n -i) such that w' • fa > w' • fa for 2 < i < k. On the other hand for every i > A; one

can choose Si € N+ so that w1 • fa + ain8i > w' • fa + ainSi.

To finish the proof set •u;n = max{(5j| k
and observe w • a\ > w • ai for i > 2.

D

Proposition 2.1.4 Lei jR = A;[x 1; . . . ,xn] be a polynomial ring over a field k and let f be a polynomial in R \ k. Then there is a change of variables Xi

= xf +yt

(i> 2)

such that /(zi.zf 2 +2/2, • . - , < " +Vn) = c0xl + dxl'1 + - - - + cr-ixi +cr,

where 0 7^ c0 € k, r > 0, and Cj £ A; [2/2, • • • ,2/n] /or « > 1. Proof. The polynomial / can be written as

where 0 ^ 6j € A; for all i. One may assume that a.\ > • • • > am are ordered

lexicographically. By Lemma 2.1.3 there is w € N" such that Xj = xWi +yi satisfies the required properties. Note r = w • a\. d

Lemma 2.1.5 If R = k [ x \ , . . . ,xn] is an affine k-algebra of dimension n, then R is a polynomial ring. Proof. One may assume R ~ B/I, where B is a polynomial ring in n variables with coefficients in the field k and / is an ideal of B. Let

/ C po C • • • C pn

be a chain of prime ideals of B of length n. If / ^ (0), then adding (0) to the chain yields a chain of length n + 1, which is impossible because

dim(.B) = n. Hence / = (0).

n

Affine and Graded Algebras_________________________ 33

Theorem 2.1.6 Let R = k[xi, . . . ,xn] be a polynomial ring over a field k and let I ^ R be an ideal of R. Then there are Zi, . . . ,zn in R such that

(a) k[x\ , . . . , xn] is integral over k\z\ , , . . ,zn], and (b) lnk[zi,...,zn] = zik[zi,...,zn] + ••• + zgk[zi,...,zn]. Proof. The proof is by induction on n. If n = 1 and 7 = ( f ( x i ) ) ^ 0, then

one sets z\ = f ( x \ ) . Assume n > 2 and 7 ^ 0 . One may assume that 7 contains a monic polynomial in x\. Otherwise take a nonzero polynomial g in 7 and apply Proposition 2.1.4 to get an isomorphism of /c-algebras

k[xi,...,xn] -A k[xi ,y2...,yn] induced by ip(x\) = x\ and ip(xi) = x™f + yi for i > 2, such that tp(g) is monic in x\. Let + • • • + Cr^iXi + CT

be a polynomial in 7 with Cj e k[x2, • • • , xn] and r > 0. Set

7i = Ir\k[x2,...,xn}. By induction there are z%, . . . ,zn such that: (i) k[x2, • • • , xn] is integral over k[z%, . . . , xn], and

(ii) 7i n k[z2, ...,zn] = (z-2, . . . , Zn). Set z\ = /. It is not hard to see that 7? is integral over k [ z i , . . . ,zn] and

k [ z i , . . . , z n ] n 7 = (zi,...,^a), as required

D

Corollary 2.1.7 If R = k [ x i , . . . ,xn] is a polynomial ring over a field k and I j^ R is an ideal of R, then dim(7?/7) =dim(7?)-ht(7). Proof. One may assume that there are Zi, . . . , zn in 7? and an integer g

such that the conditions (a) and (b) of Theorem 2.1.6 are satisfied. We will show that g is equal to the height of 7. By Lemma 2.1.5 the Zi's are algebraically independent. Hence Proposition 1.4.13 yields

ht(7) = ht(lnk[zi,...,zn]) =g. Note that there is an integral extension

k[zg+i,...,zn] ~k[zi,...,zn]/(zi,...,zg) Therefore n-g = dim(7?/7).

A k[xi,...,xn]/I. D

34_______________________________________Chapter

2

Corollary 2.1.8 (Noether Normalization Lemma) If R = k\x\ is a polynomial ring over a field k and I ^ R is an ideal, then there is an integral extension

where hi,... ,hd are in R and d = dim(_R/7). Proof. By Theorem 2.1.6 there is an integral extension

k[zg+i,...,zn] ~k[z1,...,zn]/(zi,...,zg)

A R/I.

To conclude the argument set hi = zg+i for i = 1 , . . . , d.

D

Corollary 2.1.9 If R = k[xi,... ,xn] is a polynomial ring over a field k, then R is a catenary ring. Proof. Note ht (q/p) = dim(E/p) - dim(/?/q) for any two prime ideals p C q in R. Hence by Corollary 2.1.7 we get ht (q/p) = ht (q) - ht (p). D Definition 2.1.10 Let k C L be a field extension. A subset of L which is algebraically independent and is maximal with respect to inclusions is called a transcendence basis of L over k.

Theorem 2.1.11 If k C L is afield extension, then any two transcendence basis have the same cardinality. Proof. See [191, Theorem 8.35].

n

Definition 2.1.12 Let k C L be a field extension. The transcendence degree of L over k, denoted trdeg ft (L), is the cardinality of any transcendence basis of L over k.

Corollary 2.1.13 Let k be a field and let A be a finitely generated kalgebra. If A is a domain with field of fractions L, then

dim(A) = trdeg fc (L).

Exercises 2.1.14 If A and B are affine algebras over a field k, then

dim(A <8>fc B) - dim(A) + dim(B).

Hint Use the normalization lemma.

Afflne and Graded Algebras_________________________35

2.2

Cohen-Macaulay graded algebras

In this section we will emphasize the relationship between graded CohenMacaulay rings and their homogeneous Noether normalizations. Then some useful characterizations of those rings will be given. Definition 2.2.1 Let k be a field.

A standard algebra or homogeneous

algebra is a finitely generated N-graded /c-algebra

4=0

such that j/j € 5i for all i and So = k. If we only require j/j homogeneous and deg(j/i) > 0 for all i, we say that 5 is a positively graded fc-algebra. The irrelevant maximal ideal m of 5 is defined by

t=l

Definition 2.2.2 Let A; be a field and S = k [ y \ , . . . ,yr] a positively graded ^-graded with j/j homogeneous of degree di. There is a graded epimorphism ip: R = k[xi,..., xr] —> S

given by where R is a polynomial ring graded by deg(£i) = di, the presentation of

5 is the fc-algebra R/ker(tf>). The graded ideal ker(
dim(S) = dim(S/L) + ht (L). Proof. One may assume S = R/I and L = J / I , where -R is a polynomial ring over k and J is an ideal of R containing /. Set g = ht (J/I) and r = ht (/). Let p be a prime ideal such that J C p and g = ht (p//). There is a saturated chain of prime ideals / C p o C p i C • • • Cpg = p .

Since po is a minimal prime ideal of / using Proposition 1.3.11 one obtains dim(.R//) = dim(J?/po), that is, r = ht(po). Hence there is a saturated

chain of prime ideals qo = (0) C qi C • • • C q r = po C pi C • • • C p5 = p.

36_______________________________________Chapter

2

As R is a catenary domain one obtains ht (p) = r + g and consequently ht (J) < r + g. It is not hard to verify that this inequality is equivalent to the inequality

dim(S) < ht (L) + dim(5/L). To conclude the proof recall that the inequality

dim(S) > ht (L) + dim(5/L) holds in general.

D

Corollary 2.2.4 Let R be a positively graded polynomial ring over a field k and I a graded ideal of R. If R/I is Cohen-Macaulay, then I is unmixed. Proof. It follows from Proposition 1.3.11 and Proposition 2.2.3.

D

Definition 2.2.5 Let k be a field and let S be a positively graded fc-algebra. A set of homogeneous elements 6_ = { O i , . . . , & d } is called a homogeneous system of parameters (h.s.o.p for short) if d = dim(S) and rad (0) = S+. Corollary 2.2.6 Let S be a positively graded algebra over a field k and hi,..., hci a homogeneous system of parameters for S. Then

dim S/(hi,..., hi) = dim(S) — i, forl
dim(5)
by the graded version of the dimension theorem (see Theorem 1.3.13). On the other hand if Q\,..., 9r is an homogeneous system of parameters for 5, then the sequence Si,...,9r, hi,..., hi generates an m-primary ideal of S, and applying the dimension theorem once again one has

d < r + i = dim(5) +i.

Thus dim(5) =d-i.

D

Proposition 2.2.7 Let S be a positively graded algebra over a field k and §_ = 9i,... ,9d an h.s.o.p for S. Then S is Cohen-Macaulay if and only if 9_ is a regular sequence.

Affine and Graded Algebras_________________________37

Proof. By Corollary 2.2.6 d i m S / ( 0 i , . . . , 0*) = d - i. If S is C-M, then

by Proposition 2.2.3 one has ht ( 6 1 , . . . ,$;) = i. Assume # i , . . . , 0 j _ i is a regular sequence. Next we show that Q{ is regular on A = S / ( # i , . . . , #j_i).

Otherwise if 0$ is a zero divisor of A, then #j belongs to some minimal prime p of ( $ 1 , . . . , 0j_i). Since ht (p) < i — 1 (see Theorem 1.3.14) we obtain ht ($1,... ,$,) < i — 1, which is impossible. Conversely if 9_ is a regular sequence, then depth(S) = d and S is C-M. D

Proposition 2.2.8 Let S be a positively graded algebra over afield k. If S is Cohen-Macaulay and 9_ = &i,..., 9q is a regular sequence of homogeneous elements in S+, then S/(0) is Cohen-Macaulay. Proof. It suffices to prove the case q = 1, which is a direct consequence of Lemma 1.3.10. D

Lemma 2.2.9 Let V ^ {0} be a vector space over an infinite field K. Then V is not a finite union of proper subspaces of V. Proof. We proceed by contradiction. Assume that there are proper subspaces Vi,..., Vm of V such that

1=1

where m is the least positive integer with this property. Let

vi € Vi \ (V2 U • • • U Vm) and v2 6 V2 \ (Vi U V3 U • • • U Vm). Pick m + 1 distinct nonzero scalars ko, • • • , km in K. Consider the vectors fa = V i

for i — 0, . . . ,m. By the pigeon-hole principle there are distinct vectors /3r,

/3S in Vj for some j. Since /3r - /3S £ Vj we get u2 € Vj. Thus j = 2 by the choice of v% . To finish the proof observe that /3r £ V% imply v\ 6 V^ , which contradicts the choice of vi . n Proposition 2.2.10 Let R = k[x\, . . . ,xn] be a polynomial ring over afield k with a positive grading and let I be a graded ideal of R. Then there are homogeneous polynomials hi, . . . , h<j in /?+ such that

dimR/(I, hi,. .. , h i ) = d -i,

for i = 1, . . . , d,

where dim(R/I) = d. If k is an infinite field and deg(a;j) = 1 for all i, then

hi, . . . ,hd can be chosen in RI .

38_______________________________________ Chapter 2

Proof. Assume d > 0. Let pi, . . . , pr be the set of minimal primes of I of height g = ht (/). We claim that there is a homogeneous polynomial hi not in U[ =1 pj. To show it we use induction on r. Since pi • • - p r _ i <£. pr and because pj is graded we can pick / G pi • • -p r -i l~) R^ and / 0 p r , di > 0. By induction there is g e fid2 and 5 £ U^pj, d% > 0. Assume Ri C U£ =1 pi for all i > 0, hence g £ p r . To complete the proof of the claim consider h = fd2 — gdl to derive a contradiction.

Note dim(R/I) > dim R/ (I, hi), by the choice of hi. Hence a repeated application of the claim rapidly yields a sequence hi , . . . , hs of homogeneous polynomials in R+ with s < d and such that ht (/, hi, . . . , hs) = dim(R). Therefore by Theorem 1.3.13 one concludes d = s, as required. If k is infinite and deg(rEi) = 1 for all i, then there is hi in RI and not in U[=1pi; for otherwise one has RI = U[ =1 (pj)i and since .Ri cannot be a finite union of proper subspaces (see Lemma 2.2.9) we derive R± = pi for

some i, which is impossible. Thus one may proceed as above to get the required sequence. D Theorem 2.2.11 (Noether normalization lemma) Letk be a field and let R be a polynomial ring with coefficients ink. If R is positively graded and I is a graded ideal of R, then there are homogeneous polynomials hi, . . . ,hd in R+, with d = dim(R/I), and a natural embedding

A = k [ h i , . . . , h d ] ^ R/I such that R/I is a finitely generated A-module. Moreover if k is infinite and deg(xj) = 1 for all i, then hi, . . . ,hd can be chosen in RI . Proof. Let R = k[x\ ,..., xn] and set Bi = {xa deg(x") = i}. According to Proposition 2.2.10 there are homogeneous polynomials hi, . . . , hd in m = R+ with rad (/, hi , . . . , hd) = tru Hence there is r such that

mr C (I,hi,...,hd). Note that Bi C (/, hi, . . . , hd) for i > s = r6, where 6 is the maximum of the degrees of xi, . . . ,xn. Set B = U*=1Sj. Using induction on i we now show that Bt C J = I + Bk[hi,...,hd], Vz > s. The containment is clear for i < s. Assume i > s and take xa in Bi. Let f i , • • • , fq be a homogeneous generating set for /. Since (/, hi , . . . , hd) is graded we can write 9

d

3=1

j=l

Afflne and Graded Algebras_________________________ 39

where bi,...,bg and c\, . . . , cd are homogeneous and deg(cj/ij) = i. Hence all the monomials in the support of GJ have degree less than i. Thus by induction one gets c, in J for all j and consequently xa e J. Observe that the image of B in R/I generates R/I as an A-module. Next we verify that the canonical map 9? form A to R/I is injective. Since R/I is integral over
to d.imtp(A). Using A/ker(
k [ h i , . . . , h d ] ^ 5, where hi,. . . ,hd are homogeneous polynomials in R+ and d = dim(S). Proposition 2.2.13 (Stanley) Let R = k[xi , . . . , xn] be a polynomial ring over a field k and let I be a monomial ideal with d = dim(R/I). Then

is a Noether normalization, where
Proof. It suffices to prove that S is integral over A. Let Xi = Xi + 1. Since d = dim(.R/rad(/)), one has CTJ £ rad(/) for i > d. Hence from the equality = Xn - (JiXn~l

one has

where h is nilpotent in S. Hence if hm € I, raising the last equality to the mth power yields that x^ is integral over A. D Proposition 2.2.14 Let R = k[x\,... ,xn] be a positively graded polynomial ring over a field k and let I ^ R be a graded ideal of R. If S = R/I is a Cohen-Macaulay ring and

A = k[hi,. ..,hd]<-*S is a homogeneous Noether normalization of S, then

S = Ax01 0 • • • 0 Ax13™ for any set of monomials B = {x^1,..., x@m } whose image in the Artinian ring S = R/(I, h\,..., hd) is a k-vector space basis of S.

40_______________________________________ Chapter 2

Proof. Set Mi = {xa e R\ deg(xa) = i}. First we show that the image of B generate 5 as an ,4-module. It suffices to prove M; C J = / + AB for all i > 0. Let xa G Mj. There are homogeneous polynomials fj,i, . . . , //^ in R and AI , . . . , \m in k such that

j=l

3 =1

where / € -RjD/ and deg(yUj/ij) = i if /Lij ^ 0. Since deg/itj < i, by induction one obtains that \ij € J for all j and consequently xa € J.

We claim that if c\x$l +• • • + cmx/3m belongs to (/, hi, . . . , hi-i) for some 2 < i < d and ci , . . . , cm in k[hi, . . . , /id], then a/ = 0 for all j. To prove the claim note that hi, . . . , hd is a regular sequence (see Proposition 2.2.7) and use descending induction on i starting with i = d.

Next we show 5 = Ax01 0 • • • Assume

m

OiZ^ 6 /,

for some QI, . . . , a m in A. If a^ / 0 for some £, write

where o,j are in fc[/i2> • • • > ^d]- One may assume a r o 7^ 0 for some r; otherwise we may factor out hi to some power and apply that hi is regular on S. Hence a-iox^1 + • • • + am0x^m is in (7, hi), and applying the claim with i — 2 we derive a r o = 0, which is a contradiction. Therefore a, = 0 for all j, as required. D Lemma 2.2.15 Lei A; 6e an infinite field and let

6e a standard k-algebra. If S is Cohen-Macaulay, then there exists a h.s.o.p 9_ = {61, . . . , 9d} such that 9_ is a regular sequence and 9i € Si for all i. Proof. Let Ass(S) — {pi, . . . ,p s }. Notice that Ass(S) = Min(S), because

5 is Cohen-Macaulay. We may assume d > 0, otherwise there is nothing to prove. If 5*1 C Z(S) = U|=1pi, then Si = Uf = 1 (pj)i. Since k is infinite, by Lemma 2.2.9, we obtain that s = I and pi = S+, that is, dim(S) = 0, which is a contradiction. Hence there is 9\ G Si which is regular on S.

Afflne and Graded Algebras_________________________ 4Jt By Proposition 2.2.3 dim S/ (hi) = d — 1. Applying the depth lemma (see Lemma 1.3.9) to the exact sequence

0 —> 5-l -^> 5 —> S6i —-> 0 yields depth S/ (6>i) = d- 1. Altogether S/(#i) is C-M and the result follows by induction. D

Lemma 2.2.16 Let R be an N-graded polynomial ring over a field k and I a graded ideal of R of height g. If I is a complete intersection, then there are homogeneous polynomials f i , • • • , fg such that I = (/i, . . . , fg). Proof. Let hi, . . . ,hg be a regular sequence generating the ideal /. Using the graded version of Corollary 1.1.30 one has

r = v(I) = dim* (//ml) < g, where m = /?+. On the other hand, since / is graded, there are homogeneous polynomials fi,...,fr generating / and such that {/, + m/}£=1 is a fc-basis for I /ml. By Theorem 1.3.14 one concludes r = g. D Proposition 2.2.17 Let R be a positively graded polynomial ring over a field k and let I be a graded ideal of R. If I is a complete intersection, then R/I is Cohen-Macaulay. Proof. By Lemma 2.2.16 the ideal / is generated by a sequence fi,. . . , f g

consisting of homogeneous polynomials, where g denotes the height of /. According to Proposition 2.2.10 there is a sequence hi, . . . , hn-g consisting of homogeneous polynomials such that r a d ( / i , . . . , / s , / i i , . . . , / i n _ s ) = m = R+. Therefore {/i, . . . , fg, hi, . . . , hn-g} is a homogeneous system of parameters for R and by Proposition 2.2.7 we derive that f i , . . . , fg is a regular sequence.

Finally Lemma 1.3.10 yields that R/I is Cohen-Macaulay.

D

Lemma 2.2.18 Let /,pi, . . . , p m be graded ideals of a polynomial ring R over a field k such that pi, . . . ,p m o-fe prime ideals. If f e U^pi for any

/ G / homogeneous, then I C p, for some i. Proof. By induction on m. If m = 2 and / <£ p, for i = 1, 2, then for each i pick a homogeneous polynomial /, e / \ p, of degree Uj. Since /"2 + f%1 is homogeneous we readily derive a contradiction. Hence / C pi for some i. Let G be the set of homogeneous elements in / and suppose G C U^Ljpj. One may assume pi <£ pj for i ^ j. If / <£_ pi for all i, then by induction there is /i in G \ U^1pi. On the other hand since /pi • • - p m _ i
42_______________________________________ Chapter 2 Proposition 2.2.19 Let R be a polynomial ring over a field k and let I be a graded ideal of height r, then there is a regular sequence j\ , . . . , fr in I of homogeneous polynomials. Proof. We will proceed by induction. Assume that f i , • • • , fs is a regular sequence of homogeneous polynomials in / , where s < r. Note that the ideal (/i, . . . , /«) is Cohen-Macaulay by Proposition 2.2.17, and hence it is unmixed by Corollary 2.2.4. If all the homogeneous polynomials in / belong

to Z(R/(fi, . . . ,/ s )), then by Lemma 2.2.18 the ideal / must be contained in an associated prime of (/i, . . . , fs) and consequently ht (I) < s, which is impossible. Thus there is a homogeneous polynomial /s+i in / such that /s+i is regular modulo (/i , . . . , /«). a Tensor product of affine algebras

Let A, B be two affine algebras over a field k and consider presentations A ~ Ri/Ii, B ~ R^/h, where RI = fe[x], R-2 = k[y] are polynomial rings in disjoint sets of variables and Ii is an ideal of Ri. Set R = fc[x,y] and / = /i + 1%. The map

R i—> Ri/h <8> induces a fc-algebra homomorphism:


(J(x)g(y))

= 7(x)

On the other hand there is a fe-bilinear map

^. R^h xR2/I2^ R/I, given by multiplication ij)(f,g)

= fg. By the universal property of the tensor product there is a map ip that makes commutative the following diagram

u R/I where is the canonical map and ^ = V"^- As inverse of ip. Thus we have proved:

a

consequence tjj is the

Proposition 2.2.20 If A and B are two affine algebras over a field k with presentations A ~ k\x\/I\ and B ~ fc[y]//2,

Affine and Graded Algebras_________________________ 43

Theorem 2.2.21 If A and B are two standard algebras over a field k, then depth(>l <S>fc B) = depth(A) + depth(B).

Proof. Pick a regular sequence g = g\,...,gr (resp. h — hi,.,.,hs) on A = fc[x]//i (resp. B = k[y\/ 1%) with g C (x) (resp. h C (y)) and such that

g (resp. h) consist of forms, where r is the depth of A and s is the depth of ~B. As B is a faithfully flat fc-algebra, applying the functor (•) <8>fc B to the injective map 0 —> /c gives a natural commutative diagram

such that the map in the first row is injective. Since the vertical arrows are natural ismorphisms by Proposition 2.2.20, one concludes that the map in the second row is also injective. As a consequence g is a regular sequence on fc[x,y]/(/i,/2). Similarly if we tensor the injective map

0 —>• fc[y]/(I 2) /ii, . . . , / i i - i ) A fc[y]/(/ 2) fci, - . . , /K-i)

with the faithfully flat fc-algebra fc[x]/(/i,p) it follows rapidly that /i is a regular sequence on fc[x,y]/(/i,/2,g).

Altogether g,his a regular sequence on A;[x,y]/(/i,/ 2 ). To finish the proof note that (x, y) is an associated prime ideal of fc[x, y]/(/i ,h,g, h) by Exercise 2.2.27, thus r + s is the length of a maximal regular sequence in (x,y) and by Proposition 1.3.7 one obtains depth(A ®k B) = r + s. For an

alternative proof see [115].

D

Corollary 2.2.22 If A and B are two standard algebras over a field k, then A (5>k B is Cohen-Macaulay if and only if A and B are Cohen-Macaulay. Proof. As the dimension is always greater or equal than the depth by Lemma 1.3.6, the result follows at once using Theorem 2.2.21 together with Exercise 2.1.14. D

A similar statement holds if one replaces Cohen-Macaulay by Gorenstein; indeed the type of the tensor products of two Cohen-Macaulay standard

algebras is the product of their types [115].

44

Chapter 2

Exercises 2.2.23 Let V be a vector space over an infinite field k and W,Vi,...,Vm

vector subspaces of V. If W ^ Vi for all i, prove W <£. U^Vi. #mi If VF C U^l/i, note W = \J^(W n Vi) and use Lemma 2.2.9. 2.2.24 Let /, /i, . . . , Im be ideals of a ring R and let k be an infinite field. If / C U^Lj/i and k is a subring of R, then I C I, for some i.

Note that J, I\ , . . . , Im are vector spaces over k. 2.2.25 Let R be a polynomial ring over an infinite field k and / a graded ideal of height r. If / is minimally generated by forms of degree p > 1, prove that there are forms fi,...,fm of degree p in / such that fi,...,fr is a regular sequence and / is minimally generated by f i , . . . , fm.

Hint Find a regular sequence /i , . . . , /r in 7P = Rp n I and extend this sequence to a fc-basis of Ip. 2.2.26 Let R = fc[x] be a polynomial ring over a field k and / C (x) an ideal. Note that k is an .R-module with multiplication f ( x ) • a = /(0)a, for f ( x ) G R and a & k. When is the canonical map if. k —> R/I,

a i—> a + I

a homomorphism of /Z-modules?. 2.2.27 Let A = fc[x]//i and B = k[y]/I2 be affine algebras over a field k. If (x) (resp. (y)) is an associated prime of fc[x]//i (resp. fc[y]// 2 ), then (x,y) is as associated prime of fc[x, y]/(/i, I z ) .

Hint there are embeddings A;[x]/(x) ® fc k[y]/(y)

M- fc[x]/(x) ®k B and fc[x]/(x) ® fc B <-> ^ ®fc S.

2.2.28 Let fc[x] be a polynomial ring and / an ideal (resp. graded). If k C K is a field extension, then there is a natural (resp. graded) isomorphism of K -algebras: 2.2.29 Let A be an affine algebra over a field k and let k C K be a field extension. Prove:

(a) dim(yl) = dim(A ® fc K). (b) depth(A) = depth(A 0^ Jf) if A is a standard algebra. 2.2.30 Let K be a field and let A be an affine K -algebra. Prove that A is Artinian if and only if dim^(^4) < oo.

Affine and Graded Algebras_________________________45

2.3

Hilbert Nullstellensatz

Let k be a field and let R = k[xi,..., xn] be a polynomial ring, we define the affine space of dimension n over k, denoted by A£, to be the cartesian product kn = k x • • • x k. Given an ideal I of R define the zero set or variety of I as

V ( I ) = {aekn\f(a)=Q,VfeI},

by the Hilbert basis theorem V(I) is the zero locus of a finite collection of polynomials. Conversely for any X C kn define I ( X ) , the ideal of X, as the

set of polynomials that vanish on X. An affine variety is the zero set of an ideal. The dimension of a variety X is the Krull dimension of its coordinate ring R/I(X).

Proposition 2.3.1 If R = fc[x] is a polynomial ring over a field k, then

(a) V(I H J) = V(I) U V(J) = V ( I J ) , for all ideals I and J of R. (b) nV(/ a ) = V(U/ a ), where {Ia} is any family of ideals of R.

(c) V(l) — 0 and V(0) = A£; where n is the number of variables of R. Proof. It is left as an exercise.

D

By the previous result the sets in the family

T = {A%\V(I)\ICR} are the open sets of a topology on A£, called the Zariski topology. Definition 2.3.2 An affine variety X C kn is reducible if there are affine

varieties X\ ^ X and Xi ^ X such that X = X\ U X%, otherwise X is irreducible. Theorem 2.3.3 Let k be a field and X an affine variety of A£, then X is irreducible if and only if I ( X ) is a prime ideal. Proof. See [120, Chapter 1].

D

Proposition 2.3.4 If X C A£ is an affine variety over a field k, then there are unique irreducible affine varieties Xi,... ,Xr in A£ such that T

X =\JXi and Xi
46_______________________________________ Chapter 2

Proof. See [141, Proposition 1.5].

D

The irreducible afBne varieties Xi, . . . , Xr of Proposition 2.3.4 are called the irreducible components of X. Theorem 2.3.5 Let R = k[xi, . . . ,xn] be a polynomial ring over a field k and let m be a maximal ideal of R. If k is algebraically closed, then there are ai , . . . , an G k such that

m=(xi-ai,...,xn -an). Proof. According to Corollary 2.1.8 there is an integral extension

where d = dim(E/m) = 0. Hence the canonical map tp: k -> R/m is an isomorphism because k is algebraically closed. To complete the proof choose dj G k so that
Proof. =£•) Let / be an element in \/7. If (/, 1 - tf) ^ R[t], take a prime ideal p containing (I, 1 — tf). Since / must be in p, because /" is in / for some n > 0, one concludes 1 G p which is impossible. «=) If 1 = ai/i + • • • + aqfq + a,+i(l - t f ) , where ft £ I, f £ R and

fli G /?[£]. Set z = 1/i and note 1 — tf = (z — /)/t, multiplying the first equation by zm, with m large enough, one derives an equality

zm = 61/1 + --- + bqfq + bq+i(z - /), where 6j G fi[^]. As /?[z] is a polynomial ring over R, making z = f gives fm G / and / G v7/n Theorem 2.3.7 (Hilbert Nullstellensatz) Let R be a polynomial ring over an algebraically closed field k and let I be an ideal of R, then

i(v(i)) = Vi. Proof. One clearly has / C I ( V ( I } } , hence v7/ C I ( V ( I } ) because I ( V ( I ) ) is a radical ideal. For the other containment take / G I ( V ( I ) ) and consider the ideal J = (/, 1 — ft) of the ring R[t], where R = k [ x i , . . . ,xn] and t is a new variable. Next we show the equality J = R[t]. If J ^ R[t] by

Theorem 2.3.5 one has (-?", 1 -/*) C (xi -ai,...,xn -an,t-an+i), a{ G k. Hence /? G l / (/), where /3 - (ai,...,an). Using that / G I(V(I) gives /(/?) = 0, but this is impossible because 1 — /(/?)a n +i = 0. Hence J = R[t]. Thus / G v7 by Proposition 2.3.6. D

Affine and Graded Algebras_________________________47

Corollary 2.3.8 Let X = V ( / i , . . . , / ? ) be an affine variety, defined by polynomials / i , - - - , / ? in n variables over an algebraically closed field k. Then q > n —

Proof. By the Nullstellensatz I ( X ) = rad ( / i , . . . , /9). Let p be a minimal prime of ( / i , . . . , /?) of height n — dim(X). Applying Theorem 1.3.14 we get n — dim(X) < q. O Proposition 2.3.9 Let R = k[xi,..., xn] be a polynomial ring over a field

k and let m be a maximal ideal of R. If k is uncountable, then R/m is a finite extension of k. Proof. Assume Xj is not algebraic over fc, where, as usual, f denotes r + m for any r in R. For simplicity we set i = 1. As R/m has a countable basis consisting of monomials, one has that the (uncountable) set defined by

A=

- c

is linearly dependent. Thus 3 c, c i , . . . , c g in k distinct and A i , . . . , A 9 in k \ {0} such that:

Consider the polynomial ? 9 f(x) = l[(x - Ci) - (x - c) i=l

j— 1 i^j

Since /(xi) = 0 and xi is not algebraic over k we conclude that / is the zero polynomial, which is impossible because /(c) = nl=i( c ~ ci) ^ 0. n

Exercises 2.3.10 Let / and J be two ideals in a polynomial ring R over a field k. If rad (/) = rad (J), prove that V(I) = V(J). 2.3.11 If /c is a field and X C A£, then X C l / (/(^)) with equality if X is an affine variety. 2.3.12 Let J be an ideal of a polynomial ring R over a field k. Prove that J C /(V(J)) with equality if J is the ideal of an affine variety. 2.3.13 Let k be an algebraically closed field. Prove that there is a one to one correspondence between affine varieties (resp. irreducible varieties) in kn and radical ideals (resp. prime ideals) in the polynomial ring k[xi, . . . ,xn].

48_______________________________________ Chapter 2

2.4

Grobner bases

In this section we review some basic facts and definitions on Grobner bases. The reader may consult [1, 74, 243] for a detailed discussion of Grobner bases and for the missing proofs of this section. Let k be a field and let R = k[xi , . . . , xn] be a polynomial ring. We set xa = z?1 • • • < " for a= ( a i , . . . , an) 6 N". The monomials in

are called the terms. Definition 2.4.1 A total order > of Pn is called a monomial order or term order if (a) p> I for a l l p e Pn, and (b) for all p, q,r 6 Pn, q > p implies qr > pr.

Two examples of monomial orders in Pn are the lexicographical order or lex order defined as xb > xa iff the last nonzero entry of b — a is positive, and the reverse lexicographical order or revlex order given by xb > xa iff the last nonzero entry of b — a is negative. In the sequel we assume that a monomial order for Pn has been fixed. Let / be a nonzero polynomial in R. One can write

f =

with aj e k* = k \ {0}, Mi 6 P" and MI > • • • > Mr. The fcadz'm? term MI of / is denoted by lt(/). The leading coefficient a\ of / and its leading monomial a\M\ are denoted by lc(/) and lm(/) respectively.

Definition 2.4.2 Let / be an ideal of R. The initial ideal of /, denoted by in(/), is given by in(/) = ({lt(/)| / € / } ) . Lemma 2.4.3 (Dickson) // {Mi}f^l is a sequence in Pn, then there is an integer k so that Mj is a multiple of some term in the set {Mi, . . . , M^} for every i > k. Proof. We consider the ideal / C fc[xi, . . . ,xn] generated by {M,}?^. By the Hilbert basis theorem / is finitely generated. It is readily seen that / can be generated by a finite set of terms MI, . . . , M*. Hence for each i > k, there is 1 < j < k such that Mj is a multiple of Mj . D

Afflne and Graded Algebras_________________________ 49

Definition 2.4.4 Let F = { f i , . . . , }q] C R \ {0} be a set of polynomials in R. One says that / reduces to g modulo F, denoted / —>p g , if

g = f-(au/lc(fi))fi for some /, e F, u £ Pn, a e k* such that a • u • lt(/») occurs in / with coefficient a.

Proposition 2.4.5 The reduction relation "—ip " is Noetherian, that is, any sequence of reductions

9i —>F 92 —>F • • • —>F 9i —>F • • • is stationary.

Proof. Notice that at the ith step of the reduction some term of 5, is replaced by terms of lower degree. Therefore if the sequence above is not

stationary, then there is a never ending decreasing sequence of terms in Pn, but this is impossible according to Dickson's lemma.

D

Proposition 2.4.6 (Division algorithm) ///, f i , • • • , fq are polynomials in R, then f can be written as

f = a i f i +••• + aqfq + r, where a,i,r £ R and either r = 0 or r ^ 0 and no term of r is divisible by one of lt(/i), . . . ,lt(/ g ). Furthermore if aifi ^ 0, then lt(/) > lt(o,j/j).

Definition 2.4.7 The polynomial r in the division algorithm is called a remainder of / with respect to F = {/j, . . . , f q } . Definition 2.4.8 Let / ^ (0) be an ideal of R and let F = { f i , . . . , fr} be a subset of /. The set F is called a Grobner basis of 7 if

Definition 2.4.9 A Grobner basis F = {/i, . . . , fr] of an ideal / is called a reduced Grobner basis for / if: (i) lc(/i) = 1 Vz, and

(ii) none of the terms occurring in /, belongs to in(F \ {fi}) Vi. Note that by the Hilbert basis theorem a Grobner basis of / always exist. A nice feature of a reduced Grobner basis is its uniqueness. Definition 2.4.10 Let f , g £ R. The S-polynomial of / and g is given by

where [f,g] = \cm(f,g).

50_______________________________________Chapter

2

Given a set of generators of a polynomial ideal one can determine a Grobner basis using the next fundamental procedure: Theorem 2.4.11 (Buchberger) If F = { / i , . . . , fq} is a set of generators of an ideal / of R, then one can construct a Grobner basis for I using the following algorithm:

Input: F Output: a Grobner basis G for / Initialization: G := F, B := {{/;, fj}\ /; ^ /,- 6 G}

while B ^ 0 do pick any {/, g} e B B:=B\{{f,g}} r := remainder of S ( f , g ) with respect to G if r ^ 0 then B:=Bu{{r,h}\h£G} G:=GU{r} Proof. See [1, 53, 243].

D

Proposition 2.4.12 Let I be an ideal of R and let F = {/i,..., fq} be a generating set for I. If

B = {u u e Pn and u $ (lt(/i),... ,lt(/ n ))},

then B is a basis for the k-vector space R/I. Proof. First we show that B is a generating set for R/I. For that take / e R/I. Because "—^" is Noetherian we can write

1=1

1=1

where Ai 6 k* and such that every m is a term which is not a multiple of any of the terms lt(/j). Accordingly u, is in B for all i and / is a linear combination of the ilj's. To prove that B is linearly independent assume h = ]Ci=i ^« u z ^ ^> where u, e B and Aj € ft. We must show /i = 0. If /i ^ 0, then we can label the itj's so that Ui > • • • > us and AI ^ 0. Hence lt(/i) = \iu\ £ in(7), but this is a clear contradiction because in(/) = (lt(/i),... ,lt(/ ? )). Therefore h = 0, as required. D Corollary 2.4.13 (Macaulay) /// is a graded ideal of R, then R/I and R/ in(/) have the same Hilbert function.

Afflne and Graded Algebras

Lemma 2.4.14 Let f and g be two polynomials in R and let F = {/, g}. If lm(/) and lm(p) are relatively prime, then S ( f , g ) ->F 0. Proof. Let / = M! + • • • + Mr and g = NI + • • • + Ns, where M; and Ni are monomials such that MI > • • • > Mr, NI > • • • > Ns. Set

The monomials

{JViM2, . . . , JViMr.M^, . . . .MiW,}

are distinct because MI and JVi are relatively prime, hence S(f,g) Since JViM r _i_i > MkNm for A; > r - i and m > 2, from the equality

-

MrN2

-

••• -

MrNs

we obtain S(f,g) -^F h0 ->^ fti ->^ • • • ->p /i r -2- To complete the proof notice that MiA^+i > M/t-/Vm, for m > i + 1. Hence using the identity

-

MrNi+1

-

••• -

-

M2Ni+l

-

... -

MrNs M2NS

we obtain hr-2 ^F hr-2 + N2f ->F • • • ->F ft r _ 2 + A^/ + • • • + NJ = 0,

that is, 5(/,0) -»F 0.

D

The next criterion of Buchberger will be used at several places to prove that a given set of polynomials form a Grobner basis. Observe that this criterion is important in Buchberger algorithm. Theorem 2.4.15 ([53]) Let I be an ideal of R and let F = {/i, ...,/,} be

a set of generators of I, then F is a Grobner basis for I if and only if S(fi,fj)

-^F0 for all i^j.

Definition 2.4.16 Let / be an ideal of R generated by F = {/i, . . . , f r }

and consider the homomorphism of -R-modules ip:Rr —> I ,

Cj i—> fi,

where &i is the ith unit vector. The kernel of y, denoted by Z ( F ) , is called the syzygy module of / with respect to F.

52_______________________________________ Chapter 2

Theorem 2.4.17 Let G = [g\, . . . ,gr} be a Grobner basis and write

where a^fe G R andlt(S(gi,gj}) ]

> \t(aijk9k) f°r alli,j,k. Then the set

[lt(gi),lt( gj -)] ~ —— —-— —— •

generates the syzygy module of I = ( g i , . . . , gr) with respect to G. Proof. See [1, Theorem 3.4.1] and [75, Theorem 3.2].

D

Elimination of variables Let k[z\, . . . , zn] be a polynomial ring over a field k. A useful monomial order is the elimination order with respect to the variables z\ , , . . , Zi . It is defined as

zb > za if and only if the total degree of zb in the variables z\ , . . . , zi is greater than that of za, or both degrees are equal, and the last nonzero component of

b — a is negative. Theorem 2.4.18 Let S = k[xi, . . . , xn, ti, . . . , t q ] be a polynomial ring over a field k with a term order such that terms in the Xi 's are greater than terms

in the ti 's. If I is an ideal of S with a Grobner basis G, then Gr\k[ti, . . . , tq] is a Grobner basis of I n k[ti , . . . , tq] . Proof. Set B = k[t1:. . . , t q ] and / c = / n B. Assume M e in(/ c ). There is / e Jc with lt(/) = M. Hence M = Alt( t® for all a and /3 we obtain g e G n B, that is, M 6 (in(G n B)). Thus

as required.

D

Next we illustrate the previous result using the elimination order on S with respect to x\, . . . , xn. Thus this order is given by

xatc > xbtd if and only if deg(x a ) > deg(a; i) ), or both degrees are equal and the last nonzero entry of (a, c) — (b, d) is negative.

Afflne and Graded Algebras_________________________ 53

Example 2.4.19 Let us assign deg(tj) = 2 and deg(rrj) = 1 for all i,j. Using Macaulay [15] one rapidly obtains that the reduced Grobner basis of the ideal I = (ti - XiX2,t2

- XiX3,t3 - XiX4,U

- X2X3,t5

- X2X4,t6

- X3X4)

is equal to:

t2t5-tite,

xlt4

X3t3 - X4t2,

X3X4~te,

X33ti-t2t4,

X2t2 — X3ti,

X2X4—tt,,

X2X3

X2t6 -

—

XiX3-t2,

XiX2-t1:

Hence /n k[ti, . . . , t e ] = (t3i4 -tit6,t2t5 -tit6) by Theorem 2.4.18. Proposition 2.4.20 /// and J are two ideals of a polynomial ring R over

a field k, then Ir\J=(t-I,(l-t)-J)r\R, where t is a new variable. Proof. It is left as an exercise.

D

Definition 2.4.21 Let / and J be two ideals of a ring R. One defines the saturation of / with respect to J as:

(I:J°°) = \J(I:RJi). i>i Proposition 2.4.22 /// is an ideal of a ring R and f G R, then

where t is a new variable. Proof. Let g e (1,1- tf) n R. One can write g - aifi + ••• + a q f q + aq+i(l -tf), where fa £ / and a, G R[t\. Making t = I// in the last equation and multiplying by fm, with m large enough, one derives an equality

gfm = bifi + --- + bqfq, where 6, e R. Hence gfm 6 / and g £ (/: /°°). Conversely let g e (/:/°°), hence there is m > I such that gfm 6 /. Since one can write

g=(l- tmfm)g + tmfmg and 1 - tmfm = (1 - tf)b, for some b £ R[t], one derives g e (/, 1 - tf) n R.

D

54_______________________________________ Chapter 2

Lemma 2.4.23 ([98]) Let R be a ring and let x\ , . . . , xn 6 R. If I is an ideal of R, then the radical of I satisfies

Proof. The left hand side is clearly a subset of the right hand side. To prove the other containment take a prime p containing /, it suffices to prove that p contains one of the ideals on the right hand side. If i/(/: (x\ • • -Xn)00) <£ p, there is / € R \ p such that f ( x i • • • xn)s £ / C p for some integer s > 1. Hence Xi 6 p for some i and p contains \ J ( I , X i ) , as required. d Proposition 2.4.24 Let p be a prime ideal of a ring R and let x\,... ,xn be a sequence in R \ p. // I C p is an ideal, then \I = p if and only if (a) p = A/(/: (xi • • • X n ) 0 0 ) , and (b) \/(I,Xi) = \/(p,£»), for all i. Proof. If J is any ideal of R and yi,... ,yn € R, then by Lemma 2.4.23 the radical of J satisfies: rad(J) = rad( J: (y\ • • • yn)°°) H rad(J, j/i) D • • • fl rad( J, yn).

Hence the result follows by applying this formula to / and p, together with the fact that x\ • • • xn is regular on -R/p. d The homogenization of an ideal

Let R = k[xi,..., xn] and S = R[w]

be two polynomial rings over a field k. For / £ R of degree e define

that is, fh is the homogenization of the polynomial / with respect to w. The homogenization of an ideal J C -R is the ideal Ih of R[w] given by

Proposition 2.4.25 /// is an ideal of R, then vl^ = \fl Proof. For the inclusion \fl

for / e R and m > 0.

.

C vl^, we need only observe the equality

Afflne and Graded Algebras_________________________ 55

Conversely let g £ v/^. As v/'* is a graded ideal one may assume that g is homogeneous. Write g = wsg^, where g\ £ R. Since gm is in Ih, it follows that gi £ \/7, as required. D

Let > be the elimination order on the monomials of R[w] with respect to xi,...,xn,w, this order extends the elimination order with respect to x\ , . . . , xn on the monomials of R. Proposition 2.4.26 // / is an ideal of R generated by F = {/i, . . . , f r ] , then { f i , ...,//*} is a Grobner basis if and only if F is a Grobner basis.

Proof. =») Assume {/f, . . . , ft} is a Grobner basis. Set J = (/f, . . . , /rh). It suffices to prove in(7) C (lt(/i), . . . , lt(/ r )). Let MI be a monomial in the

initial ideal of /. There is / G / C R of degree e such that lt(/) = MI. We write f = 9ifi + "-+9rfr and consider fh _ ^ (Xi xn\ fxi xn — r — z/—', 9i \\ w> • • • i w )I Ji \ \wi • • • i w u>e t=i multiplying both sides by w1 for I ^> 0, one has nits

fh _

W J

L

j/i i

i i,

//i

— All/! + • ' ' + ft r / r

for some s. Thus u//'1 e J. Since lt(/ /l ) = lt(/) we havelt(u//'') = wsMl. As {//"•, . . . , //*} is a Grobner basis one obtains

Using lt(/i) = lt(/f ) yields MI £ (IttfO, . . . ,lt(/ r )). _

4=) Assume {/L, . . . , /r} is a Grobner basis. Set I =_(/1/l, ...,/,?). We

now show that in (7) C (lt(/1/l), . . . ,lt(/ r ft )). Let M e in(7), there is / g 7 such that M = lt(/). We may assume that / is a homogeneous polynomial.

We can write

/ = wp(Mi + M2u/2 + • • • + Mswe'),

where MI , . . . , Ms are monomials in R such that MI

Using that MiWei all have the same degree we obtain 0 < e 2 < • • • <es. It follows that /' = f ( x i , . . . , xn, 1) belongs to / and lt(/') = MI. Therefore

MI belongs to in(7) - (lt(/i), . . . ,lt(/ r )). Since lt(/j) = lt(//1) we obtain that M belongs to the ideal (\t(ff), . . . , lt(/rh)). D

56_______________________________________Chapter

2

Corollary 2.4.27 Let / = ( / i , . . . , fm) be an ideal of R and let Ih = ( { f h f € /})

be the homogenization of I. If / f , . . . , fa is a Grobner basis, then rh

1

-

i fh, fh \ l/l , • • • ,Jm)-

Proof. It is enough to show {fh\ f e /} C ( f t , . . . , f a ) . Since fi,...,fm is a Grobner basis any / G / can be written as i- • • • +gmfm,

where lt(/) > lt( deg(0i/i) = deg(^) + deg(/;), hence from: ,,/j ^_____

m ————

\

fl

:

\

______

_______

1

/.

I

____I_

—T e — / , 91 I —— j • • • i —— I / z \ —— i • • • j

ui

-^

\ui

KJ /

Vu;

J

I

u; /

we get (.u n i •„ f h r (fh

fh\

riu

mat is, j t u i j - - - ) / m ^ -

Projective closure Let k be a field. We define the projective space of dimension n over fc, denoted by 1P£, to be the quotient space

(fc"+1 \ {O})/ ~ where two points a, /? in kn+l \ {0} are equivalent if a = c/? for some c £ k. It is usual to denote the equivalence class of a by [a]. Given a homogeneous ideal / of 5 = k[xi,...,xn, w} define the zero set of / as

V(I) = {[a] £ P£ /(a) = 0, V homogeneous polynomials / e /} , conversely for any AT C P£ define I ( X ) , the z'dea/ of JT, as the ideal generated by the homogeneous polynomials in 5 that vanish on X, A projective variety is the zero set of a homogeneous ideal. It is not difficult to see that the members of the family

are the open sets of a topology on P£, called the Zariski topology.

Afflne and Graded Algebras_________________________ 57

Definition 2.4.28 Let Y C A™. The projective closure of Y is denned as Y := (Y), where p is the map

A —> P ,

a —> a , l ,

and (Y) is the closure of ip(Y) in the Zariski topology of P£.

Proposition 2.4.29 If Y is a subset of the affine space A£ over a field k, then the projective closure YofY is given by

Proof. It is left as an exercise, see Proposition 7.1.14 for the affine case. D Proposition 2.4.30 Let Y C AjJ and let Y C P£ 6e iis projective closure. U /i i • • • i /r *5 fl Grobner basis of I(Y), then

I ( Y ) = (f ?,..., frh). Proof. Note that 7(F) = /(y) h and use Corollary 2.4.27.

D

Corollary 2.4.31 Let Y C A£ and /ei F C P£ be its projective closure. Then the height of I(Y) in R is equal to the height of I(Y) in S.

Exercises 2.4.32 Let / be an ideal of a ring R and / a nonzero element of R, then ( / ) ( / : / ) = i n (/).

2.4.33 Let R = k[x] be a polynomial ring, where A; is a field. Recall that / is a binomial in R if / = xa — x® for some a, j3 in N n . Use Buchberger's algorithm to prove that an ideal of R generated by a finite set of binomials has a Grobner basis consisting of binomials. 2.4.34 Let / be an ideal of a ring R and / € R. Prove that the following statements are equivalent: (a) / is regular on R/I. (b) (/:«/) = !. (c) / = ( / , ! — tf) n R, where i is a new variable. 2.4.35 If / is a prime ideal of a polynomial ring R, prove that Ih is also a prime ideal.

58________________________________________ Chapter 2

2.4.36 Let N™ be the set of n-tuples of non-negative integers endowed with the partial order given by

if and only if QJ > bi for alH. If A C N", prove that A has only a finite number of minimal elements.

Hint Use Dickson's lemma.

2.4.37 Let Y = {(t3,t2,t)\ t 6 k} C A| and let Y be its projective closure. If k is an infinite field prove that =

2.5

x -

Minimal resolutions

The aim of this section is to study homogeneous resolutions of positively graded modules over polynomial rings. In the sequel we shall be interested in the numerical data of those resolutions and in particular in the Betti

numbers of those modules. We begin with a result which is a consequence of the graded Nakayama's lemma (see Lemma 1.2.6). Proposition 2.5.1 Let R = fc[x] be a polynomial ring over a field k and let M be an N-graded R-module. If F = { f i , . . . , fq} is a set of homogeneous elements of M and m = (x), then F is a minimal set of generators for M if and only if the image of F in M/mM is a k-basis of M/mM . Corollary 2.5.2 Let R = k[x] be a polynomial ring over a field k and let

m = (x). If M is an N-graded R-module, then v(M] = dim,, (M/mM),

where v(M) is the minimum number of generators of M. Lemma 2.5.3 Let R = k[x] be a polynomial ring over a field k and let M be an 'N-graded module over R with a presentation 0 —> K = ker

0,

et H^> /».

// m = (x) and F = {/i, . . . , fq} is a set of homogeneous elements, then F is a minimal set of generators for M if and only if K C mR9 .

Affine and Graded Algebras__________________________ 59

Proof. •£=) Assume K C mRq. By Proposition 2.5.1 we need only show that the image of F in M/mM is linearly independent over k. If

i

Y^ Aj/i e mM, for some \i & k, then

( A 1 , . . . , A ( , ) - ( a l j . . . , a g ) € K C mR", for some ai , . . . , aq in m. Hence (Ai , . . . , A a ) is in mf?9 and \j = 0 for all j.

=>) If .ff £ ml?", pick z = (zi, . . . , z g ) € # such that z £ mRq . Then Zj is not in m for some i. Since the /j's are homogeneous this readily implies that fi is a linear combination of elements in _F \ {/,}, a contradiction. D Definition 2.5.4 Let R = G)^-0Ri be a positively graded ring and a € N. The graded 7?-module obtained by a shift in the graduation of R is given by

i=0

where the ith graded component of R(—a) is R(—a)i = R-a+i-

Proposition 2.5.5 Let R = ®^0-^i be a positively graded polynomial ring

over a field k with maximal irrelevant ideal m = R+ and M an N-graded R-module. Then there is an exact sequence of graded free modules

F.

• • • ->

R(-dkl) ±i> . . .

where (p^ is a degree preserving map and \m((pk) C mRbk~1 for k > 1. Proof. Let /i, • • • ,/ g be a set of homogeneous elements that minimally generate M. Set dj = deg(/j). There is an exact sequence

0 — > Zi = ker(^o) A

1Z(-di) ^> M — > 0,

where y>o is a degree preserving homomorphism such that <po(ei) = fi for all i. Note that Z\ C m^?9 by Lemma 2.5.3. Since Z\ is once again a finitely generated graded .R-module one may iterate the process to obtain the required exact sequence F. D

60_______________________________________ Chapter 2

Theorem 2.5.6 (Hilbert syzygy theorem) Let R = k[x\, . . . ,xn] be a polynomial ring over a field k and let M be an 'H-graded R-module. Then M has a graded free resolution of length at most n. Proof. Set m = R+. First notice that Torf (M, R/m) - 0 for i > n + 1, because the ordinary Koszul complex K*(x) is a graded free resolution of

R/m = k of length n. On the other hand assume that

-p —— y*\> • • • _ p —— yi,> PQ 77 _ •••_ ——.} r/t ——,> ri ——,> M M __ ——v> nU is a graded free resolution of M as in Proposition 2.5.5. Applying the functor

(•) ®.R R/m yields the complex

Fk R/m

v

-

• • • —> Fl <8> R/m ^ F0 ® R/m —> M <s> R/m —> 0.

Using im(y>fc) C mF^-i for all k > 1 one obtains that all the maps are zero. Hence Torf (M, R/m) ~ Ft ® fl/m

for i > 1. In particular Ft ® R/m ~ Fi/mFi = 0 for i > n + 1 and by Nakayama's lemma one obtains Fi = 0 for all i > n + 1. This proof was adapted from [92]. D Corollary 2.5.7 Let R = ®^0Ri be a polynomial ring of dimension n over a field k and let M be an N-graded R-module. Then there is a unique (up to complex isomorphism) exact sequence of graded modules b

9

bk

i=\

4=1 hi

fro

4=1

4=1

such that (a) g = supji | Torf (M, k) ^ 0} < n, and

(b) im(^ ft ) C mR1"1-1 for all k>l, where m = R+. Proof. Notice the following isomorphisms: bi

Torf (M, A) ~ 4=1

where Fj is the jth free module in the resolution of M. Hence the bj's and the dji's are uniquely determined and so is the length of the resolution. D

Affine and Graded Algebras

Remark 2.5.8 The entries of the matrices
dn < dzi < • • • < dgi.

Indeed if where r^ € m and r,j homogeneous, then deg(<^(ej)) > d^-i)i because r^

have positive degree. Hence one obtains d^\ > d^-i)iDefinition 2.5.10 The integers bo, . . . ,bg are the Betti numbers of M. The

dji's are the twists, they indicate a shift in the graduation. In general the Betti numbers and twists of M may depend on the base field k, see for instance [108, Example 2.10] and [237, Remark 3]. Definition 2.5.11 The .R-module Zk = ker(<^fc-i) is called the k-syzygy module of M. Definition 2.5.12 The homogeneous resolution or minimal resolution of M is the unique graded resolution of M by free .R-modules described in

Corollary 2.5.7.

If M has a minimal free resolution as above, then note that pdR(M), the projective dimension of M, is equal to g. In our situation the notions of "free resolution" and "projective resolution" coincide because of the theorem of Quillen-Suslin: if k is a principal ideal domain, then all finitely generated projective k[xi, . . . ,xn]-modules are free [250]. Theorem 2.5.13 (Auslander-Buchsbaum) Let M be an R-module. If R is a regular local ring, then

pd fl (M) + depth(M) = dim(E). Proof. See [109, Theorem 3.1] or [193].

D

Corollary 2.5.14 If R is a polynomial ring over a field k and I is a graded ideal, then

pdfl(/2/7)>ht(/), with equality if and only if R/ 1 is a Cohen-Macaulay R-module. Proof.

The result follows from an appropriate graded version of the Auslander-Buchsbaum formula. D

62_______________________________________Chapter

2

Theorem 2.5.15 (Hilbert-Burch) Let R be a polynomial ring over a field k and let I be a graded Cohen-Macaulay ideal of height two. If

0 —)• Ri-1 -A R" —> R —> R/I —> 0 is the minimal resolution of R/I, then I is generated by all the minors of size q — 1 of the matrix (p.

Example Let R =
Let us construct the minimal resolution of R/I. Consider the exact sequence

0 —> ker(^) = Zl A R(-2) © R2(-3) ^ / —> 0,

et &> f t .

Here Z\ is the module of syzygies of /. Using Macaulay we find that Z\ is generated as an l?-modulo by the column vectors, i/>i, fa, of the matrix: ZW

X

y

z

-x

-y

A=

Notice that R(-2) © 7?2(-3) is graded by

Hence f a , fa G [fi(— 2) © 7i!2(— 3)] 4 . Next consider the exact sequence

0 — > ker(v> 2 ) = Z2 A ^ 2 ( - 4 ) ^ > Z ! ^ 0 ,

&i^fa.

Since i/'i/J + ^2-R is a free -R-module, we have Z-2 = (0). Altogether the minimal homogeneous resolution of R/I is:

0 —» # ( - 4 ) - > E(-2) ©

(-3)

R—

In this example /?// is Cohen-Macaulay because pdR(R/I) = 2 = ht (/). Pure and linear resolutions

Let R = ®^0Ri be a polynomial ring over

a field fc with its usual graduation and / a graded ideal of R. Let 0

-»

#(-<W ->• • • • -> 8=1

^(-dii) ~^R^S = R/I->0

(2.1)

j=l

be the minimal graded resolution of S by free /^-modules. The ideal / (or the algebra 5) has a pure resolution if there are constants

di < d-2 < • • • < dg

Affine and Graded Algebras_________________________63

such that dn = di, . . . , dgi = dg for all i. If in addition

di = di + i — 1 for 2 < i < g the resolution is said to be di-linear . Theorem 2.5.16 (Herzog-Kiihl) If S is a Cohen-Macaulay algebra with

a pure resolution, then

Proof. See [154] for the proof and for a generalized version of this result valid for graded modules. D Theorem 2.5.17 (Huneke-Miller) Let S be a Cohen-Macaulay algebra. If S has a pure resolution, then the multiplicity e(S) of S is given by

e(S) = Proof. See [185] for the proof and for a generalized version of this result valid not only for cyclic modules. For another simple proof see [158]. D Since the Betti numbers and the twists are positive integers note that these formulae impose severe restrictions on the numbers di,..., dg that can occur as the resolution type of a Cohen-Macaulay standard algebra with a pure resolution.

Exercises 2.5.18 Let R be a polynomial ring and let 7 be a graded Cohen-Macaulay ideal of height 3 with a pure resolution:

Q-*Rb3(-(d + a + b)) ->Rb2(-(d + a)) -> Rbl(-d) -> I-> 0, where a, b > 1. Assume &i = 6 and 6 = 1 . Show that x = a + 1 and y = d + a + lisa, solution of the diophantine equation

&x(x- 1) =y(y - 1). Then prove that the integral solutions of this equation are given by

x = (X + l)/2 and y = (Y + l)/2, where \/QX + Y = ±(1 ± V6)(5 + 2\/6) n , n 6 Z.

64_______________________________________Chapter

2

2.5.19 Let R — k[xi,... ,xn] be a polynomial ring over a field k and let / be a graded ideal, then (xi,... ,xn) is an associated prime of / if and only i f p d R ( R / I ) =n. 2.5.20 Let R = k[x, y, z] and let

/ = (xn + yn - zn n> 2). If char(/c) ^ 2, prove the equality 7=(a:2+j,2-^)a;3+y3-s3>a;V).

2.5.21 Let R = k [ x , y , z ] and let / = (xn + yn - zn n> 2).

If char(/c) ^ 2, prove that

I - qi n q 2 n q 3 is a primary decomposition of /, where

qi = (y2,x-z), q 2 = (x2,y - z), and 2 2

y ,x2 +y2 - z1}.

Chapter 3

Rees Algebras and Normality In this part a detailed presentation of complete and normal ideals is given. The systematic use of Rees algebras and associated graded algebras will make clear their importance for the area. One of the outstanding references for blowup algebras is [296]. An important reference for the normality of Rees algebras is [40],

3.1

Symmetric algebras

Let M be an /^-module. Given n > 0 we define Tn(M) = M ® - - - ® M and T°(M) = R. n— times

Recall that the tensor algebra T(M] of M is the non commutative graded algebra n=0

where the product is induced by juxtaposition. The symmetric algebra of M, denoted by Sym^j(M) or simply Sym(M), is defined as the quotient algebra Sym(M) = T(M)/J

where J is the two sided ideal generated by the elements xy-yx = x®y-y®x£ T 2 (M)

65

66_______________________________________ Chapter 3

with x and y running through M. Observe that Sym(M) is commutative. Since J is a graded ideal generated by homogeneous elements of degree two, the symmetric algebra is graded by

SymJM) = Tn(M)/J

R T"(M)

and Sym 0 (M) = R.

Note Sym 2 (M) - M ® M/(x ® y - y ® x), with x and y running through M.

Exercises 3.1.1 If M is a free R-module of rank n prove that the symmetric algebra of M is a polynomial ring in n variables with coefficients in R.

3.2

Rees algebras and syzygetic ideals

Let 7 be an ideal of a ring R generated by f i , . . . , f q . The Rees algebra of I, denoted by K(I) or R[It], is the subring of R[t] given by

K(I)=R[fit,...,fqt]cR[t], where t is a new variable. Note

U(I) =R®It®---® Intn © • • • c R[t\. There is an epimorphism of /^-algebras

where B is a polynomial ring in the indeterminates ti, . . . ,tq over the ring

Ji. The kernel of , denoted by J, is the presentation ideal or toric ideal of 7i(I) with respect to f i , . . . , fq. Notice that J is a graded ideal in the tj-variables: 00

J7

- U7 ffi Jl7> ~ 1=1

where B has the standard grading induced by deg(ij) = 1. The mapping il>: Rq —> I given by

Rees Algebras and Normality________________________ 67

induces an .R-algebra epimorphism

Thus the symmetric algebra of / is:

Sym f l (/)~.R[ti,...,*,]/ker09),

where ker(/3) is an ideal of R[t] generated by linear forms:

ker(/3) = I <j y

=i

biU Y^ Z_/&,;/,; = 0 and &,; 6 R !

On the other hand the kernel of ip is generated by all forms F(ti , , . . , tq) such that F ( f i , . . . , f q ) = 0. In particular, one may factor (p through Sym / j(J) and obtain the commutative diagram:

SymR(I} We say that / is an ideal of linear type if a is an isomorphism. An important module-theoretic obstruction to "SymR(I} given by the following result.

~ 72,(/)" is

Proposition 3.2.1 (Herzog-Simis-Vasconcelos) Let I be an ideal of a

ring R. If SymR(r)~K(I), then for each prime p containing I, /p can be generated by ht (p) elements.

Proof. See [155]. Syzygetic ideals

D Let / be an ideal of a ring R and let .Hi (I) be the

first homology module of the Koszul complex H*(x^,R) associated to a set x_ = xi , . . . xn of generators of /. In [260] it is pointed out that HI (I) is related to I/I2 , the conormal module of /, by the following exact sequence:

ffi(J) -A Rn ® (R/I) A I ® (R/I) = I /I2 —* 0. Here f([z}} = z ® I and h(e{ ® 1) = x{ ® 1. Set 6(1} = ker(/). Definition 3.2.2 The ideal I is called syzygetic if 6(1} = 0.

(*)

68_______________________________________ Chapter 3

Although H\(I) may depend on the set of generators for /, 8(1} depends only on /. Indeed A. Simis and W. Vasconcelos [260] proved that where Sym 2 (/) —t I2 is the surjection induced by the multiplication map.

Definition 3.2.3 An ideal / of a ring R is said to be generically a complete intersection if IRV is a complete intersection for all p G A.ssn(R/I). Remark 3.2.4 If / is unmixed and generically a complete intersection,

then the J?//-torsion of HI (I) equals 6(1). To prove this, notice that /p is syzygetic for all p € Ass/? (R/I), consequently 6(1) is a torsion module. Computing the toric ideal of a Rees algebra Let / be a graded ideal of a polynomial ring R over a field k and /i, . . . , fq a generating set for /.

Consider the presentation of the Rees algebra:

The kernel J of (p can be obtained as follows: J

= ( t l

- t f l , . . . ,

tg-tfg)nB.

Since J is a graded ideal in the ij-variables, J = ®»>i J». The relationship between J and the Koszul homology of / is very tight. The exact sequence of Eq. (*) can be made precise:

0 —> J 2 /BiJi = 6(1) —> #i(7) —»• (7Z/7)' —-> I/I2 —> 0. In particular one can decide whether I is syzygetic - that is, J% = £?i Ji-or of linear type-that is, J = J\B.

Exercises 3.2.5 Let /? = A[x] be a polynomial ring over a ring A and / i , . . . , fq forms

of degree d > 1. Prove that there is a graded isomorphism of A-algebras

tp:A[fi,. ..,/<,] —» A[i/i,. - • ,tfq],

with y>(/j) = i/j,

where Ms a new variable and both rings have an appropriate grading such that deg(/j) - 1 and deg(i/j) = 1. 3.2.6 Let / = I%(X) be the ideal of 2 x 2-minors of the symmetric matrix: •y

X\

Xi

£3

X3

X5

X6

__

where the entries of X are indeterminates over a field k. Prove that / is a syzygetic ideal that satisfies sliding depth. See [161].

Rees Algebras and Normality________________________ 6!)

3.3

Complete and normal ideals

Let us recall the notion of integrality of ideals. Let R be a ring and / an

ideal of R, an element z G R is integral over / if z satisfies an equation

the integral closure of / is the set of all elements z G R which are integral over I. This set will be denoted by I or Ia. Definition 3.3.1 If / = /, / is said to be integrally closed or complete. If all the powers Ik are complete, the ideal / is said to be normal.

Lemma 3.3.2 Let R be a ring and I an ideal of R. An element a e R is in the integral closure of I if and only if there is an ideal L of R such that

(i) aL C IL, and

(ii) a r ann(7/) = (0) for some integer r > 0. Proof. =>) As a is in integral closure of 7 there is an equation:

an = a\an~l + • • • + an-\a + an, where QJ € I1, and n > 1 is some integer. The required L is obtained by setting

L = Ran-1 + 7a"-2 + ••• + 7™- 2 a + In~l . Indeed from the equation above

Ran C la71"1 + 72a"-2 + • • • + In~la + 7" = IL, and if i > 1 one has a(7*a™- i - 1 ) = /'a"-' = 7(7^1a"-i) C IL. Hence aL C IL. To finish this part of the proof note a n ~ 1 ann(L) = (0). •<=) Since 7? is Noetherian L = (/i, . . . , /„). One can write

where 6y G 7. Set B = (6^), C = B — aln and / = (/i, . . . , /„). Here In denotes the identity matrix. As C f l = 0, one can use the formula

C-adj(C') =det(C)7 n to conclude /jdet(C) = 0 for all i. Therefore det(C) is in ann(L) and by hypothesis ar det(C') = 0 for some r. Expanding in powers of a shows that a is integral over 7. D

70_______________________________________Chapter

3

Proposition 3.3.3 /// is an ideal of a ring R, then I is an ideal of R. Proof. Let on £ I, i — 1, 2. By the proof of Lemma 3.3.2, there is an ideal Li such that

(i) ctiLi C ILi, (ii) a[ i ann(Lj) = (0) and ITi C Li for some r-j > 0. Since on is integral over /, one has a™; 6 / for some n, > 1. Therefore for

n large enough /3™ is in Z/iI/2, where /? = ai + a2, thus /?™ann( J L 1 L 2 ) = (0). On the other hand one clearly has the inclusion flL\Li C IL\Li. Hence by Lemma 3.3.2 one concludes ft £ /. To finish proving that / is an ideal note xa\ £ / for x £ R, this follows at once from the definition of I. D

Proposition 3.3.4 /// is an ideal of a ring R and S is a multiplicatively closed subset of R, then Proof. To prove S^fT) C S1"1^) take / £ S'1^). One may assume

/ = x/1 with x £ R because the integral closure of an ideal is again an ideal. There is an equation r+

^.r-1 + ... + ^i/ + ^ o Si

Sn^i

Sn

I

where a,/Si £ S~1(I)'t = S~l(P). Thus one may assume o,j £ P and Sj € S for all i. Clearing denominators and multiplying by an appropriate element of S one has an equality (in the ring R) of the form

sxn + tiaixn~l + • • • +tn-ian^ix + tnan = 0 (s,ti 6 S), multiplying both sides of this equation by sn~l yields sx is in J. Therefore

/ = ( x s ) / s is in 5'~1(/). The other containment follows rapidly using that localizations commute with powers of ideals.

D

Lemma 3.3.5 Let A be a domain and x 6 A \ {0} such that Ax is normal. Then A is normal if and only if (x) is a complete ideal. Proof. Assume A is normal. Let z £ (a;). Since z satisfies a polynomial equation m 1

- + ••• + (am^xm-l)z + (amxm) = 0 , a{ £ A,

Rees Algebras and Normality________________________ 71

dividing by xm yields that z/x is integral over A and z 6 (x).

Conversely assume (x) is complete. Let z be an element of the quotient field of A which is integral over A. As A and Ax have the same field of fractions and Ax is normal one may assume z = a/xr , for some a G A and r > 1. To show z e A it suffices to verify a e (x). There is an equation

zm + blzm~1 + --- + bm-lZ + bm=0, he A, multiplying by xrm yields the identity

aro + (blXr}am-1 + ••• + (bm^xr(m-V)a + (bmxTm) = 0, hence a 6 (x) = (x), as required.

Proposition 3.3.6

D

If A is a domain and x e A \ {0}, then

Proof. Note that the left hand side of the equality is contained in the right

hand side because A is a domain. Conversely take an element z G Ax and z G Ay, for all p 6 AssA/(x). Write 2 = a/xn, a & A and n > 1. It is enough to prove that a e (x). If a is not in (x) note that

((x):a)cZ(AI(x}). Hence ((a;): a) C p, for some p 6 AssA/(x). Since z 6 A9 one may write z = a/xn — b/s, b 6 A and s 0 p. Therefore s £ ((x):a) C p, which yields a contradiction. D

Proposition 3.3.7 If A is an integral domain and S is a multiplicatively closed subset of A, then

Proof. Note that A and_5~1(yl) have the same field of fractions K. First we prove S~1(A) C S1"1^). Take any x in K integral over S~1(A). There is an equation Si

S n -l

Sn

1

where Oj G A andj>, e 5 for all i. _Set s — Si • • • sn. If we multiply by sn, it

follows that sx e A and x e S^_l(A).

Conversely take x e S~l(A). There is s e S such that sx is integral over A. Hence sx satisfies (sx)™ + ai(sx)"" 1 + • • • + a n _!(sx) + an = 0,

72_______________________________________Chapter

3

for some a i , . . . , an in A, dividing by sn immediately yields that x is integral over S~l(A), as required. D

Corollary 3.3.8 If A is a normal domain and is S a multiplicatively closed subset of A, then S~l(A) is a normal domain. Corollary 3.3.9 Let A be a domain and x £ A \ {0}. Then A is normal if and only if Ax and Av are normal for every p 6 AssA/(x). Proof. If A is normal, then by the corollary above Ax and Af are normal

for every x £ A \ {0} and p e Spec(yl). For the converse use Proposition 3.3.6 and observe that Ax and A$ have the same quotient field as A. D To state a useful criterion of normality of Rees algebras, it is convenient to introduce the extended Rees algebra of an ideal / in a ring R:

A = R [ I t , u ] , u = t~1. Part of its usefulness derives from the equality Au = R[t,t~1]. Proposition 3.3.10 Let I be an ideal of a ring R and A = R[It,t~l] its extended Rees algebra. If R is a normal domain, then A is normal if and only if Ap is normal for each associated prime p of t~1A.

Proof. Note At-i = R[t, t ~ 1 ] . Hence At-i is a normal domain and one can use Corollary 3.3.9. D Lemma 3.3.11 Let A be a ring and x a regular element of A. If A/(x) is reduced and p e Ass^ A/(x), then Af is a normal domain. Proof. Let Ass^ A/(x) = {pi,... ,p r }. Since R/(x) is reduced

(x) = p 1 n - - - n p r , hence (x)AVi = piAVi for all i. Note that ht (pi) = 1, by Krull's principal ideal theorem. Therefore the maximal ideal of A9i is generated by a system of parameters, that is, AVi is a regular local ring and consequently Afi is a normal domain by Theorem 1.4.16. D

Proposition 3.3.12 Let A be a ring and x a regular element of A. If Ax is a normal domain and A/(x) is reduced, then A is a normal domain.

Proof. As Ax is a domain and a; is a regular element on A one obtains that A is a domain. By Lemma 3.3.11 Af is normal for all p G A.SSA A/(x), hence A must be normal according to Corollary 3.3.9. D

Rees Algebras and Normality________________________ 73

There is another graded algebra associated to an ideal / of a ring R whose properties are related to the normality of the Rees algebra of /, it is called the associated graded algebra of / and is defined as:

= R/I

® I/I2 © • • • ffi P/Ii+1 ffi • • • ,

with multiplication

(a + P)(b + /•»') = ab + Ii+j~l

(a 6 P~\ b e I*-1).

Given a generating set f\ , . . . , fq of /, it is not difficult to verify that

gij(R) is a graded algebra over R/I generated by the following elements of degree one:

7i = /i + / 2 ,...,7, = /9 + /2.

Thus one has Lemma 3.3.13 // / is an ideal of a ring R, then

R[It]/IR[It]

~ grj(R)

and A/t~lA ~ gr 7 (#),

where A = R[It,t~l] is the extended Rees algebra of I. Proof. It is left as an exercise.

D

Theorem 3.3.14 Let I be an ideal of a ring R. If I is generated by a regular sequence f i , . . . , fq, then the epimorphism of graded algebras:

is an isomorphism, where ti, . . . ,tq are indeterminates over R/I. Proof. The proof is by induction on q, the case q = 1 is easy to show. Let J be the ideal (/i, . . . ,/g-i) and consider the epimorphism:

V'(^)

=

ft

=

fi + ^ 2 - Note (J:fq)

= J because fq is regular on R/J,

moreover since
( J m : f q ) = Jm for all m> 1. Let F € R[t] = R[ti, . . . , tq] be a homogeneous polynomial of degree d such that its image in (.R//)[t] is in the kernel of ip, that is,

74_______________________________________ Chapter 3

As if is graded it suffices to prove F € 7/?[t]. To show F 6! IR[t] we proceed by induction on d, the case d = 0 is clear. There is W £ R[t] of degree d+1 with F(f) = W(f), write

where Wi = 0 or P7j is a homogeneous polynomial in R[t] of degree d. There

are polynomials G in E[ii, . . . , i ? -i] and JEf in J?[t] of degrees d and d - 1 respectively such that i

F1 = F - £ /fWi = G + ig#.

Observe that F'(/) = 0, hence H(f) 6 (Jd: fq) = Jd C Id and by induction on d one concludes that H has coefficients in /. It only remains to prove that G has coefficients in /. There is a polynomial H1 e R[ti, . • • , * 9 -i] of

degree d with H(f) = H'(f).

Set

noting F " ( f ) = F'(f] — 0 and using that if>' is an isomorphism one derives that F" has coefficients in J, which implies that G has coefficients in / as required. D Theorem 3.3.15 Let I be an ideal of a ring R and A — R[It,t~1} its

extended Rees algebra. If R is a normal domain andgTj(R) is reduced, then A is normal. Proof. As A/t~lA ~ gr7(/?) is reduced and At-\ = R[t,t~l] is a normal domain, one may use Proposition 3.3.12. D

Definition 3.3.16 Let A be an integral domain and K its field of fractions. An element x £ K is almost integral over A if there is 0 ^ a € A such that axn e A for all n > 0. Proposition 3.3.17 Let A be an integral domain and let K be its field of fractions. Then an element x G K is integral over A if and only if x is

almost integral over A. Proof. Let x = c/d, where c, d are in A and d ^ 0. If a; is integral over A, then there is an equation xm + blXm~l + ••• + bm,lX + bm = 0, bi e A. Setting a = dm one has axn £ A for all n > 0. Conversely assume there is 0 ^ a £ A such that axn 6 A for all n > 0.

Since a~l A is a Noetherian yl-module and A[x] C a~lA one has that A[x] is a finitely generated ^-module. As A is a subring of A[x] by Proposition 1.4.2 one derives that x is integral over A.

d

Rees Algebras and Normality

75

Theorem 3.3.18 ([157]) Let I be an ideal of a normal domain R. Then the following are equivalent:

(a) / is a normal ideal of R. (b) The Rees algebra R[It] is normal. (c) The ideal IR[It] C R[It] is complete. (d) The ideal (t"1) C R[It,t~l] is complete. (e) The extended Rees algebra R[It,t~l] is normal.

Proof, (a) => (b) Set A = R[It]. Let z e A C R[t] and write

It suffices to prove bsts € A. First we prove that bsts e A. As z is almost integral over A there is 0 ^ f e A such that fzn e A for all n > 0. Hence there is 0 ^ fm 6 Im such that (fmtm)(bsts)n is in A for all n > 0, that is, bsts is almost integral over A. As R is a Noetherian integral domain using Proposition 3.3.17 one derives that bsts is integral over A. Thus there is an equation

(M') m + ai(bsts)m~l + ••• + am-i(bsta) + am = 0,

where a, = ^j=o a u'* J anc' a y ^ ^ J > Grouping all the terms of i-degree equal to sm one has the equation

Thus bs is integral over 7s and consequently bs 6 7 s , as required. (b) =>• (c) Let z = bo + bit + • • • + bsts be an element of R[It] which is integral over IR[It]. Then z satisfies an equation of the form

zm + alZm~l + ••• + am-lZ + am = 0, at 6 F R [ I t ] , multiplying by tm one obtains that tz is integral over R[It]. tz € R[It], which proves that z 6 IR[It].

Therefore

(c) => (d) Set B = R[It,t~1}. Let z be an element of B integral over 1

t~ B. As the negative part of the Laurent expansion of z is already in t~lB one may assume z = ^i=o ^it1, s > 0 and 6j € 7l for all i > 0. By descending induction on s it suffices to prove that bs G 7S+1. There is an equation

zm + aizm~l + • • • + a m _i^ + am = 0, a{ 6 (i"1)^,

76________________________________________ Chapter 3

hence zt is almost integral over B. It follows rapidly that bsts+1 is almost integral over B and thus bsts+1 is integral over B. A straightforward calculation yields that bsts+l is integral over IR[It], which shows that bs is in JS+l

(d) =>• (e) Set u = t~l and note R[It,u]u — R[t,u] is a normal domain. Therefore by Lemma 3.3.5 one concludes that R[It,u] is normal. (e) =>• (a) If z 6 7 r , then 2 satisfies a polynomial equation

multiplying by f r m yields that zir is integral over the ring R(It,t~1}. Hence

z e r.

n

Corollary 3.3.19 -Lei / be an ideal of a normal domain R and R[It] its Rees algebra. If gij(R) is reduced, then R[It] is normal.

Proof. Use Theorem 3.3.15 and Theorem 3.3.18.

D

An ideal I of a ring R is a radical ideal if / = rad (/). Note: (i) a proper ideal / is radical if and only if / is an intersection of finitely many primes, and (ii) / C rad (/) and equality occurs if / is a radical ideal.

Corollary 3.3.20 Let I be a radical ideal of a normal domain R. If I is generated by a regular sequence, then grf(R) is reduced and R[It] is a

normal domain. Proof. It follows from Theorem 3.3.14 and Corollary 3.3.19.

D

Example 3.3.21 Let R = Q[x,y] be a polynomial ring in two variables over the field of rational numbers and / = (x2,y2). Observe the equality 7= (x2,y2,xy), thus R[It] is not normal because / is not even complete.

Definition 3.3.22 Let / be an ideal of a ring R and pi , . . . , p r the minimal primes of I. Given an integer n > 1, the nth symbolic power of / is defined to be the ideal

/(«) = q i n - - - n q r , where q, is the primary component of In corresponding to p,. Let (R, m) be a regular local ring and / an unmixed ideal. An interesting problem is whether J^ 2 ' is contained in ml, although the answer is negative in general, the problem remains open in characteristic zero. For some insight

into this problem see [97, 186].

Rees Algebras and Normality________________________ 77

Proposition 3.3.23 Let I be a proper ideal of a ring R and S = -R\U[ =1 pj, where pi, . . . ,p r are the minimal primes of I. Then

/(») =S~1InnR forn> 1. Proof. Let

/™ =

qi

n • • • n q r n q r+ i n • • • n q s

be a primary decomposition, where q^ is the primary component of p, for i < r and qi is a primary component of the embedded prime pi for i > r.

Since pi n 5 ^ 0 for i > r one has S~lqt = S~~1R. Hence

To finish the argument note that 5-1q, n R = (\i and intersect with R the equality above. D

Proposition 3.3.24 Let I be a radical ideal of a ring R and pi, . . . ,p r the minimal primes of I. Then

/(») = p < n ) n - - - n p f n ) forn> 1. Proof. Let

/" = qi n - - - n q r n q r + i n - - - n q s be a primary decomposition of In, where q, is pj-primary for i < r and q^ is an embedded primary component of In for i > r. Localizing at pi yields I"Rfi = qi-Rp; and from / = pi n • • • PI pr one obtains:

Thus p"-Rpi = qi-Rp; and contracting one has p^

= qj, as required.

D

An aspect of symbolic powers that has attracted attention is to describe when the symbolic and ordinary powers of a given ideal / coincide, see [173] and [230] for a detailed discussion. Next we present a few cases where equality of symbolic and ordinary can be described in terms of properties of the associated graded ring. Definition 3.3.25 An ideal / of a ring R is called normally torsion free if Ass(R/Il) is contained in A.ss(R/I) for alii > 1 and / ^ R. In [38] M. Brodman showed that when R is a Noetherian ring and / is an ideal of R the sets Ass(R/In) stabilize for large n. If / is a radical ideal which is normally torsion free, then Ass(R/In) = Ass(R/I) for all n > 1, that is, in this case the notion of normally torsion free is a strong form of stability.

78_______________________________________ Chapter 3

Proposition 3.3.26 Let I be an ideal of a ring R. If I has no embedded primes, then I is normally torsion free if and only if I" = /(") for all n > 1. Proof. =>) Note Ass(R/In) = Ass(7?/7) for n > 1, because any associated prime p of / is a minimal prime of In and thus p £ Ass(7?/7 n ). As In has no embedded primes one concludes that 7™ has a unique irredundant minimal primary decomposition and 7" = /(") for n > 1.

4=) Let pi, . . . ,p r be the associated primes of 7. Since pi is a minimal prime of I for all i, one derives

Ass(7?/7") = Ass(fl//M) = {Pi, . . . ,p r }, as required.

D

Proposition 3.3.27 Lei 7 fee an idea/ o/ a Cohen-Macaulay ring R. If I is generated by a regular sequence, then In = 1^ for n > 1.

Proof. Let /i, . . . , fr be an 7?-regular sequence such that 7 = (/i, . . . , / r )By Krull's principal ideal theorem ht (7) = r, hence 7 is an unmixed ideal by Theorem 1.3.23. From Theorem 3.3.14 there is an isomorphism of graded rings

v. B = where B is a polynomial ring with coefficients in R/I. Therefore P/P+l is a free 7?/7-modulo because it is isomorphic to Bi, the ith graded component of B. Thus

Ass fl (7Vr +1 ) = AssR(R/I). Using the exact sequences 0 — * 7V7i+1 —•> 7?/7i+1 —> R/F —)• 0, it follows by induction that Assfl(7?/7") C Assn(7?/7) for n > 1. To finish the proof one may apply Proposition 3.3.26 to conclude the equality between the ordinary and symbolic powers of 7. D Definition 3.3.28 A proper ideal 7 of a ring R is said to be locally a complete intersection if 77?p is a complete intersection for all p £ V(I).

Proposition 3.3.29 Let R be a Cohen-Macaulay ring and I a prime ideal. If I is locally a complete intersection, then 7" = 1^ for all n > 1.

Proof. By Proposition 3.3.26 it suffices to prove Ass(7?/7") C Ass(7?/7) for n > 1. If p is an associated prime of R/In, then p7?p is an associated prime of 7?p/7^. Using Proposition 3.3.27 yields that 7P is normally torsion free, that is, the only associated prime of 7^/7™ is 7 p . Therefore p7?p = 7P and p = 7.

D

Rees Algebras and Normality

79

Proposition 3.3.30 Let p be a prime ideal of a ring R such that p/?p is a complete intersection, then p(") = p™ for all n > 1 if and only if giy(R) is a domain. Proof. =>) Since gr p (J t ?) = R[pi\/pR[pt}, it is enough to show that $R[pt] is

a prime ideal of R[pt]. Let x = a0 + ait + • • • + artr be an element in R[$t]

and x' = (a 0 /l) + (ai/l)i+ • • • + (ar/l]tr the image of a; in # p [p.R p t]. Note that aj is in pi+1 if and only if (a»/l) is in p*+1^?p, because the ordinary and symbolic powers of p coincide, thus x is in p.R[pi] if and only if x1 is in pRp[pRpt]. As a consequence p.R[p£] is prime if and only if p.Rp[p.Rpi] is prime. To finish the argument use Theorem 3.3.14 and the hypothesis that p.Rp is generated by a regular sequence to get that gr pfl (Rp) is a domain. •4=) First we prove Ass#(p l /P* +1 ) = {p} for i > 1. Let pi be an associated prime of p*/pl+1, that is, pi = ann (x + pt+1), for some x in pz \ pl+1. Note a 6 pi iff ax 6 p z+1 , and since grp(7?) is a domain, one readily concludes that a 6 pi iff a e p. Hence p! = p. There is an exact sequence:

0 — » pVp i+1 —> fl/p i+1 — > R/tf —-» 0. Hence using induction on i > 1 and Lemma 1.1.16, one rapidly derives, from the exact sequence above, that Ass(_R/p*) C Ass(_R/p) = {p}. Thus p is normally torsion free and by Proposition 3.3.26 one has p" = p( n ) for all n > 1. D

Theorem 3.3.31 ([262]) Let R be a normal domain and I a radical ideal

which is generically a complete intersection. If I is normally torsion free, then its Rees algebra Tl(I) = R[It] is a normal domain. Proof. By Corollary 3.3.19 we only have to show that the associated graded algebra gr7(.R) = R[It]/IR[It] is reduced. Let / = a0 + ait + • • • + asts + IR[It] be a nilpotent element of gr 7 (_R). Hence (asts)m is in IR[It] for some m > 1, by induction it suffices to verify that asts is in I R [ I t ] . Let pi , . . . , pr be the minimal primes of /. Since IRfi = pi-Rp ; is a complete intersection one derives that gr p . flp (/? P; ) is reduced (see Corollary 3.3.20). Therefore the

image of asts in gr p ./^ (Rf>f) is zero for all i, and one readily concludes that as belongs to the following intersection:

(R n p^ +1 # Pl ) n • • • n (R n psr+lRpr) = p[s+1} n • • • n p( s+1 >. As / is normally torsion free one can write In — C|i fl • • • fl q r , where q^ is pj-primary, localizing yields /"-RP; = p"-RP; = qi-Rp,- and consequently q; = p' n '. Making n equal to s + I proves that as is in 7S+1. D For ideals, in polynomial rings, with small data there are effective computational tests of normality, see Section 3.5, [40] and [296, Section 10.6].

80_______________________________________Chapter

3

Descent of normality Let A C B is an extension of rings such that B is normal, in general the normality of A is not inherit from B. We discuss some sufficient conditions for A to be normal.

Lemma 3.3.32 Let A C B be an extension of rings. If B = A ® C (as A-modules), then IB (~1 A = I for every ideal I of A.

Proof. Let z e IB D A and write z = X^=i bifi, where hi € B and /» € /. By hypothesis bi = a* + ci; a, e A and c» e (7. Since z e ^4 it follows that z = X^i=i aifi £ I- This proves the containment IB r\ A C I, the reverse containment is clear. D Proposition 3.3.33 Let A C B be two integral domains with field of fractions KA and KB respectively. If B — A © C (as A-modules), then

KAr\BcA. In particular if B is normal, then A is normal.

Proof. By Lemma 3.3.32 one has IB n A = I for every ideal / of A. Let b = a/c G B, a,c £ A, then a € (c)B n A = (c), hence a = Ac = be, with A £ A. Therefore b = A 6 A. D

Exercises 3.3.34 Let .R = A;[ZI, . . . ,xn] be a polynomial ring over a field k and KR its field of fractions. If R' = k[x^1,... , x^1] C KR is the ring of Laurent polynomials, prove that R1 is a normal domain.

Hint R' is the localization of R at the multiplicative set of monomials. 3.3.35 Let -R be a ring and /, J ideals of R. If J and J are integrally closed, prove that / H J is integrally closed. 3.3.36 Let p be a prime ideal of a ring R. Prove that the symbolic power p(") is a primary ideal for n > 1.

Hint Note that p"-Rp is a primary ideal because its radical is maximal. 3.3.37 Let R be a ring and S a multiplicatively closed subset of R. If p is a prime ideal such that S n p = 0, then (S-ip)(«) = 5-ipW for n > 1. 3.3.38 Let / be an ideal of a Cohen-Macaulay ring R. If / is generated by

a regular sequence, then

Ass(E// n ) = Min( J R/7) for n > 1.

Rees Algebras and Normality

3.3.39 Let / be an ideal of a ring R generated by a regular sequence. Prove that P/P+1 is a free ^//-module for all i. Hint Use Theorem 3.3.14. 3.3.40 Let / be a radical ideal of a ring R. Prove that / is normally torsion free if and only if gr/(E) is torsion-free as an ^//-module (see [188] for other equivalent conditions).

Hint If gr^ (R) is torsion-free show

AssR(r/r+l) C Ass(R/I) for i > 1 and use the exact sequence

0 —> /V/ i+1 — > R/Ii+1 — >• R/P —+ 0. 3.3.41 Let R be a ring and T = {/i}j6N a family of ideals in R. It is said that T is a filtration of R, if Ij+i C Ii, /o = ^R, and /j/j C /i+j for all i,j G N. If / is an ideal of .R, prove that the following are filtrations of R: (a) I, = /*, the ordinary powers of /, (b) Ii = /% the integral closure of P , and

(c) Ii = /W , the symbolic powers of /. Hint For (b) use the method of proof of Proposition 3.3.3. 3.3.42 Let R be a normal domain and f = {/i}jgN a filtration of R. If Ii is integrally closed for all i, prove that the Rees algebra

of the filtration T is integrally closed. See [131] for a study of the CohenMacaulay and Gorenstein property of 3.3.43 Let R be a domain and / an ideal, then

W C R[It] i=0

with equality if R is a normal domain.

Hint To show equality use the very first part of the proof of Theorem 3.3.18.

82________________________________________ Chapter 3

3.3.44 Let R = k[x\, . . . ,xn] be a polynomial ring over a field k and m

the maximal ideal (xi, . . . ,xn). If / = md and J = (x%, . . . ,x%) for some positive integer d, then ___ U(J) = n(I}.

3.3.45 Let / and J be two ideals of a normal domain R. Define the multi Rees algebra of / and J as

here u,v are new variables. Prove that

where 7™ = 7" and J"1 = J

3.4

A criterion of Jiirgen Herzog

In this section we present an elegant and useful Cohen-Macaulay criterion (see Theorem 3.4.12). Once more we recall that all rings considered in this book are Noetherian and modules are finitely generated. Quasi-regular sequences

It is useful to generalize Theorem 3.3.14 to

modules by introducing polynomial with coefficients in a module. Let M be an 7i-module and R[ti, . . . , t q ] a polynomial ring over the ring R. Set M[t] = M [ t 1 , . . . , t g ] = M®RR[ti,...,tq], and note that an element of M[t] can be regarded as a polynomial in the tj variables with coefficients in M , thus M[t] is naturally graded. If /i, . . . , fq

is a sequence in R and 7 = (/i , • • • , / g ), there is a degree preserving map of additive groups >: (M/IM) f a , . . . , t,] —> gr, (M) = ® J i=0

such that for all F(t) in M[t] homogeneous of degree i, where the notation F(t) means reducing the coefficients of F(t) modulo IM. It is not hard to verify the equivalence between the following two conditions: (a) (p is injective.

(b) For every homogeneous polynomial F in M[t] of positive degree n such

that F(f) € 7 n+1 M, one has F <E 7M[t].

Rees Algebras and Normality________________________83

Definition 3.4.1 If the map ip is an isomorphism and IM ^ M, the sequence /!,...,/, is called an M-quasi-regular sequence. Theorem 3.4.2 Let M be an R-module and f = fi,..., fq a sequence in R. I f t is an M-regular sequence, then f is an M-quasi-regular sequence. Proof. It follows adapting the proof of Theorem 3.3.14.

D

Theorem 3.4.3 (Krull Intersection Theorem) Let M be an l?-module and / an ideal of R. Then there is a £ R such that

C

oo

\

P) PM J = 0. i=i /

Proof. See [218, Theorem 8.9] or [298, Appendix A].

D

Lemma 3.4.4 Let (R,m) be a local ring (resp. N-graded ring) and M an R-module (resp. N-graded module). If N is a submodule of M (resp. submodule) and I C m is an ideal of R (resp. graded ideal), then oo

N = Proof. Using Krull's intersection theorem one has

n PM'=(o) CO

(this equality holds also in the graded case), where M' = M/N. Hence the D

required equality follows at once.

Proposition 3.4.5 Let (R,m) be a local ring (resp. N-graded ring) and M an R-module (resp. N-graded module). If f = fi,... , f q is an M-quasi

regular sequence of elements in m (resp. homogeneous elements in m = R+), then f is an M-regular sequence. Proof. Fix an integer 1 < r < q and set / = ( / i , . . . , fq). To begin with we split the ideal / as / = J + L, where J = ( / i , . . . , fr-i) and L = ( f r , . . . , fg), it is convenient to set J = (0) if r = 1. It suffices to show the equality (JM'.M fr) = JM, because this is equivalent to prove that fr is not a zero divisor of M/JM. Let m 6E (JM\M f r ) , according to Lemma 3.4.4 one has oo

P) (JM + LnM) = JM, n=l

84_______________________________________ Chapter 3 thus the proof reduces to proving by induction on n that m £ JM + LnM for n > 1. Since m 6 (JM\M / r ), there are mi in M such that

mfr = mi/i + ••• + m r _ i / r _ i

(m,mi e M).

(3.1)

Consider the polynomial • • • - TO r _i£ r __i e M[t]. Note deg t (F) = 1 and F(f)

= 0 . As f is an M-quasi regular sequence we

get m e IM = JM + LM. By induction assume m e JM + L"M, that is, there is G € M[tr, . . . , tq] homogeneous of degree n such that

...,/(7) Combining Eq. (3.1) and Eq. (3.2) frG(fr,

(6i€M).

(3.2)

yields

. . . , / , ) = Ol/l + • • • + 0,-l/r-l

Using that f is an M-quasi regular sequence one may assume (after an induction argument) that Oj 6 /"M, and consequently trG(tr, . . . , tq) is in /M[t], that is, G(tr, . . . , tq) is in 7M[t]. One can write G = GI + G 2 , where GI, G% are homogeneous polynomials in M[t] of degree n with GI G JM[t] and G2 £ LM[t]. Since Gi(/ r , ...,/,) is in JM and G 2 (/ r , . . . , /„) is in L"+1M, from Eq. (3.2) we get that m is in JM + Ln+1M, as required. D

Multiplicities Let (R, m) be a local ring, M a finitely generated Rmodule and q an ideal with rad(q) = m. As gr q (M) is a finitely generated gr q (jR)-module and AQ = R/q is Artinian, by Theorem 1.2.8 there exists a polynomial P(t) £ Q[t] of degree equal to d — 1, where d — dim(gr q (M)), such that

P(i)=lAo(qiM/qi+lM),

fori>n0.

By Lemma 1.2.10 there are integers ao, . . • ,o,d-i such that d-l

,.

foralH>0. 3=0

From the short exact sequences

—> M/qi+1M —> M/cfM

0 —> tfM/^M

—s- 0,

one derives £(M/q i+1 M) - £(M/q*M) = P(i) for i > n0. Hence using the identity + d - 1N

d

^\ d

3=0

Rees Algebras and Normality________________________ 85

we get

j=n0

for i > TIQ. Altogether the function

is a polynomial function of degree d, called the Samuel function of M with respect to q. The integer a^-i, is the multiplicity of M with respect to q and is denoted by e(q,M). Note that e(<\,M)/d\ is the leading coefficient of A W -

We recall the following result in dimension theory that relates the degree of the Samuel function with the dimension of the module. Theorem 3.4.6

The Samuel function q X M(i)

= l(M/ql+lM)

is a polynomial function for i 3> 0 of degree equal to the dimension of M .

Proof. See [218, Theorem 13.4].

D

Definition 3.4.7 Let (R, m) be a local ring of dimension d, M an ^-module and q an ideal of R with rad (q) = m. We define e P( M\d(ttq,M)-

.f

In order to show how the "multiplicity" behaves under short exact sequences we need to recall a result of E. Artin and D. Rees. Theorem 3.4.8 (Artin-Rees lemma) Let M be a module over a ring R. If N is a submodule of M and I an ideal of R, then there is a positive integer c such that

InM n N = In~c(rM n TV), Vn > c. Proof. See [218, Theorem 8.5].

D

86_______________________________________ Chapter 3

Proposition 3.4.9 Let (R,m) be a local ring of dimension d and let q be an m-primary ideal of R. If

0 —> N —> M —> N' —> 0

is an exact sequence of R-modules, then

Proof. By Proposition 1.1.32 one has

dim(M) = max{dim(TV),dim(TV')},

hence one may assume d = dim(M). Tensoring with R/qn+l the exact sequence above yields an exact sequence

0 ->• (N n q"+1M)/q"+1TV -> TV/qn+1TV -> M/q"+1M -» N'/qn+lN' ->• 0. Taking lengths with respect to R/q gives

By the Artin-Rees lemma TV n q"+1M C q" +1 ~ c /V, for some integer c > 0.

Hence

^(TV n qn+1M/qn+1N) < l(qn+1-cN/qn+1N) = XqN(n) - x%(n ~ c). as x^v(n) — XN(H — c) is a polynomial function of degree at most d — 1, the result follows by dividing Eq. (*) by nd and taking limits when n goes to infinity. n

Proposition 3.4.10 Let (R, m) be a local ring of dimension d and q an m-primary ideal. If M is a finitely generated R-module and A is the set of all prime ideals p of R with dim(J?/p) = d, then

Proof. Set 6 = ,4nSupp(M). If B = 0, then dim(M) < d and e d (q,M) is equal to 0, thus in this case the identity above holds. Hence we may assume B ^ 0, this yields the equality dim(M) = d. By Theorem 1.1.15 there are prime ideals pi , . . . , pra of R and a filtration of submodules: (0) = Mo C Mi C • • • C Mn = M,

such that Mi/Mi-i ~ -R/p, for all i. Observe

dim(R/fi)

= d},

Rees Algebras and Normality________________________ 87

to show this equality note

Ass(M) C { p i , . . . , p n } cSupp(M), see Corollary 1.1.17 and its proof, and recall that the minimal elements of Supp(M) are in Ass(M). Using Lemma 3.4.9 and the exact sequences

0 — > M,_i —-» Mi —* R/pi — ». 0

(i = l , . . . , n ) ,

we get

e(q,M) = Y^e(q,R/fi), where the sum is taken over all pi in the multiset {pi, . . . ,p n } such that

dim(.R/pi) = d. Let p e B, then if .f

P ^ P. _i

p=p

(0^ (3.3)

Since the second module is a field we get that the length of Mp is equal to the number of times that p occurs in the multiset {pi, . . . ,p n }. Therefore e ( q , M ) = ^i(MVi)e(q,R/pi), ViGB

as required. This proof was adapted from [44].

D

Proposition 3.4.11 Let (S, m) be a local ring and M an S-module with a positive rank r = rank(M) . If q is an m-primary ideal, then e(q,M) = e(q,S)rank(M). Proof. First note d — dim(M) = dim (5) by Lemma 1.1.37. Let A be the

set of all prime ideals p of 5 with dim(5/p) = d. Since M has rank r one has Mp ~ (Sf)r for any associated prime p of 5, thus Mp ~ (Sp)r for any p 6 A. Therefore using Proposition 3.4.10 one obtains e d (q,M)

= pe*

=

^ rt(S»)ed(q, 5/p) = rank(M)e d (q, S).

As e(q, M) = e^(q, M) and e(q, S) = e d (q, 5), the proof is complete.

D

In Chapter 4 we use the following important Cohen-Macaulay criterion of Jiirgen Herzog to study the Koszul homology of graded ideals.

88

Chapter 3

Theorem 3.4.12 ([148]) Let (S,m) be a Cohen-Macaulay local let M be a finitely generated S-module with a well defined and positive rank. If y = Vi ) • • • > Vd is a system of parameters of S , then

t ( S / ( y } } -rank(M) < l ( M / ( y ) M ) . Furthermore equality holds if and only if M is Cohen-Macaulay.

Proof. By Lemma 1.1.37 one has d = dim(M) = dim(S'). Let q be the m-primary ideal generated by y. There is a (graded) epimorphism oo

q*M/q' +1 M

: (M/qM)[t] = (M/qMH*!, . . . , td] -+ gr q (M) = 0

(3.4)

i-O

such that
of degree i, where F means reducing the coefficients of F modulo qM. By restriction of
i(M/qM}(t+,d~l\>l(qiM/(\i+1M) \ d - l j

for i > 0,

(3.5)

observe that both sides of the inequality are polynomial functions of degree d - 1 and consequently l(M/qM) > e(q,M). On the other hand y is a regular sequence, because S is Cohen-Macaulay (see Proposition 1.3.17), this forces

At this point one should observe that, by the arguments above, the proof reduces to show that l(M/qM) = e(q, M) if and only if M is CohenMacaulay. Next we show both implications. =>•) Taking into account Proposition 3.4.5 and Proposition 1.3.17 it suffices to prove that (p is injective; because this implies that y is an M-regular sequence and hence M is Cohen-Macaulay. Using Eq. (3.4) we obtain an exact sequence 0 — > ker(^)i — )• ((M/qM)[t])< — >• q*M/q + 1 M — > 0, hence we get *(ker(
for i > 0.

As the polynomial functions on the right hand side have both degree d — 1 and their leading terms cancel out, one concludes that £(ker(
Rees Algebras and Normality________________________ 89

Let F = Y^ mat" be a homogeneous polynomiaHn M[t] of degree p,_ with ma in M, and denote ma = rna + qM. Assume -F is in ker(^) and F ^ 0. Since mfc C q, there is an s > 1 such that m s-1 F ^ 0 and ms.F = 0, thus one may assume mF = 0 and F ^ 0. Hence there is a graded epimorphism ijj induced by multiplication by F V:(S/m)[t](-p)— >(S/q)[t]F.

We now show that ^ is injective. Let g — Y, bpt13 be an element in the kernel of V, where 6/3 € 5 and 10 = bp + m. Thus (E^^KEm^) = °; where bjj = bp + q. One may assume that ~bpt® and mata are the leading terms of g and__F respectively, w.r.t the lexicographical ordering of the ij variables. Thus bpmata+P is the leading term of gF and bprna 6 qM. Hence 6^ + q is a zero divisor of M/qM and since Ass fl / q (M/qM) = {m/q}, one has that &g + q is in m/q, which proves ba € m and fyg = 0. Altogether one derives g = 0 and ^ is injective. Using that ip is a graded isomorphism yields

which is a contradiction because the left hand side is a polynomial function of degree d — 1. Therefore F = 0 and (f is injective, as required. <=) As M is Cohen-Macaulay, the sequence y is an M-regular sequence, an application of Theorem 3.4.2 yields that

3.5

Jacobian criterion

Here we introduce the Jacobian criterion and present some applications and

examples to illustrate its use. Along the way some facts about regular local rings are discussed.

Let (R, m) be a local ring. By Nakayama's Lemma it follows rapidly that R is a regular local ring if and only if m is generated by a system of parameters of R. If R is a polynomial ring over a field k, then R is a regular ring, that is, Ry is a regular local ring for all p in Spec(-R) or equivalently -Rm is a regular local ring for any maximal ideal m of R (see [218, Theorem 19.5]). Proposition 3.5.1 If R is a regular local ring, then R is Cohen-Macaulay.

Proof. Let m be the maximal ideal of R and let k = R/m be its residue field. Assume x\,... ,Xd is a set of generators of m, where d = dim(R). If

90_______________________________________Chapter

3

k[t\,..., td] is a ring of polynomial over the field k, there is an epimorphism of graded fc-algebras oo

gr m (7?) = 0mVm <+1 ,

(3.6)

i=0

induced by
If 7 = ker(y>) ^ 0, pick a homogeneous polynomial / in 7 of degree s > 1. Using the isomorphism k[t](-8)

~ fk[t]

together with Eq. (3.6) one concludes:

dimfc(mYna* +1 ) = dim/; k[t]i — dim/; 7j <

d-1 J

\

d-1

By Theorem 3.4.6 the left hand side is a polynomial function of degree d— 1 while the right hand side is a polynomial function of degree d — 2. Therefore 7 must be zero and if is injective. D

Corollary 3.5.2 If (R,m) is a regular local ring, then R is a domain.

Proof. By Proposition 3.5.1 m is generated by a regular system of parameters. Thus grm(7?) is a domain. Let x,y G 7? such that xy = 0. If x ^ 0 and y ^ 0, then by Theorem 3.4.3 there are r, s 6 N with x £ mr \m r + 1 and y G ms \ ms+1. Multiplying the images x,y of x,y in grm(7i), one obtains xy = 0. Hence x = 0 or y = 0, which is impossible. This proves 7? is a

domain.

D

Proposition 3.5.3 Let (R,m) be a regular local ring and I / R an ideal of R. If R/I is a regular local ring, then I is a complete intersection.

Proof. Let xi,... ,Xd be a generating set of in = m/7, where x, = x^ + I and d is the dimension of R/I. Note that the set of images of x\,..., x^ in m/m 2 is a basis for m/m 2 as a vector space over k — R/m. Hence

xi + m 2 , . . . ,Xd + m2 are linearly independent in m/m 2 . Set n = dim(7t!) and m' = (xi,... ,Xd). As 7? is a regular local ring n = dim^m/m 2 ), hence using the proof of Corollary 1.1.30 and the equality m = m' + 7 one derives m = m' + J, for

some ideal J = (xa+i, • • • , xn) contained in 7. By Proposition 3.5.1 xi,... ,xn is a regular sequence, thus it suffices to show J = 7, to prove this equality observe that R/ J is a regular local ring

Rees Algebras and Normality________________________ 91 of dimension d, and therefore R/J is a domain by Corollary 3.5.2. Since R/I has also dimension d, we see that the canonical homomorphism

> R/I is an isomorphism, thus / = J, as asserted.

D

Definition 3.5.4 Let R = k [ x i , . . . xn] be a polynomial ring over a field k and / = (/i, . . . , fq) an ideal. The Jacobian matrix of / is the matrix

We denote the Jacobian matrix of / taken modulo an ideal P by

Theorem 3.5.5 (Jacobian criterion) Let B = R/I be a quotient ring, where R = k[xi,...,xn] is a polynomial ring in n variables over a field

k, and let I = (/i, . • . ,/ ? ) C R be an ideal. If P c R is a prime ideal containing I and p = P/I, one has:

(a) ra.nk(dfi/dxj)(P)

< ht(/ P ).

(b) If r a n k ( d f i / d x j ) ( P )

= ht(/p), then Bf is a regular ring.

(c) // k is a perfect field and Bf is regular, then

= ht(/p). Proof. See [217, p. 213].

D

Lemma 3.5.6 Let R be a ring and I an ideal. If I is height unmixed and P is a prime ideal such that I C P, then IP is also height unmixed and ht(/) = ht(/p).

Proof. Let be a primary decomposition of J, we may assume y/q7 C P for i = 1 , . . . s. Localizing at P gives a primary decomposition s

IRp = p) q,RP. Hence the associated primes of IP are y/qT-Rp, where i < s. To finish the proof note ht(y/q7-Rp) = ht(y/q7) = ht(/) for i < s. O

92_______________________________________ Chapter 3

Definition 3.5.7 Let k be a field and / = (/i , . . . , /g) an ideal of height g of k[xi, . . . ,xn}. The Jacobian ideal J of / is the ideal generated by the g x g minors of the Jacobian matrix 3 = Corollary 3.5.8 Let R = k[xi , . . . , xn] be a polynomial ring over a perfect field k and I = ( f \ , . . . , /?) C R an unmixed ideal. If P is a prime ideal of R containing I and p = P/I, then (R/I)f, is regular if and only if J /I fjL p, where J is the Jacobian ideal of I.

Proof. =>) By Theorem 3.5.5 and Lemma 3.5.6 one has

hence there is / e J \ P and J/I <£ P/I. <=) If ( R / I ) f is not regular, then by Theorem 3.5.5 one has the inequality Tank(dfi/dxj)(P) < ht(I), hence J/P = (0) and J/I C P/I, which is impossible. Note that this part of the proof is valid for any field k.

D

Proposition 3.5.9 Let R be a polynomial ring over a field k and I an unmixed ideal of height g. If J is the Jacobian ideal of I and ht ( J, /) > g + 2, then (R/I}9 is regular for any prime p of R/I such that ht (p) < 1. Proof. Let B = R/I and p = P/I, where P is a prime ideal of R. If Bf is not regular, then according to Corollary 3.5.8 one has J/I C p and

hence (J, /) C P. Since R is a catenary ring and / is unmixed one obtains ht(P) = g if ht(p) = 0 and ht (P) = g + I if ht (p) = 1. Therefore ht ( J, I) < ht (P) < g + 1, which is impossible. Thus Bf must be regular, as required. D Another consequence of the Jacobian criterion is the following radical test of [294]. Let k be a field and / = (/i, . . . , / 7 ) C R = k [ x i , . . . ,xn] an ideal of height g. Let J be the Jacobian ideal of / generated by the g x g minors of the Jacobian matrix 3 = Theorem 3.5.10 (Vasconcelos) If k is a perfect field and I is unmixed, then I is a radical ideal if and only if there is an element f in J \ I such that (I: f) = I.

Proof. =>) Let PI, . . . ,Pr be the associated primes of /. If J C Z(R/I), then J C Pi for some i because Z(R/I) = U^P,. Hence by Corollary 3.5.8 one derives that (R/I)pi is not a regular ring, which is a contradiction because (R/I)p, = RpJPiRpi is a field. •4=) Let I = qi fl • • • D q r be an irredundant primary decomposition of /

and Pj = ly/cjT an associated prime ideal of /. Note (R/ I)pi ~ (R/I)Vi and Ipi = (qi)p; , where pj = Pj/ 1. First we observe that (R/I)9i is a regular ring; otherwise by Corollary 3.5.8 one has J/I C Pi/I, hence / 6 Pj, which

Rees Algebras and Normality is impossible because / is regular modulo /. Therefore (Rf ' I)pi is a regular local ring and thus an integral domain by Theorem 1.4.16. As Ipi = ((\i)pi

must be prime, one concludes that q, is also prime and consequently q^ = Pi. Note that here the hypothesis k perfect is not being used. d Example 3.5.11 Let R = k[xi, . . . ,xj] and / = (x\ XT — where A: is a field. We now show that B = R/ 1 is a normal ring using Theorem 1.4.15. The Jacobian matrix of / is:

^ _ ( x^ -xg \ 0 0

0 x5

0 0 -x? -x^ x3 0

x\ —X4

Note that the monomials x7, X5x6, £ 3 £ 6 , X^XQ, X2x5, Xix5, X2x4

are in the Jacobian ideal J of /. Hence ht (J,I) > g+2, where g = ht (/) = 2. By Proposition 3.5.9 one has that B satisfies (Ri) and since / is a complete intersection it follows by Proposition 2.2.17 that B satisfies (52). Thus by Serre's normality criterion the ring B is normal. Lemma 3.5.12 Let R be a ring and I a prime ideal. If f G R \ / and Ij.n> = /" for some n > 1, then

Proof. If z £ (7":/ 00 ), then z f k 6 /" for some k > I , hence z 6 I(n) . Conversely if z e I^n\ then sz £ In for some s $ I. Since s/1 ^ // and ( s / l ) ( z / l ) £ 7™, then z/l is in the nth symbolic power of If. hypothesis z/l G 7" and this means that z € (7™: /°°).

Thus by D

There are a few methods to compute symbolic powers of prime ideals in polynomial rings, see [257] and [298, Chapter 3]. The following is a subtle

application of the Jacobian criterion due to Wolmer Vasconcelos. Proposition 3.5.13 Let R be a polynomial ring over a field k and 7 a prime ideal. If / is an element in the Jacobian ideal J of I which is not in the ideal 7, then j(n)

=

(jn : /°o)

f o r n > 1,

and such element / exist if k is a perfect field. Proof. Since the localization Rf is Cohen-Macaulay, by Proposition 3.3.29 and Lemma 3.5.12 it is enough to prove that If is locally a complete inter-

section. Let P be a prime ideal such that 7 c P and / 0 P. Set p = P/I. One has

(Rf/If)p}

~ (Rf)pt/(If)p,

- Rp/Ip ^

94_______________________________________ Chapter 3

As J/I $_ p, using Corollary 3.5.8 we conclude that Rp/Ip is a regular ring and consequently (If)pf is a complete intersection (see Proposition 3.5.3). The last assertion follows from Theorem 3.5.10. D Example 3.5.14 Let R = Q[z,y,z] and

/ = (x3 -yz,y2 -xz,z2 -x2y). The ideal / is prime because it is the kernel of the homomorphism:

:t$x,y,z] — KQ[*], where
where / = 2y2 + xz and A = -x5 + 3x2yz - xy3 - z3. See [199].

Exercises 3.5.15 Use the following procedure in Co Co A [60] to verify the formula for the second symbolic power of Example 3.5.14 Fl := x~3 — y * z; - - variables must begin with capital letters F2 := y~2 - x * z; F3 := z^2 - x2 * y; Jac := Jacobian([Fl,F2,F3]); J := Minors(2,Mat(Jac)); - - 2 x 2 minors of the Jacobian matrix I := Ideal(Fl,F2,F3);

H := Elim(t, T2 + Ideal(l - t(2 * y"2 + x * z))); Print(H);

3.5.16 Let R be a Cohen-Macaulay ring and / an ideal of R which is height unmixed. If / is a complete intersection, then / is locally a complete intersection.

Hint Use Lemma 1.3.18. 3.5.17 Let R = Q[x,y,z] and / the prime ideal generated by the set of binomials /i = x3 - yz, /2 = y2 - xz, /3 = z 2 - x2y. Prove:

(a) Ix = (/i,/2), that is, Ix is a complete intersection, and (b) /(") = (In:x°°) for n > 1.

Rees Algebras and Normality________________________95

3.5.18 Let R be a Cohen-Macaulay ring and / a prime ideal. If / g R \ I and // is a complete intersection, then /(") = (I":/00) for n> 1.

3.5.19 Let R = Q[XI ,..., x7] and let / be the principal ideal generated by the binomial / = X\X3X^X"j

— X2X^Xg.

Prove that R/I is a normal domain. Give a description of the irreducible binomials / such that R / ( f ) is normal. 3.5.20 Let R be a Noetherian ring and a an element in the Jacobson radical of R. If a is regular and R/(a) is an integral domain, prove that R is an integral domain. Recall that the Jacobson radical of R is the intersection of all the maximal ideals of R.

3.5.21 Let / = I i ( X } be the ideal of 2 x 2-minors of the symmetric matrix: X =

EI

x-2

-2

X4

x3 £5

3

X5

XQ

where the entries of X are indeterminates over a field k. Prove that IXl is a complete intersection. Find a set of generators for 1^ and prove that depth(Hi) = 1, where HI is the first Koszul homology module of /.

Chapter 4

Hilbert Series Hilbert series of graded modules and algebras are introduced and studied in this chapter. The /i-vector and a-invariant of a graded algebra S are defined through the Hilbert-Serre's theorem, when 5 is Cohen-Macaulay we

present some of their properties. Some characteristics of Cohen-Macaulay and Gorenstein extremal algebras will be presented.

4.1

Hilbert-Serre's Theorem

Unless otherwise stated we shall always assume that modules are finitely generated and N-graded. Let R = k[xi, • • • , xn] be a polynomial ring over a field k with the grading induced by deg(xi) = di, where di is a positive integer. If

i=0

is a finitely generated N-graded module over R, its Hilbert function and Hilbert series are defined by 00

H(M,i) = l(Mi) and F(M,t) = ^H(M,i)ti i=0

respectively, where l(Mi) denotes the length of Mi as a /c-module, in our

case (.(Mi) = dim fc (Mj). If d 6 N, then M(-d) is the regrading of M obtained by a shift of the graduation of M, more precisely

97

98_______________________________________ Chapter 4 where M(—d)i = M_d+i. Note that we are assuming Mi = 0 for i < 0. In this way M(—d) becomes an N-graded fl-module. Lemma 4. 1.1 F(M(-d),t)

= tdF(M, t).

Proof. Since M(—d}i = Mj_<; one has: 00

d

F(M(-d),t) = t ^l(Mi.,d)ti-d =tdF(M,t), i=d

where the first equality follows using that M;_d = 0 f o r i = 0 , . . . , d - l . D

Lemma 4.1.2 Let M be a graded R-module. If z € Rd, then there is a degree preserving exact sequence

0 —> (M/(0: z ) ) ( - d ) -^ M -^ M/zM —> 0 ((p(m) =m + zM),

where (0: z) = {m e M\ zm = 0} and the first map is multiplication by z. Proof. As the map ty: M(—d) —>• M, given by i/j(m) = zm, is a degree zero homomorphism one has that (Q:z)(—d) is a graded submodule of M(—d). The exactness of the sequence above follows because -0 induces an exact sequence

0 — » (0: z)(-d) -^ M(-d) A M -^ M/zM —> 0, where i is an inclusion.

D

Theorem 4.1.3 (Hilbert-Serre) Let k be a field and R = k[xi, . . . ,xn] a

polynomial ring graded by deg(xi) = di £ N+ . // M is a finitely generated 'N-graded R-module, then the Hilbert series of M is a rational function that can be written as F(M,

t) = „

——

for some h(t) e Z[t].

1=1

In particular if di = 1 for all i, then there is a unique polynomial h(t) 6 Z[i]

such that and h(l) / 0. Proof. If n = 0, then

(

m

\

m

ff^ Mi I = yjdinis;(Mi) < dim^M) < oo, for all m. i=o / <=o

Hilbert Series____________________________________99

Hence the Hilbert function of M is zero for i 3> 0 and F(M, t) is a polynomial. If n > 0, consider the exact sequence of graded modules

0 —» (Q:xn)(-dn) —> M(-dn) -^ M —> M/xnM —> 0. Since the ends of this exact sequence are finitely generated modules over the polynomial ring k[xi,... ,xn-i] the proof follows readily by induction on n. D

Remark 4.1.4 The conclusion of the Hilbert-Serre theorem is valid if we assume that R = k[xi,..., xn] is an N-graded algebra over an Artinian ring k. See [9, Chapter 11] for details. The integer d in the Hilbert-Serre's theorem will be denoted by d(M); this integer will turn out to be equal to the Krull dimension of M, see Proposition 4.1.8 and its proof. Definition 4.1.5 The degree of F(M,t) as a rational function is denoted by a ( M ) , it is called the a-invariant of M. The Hilbert polynomial One interesting invariant of a standard algebra is its degree or multiplicity, here we introduce multiplicities via Hilbert polynomials and explain how the a-invariant is useful to measure the difference

between the Hilbert polynomial and the Hilbert function. For the rest of this section we shall always assume that R is a polynomial ring over a field k with the usual grading.

Lemma 4.1.6

Let n > 1 be an integer, then

Proof. As the case n = 1 is clear, by induction on n one has:

i

r i v i n

n(l~t) \ CO

00

n n

n Using the identity

m + 1 (m + n\ n \m + IJ the induction process is complete.

fm + n \ m D

100

Chapter 4

Observe that Lemma 4.1.6 also follows rapidly from the classical binoexpansion: k>0

Indeed recall that if r 6 ffi and k e N, then

Using the binomial expansion for (1 — t)~n one has

1

n m l+ l + v r-ir l l 2m (m +n-1 -1 (~ ] t -1 + v f-i) _f n+2^( > (mjt 2^( > ( m (1 ) 1

\

/

°°

/

\

°°

/

^f,—1

\

/

m— 1

\

,

1

Lemma 4.1.7 If R is a polynomial ring in n variables with coefficients a field k, then

in

and

>=

Proof. Let Xi be one of the variables. From the exact sequence

we obtain

F(R, t) = F(R(-l),t) + F(R/(x1),t) Hence the formula for the Hilbert series follows by induction on n. To show

the second equality use Lemma 4.1.6.

D

Proposition 4.1.8 Let R be a polynomial ring over a field k with the usual grading, and M a graded module over R. Then there are integers

a-d,..., a_i so that j_ -i /l +d

~'1~1^,Vi> o(M)

where d = d(M) is the dimension of M and a(M) is the degree of F(M, t) as a rational function. Proof. By the Hilbert-Serre's Theorem there is a polynomial h(i) 6 Z[i] such that

Hilbert Series

101

and h(l) ^ 0. If d = 0, then a(M) is equal to deg(/i) and H(M,i) = 0 for i > a(M) + 1. Thus one may assume d > 0. Observe that by the division algorithm we can find e(t) e Z[t] so that the Laurent expansion of

F(M,t) — e(t), in negative powers of (1 — t), is equal to

j=o

v

'

Next we expand (1 — t)d~:' in powers of t to obtain oo

i-l

v—r

=e(t)

a 3=0

-
i

-j- 1 i=0

observe that (p(i) = H(M,i) for i > deg e(i) + l, where the degree of the zero polynomial is set equal to — 1. To complete the proof note that a_d, . . . , a_i

are integers by Lemma 1.2.10, and that d = d(M) is the dimension of M D

by Theorem 1.2.8.

Corollary 4.1.9 If M is a graded R-module of dimension d, then there is a unique polynomial a(M) + l. Proof. Is an immediate consequence of Proposition 4.1.8.

D

Definition 4.1.10 The polynomial

¥>M W = cd-itd~l + • • • + cit + c0 is the Hilbert polynomial of M and the integer Cd-i(d — 1)! is the degree or multiplicity of M and is denoted by e(M}. If
where h(t) is the polynomial F(M,t)(l - t)d. If d = 0, then =i(M) =dim f t (M). Corollary 4.1.12 Let M be a graded R-module. Set

r0 = min{r 6 N\H(M,i) = VM(«)> V i > r}. Tften TO = 0 if a(M) < 0 anrf TO = a(M) + 1 otherwise. Corollary 4.1.13 Let S = R/I be a standard graded algebra of dimension d over a field k. If S is Cohen-Macaulay and h = {hi, . . . , h^} a system of parameters of S consisting of linear forms, then

e(S) = i(S/hS).

Chapter 4

102 Proof. By Proposition 2.2.7

{hi,..., hd} is a regular sequence. Thus there

are exact sequences of k-vector spaces 0

0.

Therefore one has the relation s(i) ~ fs(i — 1) between Hilbert

polynomials for i > 0, where 5 = S/h\S. Let

Now,

observe that thanks to <j>s(i) = fs(i) — fs(i — 1) one derives:

_ (a — 1).

consequently e(S)

= e(5).

,lower order terms,

Since 5 is Cohen-Macaulay of dimension d - 1,

see Lemma 1.3.10, the result follows by induction.

D

Proposition 4.1.14 If A and B are two standard algebras over a field k, then

dim(A ®k B) - dim(A) + dim(5). Proof. Since r=0

where

one has the next equality between Hilbert series:

F(A,t)F(B,t)

= .i=0

.i=0

z=0

Thus Theorem 4.1.3

yields the asserted equality.

D

Making use of the Noether normalization lemma it is not difficult to prove that the result above holds for arbitrary affine algebras over a field k.

Hilbert Series___________________________________ 103

Computation of Hilbert series Let R be a polynomial ring over a field k and / C R a graded ideal. Since R/I and 7?/in(7) have the same Hilbert function (see Corollary 2.4.13), the actual computation of the Hilbert series of R/I is a two step process:

• first one finds a Grobner basis o f / using Buchberger's algorithm, and

• second one computes the Hilbert series of R/ in(7) using elimination of variables (see [16, 21]). Another approaches to the computation of Hilbert series are through minimal resolutions and Stanley decompositions (see [276]). An alternative method to compute Hilbert series of Cohen-Macaulay N-graded algebras is given by Proposition 2.2.14. Some exemplifications are given below. Example 4.1.15 Let R = k [ x i , X 2 , x 3 ] be a polynomial ring and T — 1 -* ~ x

Let us compute the Hilbert series of R/I using elimination. Pick any monomial involving more than one variable, say x\x%. The idea is to eliminate x\ from the monomials containing more than one variable. From the exact sequence of graded modules:

0 — >• R/(xlx3,x2)(-2) 4- R/I-+R/(xl,x*) — > 0, and applying Exercise 4.1.21, we get

, a-* 2 ) 2 _ -t 4 +2t 2 + 2 *+i

Example 4.1.16 Let R - Q[x,y,z,w] a n d / = (/i,/ 2 ,/ 3 ), where /i = V2 - xz, /2 = x3 - yzw, /3 = x2y - z2w.

Using Macaulay one finds that the minimal resolution of R/I is: 0 —> R2(-4) -^» R(-2) 0 J^ 2 (-3) -^ R —> R/I —> 0. Let us compute the Hilbert series of R/I using its minimal resolution: i2

1

~

(1 — t)

4

(1 — i)

2t 3 4

(1 ~ ^) 4

1.04______________________________________ Chapter 4 Example 4.1.17 Let R = k[x\, . . . , x 5 ] and / = (xiX-2,x3X4X5). / is a complete intersection, hence by Exercise 4.1.21 one has

Note that

The Laurent expansion of F(R/I,t) + 1, in negative powers of (1 — t) is

5t2 - t + 2 _

(i-*)

3

=

q_ 3

+

0-2

+

Q-I

(T^F (T^ (T^) '

where o_i = 5, a_2 = —9, a_s = 6. Hence the Hilbert polynomial of S = R/I is

Exercises 4.1.18 Let k be a field and R — k[xi,... ,xn] a polynomial ring graded by deg(a;j) = di e N+. Then the Hilbert series of R is

F(R,t) = ^—^———.

4.1.19 Let / i , . . . , fg be a regular sequence of forms of degree p in a polynomial ring R in n variables and / = ( / i , . . . , / ff ). If 0 < i < p show:

TT, .. H(p + i) H(R/I,p + i)

(i + n — 1\

= g( \ n — J.

and

-

4.1.20 Let J? = k [ x , y , z ] be a polynomial ring over a field A; and

/ = (xa,yb,zc,xaiyblzci).

If a > ai, b > bi and c > ci, then

e(R/I) = dimk(R/I) - abci + (c - Ci)(oi6 + (a - ai)6i). 4.1.21 Let ^? be a polynomial ring in n variables over a field and fi,... , f r a homogeneous regular sequence in R. If a» = deg(/j), then

(1

Hilbert Series___________________________________ 105

4.1.22 If R = k[xo, • • • , xn] is a polynomial ring over a field k and

B = {x\Xj - Xix] 0 < i < j < n}, then B is a Grobner basis for / = (B) w.r.t the reverse lex ordering. 4.1.23 In Exercise 4.1.22 show that the a-invariant of R/I is n(q — 1) by

proving that the Hilbert series F(t) — F(R/I, t) can be written as: F(t]

w + • • • + ___ wn^) —————————— + wn~l (w + ti) _ —————————— i

where w = 1 + t + - - - + tq~l.

4.1.24 Let R = k[xi,X2,xs] be a polynomial ring and

/ = (xl,x%x3,xl). Prove that the a-invariant of R/I is 3 and H(R/I, i) = 4 for i > 4. 4.1.25 Let A, B be two standard algebras over a field k and define their

Segre product as the graded algebra

S = A ®s B = (A0 ®k B0) ffi (Ai ®jt Bi) ® • • • C A ® fc B, where

(A®kB)p= i+3=P

Use Hilbert functions to prove

dim(S) = dim(A) + dim(B) - 1.

Note dim(Ai ®fc Bi) 4.1.26 Use the following procedure in CoCoA [60] to verify the formulas for the Hilbert series and Hilbert polynomial found in Example 4.1.17 Use R : : = Q[x[1..5]], Lex; Q := Ideal(x[l] * x[2], x[3] * x[4] * x[5]); Poincare(R/Q); Hilbert(R/Q);

106______________________________________ Chapter 4

4.2

a-invariants and h- vectors

Our goal here is to relate the a-invariant of a Cohen-Macaulay N-graded algebra with its minimal resolution and to prove that /i-vectors of such algebras are non negative. We include Stanley's characterization of CohenMacaulay algebras in terms of Hilbert series. We begin with a general inequality relating the a-invariant of a graded algebra with the shifts of its minimal resolution. Proposition 4.2.1 Let R = k[x\, . . . ,xn] be a positively graded polynomial ring over a field k and I a graded ideal. If

F* : 0 ->

R(-dgi) v-4 • • • -+

R(-dki] £ * . . . - + 0

R(-du) ^ R

is the minimal resolution of S = R/I, then a(S) < maxfdfcal 1 < k < g and 1 < i < bk} + a(R).

Proof. From the resolution of 5 one can write the Hilbert series of 5 as:

F(S,t) = 8=1

hence a(5) < maxjd^j} + a(R).

D

Definition 4.2.2 Let R = k [ x i , . . . ,xn] be a positively graded polynomial ring over a field k and / a Cohen-Macaulay graded ideal of height g. The canonical module of 5 = R/I is defined as

where U>R = R(—S) and 5 = J^™=1 Proposition 4.2.3 Let R = k [ x i , . .. ,xn] be a positively graded polynomial ring over a field k and I a Cohen-Macaulay graded ideal. If F* : 0 ^ 0 R(-dgi) * * . . . - > 0 R(-dki) ^ . . . -> 0 R(-dlt] ^ R is the minimal resolution of S = R/I, then maxjjdii} < • • • < and the a-invariant of S is given by

a(S) = max {cL,} + a(R) = -minii I (ws)j ^ 0}. l
Hilbert Series

107

Proof. One may order the shifts as d^i < • • • < dkbk for k = 1 , . . . , g. From the resolution above one can express the Hilbert series of S as:

In order to make use of this formula we first need to prove the assertion about the behaviour of the shifts d^iSince S is Cohen-Macaulay, the height of / is equal g, this follows from Corollary 2.5.14. Recall that by Proposition 2.2.19 there exists a regular sequence /i, . . . , fg inside /, hence using Proposition 1.3.7 yields:

Ext^(S,uR) = 0 for i < g.

Set S = X^r=i deg(xj) and UJR = R(-6). Therefore dualizing F*, with respect to the canonical module of R, one obtains the (exact) complex: Hom(F*,w fl ) : 0 ->• Rom(R,uR) -> • • • ^ Horn

Q)R(-dgi),uR

-» 0.

Observe

Horn

R(-dgi),R(-S)

~

Horn

R(-dgi),R

(-6)

Since Hom(F*,WR) is a minimal resolution of Ext^ (5, LOR), by Remark 2.5.9, one has du,1 < c?262 < • • • < dgbg . As a consequence, from the expression for F(S,t) given above, one concludes the equality a(5) = dgt,g + a(R). In particular one has a surjection ,^(-<5)) = us -)• 0.

To complete the proof note that since <5 — dsi9 < • • • < S — dgi, one has the equalities min{z| (w s )i ^ 0} = 6 - dgbg = -a(S). D

108______________________________________ Chapter 4

h- vectors Next we introduce the notion of h- vector of standard algebras which is based on the knowledge of the Hilbert-Serre theorem for Hilbert series (see Theorem 4.1.3). Definition 4.2.4 Let S be a standard algebra and

h(t) = h0 + hit + • • • + hrtr the (unique) polynomial with integral coefficients such that h(l) ^ 0 and satisfying

where d = dim(S). The h-vector of S is defined by h(S) = (ho, . . . , hr). Theorem 4.2.5 ([269]) Let S be a standard algebra and 9i,. . . ,0d a homogeneous system of parameters for S with Oj = deg(0j). Let A be the quotient ring S/(9\, . . . , 9 d ) . Then S is Cohen-Macaulay if and only if

F(S,t) =

'

.

(4.2)

1=1

Proof. =>) Assume 5 is Cohen-Macaulay. It is enough to prove the equality

F(S/(6l,...,e.),t) = for 1 < s < d. We proceed by induction on s. Assume s = 1. From the exact sequence of graded 5-modules

0 —¥ 5[-oi] A 5 —-> S/(0i) —> 0 we obtain

Since oo

F(S[-ai],f)

= t=0

i= oo

—

~

+01 V^ c. >

|i-ai

Oj_ail

taiF(S,t)

Hilbert Series

109

we conclude which proves the case s = I. Next assume s > 1 and that the equality (4.3) is true for s — I. From the exact sequence

0 ->• (S/(01, ..., 0 S _!)) [-a.] -

S/(0lt. . . , 0 g _i) —> S/(0i, . . . , 6.) -> 0

we get

F(S/(0i, . . . A), t] = (1 - ta')F(S/(6lt. . . , 0,_i), t), and the induction hypothesis yields the required equality.

•4=) Let /(i) = X^i>o fli ^ and ffW ~ St>o ^* k e two f°rmal power series, we say that f ( t ) > g(t) if a, > 6j for all i. From the exact sequence 0 —> (5/ann (
5 —> 5/(0i) —> 0

we obtain

Hence the power series F(S/(9i),t) is greater or equal than (1 -tai)F(S,t), and by induction it follows that r

F(S/(61,...er),t) > n(l-tai)F(S,t)

for 1
Set S' = S/(9i, . . . ,6d-i). Using the exact sequence 0 -^ (5'/ann (6d))[-ad] ^ S' -+ S'/(8d) =

we get

F ( S ' / ( 9 d ) , t ) = (1 - trdF(S', t) + *a' On the other hand using the hypothesis we have d

1=1

Therefore

- t)a*(F(S',t) >o

>o

= -tadF(aim(Od),t), and jF(ann(^d),t) = 0, which shows that 64 is a regular element on 5'. By induction it follows readily that 6\ , . . . , 9d is a regular sequence. This proof was adapted from [269]. D

110______________________________________ Chapter 4

Theorem 4.2.6 Let S be a Cohen-Macaulay standard algebra over a field k and h(S) = (hi) the h-vector of S, then /i, > 0 for all i.

Proof. One may assume k infinite, otherwise one may change the coefficient field k using the functor (•) ®k K, where K is an infinite field extension of k. There is a h.s.o.p y = {j/i, . . . , yd} for S, where each ?/j is a form in S of

degree one. Set 5 = 5/(y)5. From Theorem 4.2.5 we derive

for all i and consequently hi > 0.

D

Exercises 4.2.7 Let R be a polynomial ring over a field k with the usual grading and M a finitely generated N-graded module. If M is Cohen-Macaulay and h(t] = ho + hit + • • • + hrtr the polynomial in Z[i] so that h(l) ^ 0 and

F(M,t) = n _ \ d , where d = d i m ( M ) . Prove that hi > 0 for all i.

4.3

Extremal algebras

The concept of an extremal Cohen-Macaulay or Gorenstein standard algebra (resp. ideal) came up in the works of J. Sally [251] and P. Schenzel [252]. Those algebras (resp. ideals) have the smallest possible reduction number (resp. smallest possible number of generators of least degree). We will study with certain detail those two distinguished classes of algebras.

The Cohen-Macaulay case Let R be a polynomial ring over a field k with its usual grading and

i—p

a graded ideal with Ip ^ (0) . The integer p is the initial degree of J. As usual v ( I ) denotes the minimal number of generators of /, and v(Ip) stands for the minimal number of generators of / in degree p. The next result, when applied to a Cohen-Macaulay ideal / generated by forms of the same degree, provides a bound for the minimum number of generators of /.

Hilbert Series___________________________________ 111

Proposition 4.3.1 Let R be a polynomial ring over a field k and I a graded ideal in R of height g and initial degree p. If I is Cohen-Macaulay, then

Proof. We may assume that k is infinite, since a change of the coefficient field can be easily carried out using the functor (•) ®/c K , where K is an infinite field extension of k. This change of the coefficient field preserves the hypothesis on / and leaves both the height of I and the dimensions of the vector spaces of forms of a given degree in / unchanged. We set 5 — R/I. By Lemma 2.2.15, there is a homogeneous system of parameters y = {yi, . . . , y d } for 5, where each j/j is a linear form in R. If we tensor the minimal resolution of 5 with R = R/(y) it follows that

where S = S/(y)S. To get the desired inequality it suffices to observe that R is a polynomial ring in g variables over the field k.

D

Definition 4.3.2 If S is an Artinian positively graded algebra the socle of S is given by

Soc(S) = (0:s 5+). Lemma 4.3.3 ([118]) Let S = R/I be a quotient ring of a polynomial ring R over a field k modulo a graded ideal I. Let

be the minimal resolution of S. If S is Artinian, then there is a degree zero isomorphism of graded k-vector spaces:

1=1 Proof. Let R = k[xi, . . . , xg] and x_ — {xj, . . . ,xg}. The ordinary Koszul complex K* associated to the regular sequence x is an .R-free resolution of k — R/(x) and thus

On the other hand

112______________________________________ Chapter 4

(see the proof of Corollary 2.5.7). To complete the argument note

Hg(K* ®R S)[g] ~ Za(K* ®R S)[g] a

=

< ze\ A • • • A e

~

Soc(S).

A • • • A BJ A • • • A eg = 0 \ [g]

Altogether one has l>9

Soc(S)~Q)k[g-dgi], as required.

D

Proposition 4.3.4 If S a Cohen-Macaulay positively graded k-algebra over a field k with presentation R/I, then the type of S is equal to the last Betti number in the minimal free resolution of R/I as an R-module.

Proof. By making a change of coefficients using the functor (•) (gi^ K, where K is an infinite field extension of k, one may assume that k is infinite. By Theorem 2.2.11 there exists a system of parameters y = {yi, . . . , yd} for 5 = R/I with each j/j a form of degree one of R. Since Torj(5,-R/(y)) = 0 for i > 1, the minimal resolution of S/(yS) as an /?/(y)-module has the same twists and Betti numbers as the minimal resolution of 5 over R. Since is Artinian the result follows from Lemma 4.3.3. D Corollary 4.3.5 // R is a polynomial ring over a field k, then a graded ideal I of R is Gorenstein iff R/I is Cohen-Macaulay and the last Betti number in the minimal graded resolution of R/I is equal to 1 . Corollary 4.3.6 Let S be a Cohen-Macaulay positively graded k-algebra over a field k and u>s its canonical module. Then

type(S) = where v(u>s) *s the minimum number of generators ofuisProof. Use the proof of Proposition 4.2.3 and Proposition 4.3.4.

D

Theorem 4.3.7 ([61]) Let R be a polynomial ring over a field k and I a graded ideal in R of height g and initial degree p. If S = R/I is a CohenMacaulay ring, then S has a p-linear resolution if and only if the following equality holds ~

(4.4)

Hilbert Series

113

Proof. Let 61

R(-dgi) -»• ' ' ' ->• 0 ^(-diO -»• fl -> ^ = fl/I -»• 0,

(4.5)

be the minimal resolution of S. We order the shifts so that dki < • • • < dkbk for A; = 1, . . . , g; by the minimality of the resolution d\\ < cfei < • • • < dg\ . Observe that if the resolution of S is linear then the Herzog-Kiihl formulas of Theorem 2.5.16 give the required equality. Conversely assume the equality above. Since we may assume that k is infinite, there is a h.s.o.p y — {yi, . . . ,yd] for S, where each j/j is a form of degree one in R. Set 5 = 5/(y)5 and R = fl/(y). From Theorem 4.2.5 we derive that the /i-vector of 5 satisfies hp = 0, and also get /i, = H(S,i). Therefore hi = 0 for all i > p. Using Lemma 4.3.3 one obtains a degree zero isomorphism

of graded A:-vector spaces, in particular the socle of 5 can only live in degrees dgi — g, hence we conclude the inequality dgi — g < p — 1. On the other hand the minimality of the resolution of 5 gives p + (g — 1) < dgi for

all i. Altogether we have dgi = p + g - 1. Because / is C-M, then by Proposition 4.2.3 we must have p < dn>1 < G?2&2 < • • • < dgbg = p + g - 1, which implies that the resolution is p- linear, as required. D Example 4.3.8 ([237]) Let R = k[a, . . . , / ] be a polynomial ring over a

field k and / = (abc, abd, ace, adf, aef, bcf, bde, bef, cde, cdf).

Then R/I is C-M if char(fc) ^ 2 and non C-M otherwise. Thus in the first case R/I has a 3-linear resolution. Other examples of Cohen-Macaulay algebras with linear resolutions include rings of minimal multiplicity [251], the coordinate ring of a variety defined by the submaximal minors of a generic symmetric matrix [194], the coordinate ring of a variety defined by the maximal minors of a generic matrix [89], and some face rings [114]. Cohen-Macaulay rings with linear resolutions have been studied by Sally [251] for the case p = 2, and by Schenzel [252] for the general case; more general rings with linear resolutions have been examined in [93, 155].

114______________________________________ Chapter 4

The Gorenstein case To give a sharp bound for the number of generators in least degree of a graded Gorenstein ideal is harder than the Cohen-Macaulay case, we deal with this problem in the next section. The aim here is to introduce and characterize extremal Gorenstein rings in terms of their minimal resolutions and to determine their Betti numbers. Those Betti numbers are our "natural candidates" to bound the initial Betti numbers of graded Gorenstein ideals. Proposition 4.3.9 If S = SQ + • • • + Ss is a positively graded Artinian Gorenstein algebra over a field k, then the multiplication maps

Si x Ss-i -» Ss = Soc(S) ~ k are a perfect pairing, that is, the homomorphisms:

(fi'.Si^ Eomk(Ss-i,Ss) ~ Ss-i,

(pi(r)(x) = rx

are isomorphisms of k-vector spaces. Proof. To show that ) and assume r ^ 0 and i < s. Since r is not in the socle of 5, one may pick z € S homogeneous of maximal positive degree such that zr ^ 0. Note that 0 < deg(^) < s — i, because rx = 0 for x g Ss-i- Hence zr has degree less than s, using that the socle of S lives only in degree s one obtains a homogeneous element w with deg(w;) > 0 such that wzr = 0, which contradicts the choice of z. Therefore r = 0 and

Proposition 4.3.11 Let R be a polynomial ring over a field k with the standard grading and I a graded Gorenstein ideal of initial degree p and height g. If the minimal resolution of S = R/ 1 by free R-modules is: F* : 0 -

then ba — k

bk

R(-(dg - d(g_-k]i)} ~ 0 R(-dki) for 1 < k < g - 1.

(4.6)

Hilbert Series___________________________________ 115

Proof. One may always order the shifts so that dki < • • • < dkbk for k = l,...,g. As bg = 1 and dgi = dg, by Proposition 4.2.3 one has that the a-invariant of 5 is a(S) = dg — n, where n is the number of variables of R. Since Hom(F*,WR) is a minimal resolution of

and using

LUS — S(dg - n), one obtains that F* is self dual.

D

With the same assumptions and notation of Proposition 4.3.11 one has:

Proposition 4.3.12 Let h(S) = (h0,...,hs) be the h-vector of S. Then s •> 2(p — 1) with equality if and only if the minimal resolution of S is of the form

0 -> R(-(2p + g - 2)) -)• Rb'-i (-(p + g - 2)) -> • • • - » Rb2(-(p+l)) ^Rbl(-p) -+ R. Proof. As in the proof of Theorem 4.2.6 one may assume S Artinian and hi equal to H(S,i). Note that the socle of 5 lives in degree dg — g by Lemma 4.3.3, hence s > dg - g. By Proposition 4.3.11 and the minimality of the resolution one concludes

s > dg - g - dibl + d(g_i)i - g > p - g + d( 5 -i)i > 2(p - 1), where the last inequality uses dk\ > p + k — 1 for 1 < k < g — I . For the second assertion use the chain of inequalities above and compute the ft-vector of S using its resolution. n Definition 4.3.13 If s = 2(p — 1), the ring S is called an extremal Gorenstein ring and its resolution is called pure and almost linear. Proposition 4.3.14 ([252]) Let R be a polynomial ring over a field k and I a graded ideal of initial degree p and height g. If S = R/I is an extremal Gorenstein algebra, then for 1 < i < g the ith Betti number of S is given by

p-l

i

p-l

116______________________________________ Chapter 4

Proof. As in the proof of Theorem 4.3.7 one may assume S Artinian and R a polynomial ring in g variables. Set s = 1(p — 1). From the minimal

resolution of S ( B

\

5-1

/

i=l

£ hj (i - zY = i + ]T(

\i=0

where /ij = H(S,i) = /i s _j. Therefore p+m-l

Using the identity

m-l

\m-l

v

;

and the symmetry of the /i-vector of S one has

Note p+m-l

i=p m—

+

since the terms in the right hand side are zero for i > p - 1 the last summation reduces to

g \ m. -

i=0

'0-1 + A ,=Q

^-m+p-l-i)\ m

J\

p-1

J

\mj\

i

J

fg\fg + p-2 \mj \ p-1

p-1 J'

the last equality follows from Eq. (4.7) making m equal to g — m.

D

Hilbert Series

117

Exercises 4.3.15 Let -R be a polynomial ring over a field k and / a Cohen-Macaulay graded ideal in R of height g and initial degree p. Let (ho, . . . , hs) be the h- vector of R/ 1. Prove that s > p — 1, with equality if and only if the minimal resolution of R/I is p-linear. If s — p — 1 the algebra R/I is called an extremal Cohen-Macaulay algebra.

4.3.16 Let R be a polynomial ring over a field k and / a C-M graded ideal in R of height g and initial degree p. Prove that the Hilbert series of S = R/I can be written as

F(S,t) = ^—.———dr-j ——— , where d = dim S, (l-t) if and only if 5* has a p-linear resolution. 4.3.17 Let f? be a polynomial ring over a field k and / a Cohen-Macaulay graded ideal in R of height g and initial degree p. If 5 has a linear resolution, show that the Betti numbers of S = R/I and its multiplicity are given by

bk =

fp + k-2\ (p + g-l\ , , • , and e(S) = \ k - l j \ g - k j

4.3.18 Let 5 = R/I be a standard Gorenstein algebra so that / has initial degree p and codimension g. Show that the multiplicity of S satisfies:

g with equality if and only if 5 is an extremal algebra. Hint The h- vector h(S) — (h0,. . . ,hs) is symmetric and s > 2(p - 1). Then compute explicitly p-2

«=o

4.3.19 Let R be a polynomial ring over a field k and / a graded Gorenstein ideal of height 4 generated by forms of degree p. Let h(S) = (ho, . . . , hs) be the /i-vector of S = R/I. Prove that the minimal resolution of S has the form: s-2p+3

where r = s — 2p + 3 and a» = a r _j+i for i > I.

118

Chapter 4

4.3.20 If x > 1 is an integer, prove that x(x + 8) is never a perfect square.

4.3.21 Let / be a graded Cohen-Macaulay ideal of a polynomial ring R with a minimal resolution:

(p + a)) -> Rbl(-p) -^ R^ R/I. If &i = 9, use the Herzog-Kiihl formulas to prove that / is Gorenstein.

Hint Note 63 - 62 + &i - 1 = 0 and prove bib2b3 = 96 3 (& 3 + 8) is a perfect square, then use Exercise 4.3.20.

4.4

Initial degrees of Gorenstein ideals

Assume R is a polynomial ring over a field and / a homogeneous Gorenstein

ideal of codimension g > 3 and initial degree p > 2. Let

A consequence of the symmetry of the /i-vector of R/I is that the multiplicity of R/I satisfies e(R/I)>e(g,p), see Exercise 4.3.18. Given codimension g and initial degree p, a graded

Gorenstein algebra R/J with multiplicity e ( R / J ) = e(g,p) is extremal. This has strong structural implications: the minimal free resolution of R/J must be pure and almost linear; consequently all the Betti numbers bi(R/J) are determined, and in particular J is generated by z/o forms of degree p (see Section 4.3). Results in the literature dealing with Cohen-Macaulay ideals (such as [106, 249]) give upper bounds for the minimal number of generators v ( I ) of / in terms of codimension, initial degree, and multiplicity e(R/I). One of the aims below is to elucidate the multiplicity information that is already determined by the codimension and initial degree (see Theorem 4.4.3). In general one expects no such conclusion, but for graded Gorenstein algebras the symmetry of the /i-vector h(R/ 7) can be effectively exploited.

Macaulay's Theorem First we introduce the binomial expansion, and then state a famous result of Macaulay on the growth of Hilbert functions that will be needed later.

Lemma 4.4.1 Let h, j be positive integers. Then h can be uniquely written

J where a, > a^_i > • • • > a» > i > 1.

Hilbert Series___________________________________119

Proof. By induction on j. If j = 1, there is a unique aj such that (a.3') < induction / \ / \ / r = h- \ 3 } = ( J ~ 1 ) + • • • + (

u/

u-V

then h = ( j ) . Assume j > 1, then h < (a'+ ). Let r — h — ("?). By

\ | where a,-_i > • • • > a; > i > 1.

y/

From the inequality -

we obtain (°tTi) < (• < ^ 1 ) and this implies aj_i < aj. Let us show the uniqueness. Assume /i can be written as

. \:-: where bj > bj-i > • • • > bk > k > 1. By induction on j one has

Since aj is unique we derive aj = bj. Hence by induction hypothesis k = i and BJ-I = bj-i, . . . , a; = 6j. D

The unique expression for h in Eq. (4.8) is called the binomial expansion of h in base j. We set

sometimes /i^ is refer to as the Macaulay symbol. Theorem 4.4.2 (Macaulay) Let h: N —>• N 6e a numerical function and k a field. The following are equivalent (a) there exists a homogeneous k-algebra S with H(S,i) = h(i) for i > 0.

(6) /i(0) = 1 and h(i + 1) < h(i)^ for all i > I . Proof. See [298, Appendix B].

D

Theorem 4.4.3 ([222]) If I is a graded Gorenstein ideal of codimension g > 3 and initial degree p > 2, then v(Ip)


+ 9- A 5-1

and I is itself extremal if equality holds.

P +9 - 3

120______________________________________ Chapter 4

Proof. If either v(Ip) > VQ, or v(Ip) = VQ and / is not extremal, then by the symmetry of the /i-vector h(R/I) there is some j > p so that

h ( R / I ) = (ho, hi, . . . , hp-i, hp, . . . , h j , hp-i, . . . ,hi, ho), where

The idea of the argument is to use the Macaulay bound h^ for /ij+i to see that such a small value of hj cannot grow to such a large value of -«

Recall that this estimate is calculated from the binomial expansion for hj\

where a,j > a,j_\ > • • • > a$ > i > 1. Then

We may assume that /ij > j, for if not, then hj < j would imply that at—i for all I, and hence hj > /ij+i, which contradicts our assumption. Notice that by grouping the terms of (4.11) according to the value of at — t the binomial expansion for hj can be written as

h3

jn ~ 1

= £

+ Jn

n=0

"Ti

In ~ t"n H~ fan

.

.

i ,, ,

7n - « » - ! / ]

(4.13)

where a, - j = k0 > ki > • • • > kr > 0, j = jo > ji > • • • > jr, jr - ir = i,

and jn = jn-i — in-i — 1 for 1 < n
it follows that k < g - 2. From (4.11), (4.12), and (4.13), together with Pascal's identity and

b+l

J

a + 1 /a 6+1V

b

121

Hilbert Series

we have

Jn

n=0

kn

On the other hand from the upper bound on hj and hj^i = hp-i we get

Since (3 - l ) / ( p - 1) > (k + l ) / ( j + 1) it follows from (4.13) and the last

two inequalities that FQ < 0, where for 0 < s < r we set

= E

js + 1

in

kn n -in

To derive a contradiction we are going to show the following inequalities

Jr - V

Jr

Assume 1 < s + 1 < r. Notice that

n-

kn +i

n

js -is+ks s

-i

s

jT - ir + kr

jn

-l

'n - in + kn Jn

Therefore n

+ kn + l

jn - in + k n=s+i

n= S +l

Let ^4S denote the first summation and Bs the second in this last inequality; clearly As — Bs > 0. Also note that

ks + 1 k -.——- > -..

Js + 1

and

l k - > -rs+1 js - Is

3s+l +

122______________________________________Chapter

4

Putting these together, we compute F — F s+l

F .f

-L

f

' ks + 1

ks + 1 \ js - is + k.

js-«s

3s +

, + 1

- is

fc s+ l 1-1 \

/ Ka+l

ja+l +

Hence FQ > Fr > 0, which contradicts the observation that FQ < 0.

d

One might hope for even stronger estimates than suggested by the result above, for instance that v(I) < i/0. Next we present a counterexample showing that Theorem 4.4.3 cannot be extended to bound the number of generators in all degrees. Example 4.4.4 Let

/ = ((xl,x%,xl, Using Macaulay one readily obtains that / is given by: < 2 2 3 2 4 4 3 \X^, X ^ X ^ X ^ ~r3/3^C4) ^3) X ^ X ^ ^ X^t X^, X ^ X ^ ^

2 3 3 2 3 3*\ X ^ X ^ X ^ ~~r X ^ X ^ ^ X ^ X ^ ^ X ^ X ^ j .

Then R/I is a Gorenstein Artin algebra with h-vector (1,4,9,13,13,9,4,1) and Betti sequence (1,10,18,10,1), whereas the /i-vector for an extremal

Gorenstein algebra of codimension four and initial degree two is (1,4,1) and the Betti sequence is (1,9,16,9,1); notice that the multiplicity e(R/I) = 54 is far greater than the minimal value of six exhibited by an extremal algebra. In height three one can bound the Betti numbers, thanks to the structure theorem of Buchsbaum and Eisenbud. For initial degree p, the extremal Gorenstein algebra of codimension three has bi = &2 = 2p + 1. Theorem 4.4.5 Let R be a polynomial ring over a field and I be a homogeneous Gorenstein ideal of height three. If p is the initial degree of I, then

v(I) <2p+l

and b2(R/I) <2p+l.

Hilbert Series

123

Proof, (sketch) We may assume, without loss of generality, that R is equal to k[xi,X2,X3\ and A = R/I is Artinian and local with socle(A) = Aa. Then by [54] the minimal free resolution of A has the form V

V

0 -» R(-a - 3) -» 0 R(-nj) A 0 R(-rm) -» R, j=i z=i where Y is an alternating matrix, and the generators /!,...,/„ of / are the maximal pfaffians of Y. If the theorem fails every generator has degree at least (1p + l)/2 > p, which is a contradiction since at least one minimal generator has degree p. D Conjecture 4.4.6 (Miller- Villarreal) Let I C R be a graded Gorenstein ideal of initial degree p and height g. If 1 < i < g and 7, is the ith initial Betti number of R/I, then

i-l p + g - 1\ p + g - i - 2\ 1

I

.

II

P-I

1

)

g\ p + g - 2 J

V« \

Proposition 4.4.7 Let n be a given positive integer and /, • N the numerical functions given by

g(x)

= n^

and f(x) = x^,

where h^ is the Macaulay symbol. Then g(x) is non increasing and f ( x ) is strictly increasing. Proof. See [22] and [44, Lemma 4.2.13] respectively.

D

There is a short and elegant argument, based on the behaviour of the combinatorial functions g(x) = n^ and f ( x ) — x^n\ to show the inequality

of Theorem 4.4.3. We include the details of this argument in the proof below. Corollary 4.4.8 // R is a polynomial ring over a field k and I a graded

Gorenstein ideal of R of codimension g > 3 and initial degree p > 2, then + g-l\

fp + g-3

Proof. First we make a change of coefficients using the functor (•) g>& K, where K is an infinite field extension of k. Hence we may assume that k is infinite. Let A be an Artinian reduction of 5 = R/I, that is,

124______________________________________ Chapter 4

where h = hi , . . , , hj is a system of parameters of S with /ij_a form of

degree 1 for all i. Note that one can write A =^R/I, where R = R/(h) is a polynomial ring in g variables and / = IR is a graded Gorenstein ideal of R of codimension g and initial degree p. By Proposition 4.3.9 the Hilbert function of A is symmetric. Since the /i-vector h(S) = (hi) satisfy hi = dimfc(-Ai) there is some j > p so that

h(S) = (ho, hi,... ,hp-i,hp, . . . ,hj,hp-i, . . . ,hi,h0), thus

"j = hp

_ i

(1)

J

The idea of the argument is to use the Macaulay bound

hj+l < ftf ,

(2)

see Theorem 4.4.2. From the inequalities:

,

(D

, i-2

'< p-2)

we conclude

Using that /(a;) is non decreasing yields the required inequality.

D

Exercises 4.4.9 Let / be a graded Gorenstein ideal of height 4 and initial degree p, prove f ( I p ) < (p + I) 2 . If p = 4 show the values 21 < v(I^) < 24 do not occur.

4.4.10 Let / be a graded Gorenstein ideal of height 4 generated by forms of degree p and R/ J an extremal algebra of the same codimension and initial degree, prove that their Betti numbers satisfy

bi(R/I) < bi(R/J)

for all i.

Hilbert Series

125

4.4.11 Let R = k [ x i , . . . ,xg] be a polynomial ring over a field k and / a graded ideal of height g generated by forms of degree p. Prove that the Hilbert function of / satisfies:

H(I,p + i) >H(I,p)+i(9}

4.5

for all i > 0.

A symbolic study of Koszul homology

The first Koszul homology module of a Cohen-Macaulay graded ideal is studied here using the Cohen-Macaulay criterion of Jiirgen Herzog introduced in Section 3.4, our approach is symbolic and makes use of Hilbert functions. Lemma 4.5.1 Let p and g be positive integers and let

I f p > 2 andg>6 then i/j(p) > 0. Proof. Notice the equality

which is certainly positive for g > 6. We proceed by induction on p. Assume ip(p) > 0. It is easy to check that ifj(p + 1) is greater than 3

+ 9\ \ (P + 1) (4(5 - 2)p2 + (3g2 -3g- 8)p + g2 - 3g - 2) I —————————————————————————————— 9 / 1 (P + 1) (2p + l)(2p + 2)

Since the right hand side of this inequality is positive for p > 2 and g > 3, the induction step is complete. D

Theorem 4.5.2 ([240]) Let R — ®^L0Ri be a polynomial ring with its

usual graduation over an infinite field k and I a Cohen-Macaulay graded ideal of R of height g with a p-linear resolution

If I is generically a complete intersection with p > 2 and g > 3, then Hi(I), the first Koszul homology module of I, is not Cohen-Macaulay. Proof. The module HI = HI(!) has a well defined rank equal to 61 — g. Thus by Theorem 3.4.12, it suffices to prove that for some homogeneous

system of parameters y of S = R/I, one has >rank(ffi(I))-*(S/yS).

126______________________________________ Chapter 4

We start by making a specialization to the case of a polynomial ring of dimension g. Since k is infinite according to Theorem 2.2.11, there exists a system of parameters y = {yi , . . . , yd} for 5 with each yi a form of degree one of R. We make two observations: (i) Because Torj(5, R/(y)) = 0 for i > 1, it is clear that the minimal resolution of 5 = S/(yS) as an R (= -R/(y))-module has the same twists and Betti numbers as the minimal

resolution of 5 over R. (ii) With HI Cohen-Macaulay it is easy to see that the first Koszul homology module of / <S> R over R is precisely HI (/) ® R. We may then in the sequel assume that S is zero-dimensional.

We will compare the integer r=rank(original H i ( I ) ) - i ( S ) , with partial Hilbert sums contributing to t(Hi). First notice (see Exercise 4.3.17) that the length of S and its Betti numbers can be calculated using:

From the Koszul complex we obtain the exact sequences 0 -» BI ->• Zi -)• Hl -> 0,

0 ->• Z2 ->• R(b^(-1p) -)> Bl -» 0, where Zj and 5$ are the modules of cycles and boundaries defining Hi (I) . To simplify notation we set t(M)i = H(M,i), the dimension of the ith component of the graded module M. One may write

and therefore

i=0

j=0

i=0

From the minimal resolution of 5 we obtain

i > 0. Hence t(Zv\ > 6i£(E(-p))i - t(R}i and ^(Zi)i = 0 for 0 < i < p, which leads to 2p

6

2p

E^i)^ i E ^(-p))*i=0

i=p+l

Finally by (4.15) and (4.16) we have

2p

t=p+l

Hilbert Series

127

The proof now reduces to showing that the right hand side of (4.17) is greater than rank(.Hi (/)) • l(S) = (61 — g) • l(S), that is, we must show that the following inequality holds for p > 2 and g > 3

f(p,g) = &i E t(R(-p»i - E l(R^ ~ (a1) - ( b i - 9 ^ t ( S ) > o. i=p+l

i=p+l

^

'

It is not hard to see that f(p,g) simplifies to

Observe that from this equality we obtain the inequality:

It is easy to check that f ( p , g) > 0 for g e {3, 4, 5} and p > 2. The required

inequality is now a direct consequence of Lemma 4.5.1.

D

The result above complements one of Bernd Ulrich [286] in which the twisted conormal module ws ®fl I = <^s ®fi I /I2 was shown not to be Cohen-Macaulay; either result implies that 7 is not in the linkage class of a

complete intersection. Corollary 4.5.3 Let I be a graded Cohen-Macaulay ideal of height g > 3 in polynomial ring R. If I is generically a complete intersection and has a p-linear resolution with p > 2, then I is not in the linkage class of a

complete intersection. Proof. If / is in the linkage class of a complete intersection, then / is a strongly Cohen-Macaulay ideal, a contradiction because HI (I) is not CohenMacaulay. See [179] and [296, Corollary 4.2.4]. D

Chapter 5

Monomial Ideals and Stanley-Reisner Rings This chapter is intended as an introduction to the study of combinatorial commutative algebra, some of the topics were chosen in order to motivate the subject. Part of our exposition was inspired by [25, 271]. The main objects of study in this chapter are monomial ideals, face rings and simplicial complexes. Here we emphasize the connection between

algebraic properties of face rings and the reduced simplicial homology of the corresponding Stanley-Reisner complex. An understated goal here is to highlight some of the work of M. Hochster, G. Reisner and R. Stanley. It turns out that some notions encountered while studying graded algebras, such as Cohen-Macaulay rings, Gorenstein rings and Hilbert series are useful to solve combinatorial problems. Some key results leading to the proofs of the upper bound conjectures for convex polytopes and simplicial spheres are presented, most of them without detailed proofs but giving the

appropriate references. Our hope is to give a flavor of the methods and ideas of the area and to trace some routes for further reading.

5.1

Primary decomposition

General monomial ideals will be examined first, subsequently we specialize to the case of square-free monomial ideals and present some of their relevant

properties. Let R = fc[x] = k [ x i , ...,xn] be a polynomial ring over a field k. To make notation simpler we set xa = x"1 • • • x™" for a = ( a i , . . . , an) £ N™.

129

130______________________________________Chapter

5

Definition 5.1.1 An ideal / of R is called a monomial ideal if there is A C N™ such that / is generated by {xa a e A}. If / is a monomial ideal the quotient ring R/1 is called a monomial ring. Note that a monomial ideal is always generated by a finite set of monomials by Dickson's lemma.

Definition 5.1.2 A face ideal is an ideal p of R generated by a subset of the set of variables, that is, p = (x^,,.. , x t k ) for some variables Xij. Proposition 5.1.3 Let R = k[xi,... ,xn] be a polynomial ring over a field k. If I C R is a monomial ideal, then every associated prime of I is a face Proof. By induction on the number of variables that occur in a minimal

generating set of I consisting of monomials. Set m = (x\, . . . , xn). Let p be an associated prime of /. If rad(J) = m, then p = m. Hence we may assume rad(7) ^ m. Pick a variable x\ not in rad(7) and consider the ascending chain of ideals

70 = 7 and In+i = (In: Xi)

(n > 0).

Since R is Noetherian one has 7^ = (7^:xi) for some k. There are two cases to consider. If p is an associated prime of (7 n ,xi) for some n, then

by induction p is a face ideal because one can write (In,xi) = (-C> x i)i where I'n is an ideal minimally generated by a finite set of monomials in the variables X2,...,xn (cf. Exercise 1.1.45). Assume we are in the opposite case. By Lemma 4.1.2 for each n there is an exact sequence

0 —» R/(In:xi) ^ R/In —> R/(In,X!) —> 0, hence making a recursive application of Lemma 1.1.16 one obtains that p is an associated prime of In for all n. Since x\ is regular on R/Ik one concludes that 7t is an ideal minimally generated by monomials in the variables z 2 , • • • , xn, thus by induction p is a face ideal. D Definition 5.1.4 A monomial / in R is called square-free if

for some \
Corollary 5.1.5 Let R = k[xi, . . . ,xn] be a polynomial ring over a field k. If I is an ideal of R generated by square-free monomials and pi , . . . , ps are the associated primes of I, then

Monomial Ideals and Stanley-Reisner Rings_____________ 131

Proof. Set J = nf =1 pi and note J = rad (/). We only have to check J C I

because / is contained in any of its associated primes. Take a monomial / in J = rad(7) and write / = x"1 • • • a;"', where i\ < • • • < ir and o» > 0 for all i. Since fk 6 / for some k > 1, and using that I is generated by square-free monomials, we obtain x^ • • • Xir 6 7. Hence / € /. To finish the proof observe that J is a monomial ideal because the intersection of

monomial ideals is again a monomial ideal (see Exercise 5.1.24).

D

Definition 5.1.6 Let R = k[xi, . . . ,xn] be a polynomial ring, the support

of a monomial xa — x^1 • • • xfj1 in R is given by supp(x a ) = {x{ a* > 0}.

Proposition 5.1.7 ([113]) Let I be a monomial ideal in a polynomial ring R over a field k and S = R/I . Then there is a polynomial ring R and a square-free monomial ideal I' of R such that S = S' /(h), where S' = R /I' and h is a regular sequence on S' of forms of degree one.

Proof. Let R = k[x] and let F = { f i , . . . , fr} be a set of monomials that minimally generate /. Assume that one of the variables, say xi, occurs in at least one of the monomials in F with multiplicity greater than 1. Thus one may write /i = %i1gi, • • - , / , , = xl'gs,

where a$ > 1 for all i, a-\_ > 2, x\ (£. supp(^) for alii < s and x\ <£ supp(/i)

for i > s. Set

/' = (xQx^~lgi, . . .,xQxl'~lgs,fs+i,. ..Jr)cR' = R[x0], where XQ is a new variable. We claim that XQ — x\ is a nonzero divisor of S' = R' /I'. On the contrary assume that XQ — x\ belongs to an associated prime p of /' and write p = (I':h), for some monomial h. Since p is a face ideal, one has Xih € /' for i = 0,1. Hence using h £ /' gives x\h = l gihi for some i and consequently h = Xox^i~2gihi. Hence or „

, ,

fjM

> for some j > s. ,,

In both cases one obtains XQ G supp(M) and h 6 /', a contradiction. Since

S' (XQ - xi) = S one can repeat the construction to obtain the asserted monomial ideal /'. This proof is due to R. Froberg. D

The ideal /' constructed above is called the polarization of /. Thus any monomial ring is a deformation by linear forms of a monomial ring with

square-free relations. Note that / is Cohen-Macaulay (resp. Gorenstein) iff /' is Cohen-Macaulay (resp. Gorenstein).

132______________________________________Chapter

5

Proposition 5.1.8 Let q be a monomial ideal of R = k[xi,... ,xn]. Then q is a primary ideal if and only if, after permutation of the variables, q has the form: n

_ /™Qi

1 ~~ (Xl

ar

,---,-i

r

,X

61

,...,£

b,\

),

(i

where flj > 1 and U|=1supp(o; ' ) C {xi,... ,xr}.

Proof. If Ass(R/q) = {p}, then by Proposition 5.1.3 one has that p is equal to ( x i , . . . ,xr). Since rad (q) = p the ideal q is minimally generated by a set of the form: fTai \^-^ , . . . , ^™ir r , ^ bi , . . . , x b,}j.

Let Xj €. supp(a; 6i ), then xbi = XjN, where N is a monomial not in q. Since q is primary, a power of Xj is in q. Thus Xj 6 (xi,... ,xr) and consequently 1 < 3 < r! as required. For the converse note that any associated prime p of R/q can be written as p = (q: /), for some monomial /. It follows rapidly that ( z - i , . . . , xr) is the only associated prime of q. D Corollary 5.1.9 //p is a face ideal, then p™ is a primary ideal for all n. Proposition 5.1.10 If I is a monomial ideal of R — k[xi,... ,xn], then I has an irredundant primary decomposition I = qi n • • • n qr, where qj is a primary monomial ideal for all i and rad (qj) ^ rad (qj) if i ^ j.

Proof. Let / i , . . . , fq be a set of monomials that minimally generate /. We proceed by induction on the number of variables that occur in the union of the supports of / i , . . . , fq. One may assume that one of the variables in uf =1 supp(/j), say xn, satisfy xln <£: I for all z, otherwise / is a primary ideal and there is nothing to prove. Next we permute the /j in order to find integers 0 < a\ < • • • < a ? , with

aq > 1, such that fi is divisible by x^ but by not higher power of xn. If we apply this procedure to ( I , X n q ) , instead of /, note that one must choose a

variable different from xn. As (I:xnq) is generated by monomials in less than n variables and because of the equality

/=(/,<<)n (/=<') one may apply the argument above recursively to both monomial ideals occurring in the intersection and use induction to obtain a decomposition of / into primary monomial ideals q(,..., q's. Finally we remove redundant primary ideals from q(,..., q's and group those primary ideals with the same

radical.

D

Even for monomial ideals a minimal irredundant primary decomposition is not unique, what is unique is the number of terms in such a decomposition and also the primary components that correspond to minimal primes.

Monomial Ideals and Stanley-Reisner Rings_____________ 133

Example 5.1.11 If / = (x2,xy) C k [ x , y ] , then

/ = 0) n (x2,xy,y2) = (x) n ( x 2 , y ) , are two minimal irredundant primary decompositions of /.

The computation of a primary decomposition of a monomial ideal can be carried out by successive elimination of powers of variables, as described in the proof of Proposition 5.1.10. Now we illustrate this procedure with a specific ideal. Example 5.1.12 If R = k [ x , y , z ] and / = (yz2,x2z,x3y2).

= (z2,x3,x2z) Thus / = (z, y2) n (x2, y) n (z2, x3, x2z).

Corollary 5.1.13 Let R = k[xi,... ,xn] be a polynomial ring over a field k. If I is a monomial ideal, then I has a primary decomposition

I = qi n - - - n q m , such that qi is generated by powers of variables for all i. Proof. Let q be a primary ideal minimally generated by the set of monomials x " 1 , . . . , x°^, / i , . . . , fs and such that s

supp(/i) C { X I , . . . , X T } ,

where o,j > 0 for all i. Note that if f\ = x\l • • • x\T and b\ > 0, then ai > b\ and one has a decomposition: q — \3 • • • > ^r ' *^2 '

x

r 5 /2 ; • • • j /s J 5

where in the first member of the intersection we have lower the degree of a;"1 and have eliminated f i , while in the second member we have eliminated the variable x\ from /i. Applying the same argument repeatedly it follows that one can write q as an intersection of primary monomial ideals such that for

each of those ideals the only minimal generators that contain x\ are pure powers of x\. Therefore by induction q is the intersection of ideals generated by powers of variables. Hence the result follows from Proposition 5.1.10 and

Proposition 5.1.8.

D

134______________________________________Chapter

5

Proposition 5.1.14 Let I be an ideal of k[xi,... ,xn] generated by monomials in the first variables X i , . . . ,xr. If

i = /! n • • • n 7S is an irredundant decomposition of I into monomial ideals, then none of the Ii can contain a monomial in k[xr+i,.., ,xn].

Proof. Set X = {xi,... ,xr} and X' = X \ {x\,... ,xr}. Assume some of the Ii contain monomials in /c[X']. One may split the 7^'s into two sets so that 7 i , . . . , Im do not contain monomials in k[X'] (note m could be zero), while Im+i,..., Is contain monomials in the set of variables X1. For i > m + I pick a monomial gi in 7; whose support is contained in X'. Since the decomposition of 7 is irredundant there is a monomial / G H^L17j and / £" nf_ m+1 7j, where we set / = 1 if m = 0. To derive a contradiction consider j\ = fgm+i''' 9s- As f\ G 7, we get / G 7 which is absurd. d Corollary 5.1.15 Let I C k[x\,... ,xn] be an ideal generated by monomials in the first variables x\,..., xr. If

I = qi n - - - n q s is an irredundant primary decomposition into monomial ideals, then q, is generated by monomials in k[xi,... ,xr].

Proof. It follows from Proposition 5.1.8 and Proposition 5.1.14.

D

Proposition 5.1.16 If R = k[xi,... ,xn] is a polynomial ring over a field k and I a monomial ideal, then I is irreducible if and only if up to permutation of the variables I can be written as

where a; > 0 for all i.

Proof. =$>) Since 7 must be primary (see Proposition 1.1.21) from the proof of Corollary 5.1.13 one derives that 7 is generated by powers of variables. <=) If 7 is reducible, then by Corollary 5.1.13 there is an irredundant decomposition 7 = qi n • • • n q s , such that q^ ^ I and q, is generated by some powers of the variables x\,..., xr, for all i. Let xc^. G q^ \ 7, where i < a,'... Set

Note that one can write / = x^1 • • • xe£k, where &i < ami and m i , . . . , m / t distinct, but this is impossible since / G 7, thus 7 is irreducible. D

Monomial Ideals and Stanley-Reisner Rings_____________135

Theorem 5.1.17 If I is a monomial ideal in a polynomial ring R over a field k, then there is a unique irredundant decomposition

I = qi n - - - n q r such that qi is an irreducible monomial ideal. Proof. The existence follows from Corollary 5.1.13. For the uniqueness assume one has two irredundant decompositions:

qin---nqr = qin---nqi, where q; and q^ are irreducible for all i,j. Using the arguments given in the proof of Proposition 5.1.16 one concludes that for each i, there is o~i such that (\ai C q^ and vice versa for each j there is -KJ such that q'v. C (\j. Therefore r = s and q, = q'p. for some permutation p. Q

There are various routines for Macaulayl developed by S. Popescu [235] for dealing with monomial ideals, one of them computes the irredundant decomposition into irreducible monomial ideals of any monomial ideal. For further information on the primary decomposition of more general monomial ideals, e.g. monomial ideals obtained from a regular sequence or with coefficients in a ring other than a field, consult [88, 143, 144] and [307]. Example 5.1.18 Let / = (xy2,x2y) C k[x,y] and V(I] the affine variety defined by /. From the primary decomposition

I=(x)r\(y)n(x2,y2) one has:

V(I] = {(0,0)} U V(x) U V(y) = V(x) U V(y) C A 2 , where (x 2 ,?/ 2 ) corresponds to {(0,0)} which is embedded in V(x) U V(y). For this reason (x2,y2) is said to be an embedded primary component. If k is infinite, then the coordinate axes V(x) and V(y) are the irreducible components of V(J).

Monomial ideals form a lattice Let L be a lattice, that is, L is a partially ordered set or poset in which any two elements x, y have a greatest lower bound or meet x A y and a lowest upper bound or join x V y. The lattice L is called distributive if (a) x A (y V z) = (x A y) V (x A z), and

(b) x V (y A z) - (x V y) A (a; V z ) , Vx,y,z in L. Proposition 5.1.19 If L is the family of monomial ideals, order by inclusion, of a polynomial ring R over a field k, then L is a distributive lattice under the operations I A J = / D J and I V J = / + J. Proof. Use Exercise 5.1.24 and Exercise 5.1.25.

d

136______________________________________ Chapter 5

An interpretation of the minimal primes Let F be a set of monomials in the ring k [ y i , . . . ,yn] and / = I ( F ) the ideal generated by F. Consider the hypergraph

T(F) = {{j/i 1 ,...,2/i T ,}|(2/i 1 ,...,2/iJ is a minimal prime of /(F)},

whose vertex set is [J supp(/). An element in T(F) is called a minimal vertex cover of F. In order to relate the set of minimal primes of / with the set of monomials occurring in a certain geometric link of /, consider a copy F\ of F in the polynomial ring k\x\ ,..., xn] in a new set of variables x\ , . . . , xn. Proposition 5.1.20 ([3]) Let TI be the set of minimal vertex covers of

FI. If L is the ideal L= ({xit •••xir\{xil,...,xir} e TI}), then ((z):/(F)) = (z,L), where z = {xiyi, . . . ,xnyn}. Proof. Take {x^ , . . . , X i r } £ Tj and ya £ F. Hence x\,i divides xa for some 1 < j < r and we obtain x^ • • • xirya £ (z). This shows (z, L) C ((z): I ( F ) ) .

Conversely, assume M is a monomial in ((z):/(F)) \ (z). Let ya be any monomial in F. Since Mya — xtytM\ and M is not in (z), one has that xt divides M and yt divides ya . Hence supp(M) n supp(z a ) 7^ 0 for any xa in eL. D

Exercises 5.1.21 If f = /i, . . . , fr is a sequence of monomials in R, then f is a regular sequence if and only if supp(/j) n supp(/j) = 0 for i ^ j.

5.1.22 If / is a monomial ideal of R, then any associated prime p of R/I can be written as p = (/: /), for some monomial /. 5.1.23 If L is a lattice, then the following conditions are equivalent:

(a) x A (y V z) = (x A y) V (x A z), Vx, y, z in L. (b) x V (y A z) = (x V y) A (x V z), Vx, y, z in L. 5.1.24 Let / and J be two ideals generated by finite sets of monomials F

and G respectively, prove that the intersection / n J is generated by the set

Monomial Ideals and Stanley-Reisner Rings_____________ 137

5.1.25 If /, J, K are monomial ideals, prove

/ n (J + K) = (i n J) + (i n K). 5.1.26 If / and J are two monomial ideals, then (/: J) is a monomial ideal.

5.1.27 Let q be a primary ideal of k [ x i , . . . ,xn] generated by s

{x?,...,x?,xbl,...,xb-}, where |Jsupp(z 6i ) C {xi,...,xr}, i=i

and a,i > 1 for all i. If x° = x^ • • • xcrr is a monomial of least degree in the variables xi, . . . ,xr such that xc ^ q, then (xi, . . . ,xr) = (q: xc}. 5.1.28 If q is a primary monomial ideal, then q™ is a primary ideal and q n = q (n) for n > L

Hint Use Proposition 5.1.8. 5.1.29 If qi, . . . , q r are primary monomial ideals of R with non comparable radicals and / = q1 n • • • n q r , then

/(») = cftn---ntf. Hint Proceed as in the proof of Proposition 3.3.24. 5.1.30 Let / = (x\, x\x\, x\x\xz), note that / = qi n q2 is a primary decomposition of /, where qi = (x\,x\) and q2 = (x\,xix^,x\). Prove Hint If z — x\x\x\ and a = xi + x\ + x3, then az £ /3 and z & I3.

5.1.31 Use the next procedure in CoCoA to verify Exercise 5.1.30. Use R ::— Q[x[1..3]]; - - DegRevLex is the default order QI := Ideal(x[l]~3,x[2]~2); Q2 := Ideal(x[l]"3,x[l] * x[3], x[3f2); I := Intersection(Ql,Q2); 32 := Minimalized(Intersection(Qr2,Q2'~'2)); 12 := Minimalized(l^2);

Len(S2); - - checking number of generators S3 := Minimalized(lntersection(Ql"3,Q2"3)); EqSet(Gens(S2),Gens(I2)); - - checking equality G := Gens(S3); G[4] Isln ]T3;

138______________________________________Chapter

5

5.1.32 Let / = (2/1^2.2/12/2,2/12/1.^12/3,2/22/3-2/22/3,2/12/22/s). Find a primary

decomposition of /. Note (2/1,2/2,2/3) is an associated prime of /. 5.1.33 Use Proposition 5.1.20 to find the minimal primes of the ideal / generated by 2/5*2/33/5, 3/5*3/32/6. 2/5*3/22/3, 2/5*2/32/4. 3/i2/2> 2/i3/4> 2/i 2/22/3. 3/22/^5, 2/23/32/15.1.34 Show that / = (3/1,2/13/2) is a non primary ideal with a prime radical.

5.1.35 Let / be a monomial ideal in k [ x i , . . . , xn] and X = V(I) C An(k) the associated monomial variety. If k is infinite, prove that X is irreducible if and only if X = V(x^,..., x^). Hint Note I(V(xi,... ,xr)) = (xi,... ,xr).

5.2

Simplicial complexes and homology

A finite simplicial complex consists of a finite set V of vertices and a collection A of subsets of V called faces or simplices such that (i) If v £ V, then {v} e A.

(ii) If F <E A and G C F, then G € A.

Let A be a simplicial complex and F a face of A. Define the dimensions of F and A by dim(F) = \F\-1 and dim(A) = sup{dim(F)| F <E A}

respectively. A face of dimension q is sometimes refer to as a g-face or as a g-simplex. Example 5.2.1 Take the following triangulation of the disk with a whisker

attached.

0-simplices: 1-simplices: 2-simplices: #2 £3 £4

Let F be a g-simplex in A with vertices VQ , vi,..., vq. We say that two total orderings of the vertices

vio < • • • < viq and vjo < • • • < vjq

Monomial Ideals and Stanley-Reisner Rings_____________ 139

are equivalent if (IQ, . . . ,iq) is an even permutation of (jo, . . . ,jq). This is an equivalence relation because the set of even permutations form a group, and for q > 1, it partitions the total orderings of VQ, . . . , vq into two equivalence classes. An oriented q-simplex of A is a g-simplex F with a choice of one of these equivalence classes. If VQ, . . , ,vq are the vertices of F, the oriented simplex determined by the ordering VQ < • • • < vq will be denoted by [v0,...,vq].

Suppose A is a commutative ring with unit. Let (7, (A) be the free Amodule with basis consisting of the oriented g-simplices in A, modulo the relations [va,Vi,V2,...,Vq]

+ [Vl,V0,V2,...,Vq].

In particular (7g(A) is defined for any field k and dim^(7 9 (A) equals the number of g-simplices of A. For q > 1 we define the homomorphism

induced by dq([v0,Vi,...,Vq])

=

i=0

where £$ means that the symbol Wj is to be deleted. Since dqdq+i = 0 we

obtain the chain complex C(A) = {Cq(^),dq}, which is called the oriented chain complex of A. The augmented oriented chain complex of A is the complex

0 — > C d (A) -^» Cd-i(A) — > • • • — > • C 0 (A) -i> C_i(A) = A — > 0, where d = dim(A) and e(v) = 1 for every vertex v of A. This augmented chain complex is denoted by ((7(A),e). Set 80 = e and C'_i(A) = A.

Let Z ? (A, A) = ker(<9 ? ), Bq = im(9 9 +i) and fl-,(A;A) = Z,(A,A)/B,(A,A) ) for q > 0. The elements of Z 9 (A,A) and Bq(A.,A) are called cycles and boundaries respectively and Hq(A;A) is the qth reduced simplicial homology group of

A with coefficients in A. Remark 5.2.2 Note that if A ^ 0, then #i(A; A) = 0 for i < 0. Proposition 5.2.3 //A is a nonempty simplicial complex, then Ho(A;A) is a free A-module of rank r — I , where r is the number of connected components of A.

140______________________________________ Chapter 5

Proof. Let V = {xi, . . . ,xn} be the vertex set of A and AI, . . . , Ar its connected components. Denote the vertex set of Aj by Vj and pick a vertex in each one of the V*, one may assume x^ e Vi for i = 1, . . . , r. Set j3i = Xi — xr and /3i = fa + im(<9i) for i = 1, . . . , r — 1. We claim that the set Z? ={/?!,... ,/? r _i} is an ,4-basis for H0(A;A) = ker(<9 0 )/im(<9i). First we prove that B is linearly independent. Assume X^=: Qjft is in im(di) for some Oj in A, by making an appropriate grouping of terms one can write r-l

• •+

for some fry in A. Since Co (A) is a free ^.-module with basis V and the V; are mutually disjoint one has "

bij(xi — Xj) for 1 < A; < r — 1,

applying do to both sides^of this equality we obtain a/, = 0 for all k. Next we show that B is a set of generators. Let x^ and Xjl be two vertices of A, it suffices to prove that z^ — z^ + im(<9i) is in the submodule generated by B because ker(do) is generated by the elements of the form Zj — Xj. Assume z^ € Vm and Xj1 € Vs. There is a path z ^ , . . . , z$,, z m in A m and a path z ^ , . . . , ZJP , zs in A s such that every two consecutive vertices of those paths form a 1-face of A. From the equalities

\*^jl

*^J2)

'

\^J2

*^J3 / ~T ' ' ' T t,«Ejp

«£sj

—

^Jl

-^5)

one concludes Zj t - Xj1 + im(9i) = (zm - z r ) - (zs - z r ) + im(9i).

D

Definition 5.2.4 Let A and AI be simplicial complexes on disjoint vertex

sets V and W respectively. The join

A * AI is the simplicial complex on the vertex set V U W with faces FUG where F £ A and G £ AI . The cone cn(A) = w * A

of A is the join of a point {w} with A. Proposition 5.2.5 / / A is a simplicial complex and cn(A) = w * A its cone, then ^ # p (cn(A);,4) = 0 for all p.

Monomial Ideals and Stanley-Reisner Rings_____________ 141

Proof. If a = [VO,...,VP] is an oriented p-simplex, set [w,cr] equal to [W,VQ, . . . ,vp}. In general if cp = Y^niai ig a p-chain, set [u>,c p ] equal to Y^ni[w>ai}- This bracket operation is a homomorphism from C P (A) into Cp+i(w * A). If
9f7]

if dim cr = 0, > 0.

if dim ff

As a consequence one has the more general formulas:

d[w,co] d[w,cp]

= c0-e(c0)w, = cp-[w,dcp] i f p > 0 .

(5.1) (5.2)

By Proposition 5.2.3 we obtain HQ(W * A) = 0 , because w * A is connected. Assume p > 0. Let zp e Z(w * A) be a p-cycle of w * A, we will show that zp is a boundary. Write

where cp consist of those terms of zp with support in A, and dp-\ is a chain. It is enough to show that

zp - d[w, cp] — 0.

Using the identities (5.1)

and (5.2)

we obtain

zp - d[w,cp] = cp + [w,dp-i] -cp + [w,dcp] = [w,ep-i], where e p _i = dp-\ + dcp is a chain in A. Applying d to the equality above

and using that zp is a cycle yields d[w,ep-i] = 0. By identities (5.1) (5.2) one has

and

Now, the part of this chain with support in A is e p _i. Therefore ep-\ = 0 and [w, dep-i] = 0, which shows zp = 9[w, cp]. O

Definition 5.2.6

The reduced Euler characteristic x(A) of a simplicial

complex A of dimension d is given by

X(A) = - 1 + V(-l)Vi =-1 + X(A), where fi is the number of i-faces in A and x(A)

is its Euler characteristic.

142______________________________________ Chapter 5

Exercises 5.2.7 Let A be a simplicial complex of dimension d and /; the number of

z-faces in A. If A; is a field, then d

d

]T (-l^rankJ^A; k) = -1 + £)(-l)Vi. i=-l

5.2.8

i-0

Let A be a simplicial complex, prove that there is an exact sequence

Q—>H0(A;A) ^H0(&;A) —> A — j. 0 and conclude

_

Ho(&;A)~H0(&,A)®A. 5.2.9

Let A be a simplicial complex with vertex set V = {x\, . . . ,xn} and

• • • A C7o(A) A CU(A) = A-+Q, the right end of the augmented chain complex over a ring A. Prove that ker(<%) is generated by the cycles of the form Xi — Xj. 5.2.10 Let A be a simplicial complex and fi the number of z-faces of A. If

is the augmented chain complex of A over a field k, then dim/^Zi) is equal to /i - /o + r; where r is the number of connected components of A and

5.3

Face rings

Let R = k [ x i , . . . ,xn] be a polynomial ring over a field k. If / is an ideal of R generated by square- free monomials, the Stanley-Reisner simplicial complex A/ associated to / has vertex set V = {xi X{ <£ 1} and its faces are defined by A/ = {{xi, , . . . , xik }\ ii < • • • < ik , x^ • • • xik & I}.

Conversely if A is a simplicial complex with vertex set V contained in {xi, . . . ,xn}, the Stanley-Reisner ideal /A is defined as

^A = ({ZM •••xir\ii < • • •
Monomial Ideals and Stanley-Reisner Rings_____________143

Definition 5.3.1 Let / be an ideal of R generated by square-free monomials. The quotient ring R/I is called a face ring. Example 5.3.2 A simplicial complex A of dimension 1 and its StanleyReisner ideal:

/A = (xiX3,XiX2X4,X2X3Xi)

'X4

Definition 5.3.3 Let A be a simplicial complex and F € A, define lk(F), the link of F, as

lk(F) = {H e A| H n F = 0 and H U F e A}. Definition 5.3.4 Let k be a field. A simplicial complex A is said to be Cohen-Macaulay over k if the Stanley-Reisner ring A; [A] is a CohenMacaulay ring. Theorem 5.3.5 (Reisner) Let A be a simplicial complex. If k is a field, then the following conditions are equivalent:

(a) A is Cohen-Macaulay over k.

(b) Hi(lk F; k) = 0 for F e A and i < dim Ik F. Proof. See the original paper [237], or [44] for a detailed proof that combines a vanishing theorem of Grothendieck [137] with a formula of M. Hochster for the Hilbert series of the local cohomology modules of A;[A] with respect to the fine grading. D We clarify that in particular a Cohen-Macaulay complex A, being the link of its empty face, must satisfy .Hi(A; A;) = 0 for i < dim A.

In general the Cohen-Macaulay property of a face ring may depend on the characteristic of the base field, classical examples of this dependence are triangulations of the projective plane (see Exercise 5.3.31). Example 5.3.6 If A is a discrete set with n vertices

»x5 X-j

(

9X3

then A is Cohen-Macaulay and /A = (xiXj\l
144______________________________________ Chapter 5

Corollary 5.3.7 If A is a 1-dimensional simplicial complex, then A is connected if and only if A is Cohen-Macaulay.

Proof. By Reisner theorem A is Cohen-Macaulay iff .Ho (A; k) = 0, because the link of a vertex of A consists of a discrete set of vertices. To complete the argument apply Proposition 5.2.3. d Proposition 5.3.8 ([174]) If k is afield, then a simplicial complex A is Cohen-Macaulay over k if and only if lk(F) is Cohen-Macaulay over k for allF £ A.

Proof. =>) Let F be a fixed face in A and A' = \k(F). If G e A', using the equality

lk A '(G) = and Reisner criterion one infers Hj(lk A ' ((?),/:) = 0 for i < dimlkA'(G),

thus A' is Cohen-Macaulay. •4=) Since all the links are Cohen-Macaulay, it suffices to take F = 0 and observe A = lk(0) to conclude that A is Cohen-Macaulay. D Definition 5.3.9 A face F of a simplicial complex A is said to be a facet if F is not properly contained in any other face of A.

Proposition 5.3.10 // A is a simplicial complex with vertices xi, . . . ,xn,

then the primary decomposition of the Stanley-Reisner ideal of A is:

where the intersection is taken over all facets F of A, and Pp denotes the face ideal generated by all x^ such that X{ (£. F. Proof. Let F be a face of A and PF the face ideal generated by the Xi

such that Xi 7 F. It is not hard to prove that PF is a minimal prime of /A if and only if F is a facet. Therefore the result follow from Corollary 5.1.5. D

Corollary 5.3.11 Let A be a simplicial complex of dimension d on the vertex set V = {x\ , . . . , £ „ } and k a field. Then dim fe[A] — d + I — maxjs x^ • • • Xia ^ /A and i\ < • • •
Monomial Ideals and Stanley-Reisner Rings_____________ 145

Corollary 5.3.12 A Cohen-Macaulay simplicial complex A is pure, that is, all its maximal faces have the same dimension.

Proof. Use Proposition 5.3.10 and Corollary 2.2.4.

D

Definition 5.3.13 The q-skeleton of a simplicial complex A is the simplicial complex A 9 consisting of all p-simplices of A with p < q.

Proposition 5.3.14 Let A be a simplicial complex of dimension d and A 9 its (/-skeleton. If A is Cohen-Macaulay over a field k, then A 9 is CohenMacaulay for q < d. Proof. Set A' = A" and take F e A' with dim(F) < q < d. Note

Since A is pure there is a facet F' of A of dimension d containing F, hence F' \F is a face of lk&(F] of dimension d— \F\. It follows that the dimension of lk&i(F ) is q — \F\. Using Proposition 5.3.8 we get Hi(lkA(F); k) = 0 for i < dim(lk A (F)) = d - \F\, and consequently F

\;k)=Hi(\k&(F);k)

= Q for i < q - \F\ < d - F .

Observe that the equality between the homology modules follows simply because the augmented oriented chain complexes of (lkA(^)) ? ~' F ' and IkA(-F) are equal up to degree q - \F\ and thus their homologies have to agree up to degree one less. Hence by Reisner criterion A 9 is Cohen-Macaulay. D Example 5.3.15 ([114]) Let R = k[x\, ... ,xn] be a polynomial ring over

a field k and A the (p — 2)-skeleton of a simplex of dimension n — 1, where p > 2. Then /A is a Cohen-Macaulay ideal and the face ring

= R/({xtl • • • x has a p-linear resolution by Theorem 4.3.7. Face rings with pure and linear resolutions have been studied in [46] and [114, 116]; special results are obtained when the corresponding StanleyReisner ideal is generated in degree two.

Proposition 5.3.16 //Ai and A2 are simplicial complexes, then their join AI * A2 is Cohen-Macaulay if and only if Aj is Cohen-Macaulay for i = 1,2.

146______________________________________ Chapter 5

Proof. Since one has the equality ^A!*A 2 = /Ai + ^A 2 >

there is a graded isomorphism of fc-algebras

fc[Ai * A 2 ] ~ fc[Ai] fc &[A 2 ], see Proposition 2.2.20. To complete the proof use Corollary 2.2.22.

n

Given a simplicial complex A and AI, . . . , A s sub complexes of A, we define their union

as the simplicial complex with vertex set Uf-jVj and such that F is a face of U|=1Aj if and only if F is a face of Aj for some i. The intersection can be defined similarly.

Definition 5.3.17 A pure simplicial complex A of dimension d is shellable if the facets of A can be order Fi,...,F8 such that

is pure of dimension d — 1 for alH > 2. Here

If A is shellable, Fi,...,Fs is called a shelling. Theorem 5.3.18 Let A be a simplicial complex. //A is shellable, then A is Cohen-Macaulay over any field k. Proof. Our proof is based on a refined argument of T. Hibi [162]. Let V = {xi, . . . ,xn] be the vertex set of A and R = fcjx] a polynomial ring over a field_fc. Assume that FI, . . . , Fs is a linear order of the facets of A such that Fi n (Ul,j2\Fj) is pure of dimension d - 1 for alH > 2, where d is the dimension of A. Set <7 =

I

-

-

, ^

-j

" I

I \

\3 = 1

one may assume a — {xi,... ,xr}. There is a short exact sequence 0 —> R/(I: /) -4 R/I —> R/(I, /) —)• 0,

Monomial Ideals and Stanley-Reisner Rings_____________ 147

where / = x\ • • • xr and / = /A is the Stanley-Reisner ideal of A. First we show the equality R/(I:f)=k[Fs], note that k[Fs] is a polynomial ring in d+l variables. By Proposition 5.3.10

one can write I = nf =1 pi, where p, is generated by V \ Fi for all i. Hence ( / : / ) = P| (p-/) = p| ( P i :/). i=l

aCFi

Observe that if a C Fi for some i, then i = s; otherwise if i < s, then a belongs to Fs n (U^lJ-Fj), but since this simplicial complex is pure of dimension d — 1, there is v 6
Next let us show

Note that / 6 pi for i = 1, . . . , s - 1; otherwise if / ^ pj for some i < s, then a C Fi and the previous argument yields a contradiction. If p is a minimal prime of (/, /) different from pi, . . . ,p s -i, then ps C p and Xi G p for some x^ G cr, thus by construction of a one has Fs \ {x,} C F& for some

fc < s. Therefore V\Fk C {xi} U (V \ Fs) C p, and by the minimality of p one derives p& = p s , which is impossible. Altogether we conclude that pi, . . . , p s _i are the minimal primes of (I, /) and s-l

as required.

Using induction on s and the exact sequence (*) it follows that R/I is Cohen-Macaulay by a direct application of the depth lemma. To apply the depth lemma one uses that the extremes of the exact sequence (*) are both Cohen-Macaulay -R-modules of dimension d+l. D

Exercises 5.3.19 Let R be a polynomial ring over a field k. Prove that the family of

ideals of R generated by square-free monomials is a sublattice of the lattice of monomial ideals.

5.3.20 A monomial ideal /, with coefficients in a field, is generated by square-free monomials iff any of the following conditions hold: (a) I is an intersection of prime ideals.

148______________________________________Chapter

5

(b) rad (/) = /. (c) A monomial / is in I iff x\ • • -xr G /, where supp(/) = {xi}ri=l. 5.3.21 If A is the simplicial complex

'X4

prove the equality

/A = (xi, x-i) n (x\, xz) n (xi, £4) n (3:2, xs) n (xs, X4) by using the correspondence between facets and minimal primes.

5.3.22 Let A be a Cohen-Macaulay simplicial complex and F G A. is not a discrete set of vertices, prove that lk(F) is connected.

5.3.23 Let A be a simplicial complex on the vertex set V and / its StanleyReisner ideal. Take x G V and set J = (/: x). (a) Prove that J = r\£ p; pi, where Ass(/) = { p i , . . . , p r }. (b) Find a relation between A and Aj. 5.3.24 Let R be a polynomial ring over a field k and / an ideal minimally

generated by square-free monomials / i , . . . , fq. If R/I is Cohen-Macaulay and x is a variable not in / such that x G U' =1 supp(/j), then R/(I:x) is Cohen-Macaulay. 5.3.25 Let A be a pure simplicial complex on the vertex set V and / = /A. Assume that x is a variable in U' =1 supp(/j), where the /$ are monomials that minimally generate I. If (/: x) and (/, x) are Cohen-Macaulay, then / is Cohen-Macaulay. Hint Show ht(7) = ht(/:x) = ht(J,x).

5.3.26 Let A a simplicial complex and a G A, define the star of a as star(a) = {G G A| a U G G A}. Prove that the facets of star(cr) are the facets of A that contain a. 5.3.27 Let R = k\x\ be a polynomial ring over a field k and A a simplicial complex with vertex set x. If a = { x i , . . . , xr} G A, prove the equality

here / = Hi=i ^i

an

d star(cr) is the star of a.

Monomial Ideals and Stanley-Reisner Rings_____________149

5.3.28 If A is a Cohen-Macaulay complex and F a face of A, then star(F) is a Cohen-Macaulay complex. Hint star(F) = F * lk(F). 5.3.29 A pure simplicial complex A is shellable if and only if the facets of

A can be listed FI, ..., Fs such that for all 1 < j < i < s, there exist some v £ Fi \ Fj and some k 6 {1,..., i - 1} with F, \ Fk = {v}. 5.3.30 Let A be a pure simplicial complex. Prove that A is shellable if and only if the facets of A can be ordered FI ,..., Fs such that for all 1 < J < i < s, there is v € Fi\Fj and k < i with F^Fj C F^Ff. = Fi\{v}. The next exercise is a subdivision of the minimal triangulation of the projective plane given by G. Reisner [237, Remark 3].

5.3.31 (N. Terai) If A; is a field and A is the following triangulation of the real projective plane F2

then the link of any vertex is a cycle and fc[A] is Cohen-Macaulay if and only if char(/s) ^ 2.

Hint Hi(&;k) ~ fl"i(P 2 ;fc) = k/2k and use Theorem 5.3.5.

5.3.32 Let R be a ring and /i, 72 ideals of R. Prove that there is an exact sequence of .R-modules: r\ __,

r-> / / r

r~i 7" \

<, K* / T

ff\

Z? / T

\ /? / { T

_l_ 7" ^ __\ fl

where ip(r) = (r, — F) and 0(Fi,F 2 ) = r\ + r 2 . 5.3.33 ([172]) Let R = k[x] be a polynomial ring over a field k and A!, A2 simplicial complexes whose vertex sets are contained in x. Prove that there is an exact sequence of /^-modules:

0 —> fc[Ai U A 2 ] —> fc[Ai] © fc[A2] —> Jfc[Ai n A 2 ] —>• 0.

150______________________________________ Chapter 5

5.4

Hilbert series of face rings

For an Stanley- Reisner ring A; [A] there are explicit formulas for its Hilbert function and Hilbert series in terms of the combinatorial data of the complex, we begin with a direct approach to compute these formulae. Definition 5.4.1 Let k and n be two positive integers. A k-partition of n is an element a = (ai, . . . ,0^) in N+ so that n = QI + • • • + cc^.

Lemma 5.4.2

The number c(n,k) of k-partitions of n is equal to

Proof. If a = (ai, . . . , afc) is a fc-partition of n define ifj(a) by

tp(a) = {0-1,0:1 + a 2 > . .. ,ai + ••• + otk-i} C {!,..., n}.

To finish the proof note that the map ip induces a bijection between the set of ^-partitions of n and the set of subsets o f { l , . . . , n — 1} with fc - 1 elements. D Definition 5.4.3 Given a simplicial complex A of dimension d its f -vector is the (d + l)-tuple:

/(A) = ( / „ , . . . , / „ ) , where fi is the number of i-faces of A. Note /_i = 1. Proposition 5.4.4 // A is a simplicial complex of dimension d and (fi) its f -vector, then the Hilbert function of A; [A] is: d

/ .

.. \

H(*[A],j) = W J 7 U, forj>landH(k[&\,0) i=o ^ l '

= l.

Proof. Let x\, . . . ,xn be the vertices of A. Given xa = x"1 • • • x^71 we denote its image in A; [A] by xa. The face ideal /A is generated by squarefree monomials, hence a monomial xa is in /A if and only if supp(xa) = {xi ai > 0} is not a face of A. Therefore the set of all monomials xa such that supp(a;Q) is a face in A form a basis for fc[A] as a vector space over fc. If {x^ , . . . , Xir } is a face of A of dimension r — 1, we can use Lemma 5.4.2 to count the number of monomials of the form x i\f 1 • • • x ifrr such that (ai, . . . , a r ) is an r-partition of j. The desired formula follows at once. D Corollary 5.4.5

If A is a simplicial complex and /(A) = (/o, . . . , fd) its

f -vector, then the Hilbert series of fc[A] is given by F(k[*\,z)= ^

fi

*l+*, d = dim(A).

Monomial Ideals and Stanley-Reisner Rings

151

Proof. By the proposition above we have \

rf

Hence using the identities

the formula is established.

D

^4 result of Kruskal-Katona Kruskal and Katona [202] showed that (/o, /i, • • • , /d) G Z d+1 is the /-vector

of some (i-dimensional simplicial complex if and only if

where f^1

is defined according to Equation 4.9.

Simplicial complexes and their /i-vectors As an application we derive formulas for /i(A), the h- vector of A; [A]. If A is Cohen-Macaulay, we present an important numerical constraint for h(A) which is basic for some

applications of Commutative Algebra to Combinatorics.

Theorem 5.4.6 If A is a simplicial complex of dimension d and (hi) the h-vector of k[A], then hk = 0 for k > d + 1 and hk =

(-1)*-' i=o

\

K

_

l

fi-! for 0
(5.3)

J

Proof. The idea is to write the Hilbert series of fc[A] in two ways. By Corollary 5.4.5 we have:

i=-l

v

'

On the other hand by the Hilbert-Serre Theorem 4.1.3 there is a (unique) polynomial h(z) = ho + h\z + • • • + hrzr with integral coefficients so that h(l) ^ 0 and satisfying 0 = 7i——r^TT-

Comparing (5.4) and (5.5) yields the asserted equalities.

(5-5)

D

152

Chapter 5

Definition 5.4.7 The h-vector of a simplicial complex A, denoted by /i(A), is defined as the h-vector of the standard graded algebra A;[A]. Computation of the h-vector of face rings To compute the h-vector (of face rings) the following procedure of R. Stanley [271] can be used. Consider the 3-dimensional convex polytope P consisting of two solid tetrahedrons joined by a 2-face:

one has /(P) = (5,9,6). Write down the /-vector on a diagonal, and put a 1 to the left of /o.

Complete this array constructing a table, by placing below a pair of consecutive entries their difference, and placing a 1 on the left edge:

1 1

h(P)=

(1

2

1 5 4 9 3 5 6 1)

After completing the table the next row of differences will be the ^-vector. Hence the boundary simplicial complex A = A(P) is 2-dimensional, A; [A] has multiplicity 6, and IA =

Now we present a result that plays a role in the proof of the upper bound conjecture for spheres.

Theorem 5.4.8 Let A be a simplicial complex of dimension d with n vertices. //A; [A] is Cohen-Macaulay and k is an infinite field, then the h-vector of A satisfies 0

i + n - d- 2

for 0
(5.6)

Monomial Ideals and Stanley-Reisner Rings_____________153

Proof. Set S = fc[A] = R/I, where R = k [ x i , . ..,xn] and I = I/\. By Lemma 2.2.15 there exists a regular homogeneous system of parameters

9_ = $ 1 , . . . , 9^+1 for S such that each $j has degree one. Since A = S/(0_)S is Artinian, in this case Theorem 4.2.5 says that /ij(A) — H(A,i). Note that S/(d_)S ~ R/I, where R = R/(&} is a polynomial ring in n — d — 1 variables and / is the image of / in R. Therefore we have

/ii(A) = H(A,i) = H(R/l,i) < H(R,i) = as required.

D

Example 5.4.9 Let A be a 2-dimensional complex with /(A) = (5,6,2). Thus /c[A] is not Cohen-Macaulay because its h vector has a negative entry: 1

1 1 /i(A) = (1 A shellable complex

4 3

2

5

6 2

-1

2 0).

Let R = k[xi , . . . , xn] be a polynomial ring over a

field fc and A a simplicial complex with vertices xi, . . . ,xn. Associated to the Stanley-Reisner ideal /A we define another ideal

/'(A) = (I&,xiyi,...,xnyn) C k [ x i , ... ,xn,yi, .. . , y n ] , and denote by 5(A) the Stanley-Reisner complex corresponding to this ideal /'(A). Proposition 5.4.10 The simplicial complex 5(A) is shellable.

Proof. To prove that 5(A) is shellable we will use the criterion of Exercise 5.3.29. Let us begin by describing the facets of 5(A). It is not hard

to show that 5(A) is pure. Notice that the facets of 5(A) are of the form F' = {yri , . . . , yrk } U F, where F = {xsi ,..., xse } is a face of A so that k + I = n, {TI , . . . , ffc } n {si , . . . , si] = 0. Therefore F i-^ F' defines a one to one correspondence between the faces of A and the facets of 5(A). To construct a shelling for 5(A) we first linearly order the faces of A according to its dimension, and then linearly order the facets of 5(A) in such a way

that if F, G are faces of A and F < G then F' < G". Let F', G' be facets of S(A), with F' < G', there is Xj € G'\F', setting H1 = (G1 Ufe}) \ { X j } we obtain H' < G' and G' \ H' = { x j } , as required.

D

Corollary 5.4.11 5(A) is a Cohen-Macaulay complex and the f -vector of A is related to the h-vector of 5(A) by

/i(A)=/i i + 1 (S(A)) for all i.

154______________________________________Chapter

5

Proof. The first assertion follows from Theorem 5.3.18 and the previous result. For the other claim set J = (/A, x\,..., x^) and note that the face ring A;[5'(A)] = k[x,y]/I'^ reduces to k[x]/J modulo z = {yi — Xi,... ,yn — xn}. Thus z is a system of parameters for fc[5(A)]. Since 5(A) is CohenMacaulay, z must be a regular sequence on fc[5(A)] according to Proposition 2.2.7. Using Theorem 4.2.5 one has that the /i-vector of 5(A) is then equal to the /z-vector of k[x]/J, which is nothing other than /(A) shifted by one. D

Exercises 5.4.12 Let A be a simplicial complex of dimension d and /i(A) = (hi) its

/i-vector. Prove hd+i = (-l) d x(A),

where x(A) is the reduced Euler characteristic of A. 5.4.13 Let 5 = A;[A] be the Stanley-Reisner ring of a simplicial complex A over a field k and (p(t) the Hilbert polynomial of S. Prove: (a) ip(i) = dimfc(Si) for all i > 1, and (b) <^s(0) — 1 if and only if x(A) = 0. 5.4.14 Let A be a simplicial complex of dimension d and ( / o , . . . , fd) its

/-vector. Prove that fd is equal to e(fc[A]), the multiplicity of /c[A]. 5.4.15 Let S — fc[A] be the Stanley-Reisner ring of a simplicial complex A with vertices xi,... ,xn and A = S/(x\,... , x ^ ) - Prove that A is pure if and only if all the nonzero elements of the socle of A have the same degree.

5.5

Upper bound conjectures

It is our intention here to give flashes of some of the steps in the proof of

the upper bound conjecture for simplicial spheres. An excellent reference for a detailed proof of this conjecture is [44]. Convex polytopes

A convex polytope P C M s , or simply a polytope, is

the convex hull of a finite set of points x\,..., xn in E s , that is, P is the set of all convex combinations of those points: P = {aiXi + • • • + anxn Oj > 0, 01 + • • • + a n = 1, Oj e E}.

For A C fi s , let aff(A) be the affine space generated by A. Recall that aS(A) is the set of all affine combinations of points in A: aff(A) = {aipi + ••• +arpr\pi e A, a: + • • • + a r = 1, a» € E}.

Monomial Ideals and Stanley-Reisner Rings_____________ 155

One can represent aS(A) as aff(A) = XQ + V, where xo € Es and V is a (unique) linear subspace of Es . The dimension of A is defined as dim(A) = di Given y £ Rs \ {0} and a 6 ffi, define the hyperplane

,a) = {xeW.s\(x,y)=a},

where { , ) is the usual inner product in Es . The two closed half spaces bounded by H(y,a) are H+(y,a) = {x<E M.s\(x,y) > a} and H-(y,a) = H+(-y, -a). Definition 5.5.1 A proper face of a convex polytope P is a subset F C P such that there is a supporting hyperplane H(y,a) satisfying:

(b)P<£H(y,a)andPcH-(y,a).

The improper faces of a polytope P are P itself and 0. It is usual to call vertex to a face of P of dimension zero; it is a result of polyhedral geometry that P is in fact the convex hull of its vertices, see [39, 315].

Proposition 5.5.2 // P C Es is a convex polytope, then P is a compact

convex subset of Es with a finitely many faces and any face of P is a convex polytope. Proof. See [39].

D

A set of points 7/1, . . . , ym G Es is affinely

independent if the relation

ajj/j = 0 with ai + • • • + a m = 0 and aj 6 can hold only if a» = 0 for all i. Definition 5.5.3 A q-simplex is a polytope generated by q + 1 affinely independent points. A polytope is called a simplicial if every proper face is a simplex.

Let P be a polytope of dimension d+1 and denote by /, the number of its faces of dimension i, thus /o is the number of vertices of P. The f -vector of P is the vector /(P) = (/o, . . . , /d) and we set /_i = 1.

156______________________________________ Chapter 5

Simplicial spheres Let A be a finite simplicial complex with vertices X i , . . . ,xn and ei the ith unit vector in W1 . Given a face F 6 A set

|P| =conv{ei\xi e P}, where "conv" denotes convex hull. Define the geometric realization [A| of A as

IAI= u in FeA

thus A is a topological space with the induced usual topology of W1 . The boundary complex A(P) of a simplicial polytope of dimension d+ 1 is by definition the "abstract simplicial complex" whose vertices are the

vertices of P and whose faces are those sets of vertices that span a proper face of P, thus A(P) has dimension d. Note that P and A(P) have the same /-vector and accordingly one defines the h-vector of P as the /i-vector of A(P). The geometric realization |A(P)| is therefore homeomorphic to a of-sphere Sd. This suggest the following more general concept. Definition 5.5.4 plicial sphere if

A simplicial complex A of dimension d is called a sim-

|A|~5d

and in this case A is said to be a triangulation of Sd.

Theorem 5.5.5 then

Let A be a simplicial complex of dimension d. //|A| = Sd,

Proof. See [44, Chapter 5].

D

Proposition 5.5.6 //A is a simplicial sphere, then A is Cohen-Macaulay and the following equality holds x(lkP) = ( - l ) d i m l k f for F G A .

Proof. Use Reisner's criterion and Theorem 5.5.5.

D

Remark 5.5.7 If P is a simplicial
*(P) = x(A(P)) = £) (-l)Vi = (-1)""1 = -1 + X(P), i=-l

where / = (/o, . . . , f d - i ) is the /-vector of P. The Euler formula is valid for any d-polytope, not just the simplicial ones (see [39, Theorem 16.1]).

Monomial Ideals and Stanley-Reisner Rings_____________ 157 Corollary 5.5.8 //A is a simplicial sphere, then A; [A] is a Gorenstein ring. Proof. It follows from [274, Theorem 5.1] and Theorem 5.5.5.

D

Theorem 5.5.9 / / A is a simplicial sphere of dimension d, then its hvector satisfies hi = hd+i-i for i = Q,...,d+l. Proof. Since fc[A] is a Gorenstein ring, its /i-vector is symmetric according to Exercise 4.3.9. D Remark 5.5.10 If A is a simplicial sphere, then the linear relations

of its h- vector contain the Euler formula as a particular case (make i = 0) .

Those relations are called the Dehn-Sommerville equations. Corollary 5.5.11 If h = (ho, . . . , hd+i) is the h-vector of a simplicial polytope of dimension d + 1, then h satisfies

hi = hd+i-i for 0 < i < d + I. Let us mention two conjectures raised around 1993 in connection with the problem of bounding the Betti numbers of graded Gorenstein ideals [222]. Conjecture 5.5.12 (Miller- Villarreal) Let A be a simplicial sphere and I the Stanley-Reisner ideal of A. If I has initial degree p and height g, then for 1 < i < g the ith Betti number /Hi of fc[A] satisfies:

Conjecture 5.5.13 (Miller- Villarreal) Let A be a simplicial sphere and I its Stanley-Reisner ideal. If I has initial degree p and height g, then

9-1

\ 9-1

158______________________________________ Chapter 5

Cyclic polytopes In order to formulate the upper bound conjecture for simplicial spheres and the upper bound conjectures for convex polytopes we

need to introduce some results on cyclic polytopes, the reader is referred to [39] for a detailed discussion of this distinguished class of polytopes. Consider the monomial curve F C Md+1 given parametrically by

A cyclic polytope, denoted by C(n, d+ 1), is the convex hull of any n distinct points in T such that n > d + 1, The combinatorial type or /-vector of C(n, d + 1) depends only on n and d and not on the points chosen and dim C(n,d+l) = d+l. One of the relevant properties of cyclic polytope is that they are simplicial, thus the boundary 8C(n,d+ 1) defines an abstract simplicial complex

A(n,d + 1) = A(C(n,d + 1)) so that |A(n,d + 1)| = Sd. Notice that dC(n,d + 1) is a union of proper faces (in fact of facets) of C(n,d + 1). A polytope P has the highest possible number of f-faces when every set of i + I vertices is the vertex set of a proper face of P. In this case it is said that P is (i + l)-neighbourly.

Theorem 5.5.14 A cyclic polytope C(n,d+ 1) is [(d+ l)/2\-neighbourly. In particular

i/

I 2

Proof. See [39].

n

Corollary 5.5.15 If h = (ho,..., hd+i) is the h vector of C(n,d+l), then

for 0 < fc <

k

Proof. Let 0 < k < [(d + 1)/2J. One has

s •i=0 n - d+ k k

- ,

,=0

,V(k-i) fk-d-2\ (n . * - i A*

The second equality uses Eq. (4.1) and the last equality follows using induction on fc, or setting a = n and c = k — d — 2 in:

, , fora,ceZ. For a proof of this formula see [39, Appendix 3].

D

Monomial Ideals and Stanley-Reisner Rings_____________ 159

Upper bound conjecture for simplicial spheres Next we present the upper bound conjecture for simplicial spheres. It was conjectured by V.

Klee and proved in 1975 by R. Stanley [268]. It was motivated by the performance of the simplex algorithm in linear programming. Conjecture 5.5.16 (UBCS) Let A be simplicial complex of dimension d with n vertices. If |A| = Sd, then

for i = Q,...,d. Proof, (sketch) Consider the following four conditions:

(a) /ii(A) < hi(A(C(n, d + 1))) for 0 < i < [(d + 1)/2J . (b) /i i (A(C(n, d + 1))) = hd+i-i(&(C(n, d + 1))) for 0 < i < d + 1. (c) ftj(A) < ( i+ "- d - 2 ) for all i. (d) /ii(A) = /i d + i_i(A) for 0 < i < d + I . Observe that (a), (b) and (d) imply the upper bound conjecture. To show it notice that if [(d+ 1)/2J
fci(A) - ft d +i_i(A) < hd+1-i(^(C(n, d + 1))) - hi(&(C(n, d + 1))), hence /li(A) < /i i (A(C'(n, d + 1))) for 0 < i < d + 1. Therefore d+l / j , -i

-\

d+1

and the proof of the observation is complete. By Corollary 5.5.15 (c) =>

(a), and by Corollary 5.5.11 (b) is satisfied because C(n,d + 1) is a simplicial polytope. As a consequence the upper bound conjecture is reduced to prove (c) and (d). To complete the proof notice that (c) follows from Theorem 5.4.8, and (d) follow from Theorem 5.5.9. D As a particular case of the UBCS one obtains the upper bound conjecture for convex polytopes, this conjecture was posed by Motzkin in 1957 and proved in 1970 by McMullen [221],

Corollary 5.5.17 (UBCP) If P is a convex polytope of dimension d with n vertices, then

fi(P) < f i ( C ( n , d ) )

for 0
160______________________________________ Chapter 5 Proof,

(sketch) By a process known as "pulling the vertices" P can be

transformed into a simplicial polytope with the same number of vertices as P, and at least as many faces of higher dimension. Hence we may assume that P is simplicial. Since P is simplicial the boundary complex A(P) is a simplicial sphere, that is, |A(P)| = 5 d ~ 1 . Using the UBCS we obtain

for i = 0, . . . , d.

n

Remark Not every triangulation of a sphere is the boundary complex of some simplicial polytope. There are examples, found by Griinbaum in 1965, with d = 4 and n = 8 satisfying |A| = Sd~1 and A not isomorphic to A(P) for every d-polytope. Moreover it follows from a result of Steinitz (a graph is the 1-skeleton of a 3-polytope in R3 if and only if it is a 3-connected

planar graph) that such an example does not exist for d = 3.

Exercises 5.5.18 If A is a connected simplicial complex having all its vertices of degree two, then A is a cycle. 5.5.19 If A is a simplicial sphere on n vertices whose geometric realization |A| is isomorphic to the unit circle S1 , then

(a) A is a cycle of length. (b) i/(J A ) = (3) - " for n > 4. 5.5.20 ([18]) Let R = k [ x i , . . . ,xn] be a polynomial ring over a field k. If / is a monomial Gorenstein ideal and rad (/) = (x\ , . . . , xn), prove that 17- _

(-.0.1

— (xl

On\

) • • • ixn I

Hint Prove that / is irreducible. See [43] and [218, Theorem 18.1]. 5.5.21 Give an example of a monomial Gorenstein ideal which is not a complete intersection.

Chapter 6

Edge Ideals Edge ideals are the most simple polynomial ideals after face ideals, they are generated by square-free monomials of degree two and they can be associated to graphs and to Stanley-Reisner complexes. At first sight the use of graph theory to study edge ideals is convenient because there are common objects of study, but in fact some remarkable applications of graph theoretical results will take place in this chapter. There are many works in the literature relating graph theory with other branches of algebra, see for

instance [17, 24, 135, 136, 164, 262, 263]. In the first section the connection between ideals and graphs is established through the notion of minimal vertex cover. Some graph theory results and terminology are recollected and various versions of a theorem due to Konig are discussed. As we learned graph theory from [28] and [139], those remain as our main references for the subject and they are highly recommended. We study some properties of edge ideals, such as the Cohen-Macaulay property, and stress that some of those properties can be better understood using graph theoretical notions. A structure theorem for Cohen-Macaulay trees is presented in Section 6.3; this leads to a deeper result showing that Cohen-Macaulay bipartite graphs can be defined recursively. The notion of "ideal of covers" of a monomial ideal is introduced in Section 6.7, and we study the connection between properties of an edge ideal and properties of its ideal of covers, the notion of triangulated graph is used to link properties of those two ideals.

6.1

Graphs and ideals

A simple graph G consists of a finite set V of vertices and a collection E of subsets of V called edges such that every edge of G is a pair {vi,Vj} for

161

Chapter 6

162

some Vi,Vj in V. Note that a simple graph G has no loops, that is, we are requiring i>j ^ Vj for all edges { v i , V j } of G. When working with several graphs it is convenient to write V(G) and E(G) for the vertex set and edge set of G respectively. Let H and G be two graphs, it is said that H is a subgraph of G if V(H) C V(G) and E(H) C E(G). A spanning subgraph is a subgraph H of G containing all the vertices of G.

Let G be a graph on the vertex set V. If z = {v$, Vj} is an edge of G one says that the vertices Vi and Vj are adjacent or connected by z, in this case

it is also usual to say that the edge z is incident with the vertex v^. The degree of a vertex v in V, denoted by deg(i>), is the number of edges incident with v. A vertex of degree one (resp. degree zero) is called an end vertex (resp. isolated vertex). If all the vertices of G are isolated, G is called a discrete graph.

A walk of length n in G is an alternating sequence of vertices and edges w = {v0,zi,vi,...,vn-i,zn,vn}, where z, = {w,_i,i>i} is the edge joining Vi-i and Uj. A walk may also be written {VQ, . . . , vn} with the edges understood, or zi,Z2,...,zn with the vertices understood. If VQ = vn, the walk w is called a closed walk. A path is a walk with all its vertices distinct. We say that G is connected if for every pair of vertices v\ and v% there

is a path from vi to v%. Note that G has a vertex disjoint decomposition

1=1 where GI , . . . , Gr are the maximal (with respect to inclusion) connected subgraphs of G, the Gi are called the connected components of G.

If z is a edge, we denote by G\ {z} the spanning subgraph of G obtained by deleting z and keeping all the vertices of G. The removal of a vertex v from a graph G results in that subgraph G \ {v} of G consisting of all the vertices in 7? except v and all the edges not incident with v. For instance:

G\{Xl}

If A is a set of vertices of G one can define G \ A recursively by removing one vertex at a time.

Edge Ideals_____________________________________163

A cycle of length n is a closed walk {VQ, ... ,vn} in which n > 3 and the vertices Vi,... ,vn are distinct. A cycle is even (resp. odd) if its length is even (resp. odd). We denote by Cn the graph consisting of a cycle with n vertices, C^ will be called a triangle, C^ a square and so on. A tree is a connected graph without cycles, and a forest is an acyclic graph. The complete graph K,n has every pair of its n vertices adjacent. For instance:

Tree

A graph G is bipartite if its vertex set V can be partitioned into disjoint

subsets V\ and Vz such that every edge of G joins V\ with VzThe distance d(u, v) between two vertices u and v of a graph G is defined to be the minimum of the lengths of all possible paths from u to v. If there is no path joining u and v, then d(u,v) = oo.

Proposition 6.1.1 A graph G is bipartite if and only if all the cycles of G are even.

Proof. =£•) Let V\ and Vz be a partition of V(G) such that every edge of G joins Vi with Vz. If {VQ, • • . , vn} is a cycle of G, one may assume VQ £ V\. Note vi e l / 2 , it follows at once that Vi is in Vi if and only if i is even, thus n must be even.

•4=) It suffices to prove that each connected component of G is bipartite, thus one may assume G connected. Pick a vertex VQ 6 V. Set

FL = {v e y(GO|d(>,t>o) is even} and V2 = F(G) \ Vl. It follows that no two vertices of Vj are adjacent for i = 1,1, otherwise G would contain an odd cycle. Therefore G is bipartite. n

Corollary 6.1.2 If T is a forest, then T is a bipartite graph. Definition 6.1.3 A set of edges in a graph G is independent if no two of

them have a vertex in common. Let G be a graph on the vertex set V(G). Given a subset U C V(G), the neighbor set of U, denote NG(U) or simply N(U), is defined as N(U) = {v 6 V(G)

v is adjacent to some vertex in U}.

Notice that if Cr is a cycle, then Cr C N(Cr).

164______________________________________ Chapter 6

Proposition 6.1.4 LetG be a bipartite graph with vertex sets V\, V? having m and n vertices respectively. If

\A\ < \N(A)\ for all A C Vi, then there are m independent edges in G. Proof. By induction on m. If m = 1, then there is at least one vertex in

Vi connected to Vi, thus there is one independent edge. Assume m > 2. Case (I): First assume \A\ < \N(A)\ for all A C Vi with \A\ < m. Let x\ e Vi and yi 6 V-z be two adjacent vertices. Consider the subgraph

H = G\{Xl,yi} obtained from G by removing x\ and y\. Note that for every A C Vi \ {x\}

one has NH(A) = N(A) if yl $ N(A) and NH(A) = N(A) \ { y i } otherwise. Hence by induction there are z2,...,zm independent edges in H, which together with z\ = {xi,yi} yield the required number of independent edges. Case (II): Next assume \A\ = \N(A)\ for some A C Vi with \A\ < m. Set r = \A\. Consider the subgraph

H = G\(Vl\A). Since Nfj(S) = N(S) for all 5 C A, by induction there are independent edges z\, . . . , zr in H. On the other hand consider F = G\(AU N(A)). lfScVi\A, then

\N(A\JS)\ > \A\JS\ = \S\ + \A\, since

N(S\JA) = N ( A ) ( J N F ( S ) and r = \N(A}\ one derives IA/F^)! > \S\. Applying induction there are wi , . . . , wm-r independent edges in F. Altogether it follows that there are m independent edges in G, as required. This proof is due to Halmos and Vaughn. D Corollary 6.1.5

LetG be a bipartite graph with vertex sets V\, V%. If there

is an integer p > 0 such that \A\ — p < \N(A)\ for all A C V\, then there are m — p independent edges in G, where m = \V\\. Proof. By Proposition 6.1.4 the assertion holds for p = 0. Consider the graph F obtained from G by adding a new vertex y to V? and joining every

vertex of Vi to y. Let A C Vi, since NF(A) = N(A) U {y} one concludes 1-^(^4)1 > l - ^ - l ~ ( P ~ l ) - Hence by induction there are m—p+1 independent edges in F, it follows that there are m — p independent edges in G. D

Edge Ideals_____________________________________165

Let G be a graph with vertex set V. A subset A C V is a minimal vertex cover for G if: (i) every edge of G is incident with one vertex in A, and (ii) there is no proper subset of A with the first property. If A satisfies condition (i) only, then A is called a vertex cover of G and A is said to cover all the lines of G.

A — {xi,X2,x4,xe} minimal vertex cover of G.

It is convenient to regard the empty set as a minimal vertex cover for a graph

with all its vertices isolated. A set of vertices in G is independent if no two of them are adjacent. Notice that a set in G is a maximal independent set iff its complement is a minimal vertex cover for G. The vertex covering number of G, ao(G), is the smallest number of vertices in any minimal vertex cover.

Definition 6.1.6 The independence number of a graph G, denoted by 0i(G), is the maximum number of independent edges in G.

Theorem 6.1.7 (Konig) If G is a bipartite graph, then /?i(G) = ao(G).

Proof. Let V = V\ U V2 be a partition of the vertex set of G such that every edge of G joins Vi with V"2- Set r = Pi(G) and m — |Vi|. Since there are r independent edges in G one obtains «o(G) > r. Note there is A C Vi such that |AT(.A)| < \A\ — (m — r), otherwise if | AT (A) | > \A\ - (m - r) for all A C Vi, then by Corollary 6.1.5 one would have r +1 independent edges which is impossible. Therefore N(A) U (Vi \ A) is a vertex cover of G with at most r elements, thus ao(G)
A pairing off of all the vertices of a graph G is called a perfect matching. Thus G has a perfect matching if and only if G has an even number of vertices and there is a set of independent lines "covering" (containing) all the vertices. There are several results equivalent to Konig theorem (see [28, 139, 223]), next we present its connection with perfect matchings; see Proposition 8.8.11 for a generalization of the marriage problem. Theorem 6.1.8 (Marriage problem) If G is a bipartite graph with vertex set V, then the following are equivalent (a) G has a perfect matching.

(b) A\ < \N(A)\ for all A C V independent set of vertices.

166______________________________________Chapter

6

Proof. Let V\, V2 be a partition of V such that every edge of G joins V\ with V-2,. Set r = /3i(G). It is clear that (a) implies (b). (b) =>• (a): Since Vi is an independent set for i = 1,2, one readily derives \ViI = 1^21- By Konig theorem there are Zi,..., zr independent edges and a minimal vertex cover A of G with r elements, hence Z{ n A has exactly one vertex for any i and A is an independent set. Using that also V \ A is an independent set of vertices and the equality N(V \ A) = A we conclude

\V\A\<\N(V\A)\ = \A\. It follows that |Vi = r, thus Zi,... ,zr yields a perfect matching.

D

The cover complexity Given a graph G, the family of its minimal vertex covers will be denoted by T(G). One may call the integer |T(G)| the cover complexity of the graph G. We are interested in algebraic ways of grasping

this important number. The best known estimate is due to E. Sperner (cf. [112, Theorem 3.1]): Theorem 6.1.9 ([265]) // G is a graph with n vertices, then

'

n

N

>/u Proposition 6.1.10 Let

jn = sup {|T(G)| I G has n vertices} . // {Fn} denotes the Fibonacci sequence given by

F0=F1 = l and Fn = F n _i + F n _ 2 , then 7n < Fn for n > 1. Proof. The proof is by induction on n. The required inequality is clearly

satisfied if n < 4. Assume 7^ < Fk for k < n and n > 5. Let G be a graph with n vertices. We write G = GI UG 2 where GI is a connected component of G with at least two vertices. Let a: be a vertex in GI and xi,...:xr the vertices of G adjacent to x. Let TI be the set of minimal covers of G that contain x and T2 = T(G) \ Tj.. If A 6 TI notice that A \ {p} is a (possible empty) minimal vertex cover for G \ {x}, hence |Ti < Fn-i. If A € T 2 , then A \ {xi,... ,xr} is a (possible empty) minimal vertex cover for G \ {x,Xi,... ,xr}, which implies [T2 < Fn-r-i. Therefore 7ra < Fn as required. D Proposition 6.1.11 Let F0 = FI = I, F2 = 2 and Fn = F n _ 2 + F n _ 3 for n>3. If Pn is the path {xi,... ,xn}, then

Fn = |T(P n )|.

Edge Ideals_____________________________________ 167

Proof. To show it note: (i) The minimal vertex covers of Pn not containing x\ are in one to one correspondence with the minimal vertex covers of the graph Pn \ {xi,xz}, and (ii) the minimal vertex covers of Pn containing x\ are in one to one correspondence with the minimal vertex covers of the graph Pn \ {xi,x2,x3}. Q

Proposition 6.1.12 If T is a forest, then |T(T)| > a0(T) + 1. Proof. The proof is by induction on the number of vertices. Assume the inequality for all forests with less than n vertices and assume T is a forest with n vertices. Let z be a vertex of degree one in T and x the vertex in T adjacent to z. Set Tz = T \ {z}. Let A e T(T Z ). If x 6 A one has A € T(T), and if x £ A, then A U {z} 6 T(T). Hence |T(T)| > |T(T Z )|. If a 0 (T) = a 0 (T z ) the inequality follows by induction hypothesis. If

ao(Tz) < a 0 (T), then ao(T^) = 0:0(3") — 1. Therefore there is a minimal vertex cover A\ of T so that | AI| = a0(T) and z e Aj,. Notice that

A2 = (A, \ {*}) U {x} is not a minimal vertex cover for Tz because AI \ {z} is a minimal cover for

TZ . As a consequence A2 is a cover of T which is not a lift of a cover of Tz . The required inequality now follows using the induction hypothesis. D Some types of graphs For convenience let us recall the notion of CohenMacaulay standard algebra. Let R be a polynomial ring over a field k with irrelevant maximal ideal m and / a graded ideal of R. The depth of R/I, denoted by depth R/I, is the largest integer r so that there is a homogeneous sequence /i, ...,/,, in m with fa not a zero-divisor of R/(I, fi, . . . , f a - i ) for all 1 < i < r, here we set /o = 0. This number r is bounded by dim R/I according Lemma 1.3.6. Definition 6.1.13 The quotient ring R/I is called Cohen-Macaulay (C-M for short) if

and the ideal / is said to be Cohen-Macaulay if R/I is Cohen-Macaulay. Let G be a graph with vertices v\ , . . . , vn and

R = k[xi,. ..,£„] a polynomial ring over a field fc, with one variable Xi for each vertex Vi, we will often identify the vertex v^ with the variable Xi.

168______________________________________ Chapter 6

Definition 6.1.14 The edge ideal I(G) associated to the graph G is the ideal of R generated by the set of square-free monomials XiXj such that Vi is adjacent to Vj, that is,

({xixj\{vi,vj}€E(G)})cR. If all the vertices of G are isolated we set I(G) = (0). Note that the non zero edge ideals are precisely the ideals of R generated by square-free monomials of degree two. Definition 6.1.15 The graph G is said to be Cohen-Macaulay over the field k (C-M graph for short) if R/I(G) is a Cohen-Macaulay ring. According to Theorem 6.4.7 the Cohen-Macaulay property of a bipartite graph is independent of the field k, but in general this property may depend

on the field k (see Example 5.3.31). The next result establishes a one to one correspondence between the minimal vertex covers of a graph and the minimal primes of the corresponding edge ideal. Proposition 6.1.16 Let R = k[xi,. . . ,xn] be a polynomial ring over a field k and G a graph with vertices x\ , . . . , xn . If p is an ideal of R generated by A = {xjj , . . . , Xir }, then p is a minimal prime of I(G) if and only if A is a

minimal vertex cover of G. Proof. Note that I(G) C p if and only if A is a vertex cover of G. Assume that A is a minimal vertex cover of G, hence I(G) C p. By Proposition 5.1.3 any minimal prime of /(G) is a face ideal, thus p is a minimal prime of /(G). The converse is clear. D Definition 6.1.17 A graph G is said to be an unmixed graph if any two minimal vertex covers of G have the same cardinality.

Corollary 6.1.18 If G is a graph and I(G) its edge ideal, then the vertex covering number ao(G) is equal to the height of the ideal I(G). Proof. It follows at once from Proposition 6.1.16.

D

Definition 6.1.19 An edge x of a graph G is critical if

a0(G\x)
and G is called vertex- critical if all its vertices are critical.

Edge Ideals_____________________________________169

Proposition 6.1.20 Let G be a graph. If ao(G \ v) < ao(G) for some vertex v of G, then ao(G \ v) = cto(G) — 1.

Proof. Set r = O.Q(G\V) and note r + 1 < ao(G). Pick a minimal vertex cover A of G \ v with r vertices. Since v cannot be an isolated vertex

A U {v} is a minimal vertex cover of G. Hence a0(G) < r + 1 and one has the asserted equality.

d

If G is a graph without isolated vertices, then the relationships between the various notions defined above are the following: C-M =£• unmixed =>• vertex-critical •<= edge-critical,

where none of the arrows can be reversed in general. We now prove the first two implications and leave the third implication as an exercise.

Proposition 6.1.21 If G is a Cohen-Macaulay graph, then G is unmixed. Proof. By Corollary 5.1.5 I(G) is the intersection of its minimal primes. Therefore I(G) is an unmixed ideal by Corollary 2.2.4, thus G is unmixed by the correspondence between minimal covers and minimal primes. D

Proposition 6.1.22 // G is an unmixed graph without isolated vertices, then G is vert ex-critical.

Proof. Let v (E V(G). Since G is unmixed there is a minimal vertex cover A of G containing v and such that \A\ = a0(G). As A \ {v} is a minimal vertex cover of G \ v, one has a0(G \ v) < aQ(G). D Definition 6.1.23 A partially ordered set or poset is a pair P = (V, <), where V is a finite set of vertices and < is a binary relation on V satisfying: (a) u < u, Vw € V (reflexivity). (b) u < v and v < u, imply u = v (antisymmetry). (c) u < v and v < w, imply u < w (transitivity).

A poset P = (V, <) with vertex set V = {xi,... ,xn} can be displayed by an inclusion diagram, as with the diagram ,12

of the divisors of 12. In such a diagram X{ is joined to Xj by raising a line if Xi < Xj and there is no other vertex in between.

170______________________________________Chapter

6

Given a poset P with vertex set V = {x\.... ,xn} its order complex, denoted by A(P), is the simplicial complex on V whose faces are the chains (linearly ordered sets) in P, thus k[A(P)}=k[V}/(xixj\xi^xj) is the Stanley-Reisner ring of A(P). Here Zj •/ Xj means that Zj is not comparable to Xj. The Cohen-Macaulay property of those rings associated to posets has been extensively studied, see [12, 26, 44, 123, 162, 274] and the references there. Those simplicial complexes that occur as order complexes have been nicely characterized by Stanley [270], see also [162]. Example 6.1.24 Let P = (V, <) be a partially ordered set with vertex set V — {xi,... ,zg} ordered according to the next inclusion diagram. The Stanley-Reisner ring of the order complex A(P) is k[V]

facets of A(P) {zi,Z2,Z4,z 5 ,z 6 ,a;8} {zi,Z2,Z4,Z5,Z7,Z8} {zi,Z3,Z4,Z5,Z6,Z8}

Simplicial complexes of edge ideals Given a graph G on the vertex set V one defines a simplicial complex Simp(G) on the same set of vertices:

Simp(G) = {F| 3 H a subgraph of G such that F = V(H] and H ~ /C r }, where K,r denotes a complete graph on r vertices. Conversely a simplicial complex A defines a graph Skel(A), the 1-skeleton of A, on the same set of vertices: Skel(A) = {F€ A| dim(F) < 1}. In general one has A c Simp(Skel(A)). Proposition 6.1.25 Let A be a simplicial complex and /A its StanleyReisner ideal. Then A = Simp(Skel(A)) if and only if /A is generated by square-free monomials of degree two.

Proof. Let V = {xi,... ,xn} be the vertex set of A. Assume /A is generated by square-free monomials of degree two. Take F € Simp(Skel(A)), for simplicity set F = {xi,... ,xr}, that is, { x i , X j } for all 1 < i < j < r. If F $ A, then x\ • • • xr is in /A, and by assumption X{Xj is in /A for some 1 < i < j < r. Hence {xi,Xj} $ A, which is a contradiction. This shows A = Simp(Skel(A)). The converse also follows rapidly. d

Simplicial complexes associated to edge ideals have been studied in the literature, they are called flag complexes, see [274, Chapter III] and the references there for a series of amusing related problems.

Edge Ideals_____________________________________171

Exercises 6.1.26 Prove that G = Skel(Simp(G)), for any graph G.

6.1.27 Let G be a graph with vertex set V = {x\,... ,xn}, prove Euler's identity:

6.1.28 (P. Hall) Let X be a finite set and S i , . . . , Sm a collection of subsets of X. If k

IK

>k

for all i i , . . . , ik distinct, prove that there are x\,..., xm in X distinct such that Xi £ Si for all i.

Hint Consider the bipartite graph with vertex sets V\ = {Si,... ,Sm} and T/2 — X, where x G X is adjacent to Sj if x € Sj. 6.1.29 Let H be a subgroup of G of index n and G/H (resp. (G/H)L) the set of right (resp. left) cosets. Then there are gi,... ,gn in G such that G/H = {Hgi,..., Hgn} and (G/H)L = {9lH,.. .,gnH}.

Hint If G/H = {Hl,...,Hn} and (G/H)L = {H[,...,H£, define St as the set of j such that Hi n Hj ^ 0 and write

where Br is equal to s=l

with ii,... ,ir distinct. Note that at least r of the Br n H[ are non empty. 6.1.30 If G is a connected graph and d the distance function, then:

(a) d(x,y) > 0, and d ( x , y ) - 0 & x = y, (b) d ( x , y ) = d ( y , x ) (symmetry), (c) d(x,y) < d(x,z) + d ( z , y } (the triangle inequality).

6.1.31 Let G be a graph. If a 0 ( G \ x ) < a 0 (G) for some edge x of G, prove that a0(G \ x) = a 0 (G) - 1.

172

Chapter 6

6.1.32 If G is an edge-critical graph without isolated vertices, prove that G is vertex-critical. 6.1.33 In [234] there is a list of all the known edge-critical graphs with

fewer than nine vertices, prove that all these graphs are unmixed. Can you give an example of an edge-critical graph which is not unmixed?.

6.2

Cohen-Macaulay graphs

Let G be a graph on the vertex set V = {xi,...,xn}. complementary simplicial complex AG of a graph G by

We define the

AG = {^4 C V\ A is an independent set in G},

notice that AG is precisely the Stanley-Reisner simplicial complex of /(G). According to Reisner's Theorem 5.3.5 the Cohen-Macaulay property of a given graph can be decided once the reduced homology of the links of AG is computed. Unfortunately the simpler the graph G the more complicated the complex AG- Following [262, 302] we will study C-M graphs by looking at both the complementary simplicial complex and the graph itself.

Construction of Cohen-Macaulay graphs One of the purposes here is to show how large classes of C-M graphs can be produced and to show some obstructions for a graph to be C-M. w

First construction

Let H be a graph with vertex

set V(H) = {xi,... ,xn,z,w} and J its edge ideal. Assume that z is adjacent to w with deg(z) > 2 and deg(w) = 1. We label the vertices of H such that Xk X i , . . . , X k , w are the vertices of H adjacent to z, as shown in the figure. The next two results describe how the Cohen-Macaulay property of H relates to that of the two subgraphs G = H \ {z,w} and F = G \ {xi,... ,Xk}. One has the equalities J = (I, X i Z , . . . , x^z, zw] and (/, xi,..., x ^ ) = (L, x\ ..., Xk),

where / and L are the edge ideals of G and F respectively. Assume H is unmixed with height of J equal to g + I . Since z is not isolated, there is a minimal prime p over / containing {xi,...,Xk} and such that ht(/) = ht(p) = g. It is not difficult to prove that k < n and > 2 for i = l , . . . , f c .

Proposition 6.2.1 If H is a C-M graph, then F and G are C-M graphs.

Edge Ideals_____________________________________173

Proof. Set A = k[xi,... ,xn] and R = A[z,w\. By Proposition 9.3.2 there exists a homogeneous system of parameters {/i, • • • , / < * } for A/1, where fi 6 A+ for all i. Because of the hypothesis and the equalities

z(z — w) + zw = z1 and w(w — z) + zw = w2,

the set {/i,..., fd, z — w} is a regular system of parameters for R/J. Hence f i , . . . f d is a regular sequence on R/I, that is, G is C-M. To show the corresponding property for F we consider the sequence „/,

0 —

where the first map is multiplication by z and ijj is induced by a projection. The exactness of this sequence follows from Corollary 5.1.5. Taking depths

w.r.t the maximal ideal of R, by the depth lemma one has n — g + I < depth R/(!,XI, ... , X k , w ) , where g = ht(J).

Since (/, x\,..., Xk, w) = (L, x\,..., Xk, w), F is C-M.

D

Proposition 6.2.2 // F and G are C-M graphs and {xi,... ,Xk} form a part of a minimal vertex cover for G, then H is a C-M graph.

Proof. Consider the exact sequence of Proposition 6.2.1. Since the ends of this sequence have R-depth equal to dim (.R/J), by the depth lemma H is a Cohen-Macaulay graph. D Corollary 6.2.3 If G is C-M and {xi,...,Xk} is a minimal vertex cover

for G, then H is C-M. Proof. Note that in this case F is Cohen-Macaulay because I ( F ) = (0). D Second construction

For the discussion of the sec-

ond construction we change our notation. Let H be a graph on the vertex set V(H) = {xi,... ,xn,z}. -x Let {x\,... ,Xk} be the vertices of H adjacent to z, x as shown in the figure. Taking into account the first construction one may assume deg(zj) > 2 for i = 1 , . . . , k and deg(z) > 2. Setting G = H \ {z} and F = G \ {xi,..., Xk}, notice that the ideals J, / and L associated to H, G and F respectively are related by the equalities J = (I,xiz,.. .Xkz) and (I,x\,... ,Xk) = (L,xi,..., £&)• Proposition 6.2.4 If H is C-M, then F is C-M.

Proof. We set R = k [ x i , ...,xn,z], A = k[xi,... ,xn] and ht(J) = g + 1. The polynomial / = z - x\ - • • • - Xk is regular on R/J because it is

174______________________________________ Chapter 6

clearly not contained in any associated prime of J. Therefore there is a sequence {/, fi, • • • , fm} regular on R/J so that { f i , • • • , fm} C A+, where m = n — g — 1. Observe that {/1; . . . , fm} is in fact a regular sequence on A/ 'I, which gives depth(A/7) > n — g — 1. Next, we use the exact sequence 0

and ht(I,xi, . . . ,Xk) = g + 1 to conclude that F is C-M.

D

Proposition 6.2.5 Assume xi,...,Xk do not form a part of a minimal vertex cover for G and ht(/, x\ , . . . , Xk) = ht(/) + 1. If F and G are C-M, then H is C-M. Proof. The assumption on {xi, . . . ,xk} forces ht(J) = ht(/) + 1. From the exact sequence

we obtain that H is C-M.

D

Corollary 6.2.6 // G is C-M and {xi , . . . , Xk-i } is a minimal vertex cover for G, then H is C-M. Connected components of Cohen-Macaulay graphs A good property of Cohen-Macaulay graphs is its additivity with respect to connected components.

Lemma 6.2.7 Let R\ = k[x] and R% = k[y] be two polynomial rings over a field k and R = k [ x , y ] . If I\ and /2 are graded ideals in R\ and R% respectively, then depth(Ei//i) + depth(# 2 //2) = depth R/(Ii + J 2 ). Proof. We give two proofs. First proof: the formula is a direct consequence of Proposition 2.2.20 and Theorem 2.2.21. Second proof: by a change of base (Ri/h)®kRi — R/hR is a flat /?i-module. The equality now follows from a general property of tensor products [217, Theorem 50]. D Proposition 6.2.8 If G is a graph and GI,.. .,Gn its connected components, then G is C-M if and only if Gi is C-M for all i.

Proof. The result follows from Lemma 6.2.7.

D

Edge Ideals_____________________________________175

Complementary complex versus the complement The complement G of a graph G has vertex set V(G) and two vertices are adjacent in G if and only if they are not adjacent in G. Along the chapter we will encounter the complement G, and witness its usefulness to encapsulate information of the edge ideal /(G). Proposition 6^2.9 Let G be a graph and AG its complementary simplicial

complex. Then G C AG, with equality if and only if G has no triangles. Proof. If x = {vi,V2} is an edge of G, then {vi,v^} is an independent set of G and x G AQ. Hence G C AQ. Assume G has no triangles. If A — {vi,..., vn} is an independent set of G, then n = 1 or n = 2; otherwise if n > 2, then G would contain the triangle {VT_, v%, vs}. Therefore A is

either a point or an edge of G, thus G = AQ. The converse also follows rapidly. D In general a Cohen-Macaulay graph G can have all the subgraphs G\ {v} failing to be C-M, for v in G; see Example 6.2.24. Under special circumstances we will study the existence and distribution of the vertices of G whose removal preserve the C-M property, the case of bipartite graphs is analyzed in Section 6.4.

Definition 6.2.10 A cutpoint of a graph is a vertex whose removal increases the number of connected components.

Proposition 6.2.11 I f u , v are two vertices at maximum distance in a non discrete connected graph G, then u,v are not cutpoints.

Proof. Assume v is a cutpoint and write G \ {v} = G\ U • • • U Gs, where G i , . . . , G s are the connected components of G \ {v}. One may assume u £ GI, since s > 2 there is w e G, for some i > 2. Note that v is in any path joining u and w, thus d(u,w) > d ( u , v ) , a contradiction. This argument is due to Harary. D

Definition 6.2.12 A set of vertices D of a graph G is called dominant if every vertex of G is adjacent with at least one vertex in D. Lemma 6.2.13 Let G be a connected graph. If G is connected and has no triangles, then

D = {v e V(G) is a dominant set of vertices of G.

G \ {v} is connected}

176______________________________________ Chapter 6

Proof. Assume G \ {v} is disconnected for some v G V(G). We set

V(G) = {v,xi,...,x where x\ , . . . , xk are the vertices of G adjacent to v. Since G has no triangles

d(xi,Xj) = 2 for k + 1 < i < j < m, where d is the distance in G. Notice that at least one of the vertices {xk+i ,... ,xm} has degree greater or equal than two in_G, say xm. We order the vertex set so that xk and xm are adjacent in G. Let GI , . . . , G^ be the connected components of G \ {v}. We may also assume that Xk and xm are in V(G\). Notice that d ( x k , b ) > 3 for all b in V(G-i). Using Proposition 6.2.11 one has that^any pair of vertices at maximum distance in G must be in D. Therefore G \ {xi} is connected for some 1 < i < k, that is, x, is a vertex in D adjacent to v in G. D Proposition 6.2.14 Let G be a connected graph and G its complement. If G is connected and has no triangles, then

D= {z£V(G)\G\{z} is C-M} is a dominant set. Proof. Let AG be the complementary simplicial complex of G. Observe that AG = G, hence by Reisner's Theorem G is C-M. The rest of the proof is an immediate consequence of Lemma 6.2.13. D

Exercises 6.2.15 Show that a cycle Cn of length n > 3 is unmixed if and only if n = 3,4,5,7. 6.2.16 Let G be a graph such that G is connected and has no triangles. Show that I(G) is Gorenstein if and only if G is a cycle.

Hint It follows from [274, Theorem 5.1] that /(G) is Gorenstein iff G is a pseudo manifold. Since G is a graph it is a pseudo manifold iff it is a cycle. 6.2.17 ([309]) Let GI, G% be two graphs with vertex set x. If their complements are triangle free and they have the same number of edges, then the Hilbert series of fc[x]//(Gi) and fc[x]//(&2) are identical. 6.2.18 Let G be a graph on the vertex set V = { X I , . . . , X P } such that ht(/(G)) =p-2. Prove the following: (a) G is Cohen-Macaulay if and only if AG is connected, and (b) if /(G) is not Cohen-Macaulay, then /(G) is unmixed if and only if AG has no isolated vertices.

Edge Ideals_____________________________________177

6.2.19 If G is a C-M graph with q edges, prove that

q<(92 where g is the height of I(G).

6.2.20 A graph G with q edges is said to be saturated'^q = g(g+l)/2, where

g is the height of /(G), and G is C-M. Let G be a graph with connected components G i , . . . , G m . Prove that G is saturated if and only if Gi is saturated for all i. 6.2.21 If G is a saturated graph, prove that at most one of its connected components has more than one vertex. 6.2.22 Let G be a Cohen-Macaulay graph over a field k. Prove that G is

saturated if and only if R/I(G) has a 2-linear resolution. 6.2.23 Prove that the following graph is Cohen-Macaulay.

6.2.24 Prove that the ideal / = /A of Example 5.3.31 is generated by XiX7,

XiXiQ,

XiXu,

X2X4,

X2X5,

X2Xn,

X3X5,

X3X6,

X3X&,

X3Xn,

X4X6,

X5X7,

X5Xg,

X5Xn,

X6XS,

X6Xg,

X7X9,

X2XS,

Let Ri — k [ x i , . . . , Xi, . . . ,xu] and Gj = G \ {xi}. Note / = /(G) for some graph G. Prove that the /-vector of A Gi is (10, 25, 15) or (10, 24, 14). Thus the ^-vector of Ri/I(Gi) is (1,7, 8, -1) or (1,7,7, -1). Use Theorem 4.2.6 to conclude that Gj is not a C-M graph for all i. 6.2.25 If G is a Cohen-Macaulay graph and X i , x% are two vertices of degree

1, prove that they cannot have an adjacent vertex in common. 6.2.26 Let G be a connected graph and {zi, . . . , zn} a path with n > 4. If deg(zi) = 1, deg(rEj) = 2 for i = 2, . . .n — 1 and deg(z n ) > 3, prove that G cannot be unmixed.

178______________________________________Chapter

6.3

6

Trees

In this section we give an effective description of Cohen-Macaulay trees and show an interesting family of graphs containing all Cohen-Macaulay

trees. The first Koszul homology module of a Cohen-Macaulay forest is also studied.

Proposition 6.3.1 Let S be a positively graded algebra of dimension d over a field k. If A = k[hi,..., hj\ M- 5 is a homogeneous Noether normalization of S, then S is a Cohen-Macaulay ring if and only if S is a free A-module.

Proof. By Theorem 2.5.13 pdA(S) + depth(S) = dim(^l), where the depth of S is taken with respect to A+. Assume S is a free A-module, then

depth(S) = d and thus the depth of S with respect to 5+ is also equal d. Therefore 5 is Cohen-Macaulay. Conversely assume S Cohen-Macaulay. Note that A is a polynomial ring. Let p 6 Spec 5 be a minimal (graded) prime over A+S. There is an integral extension

k = A/A+ <-» S/p, hence S/p is a zero dimensional domain and consequently p is a maximal ideal. Therefore rad (A+S] = S+, that is, hi,..., hd is a h.s.o.p for 5. Now

use Proposition 2.2.7 to conclude that hi,... ,hd is a regular sequence in S, that is, depthS = d. Another application of the Auslander-Buchsbaum formula yields that S is a free /1-module. D Next we apply the above characterization of Cohen-Macaulay graded algebras to give a proof, due to Wolmer Vasconcelos, of the following key result:

Proposition 6.3.2 Let R = k[xi,... ,xn,yi,.. . ,yn] be a polynomial ring over a field k and M C {(i, j)\ 1 < i < j < n}. Then the ideal

I = ({%iyi,yjyt\ i = l , . . . , n and (j,l) e M}), is a Cohen-Macaulay ideal.

Proof. We start by doing a change of variables Zi = X{ — y, for i = 1 , . . . , n. Its effect is that A = k[zi,...,zn,yi,...,yn]/I becomes an integral graded algebra over k[zi,... ,zn], where

I = V +ViZ

Edge Ideals_____________________________________ 179

Observe that the images of y^ • • • yiT (1 < i\ < • • • < ir < n) not divisible by any of the monomials yjyi form a generating set for the k [ z i , . . . , zn]module A. To complete the proof we now show that they are in fact a basis. We set z = (zi, . . . , zn) and y = (yi, . . . , yn). Assume the following equality

Using induction we will prove that fi = fi^2 — • • • = fi1...im = 0. First we show fi = 0 for 1 < i < n. Making yj = 0 for j ^ i in the equation

above we obtain fi(z) — Qi(yi,z)(yi + z^), which forces fi = 0. We set the next induction step by assuming /» = f^^ = . . . = /, 1 ...t r ._ 1 = 0. The substitutions 2/j = 0 for yj ^ {y^, . . . ,yir} and yip — -ZIP for 1 < p < r into the first equality gives

therefore f i 1 . . . i r ( z ) = 0. Altogether A is a free module over k[zi,...,zn] and this implies that / is C-M by Proposition 6.3.1. Let us give a quite different second proof pointed out by Jiirgen Herzog. Note that the polarization of the ideal

J = ({x^XjXt]

1 < i < n and (j,l) £ M})

is equal to / and apply Proposition 5.1.7.

D

Lemma 6.3.3 Let T be a tree. If v and w are two adjacent vertices of degree at least two, then there is a minimal vertex cover for T containing both v and w. Proof. Let x be the edge joining v and w. Since T \ {x} has exactly two components, say T\ and TZ, there are minimal vertex covers A and B for

TI and TZ respectively so that v € A and w € B. Therefore A U B is the required cover for T. D

Theorem 6.3.4 Let T be a tree with vertex set V and edge set E. Then T is C-M if and only if \V\ < 2 or 2 < |V| = 2r and there are vertices xi,...,xr,yi,...,yr so that deg(o;i) = 1, deg(j/j) > 2, and {xj,j/j} e E for Proof. =>) As T is a bipartite graph there are disjoint sets Vi and V% such that V = Vi U Vi and every edge of T joins V\ with Vi- Since V\ and Vi are minimal vertex covers for T, and T is unmixed, we have

r = \V1 =\V2\=a0(T).

180______________________________________ Chapter 6

By Theorem 6.1.7 there are r independent edges. Therefore we may assume Vi = {zi,...,zr}, V-2 = {wi,...,wr}, and {zi,Wi} € E for i - l , . . . , r . To complete the proof is enough to show that for each i either zi or Wi has degree one. Assume deg(zj) > 2 and deg(wi) > 2 for some i. By Lemma 6.3.3 there is a minimal vertex cover for T, say A, containing Zi and Wi. Since {zj,Wj} H A ^ 0 for j ^ i we conclude |.A| > r + 1, which is a contradiction. •4=) The sufficiency follows from Proposition 6.3.2. D Corollary 6.3.5 If G is a tree, then G is C-M if and only if G is unmixed.

Proof. If G is unmixed, using the proof above it follows that /((?) is a C-M ideal of the kind described in Proposition 6.3.2. D Corollary 6.3.6 The only C-M cycles are a triangle and a pentagon.

Proof. Let Cn — {XQ,XI, . . . ,xn = XQ} be a C-M cycle of length n. The cases n = 4, 6, and 7 can be treated separately, so we assume n > 8. By Proposition 6.2.4 Cn \ {XQ,XI, x n -i} is a C-M path of length n — 3 which is impossible by Theorem 6.3.4 D Proposition 6.3.7 If I = I(T) C R is the edge ideal of a Cohen-Macaulay forest with In vertices and without isolated vertices, then

Proof. By Theorem 6.3.4 there are vertices X i , . . . ,xn, yi, . . . ,yn with deg(xj) = 1, deg(j/j) > 1 for all i, such that / has a generating set:

{xiyi,...,xnyn,yjye\

(j,€) € M},

where M C [l,n] x [l,n].

Set R = fc[x, y], where k is a field. We proceed by induction on n. If / is a complete intersection or 1 < n < 2 the result follows from Theorem 3.3.14 together with the exact sequence 0 —> I /I2 —> R/I2 —» R/I —> 0. Notice that if / is not a complete intersection, then T has a vertex of degree two, its existence follows from Theorem 6.3.4 and the fact the sum of the degrees of the vertices of a graph is twice the number of its edges. We order the variables so that deg(y n ) = 2 and yn-iyn G / and assume the result true for CM forest with less than 2n vertices. The ideal of R obtained by evaluation of / at yn = 0 will be denoted by J, while L will denote the ideal in R obtained by evaluation of J at yn-i = 0. We also set K - (J2,yn-iyn,xnyn).

Edge Ideals_____________________________________ 181

In the sequel we shall also make use of the fact that, if / is a monomial ideal, and / is a monomial, then (/: /) is a monomial ideal. Consider the complex (easily shown to be exact by the preceding remark):

0 — >• RKL^y^) ^ R/(J2,yn_iyn) —> R/(J2,yn) —> 0.

(1)

By an appropriate induction hypothesis

depth(.R/(L 2 ,2/ n _i)) > n and depth(R/(J2, yn)) >n-l. Hence from (1) depth(jR/(J 2 ,2/ n _i2/ ra )) > n - 1, which together with the exact sequence

O ^ E / ( L 2 , <;„_!) *"-^ R/(j^yn_iyn)^R/K^Q

(2)

gives depth(R/K) >n — l. Because of the equality K — (I2, y n _iy n , xnyn) we shift our attention to the exact sequence

0 — >• R/(J, yn-iyn, xn) """-^

R/(I2, xnyn) — y R/K — > 0.

(3)

To estimate the depth of the module on the left, we use the exact sequence

0 —-> fl/(L,y n _i) -^ R/(J,yn-iyn) — »• #/(,/, y n ) —> 0

(4)

Taking into account that L and J are attached to CM forests, we have that the left end of the last sequence has depth equal to n + 1 while its right end has depth equal to n. Hence a combination of (3) and (4) gives depth(fl/(/ 2 ,a; n j/ n )) >n-l. To complete the proof observe that from the sequence

O^R/IXn^ we get depth(#// 2 ) > n - 1.

R/I2—>R/(I2,xnyn)—+0

(5) D

Exercises 6.3.8 Let T be a tree with 2r vertices. If r > 2, prove that T is CohenMacaulay iff ht /(T) = r and T has exactly r vertices of degree 1.

6.3.9 Let G be a graph with vertex set V = {x\ , . . . , xn, y\ , . . . , yn} and edge set X = { { x k , y k } , {y«, Vj}\ k = 1, . . . , n a n d l
182______________________________________Chapter

6.4

6

Bipartite graphs

Let G be a Cohen-Macaulay bipartite graph and / = /(G) its edge ideal.

We prove that G \ {v} is Cohen-Macaulay for some vertex v in G. Then we show that the Stanley-Reisner simplicial complex of / is shellable. Cohen-Macaulay Bipartite graphs

A graph G is a complete bipartite

graph if its vertex set V can be partitioned into disjoint subsets Vi and V% such that G contains every edge joining Vi and V2. If V\ and V2 have m and n vertices respectively, we denote such a complete bipartite graph by /C m>n . In the following illustration Vi (resp. V 2 ) is the set of vertices in the top (resp. bottom) row.

Lemma 6.4.1 The complete bipartite graph /Cm,n is Cohen-Macaulay if and only if m = n = 1. Proof. By Exercise 5.3.22 it is enough to observe that the complementary complex of TCm>n is disconnected for m + n > 3. D

Lemma 6.4.2 Let G be an unmixed bipartite graph and I(G) its edge ideal. If I(G) has height g, then there are disjoint sets of vertices Vi = {xi,... ,xg}

and ¥2 = {yi,...,yg} such that: (i) {xi,yi} is an edge of G for alii, and (ii) every edge of G joins V\ with V 2 .

Proof. The statement follows rapidly if one uses that g is equal to the maximum number of independent edges of G, see Theorem 6.1.7. D

Definition 6.4.3 Let G be a graph and v £ V(G). of v is the set of vertices of G adjacent to v.

The neighbor set N(v)

Theorem 6.4.4 Let G be a Cohen-Macaulay bipartite graph. If G is not a discrete graph, then there is a vertex v € V(G) such that deg(t>) = 1.

Proof. We may harmlessly assume that G has no isolated vertices. Set g = ht/(G). Let Vi and V2 be as in Lemma 6.4.2. We proceed by contradiction. Assume deg(w) > 2 for all v € Vi. Let v 6 Vi be a vertex of minimal degree. We may assume v = x\. If deg(i>) = g , then G = /C 5jff is a complete bipartite graph, but this is impossible by Lemma 6.4.1. Therefore we may also assume 2 < deg(zi) < g - 1. Set r = deg(a;i). Let N(XI) be the set of vertices in V adjacent to x\, say N(x\) = {3/1,... ,yr}.

Edge Ideals_____________________________________ 183

We claim that deg(xi) = r, for alH = 1, . . . , r. For simplicity we assume i = 2. If deg(z2) > 71, then x-2 is adjacent to yj for some r + 1 < j < g.

Let G" = G \ {xi,yi, . . . ,yr}. By Proposition 6.2.4 we derive that G" is C-M, and hence G' is unmixed. Notice that {yr+i, . . • ,yg} is a minimal vertex cover for G' and there is also a minimal vertex cover for G" containing {x2, xr+i . . . , xg}, which is a contradiction since G" is unmixed. Thus deg(x2) = r. The previous argument also shows that N(XI) = {yi, . . . ,yr},

for all 1 < i < r.

Hence N(yi) is a subset of {xr+i, . . . ,xg}, for all

r + 1 < i < g. Consider the graph

H = G\ ({yr+i,. • • ,ya} U N(yr+l) U • • • U N ( y g ) ) .

To complete the proof observe that a repeated use of Proposition 6.2.4 shows that H is a C-M bipartite graph. This is impossible since H = K,r,r is a complete bipartite graph and r > 2. D Corollary 6.4.5 If G is a Cohen-Macaulay bipartite graph, then G \ {v} is C-M for some vertex v in G.

Proof. Let {v,z} be an edge of the graph G such that deg(w) = 1. Then by Proposition 6.2.1 we obtain that G \ {z} is C-M. D The results above yield a recursive algorithm to construct all the C-M bipartite graphs with a given number of vertices, see exercises.

Shellability The vertex set and edge set of a graph G are denoted by V(G) and E(G] respectively. Recall that the faces of AG, the complementary simplicial complex of G, are precisely the independent sets of G (together with the empty set). Notice that the facets (faces of maximal dimension) of AG are the maximal independent sets of G. Let G be a graph on the vertex set V — {zi, . . . ,zn,x,y}. Assume that x is adjacent to y and deg(t/) = 1. We label G so that y, Zi, . . . , Zk are the vertices of G adjacent to x.

Lemma 6.4.6 Let H = G\ {x,y} and A c V. If G is Cohen-Macaulay, then A is a facet of AG if and only if A can be written as

(a) A = F U {y}, where F c V(H) is a facet of A//, or (b) A = F U {x}, where F C V(H) is a facet of A// and z\ , . . . , z^ are not in F.

Proof. Let A be a facet of A G . Suppose x £ A, then y e A. Set W = V(H) and F = A \ {y} C W. We must show that F is a facet of A//. First notice that F is an independent set of H, because E(H) C E(G). Next let us show that F is not contained in any other independent set of H. Let v G W \ F.

184______________________________________Chapter

6

Since A is a facet of AG, we obtain that there is an edge L in E(G) so that L C A U {v}. As x £ A and degQ/) = 1, we have L e E(H) and L C F ( J { v } . Hence F is a facet of AH and we are in case (a). On the other hand if x € A, then we have y,z\,... ,Zk $ A. Set F = A \ {x}. Let C be a facet of AH containing F. Since C U {y} is an independent set of G, and G is C-M, we obtain that |G| + 1 < |A| = |F| + 1. Hence C = F and we are in case (b), as required. The converse also follows readily using similar arguments. D Theorem 6.4.7 Let G be a Cohen-Macaulay graph. If G is bipartite, then the Stanley-Reisner simplicial complex AG of /(G) is shellable. Proof. We will use the shellability criterion of Exercise 5.3.29. One may

assume that G has at least one vertex of positive degree, for otherwise AG is a simplex with a unique facet. Set g = ht/(G). We argue by induction on g. Let V\ and 1/2 be as in Lemma 6.4.2. By Theorem 6.4.4 we may assume that deg(yi) = 1. Let us label the vertices of G so that y2, • • • ,Vt be the vertices of G (if any) adjacent to x\. According to Proposition 6.2.1 the graph H = G \ {zi,2/i} is C-M, hence by induction AH is shellable. Let FI, ... ,Fm be a shelling of AH. By Lemma 6.4.6 there are integers 1 < TI < • • • < rs < m so that the facets of AG can be linearly ordered as follows F1\J{yi},...,Fm\J{y1},FriU{xi},...,Fr,

U{zi},

where Fri,..., FTa are the facets of H not containing y % , . . . yt (or equivalently containing x2,...,xt). We claim that the subcomplex A' of AG generated by Fri U {xi}, ...,Fri U {xi} is shellable, with shelling given by Fn U {zi}, • • . , F T f U {xi}. To show the claim assume 1 < j < i < s. Since AH is shellable there is v £ Fri \ Frj and I < r» so that Fri \Ff = {v}. It follows that X2, • • • , xt € FI. Hence FI = Frk, for some 1 < k < i. Therefore A' is shellable. It now follows readily that the linear ordering of the facets of AG given above is a shelling for AG. D

Computing the type Let G be a C-M bipartite graph on the vertex set V. Assume G has no isolated vertices. There are disjoint sets Vj and 1/2 so that V = l/i U V-2 and every edge of G joins Vi with 1/2- Since V\ and 1/2 are minimal vertex covers for G, and G is unmixed, we have 9 = \Vi\ = \Vi\ = height/(G). By Konig's theorem there are g independent edges. Therefore we may assume

Vi = {yi,...,yg}, 1/2 = {wi,...,wg}, and y,Wi € /(G) for all i. Notice that w = {yi — Wi}3i=i is a regular system of parameters for _R//(G), where R = &[y,w], and R/I(G) modulo (w) reduces to:

Edge Ideals

185

where H is a graph on the vertex set Vi. The algebra A is generated, as a k-vector space, by the image of 1 and the images of all the monomials 2/ii • • • yir such that {y^,..., yir} is an independent set of H. Observe that the type of R/I(G) is equal to |T(.ff)|, because Soc(A) is generated by the images of all the monomials y^ • • • yir such that {y^,..., j/j r } is a maximal independent set of H. Example 6.4.8 Let G be the C-M bipartite graph whose edge ideal is: I(G)

= (2/1^1,2/1^2,2/1^3,2/1^4,2/2^2,2/2^3,^2^4,2/3^3,2/4^4).

The graph H obtained by reduction has edges 2/12/2,2/13/3,2/12/4,2/22/3,2/22/4Since this graph has only three minimal vertex covers, the type of R/I(G) is equal to three.

Exercises 6.4.9

Let G be a C-M bipartite graph. Show that the /i-vector of R/I(G)

has length at most g = htI(G). Can we replace C-M by unmixed? 6.4.10 Prove that the following is the full list of Cohen-Macaulay connected bipartite graphs with eight vertices, in each case determine the type and the multiplicity of the edge ring.

6.4.11 Let / = (zjj/j'l 1 < i < j' < n), draw a picture of the bipartite graph G such that / = I(G) and prove that / is a Cohen-Macaulay ideal. 6.4.12 Let G be a Cohen-Macaulay bipartite graph. If G is not a discrete graph, prove that G has at least two vertices of degree one.

6.5

Links of some edge ideals

In this section we give graph theoretical interpretations for the links of

certain ideals attached to graphs.

186______________________________________ Chapter 6

Let GO be a graph on the vertex set V = {y\, . . . ,yn} and / = /(Go). Take a new set of variables x\ , . . . , xn and define 5 (Go) to be the new graph obtained by attaching to each vertex j/j a new vertex Xi and the edge {xi, yi}. We set K = /(5(G 0 )) = ( x i y 1 } . . . , xnyn,I) CR = fc[x, y], where k is a field. The 5 in the notation is intended for suspension or attaching whiskers to the base graph. The significance of this construction lies partly in the following restatement of Theorem 6.3.4: Theorem 6.5.1 If G is a graph, then G is a Cohen-Macaulay tree if and

only if G = S(G0) for some tree G0. Of course, all the structure of S(Go) lies in the graph GQ. The following simple computation conforms to this view.

Proposition 6.5.2 Let GO be a graph and G = S(Go) its suspension. Then

the multiplicity of R/I(G) is d

j_£/«, _^ where /» is the number of i -faces of the Stanley-Reisner complex of I (Go).

Proof. Let yi,...,yn be the vertices of G 0 . By Proposition 6.3.2, the set of linear forms {j/j - Xi}f=1 is a regular system of parameters for R/I(G). The reduction of R/I(G) modulo these forms is

whose k- vector space basis is formed by the standard monomials, that is, the faces of the attached simplicial complex. D The next proposition describes the first link of S(Go) relative to the regular sequence z = {xij/i, . . .,xnyn}.

Proposition 6.5.3 Let GO and G\ be two copies of the same graph on disjoint sets of vertices {yi,..-,yn} and {x±, . . . ,xn} respectively. Let T be the set of minimal vertex covers of G\ and

L = ({xil • • • X i , {xil ,...,xia} 6 T(Gi)}). Then ( ( z ) : K ) = (z,L), where K = I(S(G0)).

Proof. Take {x^, . . . , X i r } G T and yuyt £ /• Since is £ {k,l} for some s we obtain x^ • • -x^y^yi € (z). This shows (z,L) C ( ( z ) : K ) . Conversely, assume M is a monomial and M € ((z): K)) \ (z). Let {yk,Ve} be any edge in GO- Since My^yi = xtytMi, either Xk divides M or xi divides M; in either case, M e L. D

Edge Ideals_____________________________________ 187

Corollary 6.5.4 The type ofR/I(S(G0))

is equal to |T(G 0 )|.

Proof. Since R/K is a Cohen-Macaulay ring and z is a maximal regular sequence in K, where K = /(5(Go)), using Proposition 1.3.7 one has:

ExtnR(R/K,R) ~HomR(R/K,R/(z)) Thus by Corollary 4.3.6 one obtains type(R/K) assertion follows from Proposition 6.5.3.

~ ((z):^)/(z). = z/(((z):A')/(z)) and the O

Observe that these results can be used to compute the primary decomposition of an edge ideal. Example 6.5.5 Let /(G 0 ) = (yiys,yiy4, 3/22/4, 2/22/5, 2/32/5,2/32/6,2/42/6)- Note that the ideal ((z): /(S(G 0 )) is generated by z = {ziyi, . . . , xey6} together with: and these monomials correspond to the minimal primes of /(Go)- The type

of R/I(S(Go)) is 5 and the multiplicity of the face ring of /(Go) is 2. Definition 6.5.6 A star on n vertices is a complete bipartite graph of the form /Ci ] n _i and is denoted by star(n).

Corollary 6.5.7 If GQ = star(n) is the star on n vertices, then I(S(G0)) lies in the linkage class of a complete intersection. Proof.

Let y i , . . . , j / n be the vertices of Go and yi its center; then the

only minimal vertex covers of GI are {xi} and {x2, ..., xn}, so that, in the notation above, the first link is J

= ((z = (xi,x2y2,...,xnyn,x2---xn).

Take now its link with respect to the regular sequence formed by the first n elements:

(x\ , x2y2, . - . , xnyn): J = (xi , y2, . . . , j/ n ), which is a complete intersection.

D

Exercises 6.5.8 Let GO and GI be two copies of the same graph on disjoint sets of vertices {yi,...,yn} and { X I , . . . , X H } respectively. Let T be the set of minimal vertex covers of GI and

Then ( K , z ) and (z,L) are geometrically linked, where K = I(S(Go)).

6.5.9 Let / = (xiyj\ l
188______________________________________ Chapter 6

6.6

First syzygy module of an edge ideal

Let G be a graph and I(G) C R its edge ideal. An interesting problem is

to express some of the first initial Betti numbers of I(G) in terms of the graph theoretical data of G; below we give an indication that this might be possible, see [103] for further information.

Let R be a polynomial ring over a field K with the usual grading and / a graded ideal of R. The minimal graded resolution of R/ 1 by free ^-modules

can be expressed as: (-dgi)

«!> . . . __> 0

R^ (_ d l .) J^ R __,. R/I -^ 0,

where all the maps are degree preserving and the dij are positive integers.

One may assume dji < • • • < djCj for all j — 1, . . . , g. The integer g is equal to pdR(R/I), the protective dimension of R/I. To simplify notation set

1=1 The rank of Fj is the jth Betti number of /. From the minimality of the resolution du < • • • < dgi, see Remark 2.5.9. We define the jth initial Betti

number of / as j j = bj\ and the jth initial virtual Betti number of I as

Note that some of the initial virtual Betti numbers may be zero.

For a general monomial ideal / there is an explicit and elegant description of a minimal set of generators for the first module of syzygies of / due to S. Eliahou [101]; see also [43].

Definition 6.6.1 The edge graph of G, denoted by L(G), has vertex set equal to E — E(G) with two vertices of L(G) adjacent whenever the corresponding edges of G have exactly one common vertex. Proposition 6.6.2 If G is a graph with vertices Xi,...,xn and edge set E(G), then the number of edges of the edge graph L(G) is given by

\E(L(G»\ =

= -\E(G)\ i=l \

'

,

+

(6.1)

i=l

Proof. To prove the first equality note that each vertex Xi of G of degree

di contributes with (^) edges to the number of edges of L(G).

Edge Ideals_____________________________________ 189

The second equality follows from the Euler's identity

see Exercise 6.1.27.

d

Let us now present a formula, in graph theoretical terms, for the second initial Betti number of I(G). Proposition 6.6.3 ([103]) Let I = I(G) C R be the edge ideal of a graph G and V the vertex set of G. If

• • • — > Rc(-4] © Rb(-3) — > R"(-2) A R — > R/I — > 0 is i/ie minimal graded resolution of R/I. Then b = \E(L(G))\ — Nt, where Nt is the number of triangles of G and L(G) is the edge graph of G.

Proof. Assume / — (/i, • • • ,/ 9 ) and ^(e,) = /j. Let Z[ be the set of elements in ker(?/>) of degree 3. We may regard the fi as the vertices of L(G}. Every edge e = {fi, fj} in L(G) determines the syzygy syz(e) = XjSi — XiBj, where ji — XiZ and fj — XjZ for some z , X i , X j in V. According to Theorem 2.4.17 the set of those syzygies generate Z [ . Given a triangle G3 = { z i , Z 2 , z z } in G we set

where fi = ziz2, fj = z2z3, and fk = ziz3. Notice that (f>(C3) Pi (j)(C3) = 0 if C*3 ^ C3. For every triangle C3 in G choose an element p(C3) 6 4>(C3). It is not hard to show that the set B = {syz(e)| e e L(G)} \ {p(C3)\ C3 is a triangle in G} is a minimal generating set for Z[ .

D

It is a fact that the second Betti number of a Stanley- Reisner ideal (resp. monomial ideal) is independent of the field [169] (resp. [43, 101]), moreover for edge ideals the third and fourth Betti numbers are also independent of the field [168]. An application to Hilbert series Let A be a simplicial complex of dimension d and / = (/o, • • • , fd) its /-vector, where /j is the number of

faces of dimension i in A and /_i = 1. By Corollary 5.4.5 the Hilbert Series of the Stanley-Reisner ring S = R/I& can be expressed as

We now use this formula to compute a particular instance.

190______________________________________Chapter

6

Corollary 6.6.4 Let G be a graph with q edges, V = { x i , . . . , xn] its vertex set and F(z] the Hilbert series of S = R/I(G). If S has dimension 3, then F(z)(l — z)3 is equal to

2 J-q)*~+(~

"

~———6——^———:———)Z3

where g is the height of I(G), N-t is the number of triangles of G, and

Proof. Let A = AG be the Stanley- Reisner complex of / = /(G). The first entries of the /-vector of A are given by

f =i 1, /-i

tfo- n,

n n ( -~—— !) - q. fti = ——

To compute /2 notice that the resolution above yields dim( J R//) 3 = (n + f) - qn + b \n ij and /2

=

™

-

where the last expression for /2 uses Proposition 5.4.4. The desired formula follows by substitution of the /, in Eq. (6.2) and using Eq. (6.1). D The number of triangles of a graph

Let G be a graph with q edges

and vertex set V = {xi ,... ,xn}. As the number of edges of the edge graph L(G) of G is given by the formula:

observe that Proposition 6.6.3 provides a method to compute the number of triangles of a graph G by computing syzygies with Macaulay or Co Co A. An alternative method using linear algebra is recalled next. The adjacency matrix of G is the n x n matrix A = (a^) whose entries are given by _ J 1 if Xi is adjacent Xj, l i \ 0 otherwise. It follows directly that A is symmetric and that its trace is equal zero. If

f ( x ) =xn + cixn~l + c 2 x n ~ 2 + c 3 x"~ 3 + • • • + cn,

Edge Ideals_____________________________________ 191 is the characteristic polynomial of A, then c\ = 0, — c-2 is the number of edges of G, and — 03 is twice the number of triangles of G. For an interpretation of all the Cj 's consult [24] . Let G be the graph whose complement is the union of two disjoint copies of a path of length two. The edge ideal of G is equal to: /(G) = (xi £4,0:1 x5, ZiZ6,Z2Z3,Z2Z4,Z2Z5,Z2Z ; 6,Z3Z4, £32:5, X3x6,x5x6), and part of its minimal resolution is:

0. Hence the number of triangles Nt of G is equal to 6. Of course the same number is obtained noticing that f ( x ) = x6 - ILr 4 - 12x3 + 3a;2 + 4x - 1

is the characteristic polynomial of the adjacency matrix of G.

Exercises 6.6.5 Let G be a graph with vertex set x and FG(Z) the Hilbert series of fc[x]//(G), where A; is a field. For any vertex x € V(G) one has:

FG(z) = FG\{x}(z) + -^——;FG\(N(X)(J{X})(Z),

with F$(z) - 1.

6.6.6 If Kr^s is a complete bipartite graph, then

6.6.7 ([309]) Let Pn be a path graph with vertices x = x\, . . . ,xn. Prove that the Hilbert series of fc[x]//(P n ) can be express as:

6.6.8 Let Pn be a path graph with vertices x = x\ , . . . , xn. Prove that the Hilbert function of k [ x ] / I ( P n ) is given by

m-3

Notice that for m > L^^-J this is the Hilbert polynomial.

192______________________________________ Chapter 6

6.6.9 Let G = CT be an heptagon. Prove the formula:

-3

2

What is the number of minimal primes of /(G)? 6.6.10 Let G = Cf be an heptagon and AQ its complementary simplicial

complex. Prove that AG is a triangulation of a Mobius band whose reduced Euler characteristic is X^-G — — 16.6.11 Let / = I(G) C R be the edge ideal of a graph G. If

• • • — > fl c (-4) © Rb(-3) — > E»(-2) A E —> fl/J —> 0 is the minimal graded resolution of .R//. Prove that c is equal to the number of unordered pairs of lines {/;, fj} such that /; and /j are independent lines that cannot be joined by an edge.

6.7

Edge rings with linear resolutions

A noteworthy use of graph theoretical methods in commutative algebra is implicit in Lyubeznik thesis [212] and explicit in the work of R. Froberg [116]. In both cases the central notion is that of a triangulated graph. It will turn out that this is the right notion to express some homological properties

of edge ideals and the Cohen-Macaulay property of unmixed ideals of height two generated by square-free monomials. Given a graph G an interesting duality between /(G) and IC(G) , the ideal of covers of G, will take place in this section. Under special circumstances /(G) reflects properties of IC(G) and viceversa. The Cohen-Macaulay case of dimension two

Let us show a family

of flag complexes where shellability is equivalent to Cohen-Macaulayness. We begin with:

Lemma 6.7.1 Let A be a simplicial complex of dimension 2 on the vertex set V . If A is Cohen-Macaulay, then there are facets FI, , . . ,FS such that

(i)

V = Ui=1Fi,

(ii) \Fi U • • • U Fi\ = 2 + i for 1 < i < s, and

(iii) (Fi U • • • U Fk-i) n Fk is a face in A of dimension 1 for k > 2.

Edge Ideals______________________________________ 193

Proof. We will proceed by induction on k. Assume that there are facets

F1,...,Fk in A with |Fi U • • • U F = 2 + i for 1 < i < k, and such that (Fi U • • • U Fi-i } n Fi has dimension 1 for 2 < i < k. Set Ui = FI U • • • U FJ. Assume Uk ^ V, otherwise the proof is complete. Choose F <£. Uk so that |F n Uk\ has maximal dimension. Set G = F D Uk and r — \G\. We will show r = 2. Assume r = 1, i.e., G = {p}. Notice that p e Fi C Uk for some i < k. We pick z0 e F \ Uk and x0 £ Fi, x0 ^ p. Observe that z0 and x0 are both in Ik (G) and that Ik (G) is connected, hence there is a path [ZQ = W0,...,Wm = X0}

in Ik (G) joining ZQ and XQ. We pick j so that uij ^ Uk and ifj+i G Z7fc. Since {uij,Wj+i} 6 Ik (G) we obtain that F' = {WJ,WJ+I,P} is a facet of A so that F' <£_ Uk and F' n Uk — {p, Wj+i}, which contradicts the choice of F. Thus r = 2, and we can set F = Ft+i to complete the induction. D

Proposition 6.7.2 ([116]) Let A be a simplicial complex of dimension

2. // the Stanley-Reisner ring A; [A] has a 2-linear resolution, then A is a Cohen-Macaulay complex if and only if A is shellable. Proof. Assume that A; [A] is a Cohen-Macaulay ring with a 2-linear resolution. The /-vector of A is given by /o = n, f\ = In — 3 and /2 = n — 2,

where n is the number of vertices of A (see Exercise 6.7.17). By Lemma 6.7.1 there are facets F I , . . . ,FS in A so that V = Uf =1 Fj, \Fi U • • • U Ft\ = 2 + i for 1 < i < s, and (Fx U • • • U Fk-i) n Ft is a face of dimension 1 of A for 2 < k < s. Notice that in our case s = n - 2, hence FI , . . . , Fs are precisely the facets of A. Set

Ui = F! U • • • U Fi and A; - FI U • • • U F;, where Fj = {a 6 A|
Using f i ( U s ) - 2n - 3 yields_/i(!7;) = /i(A^) = 2i + I for all i. A similar argument shows fz(Ai) = fi(Ui) for all i. To complete the proof notice that

G = Fkn c/ fc _i e Fk n A 4 = F, n !/*_!, accordingly G C Fk n Fj for some j < k, using that |G| = 2 we get

Fk n F.,- = Fk \ {a;} for some x E V. Thus A is shellable by Exercise 5.3.29. The converse is true quite generally (see Theorem 5.3.18). D

194______________________________________Chapter

6

Corollary 6.7.3 If k[A] is a Cohen-Macaulay ring of dimension 3 with a 2-linear resolution over a field k, then K[A] is a Cohen-Macaulay ring with a 2-linear resolution for any other field K. Example 6.7.4 Let R = k[a,...,/] be a polynomial ring over a field k and A the following triangulation of the real projective plane IP2:

Note that /A = (abc, abd, ace, adf, aef, bcf, bde, bef, cde, cdf). The face ring fc[A] was studied by Reisner [237]. One has that A;[A] is Cohen-Macaulay and has a 3-linear resolution if k has characteristic other than 2, and is not Cohen-Macaulay and has a non linear resolution otherwise (compare with

[114, Example 3]). Hence A is not shellable and dim A = 2. The general case Let G be a graph and G its complement. The graph G is said to be triangulated or chordal if every cycle Cn in G of length n > 4

has a chord in G. A chord of Cn is an edge joining two non adjacent vertices

ofC n . The following result provides a graph theoretical characterization of the rings A; [A] having a 2-linear resolution.

Theorem 6.7.5 ([116]) Let G be a graph and A the Stanley-Reisner complex of I(G). Then A; [A] has a 2-linear resolution if and only if G is a triangulated Next we introduce the notion of d-tree and state an interesting consequence due to R. Froberg. Definition 6.7.6 A d-tree is defined inductively: (i) a complete graph K^+i is a d-tree, (ii) let G be a d-tree and H a subgraph of G with H ~ K^, if v

is a new vertex connected to all vertices in H, then G U {v} is a d-tree. Theorem 6.7.7 ([116]) Let G be a graph and A the Stanley-Reisner complex of I(G). IfG is Cohen-Macaulay over a field k, then A;[A] has a 2-linear resolution if and only if G is a d-tree, where d = dim A.

Edge Ideals_____________________________________ 195

Corollary 6.7.8 Let G be a connected Cohen-Macaulay bipartite graph with q edges. If q is equal to g(g + l)/2, where g = htl(G), then

Proof. First notice that R/I(G) has a linear resolution by Theorem 4.3.7.

Hence, according to Theorem 6.7.7, the 1-skeleton of AQ is a (g — l)-tree, which implies that there is a vertex w of G of degree g. We now use the notation (and similar arguments) of the proof of Theorem 6.4.4 to conclude that deg(xi) = 1 and deg(t/j) = 1, for some i, j. To finish the proof consider

the graph G \ {w} and use induction.

d

The more general problem of determining the simplicial complexes A whose Stanley-Reisner ideal /A has a pure resolution has been addressed by W. Bruns and T. Hibi [46, 47], where a very concrete solution can be found

if /A is generated by square-free monomials of degree two. Edge ideals and their ideals of vertex covers

Triangulated graphs

have been extensively studied, they can be constructed according to a result of G. A. Dirac, see [84, 285]. Those graphs have appeared in the commutative algebra literature in the works of R. Froberg [116] and G. Lyubeznik [213], here we point out a connection between these two works that leads to

an interaction between some properties of an edge ideal and some properties of the ideal of minimal covers of the graph.

Lemma 6.7.9 Let G be a connected graph with vertex set V(G) and

d = min{deg(v)\v € If G is not a complete graph, then there is S C V(G) such that G \ S is disconnected and d = \S\.

Proof. Let v be a vertex of G of degree d and S = N(v) the neighbor set of v. Since G is not a complete graph and v has minimum degree there is a vertex w $ S U {v}. Thus G \ S has at least two components. D Definition 6.7.10 Let G be a graph and S a set of vertices, the induced subgraph <S> is the maximal subgraph of G with vertex set S. Thus two vertices of S are adjacent in <S> if and only if they are adjacent in G.

Proposition 6.7.11 ([84]) If H is a triangulated non complete graph, then it can be constructed out of two smaller disjoint triangulated graphs HI and HI by identifying two (possibly empty) complete subgraphs of the same size

in HI and H%.

196______________________________________Chapter

6

Proof. One may assume that H is a connected graph with n vertices. Since

H has at least one vertex of degree less than n — 1, by Lemma 6.7.9

there is

a minimal set of vertices S such that H \ S is disconnected and \S\ < n — 2.

Hence we can write H\S = FI UF2, where FI and F2 are disjoint subgraphs of H. Set

H1=

and H2 = ,

where < A > is the induced subgraph with vertex set A. Note HI U H2 = H

and HI n H2 = < £ > . Next we show that <5> is a complete graph. Assume there are vertices xi and x2 in S not connected by an edge of H. Since H \ (S \ {x^) is connected for i = 1,2, there are paths pi and p2 in HI and H2 respectively joining x\ with x2, whose only vertices in S are xi and x2. If pi and p2 are the shortest paths with this property, then pi U p2 is a cycle of length greater or equal than four without a chord, a contradiction. D Lemma 6.7.12 ([285]) Let H be a triangulated graph and K a complete subgraph of H. If K ^ H, then there is x $ V(K) such that the subgraph

induced by the neighbor set N(x) of x is a complete subgraph. Proof. By induction on V(H). If H is complete, any x $ V(K) would satisfy the requirement. Otherwise, using Proposition 6.7.11 H = HI U H2, where HI , H2 are triangulated graphs smaller than H such that HI n H2 is

a complete subgraph. It follows that AT is a subgraph of Hi for some i, say i = 1. As HI r\H2 is a proper complete subgraph of H2, by induction x can be chosen in H2 \ (Hi n H2). D Let R = k[xi,... xn] be a polynomial ring over a field k and / a monomial ideal of /. The ideal of minimal covers or simply the ideal of covers of /, denoted Ic , is the ideal of R generated by all the monomials x^ • • • Xik such that the ideal (x^,..., Xik) is an associated prime ideal of /. This terminology is intended to emphasize that a minimal prime "covers" or contains all the monomials in /. If / is a Stanley-Reisner ring one has a duality:

j _

(x- •••x-

) = (x-

Z• ) n • • •

Let G be a graph and E = E(G) its edge set, in this case we denote the ideal of minimal covers of I(G) by IC(G). One has

IC(G}=

Thus IC(G) is an unmixed ideal of height two. The Cohen-Macaulay property of this ideal was successfully studied by G. Lyubeznik.

Edge Ideals_____________________________________ 197

Theorem 6.7.13 ([213]) Let G be a graph and G its complement. Then the ideal of minimal covers / C (G) is Cohen-Macaulay if and only if G is a triangulated graph.

Proposition 6.7.14 Let G be a graph and IC(G) its ideal of covers. If IC(G) is Cohen-Macaulay, then

where i/(/ c (G)) is the minimum number of generators of IC(G) and g is the

height of I(G). Proof. By the Hilbert-Burch Theorem 2.5.15, / C (G) is generated by the m — 1 minors of an m x (m — 1) matrix A with homogeneous entries, where m = z/(/ c (G)). Take / in IC(G) of degree g, since any m — 1 minor of A has

degree at least m - 1, one obtains deg(/) = g > m - 1.

d

Proposition 6.7.15 Let G be a graph whose complement G is a triangulated graph. Then I(G) is Cohen-Macaulay if and only if IC(G) has a linear resolution. Proof. =>•) By the previous results one has that IC(G] is Cohen-Macaulay and the following inequality holds

v(Ic(G)) < h t / ( G ) + l, according to Theorem 4.3.7 it suffices to prove that one has equality. Note that z/(7 c (G)) is equal to the multiplicity of R/I(G), because all the minimal primes of /(G) are of height g = ht/(G). On the other hand the multiplicity of R/I(G) is equal to g + 1 (see Exercise 6.7.17). •£=) By induction on the height of /(G). One may assume g = htI(G) >1, otherwise /(G) = 0. Note that the hypotheses imply that G is an unmixed graph such that the multiplicity of R/I(G) is equal to e = g + 1. Hence any two maximal complete subgraphs of G are isomorphic. Note d+l =n — g,

where n is the number of vertices of G and d+ 1 is the dimension of R/I(G). Let K be a proper complete subgraph of Gon d+l vertices, by Lemma 6. 7. 12 there is x £ V(K) such that the subgraph induced by the neighbor set N(x) of £ is a complete subgraph. Since x belongs to a complete subgraph L of G on d + 1 vertices, it follows rapidly that deg^(z) = d or equivalently deg G (x) = g. Hence by induction /(G \ {x}) is C-M of height

5-1. Using Proposition 6.2.5 one concludes that J(G) is C-M.

D

198______________________________________ Chapter 6

One can use a duality criterion described next to see that this result is valid in a more general setting and without the hypothesis that the graph G is triangulated (see Exercise 6.7.21). Definition 6.7.16 Let A be a simplicial complex on the vertex set V, the Alexander dual A* of A is the simplicial complex given by

A* ={GcV\V\Gi A}. According to [90] the Stanley-Reisner ideal of a simplicial complex has a linear resolution if and only if its Alexander dual is Cohen-Macaulay. This has been recently generalized [151] replacing linear resolution by the notion of componentwise linear ideal, and Cohen-Macaulay by sequentially Cohen-Macaulay.

Exercises 6.7.17 Let A be a simplicial complex of dimension d with n vertices and / = (/Oi • • • j /d) its /-vector. If A; is a field and A; [A] is a Cohen-Macaulay ring with a 2-linear resolution, then

6.7.18 Let A be a simplicial complex and k a field. If the ring A; [A] is Cohen-Macaulay and has a 2-linear resolution, then the reduced Euler characteristic x(A) is equal to 0 and -Hd(A) = 0, where d = dim A. 6.7.19 Let A be a simplicial complex. If fc[A] is a Cohen-Macaulay ring with a 2-linear resolution over a field k, prove that A is shellable.

Hint Use Theorem 6.7.7. 6.7.20 Show that the complement of a triangulated graph cannot have a chordless cycle with five or more vertices. 6.7.21 Let A be a simplicial complex on the vertex set V = [x\, . . . ,xn} and A* its Alexander dual. Prove that the Stanley-Reisner ideal of A* is equal to the ideal of minimal covers of / = /A, that is: JA* = ({x^ • • • Xir (x^ ,... , X i r ) is a minimal prime of I}). 6.7.22 Prove that the third Betti number of the ideal

/ = (abc, abd, ace, adf, aef, bcf, bde, bef, cde, cdf)

depends on the base field, and show that I is equal to its ideal of vertex covers. See Example 6.7.4.

Edge Ideals_____________________________________199

6.7.23 Let k be a field of characteristic zero or two. UseMacaulay to verify that the initial Betti numbers and initial degrees of the Stanley-Reisner ideal / of Exercise 6.2.24 are as follows: 10: 11 78

9: 80 77

8 : 245 76

6: 6 75

5: 40 74

4: 95 73

3: 80 72

2: 25 71

where the number before the colon indicates initial degree. What are the initial virtual Betti numbers of I?

Chapter 7

Monomial Subrings Two of the main objects of study here are subalgebras generated by monomials and the corresponding Rees algebras. In general one would like to relate properties of those two algebras, such as the normality property or the Cohen-Macaulay property. Some of the relevant topics of the chapter are the descriptions of the integral closure of ideals and monomial subrings and how the various notions of normality for algebras and ideals relate to each other. Special attention is given to showing the relation between the normality of the Rees algebra and properties of other related objects. We present some degree bounds for the generators of the integral closure of Rees algebras and monomial subrings. Upper bounds for the a-invariant of some subrings will also be given.

7.1

Basic properties

Let R = k[x] = k[xi,..., xn] be a polynomial ring over a field k in the indeterminates X I , . . . , X H and F = {/i,... , f g } a finite set of distinct monomials in R such that j{ ^ I for all i. For c 6 N™ we set xc = x{1 • • • < " . The monomial subring spanned by F is the /c-subalgebra

k[F] = k [ f l i . . . , f q ] c R . The exponent vector of /j = xai is denoted by log(/j) = cc, and log(F)

denotes the set of exponent vectors of the monomials in F. To try to avoid cumbersome notation we set fc = /f1 • • • fqq if c 6 W.

Note that k[F] is equal to the semigroup

k[C] = k[{xa\aeC}},

201

202______________________________________ Chapter 7

where C = Nlog(/i) + - • - + Nlog(/ ? ) is the subsemigroup of N™ generated by log(F). Thus as a fc-vector space k[F] is generated by the set of monomials of the form xa , with a 6 C. An important feature of k[F] is that it is a graded subring of R with the grading k[F]i = k[F] n #,. There is a graded epimorphism of fc-algebras:

¥>: B = % , . . . , * , ] — ) • k[F] — )• 0, induced by pfc) = /<, where B is a polynomial ring graded by deg(ij) = deg(/j). Note that the map ip is given by
The kernel Pp of
Definition 7.1.1 A binomial f in B = k[t\, . . . ,tq] is a difference of two monomials, that is, / = ta - 113, for some a,/? in N 7 . An ideal generated by binomials is called a binomial ideal. Proposition 7.1.2 The presentation ideal Pp of k[F] is a graded prime ideal generated by a finite set of binomials. Proof. Observe B/Pp ~ k[F] and that k[F] is a Noetherian domain. If a polynomial ho + • • • + hs is in Pp with hi G Bi, then y(/ij) = 0 because f

is graded. Hence hi £ (Pp)i f°r all i, and Pp is a graded ideal. Consider the ideal L of B generated by the binomials in Pp . Since Pp has a finite basis of homogeneous polynomials, to show L = Pp it suffices to prove Bd n Pp C L for all d. Let h G Bd n PF , write h=

with a7i 6 k and tji G B,j. We show that h G L by induction on r. r > 3. Making the substitution Xi = t\ in the equality Y^i=i a-nf~ti obtains (^i_i o,^)tf = 0 and ]Ci=i a7; — 0. One may assume fji for some i ^ j, otherwise aTi = 0 for all i. For simplicity assume i j = 2. Since a72 = — oTl + /?, with ^ € k one has

By induction one gets g £ L and h & L, as required.

Assume — 0 one = f7' = 1 and

D

Definition 7.1.3 If F = {/i,..-,/ ? } is a set of monomials in fc[x], the associated matrix of fc[F] is the n x q matrix M whose columns are the

exponent vectors log(/i), . . . , log(/ ? ).

Monomial Subrings_______________________________203

Recall that the support of a vector j3 e K.g is supp(/3) = {i \ /3, ^ 0} . Note that any vector a in Rn can be written as a = a+ — a _ , where a+ and a_ are two non negative vectors with disjoint support. Corollary 7.1.4 If Pp is the toric ideal of the monomial subring k[F] and M is the associated matrix of k[F], then

PF = ({ta+ -ta~ a € Zq and Ma = 0}). Proof. Let ta - t@ be a binomial in Pp. If supp(a) l~l supp(/3) ^ 0, then we can write /<* - f0 = f(fa' - f 0 ' } = 0, where 7* = min{a;,,S;}. Note that a' and /3' have disjoint support and fa - f13 = 0 iff M(a - /?) = 0. To complete the argument use that Pp is generated by binomials. D

Corollary 7.1.5 The toric ideal Pp of a monomial subring k[F] has a Grobner basis consisting of binomials with respect to any monomial ordering of the polynomial ring B. Proof. Let E be a finite set of generators of Pp consisting of binomials and let /, g e E. Since the 5-polynomial 5(/, g) is again a binomial and

the remainder of S(f, g) with respect to E is also a binomial, it follows that the output of the Buchberger algorithm (see Theorem 2.4.11) is a Grobner basis of Pp consisting of binomials. D Definition 7.1.6 Let R = K[XI, ... ,xn] be a polynomial ring over a field K and k > I an integer. The monomial subring of k-products or simply the subring of k-products is:

K[Vk] C R, where Vk = {xh • • • xik\ I < ii < • • • < ik < n}. The kth Veronese subring of R, denoted by R^, is by definition:

i=0

where an element in Rik is said to have normalized degree i. Thus is again a standard algebra generated by elements of normalized degree 1. Sometimes the subring K[T4] is called the square-free kth Veronese subring of R, because it is contained in R^.

Corollary 7.1.7 ([111]) Let R = K[XI, ... ,xn] be a polynomial ring over a field K and let P be the toric ideal of the subring of k-products K[Vk]. If k > 2, then P is generated by homogeneous binomials of degree two.

204______________________________________ Chapter 7

Proof. Let 14 = {x^ • • • Xik \ 1 < i\ < • • • < i^ < n} be the set of squarefree monomials of degree k in R. To begin with consider a polynomial ring:

with one variable t^...ik for each monomial x^ • • • Xik in 14. Let
(p:B —> K[Vk], induced by til...ik H-^ x^ • • • x i k ,

where B has the usual grading and AT [14] has the normalized grading. Thus P = ker(2. We proceed by induction on s. Assume

s > 3 and that all the binomials in P2, . . . , P s _i are in P2. Let T= {^...ijl
where ti, t\ £ T for all i. If (^) = f t , then after reordering the variables we may write f i — x\- • • xrxr+\ • • • xk, where xj.,...,xr are in nf =1 supp(/,) = nf =1 supp(/i) and xr+i,. ..,xk are not in nf =1 supp(/;). Set h equal to x\ • • -xr and write f[ = hxir+1 • • - X i k . If xr+i £ supp(/{), then fp = hxr+iXpr+2 • • • xpk for some p > 2. By our assumption there is

f'( so that Xir+l is not in supp(/^) for some (. > 2. First consider the case xr+i e supp(/^). Write $[ = hxr+iX(r+2 • • -xtk. Note that j[j't = f"f", where f{' = (xr+if[)/xir+l and f'J - (xir+lf'e)/xr+i. Since f{' and j'l are in 14, there are t" and t"t in T mapping under

' ^^2 * "

^ _ 1 /-f-1 * * * s

we get F £ (P2) if and only if FI e (P 2 ). On the other hand if xr+i is not in supp(/^), write j't = hxtr+1 • • • x e k . Note that {a;^,,+1 , . . . , xtk } is not contained in A = {xpr+2,. . . ,xpk}, say xtr+l is not in A. Setting j'l = (xr+1fe)/xlr+l and /; = (xtr+1fp)/xr+1, we get G2 = t'et'p-t'{t'p £ P2, where t", i^ map under

we obtain F 6 (P2) if and only if F2 e (P2). Set /^" = (fe'xir+1)/xr+1, f i = (fixr+i)/Xir+T. and G3 = t(tf - t"t'" . From the identity

PS

=

FI + G3(t'2 • • • ^_i^ +1 • • • t'p-itpt'p+l • • • t's)

Monomial Subrings_______________________________205

we obtain P2 & (P2) if and only if FS € (P?)- Hence from the outset one may assume

/i = hxr+i • • • xk and f{ = hxr+ixir+2 • • • x i k . Using the same argument as above with xr+z and Xir+2 playing the role of xr+i and Xir+1, and so on, we derive that there is a binomial Fm e Ps of the form Fm = ti • • • ts - tiu-i • • -us, with u 2 , • • • , u s € T, so that F 6 (P2) if and only if Fm e (P 2 ). To complete the proof notice that Fm is in (P2) by the induction hypothesis. D

Corollary 7.1.8 Let R = K[XI, . . . ,xn] be a polynomial ring over a K and P the toric ideal of the Veronese subring R^ . If k > 2, then P is generated by homogeneous binomials of degree two. Proof. Consider a polynomial ring:

B = K[{ta\ a <E N™ and |a| = k}}, with one variable ta for each monomial xa of degree k. Let ip be the graded homomorphism

y.B —> R(k\ induced by ta ^ xa, where B has the usual grading and R^ has the normalized grading. By Proposition 7.1.2 the toric ideal P — ker(y) is a graded ideal P = ®^_ 2 -^ generated by homogeneous binomials. Hence we need only show that all the binomials in Ps are in (P2) for s > 2. We proceed by induction on s. Let T = {ta\ a G If1 and a = k} and F = ti • • • ts - HI • • • us e Pg,

where ti, Ui G T for all i. If
g\. Note that g\g^ = /ii/i 2 , where h\ = xigi/xt and /i2 = x^g^/xi. Since hi and ft 2 have degree k, there are u'j and u2 in 7" mapping under ip to /ij and /i2 respectively. From the equality PI

=

(ti •

we conclude that F e (P2) if and only if PI £ (P 2 ). Case (b): x\ divides g\- • • gs for some z, say i = 1. By a repeated use of case (a) one may assume g\ = x\. Hence by symmetry one may also assume /i = £*, that is, P = ti • • • ts — t\u
206______________________________________Chapter

7

Case (c): x\ does not divide g\ • • • gs for all i. Let

Note Ci; < k for all j. Thus using case (a) one may assume g\ = x^ • • • xc£, where x^ £ u^isupp^). A repeated application of case (a) yields that one may assume
(p: B = k [ t i , . . . , t < , ] —> k [ f i , . . . , f q ] ,

induced by ip(ti} = /»,

is the ideal (t\ — fi,..., tq — fq) fl B. Proof. Let F G ker((^). Making ij = (ti — fi) + fi we can write F as:

Since F ( f i , . . . , f v ) = 0 one derives J]^ apf® = 0, and consequently F is in (ti — /i j • • • i tq — fq) H B. The other containment also follows readily. D

Let us illustrate a technique used to compute the toric ideal Pp of a monomial subring k[F] using Grobner bases and elimination of variables. Example 7.1.10 Let F = {xiXj 1 < i < j < 4} and k[F], the graded epimorphism of fc-algebras with y(%) = a;ja;j and deg(i^) = 2. By Proposition 7.1.9 one has

PF = ker(v) =Ln fc[%], L = (ttj - fa l
£4*23 — *24*34)

£4

- £4*23,

^3*12 - *13*23,

2:22:4 -*24,

2^2:3 -i 2 3 ,

2:1*34 - X 4 t i 3 ,

2:1*24-2:4^12,

2:1*23-2:3*12,

XiX4-ii4,

^i^s-iis,

X i I 2 -£12,

2:3*12*34 - 2;4*13*23-

Therefore Pp = (*i4*23 - *12*34, *is*24 - *12*34) by Theorem 2.4.18.

Monomial Subrings_______________________________ 207

Another method to compute Pp can be found in [83]; this method uses Buchberger's algorithm [53] adapted for the specialized situation of toric ideals. In [23] some algorithms are devised, that in many respects improve existing algorithms for the computation of toric ideals. For a general study of binomial ideals we refer the reader to [98] . Lemma 7.1.11 Let f be a polynomial in k[xi, . . , ,xn]. If k is an infinite field and /(a) = 0 for all a £ kn , then f is the zero polynomial.

Proof. By induction on n. The case n = 1 is clear because any nonzero polynomial in k\x\} of degree r has at most r roots. If n > 2 write / = /O + flXn + ' • • + frXrn,

where /$ G k[xi, . . . ,xn-\] for all i. Fix c = (ci, . . . ,c n _i) in kn~1 and consider the polynomial d(xn) = f(c,Xn]

= / 0 (c) + fl(c)Xn

+ • • • + fr(c)xrn.

As g vanishes on k and g is in k[xn] one obtains that g is the zero polynomial. Hence /j(c) = 0 for all i. Since c was an arbitrary point in the affine space kn~1 by induction one has that fi is the zero polynomial for all i, thus / is the zero polynomial. CI

Corollary 7.1.12 Let R — k [ x i , . . . ,xn] be a polynomial ring over a field k and X C kq a set given parametrically by ti = fi(x\, . . . ,xn), where F = {/i, . . . , fq} C R. If k is infinite, then the ideal I ( X ) of polynomials f(ti, . . . ,tq) that vanish on X is equal to Pp , the. presentation ideal of k[F]. Proof. As Pp = (ti - /i,. . . ,tq - fq) H k [ t i , . . . ,tg], one has Pp C I ( X ) . Conversely take / £ I ( X ) and consider the revlex ordering such that ti > Xj for all i,j. By the division algorithm f(ti,..,,tq) = ]Cj5«(^"~/«) + r; where r is a polynomial in R. Hence for any a G kn one has

/(/i («),...,/») =

) - fi(xi,...,xn)) + r(xi, , . . ,xn) = 0 . This proves r(a) = 0 for any a 6 kn, thus r is the zero polynomial because

k is an infinite field (see Lemma 7.1.11).

D

Definition 7.1.13 Given a subset X at the affine space A£ over a field k, the Zariski closure of X_, denoted by X, is the closure of X in the Zariski topology of A£ . Thus X is the smallest affine variety of A£ containing X. Proposition 7.1.14 If k is a field and X a subset of the affine space A£, then X = V ( I ( X ) ) , where X is the Zariski closure of X .

208______________________________________Chapter

Proof. AsV(I(X))]aclosedwadXcV(I(X)),onehasXcV(I(X)).

7

For

the other containment note X = V(J), where J is an ideal of a polynomial ring R over k. From X C V ( J ) one derives I(V(J)) C I ( X ) , and since one also has the containment J C I ( V ( J ) ) by transitivity J C I ( X ) . Therefore V ( I ( X ) ) C V ( J ) and consequently V ( I ( X } ) C~X. D

Definition 7.1.15 An affine toric variety V is defined as the zero set of a

toric ideal, that is, V = V(Pp) for some toric ideal Pp. Corollary 7.1.16 Let R — k[x\, . . . ,xn] be a polynomial ring over a , k and X C kq a set given parametrically by ti = f i ( x i , . . . ,xn), where F = {/i, . . . , fq] is a subset of R. If k is infinite, then the Zariski closure of X is equal to V(Pp). Proof. By Corollary 7.1.12 and Proposition 7.1.14 one has the equalities

I(X)=PF andX: = V ( I ( X ) ) .

D

Proposition 7.1.17 If k[F] is a monomial subring over a field k generated by a finite set F of monomials and M is its associated matrix, then dim k[F] = rank(M),

where dim k[F]

is the Krull dimension of k[F].

Proof. Let F = {/i, . . . , fq} and fa = xai , where a; = (an,. . . ,ain) for

i = 1, . . . , q. Let M = (a^) be the n x q matrix of k[F]. Set r = rank(M). By the Noether normalization lemma dim k[F] = trdeg fc k[F], therefore it suffices to prove the equality r = trdeg^ k[F]. We may assume a\ , . . . , ar is a basis for the column space of M. We now show that /i, • • . ,/ r are algebraically independent over k. If /i, . . . , fr satisfy a polynomial relation, then there are

a= ( a i , . . . , a r , 0 , . . . , 0 ) 6 N9 and c = (ci, . . . ,c r ,0, . . . ,0) e N9

such that fa = f°. Hence M (a) = M(c) and 71 a i + • • • + jrar = Q, where 7i = o-i ~ ^. Therefore a$ — c,, that is fa = fc is a trivial relation. To show that fr+i is integral over the quotient field k ( f i , . . . , / r ) pick s € N such that sar+i = AI«I + • • • + \rar, where Aj e Z for all j. Therefore fr+i = /iAl • • • fr" , as required. D

Exercises 7.1.18 Let k = Z2 and X C k3 the space curve given parametrically by t\ = x[, t? = x\ , ts = x\. Prove that

PF = (ti ~ ^2*3, ^2 ~ tits, t\t-2 — *s) and PF ^ I ( X ) .

Monomial Subrings_______________________________ 209

7.1.19 If P is the toric ideal of the subring of /c-products SHik = K[Vk] and n > 2k > 4, then the minimal number of generators of P is given by:

"(p) = y(n}(n~i}\(2(k~i}~

^ W U * - * ) / [\ k - i - l

7.1.20 Use Theorem 2.4.18 together with Proposition 7.1.9 to give a second proof of the fact that toric ideals of monomial subrings have Grobner bases consisting of binomials. 7.1.21 Let R = K [ x i , X 2 , y i , • • • ,yi\ be a polynomial ring over a field K and 5 = K[F] the monomial subring spanned by F = { x i y i y j , x 2 y i y j \ 1 < i < j < 5}. Prove that (xiyiy2:s X2y3y4) =

7.1.22 Let R = k [ x i , X 2 , x 3 ] and R^ the second Veronese subring. If

:*[*«]— > f l < 2 >

induced by ip(tij) = XiXj, prove that the toric ideal P = ker(y>) is generated by the 2-minors of the symmetric matrix il

7.1.23 Let R = k[xi,X2,xs] and let 5 = R^ be the second Veronese subring endowed with the normalized grading 00

Q 0

- ff\

~ VL7 i=0

where Si = R-H. Prove that the Hilbert series of R^ is equal to

7.2

Integral closure of subrings

Here we present a description of the integral closure of a monomial subring, which links polyhedral geometry and integer programming techniques with the problem of finding integral closures of monomial subrings. The problem

of the effective determination of the integral closure of a monomial subring has a satisfactory answer given by an algorithm of W. Bruns and R. Koch;

see [48] for details on this algorithm.

210______________________________________ Chapter 7

Polyhedral sets and cones An affine space or linear variety in E™ is by definition a translation of a linear subspace of K™ . Let A C En . Let us recall that aff(A), the affine space generated by A, is the set of all affine combinations of points in A:

&S(A) = {aipi + • • • + arpr\pi & A, ai + • • • + ar — 1, a* 6 There is a unique linear subspace V of E™ such that

for some x0 € E n . The dimension of A is defined as dim A = dim R (F). A point x G E™ is called a convex combination of pi, . . . ,pr E En if there are non negative real numbers a\ , . . . , ar such that

x — aipi + • • • + arpr and ai + • • • + ar = 1.

Let A C K". The convex hull of A, denoted by conv(A), is the set of all convex combinations of points in A. If A — conv(A) we say that A is a convex set. Definition 7.2.1 A convex poly tope P C En is the convex hull of a finite set of points p\, . . . ,pr in R™, that is, P = conv(pi, . . . ,pr).

The inner product of two vector x = (xi , . . . , xn) and y = (yi, • • • ,yn) in R" is defined by ( x , y ) = xiyi + ••• + xnyn. Given a 6 R™ \ {0} and c £ E, define the hyperplane

H(a,c) = {x e K n |(z,a) = c}. Note that every hyperplane of K" is of this form. The two closed halfspaces bounded by H(a,c) are

H+(a,c) = { x 6 R " | ( x , a ) > c} and F"(a,c) =F + (-o,-c).

Definition 7.2.2 A polyhedral set or convex polyhedron is a set Q in Rn which is the intersection of a finite number of closed halfspaces of E™ . Since the intersection of an arbitrary family of convex sets in En is convex, one derives that any polyhedral set is convex and closed. It is a fact that P is a convex polytope in R" if and only if P is a bounded polyhedral

set; see [310, Theorem 3.2.5]. Definition 7.2.3 If C C E" is closed under linear combinations with non negative real coefficients, we say that C is a convex cone or simply a cone. A polyhedral cone is a convex cone which is also a polyhedral set.

Monomial Subrings_______________________________211

If A is a set of points in E", the cone generated by A, denoted by E+ A

or coue(A), is defined as Oj € K+ and ft £ A, for alii > ,

where E+ is the set of non negative real numbers. We say that C is a finitely generated cone in Era if (7 = E+^L, for some finite set of points A C K n . Theorem 7.2.4 If C C E", i/zen C is a polyhedral cone in Kn if and only if C is a finitely generated cone. Proof. See [310, Theorem 4.1.1].

D

One of the fundamental results on polyhedral geometry is the following remarkable representation theorem. Theorem 7.2.5 (Finite basis theorem) // Q is a set in E™, then Q is a polyhedral set if and only if Q can be expressed as Q = P + C, where P is a convex polytope and C is a finitely generated cone.

Proof. See [64, Theorem 16.2] and [310, Theorem 4.1.3].

D

Thus according to the finite basis theorem a convex polyhedron has two representations. We have found it very convenient to use the computer program Polyhedron Representation Transformation Algorithms (PORTA for short) to switch between the two representations. See [63]; for the case of polyhedral cones see [48] . Definition 7.2.6 Let Q be a closed convex set in E™. A hyperplane H of

E" is called a supporting hyperplane of Q if Q is contained in one of the two closed halfspaces bounded by H and Q fl H ^ 0. Definition 7.2.7 A proper face of a polyhedral set Q is a set F C Q such that there is a supporting hyperplane H ( y , c ) satisfying the conditions:

(b) Q £ H(y,c), and Q C H+(y,c) or Q C H~(y,c). The improper faces of a polyhedral set Q are Q itself and 0. Proposition 7.2.8 If Q is a polyhedral set in E™ and FI, F% are faces of Q, then their intersection F = FI n F% is a face of Q.

212______________________________________ Chapter 7

Proof. Let Hi = H(a,i,ci) be supporting hyperplanes of Q, where Cj G M

and 0 ^ a, 6 K™, such that F{ = Q n #; and Q C ff+ for i = 1,2. Let us prove the equality +a2,ci +c2). The left hand side is clearly contained in the right hand side. On the other hand if x e Q PI #(ai + 0,3,01 + c 2 ), then using (x,a,i) > Cj one has:

ci + c2 = (o;,ai + a 2 ) = (x,ai) + (x,a2) > GI + c 2 , hence (x, a;} = Cj for i = 1, 2 and x 6 -F. Since

Q C ff + (ai + a 2 , c i +c 2 ) the set F is a face of Q.

D

Definition 7.2.9 Let Q be a polyhedral set and x0 e Q. The point x0 is called a vertex or an extreme point of Q if {XQ} is a proper face of Q. Definition 7.2.10 Given x = (xi,.. ., xn) 6 Kn its Euclidean norm is

||x|| = \/(x,x}. Lemma 7.2.11 Let Q be a convex polyhedron in E™ and XQ € Q. Then XQ is a vertex of Q if and only if Q\ {XQ} is a convex set.

Proof. =>) Let H(a,c) be a proper supporting hyperplane of Q such that {XQ} = Q n H(a,c) and Q C H+(a, c). Take x,y £ Q\ {x0} and consider z = iy + (1 — t)x with 0 < t < 1. Note that x, y are not in H(a,c), thus (z,a) = t ( y , a ) + (1 -t)(x,a) > tc+ (1 - i)c = c. Hence (z, a) > c, that is, z ^ XQ and z G Q, as required.

<=) Let <5 = n[ =1 7f + (ai, Cj) be a decomposition of Q as an intersection of closed halfspaces, where 0 ^ a, 6 R™ and Cj 6 K for all i. First observe that x0 is in H(a,i,Ci) for some i, otherwise if (x 0 ,aj) > Cj for all i, there is an open ball BS(XQ) in R™, of radius <5 centered at XQ, whose closure lies in Q. Hence taking two antipodal points xi,x^ in the boundary of BS(XQ) one obtains XQ € Q \ {XQ}, a contradiction. One may now assume there is k > 1 such that XQ € H(a,i, Cj) for i < k and (XQ, QJ) > Cj for i > k. Set A = ff (ai, ci) n • • • n H(ak,ck). We claim that A = {XQ}.

If there is xi & A\ {XQ}, pick Bg(xo) whose

closure is contained in H+(ai,Ci) for all i > k. Since z = txo + (1 — t)xi is in A for all t € K, making i = 1 + (J/ll^i ~ ^oll one derives z G A and ||z — xo|| = 5, thus z G Q. Note that if k = r, then x\ is already in Q

Monomial Subrings_______________________________ 213 and in this case we set z = x\. Making t = — 1 in z\ — tz + (1 — t)xo one concludes z\ € A and \\z\ — XQ|| = 6. Altogether z,z\ are in Q \ {a;o} and

XQ = (z + Z i ) / 2 , a contradiction because Q \ {XQ} is a convex set. From the equality {XQ} = A = Af~}Q one has that {XQ} is an intersection of faces. Using Proposition 7.2.8 yields that {x0} is a face, as required. D Proposition 7.2.12 Let Q be a polyhedral set in E™ which is not an affine space and t=i

where 0 ^ aj 6 E" and c; 6 E /or aH i. // XQ G E", £ften XQ is a vertex of Q if and only if (a) {XQ} — r\i£iH(ai,Ci) for some I C {!,..., r}, and

(b) (x0,ai) > Ci for all i = 1, . . .,r. Proof. =£•) This direction follows at once from the proof of Lemma 7.2.11.

•<=) Note XQ E Q. Since the intersection of faces of Q is a face, see Proposition 7.2.8, one has that {XQ} is a face. D If a € E™, a ^ 0, then the set Ha will denote the hyperplane of E™ through the origin with normal vector a, that is, Ha — {x € En | (x, a) =0}.

This hyperplane determines two closed halfspaces denoted by

H+ = {x e Kn (x, a) > 0} and H~ = {x € En | (x, a) < 0}. Proposition 7.2.13 I f C is a dosed convex cone in En and H a supporting hyperplane of C , then H is a hyperplane passing through the origin.

Proof. LetH = H(a,c). Since 0 e C C H+, one has c < 0. If CnH = {0}, then c = 0 as required. Assume C D H ^ {0} and pick 0 ^ z € C1 such that (z, a) = c. Using tz G C C H+(a, c) for all t > 0, one derives

ic = t(z,a) = (tz,a) > c,

for every i > 0. Thus c = 0.

D

Proposition 7.2.14 Let C be a cone in E™ and /ei

C = H+ n • • • n H+ be a representation of C. If aj € Q™ \ {0} /or a// i, then C is a finitely generated rational cone, that is, there are fa € Q™ such that

214______________________________________ Chapter 7 Proof. Consider the polyhedral set P=

#-0,1,

where

A = {eiei + • • • + enen\£i e {1, -1} for all i}. Note that P is a bounded set and the origin is an interior point of P. Hence P is a polytope and consequently the set Q = C n P is also a polytope. Thus by [39, Theorem 7.2] one can represent Q as Q = conv(/?i, . . . ,/3s), where /?i, . . . ,fa, are the vertices of Q. By Proposition 7.2.12 each vertex of Q is the unique solution of a system of linear equations with rational coefficients. Thus all the fa have rational entries.

Since fa e C for all i one has K+ /?i + • • • + E+ J3S C C. Conversely if x 6 C, there is an scalar t > 0 such that tx G Q, and one rapidly obtains that x is in the cone generated by /3i , . . . , f3s . D Definition 7.2.15 A proper face P of a polyhedral set Q C En is called a facet of Q if dim(F) = dim(Q) - 1. Theorem 7.2.16 Let Q be a polyhedral set in E™ which is not an affine space such that

Q = aff(Q) n H+(ai, ci) n • • • n H+(ar,cr)

where
Definition 7.2.17 If a polyhedron Q in E" is represented as

and satisfies

for all j, we say that (*) is an irreducible representation of Q.

D

Monomial Subrings_______________________________ 215

We shall always assume that an affine space in E" has the topology induced by the usual topology of E" . By the relative interior of a set X in E", denoted by ri(^T), we mean the interior of X in aS(X). Theorem 7.2.18 Let Q be a polyhedral set in E™ with dim(Q) = n and such that Q ^ W1 . Let

be a representation of Q with H+(ai, c\ ) , . . . , H+(ar,cr) distinct, where a,i is in K" \ {0} for all i. Set Fi = Q n H(a{, a), for i = 1, . . . , r. Then (a) ri(Q) = {x&En\(x,ai) > Cll. . . ,(x,ar) >cr}. (b) Each facet F of Q is of the form F = F \ for some i. (c) Each Fi is a facet of Q if and only (*) is irreducible.

Proof. See [39, Theorem 8.2] and [310, Theorem 3.2.1].

D

Corollary 7.2.19 If Q ^ E™ is a polyhedral set of dimension n, then Q has a unique irreducible representation

as an intersection of closed halfspaces. Proof. Follows from parts (b) and (c) of Theorem 7.2.18. Note that if the set F = Q H H(a, c) is a facet of Q, then aff(F) = H(a, c). D Corollary 7.2.20 // C ^ E" is a polyhedral cone of dimension n, then there is a unique irreducible representation

C = tf+ n • • • H tf+ , where at e Kn \ {0}. Proof. Note that according to Proposition 7.2.13 a set F is a proper face of C if there is a supporting hyperplane Ha of C such that

F = Cr\Ha ^ 0 and C £ Ha. In particular the facets of C are defined by hyperplanes through the origin, therefore the irreducible representation of C has the required form. n

The following two results are quite useful for determining the facets of a polyhedral cone. Proposition 7.2.21 Let A be a finite set of points in TLn. If F is a nonzero face ofR+A, then F = K+.4' for some A' C A.

216______________________________________ Chapter 7

Proof. Let F = ffi+,4 n Ha with E+A C H~ . Then F is equal to the cone generated by the set A' = {a 6 A\ (a, a) = 0}. D Corollary 7.2.22 Let A be a finite set in 1n and F a face ofE+A.

(a) If dim(F) = 1 and A C N™, then F = R+a for some a € A. (b) J/dim(IR+.4) = n and F is a facet defined by the supporting hyperplane Ha, then Ha is generated by a linearly independent subset of A.

Definition 7.2.23 Let Q be a polyhedral cone in E™ with dim(Q) = n and such that Q ^ K™ . Let

be the irreducible representation of the cone Q. If Oj = (oa, • • • , ^m) we shall call a set of equations of the cone Q. Proposition 7.2.24 // Q is a proper cone in W1 that can be written as

with a.i 6 Qn for all i and dim(Q) = n, then there are unique (up to sign) ai, . . . ,ar in TLn with relatively prime entries and such that

is the irreducible representation o f Q . Proof. First note that if HI, is a supporting hyperplane of Q generated by a set of n - 1 linearly independent vectors in {cti,...,aq}, then by the Gram-Schmidt process HI, has an orthogonal basis of vectors in Q™ , and consequently there is a normal vector a to HI, such that a 6 Qn and Ha — HI,. Hence multiplying a by a suitable integer and then dividing by the greatest common divisor of the entries, one may assume Ha — HI,, where a is in Z" and has relative prime entries. Observe that a, b are linearly dependent because the orthogonal complement of Ha is one dimensional. It is readily seen that a is uniquely determined up to sign. To complete the proof use Corollary 7.2.22 (b) to see that any facet of Q is defined by a supporting hyperplane HI, as above. D

217

Monomial Subrings

Definition 7.2.25 Given a = (GI , . . . , an) € K" , we set

= K| + •• • + \an\. Theorem 7.2.26 Let A be a finite set of nonzero points in N™ and A' the set of a € A such that R+a is a face of K^ A of dimension 1. Then

Proof.

Set A = {ai,...,aq} and S = conv(ei, . . . ,en). Consider the

convex polytope P = S r\ ffiL+A Note that P = conv(/?i, . . . ,/3q), where /3i = a.il\a.i . Therefore by [39, Theorem 7.2] the set of vertices V(P) of P is contained in {/?i, . . . , /3q} and P is the convex hull of V(P). Since K+.4 = E+V(P) it suffices to show that E^/3 is a face of the cone spanned

by A for every /3 in V(P). Let /3 £ V(P). There is a hyperplane H = {x £ Rra (a;, a) = c] such that P n H = {13} and P C H~ . Set a = (ai, . . . , a n ). To complete the proof we now show that

R+AcH^ and R+An Hb = R+/3, where b = (0,1 — c, . . . , an — c). Since ai/\ai\ £ P C H~ for all i one has


a

i

\ i\

Hence j3 £ HI,. To prove the reverse inclusion take 0 ^ x € IR+.4 n Then I x \ ( x , a ) = ( x , ( c , . . . , c ) ) =c\x => ( — ,a}=c. \\x I Thus x/\x\ € H. Observe that x/\x\ £ 5 and x/\x &R+A, hence

Altogether

this proves x £

x \M Jj

n

218

Chapter 7

Notation Given x = (x\,..., xn) 6 R™, we write x > 0 if z, > 0 for all i. Theorem 7.2.27 (Farkas) Let A an n x q matrix with entries in a field K and z e Kn. Assume K — Q or K = K. Then either there exists x £ Kq with Ax1 = z1 and x > 0 or there exists v G Kn with vA > 0 and vzi < 0, but not both. Here x1 denotes the transpose of x. Proof. See [315, Proposition 1.8] for the case K = E and [207, 216] for the

case K = Q.

D

Normality of monomial subrings One of the issues to be discussed in first place is a description of the normalization of a monomial subring. For convenience recall that if A is a domain with field of fractions KA , the normalization or integral closure of A is the subring A consisting of all the

elements of KA which are integral over A. Theorem 7.2.28 Let R = k[x\,..., xn] be a polynomial ring over a field k and let F be a finite set of monomials in R. Then

k[F] = k[ where A — log(-F) is the set of exponent vectors of the monomials in F.

Proof. Set B = k[{xa

First we show k[F] C B. Take z e k[F], since k[F] C R and R is normal we get k[F] C R and z G R. Hence one can write

where d; £ k \ {0} and 71,..., jr distinct nonzero points in N n . Write 7i — rii/3i, with m equal to the gcd of the entries of ji (observe that f 3 i , . . . , ( 3 r are not necessarily distinct). Consider the cone C spanned by A and {/3i, . . . ,/3r}. One must show ft e 1+^4 for all i. Assume on the contrary that C is not equal to K+ A By Theorem 7.2.26 one may assume that M+/?i is a face of C not contained in K+.4. Let Ha be a hyperplane so that Ha n C = ffi+/3i and C C H~ . There is 1 < I < r such that

nt = sup {rijl-jj =nj/3i}. Since z is integral over k[F] it satisfies a monic polynomial / of degree m with coefficients in k[F]. To derive a contradiction we claim that the term

Monomial Subrings_______________________________ 219

xmje occurs only once in the expansion of f ( z ) as a sum of monomials. Assume the equality

where m^. > 0 for all j, m > XL=1 tn^ and a 6 ffi+A As jf = nePi, one has (mryt,a) = 0 and from this equality one rapidly derives (0,0) = (#1,0) = ••• = (&,, a) =0.

Hence a, /3i; , . . . , fiit e Har\C = !+/?].. Note /?i £ K+A Thus a = 0 and ft.,- = /?i f°r all J- Therefore t = 1 and 7^ = 7^, as claimed. Altogether we get f ( z ) ^ 0, which is impossible. This proves that 7, € R+vA for all i. On the other hand using that z belongs to the field of fractions of k[F] it follows that 7j G 1iA for all i. __

Let us now show the other containment B C k[F]. Note the equality

which follows from Theorem 7.2.27. An straightforward calculation rapidly shows that xa is in the field of fractions of k[F] if a £ ZA, and xa is an integral element over k[F] if a £ Q+ A. Hence B C k[F}. D Corollary 7.2.29 Let R be a polynomial ring over a field k and F a finite set of monomials in R. Then k[F] is normal if and only if

NA = where A = log(F). Proof. It suffices to observe the equality

k[F] = k[{xa\a € NA}] and to use the description of k[F] given above.

HI

Theorem 7.2.30 (E. Noether) Let B = k[ti,... tq] be a polynomial ring over a field k andtp a prime ideal of B. Then the integral closure of B/f is a finitely generated B/p module.

Proof. See [92, Corollary 13.13].

D

In general it is hard to compute the integral closure of an affine domain; see [295] and [298, Chapter 6].

Corollary 7.2.31 Let R = k[x\,... ,xn] be a polynomial ring over a field k and F a finite set of monomials in R. Then

k[F] = ^k[F] for some x^1,..., x@m in k[F].

220______________________________________Chapter

7

Proof. Use Theorem 7.2.28 and the fmiteness of the integral closure of an afflne domain, see Theorem 7.2.30. This result also follows from the next lemma. D Lemma 7.2.32 Let A = { c c i , . . . , aq} be a finite set of points in TLn. Then there exist 7 1 , . . . , jm in 1in such that

and ji e — q max ai |, q max at

/or a// i.

Proof. Set M = q max |at| and TV = — q max \a~ . Recall the equality

Let /3 e (7. One can write

where x, G N and 0 7^ j/j G N. By the division algorithm there are non negative integers TJ, n, such that Xj = n^ + r» and 0 < r, < y^. Therefore one can write

where n» 6 N and at £ [0, 1] n Q. Since ^Li a*ai is in (7 n [N, M]n one rapidly obtains that A U ((7 n [JV, M]n) is a generating set for (7 with the required property. Our argument was based on the proof of Gordan's Lemma given in [44]. D Definition 7.2.33 Let R be a polynomial ring over a field k and F a finite set of monomials in R. A decomposition

of the fc-vector space k[F] is an admissible grading if k[F] is a positively graded fc-algebra with respect to this decomposition and each component k[F]i has a finite fc-basis consisting of monomials.

Monomial Subrings_______________________________ 221

Theorem 7.2.34 (Danilov, Stanley) Let R = k [ x i , . . . ,xn] be a polynomial ring over a field k and F a finite set of monomials in R. If k[F] is normal, then the canonical module U>I.[F] ofk[F], with respect to an arbitrary admissible grading, can be expressed as

(7.1) where A — log(F) and ri(R±A) denotes the relative interior o/K_|_.4.

The formula above represents the canonical module of k[F] as an ideal of k[F] generated by monomials. Soon it will become clear how to apply it to estimate some a-invariants. For a comprehensive treatment of the Danilov-Stanley formula see [44, Theorem 6.3.5] and [76]. Theorem 7.2.35 (Hochster) Let R be a polynomial ring over a field k and F a finite set of monomials in R. If k[F] is normal, then k[F] is Cohen-Macaulay.

Proof. See [30] and [172].

D

Theorem 7.2.36 Let R = k[xi, . . . ,xn] be a polynomial ring over a field k and F a finite set of monomials in R. If

is a homogeneous Noether normalization of k[F], then k[F] is a free Amodule whose generators have degree at most X)i=1 deg(/i$). Proof. Let S = k[F] endowed with the induced grading 5, = S r\R,. Note that the composition

A^f S^-S is a homogeneous Noether normalization. Since 5 is Cohen-Macaulay by Theorem 7.2.35, one can then use Proposition 2.2.14 to obtain a direct sum decomposition S = Axfil © • • • e A z / 3 m . Therefore the Hilbert series of S can be expressed as

F(S, *) =

z=l

Applying Theorem 7.2.34 one derives that a(5), the a-invariant of S, is negative. Using that a(S) is equal to the degree of F(S,t) as a rational function (see Proposition 4.2.3) one concludes degx/3i < X^i deg/ij. D

Chapter 7

222

An algorithm to compute normalizations Let us describe some steps of an algorithm of Winfried Bruns and Robert Koch [48] that effectively computes normalizations. Let A be a finite set of points in N™. First one finds Wi, . . . ,uid in Zn such that

There exists an isomorphism of Z-modules

T:ZA —>Zd with T(WI) = e». Because of Lemma 7.2.32 there are ft, . . . j3r in a bounding

box [N, M]d such that

The determination of the fa's can be accomplished by finding the irreducible representation of the cone R_|_T(^4) as an intersection of closed half spaces (see Corollary 7.2.20), and then testing which elements in

lie in the cone spanned by T(A). To finish the description note

Z,Ar\ E+A = NT'1 (ft) + • • • + NT"1 (ft). Observe that the computation of the irreducible representation of M+T(A) can be performed using Corollary 7.2.22; in several specific examples we have used [48] or PORTA [63] to rapidly determine such representation. Example 7.2.37 Let A C N7 be the set of points

ai = ei + e 2 , ai — 62 + e 3 , 0:3 = ei + e3, 04 = e3 + e 4 , a5 = €4 + e 5 ,

ae = e5 + ee,

0:7 = e6 + e 7 ,

Note 0 •••0

where the itij's are the row vectors of the matrix:

P=

1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0

0

0

0

0

0

a8 = e5 + e 7 .

Monomial Subrings

223

Next we consider the affine transformation T determined by T(u)i) = &i. An straightforward computation shows that the points in T(A) correspond to the rows of the matrix:

"1

0 0 0 0 0 0 " 1 0 0 0 0 0 1 -1 2 -2 2 -2 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 1 -1 1 0

Using [48] or PORTA [63] we obtain that a point x = (xi,...,x7) is in if and only if x satisfies the system of linear inequalities

-2x7 -X 5

-2x5

-2xe

-x3 -X3

-2x7 +2x 7 -2x 7

-Z

+X 7 ~X2

-X 7

~X2

+ X7

-2X2

-x6

-Xi

-x7

-X4


-Xi


= l,3,7.

It follows that

ZdnE+T(A) = HT(A) + where ft = (1,0, 1, -1,2, -1, 1). Hence

The linear inequalities shown above define the facets of the cone spanned

by T(A). Homogeneous subrings

The class of homogeneous monomial subrings

has been studied with detail in [278]. Examples that fall into this class are Rees algebras of ideals generated by monomials of the same degree, subrings generated by monomials of the same degree, and polytopal subrings.

224______________________________________ Chapter 7

Definition 7.2.38 Let F — {xai, . . . ,xa*} be a set of monomials in the ring of polynomials R — K[x\, . . . ,xn], where K is a field. The monomial

subring K[F] is said to be homogeneous if there is v 6 Q™ such that (ai,v) = 1 for i = l,...,q. Proposition 7.2.39 Let F = {xai ,..., xaq } be a set of monomials in the ring R = K[x\, . . . , xn], where K is a field. Then K[F] is a homogeneous subring if and only if K[F] is a standard graded algebra with the grading

K[F]i = Y^ K(xai)Cl • • • ( x a * ) c « ,

where \c = Cl + • • • + cq.

\c\=i

Proof. =>•) Let v e Q" such that (cti,v) = 1 for all i. It suffices to prove

because one clearly has

C K[F]i+j

for all

First let us prove that K[F]i n K[F]j = {0} for i ^ j. If this intersection is not zero, one has an equality

with |a| = i and c| = j. Hence q = /( ^2atai,v\\ =

a contradiction. To finish assume /^ + • • • + /»r = 0, with /^ £ K[F]ij and ii < • • • < ir. If fir ^ 0, using that the monomials of R form a JC-basis, one has K[F]ir n K[F]ii ^ {0} for some j < r, a contradiction. Thus fir = 0 . By induction all the jij must be zero, as required. •4=) Set A — aff(Q)(ai, . . . ,aq) and r = dim(A), To begin with we claim that 0 is not in A. Otherwise if 0 € A, write 0 = 77777 + • • • + aqaq with

7j £ Q for all z and ^'=1 ai = l- There is 0 ^ p e Z such that pa^ e Z for all i. One may assume, by reordering the a*, that pa^ > 0 for i < s and poi < 0 for i > s. As -^[-F] is graded, from the equality

(xai)pai •••(xa')pa' = one derives

a s ) = -p(as+i

Monomial Subrings_______________________________ 225

that is, pY^1=iai = 0, a contradiction. Thus 0 ^ A, and consequently r < n. It is well known that a linear variety A in Q" of dimension r can be written as an intersection of hyperplanes

A= where 0 ^ y{ <E Q" and c, £ Q for all i; see [310, Corollary 1.4.2]. Since 0 ^ A, one has Cj ^ 0 for some j. To finish it suffices to set v = yj/Cj. O Corollary 7.2.40 Let F C K[x\ be a finite set of monomials and Pp the toric ideal of K[F}. Then PF is a graded ideal with respect to the standard grading if and only if K[F] is a homogeneous subring. Proof. As Pp is generated by a finite set of binomials, the result follows rapidly. Q Proposition 7.2.41 Let F = { f i , . . . , f q } be a set of monomials in the polynomial ring R = K[x] over a field K and Ti, . . . ,Tq,T a new set of variables. If
K[TF]

Tfi

then there is a unique an epimorphism if> such ijj = ip) is a binomial ideal. Note that ker(y) and ker(^) are both prime ideals. The map tp is an isomorphism if and only if ker(y>) = ker('0), thus to finish we need only observe that the last equality holds if and only if K[F] and K[TF] have the same dimension. n

With the notation above one has Corollary 7.2.42 The map ip:K[TF] —>• K[F] is a isomorphism if and only if K[F] is a homogeneous subring. Proof. If i/) is an isomorphism, then by the proof above ker(^) = and K[F] is homogeneous because ker(y) is homogeneous with the standard grading of K[Tlt. . .,Tq]. Conversely assume ker('0) homogeneous with the standard grading. Note that any homogeneous binomial in ker(^i) is also in ker(y>), thus one has ker(V») = ker(yi) and i/j is an isomorphism. D

226______________________________________ Chapter 7

Ehrhart function In this paragraph we consider a fixed set of distinct monomials F = {xai , . . . , xai} in a polynomial ring R = K\x\ . . . , xn] over

a field K. Let P = conv(.4)

be the convex hull of the set A = {a-i, . . . ,aq}. The normalized Ehrhart ring of P is the graded algebra oo

AP = 0(^)i C R[T\, z=0

where the ith component is given by

KxaTi

-

Note that because of the convexity of P one has • (Ap)i(Ap)j C (AP)i+j and . (AP)i n (AP)j = {0} for i jt j.

Thus Ap is in fact a graded /^-algebra. Proposition 7.2.43 K[FT] C Ap, with equality if K[F] is homogeneous. Proof. Set A1 = {(«!,!), . . . , (aq, 1)} C Z n+1 . First we show:

ZA'nS+A' C {(a,i)\a e Z,4niP and i € N}.

(*)

Take z = (a, i) in 1A' n M+ .-4' and write

z

= =

n i ( a i , l ) + • • • +ng(ctg, 1) A i ( a i , l ) + ••• + X q ( a q , l ) ,

where n^ e Z and \j > 0 for all j. Note a = 0 if i = 0, and a/i 6 P if i > 1. Hence z is in the right hand side of (*). By Theorem 7.2.28 one concludes the required inclusion.

Assume K[F] is a homogeneous subring. Let v be a vector in Qn such that (cej,v) = 1 for all j. It suffices to prove that equality holds in (*). Take z = (a,i) in the right hand side of (*) and write a = mon + • • • + nqaq = z(Aio:i + • • • + A ? a ? ),

where nj £ Z, \j > 0, and AI + • • • + Xq = 1. Hence (a,v) z = (a,i)

= ni+-'- + nq = i, and = =

ni («!,!) + ••• + nq(aq,l) i\i(ai,l) + • • • + i X g ( a q , l ) .

Thus z £ "LA1 n 1+.4', as required.

D

Monomial Subrings_______________________________ 227

Corollary 7.2.44 // K[F] is homogeneous, then K[F] is normal if and only ifK(FT] = AP. Proof. Use Corollary 7.2.42 and Proposition 7.2.43.

D

The normalized Ehrhart function of P is defined as

EP(i) = dimK(AP)i = \ZAniP\. Corollary 7.2.45 // K[F] is a homogeneous subring of R and h is the

Hilbert function of K[F], then (a) h(i) < Ep(i) for all i>0, and (b) h(i) = Ep(i) for all i > 0 if and only if K[F] is normal. Proposition 7.2.46 // K[F] is a homogeneous subring of dimension d, then Ep is a polynomial function of degree d — 1 . Proof. As the Ehrhart ring AP is equal to K[FT], by Corollary 7.2.31 one has that Ap is a finitely generated K-algebra. Hence Ap is a finitely generated graded K[FT]-module. Using Theorem 1.2.8 and Proposition 1.4.12 one concludes that the Hilbert function of Ap is a polynomial function of degree d — 1. D

Corollary 7.2.47 // K[F] is normal and homogeneous, then the Hilbert polynomial of K[F] is equal to the normalized Ehrhart polynomial of P. Example 7.2.48 Let F = {x4, x3y, xy3, y 4 } and P = conv(_4), where A = log(.F). The normalized Ehrhart ring is

AP = K(x4T, x3yT, xy3T, y*T, x2y2T}. The subrings K[F] and Ap have different Hilbert functions, but they have the same Hilbert polynomial, which is equal to 4£ + 1.

Proposition 7.2.49 ([278]) If K[F] is a homogeneous monomial subring whose toric ideal has a square-free initial ideal with respect to some term order, then K[F] is normal. Proof. See [278, Proposition 13.15].

Normal polytopes

D

Some special types of monomial subrings associated

to polytopes are introduced next. Our exposition has been influenced by the beautiful paper [41], that presents an appealing study of normal polytopal subrings.

Chapter 7

228

Let P be a lattice polytope in K™, that is, P has all its vertices in Z". For simplicity we assume that the vertices of P are in W1 . Given a field K the monomial subring

K[P] = K[{xaT\ a

C K [ X l j . ..,xn,T]

is called the polytopal subring of P. The polytope P is said to be normal if K[P] is normal. Proposition 7.2.50 If P is a lattice polytope in W1 and K a field, then

dimK[P] =dim(P) + l. Proof. Let cei,...,aq 6 N™ such that P = conv(ai, . . . ,aq). First we observe the equality

aS(P}=aS(ai,...,ag)=a1+V, where V = R(a2-a1) + -•- + R(aq-ai). Thus one has dim(P) = dim R (V). On the other hand using the identity 0:2-0:1

Oil

0

— rank

rank

aq

1,

1

the asserted formula follows from Proposition 7.1.17.

D

Proposition 7.2.51 Let AQ, . . . , An be a set of affinely independent points in W1 and A = conv(Ao, . . . , An}. Then the volume of the simplex A is

A

det vol(A) =

(° M : \An

: l)

det

I Al - A0 \ : \An-A0) n\

Proof. The result follows using linear algebra or applying the change of variables formula. See [206, Chapter XII] or [310, Chapter 6]. d

Let A be an n-dimensional lattice simplex in K", that is, A is the convex hull of a set of n+1 affinely independent point in Z n . The normalized volume of A is defined as n!vol(A). From the result above one has vol(A) > I, recall that A is called a unimodular simplex if equality holds.

229

Monomial Subrings

Proposition 7.2.52 Let A be an n-dimensional lattice simplex in En with vertices AQ, . . . ,An in Z™. Then A is unimodular if and only if either of the following two equivalent conditions hold

(a) 1n+l = Z(Ao, !) + • • • + Z(An, 1). (b) Z" = Proof. Recall that a set of vectors ai,...,an in Z" generate Z™ as a Z-module if and only if the n x n matrix A whose rows are the a, has determinant ±1. Thus the result follows from Proposition 7.2.51. D Proposition 7.2.53 // A is a unimodular simplex in E™ with vertex set

A = {QO, . . . , an}, then A n Z" = A. Proof. It suffices to show that A n Z" is contained in A. Let a e A n Z™. There are Aj in K+ such that a = \oao + ••• + \nan and ElLo Aj = 1. If A0 > 0, from the equalities

1 < det

/ a ai

1\ 1 ^

^ an

det

1/

= A 0 n!vol(A) = A 0 , \

I

one derives AQ = 1, and consequently Aj — 0 for all? > 1. Hence a = a^. Similarly if Aj > 0 for some i > 0, then a = a». D Corollary 7.2.54 Let K be a field and P a unimodular simplex in En with

vertex set A = {OQ, . . . ,an}- Then K[P] = K[xa°T,...,xanT}

and K[P] is a polynomial ring. Proof. The asserted equality is a direct consequence of the result above. As K[P] has dimension n + 1, using Lemma 2.1.5 gives that K[P] is a polynomial ring. D Proposition 7.2.55 ([41]) Let {C,Ci} be a family of finitely generated subsemigroups of N™ such that C = UjCj. // ZC = ZCj and K[Ci] is normal for all i, then K[C] is normal. Proof. By Corollary 7.2.29 it suffices to show that ZCn l+<7 is contained

in C. Take z e ZCnR+C. There is an integer s > 1 such that sz 6 C, thus sz £ Ci for some i. As ZC = ZCj and K[Ci\ is normal we get z G Ci C C, as required. CI

230______________________________________ Chapter 7

Definition 7.2.56 Let P be a convex polytope in K". A collection {A;} of unimodular lattice simplices in K" is called a unimodular covering of P if P = U;Ai. Theorem 7.2.57 ([41]) Let P be a lattice polytope in K™ of dimension n with vertices in W1 . If P has a unimodular covering and K is a field, then K[P] is normal.

Proof. Let { A^} be a unimodular covering of P and ai,...,aq the vertices of P. By Proposition 7.2.55 we need only show that C = UfCi, where C and Ct are the subsemigroups of W+l denned as:

,l)|a e Z " f l P }

and Ct = N{(a, 1)| a £ Z"

One clearly has the containment UfCi C C, because P — U^A^. Conversely take z £ C 2 r = ( / 3 , a ) = n i ( / 3 i , l ) + -" + n r ( j 8 r ) l), where n, € N and ft € Z" n P for alH. There are scalars ^ > 0 such that

ft = Mil "I + • • • + Piqdq and Mil + • • • + Mig = 1 Vi. Note that /3/s is in P, where 0 = Y^i n »ft anc^ s = ]Ci n «- Hence /3/s is in Af for some £.. We can write A^ = conv(7o, . . . ,7n), with {70, . . . ,jn} the vertex set of A^. Thus

/3 for some fa > 0 satisfying ^^ //, = !. Hence Z = (ft s) = SyUo(70, !) + • • • + S^n(7n, 1) 6 IK+ {(TO, 1), • • . , (in, 1)},

by the unimodularity of A^ one has

Therefore using that jK"[A^] is normal (see Corollary 7.2.54) yields z 6 Ct, as required. This proof is due to W. Bruns, J. Gubeladze and N. Trung. n

Exercises 7.2.58 Let A C W1 . Prove that A is an affine variety in Kn if and only if A! A + X2A C A for all AI, A2 in K such that AI + A 2 = 1. 7.2.59 If K[F] is a homogeneous monomial subring over a field K, use the Danilov-Stanley formula to show a(K[F}) < 0.

Monomial Subrings_______________________________ 231

7.2.60 Let o
(b) k[C] = k[{xa\a e C}] C fcjx] is a normal domain, where k is a field. 7.2.61 Let A C W1 . Prove that A is an affine variety in B™ if and only if

\iA + X2A C A for all AI, A 2 in B such that AI + A 2 = 1. 7.2.62 Determine the facets of the convex polytope

P = conv(±ei , . . . , ±e n ) C E" . 7.2.63 Let A = {a^...,aq} C N™ and C = No-i + • • • + Ncc, . Prove that the semigroup ring k[C] is normal if and only if any element a 6 TLA satisfying ra £ C for some integer r > 1 belongs to C.

7.2.64 If P is a unimodular lattice simplex in B™ with vertices CHQ, . . . , an, prove that the polytopal subring K[P] is not isomorphic (as a ring) to the monomial subring K[xa° , . . . ,xan] C K[x], where K is a, field. 7.2.65 Let 1 < k < n be two integers and A = {e^ + - - - + eik 1 < ii < • • • < ik < n},

where &i is the ith unit vector in E™. If P — conv(ai, . . . ,a ? ), for some ai,...,otq £ A, then P n Z™ = {ai, . . . ,a,}. 7.2.66 Let F = {xai , . . . , xa« } be a finite set of square-free monomials of degree k in a polynomial ring R = K[xi,...,xn] over a field K. If P = c o n v ( a i , . . . , a ? ) , then K[P] ~ K[F]. 7.2.67 Let F = {xai , . . . , z"9 } be a set of monomials in a polynomial ring

R = K[XI , • • • , xn] over a field K. Prove that K[F] is a homogeneous subring of R if and only if 0 <£ A = aff 0 (ai , . . . , aq) C Qn .

Hint A is an intersection of n — r rational hyperplanes, where r = dim( J 4). 7.2.68 Let {ai, . . . ,aq} be a set of vectors in Qn and A — affQ(a 1 , . . . , a ? ). Prove that 0 ^ A if and only if

rank

Vi o ) 7.2.69 Is the integral closure of a homogeneous subring homogeneous?

232______________________________________Chapter

7

7.2.70 Is the monomial subring K[x2,xy,y3] C K [ x , y ] homogeneous? 7.2.71 Let F = { x ™ 1 , . ..,xa"} C K[x\. Prove that K[F] is homogeneous if and only if rank |

*i : aq

1 • I = rank I

7.2.72 Let F = {xai,..., xaq} be a set of monomials in R — K[x\, • • • , xn] and consider P = conv(log(F)). If A' = {(p, l ) | p € Z ™ n P } , then

and K[FT] C K[P] C A(P).

The ring A(P) is called the Ehrhart ring of P. In the following problems we keep the notations of the previous exercise. For more details on Ehrhart rings consult [44, 163, 272]. See [48] for related algorithms. 7.2.73 Prove that the Ehrhart ring A(P) is a finitely generated Jf-algebra and a normal domain.

7.2.74 Prove that i=0

is a graded /C-algebra with its ith component given by

7.2.75 Prove that the Ehrhart ring A(P) is integral over K[FT], and that the Hilbert function of A(P} is a polynomial function of degree d = dim(P). The corresponding polynomial is called the Ehrhart polynomial of P. 7.2.76 If d = dim(P), then according to [272, Proposition 4.6.30] one has: vol(P) = lim -——-;——-. i—>oo

{"•

Thus vol(P) is the leading coefficient of the Ehrhart polynomial of P. Prove this formula for the case n = dim(P). In practice one can compute volumes

of lattice polytopes using [48].

Monomial Subrings_______________________________ 233

7.3

Integral closure of monomial ideals

As noted earlier the simplest kind of integrally closed monomial ideals are the Stanley-Reisner ideals and the powers of face ideals. Below we give a few basic properties of the integral closure of monomial ideals.

Lemma 7.3.1 Let R be a polynomial ring over a field k and let I be an ideal of R. Assume that MI, . . . , Mq is a set of linearly independent monomials over k such that

Mlji £ k, where Sj = rrij +

Mj* = bmj i¥=i

r^ > 0, tyi

rrij e N and Tji £ N for j — 1, . . . , q. If bmj is a monomial in Imi for nij > 0 and bmj £ k for rrij — 0, then M^ G I* for some I > 0. Proof. We argue by induction on q. The case q — 1 is clear. Assume q > 2.

For j > 2 we substitute M*1 into M - j S l to obtain: M^Sl-n,r,1=6^^

-Q

^i + rnr^

l?^j

Notice that in these equations MI has been eliminated. Hence applying the induction hypothesis to M%, . . . , Mg yields Mlq e I* for some I > 0. D

As a consequence of Lemma 7.3.1 we derive the following result using a short algebraic argument. Proposition 7.3.2 ([198]) Let R be a polynomial ring over a field k and let I be a monomial ideal in R. Then I, the integral closure of I, is a monomial ideal. Proof. Let z € / and let z — MI + • • • + Mq, where the Mj's are linearly independent monomials over k. There is an equation

zn + aizn~l + ••• + an_iz + an - 0

(a* <E P).

By induction it suffices to verify that M» € / for some i. If M" e In, then Mi 6 /. Hence we may assume M" ^ In for all i. Since I1 is a monomial ideal, all the monomials in the support of a, belong to P. Using the equation above we obtain (after cancelling out common terms) equations of the form

Mjj = bmj

Mp* , where Sj = rrij +

&j

Tji > 0,

ift

nij 6 N and TJJ € N for j = 1, . . . , q, and such that bmj are monomials in Imi for rrij > 0, and bmj € k for rrij = 0 . As a consequence Lemma 7.3.1

yields Mq e 7.

D

234

Chapter 7

Proposition 7.3.3 Let R = k[x\, . . . ,xn] be a polynomial ring over a k and I C R a monomial ideal. Then the integral closure of I is given by

7 = (xa xma e Im for some m > 1). Proof. If xma € Im, then xa satisfies a polynomial of the form zm + am, with am e Im, hence xa is in /. On the other hand if z = xa 6 7, then there is an equation

zr + aizr~l + ••• + ar_iz + ar = 0

(a» e P).

Since / is a monomial ideal one obtains zm e Im for some m > 1. Noting that / is a monomial ideal the asserted equality follows. D A geometric description of the integral closure

Let a €
Q_)_ is the set of non negative rational numbers. We define the upper right corner or ceiling of a as the vector \a] whose entries are given by if on e N,

a,

where \pti\ stands for the integral part of o^. Let a, = (an, . . . , Hj n ) £ N" and let conv(ai, . . . , ar) be its convex hull (over the rationals), that is, r

V—"v

v—-v

conv(ai,... ,ar) = ^ y^ Ai
is the set of all convex combinations of a i , . . . , a r . Proposition 7.3.4 Lei R = k[xi,... ,xn] be a polynomial ring over a field k. If I C R is an ideal generated by monomials xai,..., xar, then its integral closure is: T _

Proof. We set F =

a € conv (ai ,... ,ar)\ .

Let x^0^ 6 F and let a = ^[=j Ajaj be a convex combination of 0,1, . . . , ar. Since [a] > a, there is j3 € Q™, with /? > 0, such that ["«]=/? + a. Hence

there exists p > 0 so that p/3 G N" and pAj € N for all i. Therefore a-Pkl

=

^^P" = xP/3 (a; a i)P Al . . . (xa^ e P,

showing a;!""! e l and (F) C 7.

Monomial Subrings

235

Conversely let z7 € /, that is, xp"< 6 Ip for some p > 0. There are non-negative integers s, satisfying ar' — x

Hence <J A /* 7= - + > — p t^ V P

Set a = Xl[ =1 (si/p)aj. By dividing the entries of S by p, we can write
Example 7.3.6 Let / = (x 3 ,y 4 ) C k[x,y}. The integral closure of / is generated by the monomials of the points marked in black dots in the figure:

Therefore 7=

236______________________________________Chapter

7

Proposition 7.3.7 Let k be a field and R = k [ x i , X 2 \ a, polynomial ring in two variables. If I is generated by monomials xai,..., xaq of degree d, then I is also generated by monomials of degree d. Proof.

One may assume an < ••• < a ? i, where a, = (flji,aj2).

We

claim that / is generated by the set A of all xc such that c\ + c-i = d and an
c = tai + (1 - t)aq and c 6 conv(ai,a 9 ), that is, xc € /. Conversely let x13 6 /, by Proposition 7.3.4 one may assume 0 = fa] for some a in

conv(ai,... ,o g ). Set c = (\ai],d- fail), by the previous argument xc is in A. Since (3 > c one has x13 6 (.4). D Proposition 7.3.8 Let R = k [ x i , X 2 \ and I an ideal of R generated by monomials of the same degree d. If / is integrally closed, then / is normal.

Proof. The ideal / is of the form / = z J, where z is the greatest common divisor of the monomials in /. Since / is integrally closed if and only if J is integrally closed, one may assume / is m-primary, where m = ( x i , x % ) . As

Xi and ^2 belong to /, one rapidly derives / = m.

D

In polynomial rings in two variables any integrally closed ideal is normal; more generally one has the following result of Zariski.

Theorem 7.3.9 If R is a regular local ring of dimension two and Ii,... ,Ir

are integrally closed ideals, then l\- • -Ir is integrally closed. Proof. See [314, Appendix 5, Theorem 2'].

D

Example 7.3.10 Let R = k[xi,X2,X3\ and I = (x\,x<2X\).

Using the

program Normaliz [48] one can compute the integral closure of /. input and part of the output file are: example.in

The

example.out

integral closure of / 2 3 3 0 0 0 1 2 3

(number of generators) (dimension o f ambient lattice) (generators)

3 0 0 0 1 2 2 1 1

(option or mode)

Definition 7.3.11 If xa is a monomial in R = k[x] we set log(:ra) = a.

Given a set F of monomials, the log set of F, denoted log(F), consists of all log(o;a) with xa e F.

Monomial Subrings_______________________________ 237

Proposition 7.3.12 Let R = k[x\ , . . . , xn] be a polynomial ring over a field k and I C R a monomial ideal. Then the integral closure of I is given by

7=(xa a € conv(log(/)) n Z"). Proof. Set

J — (xa\ a £ conv(log/)). Let a e conv(logJ). One can write

where m > 0, £?=1 m = 1 and xf £ /. Note that Atf = (!,«)*, where A is the (n + 1) x q matrix whose columns are (1, /?;)*, i = l,...,q, hence by Farkas's Lemma one may assume Hi 6 Q. By choosing 771 6 N+ such that

mfj,i G N for all i one obtains xma 6 Im and x" e /. Conversely let us show I C J. Since / is generated by monomials for any xa e / there is m 6 N+ such that x ma e Im. It follows readily that a 6 conv(logl). D Proposition 7.3.13 Let F be a finite set of monomials of degree k in a polynomial ring R over a field K. If I = (F) is complete, then

(T)

conv(log(F)) n Zn = log(F).

Proof. Set F = {xai ,...,xa«}. Let /? e conv(logF) n Z", then

where ^ > 0 and $3?=i Mi = 1- Since |/?| = fc and x'3 is in / = / (see Proposition 7.3.12), then x13 — xai for some i and /? = c^, D Condition (P) says that log(F) is the set of all lattice points in the convex polytope spanned by itself. We may interpret it as asserting that

the integral closedness of / is not violated in the lowest possible degree. If F consists of square-free monomials, then P is certainly satisfied. Symbolic Rees algebras of monomial ideals

For a monomial ideal /

generated by square-free monomials it is possible to compare the integral closure of the powers of / with the symbolic powers of /, to make the comparison we begin with the following description of the symbolic powers.

Proposition 7.3.14 Let R be a polynomial ring over a field K and let I be an ideal of R generated by square-free monomials. Then

for all n > 1, where p 1; . . . , pr are the minimal primes of I.

238______________________________________ Chapter 7

Proof. From Proposition 5.1.8 one has pf = pj . Since / is a radical ideal the result follows from Proposition 3.3.24. D Corollary 7.3.15 Let R be a polynomial ring over a field K . If I is an ideal of R generated by square-free monomials, then /("' is integrally closed for n > 1.

Proof. Set J = /( n ). Let pi, . . . , pr be the associated primes of /. If / G J, then there is TO > 1 so that fm G p™ m for all i. By Corollary 3.3.20 pf is complete, thus / G p™ for all i. Hence Proposition 7.3.14 shows / G J. n Corollary 7.3.16 Let R be a polynomial ring over a field K and let I be a monomial ideal. If I is a radical ideal, then In C /'"' for n > 1. Proof._Let x £ 7", there is k > 1 so that xk G (/")* C (/( n ) ) f c , thus x is in G /("). Note that /(") is integrally closed by Corollary 7.3.15, hence

zG/(n).

n

Let us present the monomial version of Theorem 3.3.31, which can be proved directly in the monomial case. Proposition 7.3.17 Let R be a polynomial ring with coefficients in afield. If I is a radical monomial ideal of R which is normally torsion free, then its Rees algebra R[It] is a normal domain.

Proof. By Proposition 3.3.26 In = 7 (n) for n > 1. Hence thanks to Corollary 7.3.16 I™ is integrally closed for n > 1, and consequently R[It] is normal. D Proposition 7.3.18 ([214]) Let R = k[xi,...,xn] be a polynomial ring over a field k and I an ideal generated by square-free monomials. Then the

symbolic Rees algebra + ••• + /(">£" + • • • C R[t] is a finitely generated R-algebra. Proof. We follow Lyubeznik's original argument. Let p i , . . . , p s be the minimal primes of /. Given a monomial y = x\l • • • XT^ , set dega.j (y) — r, and degpi (y) = X^-<E P i deg (y). To every monomial y in j( m ) we associate the vector:

w(y) = (dega.1 (y), . . . , deg^ (y), deg pl (y) - m, . . . , deg pj (y) - TO). By Proposition 7.3.14 a monomial y is in /( m ) if and only if deg p .(j/) > TO for all i. Hence w(y) 6 N"+s for all y g /( m ).

Monomial Subrings_______________________________ 239

Note that the set Wi+s is partially ordered by: (a;) > (6j) if and only if a, > bi for all i. Denote by A the set of all w(y) such that y is a monomial which is part of a minimal generating set of I^ for some m > 1. By Dickson's lemma A has only a finite number of minimal elements, and for each c in A there is a minimal element d of A with c = b + d for some b e Nn+s . Let yi, . . . ,yp be the monomials in R such that u>(j/i), . . . , w(yp) are the minimal elements of A. Thus yi is part of a minimal generating set of /( m ; ), for some TOJ > 1.

We claim U S ( I ) = R[yitm\. . .,yptm"}. Let ytm 6 / (m ^ m , where y is a

monomial in I^m\ note that one can write w(y) = b + w(yj) for some j and some 6 G p*j«-+*. One has:

(i) y = xayj, and (ii)

de

SPi (y)-m> deg p . %) - m.,- for all i.

Combining (i) and (ii) yields: p.

(y) - deg p . (?/_,•) > m - TO,,- for all i.

Hence zT1-"1'' e /(m-m^m-m^ and since ytm = (xatm-m')(yjtm^ follows by induction that y is in the subring R[yitmi , . . . , yptmp}.

it O

Exercises 7.3.19 Let R = k[xi,...,xn] be a polynomial ring over a field k and m its irrelevant maximal ideal, then md is the integral closure of the ideal Td •"'I i •

7.3.20 Prove In J ^ In J, where / = (x5,y3) and J= ( y 5 , z 2 ) .

7.3.21 Let R = k[xi,... ,xm,yi, • • • ,yn] and L = I\JT + ITJ\, where Jt (resp. I s ) denote the ideal of R generated by the square-free monomials of degree t (resp. s) in the j/j (resp. Xi) variables. If r > 3, prove that L3 is not integrally closed.

Hint Set / = x\ • • • x2ry\ • • • y% and prove / e L3 \ L3.

7.4

Normality of some Rees algebras

There are several interesting examples of normal ideals generated by squarefree monomials, here we present some of them. Since the normality of an ideal /, in a polynomial ring R over a field k,

is equivalent to the normality of the Rees algebra of / (see Theorem 3.3.18), it is often the case that is much easier to verify the normality of the ideal than the normality of its Rees algebra.

240______________________________________Chapter

7

An advantage of proving normality of the Rees algebra of / occurs when the ideal is generated in a single degree p, in this case one automatically obtains the normality of the subring k[Ip], by the next general result on descent of normality. See Example 8.7.14 for a counter-example if / is

generated in different degrees. Proposition 7.4.1 Let R be a polynomial ring over a field k and I an of R generated by homogeneous polynomials f i , . . . , fq. Assume deg(/j) = d for all i. If Tl(I) is normal, then k[fi, . . . , fq] is also normal.

Proof. Let m = R+ be the irrelevant maximal ideal of the polynomial ring R and A = k[Tfi, . . . ,Tfq]. Observe that there is a decomposition of A- modules:

= R[Tf1} . . . , T f g ] = fcpVi, . . . ,Tfq] 0 As A ~ k[fi , . . . ,fq] (as rings) the result follows from Proposition 3.3.33. n Proposition 7.4.2 Let R = k[xi,...,xn] and R[x] be polynomial rings over afield k. If I is a normal ideal of R generated by square-free monomials of the same degree t >2, then the ideal J = I + (xxi, . . . , xxn) is normal. Proof. Set V = {xi, . . . ,xn} and J = I+(xxi, . . . ,xxn). Here to make the notation simpler we denote the integral closure of Jp by Jp. By induction

on p we will show that Jp = Jp for all p > 1. If p = 1 then J, being an ideal generated by square-free monomials, is complete (see Corollary 5.1.5). Assume Jla = J"1 for i < p and p > 2. Using Proposition 7.3.3 we have

Ja — ({y\ 2/ is a monomial in R[x] and ym £ Jmp for some m > 1}). Let y be a monomial in J%, then ym € Jmp, m > 0. Let us show y G Jp. Since Jp C Jp~l = Jp~l we can write y = xiM(xwi)---(xws}fi

••• fp-s-i,

for some 0 < s < p — 1, where M is a monomial with x ^ supp(M), the fi's are degree t monomials in J with x £ supp(/^) for all i, and Wi € V for 1 < i < s. Likewise we can write

ym = N(xzi) • • • (xzr)gi • • • gmp-.r, for some 0 < r < mp, where TV is a monomial, the ^'s are monomials of

degree i in J with x £ supp(^) for all i, and Zi £ V for all 1 < i < r. To clarify notation set XWQ = XZQ = 1 and /o = go = 1- Altogether we obtain:

=

Nxrzi • • • zrgi • • -gmp-r,

(7.2)

Monomial Subrings_______________________________ 241 making x = 1 and taking degrees one readily derives the inequality m(ts + t-s)
(7.3)

where r < min{mp, ml + ms}. (a) If deg(M) — 0, the inequality above gives I > 2 since r < m(i + s), and s < p — 2 since r < mp. Therefore xl j\ e J2 and y € Jp . (b) Assume deg(M) > 0, notice that t — 0 and r < ms, otherwise y G Jp . Let j/i be the result of evaluating y at x = 1. From Eq.(7.2) we derive yf € Imv-r c jm( P - s ) ) and yi 6 JP-S = / P - S _ Using Eq.(7.2) and Eq.(7.3) we obtain

as y = xs yi by a degree argument it follows that y £ Jp.

d

The next example shows that Proposition 7.4.2 does not extend to monomial ideals with minimal generators in different degrees.

Example 7.4.3 Let R = k [ x i , . . . , xg] be a polynomial ring over a field k of characteristic zero and let /I = XiX2X3X4,

/2 = X5XeX7X$,

/3 = X

/4 = X3X4X6XS,

/5 = XiX3Xr,

/8 — X2X7XS.

If / is the ideal of R generated by /i , . . . , /e, then Ti(I) is a normal domain, whereas the ideal J = I + (xxi, . . . ,xxs) is not a normal ideal of the polynomial ring R[x] .

Let us briefly explain how to verify the assertions in Example 7.4.3. Let P be the toric ideal of K(I) with respect to /i , . . . , /e and let L be its Jacobian ideal, that is, L is generated by the g x g minors of the Jacobian matrix of P, where g is the height of P (in our case g = 5). Using the computer algebra system Macaulay [15] we readily obtain the equality

pdB(B/P) = height (P), where B = R[Ti,. . . ,T6}. Therefore K(I) is Cohen-Macaulay by Corollary 2.5.14. Again making use of Macaulay we derive that (L, P) is an ideal of height 7. Therefore the normality of Ti(I) follows from the Serre's normality criterion together with the Jacobian criterion for regularity in polynomial rings; see Example 3.5.11 for details. On the other hand observe that J3 is not complete because x/i/2 ^ J3 and (x/i/ 2 ) 2 = (xa; 5 )(xx 6 )/3/4/ 5 / 6 e J6. Proposition 7.4.4 Let R = k[xi,...,xn] and R[x] be polynomial rings over a field k. If I is an ideal of R generated by square-free monomials and I is normal, then J = (7,xxi) is a normal ideal of R[x}.

242

Chapter 7

Proof. By induction on p we will show Jp = Jp for all p > 1. If p = 1,

then Jp is integrally closed by Corollary 5.1.5. Assume J* = Jl for i < p and p > 2. Let y be a monomial in JP, then y m e J pm , m > 0. Since JP C JP'1 = J^""1 we can write y =

where M is a monomial with x ^ supp(M) and the f i ' s are monomials in J with x ^ supp(/i) for all i. By Proposition 7.3.2 it suffices to show y £ Jp. Since ym e Jpm we have

ym = xtm(XXlymzm = N(XXl)sgi • • • g m p - s ,

(7-4)

where N is a monomial, x ^ supp( 0, we may assume zi ^ supp(M), otherwise y 6 Jp. We may also assume x\ ^ supp(/j) for all i, otherwise it is not hard to see that we are back in case (a). Notice that s < rm, because xi $ supp(z). Since x\m divides ym it follows that z e Ip"r = Ip~r and y = xl(xx^)rz & Jp. D Proposition 7.4.5 Let R = k\x\,..., xn] be a polynomial ring over a field k. If It is the ideal of R generated by the square-free monomials of degree

t, then It is a normal ideal. Proof. We will proceed by induction on t. Assume that /j is a normal ideal for i < t and let I = It. We set m = (x\,... ,xn). It is enough to show (by induction on p > 0) that Ip is integrally closed for all p. Assume p > 0 and P is integrally closed for i < p. In our case the integral closure of Ip is a monomial ideal given by Ip = (y£ R\ yj e Ijp for some j),

see Proposition 7.3.2. Assume M = IP/IP is non-zero. We claim that M has finite length. If p G Ass# M\{m}, say xi £ p, then Jp = (/t_i) p , where It-i is the ideal of R generated by the square-free monomials of degree t - 1 in x % , . . . ,xn. By the first induction hypothesis we get Mp = (0), which is impossible. Hence ASS.R M = {m}. Let y be a monomial in Ip \ Ip which is conducted by m into P'. Note that, by the above description of Ip, any monomial in Ip has degree at least tp. We have y £ Ip C Ip~l = Ip~l, so we can write y = zi • • • Zp-iu, where z\,..., zp-i are monomials of degree t in / and u is a monomial with deg(u) > t. Notice that |supp(u)| < t — 1, otherwise y G Ip. We can write,

Monomial Subrings_______________________________243

after permutation of variables, u = x\l • • -xb^, bt > I all i. If x\ £ supp(^) for some i, then there is Xjl € supp(zj) \ supp(u), and we can re-write

y = zi • • • Zi-iz'iZi+i •••zp-1xbl1~lxb22 •••xb^xjl, where z( = (zifxj^)x\. We have three cases to consider. (i) The number of Zj's not containing x\ is greater or equal than &i, then we can re-write y=

Wl

- - - wp-lXbJ • • • xb^xh • • • xjbi,

where Wi are monomials in / \ ml, the variables X2, • • •,x\, X j 1 , . . . , Xjb distinct, and A + &i
y = wi •••wp-ixbli"qx\2 •••xbxxxjl •••Xjq, where xi, x 2 , • • • , x\, X j 1 , . . . , Xjq are distinct, w^ are monomials in / \ ml and x\ in supp(wj) for all i. (iii) All the z^s contain xi, which reduces to case (ii). Applying the same arguments to x%,..., x\ yields that we can write

y = yi---yP-ixail ••• 1 for all i, xtl,..., Xj r , X j l , . . . , Xjs distinct and

suppO/j). t=l Note that r + s < t — 1, otherwise y G -fp- We set Xj3 = 1 if s = 0. We substitute x^ = • • • = Xir = 1 in x^y and denote by g the monomial obtained from x^y after evaluation. To derive a contradiction observe that

while on the other hand, since x^y G Ip we have deg(^) > p(t — r). Hence r + s > t, which is impossible. D Some of the arguments given in Proposition 7.4.5 may apply for other groups of square-free monomials whose elements are of degree t or less. Next we show a result in this direction. Theorem 7.4.6 Let R = k[xi,.. .xn] and R[x] be polynomial rings over a field k and let Lt - It +x/i_ 2 , where It is the ideal of R generated by the square-free monomials of degree t > 2. Then Lt is a normal ideal of R[x].

244______________________________________Chapter

7

Proof. We use induction on t. If t = 2, then Proposition 7.4.5 yields that L 2 = /2 + xR is normal. Assume that Li is a normal ideal for i < t and set m = ( x i , . . . ,xn, x), L = Lt, I = It-2, and J — It- By induction on p > I we now show that Lp is integrally closed for all p. Recall that L is an intersection of prime ideals because L is generated by square-free monomials (see Corollary 5.1.5), hence L = La. Assume that L1 is integrally closed

for i < p and p > 2. Let us show Lp = Lp. If on the contrary Lp ^ Lp, we set (0) 7^ M = LP/LP. Observe that for any p G AssA M \ {m} one has Lp = ( L t - i ) f if Xi ^ p for some i and L p = (/j_2)p if x ^ p, therefore by induction hypothesis and Proposition?.4.5 it follows that the only associated

prime of M is m. Hence there is a monomial y G Lpa \ Lp so that my C Lp, since Lp C L^1 = Lp~1 we can write y = u(xhi) • • • ( x h s ) f i • • • fp-s-.i,

where the /ij's are monomials in / of degree t - 2 and the //s are monomials in J of degree t. Observe that if a; G supp(u), then we may re-write

y=

(XiX'U\ 3

} (xhi) • • • (xhs)

(

x

( j\ \\

——

/2 • • • fp-s-i,

for some Xi,Xj € supp(/i) and i ^ j, a repeated use of this rearrangement of the factors of y yields that either y = u(xhi) • • • (x/i s )/i • • • fp-s-i with x $. supp(u), or y = u(xh\) • • • (xhp-i) with x G supp(u). In the second case using that ym G Lmp for some m > 1, together with Proposition 7.4.5, rapidly gives y G Lp. As a consequence from the start we may assume x £ supp(it); notice that deg(w) > t. Let u = x^1 • • • x1^, 1 < A < t — 1, ai > 2. There are three cases to consider: (a) Assume |supp(u)| = t — 1. In this case p—s — l Xi G

P)

SUpp(/j),

1=1

otherwise y G Lp. We may assume xi £ supp(hi) for some I, otherwise by induction on t we obtain y G Lp. We claim that supp(ftj) C supp(u) C supp(/j), for all i,j. If the first containment is false, then pick xr G supp(/ij) \ supp(u). If x\ is

not in supp(/ij), then we re-write ILXr

\

(

-,

\

(

7

\ I ^^±'"1

\

/

,

\

/ 7 \ J * ^ 7 - 2 ?

—— I (xhi) • • • ( x f i i - i ) I ——— I (x/ij+i) • • • (xhs)j G -i/, xi / \ xr J where / = /!•• • fp-s-i, which contradicts the choice of y. On the other hand if x\ G supp(/ij), then supp(hi)
Monomial Subrings_______________________________ 245

z 6 supp(h() and z $ supp(/ij). Therefore V =

uxr

——

(xhi) • • • (xht-i)

\

z

• • • ( x h i - i ) ( ——- ) • • • ( x h 3 ) f , \ xr j where / = /i • • • fp-a-i, hence y 6 Lp, and again this is a contradiction. To show the second containment note that we can write he = x% • • • xt-\. If

supp(u) <£_ supp(/j), pick XT in supp(u) \ supp(/j) and re-write y = u(xhi) • • • • • • (xhs)f

where h'e = (xihe)/xr and /j = ( x r f j ) / x i , since xi is not in supp(/j) and supp(/j) <£_ supp(u) we get y £ Lp, and the proof of the claim is complete. Let g fee the result of evaluating y at xi = • • • = o;<-i = 1, notice that deg(i?) = p — 1, while by evaluating x\y G Lp at the same points we obtain

deg( p, which is a contradiction. Altogether in this case y € Lp. (b) Assume |supp(u)| = t - 2. If

X! e I p| supp(Aj)

p)

p| supp(/i)

i=l

then by induction hypothesis on t we derive y £ Lp, otherwise we reduce to a case similar to (a). (c) If |supp(u)| < t - 3. Since deg(/ij) = t - 2, supp(/ij) £ supp(u) and supp(/j) <£. supp(u) for all i, j. Therefore we may apply the method of proof of Proposition 7.4.5 to conclude that (after at most 0,1 + • • • + a\ rearrangements) we reduce to a case similar to (a), (b), or there is some Xj in the support of u, hi, and fj for all i, j, which implies (by induction on t) that y € Lp. D

Definition 7.4.7 Let / and J be two monomial ideals of the polynomial rings k[xQ, . . . , xm] and k[yo, . . . , yn] respectively. The join of / and J is:

where K = (zji/j 0 < i < m and 0 < j
= k[xo,...,xm] and S = k [ y o , . . . , y n ] be and let I , J be two monomial ideals of R are normal ideals generated by square-free > 1, then their join I * J is normal.

246______________________________________ Chapter 7

Proof. Set X = {x0>. . . ,xm}, Y = {y0, . .. ,yn} and L = I + J + K, where K = ( X ) ( Y ) . By induction on p we will show that Lpa = Lp for all p > 1,

where Lva denotes the integral closure of Lp. If p = I then L is a radical

ideal (see Corollary 5.1.5), hence L is integrally closed. Assume L\ = Ll for i < p and p > 2. Using Proposition 7.3.2, we have Lpa = ({z\ z is a monomial in k[X, Y] and zq e Lqp for some < ? > ! } ) .

Let z be a monomial in L£, then zq e L9P, 0. Let us show z € £p.

Since L^ c I-^"1 = I/*1"1 we can write z = Mhi • • • hsgi • • • grfi • • • / p _ r _ s _i ,

where M is a monomial, the hi's are monomials of degree two in K, the g^s and fi's are degree t monomials in J and I respectively. Likewise we can write where JV is a monomial, deg(ftj) = 2 and h\ is a monomial in K for all i, g[ and f'j are degree i monomials in J and / respectively for all i,j. From the last two equalities we have

. . / i g ) « ( f f i - - - < ? r ) 9 ( / i " - !P-r-s-i)q

(7.5)

From Eq. (7.5) one readily derives the inequality (qs-Sl)(t-2)+qt
(7.6)

We may assume M = xa or M = y a , otherwise z € L p . By symmetry we may assume M = xa, where a > 0. (a) If t > 3, then r = 0 o r p — r — s — 1 = 0, otherwise 51/1 € L3 and hence z 6 Lp.

First we treat the case r = 0. By taking degrees in Eq. (7.5) w.r.t. the variables y0, . . . ,yn one has si + tr\ < qs, which together with Eq. (7.6) yields degM > t. Let z\ be the result of evaluating z at yi = 1. From Eq. (7.5) we derive and z\ G -^o~s = Ip~s • We may write ^ = yezi and 21 = x^w, where deg(j/ e ) = s and w is a monomial in Ip~s of degree i(p — s). Since

deg(^) = deg(M) +s + t(p-s-l) we obtain deg(zi) > s + t(p — s), hence deg(x/3) > s. Altogether we derive z = yex@w € Lp. Next we consider the case r = p — s — 1 > 1, observe that

Monomial Subrings_______________________________ 247

deg(M) < 1, otherwise z 6 Lp. Therefore either z = h\- • • hsgi • • • gr, or we may rewrite

z = y&hi • • -hshs+igi • • • gr-i, where deg(/i s +i) = 2 and /i«+i € K, interchanging the x» and i/j variables

we may apply the arguments above to conclude z 6 Lp. (b) Assume t = 2. Using z ? £ L9P one rapidly obtains deg(M) > 2, hence we may assume r = 0 (otherwise z £ L p ) and the arguments of case (a) can be applied to conclude z G Lp. D

Corollary 7.4.9 ([256]) Let X = {x0,...,xm} and {yo,...,yn} b disjoint sets of indeterminates over a field k. Let I be a normal ideal of k[X] generated by square-free monomials of degree t and let L = I + K , where K = ( X ) ( Y ) . Then L is a normal ideal. Proof. Proceed as in the proof of Theorem 7.4.8 and notice that in this case r = p — s — 1. d

Exercises 7.4.10 Let R = k[xi,...,xn] be a polynomial ring and / = (/i, . . . , / , ) , where k is a field. If /j is homogeneous of degree d > 1 for all i. Prove

(a) n(I) ~ U(I)/mn(I)

® mTl(I) as K(I) modules, where m - R+,

(b) &[/i,. ..,/„]=: K(I)/mK(I)

as fc-algebras,

(c) if F = {/i,...,/ 9 } is a set of monomials, then *R.(I) ~ k[F,tx\ as fe-algebras, where x = {xi ,... ,xn}. Hint For (c) consider ip: k[x,tj] —> H ( I ) , where ^(^i) = xi> ^ ( t j ) — ^/ji

an

d

: fc[x, ij] ->• fc[F,ix], where (j>(xi) = tej, <^(^) = fj. Use Proposition 7.1.2 to show that 0 and t/> have the same kernel. 7.4.11 Let I? = jff[a;i, . . . , s m , j/i, . . . ,yn] and L — IsJt, where Is (resp. Jf) denote the ideal of R generated by the square-free monomials of degree s (resp. t) in the Xi (resp. j/j) variables. Prove that L is normal. 7.4.12 Let R = K{x] be a polynomial ring over a field K. If / is an ideal of R generated by square- free monomials of degree 2, then / and I2 are both integrally closed.

248______________________________________Chapter

7.5

7

Ideals of mixed products

Let R = K[x,y] be a polynomial ring in two disjoint sets of variables x, y over a field K. We will study ideals of mixed products

such that k + r = s + t, where Ik (resp. Jr) denotes the ideal of-R generated by the square-free monomials of degree k (resp. r) in the x (resp. y) variables. The notion of ideal of mixed products was introduced in [241].

The aim is to present a complete classification of the normal ideals of mixed products, the main result (see Theorem 7.5.8 and [241]) generalizes the normality of various previously known normal ideals and contain as a particular case the normality of generalized graph ideals associated to complete bipartite graphs. The dimension of the monomial subring K[F] will be examined, where F denotes the minimal generating set of L consisting of monomials.

Powers of ideals of mixed products Let R = K[XI, . . . , xm,yi,..., yn] a polynomial ring over a field K. Given non negative integers k,r,s,t such that k + r = s + t consider the square-free monomial ideal of mixed products given by L = IkJr + Ia Jt j

where Ik (resp. Jr) is the ideal of R generated by the square-free monomials of degree fc in the variables x\,..., xm (resp. j / i , . . . , yn). Note that L has a dual ideal L1 obtained by an interchange of the two sets of variables:

L1 = ItJs +IrJk C R' = K[xi,...,xn,yi,...,ym]. It is convenient to set /o = JQ = R. By symmetry there are essentially two cases:

(i) L = IkJr + IsJt with 0 < k < s, or (ii) L = IkJr with k > I or r > 1. One of the results that will be used throughout this section is the fact

that the powers of the ideals Ik and Jr are complete, see Proposition 7.4.5. Remark 7.5.1 (a) If A = R[yi~l\ is the Laurent polynomial ring for some

2/i, then

JtA = Ji^A,

Monomial Subrings_______________________________249

where J(_^ is the ideal of R generated by the square-free monomials of degree t — 1 in the variables j / i , . . . , 2 / i - i , yt+i, • • • ,yn- In fact one clearly has Jj C J't-ii thus JtA C J^_1A. On the other hand consider a monomial / in Jt'_1; then j/j/ e Jt and / 6 JtA. Hence J[_1A C JtA (b) If a variable j/j is not in a prime ideal p C R, then the localization of Jj at p is the same as the localization of Jj._1 at p. A similar statement holds if a variable Xj is not in p and one consider /& (resp. I'k_i) instead of Jt (resp. J[_i). We need this observation only when p is a face ideal. For use below recall that products of ideals and quotients of modules

commute with localizations.

We also recall that the integral closure of

ideals commute with localizations, see Proposition 3.3.4. For basic facts on

monomial ideals see Chapter 5. As usual a monomial / = x1^1 • • • x^y^ • • • yb^ will be denoted simply by / = xayP, where a = (ai,... am) and /? = ( & i , . . . , bn). The support of /, denoted by supp(/), is: supp(/) = {xi\ ai > 0} U {yi 6» > 0}. We begin by treating some special cases.

Lemma 7.5.2 Let R = K[XI, ...,xm,yi,...,yn] be a polynomial ring over a field K. If L=Jr + Im,

then L is a normal ideal for any r.

Proof. Let us prove directly that LP = Lp for all p > 1. Take f = xay0 in LP, where xa = x*1 • • • x%?. Thus_/* is in Lpi for some i > 1. If o, = 0 for some j, then y^ 6 jy and y& e J? = Jp- Hence one may assume that ai > 1 is the smallest value of the a,. Note that p < a\ implies xa € /^. Assume p > ai. It follows that y^ G Jl ai'. Using that Jr is normal yields y? £ J^~ a i , which together with xa £ 1% proves / 6 L". D Example 7.5.3 If m = n = 3, then L = J% + 1% is not a normal ideal because / G L3 \ L3, where / is the product of all the variables. Proposition 7.5.4 Let R = K[XI, ... ,xm,yi,... ,yn] be a polynomial ring over a field K, If L = Jr + ImJt

with r = m + t, then L is a normal ideal. Proof. By induction on r. If r = 1 or t = 0 the result is clear because of Lemma 7.5.2. Next we show by induction on p that Lp = Lp for all p > 1;

the case p — 1 is clear because L is a radical ideal. Assume L1 is integrally

250______________________________________ Chapter 7

closed for 1 < i < p and set M = LP /Lp. If M ^ (0), take an associated prime ideal p of M. As M <-» R/LP, the prime ideal p is a face ideal. Note that yt is in p for all i; for otherwise if j/j ^ p, then by Remark 7.5.1 one concludes

where and J'r_l is the ideal generated by the square-free monomials of degree r — 1 in the variables 3/1, . . . , 2/j-i, J/i+i, • • • , 2M- Thus by induction on r, the immediate consequence is that the localization of M at p is zero, that is,

Mp = (Z?/L*)P = (0), a contradiction because p is in the support of the E-module M. Hence t/j is in p for all i. As p is generated by a subset of the variables, there is a monomial / in LP \ Lp such that p = (Lp: /). Since y,/ is in Lp and p > 2, one can pick a variable yi in supp(/) such that deg yi (/) > deg y; (/) for all i. Therefore one can write 2/1 / = gwi • • • w p , (1) where W I , . . . , W P are monomials of degree r in L and g is some monomial of R. Note deg( 0, because / l € Lpl for some i > 1, and y\ £ supp(5), because / ^ Lp. Case (a): First we assume yj divides g for some j, note j 7^ 1 because / ^ I/p. Set c = deg yi (/). Since j/j +1 divides j/i/ one has that y^+1 divides w\ • • • wp , thus one may assume y\ is in the support of Wi for i = l , . . . , c + l . In order to derive a contradiction note that yj SE w^ for alH = 1 , . . . , c + 1 , this follows by observing that if yj <£. Wi, then from the equality

\yjj-

' - '•••^•••1

one derives / € Lp. Therefore y^+l divides /, which contradicts the choice of c. Case (b): Next we assume g = xa. One may order the u>; such that u>( G Im Jt for t < j and Wf € Jr for t > j, where j is some fixed integer between 0 and p. Therefore from Eq. (1) one has ;„(/) = deg^j,!/) -l=jt + (p- j)r -I.

(2)

Monomial Subrings_______________________________251

On the other hand since / l £ Lpl for some positive integer i, one can find monomials z±,... ,zpi in L of degree r such that

/* = hzi • • -zpi,

(3)

for some monomial h in R. We order the z, such that ^ £ ImJt for £ < j) and Zf £ Jr for (. > ji, where ji is some integer between 0 and pi. Using

Eq.(2) and taking degrees in Eq.(3) gives

> jit + r(pi - j i ) -ij)>i.

(4)

Hence j < p and consequently |supp(a; a )| < m, because otherwise y £ Lp. Pick a variable x\ $ supp(z a ) and note that Eq. (1) gives degxi(ft) = ij, which together with Eq. (3) yields j\ < ij, but this contradicts Eq. (4). D Proposition 7.5.5 Let R = K[XI, . . . ,xm,yi, . . . ,yn] and let

L = Jr + Is Jt • Ift>l,r

= s + t and 2 < s < m, then L2 is not integrally closed.

Proof. Since s < m it makes sense to set

In order to prove that / £ L 2 , set yo = 1 and observe that from the equalities:

belongs to /° +1 1l2s . . . ,;2s

Vi

S

,; Vt-iVt

... IIs

Vr

ysi+1 • • • yf-l y f - y r

—

S

l

S

— (11, (Vi . . .11Vr)~\ ~ (ll(yi +

-

1

. . . 1IS+11U • • - I I

Vt-iyt

yr)~\

(yi • • • yt-iyt) • •• (y\ •••yt-iy^ s+i terms in Jt

we obtain / s £ L2s , as required. By counting degrees, and using s > 2, it follows that / £ L2. D Proposition 7.5.6 Let R = K[XI, . . . ,xm,yi, . . . ,yn] and let

L = IkJr + Is Jt • I f s > k + l>2,t>l and k + r = s + t, then L3 is not integrally closed.

252______________________________________ Chapter 7 Proof. Set

/ = x\x\ • --xl^xl • --xlylyl • ••y3t_1y2t • • • y * , where XQ — J/Q = 1. To show that / is in the integral closure of I/ 3 , it suffices to observe that from the equalities. f2 J x

i ' • ' xt-ixk ' • ' xl

_ —

=

6 6. 1X2

X

6 4 _ .. k~lxk

x

4.6 6

(xi---xk 3 3 3 ( T r ---T [J-1^2 •lf,

3 3

y\---yl-iy\---yl = (vl • • • yl}(yt+z • • • yr)(yi • • • yt-\yt)(yi • • • yt-iyt+i)(y\ •••ytit follows that / 2 is in L6. As degx(f) -2s + k-l and deg J/ (/) = 2r + t-l, by a counting degree argument it follows that / is not in L3 . Q Proposition 7.5.7 Let R — K[x\, . . . ,xm,yi, . . . , yn] and let

L = IS + JS. If 2 < s < inf{rn,n}, then Ls+l is not integrally closed.

Proof. Let / = (x[~l • • • xss^ll)(yi • • • y s + i ) . To show that / lies in the integral closure of Ls+l note that from the equality:

r = ((x1---Xs+1yy-l(y1---ys+1y is in / ^ -

is in J1

it follows that fs is in I/( s+1 ) s . As s > 2, by a counting degree argument, it follows that / is not in Ls+l . D

Theorem 7.5.8 ([241]) Let R = K[XI, . . . ,xm,yi, . . . , y n ] be a polynomial ring over a field K . If is an ideal of mixed products with k + r = s + t, then L is normal if and only if L can be written (up to permutation of k, s and r, t) in one of the following forms:

(a) L = IkJr+ Ik+iJr-i, k > 0 and r > 1. (b) L = IkJr, k > 1 orr > 1. (c) L =

Monomial Subrings_______________________________253 Proof. =>) If k — s, then r = t and we are in case (b). By symmetry one may assume 0 < k < s < m. Taking into account the previous results one can readily conclude the following classification. If k = t — 0, then

{

Ji + /i, s = 1, J s + / s , s = inf{m,n}, Js + Is, 2 < s < inf{m,n},

case (c). case (c). not normal.

If k = 0 and t > 1, then

{

Jr + IiJt, Jr + ImJt, Jr + IsJt,

s = 1, s = m, 2 < s < m,

case (a). case (c). not normal.

If k > 1 and t = 0, then

{

hJi + 4+i, r = 1, IkJn + I s , r = n, IkJr + Is, 2 < r < n,

case (a). case (c). not normal.

If k > 1 and t > I , then L_

r

Ik+iJr-i,

|_ Ik Jr + Is Jt,

s = k + l,

case (a).

s > k + 2,

not normal.

Altogether one obtains that L has the required form.

•<=) (a): By induction on k + r we show that L is normal. If k + r = 1, then L = Ji +/i and L is normal. Assume k + r > 2. Next we use induction on p to prove that LP = Lp for all p > 1; the case p = 1 follows because L is a radical ideal. Assume L1 is integrally closed for 1 < i < p and set M = TP/LP. If M ^ (0), take an associated prime ideal p of M. Since M <-> R/LP, the prime ideal p is generated by a subset of the variables. Note that by Remark 7.5.1 one has Lp = (Li)p, under the following conditions

L

_ 1

IkJ'r-i + h+iJr-2,

if

Vi i P and r > 2,

I'k-iJr + I'kJr-i, J 'r-i + tiJ'r-i,

if if

%i i P and k > 1, Vi i P and k = 0,

JT-\I

if

x

4-i^i + 4 >

if

x

4,

if

i $ P and ^ — 0)

« ^ P and

r

- !> 2/j ^ p and r = 1.

Here J^_ : and J^j have the same meaning as in Remark 7.5.1. Thus if p is different from the maximal ideal m = R+ , using the induction hypothesis and the fact that Ik and J r _i are normal, one concludes that Mp = (0), a contradiction because p is in the support of M. Hence m is the only associated prime ideal of M and there is a monomial / in LP \ Lp such that

254______________________________________ Chapter 7

m = (Lp: /). Observe that the support of / contains one of the j/i variables; because if / = xa, then fl £ Lpt for some i > 1 but this can only take place if r = 1 and fl € (/*+i )p', as Ik+i is normal one has / £ If,+1 C Lp which is a contradiction. Pick a variable yi in supp(/) such that deg yi (/) > degy. (/) for all i, then one can write yif = g w i - ' - W p , (1) where w\ , . . . , wp are monomials of degree k+r in L and g is some monomial of R. Note deg( 0 because fl £ Lpi for some i > 1, and 1/1 is not in supp(g) because / ^ L p . Case (I): First assume yj divides g for some j fixed, note j ^ 1 because / ^ L p . Set c = deg yi (/). Since j/j +1 divides j/j/ one has that y^+1 divides wi • • • wp, thus one may assume y\ is in the support of Wi for i = 1, . . . , c+ 1. Note that yj is in supp(wj) for alH — 1, . . . , c+ 1; because if yj £ then from the equality yif

=

Wl

• • • (wiy-j/yi) • • • wc+1 • • • w p ( y i g / y j )

one derives / £ Lp. Hence y^+l divides /, which contradicts the choice of 2/iCase (II): Next assume g = xa and Xj divides g , for some fixed j. We divide this case into two subcases. Subcase (A): First assume that there is a term we of the form we = (x^ • • • x i k ) ( y i y J 2 • • • y j r ) ,

with yi £ supp(u^),

observe that Xj is in supp(w^), otherwise write

w1---we-i(xil • • • x i k x j } ( y h • • • y to obtain / £ Lp '. We claim that Xj ^ n^ =1 supp(wi), because if Xj is in the support of w^ for all i, then x?+ divides Xjf, but since Xjf £ Lp one readily derives that / is also in Lp, a contradiction. Then there is wq such that

(i) (ii)

(xSl---xSk+l)(ytl---ytr_1), (xai---xSk)(ytl---ytr),

Xj $ supp(w ? ), or Xj$

If (i) occurs {xsi,...,xsk+l} <£ {xh, . . . ,xik}, say xsi <£ {x^ , . . . ,xik}. From

one derives / £ LP'. If (ii) occurs {2/ tl , . . . , j/ tr } ^ {j/j-2, . . .,yjr}, say y tl is not in { 2 / j 2 , - . - , 2 / > } - From J ^ = (y1g/xj)(ytlwf/yi)(xjwg/ytl

Monomial Subrings_______________________________255

one derives / G Lp, again a contradiction. Subcase (B): Now assume all the terms u>i that contain j/i in their support (and there is at least one of such terms) are of the form wi = (xil---xik+l)(ylyh---yjr_l),

yi G supp(t^).

(*)

Consider the following two possibilities for the terms wi,... ,wq Wq

_ / (i) ( x s i - - - x s h ) ( y t l - - - y t r ) , \ (ii) (xsi •••xsk+l)(ytl • • • 2 / t r _ 1 ) ,

for some 1 < q < p, or for all 1 < q < p.

If (i) occurs and we is any term as in Eq. (*), pick

3/ti e {j/tl , . . . , ytr } \ {yj2 ,..., yj r -i } and Xil G {x{l , . . . , £j fc+1 j \ \xsi , . • • , xSk Note Xj is in supp(u> 3 ) because otherwise from

one derives f £ Lp. Assume Xj $ supp(u^) for some wi (that contains in its support) and note that from

JJ one derives f £ Lp. Hence one may now assume Xj G supp(w^) for all Wi containing yi in its support. From subcase (A) it follows that Xj £ n? =1 supp(wi). Hence Xj is not in the support of wg for some 1 < 0 < p. If wg has the form wg = x^ • • • xxkym • • • y^r , one may assume y Ml ^ { % ! » • • • »Wr-i} and use

Vif = (yi9/ to derives / e Lp. If ^^ has the form 1^9 = x^ • • •x\k_t_1yl_ll • • •yl_lr_1, choose a^Ai e {^AI , • • • , a;A fc+1 } \ {xsi ,...,xsk}, then from

= (yi9/Xj)(xxlwq/ytl)(xjWe/xXl')(ytlwi/yi)

JJ

Wi

'

one derives / G Z/ p . Thus the argument for case (i) is complete. Assume (ii), that is, all the w^ have the form w^ = xayb, where the degree of xa (resp. yb) is equal to k + 1 (resp. r — 1). Using Eq.(l) and the fact that g = xa one concludes deg y (/) = p(r — 1) — 1. On the other

256______________________________________Chapter

7

hand, since /•?' 6 L JP for some j > 1, we get j degy(f) > (r - l ) p j . Hence deg y (/) > (r — l)p which is impossible. This completes the proof of the normality in case (a). To finish the proof of the theorem note that the normality of case (b) follows readily using that /& and Jr are normal, and the normality of case (c) follows from Proposition 7.5.4. Q

Definition 7.5.9 Let G be a graph with vertices x\,... ,xn, the . graph ideal, denoted by IQ(G), is the ideal of K[x\,... xn] generated by the square-free monomials x^ • • -x^ such that the Xii is adjacent to Xij+l for all 1 < j < q - 1. Corollary 7.5.10 // G is a complete bipartite graph, then the generalized graph ideal Iq(G) is normal for all q > 2.

Proof. Let xi,... ,xm,yi,...,yn be the vertex set of G, one may assume that the edges of G are precisely the pairs of the form { x i , y j } . Therefore Jt+i + It+iJt and the result follows from Theorem 7.5.8.

Q

Segre products and a dimension formula Let us begin with two results that are useful. The first is the following dimension formula, and the second result states that certain monomial subrings of mixed products are isomorphic to Segre products.

Proposition 7.5.11 Let B and C be two standard algebras over a field K, then their Segre product:

B ®s G = (B0 ®K Co) ® (Bi ®K Ci) © • • • C B ®K C, has Krull dimension equal to dim(B) + dim(C') — 1. Proof. See Exercise 4.1.25.

D

Definition 7.5.12 Let R = K[XI, ... ,xn] be a polynomial ring over a field

K. The set of k-products in the Xi variables, denoted by V/,, is defined as: T/7

—— J T ' *k — \^"i\

1 "^ t 1
• • • T •

where k is a positive integer.

• • •

<"' 1 J *^ f~) > ^^k _: '''/)

Monomial Subrings_______________________________ 257

Proposition 7.5.13 Let Vk C K[x] and VJ, C K[y] be the sets of k and r products respectively in disjoint sets of variables. Then the Segre product of K[Vk] C K[x] and K[V^.} C K[y] is isomorphic to

Proof. We grade K"[T4], K[VJ.] and A with the normalized grading, thus

their generators have normalized degree 1. It is no hard to see that there is a well defined graded epimorphism

such that tp(fg)

= f <S> g for all / £ Vk and g £ V/, where

K[Vk] ®5 K[VA = (K[Vk}Q ®K K [yr']0) Since

= dimK(K[Vk}i ® for all i one obtains that the map ip is an isomorphism.

D

Lemma 7.5.14 Let A be the matrix with rows

e^ + • • • + eir, -ei + e 2 , -e2 + e3, . . . , -en-i + en, where ei is the ith unit vector in ffi™ . Then det(A) = r for n > 2. Proof. Expanding by the first column of A and using induction on the size n of the matrix, the result follows rapidly. D Proposition 7.5.15 // L is the ideal of mixed products in m and n vari-

ables and F the minimal generating set of L consisting of square-free monomials, then the Krull dimension d of K[F] is given by

+ n-I, i f L = IkJr , _ i m + n, if L = IkJr + IsJt

1 < k < m and 1 < r < n, 1 < k < s <m, 1 < t < r < n, and k + r = s + t.

Proof. The first formula follows at once using Proposition 7.5.11 and Proposition 7.5.13. To prove the second formula consider the linear subspace

W of Em+ra generated by the set of exponent vectors of the monomials that minimally generate L. Note that the exponent of Xi is e^ and the exponent of yi is em+i, where e^ is the ith unit vector. Recall that

=dimK(W), that is, the proof reduces to showing dim^- W = m + n.

258______________________________________Chapter

7

Give two integers 1 < i < j < m, pick integers j i , . . . , j f c _ i between 1 and m such that i,j,ji, • • • ,jk-i are all distinct, this choice is possible because 1 < k < m. Hence subtracting the vectors e

i + eji + • • • + Zjk-i + e m +i + • • • + em+r

Gj T 6jj ~r ' ' ' ~r ^jk—i + 6 m _j_i + • • • T- &m+r

we get Ei — BJ in W for 1 < i < j < m. A symmetric argument using 1 < t < n shows that &i — ej is in W for m + 1 < i < j < m + n. Similarly subtracting

• • + eia + ejl + • • • + 6jt • • + eik + Cj1 + • • • + ejr

(1 < it < m) (1 < it < m)

(m < jt <m + n) (m < ji < m + n)

yield that &ik+l + • • • + Cj 0 — £jt+1 — • • • — Cjr are in W for all possible choices,

or equivalently the vectors e^ + • • • + &ia_k — Cj1 — • • • — ejr_t are in W for all possible choices. By adding s-k

r-t

1=2

1=1

to any of these vectors we conclude (s — fe)ej 1 — (r — t)ej1 € W, but by assumption s — k = r — t, thus e^ — BJI are in W for 1 < i\ < m and TO < j\ < m + n.

Altogether a - e.j are in W for all 1 < i < j < m + n. To finish the argument apply Lemma 7.5.14 to the matrix having first row

to conclude dim^-(W / ) = n + m.

D

Remark 7.5.16 There is a notion of sequentially Koszul algebra that has been recently studied [7] and that is likely to hold for some members of the family of monomial subrings associated to mixed products. See also [152] for the related notion of strongly Koszul algebra.

Exercises 7.5.17 Let Vk C K[x] and Vr' C K[y] be the sets of k and r products respectively, in disjoint sets of variables, and

If -ftTfVfc], ^[Vr] and A have the normalized grading, then for all d one has the equality

dimK(Ad) = dimK(K[Vk]d ®

Monomial Subrings_______________________________259 7.5.18 Let L = IkJr +IsJt be an ideal of mixed products with k + r = s + t and F the minimal generating set of L consisting of monomials. Prove that dim K[F] = m + n in the following three "extreme" cases:

(a)

(6) (c)

0
0=k< s< m

and

and and

1 < t < r < n, 0 < t < r < n,

0 = t < r < n.

7.5.19 Let L = IiJ^ + I^Ji be an ideal of mixed products in the polynomial ring K[XI, . . . , £ 4 , 3 / 1 , . . . ,7/4]. Prove that the integral closure of I/3 is not generated by monomials of degree 15.

7.6

Degree bounds for some integral closures

Let S be an affine domain and 5 its integral closure or normalization. When S is graded, it is rarely possible to have estimates of the degrees of the generators of S unless there are very circumscribed situations. Here we discuss one of these when 5 is a monomial subring permitting a sharp regrading. In this section we use some techniques introduced by W. Bruns, J. Gubeladze and N. Trung [41]. Theorem 7.6.1 ([51]) Let R — K[xi,...,xn] be a polynomial ring over an infinite field K and F a finite set of monomials in R of the same degree

k. Let I be the ideal of R generated by F and oo

n = 0 rt c R[t] j=0

its Rees algebras. Then the normalization Ti is generated as an Tl-module, and thus as an Tl-algebra, by elements g £ R[t] of t-degree at most n.

Proof. Let F = {/i,...,/,} and U = K [ X I , ..., xn, fit,..., /,*]. We introduce a new grading S on R[t] by setting

6(xi) = I and S(t) = -(k-l). Note that Tl is generated as a /sT-algebra by elements of degree 1 because S(fit) = 8(fi) + 6(t) = 1. Consider the subsemigroup

C = Nlogfo) + • • • + Nlog(x n ) + Nlog(/it) + • • • + Nlog(/,*) of N"+1 generated by ei,...,e n ,log(/ 1 t),...,log(/^). Since ZC = Zn+1, according to Theorem 7.2.28 the integral closure of 7?. can be expressed as:

n = K[{xatb\(a,b) ez™ +1 nE+c 1 }],

260______________________________________ Chapter 7

where E+ C is the cone spanned by C in R"+1 . If M = xatb with (a, b) ^ 0 and (a, b) _e Z"+1 n K+C1, then it is not hard to show that 6(M) > b. Therefore 'R, is generated as a K-algebra by monomials of positive degree. There is a Noether normalization of Tl

where z\ , . . . , zn+i € Tl\ . Note that

is a Noether normalization ofJR,. By Theorem 7.2.35 K is Cohen-Macaulay. Hence by Proposition 2.2.14 'R, is a free module over A and one may write

where Mj = x ft i6i . Using that the length is additive one has the following expression for the Hilbert series

where hi = \{j \ 6(Mj) = i}\. Note that a(7?.), the a-invariant of H, is equal to s - (n + I). On the other hand by Proposition 4.2.3 one has

a(K) = -mm{i (u^i ^ 0}, where ui^ is the canonical module of 72.. An application of Theorem 7.2.34 yields a(jl) = s - (n + 1) < 0 and s < n. Altogether one has 5 (Mi) < n. An straightforward computation shows that if Mj = x/3itbi, then the i-degree of M, is less or equal than n, that is, bi < n. D

For a more general version of Theorem 7.6.1 see [51]. We give a variant of the previous theorem for the subalgebra generated by monomials of the same degree k. A homogeneous polynomial of degree ik is said to have normalized degree i. Theorem 7.6.2 ([51]) Let F be a finite set of monomials of degree k and

the monomial subring generated by F. Then the normalization S of S is generated as an S-module, and thus as an S-algebra, by elements g G R of normalized degree at most dim(S') — 1.

Monomial Subrings

261

Proof. Let i-Q

be the kth Veronese subalgebra of the polynomial ring R. Both 5 and S are contained in R^ . We grade 5* by setting Si = S fl Rik • Then we may

adapt the arguments of the previous proof.

7.7

D

Degree bounds in the square-free case

Let R = K[XI, ..., xn] be a polynomial ring over a field K and F a finite set of square-free monomials of the same degree k. One of the goals is to present bounds for the generators of the normalization of the monomial

subring K[F]. In this case there are embeddings

K[F] CSCRW, where

S = KKxh • • • xik I
take place inside of one of these algebras. Our exposition will follow closely that of [51]. Especially we want to present an explicit generating set for the canonical module u>$ of S that appears in [51]. In particular we will fully describe the algebras 5 with the Gorenstein property and compute the a-invariant of 5. The analysis below will determine the Cohen-Macaulay type of 5 but also leads to the control of degrees in K[F]. A description of the canonical module

Let us fix some of the notation that will be used throughout the rest of the section. Let n > 2k > 4 be two integers (this is not an essential restriction; see Remark 7.7.12). We set •A ~ {e^ + • • • + eik 1 < ii < • • • < ik < n},

where e.\,..., en are the canonical vectors in ffi". The affine subsemigroup of Nn generated by A will be denoted by C, that is, we have

C = T Na.

262______________________________________Chapter

7

The cone generated by C will be denoted by M+ C, thus Of € 1+, at e ,4, r e N I .

If 0 7^ a G E™ , then the set Ha will denote the hyperplane of R" through the origin with normal vector a. Thus Ha = {x £ R™| (x,a) = 0}, and Ha determines two closed half-spaces

H+ = {x e 1" | (x, a) > 0} and #~ = {a; 6 R™ | (a;, a) < 0}.

Let R — K[XI , . . . , xn] be a polynomial ring over a field K. Observe that the -fT-subring of R spanned by the set {xa\ a e .4} is equal to the affine semigroup ring K[C]. Remark 7.7.1 Let (p be the matrix whose columns are the vectors of A. By Proposition 7.1.17, we obtain

rank(y>) = dimfC[C < ] = n. Hence the vector subspace generated by A is equal to Rn and dim R+ C = n. The equations of the cone In order to describe the canonical module of S we will use the Danilov-Stanley formula. Thus the equations of the

cone R+C1 should be determined first. Lemma 7.7.2 Set NI = {-ei, . . . , -en}, N - A/i U N2 and

NI = {-ei - • • • - ej_i + (k - l)ej - ei+i - • • • - en 1 < i < n}. If H is a supporting hyperplane of the cone R+C* such that H contains a set ai, . . . ,an-i of linearly independent vectors in A, then H = Ha for some vector a € N . Proof. Let 0 ^ a = ( a i , . . . , a n ) , such that H = Ha. Let M be the (n — 1) x n matrix whose rows are the vectors a\ , . . . , an-\ . First note that if the j-column of M is equal to zero, then (a^Cj) = 0 for all i and H = Hej . On the other hand, if all the entries of the j-column of M are equal to 1, then H — Ha for some a G A^. Therefore we may assume that all the columns of M have some zero entries and also some entries equal to 1. Set J = { z ^ > 0} and J = { i | f l j < 0 } ,

without loss of generality we may assume

A=

Monomial Subrings_______________________________ 263

where I = \X\. By symmetry the proof can be easily reduced to the following four cases.

(a) Assume T = 0. Using that all the columns of M have at least one entry equal to 1 we obtain that a = 0; hence this case cannot occur.

(b) Assume 2 < i < k — 1. We claim that there exists ccj so that in the set of the first i entries of a* there is at least one entry equal to zero and at

least one entry equal to 1. Otherwise, using that rank(M) = n — 1 and rowreducing M to its normal form we obtain 1 = 2, and hence a = a\e\ — Setting

/3 = ei+e3 + • • • + ek+i and 7 = e2 + e3 we get (/3,a) = ai > 0 and (7,0) = — 01 < 0, which is impossible because Ha is a supporting hyperplane of K+ C and the

proof of the claim is complete. For simplicity we assume ai = Ct! + ei2 + • • • + eim + 6j1 + • • • + 6jr ,

e^ = ei, m + r = k, 3 < ii < • • • < im < I < ji < • • • < > < n and r,m > I.

Let us show (/3, a) > 0, where /3 = e\ + • • • + ei + e^ + • • • + €jk_e . Note that {/?, a)

>

GI + a3 + • • • + at + fljj + • • • +

> ai + aj + • • • + ai + a

+ ••

because (ai,a) = 0. As a consequence j3 € A and (/?,a) > 0, as required. If 2 < \J\ < k — 1, we may apply similar arguments to prove that there

is a 7 € A so that (7, a) < 0, which contradicts that H is a supporting hyperplane. Therefore we may further assume that \J\ is either equal to 1 or \J\ > k; in this situation we can rapidly find 7 e A so that (7,0) < 0, which again yields a contradiction. (c) Assume I = 1 and \J\ > k. Set r = \J\ and B = {at\i e J}, say B = {02, . . . , ar+i}- If r = n — 1, then using (a^ a} = 0 for all i we derive that all the entries of the first column of M are equal to 1 , hence H = Ha for some a e N%. Next we assume r < n — 2. Let aj be a vector with its first entry equal to zero. Since (cti,a) = 0 it follows that the first r + 1 entries of a, are equal to zero and n > r + k. Setting j3 = BI + er+2 + • • • + er+k and 7 = /3 - e\ + e? we obtain (/?, a) > 0 and (7, a) < 0,

which is impossible. The case 1=1 and \J\ = 1 can be treated similarly. (d) Assume i > k and \J\ > k. This case cannot occur because one can rapidly find vectors /3,7 in A so that (/?, a) > 0 and (7, a) < 0. n

264______________________________________ Chapter 7

Remark 7.7.3 The converse of Lemma 7.7.2 is also true because we are assuming n > Ik > 4, and this implies n > k + 1. Note that if n = 3 and k = 2, then the cone R+C has only three facets. Proposition 7.7.4 ([51]) A point x = (xi, . . . ,xn) G R" is in R+C1 if and only if x is a feasible solution of the system of linear inequalities

—Xi

< 0, i = I , . . . ,n

Proof. Let R+C = H^ n • • • n H^ be the irreducible representation of R+C as an intersection of closed half spaces. By Theorem 7.2.18 the set H^ n R-I. C is a facet of R+ C' . Note that Hbi is generated by a set of linearly independent vectors in A, see Corollary 7.2.22. Therefore by Lemma 7.7.2 one obtains Hf)i = Ha, for some a G N, here N denotes the set defined in Lemma 7.7.2. D A generating set for the canonical module

Let X be an arbitrary

subset of Rn . The relative interior of X, denoted ri(.X'), is the interior of X relative to aff(v\r), the afHne hull of X . Lemma 7.7.5 Let a be a vector in Cr\ri(E^.C) and set A = {i a^ > 2}. // \A\ > k and ii, . . . , i/, are distinct integers in A, then a1 = a — e^ — • • • — e.ik

also belongs to C fl ri(R + C 1 ).

Proof. Without loss of generality one may assume a\ > a^ > • • • > an, a,k > 2 and a' = a — e\ — • • • — e^- We claim that a' G ri(R+(7). First recall that R C 1 has dimension n; thus one has

ri(R+C r ) = int R+C = int [ j H~ = f ] inttf- - f ] tf- \ Hc

(7.7)

by Proposition 7.7.4. Hence, using that a G ri(R+C) we readily obtain

(k-l)(a,i - 1) < -1+ ^(aj-l)+

]T dj, for l
On the other hand, if k < i < n we have a,

>

ai + (k - l)ai + ^2 aj — °i + kdi + ^ a.,-

> kdi + n - k + 1.

(7.8)

Monomial Subrings

265

As n > Ik > 4 we obtain n

2^ a-j > kai + k + 1, or equivalently, one has k

(k - l)a; < -1 + ]T(aj - 1) + ]T aj: for k + 1 < i < n.

(7.9)

Altogether making use of (7.7), (7.8), and (7.9) we obtain a' e ri(ffi + C'). By Proposition 7.4.5 the subring K[C] is a normal domain, and therefore C = ZCn l+C. Since a' e 1C we conclude a' 6 C. Q

Let .R = ® °^.0 Ri be the standard grading of R and let

z=0

be the kth Veronese subring of R graded by

ipOh. — /?.. X,JI — •* ''/CZ • 1L

Notice that K[C] is a graded subring of R^ with the normalized grading: 00

K[C] = 0 i=0

where

In the sequel we shall assume that K[C] has the normalized grading.

Theorem 7.7.6 ([51]) Set S = K(C}. Let LJS be the canonical module of S and let 03 be the set of monomials M = x^1 • • • x°^ satisfying the following conditions:

(a) Oj > 1 and (k — l)a; < — 1 + J2j^ti aj' for

a

^ *•

(b) E ? = i O i = Omod(fc).

(c) \{i en > 2}| < k-l. If n > Ik > 4, then *B is a generating set for us-

266______________________________________Chapter

7

Proof. According to Theorem 7.2.34 we have

uis = ({xa a € Cnri(R + C')}).

Taking into account the arguments of the proof of Lemma 7.7.5 and by a repeated use of Lemma 7.7.5 it is enough to prove that 03 C u>s- Let M € 03; without loss of generality we may assume M = x°^ • • • x^'^Xk • • • xn, where 0,1 > • • • > dk~i > 1. The monomial N = x°^ • • • x°£~i can be factored as (xl ' ' ' £i)a'"~ai + 1 , z

...^j.^

if 1 < i < k - 2

K^IL

On the other hand, by the properties (a) and (b) it follows that we can write

i=k

where

~ ^ai ~ ai+l^ -1,

if l - i - k i~2 :f „• _ jL II I — K — 1,

and deg(TV') = Omod(fc). Hence

k-i 1=1

is in -ft^C1], which readily implies M € u>s-

Applications of the description of the canonical module To find the presentation of K[C] as a quotient of a polynomial ring B modulo a prime ideal P, consider the polynomial ring over the field K

with one variable T^...ik for each monomial x^ • • • Xik . Here B has the usual grading. There is a graded homomorphism of ^T-algebras ih

V>:5 —>• /^[C], induced by T^...ik i—s^ z^ - - - x ^ , the ideal P = ker(?/>) is the presentation ideal or ioric idea? of ^[C1]. Thus the Cohen-Macaulay type of the ring AT[C], denoted by type(K[C]), is the last Betti number in the minimal free resolution of B/P as a £?-module.

Monomial Subrings_______________________________ 267

Remark 7.7.7 Set 5 = K[C}. By Corollary 4.3.6 the type of 5 is equal to the minimal number of generators of the canonical module u>s of S. To compute the type of 5 notice that a monomial M = x^1 • • • x°^~^xk • • • xn is in 05 if and only if for all 1 < i < k — 1 one has k-l

Oj = mk — n + k — 1 and 1 < a» < m — 1, for some m > 2. These two conditions imply

.

k ~

Therefore, by Theorem 7.7.6, the computation of the type of S reduces to counting partitions of positive integers.

Corollary 7.7.8 Let u>s be the canonical module of S = K[C], Assume k — 1 and n > Ik. If n is odd, then

-x

1 < j < n, l
Corollary 7.7.9 // k = 2 and n > Ik, then

n(n - 3)

. if n is odd,

(n2 - 4n + 2) ———-———

., «/ n is even.

Proof. The details are left as an exercise.

D

Corollary 7.7.10 If n = 2k + 1 > 5, then

Proof. Left as an exercise.

D

Let S be a Cohen-Macaulay positively graded AT-algebra over a field K and ujs the canonical module of S. We recall that a ( S ) , the a-invariant of 5, is given by a(S) = -min{ i \ (w s )j ji 0}. Corollary 7.7.11 ([51]) Ifn>2k>4, then the a-invariant of K[C] is:

a(K(C}) = - [ = ] , where \x\ is the least integer greater or equal than x.

268______________________________________Chapter

7

Proof. Set S = K[C] and m = [f ]. It follows from Remark 7.7.7 that the degree of the generators in least degree of ws is at least m. To complete the proof we exhibit some generators of uis living in degree m. Write n = qk + r, 0 < r < k; note q > 2. If r > 1, observe that the monomials

belong to (ws) m . In particular, S cannot be a Gorenstein ring in this case. If r = 0, then the monomial M = Xi • • • xn satisfies M € (ws) m . D Remark 7.7.12 (Duality) Let R = K[xi,...,xn] be a polynomial ring over the field K, and k an integer such that 1 < k < n — 1. Consider

gn k _ xl\Xi • • • Xi

I < ii < • • • < ik < n}]

the .ftT-subring of R spanned by the x^ • • -a^'s. Observe that there is a graded isomorphism of J^-algebras of degree zero: P • Sn,k —> Sn,n-k, induced by p(xil • • - x i k ] - xjl • ••xjn_k, where { j i , . . . , jn-k} = {1; • • • ,n} \ {ii, • • • >*fc}- In particular if n < 2k, then

n —k

Because of this duality one may always assume that n > 2k. The next corollary was shown by De Negri and Hibi [81] using different methods. Corollary 7.7.13 ([51]) The subring Sn^ is a Gorenstein ring if and only if k 6 {1, n — 1} or n = 2k.

Proof. By duality one may assume n > 2k > 4. Set 5 = Sn>k- If S is Gorenstein, then by the proof of Corollary 7.7.11 we may assume n = qk. If q > 3, then x\ • • • xn and x\x\ • • • x\_vx^ • • • xn belong to (ws) ? and (u>s)q+i respectively, which is impossible. Therefore q = 2, as required. Conversely assume n = 2k. Let 05 be as in Theorem 7.7.6. Take a monomial M in 03; it suffices to verify that M is equal to x\ • • • xn. One may assume that M = x^1 • • •x^.k_~11Xk • •-xn, where a^ > aj+i > 1. By hypothesis one has k-l

,-=fc(m-l)-l,

(7.10)

for some m > 2. On the other hand one has k-l

a,j, for all 1 < i < k - 1.

(7.11)

Monomial Subrings_______________________________ 269

Next we combine (7.10) and (7.11) to obtain Oj < m — 1 for all i. Therefore using (7.10) again we rapidly derive

k(m- 1) - 1 < (k- l)(m- 1), which yields m < 2. As a consequence aj = 1 for all i, as required.

D

Proposition 7.7.14 ([51]) Let F be a finite set of square-free monomials of degree k in R and let A — K[F] be the K-subring of R spanned by F. If dim(A) = n, then

a(A] < a(K[C}), where A denotes the integral closure of A and C is the subsemigroup of N™ generated by A = {e^ + • • • + 6ik 1 < i\ < • • • < z& < n}.

Proof. Let C and CF be the subsemigroups of N™ generated by A and log(F) respectively. Set S = K[C}. Since S is normal by Proposition 7.4.5, we obtain A C S. Let

Mi = {xa\a& C^nriOR+CF)} and M2 = {xa\a<= C D ri where

Notice_K+CV - H+CV and afi(R+CF) - K™. Therefore the relative interior of K+CV equals its interior in R™. For similar reasons we have ri(ffi+C < ) = (l+C) 0 . Hence ri(!+C^) C ri(l + C ( ). Altogether we obtain MI C M2. Let xb be an element of minimal degree in M\ so that deg(a;6) = —a(A). Set r = -a(A). Since xb is in M2 and xb e Ar C Sr, we conclude -a(S) = min{deg(x a )|a £ M2} < r = -a(A).

Hence a(A) < a ( S ) , as required.

D

Corollary 7.7.15 ([51]) Let R — K\x\, . . . ,xn] be a polynomial ring over a field K and F a finite set of square-free monomials of the same degree k. If dim K[F] = n, then x

r m ~\

ifn> 2k, a(K(F}}

n —k

if n < 2k, n ^ k.

Proof. Use Corollary 7.7.11, Proposition 7.7.14, and duality.

D

270

Chapter 7

Remark 7.7.16 It is not hard to verify the equality

"<«<"> = - fsl • where R^ is the kth Veronese subring of R. Let K[C] be the subring generated by the square-free monomials of degree k. Notice that

a(K[C}}
Because of this, to keep better control of the a-invariant, it is preferable to embed K[F] into K[C], instead of R^.

Corollary 7.7.17 Let R = K[XI, . . . ,xn] a polynomial ring over a field K

and F a finite set of square-free monomials of degree k in R. If K[F] has dimension n and n > 2k > 4, then K[F] is generated as a K-algebra by elements of normalized degree less or equal than n — [^] . Proof. Proceed along the lines of the proof of Theorem 7.6.2 and use Corollary 7.7.15. D

Example 7.7.18 Let K[F] be the subring of R — K[XI, . . . ,x$] spanned by the monomials of R defining the edges of the graph shown below.

/I =. /2

=

/3 = /4

/i2 =xix 2 x 3 a;5a; 6 a;8,

=•

/14 =2;io:2X3X5X 6 a;7, /i5 = xix2o;3a;52:7a:8.

/5 = Z4

fe =x5xe, Jf

— XfjXf)

/8 = a;5a;7, J9

K[F] =

^^ X f X g ,

/io = a;5a;8) /n = The generators of K[F] can be computed using the program [48], see also [298, Section 7.3]. A Noether normalization for K[F] is given by

Monomial Subrings_______________________________ 271

where

hi = /i,

h3 = fa - /u,

h5 = /2 - /a, /IT = /5 - /? - /9 - /ii,

/l2 = /6,

/M = /9 — /10,

he = fa - /5,

hg = fl - 53j=2 /*•

Since AT[F] is Cohen-Macaulay we obtain a decomposition

-4o/73 © A>/72/ii © A>/i3o A)/12 © Ao/13 ® A>/14 ®

Therefore the Hilbert series of K[F] is equal to

H(K\F\,*)=l

+3Z+

,?*l?Z*+-

Altogether K[F] is generated as an A0-module by monomials of normalized degree less or equal than dim R + a(K[F]) = 4. However, as a /iT-algebra it is generated by square-free monomials of normalized degree at most 3. Next we show a general statement on degrees valid when S and 5 are both Cohen-Macaulay.

Proposition 7.7.19 Let S be a standard graded algebra and let M C N be finitely generated graded S-modules of the same dimension d and multiplicity

e. If M and N are Cohen-Macaulay and ' 7i———Twf>

tlN\t)

=

T:

TTT

are their Hilbert series, then deg/(i) > degg(i). Proof. Consider the exact sequence

0 - > M —> N —> P - > 0 of graded modules. If M 7^ A7", P is a module of dimension < d since M and N have the same multiplicity. Since M and TV are Cohen-Macaulay, standard depth chasing (see Lemma 1.3.9) implies that P is Cohen-Macaulay of dimension d - 1. We have the equality of Hilbert series, g(t) (l-t)d

=

f(t) (l-t)d

h(t) (I-*)*- 1 '

and therefore

The assertion follows since the /i-vectors of these two modules are positive (see Chapter 4 and [44, Corollary 4.1.10]). D

272______________________________________Chapter

7

Lemma 7.7.20 If K[F] C K[x] is a monomial subring over a field K, then there is a positive integer UQ such that (xa)n° £ K[F} for all xa £ K[F]. Proof. By Lemma 7.2.32 there are 71,..., jr £ N™ such that

For each i one can find a positive integer n, such that xniji £ K[F\. It is readily seen that the integer no = 1cm(ni,... ,n r ) satisfies the required condition, because any xa £ K[F\ can be written as xa = x aiTl • • • xarjr, for some GI , . . . , a r £ N. d Proposition 7.7.21 If A = K[F] C K[x] is a monomial subring over a field K and I is the io

1= (A:AA) = {a£ A\aAc A}, then there is a monomial x1 £ /. In particular 1^0. Proof. Let 71,..., 7r £ Nn such that A = K[xjl,..., x 7r ]. One can write !3i and xSi are in A. By Lemma 7.7.20 there is xii _ X0i jxi>i ^ where x n a 0 ^ n0 £ N_such that x ° £ A for all xa £ A. Set a;7 = a;no<51 • • • xn°s". Take a;" in A and write X Q = (x71)"1 • • • (x 7 ") a ", where a» € N for all i. By the division algorithm there are integral equations a^ = qiUQ + Cj, where 0 < Q < no for all i. From the equality

is in A

is in ^4

is in A

one concludes x^xa £ A Since xa was an arbitrary monomial in A, we get C A and a;7 £ /. D Proposition 7.7.22 If A = K[F] C K[x] is a homogeneous monomial

subring over a field K, then A is a positively graded K-algebra and the multiplicity of A is equal to the multiplicity of A. Proof. Set A = log(F) and P — conv(.A). As A is homogeneous, using Proposition 7.2.43 one has that

i=0

is a graded .ftT-algebra whose ith component is given by

Monomial Subrings_______________________________ 273

Recall that A is graded as in Proposition 7.2.39. Jl/et h and E be the Hilbert functions of A and A respectively. Since A and A have the same dimension d, one can write

h(i) = aoid~1 + terms of lower degree, E(i) = Coid~1 + terms of lower degree,

for i 2> 0. One clearly has OQ < CQ because Ai C At for all i. On the other hand by Proposition 7.7.21 there is x7 £ A of degree m such that x"1 A C A. Thus for each i there is a one to one map

Hence E(i) < /i(i + m) for all i, and consequently c0 < o 0 . Altogether OQ = CQ. Therefore

as required.

d

Corollary 7.7.23 ([51]) // .K" is a /ze/d anrf K[F] is a Cohen-Macaulay homogeneous monomial subring, then

a(K[F}) < a(K[F}). Proof. Use Proposition 7.7.19 and Proposition 7.7.22.

D

Exercises 7.7.24 Let F be a finite set of square-free monomials of degree k in the polynomial ring K[XI, . . . ,xn], K a field. If n < 2k and K[F] is not a normal domain, prove that K[F] cannot be generated as a K algebra by square-free monomials. 7.7.25 ([130, 219]) Let R^ be the kth Veronese subring of a polynomial ring R in n variables over a field K. Prove that R^ is Gorenstein if and only if k divides n. Hint The interior of the cone spanned by the exponent vectors in R^ is the "first open quadrant" of ffi™ and use the Danilov-Stanley formula.

7.7.26 Let R^ be the kth Veronese subring of a polynomial ring R in n variables over a field K. Prove that the a-invariant of R^ is given by

274

Chapter 7

7.7.27 If K[F] is a two dimensional homogeneous monomial subring, prove that the Hilbert polynomial of K[FT] is equal to the Hilbert polynomial of K[FT], where T is a new variable. 7.7.28 Let A = K[F] be a Cohen-Macaulay homogeneous monomial subring. If dim(yl) = 2 and a(A) < 0, prove that A is normal.

7.7.29 Let 5 be the kth square-free Veronese subring of a polynomial ring R = K[XI, ..., xn] over a field K. If KS is the field of fractions of S. Use the equality k+l

^1 ' ' ' Xi—iXjx^-|_i • • • Xk^.~i

\k-l

_ rk —— -t-1

1

to derive x\ G KS- Using a similar equality prove K[x^, . . . , x^] C KS, then conclude tic.degK(S) = n. 7.7.30 Let F be the set of square-free monomials of degree k in the variables X i , . . . ,xn and S — K[F] the monomial subring spanned by F endowed with the normalized grading. If n = 2k + 1 > 5, prove that the canonical module of 5 is given by

ujs = ({xa\ card({z|aj — 2}) — k - I and card({i|aj = 1}) = n - k + 1}) .

Recall that a standard Artin algebra A over a field K is called a level algebra [27] if all the nonzero elements of the socle of A have the same degree. Prove that if hi, . . . , hn is a system of parameters for S = K[C] with each hi a form of degree 1, then

is a level algebra. Consider the homogeneous subring A = K[F], where F is the set of all monomials X{Xj such that x^ and Xj are connected by a line of the graph:

Note that A is a graded subring of A, where both rings are endowed with the

normalized grading. In the following problems we indicate how to compute some invariants of A and A.

Monomial Subrings_______________________________ 275

7.7.31 Use CoCoA, and the procedures indicated below, to prove that the Hilbert series and Hilbert polynomial of A and A are given by:

144

16

18

6

0

>

In particular e(7) = e(7) = 5, a(A) = -3 and a(A) = -4. 7.7.32 Use this CoCoA procedure to compute the Hilbert series and Hilbert

function of A. Note that we are first computing the toric ideal P of A, through the usual procedure of elimination of variables. U s e R : : = Q[t[1..8],x[1..7]]; Q := Ideal(x[l] * x[2] - t[l], x[2] * x[3] - t[2], x[l] * x[3] - t[3], x[3] * x[4] - t[4], x[4] * x[5] - t[5], x[5] * x[6] - t[6], x[6]*x[7]-t[7],x[5]*x[7]-t[8]); P:=Elim(x,Q); Use R ::= Q[t[1..8]];

P := Ideal(t[2] * t[3] * t[5p2 * t[7] - t[l] * t[4f2 * t[6] * t[8]); Poincare(R/P);

Hilbert(R/P);

7.7.33 Use Normaliz to verify that A = A\X\XIX?,X§X§XT\. required for the next computation.

This step is

7.7.34 Use the following CoCoA procedure to compute the Hilbert series of A. Note that we are also computing the toric ideal P of A. Use R::= Q[t[1..9],x[1..7]]; Q := Ideal(x[l] * x[2] - t[l], x[2] * x[3] - t[2], x[l] * x[3] - t[3], x[3] * x[4] - t[4], x[4] * x[5] - t[5], x[5] * x[6] - t[6], x[6]*x[7]-t[7],x[5]*x[7]-t[8], x[l] * x[2] * x[3] * x[5] * x[6] * x[7] - t[9]); P := Elim(x, Q); Use R ::= Q[t[1..9]], WeightsQl, 1, 1, 1, 1, 1, 1, 1,3]); P := Ideal(t[l] * t[4] * t[6] * t[8] - t[5] * t[9], t[2] * t[3] * t[5] * t[7] - t[4] * t[9], t[l] * t[2] * t[3] * t[6] * t[7] * t[8] - t[9]"~2); Poincare(R/P);

276______________________________________ Chapter 7

7.7.35 Verify the assertions about A using the following input file for the program Normaliz, and running Normaliz with the option —h. bowtie. in

8 7 1

1 1 0 0 0 0 0 0

0 1 0 0 0 0 0 2

7.8

0 1 1 1 0 0 0 0

0 0 0 1 1 0 0 0

0 0 0 0 1 1 0 1

0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1

(number of generators) (dimension of ambient lattice) (generators)

(option or mode)

Some lexicographical Grobner bases

A lexicographical order of the terms in a polynomial ring K[ZI, . . . ,zr] is an order obtained by representing terms as z^ • • • z^ for a fixed permutation

TT of the variables and then using the lexicographical order. The Grobner bases obtained by lexicographical orders are called lexicographical. There are several classical examples of graded prime ideals that are minimally generated by a lexicographical Grobner basis [59, 66, 159, 276]. Below we study some special linear orders and relate them to toric ideals of subrings of fc-products. Some linear orders of the variables

Given two integers n > Ik > 4

we consider a "label set" : Tk= {[ii,...,u]|l
Let A = (ay) be a non-singular matrix of order k with entries in K and

Ui = (an, . . . ,a,ik) its ith row.

This matrix defines a linear ordering in Z fc given by

x = (xi,...,xk) >A y = ( y i , - - - , y k ) iff Ax >iex Ay, that is x >A y if the first non-zero entry of (x • u\ — y • HI, . . . , x • Uk — y • Uk) is positive. Thus we can linearly order the variables in T by:

T [ i i , . . . , i k ] > T [ j i , . . . , j k ] i f f ( Z I , . . . , H ) >A ( j i , - - - , j k ) -

Monomial Subrings

277

Example 7.8.1 If n — 5 and k = 2, then the linear ordering induced by the matrix: 0 1 A ~ ' -I 0

is T[l, 5]T[2,5]T[3,5]T[4,5]T[1,4]T[2,4]T[3,4]T[1,3]T[2,3]T[1,2]. A distinguished group of matrices order k of the form

Consider the set A of matrices of

such that the set {e^, . . . ,e, fc } is a permutation of the set of canonical vectors {ey . . . , e&} in R* . Note that ^4 is a group of order 2kkl. As the rows of any matrix in A are orthonormal, we see that A is a subgroup of the orthogonal group Finding lexicographical quadratic Grobner bases Let K be a field and R = K[XI, . . . ,xn] a polynomial ring. Let Vk be the set of square-free monomials of degree k in R. Consider the presentation:

K[Vk], induced by T(i, • • • ik] ^ xh • • • xik , thus P = ker(
quadratic Grobner basis, with respect to some term order, for the toric ideal P o f K [ V k ] , see [278].

The examples below can be verified using Macaulay. We thank Eduardo Viruena Silva for writing some Pascal programs to generate all the matrices in A, and the corresponding linear orderings of the variables. Example 7.8.3 The matrices below are precisely the matrices in A that induce a quadratic Grobner bases for the toric ideal of Q[Vfc] corresponding to k = 2 and n — 6.

1 0 1 . [1 0 1 A 0 1 * = 0 -1

, F -1 0 * = 0 -1

A

0 - 1

i

o

Chapter 7

278

Example 7.8.4 For n = 6 and k = 3 the full list of matrices in A that induce quadratic Grobner bases are: -

A,=

1 0 0 " 0 1 0 0 0 -1 _

0 " - 1 0 0 0 0 - 1

AIS =

" 0 - 1

A16 =

0 1 0 " 1 0 0 0 0 1 "0-1

" 0 1 AIS =

0 " 1 0 0 0 0 - 1

A9 =

A3S =

0 " 1 0 0 0 0 - 1

" 0 0 -1 " 0 - 1 0

1

0

0

A,! =

" 0 1 0 " 0 0 1 1 0 0

A45 =

' 0 1 0 " 0 0 - 1 1 0 0 _

A,7 =

"0-1 0 " 0 0 - 1 1 0 0

A48 =

" 0 - 1 0 " 0 0 - 1 -1 0 0 _

Example 7.8.5 For n = 8 and k — 4 the complete list of matrices in A that give quadratic Grobner bases is:

A25 =

" 0 1 0

0

1 0 0

0

0 0 1 0 _ 0 0 0 -1

^367 —

"0 0-1 0" 0 - 1 0 0 0 0 0 - 1 1 0 0 0

1 0 " 0 0 0 0 0 -1 _

" 0

" 0 1 0 89 =

0 0 0 1 0 1 0 0 0 _ 0 0 0 -1

A7g =

0 00 - 1 1 0 0 0

351

377

~

0-1 0 " 0 0 0 - 1 0 - 1 0 0 1 0 0 0

" 0 0 ~ 0 1

1 0

0 "

0 1 0 0 0 - 1

0 0

0

Conjecture 7.8.6 If P is the toric ideal of K[Vk] and n > 2k > 4; then there exists a linear order of the variables in T 50 that the reduced Grobner basis of P, with respect to the lexicographical order, is generated by quadrics.

Monomial Subrings

279

Exercises 7.8.7 If we order 3£, this induces an order on 3™_ fc . Prove that P(Vk) has a lexicographical quadratic Grobner basis if and only if P(Vn-k) does. Here P(Vk] denotes the toric ideal of K[Vk]. 7.8.8 (V. Bonanzinga) Let n > d > 1 be two integers. If 3JJ has the lex order and N[II , . . . , id] represents the position of [ii , . . . , id] with respect to this order. Prove the formula: d

n — d+

See [29] for more details about this formula.

Chapter 8

Monomial Subrings of Graphs Here we study the monomial subrings associated to graphs and their toric ideals. A full description of the integral closure of those subrings will be presented along with some connections with graph theory and polyhedral geometry. Let G be a graph. The following diagram gives an idea of the objects

associated to G that will be studied along this chapter. Graph: G

Edge subring: k[G]

Rees algebra

Toric ideal

Toric ideal

Circuits & Smith form

Grobner basis

The edge subring k[G] associated to a graph G is the monomial subring generated by the monomials corresponding to the edges of G, where k is a field. The description of the integral closure of k[G] will be given in

281

282______________________________________ Chapter 8

terms of special circuits of the graph (see Theorem 8.7.9). This description links the normality property of k[G] with the combinatorics of the graph G (see Proposition 8.8.11). The incidence matrix of G plays an important role

because it is related to the toric ideal of k[G] and its rank can be interpreted in graph theoretical terms.

8.1

The subring associated to a graph

Let G be a graph on the vertex set V = {vi , . . . , vn} and 00

R = k [ x i , . . . , x n ] = @Ri i=0

a polynomial ring over a field k with the standard grading. To simplify notation sometimes one identifies the indeterminate X{ with the vertex v^. The monomial subring or edge subring of the graph G is the fc-subalgebra

k[G] = k[{xiXj\ Vi is adjacent to Vj}] C R. If F = {/i, . . . , fq} is the set of monomials XiXj such Vi is adjacent to Vj, the elements in k[G] are polynomial expressions in F with coefficients in

k. A monomial / = x^Xj in R is said to be an edge generator of G if Vi is adjacent to Vj. As k[G] is a standard /c-algebra with the normalized grading

there is a graded epimorphism of fc-algebras

where B is a polynomial ring graded by deg(Ti) = 1 for all i. The kernel of f , denoted by P(G), is a graded ideal of B called the toric ideal of k[G] with respect to /i, . . . , fq. Definition 8.1.1 Let w = {VQ,VI, . . . ,vr = VQ} be an even closed walk in

G such that fi = Xi-\Xi. As /1/3 ' ' ' fr-l

= /2/4 ' ' ' fr,

the binomial

Tw = TI • • • Tr-i — T
Monomial Subrings of Graphs_______________________ 283

Next we identify a set of generators for P(G] corresponding to the even closed walks of G. A closed walk of even length will be called a monomial walk.

Notation For use below Xs denotes the set of all non-decreasing sequences a = (ii , • • • , ia} of length s. Let a\ , . . . , aq be a sequence and a = (i\ , • • • , is) in Is , then we set aa = a^ • • • aia .

Proposition 8.1.2 ([303]) If G is a graph and P = P(G) the toric ideal of the edge subring k[G], then (a) P = ({Tw\ w is an even closed walk}), and

(b) P = ({Tw w is an even cycle}) if G is bipartite. Proof. Let fi, . . . , f q be the edge generators of G and P the kernel of the homomorphism of fc-algebras

where B is the polynomial ring k\T\, . . . ,Tq] and (Tj) = fi. Recall that P is graded and /oo

\

P =B.(\JP.). \s=2

/

First we set B = {Tw\w is an even closed walk}. One clearly has (B) C P. To prove the other containment we use induction on s. First recall that P is a binomial ideal by Proposition 7.1.2. If s — 2, it is enough to note that the binomials in P% come from squares. Assume Ps_! C (B). To show Ps C (B), take Ta - T0 in Ps, where a = (ij, . . . ,is) and j3 = (ji, . . . , js). Define GI as the subgraph of G having vertex set Vi = {xi £ V\ Xi divides /„} and edge set

EI = {{x,y} £ E ( G ) \ f e =xy for some t e {ii, . . . ,is,ji, • . • , js}}. Note that if fa • • • fim = fjl • • • fjm for some m < s and for some ordering of the generators then Ta - Tp e (B); to prove it, notice that Ta - Tp can be written as

Ta - Tp = Tim+l • • • TJ. (Ttl • • • Tlm - Th • • • Tjm) +Th • • • Tjm (Tim+1 • • • Tia - TJm+1 • • • Tja ) and use induction hypothesis. Now we can assume /^ • • • fim ^ fjt • • • fjm for m < s and for any re-ordering of the /j's. Observe that this forces GI

284______________________________________Chapter

8

to be connected. Take VQ 6 V\. Since fa = fp, it is easy to check that after re-ordering we can write f i k — v^k-^v^k-i and j j k = V2k-i^>2k, where 1 < k < s and I'D = «2s- Therefore the monomial walk w — {i>o, • • • ,v^n} satisfies Tw = Ta — Tp and the induction is complete. This proves part (a).

Assume G is bipartite. The second part of the proposition can be proved similarly if we notice that this time the condition /^ • • • /jm ^ f j i ' ' ' fjm for m < s and for any re-ordering of the /j's implies that the graph GI is

an even cycle.

Q

Definition 8.1.3 Let F be a finite set of monomials in a polynomial ring

k[x]

and let P be the toric ideal of k[F}. A binomial fa — T@ c. p

is called primitive if there is no other binomial Ty — Ts G P such that T7

divides Ta and T5 divides T13. According to a result of [278], the set of all primitive binomials in a toric ideal P is finite and is called a Graver basis of P. Corollary 8.1.4 If G is a graph and f = Ta — T13 is a primitive binomial in P(G], then f = TW for some even dosed walk w of G. Proof. It follows from the proof of Proposition 8.1.2.

D

Corollary 8.1.5 If G is a bipartite graph and f = Ta — T13 is a primitive

binomial in P(G), then f = TW for some even cycle w of G. Proof. It follows from the proof of Proposition 8.1.2.

D

Proposition 8.1.6 Let G be a graph and let P(G) be the toric ideal of the edge subring k[G]. If f = Ta — T@ is a primitive binomial in P(G), then the entries of a satisfy ccj < 2 for all i. Proof. By Corollary 8.1.4 one can write _

rr\

_

rr\ rr\

rrt

rj-i rr\

rj~\

= J-w = J - l J - 3 - • • J-l-l — J . 2 J - 4 - • • J - e ,

where i is even, and w is an even closed walk in G of length f,: w = {v0,vi,...,vi-i,ve - v0} such that /i/s • • • fi-i = /2/4 • • • fi and /j = Vi-\Vi for all i. Note that there may be some repetitions of the variables T, and accordingly some

repetitions in the monomials /». As usual we are identifying the vertices of G with variables. It suffices to verify that VQ occurs at most three times

Monomial Subrings of Graphs_______________________ 285

(including the end vertices) in the closed walk w. If VQ occurs more than three times in w, one can write V0,V1,. . . ,Vtl = V0,Vtl+i,. . . ,Vtl+t2 = V0,Vtl+t2 + i , . ..,Vf=

VQ,

note that ti must be odd; otherwise using w\ = {VQ,VI, . . . ,v^ = VQ} one has that TWl = T^ - Td , where TT divides TiT3 • --Tt-i and Ts divides TiT± . . .Ti, a contradiction because / is primitive. By a similar argument *2 must be odd. Thus ti + £2 is even, to derive a contradiction note that the closed walk w

' = (Vti+t2 = vo,vtl+t2+i,...,vt =v0}

is of even length, and apply the previous argument.

D

Lemma 8.1.7 Let P be the toric ideal of a monomial subring K[F]. If f = Ta — T® is a binomial in the reduced Grobner basis of P with respect to some term order, then f is a primitive binomial.

Proof. See [278, Lemma 4.6].

D

Lemma 8.1.8 Let G be a graph and let P = P(G) be the toric ideal of k[G}. If f is a polynomial in any reduced Grobner bases of P, then (a) / is a primitive binomial and f — Tw for some even closed walk w of the graph G.

(b) // G is bipartite, then f is primitive and f = Tw for some even cycle w of the graph G. Proof. Let A be a reduced Grobner basis of P and take / e A. Using Corollary 7.1.5 together with the fact that normalized reduced Grobner bases are uniquely determined, we obtain / = Ta - T/3 . By Lemma 8.1.7 / is primitive, thus by Corollary 8.1.4 and Corollary 8.1.5 / has the required form. D Proposition 8.1.9 If G is a graph and P — P(G) is the toric ideal ofk[G], then the set

{Tw\ Tw is primitive and w is an even closed walk}

is a universal Grobner basis of P. Proof. It follows from Lemma 8.1.8.

D

Proposition 8.1.10 If G is a bipartite graph and P = P(G] is the toric

ideal of k[G],

then the set {Tw w is an even cycle}

is a universal Grobner basis of P. Proof. It follows from Lemma 8.1.8.

D

286

Chapter 8

Exercises 8.1.11 Let G be a graph with q edges and P(G) the toric ideal of the edge subring k[G\. If / is a primitive binomial in P(G) prove:

,

, _,. { q if q is even, and &\J ) - | g _ } otherwise.

8.1.12 Let G be a bipartite graph and k[G] its edge ideal. Prove that k[G] is a normal domain.

8.2

Rees algebras of edge ideals

We will look closely at the presentation ideal or toric ideal of the Rees

algebra of an edge ideal, and then show a few applications.

Notation For / and g in a unique factorization domain we define [f,g} = l c m ( f , g ) .

As before Is denotes the set of all non-decreasing sequences a = (i\, • • • , is) of length s. Let a i , . . . , aq be a sequence and let a = (ii, • • • ,is) £ Is, then we set ^ • • • flj o and aat = a^ • • -a^ • • • dj a .

Theorem 8.2.1 Let I be the edge ideal of a graph G and F = {/i,..., fq} the set of edge generators of I. If Bj J is the presentation of the Rees algebra of I with respect to F, then /oo

\

J = B J, + B • |J Ps \s=2

, /

where Ps = {Ta -Tp\ fa= /g, for some a, j3 £ Is}. Proof. Let xi,..., xn be the set of vertices of G and R a polynomial ring in n variables over a field k. Here we identify the vertices of G with the variables of R. Let Tl(I) — R[IT] be the Rees algebra of I. The toric ideal J of 7?-(/) is the kernel of the graded epimorphism: 1,...,Tg]

It is sufficient to show the recursive formula

Js =B1JS^1+RPS for s > 2.

Monomial Subrings of Graphs_______________________287

According to Theorem 2.4.17 Js is generated by polynomials of the form '[fctJ^rr

/o -, N

I8-1)

H/-//3]

/a

where a = (ii,... ,is), /? = ( j i , • • . ,js) are in Is. It suffices to prove that all those generators are in £?iJs-.i + Bs-iJi + RPS. Since for fa = fp

expression (8.1) simplifies to Ta - Tp 6 RPS, we may assume fa ^ fp- In this case there is a variable z G {zi,..., xn} so that fa = zah and fp = zbg, with a ^ b. Assume b > a > 0, it is not hard to see that there are integers m and t so that [fa, fp] is a multiple of f j m f a t . Notice that the generators of Js can be re-written as J? 7/3

'"lXt

/ "-Pm /

I - 'Jm /

'

\ -" J/m V

I

J?

)

T*i

The proof now reduces to showing that there are polynomials A, n, A in R and integers m and t so that fa,fft]

Pm

f

//3m

/

and Arp

A1

Jm -

I IJO" -IP1

1——7——— V

Ja

\ rp

/

_

I" I [ IJlt 1 J]ml \ rp /^ rU

-Lit -fj'((

\L r (ifitJij I IJltl JJmJ

——J.——— ) 1J^ ~ (——J———

\ V

J]m

/

\

Jit

\ rp

/

^t

To complete the proof observe that such polynomials exist if [/a,/g] is a multiple of fjm fat for some integers m and t.

D

The example below proves that Theorem 8.2.1 do not extends to ideals generated by square-free monomials of higher degree. Example 8.2.2 Let R = k[xi,... ,£7] be a polynomial ring and let / be the ideal of R generated by , /2 = X2X4X5,f3

= X5X6X-f,

/4 = X3X6XT.

Then using Macaulay [15] we get that the toric ideal of 72.(/) with respect to F = { / i , . . . , /4J is minimally generated by the binomials

288______________________________________Chapter

8

Corollary 8.2.3 Let I = I(G] be the edge ideal of a graph G. Then I is a syzygetic ideal if and only if G contains no squares. Proof. Notice that according to Proposition 8.2.1 / is syzygetic if and only

if RPz = (0).

Hence it is sufficient to observe that if / is generated by

/ i , . . . , fq then a non trivial relation of the form fifj if fifj is a square-free monomial.

= fkfi can only occur D

Corollary 8.2.4 ([303]) Let G be a connected graph and let I be its edge ideal. Then I is an ideal of linear type if and only if G is a tree or G has a

unique cycle of odd length. Proof. Let V = {xi,..., xn} be the vertex set of G and / = ( / i , . . . , /9). If / is an ideal of linear type then by Proposition 3.2.1 G is a tree or G has a unique cycle of odd length.

Conversely assume G is either a tree or G has a unique cycle of length. We claim that RPS = (0) for s > 2, we proceed by induction on n. If n < 3 this is easily verified, assume the claim true for graphs with less than n vertices. Using induction on s we now show RPS = (0) for s > 2, notice

that by Corollary 8.2.3 AP2 = (0). Assume RPs-i = (0) and RPS ^ (0), take a nontrivial relation ftl • • • fia = fj1 • • • fja. Since a tree has a vertex of degree one,

by induction hypothesis it follows that G must be a cycle.

Therefore, using induction again, we may write { f a , . . . , f a } = {gi,..., gs}, where gcd(
D

Let us show an application of Proposition 6.3.7

that lends some support

to the following open question [291]:

Conjecture 8.2.5

(Vasconcelos) Let I be a Cohen-Macaulay ideal in a

regular local ring R. If I is syzygetic and I/I2 is Cohen-Macaulay, then I is a Gorenstein ideal. Corollary 8.2.6 Let T be a tree with vertex set V and let I be its edge ideal. If I is Cohen-Macaulay and m = \V\ > 3, then depth(/// 2 ) = | - 1. Proof. First note that / has the description given in Theorem 6.3.4, here we use the notation of this theorem. Set R = fc[x,y] and z = {xt - yi}ri=1,

Monomial Subrings of Graphs_______________________ 289

where m = 2r. Since I is syzygetic (see Corollary 8.2.3), we have an exact sequence 0 —4 Fi(/) —•> (R/I)m~l — > I// 2 —» 0. If /// 2 is Cohen-Macaulay, then

Torf (I/I2, R/(z)) = 0.

Therefore if we tensor this exact sequence with R/(z) we get that the ideal

({Vi\ 1 < i < r}) + ({yjVel Xj adjacent to ye}) obtained from / by making Xj = yt is syzygetic, which is contradicted by

the relations (yjye)2 = y]y\.

Q

Corollary 8.2.7 Let G = K.n be the complete graph on n vertices and let I be its edge ideal. Then the toric ideal of Ti(I) is generated by binomials of degree two.

Proof. Let P = ker(B = k[Ti,. . .Tq] 4 k[G}), where (p(Ti) = fc. Recall that we can write /oo

\

P =B(\JP.). \s=2

/

By Theorem 4.3.7 (cf. [251]) the edge ideal / has a linear resolution, hence using Theorem 8.2.1 the proof reduces to showing P = (Pz). Using induction on s we will show Ps C PI for all s. Assume Ps~\ C P%. Let /(T) = Ta — T@ G Ps. By the induction hypothesis we may assume a and (3 have disjoint support and that f ( T ) comes from a monomial walk. In particular we may assume that 0.1,0:2, and /3s are positive integers so that f i = xiX2,fa = x^xs and /s = 0:3X4, where xi,...,x^ are distinct vertices. Note that fk = xix± for some k and g(T) = TiT2 - T3Tk € P2. Using /(T)

=

T ai

+T3 the induction assumption shows /(T) 6 Ps, as required.

D

The next result uses a formula of Huneke and Rossi [187, 261] for the Krull dimension of the symmetric algebra of a module. For another proof of this formula based on slightly different ideas, as well as for a nice survey on symmetric algebras, see [293]. Theorem 8.2.8 ([303]) Let G be a graph with n vertices and q edges and I = I(G) its edge ideal. If G is connected, then dim(Sym(/)) = sup{n + 1, q}.

290______________________________________ Chapter 8

Proof. According to the Huneke-Rossi formula we have dim Sym(J) = sup{z/(/ p ) + dim (R/p)\ p 6 Spec(f?)} , where z/(/ p ) denotes the minimal number of generators of the localization of / at p. By this formula it is clear that the Krull dimension of Sym(J)

is at least the given bound. To prove the converse, take a prime ideal p of height n — i containing / and set B = {v e V\v ^ p}. It follows that p contains a face ideal Q, that is, Q is an ideal generated by a subset of the vertices of G. We may assume Q = (A], where A is a minimal vertex cover for G. De7ne now C = {x 6 A\ x is adjacent to some vertex in B} and

Y = {{x,y} £ E(G)\ {z,2/}nC = 0}. Notice that z/(/ p ) < \C\ + \Y\. Since \B\ > i we obtain

The formula is now a consequence of Lemma 8.2.9 below.

Q

Using the notation above one has:

Lemma 8.2.9 Let G be a connected (n, q)-graph with vertex set V = V(G) and edge set E = E(G). Let A be a minimal vertex cover for G and B a subset of V\A. Then \C\ + \Y\ + \B\ < supjn +l,q}. Proof. Let Y' be the set of edges cover by C, that is, Y' = E(G)\Y. Consider the subgraph G' of G with edge set equal to Y' and vertex set

V(G') = {ze V(G)\ z lies in some edge in Y'}.

If Y = 0, then \B\ + \C\ + \Y\ < n. Assume Y ^ 0, and denote the connected components of G" by G I , . . . , Gm. Set Bi=Br\ V(Gi) and d = C D V(Gi); notice B = UBt and C = UCj. We claim that \Bi + \d\ < n{ - 1, for all i, where m = \V(Gi)\. For that, fix an edge {x,y} £ Y. If z e V(G r j), then x
This yields a path {x = XQ,XI, . . . ,xr = z}. As xr-\ ^ V(Gi), z £ Bi U Ci, which gives \Bi\ + \Ci < n^ — 1. Altogether we have m

m

\B\ + \C\ + \Y\ < ^(n, - 1) + \Y\ = ^(n, - 1) + (q - \Y'\).

Monomial Subrings of Graphs_______________________ 291

But Y' is the disjoint union of E(Gi) for i — 1, • • • , m, and thus

here
\B\ + \C\ + \Y <£("*- ^ -!) + ?

z=i Since G; is a connected (rij, g^)-graph we have n; — 1 <
\B\ + \C\ + \Y\ < q, establishing the claim.

D

A dimension formula Let us describe the cycle space of a graph G over the two element field F = Z 2 . We denote the edge set of G by

E(G) = {/i,...,/J and the vertex set of G by

V = {xi,...,xn}. Let CQ and C\ denote the vector spaces over F of 0-chains and 1-chains respectively. Recall that a 0-chain of G is a formal linear combination

of points and a 1-chain is a formal linear combination

of edges, where a^ G F and 6j G .F. The boundary operator d is the linear transformation denned by

The cycle space Z(G) of G over F is equal to ker(<9). The vectors in Z(G) can be regarded as a set of edge-disjoint cycles. A cycle basis for G is a basis for Z(G) which consist entirely of cycles, such a basis can be constructed as follows:

292______________________________________Chapter

8

Remark 8.2.10 If G is connected, then G has an spanning tree T. The subgraph of G consisting of T and any edge in G has exactly one cycle, the

collection of all the cycles obtained in this way form a cycle basis for G. See [139] for details. In particular dim/? Z(G) — q — n + 1,

see Exercise 5.2.10. Lemma 8.2.11 Let G be a connected graph and Ze(G) the subspace of

Z(G) of all cycle vectors of G with an even number of terms. Then + I1 ,. ,-, ,^-A f q -— n + dimp Ze(G) = <

[ q —n

bipartite, and ifif G is bip ,,

otherwise.

Proof. Let GI , . . . , GI , c ; + i , . . . , cm be a cycle basis for G, where c\,..., c;

are even cycles and c ; + i , . . . , cm odd cycles. It is clear that a bases for Ze(G) is given by {ci,... ,c/,c/+i + c m , . . . , c m _ i +cm}.

D

The next result shows how the Krull dimension of k[G] is related to the cycle space of its graph. Proposition 8.2.12 ([303]) Let G be a connected graph and let P be the toric ideal of the edge subring k[G}. Then

ht(P) = dim F Z e (G). Proof. Let G be a connected graph with q edges and n vertices. Assume that /(G) is minimally generated by the monomials / i , . . . , / ? . There is

an spanning tree T of G so that (after re-ordering) I ( T ) — (fi,...,/n-i). Since I(T) is an ideal of linear type dimfc[G] = tr.degj. k[G] >n-l.

If G is bipartite notice that /& € k ( f i , . . . , / n -i) for k > n; to prove it we write /& = xz and observe that the graph T U {x, z} has a unique cycle

of even length. This shows the equality dim k[G] = n — 1. If G is not bipartite, then for some /& = xz, k > n. Note that the graph T U {x, z} has a unique cycle of odd length. By Corollary 8.2.4 the ideal (fi, •••, f n - i , f k )

is of linear type, hence dimfcfG] > n; to show equality

recall that tr.deg ft k[G]

< n. To finish the proof use Lemma 8.2.11.

n

Corollary 8.2.13 // G is a connected graph with n vertices and k[G] its edge subring, then

.. /, r ,-,-,N ( n if G is not bipartite, and dim (A; , ,, v l GJ ) = <^ [ n — 1 otherwise. Proof. It follows from the proof of Proposition 8.2.12.

D

Monomial Subrings of Graphs_______________________293

The role of the cone of a graph

The cone C(G), over the graph G, is

obtained by adding a new vertex t to G and joining every vertex of G to t. Example 8.2.14 A pentagon and its cone:

C(G) Proposition 8.2.15 Let G be a graph and let C(G) be the cone over G. Then there is an isomorphism

k(C(G)]. Proof. Let V = {xi, . . . , xn} be the vertex set of G and K(I(G))=k[{x1,...,xn,tfi\l
the Rees algebra of the ideal /(G) = (/i , • . • , /?) • We assume that f i , • • • , fq are the monomials in the Xj's corresponding to the edges of G. Let k[C(G)\ = k [ { t x i , fj l
(p:Tl(I(G)) —> k[C(G)], induced by (p(xi) = txt and
D

Exercises 8.2.16 Let G be a connected graph with vertices xi, . . . ,xn and /i, . . . ,/ ? the edge generators of G. Prove that

(a) k ( f i , . . . , f q , x i ) = k(xi,...,xn), and (b) dimk[G][x2l} = n.

Here k(F) denotes the field of fractions of k[F}.

294______________________________________Chapter

8

8.2.17 If G is any non discrete graph with n vertices and C(G) its cone, prove that the edge subring k[C(G)} has dimension n + I .

8.2.18 Let G be a graph with vertices xi,... ,xn and / c R = k[x\,..-, xn] its edge ideal. If ip, tp are the maps of /o-algebras defined by the diagram

R[T^...,Tq]

k[C(G)]

fi

xt

i

where C(G) is the cone of G and f:,..., fq are the edge generators, then ker(y) = ker(^).

8.3

Incidence matrix of a graph

One of the most interesting matrices that have been associated to a graph is its incidence matrix, here we present some properties of this important matrix. The reader is encouraged to consult [136] for a thorough study of the minors of an incidence matrix.

Let G be a simple graph with vertex set V = [xi,... ,xn} and edge set E = {zi,... ,zq}, where every edge Zi is an unordered pair of distinct

vertices z^ = { x i j , X i k } . The incidence matrix MQ = [by] associated to G is the n x q matrix defined by 1

if Xi G Zj, and

Note that each column of MQ has exactly two 1 's and the rest of its entries equal to zero. If Zi = {a;^., Xik} define on = e^. + e^, where e» is the ith canonical vector in R™. Thus the columns of MQ are precisely the vectors ai,...,ag regarded as column vectors. As a simple example consider a triangle G with vertices x i , x % , X 3 . In this case: Mn = with the vectors a\ = 61+6%, a? = 63+63, and as =61+63 corresponding to the edges z\ = {xi,x2}, z? — {x2,x3}, and z3 = { x i , X 3 } . Remark 8.3.1 Let A be a square submatrix of MG- In [136] it is shown

that either det(.A) = 0 or det(.A) = ±2*, for some integer k such that 0 < k < TQ, where TO is the maximum number of vertex disjoint odd cycles in G. Moreover for any such value of k there exists a minor equal to ±2k.

Monomial Subrings of Graphs_______________________295

Recall that a graph G is bipartite if all its cycles are of even length. Thus any tree and in particular any point is a bipartite graph. The number of bipartite connected components of G will be denoted by CQ, and the number of non bipartite connected components will be denoted by c\. Thus c = CQ + Ci is the total number of components of G. Lemma 8.3.2 ([136]) IfG is a graph withn vertices and MG its incidence matrix, then rank(Mc) = n — CQ.

Proof. Let Gi,...,Gc be the connected components of G, and n; the number of vertices of Gi. After permuting the vertices we may assume that MG is a "diagonal" matrix MG = diag(M G l ,...,M G J, where MG; is the incidence matrix of Gi. By Proposition 8.2.12 the rank of MG; is equal to rij — 1 if Gi is bipartite, and is equal to HJ otherwise. Hence the rank of MG is equal to n — CQ. D Definition 8.3.3 Let U be a square matrix with entries in a commutative ring R. The matrix U is called unimodular if det(C7) is a unit in R.

Theorem 8.3.4 ([136]) // G is a graph with n vertices and MG is the incidence matrix of G, then there are unimodular integral matrices U, V such that \

0

0

where

n — c is the number if 1 :s and c\ is the number of 1 's. Proof. See [136, Theorem 3.3].

D

The matrix S is called the Smith normal form of MG and the numbers in the diagonal matrix D are the invariant factors of MG. Corollary 8.3.5 Let G be a graph with vertices x\, . . . ,xn and AG the set of all vectors Ci + e,j such that { x i , X j } is an edge of G. Then

where r — n — CQ is the rank Proof. It follows at once from the fundamental structure theorem of finitely generated modules over a principal ideal domain. See [190, Chapter 3]. D

296______________________________________Chapter

8

Exercises 8.3.6 Let G be a cycle of length n and MQ its incidence matrix. Prove i n,, \ f ±1 det(M v G) = < „ [0

if n is odd ., . if n is even.

8.3.7 Let U be a square matrix with entries in a commutative ring R. Prove that U is unimodular if and only if U is invertible.

8.4

Circuits of a graph and Grobner bases

Let G be a graph. In this section we study the toric ideal P(G) of the edge subring k[G] from the view point of Grobner bases, and examine the elementary integral vectors of the incidence matrix of G. For general results

on Grobner bases of toric ideals a standard reference is [278, Chapter 4]. The elementary integral vectors The notion of elementary integral vector occurs in convex analysis [247] and in the theory of toric ideals of graphs [278, 303].

If a e W1, its support is defined as supp(a) = {i\cti ^0}. Note that a = a+ - a_, where a+ and a_ are two non negative vectors with disjoint support. Remark 8.4.1

If a is a linear combination of j3i,..., J3r € W, then r

supp(a) C |Jsupp(/?i). i=i

Definition 8.4.2 Let N be a linear subspace of Q9. An elementary vector of TV is a nonzero vector a in N whose support is minimal with respect to inclusion, i.e. supp(a) does not properly contain the support of any other nonzero vector in N. The concept of an elementary vector arises in graph theory when N is the kernel of the incidence matrix of a graph G. The reader is referred to

[247, Section 22] for a precise interpretation of the vector in N as flows of G which are conservative at every vertex. Lemma 8.4.3 If N is a linear subspace o/Q9 and a, 0 are two elementary vectors of N with the same support, then a = X/3 for some A € Q.

Monomial Subrings of Graphs_______________________ 297

Proof. If i € supp(a), then one can write a, = Aft for some scalar A. Since supp(a — A/?) C supp(a), one concludes a — A/? = 0, as required. D

Definition 8.4.4 Let N be a linear subspace of Q1 . An elementary integral vector or circuit of N is an elementary vector of N with relatively prime integral entries.

Corollary 8.4.5 If N is a linear subspace o/Q 9 , then number of circuits of N is finite, Proof. It follows from Lemma 8.4.3.

D

Definition 8.4.6 Two vectors a = (a,) and B = (ft) in Q9 are in harmony if aift > 0 for every i.

Lemma 8.4.7 ([246]) Let N be a linear subspace of then there is an elementary vector 7 G N in harmony with a such that supp(7) C supp(a).

Proof. Let (j = (ft, . . . ,/3 9 ) be an elementary vector of N whose support is contained in supp(a). By replacing /? by —j3 one may assume cujft > 0 for some i. Consider a \ i *j - Wo . ~ J

• mm i ^ i^ _

\\ ^~ f-t.

0

Note that ^ — (on - Ajft)aj > 0 for all i. Indeed if a,ft < 0, then clearly Zi > 0, and if ctjft > 0 by the minimality of Xj one has Zi > 0. Thus the vector a — Xj/3 is in harmony with a and its support is strictly contained in the support of a. If a - Xj/3 = 0 we are done, otherwise we apply the same argument with a — Xj/3 playing the role of a. Since being in harmony is an equivalence relation, the result follows by a recursive application of this procedure. D

Theorem 8.4.8 ([246]) If N is a vector subspace of
for some elementary vectors B\,..., ft- of N with r < dim N such that (i)

/ ? i , . . . , ft are in harmony with a,

(ii)

supp(ft) C supp(a) for all i, and

(iii) supp(ft) is not contained in the union of the supports o / f t , . . . ,ft_i for all i > 2.

298______________________________________Chapter

8

Proof. By induction on the number of elements in the support of a. If a. is not an elementary vector of N, then by Lemma 8.4.7 there is an elementary vector ai G N in harmony with a such that supp(ai) C supp(a). Using the proof of Lemma 8.4.7 it follows that there is a positive scalar AI > 0 such that a — \\cti is in harmony with a and the support of a — \\oti is strictly contained in supp(a). Therefore the result follows applying induction. This proof uses the original argument given in [246]. D

Corollary 8.4.9 If N is a linear subspace of Q7, then the circuits of N generate N as a Q-vector space. Theorem 8.4.10 Let M be an n x q integral matrix and ip the Z-linear homomorphism TJj:Z
——J.Z"

given by ijj(a) = M(ot). If M has rank n, then ker('0) is generated as a Z-module by the elementary integral vectors o/ker(i/>). Proof. Set TV = ker(^). Let ai,..., aq be the column vectors of M and e,{ the ith unit vector of Zq. By the proof of [278, Lemma 4.9], the set of circuits of N is the set of vectors of the form

±1 n+l

±1

-^-X)(-1)J • d e t ( a i l , . . . , « , . _ ! , a i j + 1 , . . . , a i n + l ) • eij = — • v, .7 = 1

where 1 < i\ < • • • < in+i < (/, and h is the greatest common divisor of the entries of the vector v, if v is nonzero. For convenience we introduce the notation v = v(ii,... ,in+i) and

[ i i , . . . , i j , . . . , in+i] - d e t ( a i l , . . . , aij_1, aij+1,..,, ain+1). Set d equal to the greatest common divisor of all the nonzero integers having

the special form [ii,...,ij,...,in+i\ and < ii < ••• < in+i < q Note WjB C L C N, where L is the Z-module generated by the elementary integral vectors of N. We claim that ZB = N. First observe that L and N have both rank q - n, see Corollary 8.4.9. Since ZB and L have equal rank, one concludes that N/ZiB is a finite group. Thus to show Z£? = TV it suffices to prove that N/ZB is torsion-free or equivalently that Zq/ZB is torsion-free of rank n. Consider the matrix A whose rows are the vectors of B. By the fundamental theorem of finitely generated abelian groups the proof reduces to

Monomial Subrings of Graphs_______________________299

showing that the ideal /,_ n (A) of Z generated by the (q - n)-minors of A is equal to Z. Fix A = [ii,... ,ij,... ,in+i]/d. For simplicity of notation one may

consider the case A = [ l , . . . , n ] / d . Note that A occurs in v(l,... ,n,i)/d with a ± sign for i = n + 1 , . . . , q. Thus the matrix

A' =

on

•••

ain

A

0

0 ••• 0 \

O21

•••

d2n

0

A

0

•••

o ( ,_ n ) n

0

0

0

V o ( ,_ n ) 1

0

•••

A J

whose ith row is the vector v(1,... ,n,i)/d is a submatrix of A. Therefore A 9 ~™ belongs to Iq-n(A). If 7 9 _ n (A) ^ Z, there is a prime number p dividing A 9 ~ n for all A. Thus p divides A for all A, and this implies that pd divides d, a contradiction. D

On the circuits of a graph Let G be a graph on the vertex set V with edge set E and incidence matrix M. We set N = ker(M), the kernel of M in Q7. Note that a vector a G TV 7 Z 7 determine the subgraph Ga of G having vertex set Va = {xe V x divide fa+} and edge set Ea = {{x,y} G E | xy £ supp(/ a + ) U supp(/ a ~)},

where / i , . . . , fq are the square-free monomials corresponding to the edges of G, that is, the edge generators of G. Note that we are using the following notation: Notation For a = (ct\,..., aq) G N9 and f i , • • • , fq in a commutative ring with identity we set /" = /"* • • • /"'. The support of fa is defined as the set supp(/ a ) = {fi ai ^ 0}.

The next result gives a geometric description of the elementary vectors of the kernel of the incidence matrix of a graph G. Proposition 8.4.11 ([303]) Let G be a graph with incidence matrix M and let N be the kernel of M in Q*. Then a vector a G Z ? n TV is an

elementary vector of N if and only if

(i) Ga is an even cycle, or (ii) Ga is a connected graph consisting of two edge disjoint odd cycles joined by a path.

300______________________________________ Chapter 8

Proof. Assume a e Z9nN is an elementary vector of TV so that Ga is not an even cycle. From the equation fa+ — fa~ we obtain a walk in Ga

w = {x0,xl,...,xn = x0,xn+i,...,xe}, so that xi, . . . ,xt-i are distinct vertices, n is odd and xe is in {xi, . . . ,xe-2}. The walk w can be chosen so that XjXj+i 6 supp(/ a +) if i is even and XiXi+i e supp(/ a ~) otherwise. We claim that xg is not in {xi, . . . , x n _i}.

Assume a^ = xm for some 1 < m < n - 1. If m and £ - n are odd then the monomial walk U>i = {x0,Xi,... ,Xm,X(-i,Xe-2,.

.. ,Xn = X0}

yields a relation /* = / 7 with supp(<5 — 7) properly contained in supp(a), which is a contradiction. If m is odd and I — m is even then we can consider

the walk ^1 — \<£n> -^n-j-1 > • - • 5 ^t — ^m ? ^m-\-l 5 • • • > *^n /

and derive a contradiction, the remaining cases are treated similarly. We may now assume xt = xm for n < m < I — 2. The cycle

must be odd, otherwise Ci would give a relation f s — /7 with supp(<5 — 7)

a proper subset of supp(a). The walk w gives a relation fs = /7 with suppC/ 15 ) C s u p p ( / a + ) and supp(/ 7 ) C supp(/ a -).

Hence supp(<5 — 7) is a subset of supp(a), and since the support of a is minimal one has the equality supp(<5 — 7) = supp(a). Therefore

Va = {xi,...,xe} and Ga is as required. The converse follows from Corollary 8.2.4.

D

Definition 8.4.12 A circuit of a graph G is either an even cycle or a subgraph consisting of two edge disjoint odd cycles joined by a path. Proposition 8.4.13 If G is a graph with incidence matrix M and P is the toric ideal of k[G], then

P = ({Ta+ -Ta-\a&Zq and Ma = 0}). Proof. The result is a consequence of Corollary 7.1.4.

D 3

Let G be a graph with incidence matrix M and let N = ker(M) C Q .

The next example was computed using Macaulay [15], it shows that the set of elementary integral vectors of N does not determine the presentation

ideal of A; [G].

Monomial Subrings of Graphs

301

Example 8.4.14 Let R = k[a,..., o] be a polynomial ring over a field k and let G be the labeled graph

m

n -9—

the toric ideal P of k[G] has height three and k[G] is not Cohen-Macaulay. The toric ideal P is minimally generated by

*7*g*n*i3 — *6*8*io*i2*i4i *i*3*6*8*io*i5*i7 — £2*7*9*11*16*18 *4*9*11*13*15*17 — *5*10*12*14*16*18, *4*6*8*10*15*17 — *5*7*9*11*16*18, *2*4*9*n*13 ~ *1*3*5*10*12*14, *1*3*5*7 — *2*4*6*8,

*1*3*13*15*17 ~ *2*12*14*16*18

*1*3*5*10*15*17 — *2*4*9*11*16*18

*4*6*8*13*15*17 ~ *5*7*12*14*16*18

*1*3*5*7*9*11*13*15*17 ~~ *2*4*6*8*10*12*14*16*18-

Remark 8.4.15 By Proposition 8.4.11 there are, up to sign, exactly nine elementary integral vectors of TV. They are in one to one correspondence with the exponent vectors of the first nine generators given above. Therefore the set

{Ta+ — Ta~ | a is an elementary integral vector of N} is not a generating set for the presentation ideal of k[G]. Corollary 8.4.16 Let G be a graph with incidence matrix M. If A — (Aj) is an elementary integral vector of N = ker(M), then |A, < 2 for all i.

Proof. Let A be an elementary integral vector of N. Since G\ is an even cycle or two edge disjoint cycles joined by a path we obtain an elementary integral vector a with supp(a) = supp(A) and so that oti\ < 2 for all i. Because any two elementary vectors with the same support are dependent we obtain A = a. D

302______________________________________ Chapter 8

Theorem 8.4.17 Let G be a connected graph with q edges and let P be the toric ideal of k[G]. Then the total degree of a polynomial in any reduced Grobner basis of P is less or equal than 2qdimZe(G). Proof. Let M be the incidence matrix of G and let

N = a£Q?

Assume F is the reduced Grobner basis of P and take / = /(T) e F. Using Corollary 7.1.5 and Corollary 8.4.13, together with the fact that normalized

reduced Grobner basis are uniquely determined, we obtain /(T) = Ta — T@ . Notice that supp(a) D supp(/3) = 0, otherwise take i in the intersection

and write f ( T )

= Ti(Ta - Tb). Since Ta - Tb reduces w.r.t F it follows

that /(T) reduces w.r.t F \ {/} which is impossible. We can now write /(T) = Ta+ - Ta~ for some a e N H 1q. By Theorem 8.4.8 we can write a = X];=i ai^i f°r some ai € Q+ , and for some elementary integral vectors A; each in harmony with a and such that supp(Ai) C supp(a) for all i and

r < dim AT. Hence r

a+ = ]T] 1 for some i then CH+ and ttj_ are componentwise larger than Aj_|_ and Aj_ respectively, note that this is not possible by the previous argument.

Hence 0 < a, < 1. Using Corollary 8.4.16 it now follows that the entries of a satisfy on < 2dimZ e (G) for all i, which gives the asserted inequality. D

Let G be a graph with incidence matrix M and N = ker(M) the kernel of M in Q9 . Let AI, . . . , A s are the elementary integral vectors of N. Define the elementary zonotope of N as £?j — [0, AI] + • • • + [0, A s ].

Corollary 8.4.18 If N is the kernel of the incidence matrix of a graph G and P is the toric ideal of k[G], then

U = {Ta+ -Ta~\a££Nn 1q] is a universal Grobner basis for P.

Exercises 8.4.19 Prove that the elementary integral vector of the kernel of the matrix M are precisely the row vectors of the matrix A with a ± sign

Monomial Subrings of Graphs_______________________ 303

8.5

Edge subrings of bipartite planar graphs

It is the goal of this section to extract information about the edge subring k[G], where G is a bipartite planar graph, by studying the particular geometry of the graph G, Our exposition follows closely that of [85] , all the main results and proofs in this section are due to Luisa Doering and Tor Gunston.

Atomic cycles

Let G be a bipartite planar graph with vertex set

and

R = k[xi,...,xn] a polynomial ring over a field k. If j\ , . . . , fq are the edge generators of G, that is, the monomials that correspond to the edges of G, then the presentation ideal P(G) of the edge subring k[G] is the kernel of the homomorphism of ^-algebras induced by the assignment Ti >-)• /j. By abuse of notation, we will sometimes

refer to "the edge Tt" . Fix an embedding of the graph G in the plane. Let c be an even closed walk, c = TJJ , . . . , Tj2r, . For any such walk, write -* c — -^ ii -L is ' ' ' -L izr-i

22

*4

'

i?r '

By Proposition 8.1.2, we have that the generators of P(G) correspond to cycles in G:

P(G)

= ({Tc c is a cycle in G}).

Since G is embedded in the plane, G divides the plane into regions. If c is a closed walk that bounds a region (not including the unbounded region) we say that c is an atomic cycle (even though it need not have distinct

vertices). Let A(G) C P(G) be the ideal generated by

{Tc c is an atomic cycle in G}. As G is embedded in the plane, every cycle in G has a locus in the plane. We denote cycles by lower case letters, and use the same symbol for its locus; the corresponding upper case letter denotes the open region in the plane: a = dA. Similarly, for each atomic cycle Cj, we denote the open region it bounds by Gj.

If G can be written as G = GI U G2, where G\ and G% are subgraphs with at most one vertex in common, then

304______________________________________Chapter

8

and

k[G\ = fc[Gi] ® fc k[G2]. Therefore, we will limit ourselves to graphs which are 2-connected (that is, connected graphs with no cutvertices). Under this assumption, atomic cycles will actually be cycles. For any vector v = (vi,..., vq) we will write Tv for T"1 • • • Tgq. If u and v are vectors with non-negative components, then we define ( u , v ) to be the vector whose ith component is given by (u,v)i = max(0,Wj — Vi). In the sequel, G is a 2-connected bipartite planar graph with a fixed in the plane, n vertices, q edges and r regions. Polarizations

The concept of polarization will allow us to define term

orders which in some respects are the "right ones" to study P(G) and some homomorphic images of P(G). Definition 8.5.1 Let c be an even closed walk of G, c = T^, . . . ,Tit. A polarization of c is the set I-MI i -MS j • • • > -'•it — i J

or the set {-' J 2 i -'141 • • • > -Mi / •

Let {GI ,... ,cr} be the atomic cycles of G. A polarization P of G is a set

where Pi is a polarization of Cj for all i such that

(a) for i ^ j we have PiC\Pj = 0 , and (b) {Ti| TJ ^ PJ for j = 1, . . . ,r} is a polarization of w, where w is the

boundary of the unbounded region. Proposition 8.5.2 ([85]) G has a polarization.

Proof. Let {xi, . . . ,xn} be the vertex set of G. Form a new graph G' as follows. G" has vertex set

{zi, ...,xn,yi,... ,yr} and edge set { { x i , X j } \ { x i , X j } is an edge in G} ( _ ) { { x i , y j } \ Xi is in the atomic cycle Cj}.

Monomial Subrings of Graphs_______________________305

In other words, G' is obtained from G by adding a vertex in each bounded region and connecting it to each vertex in the boundary of that region. We

view G as a subgraph of G'. It is clear that G has a polarization if the plane map determined by G' can be 2-colored. In order to verify this, we let G" be the graph with vertex set {vi\ Vi is a bounded region of G'}

and edge set { { v i , V j } \ Vi is adjacent to Vj in G'}

We claim that G" is bipartite. G" has an embedding in the plane induced from the embedding of G, so we need only check that every atomic cycle of G" is even. The atomic cycles of G" are of two sorts: those corresponding to a vertex Xi and those corresponding to a vertex y ^ . A cycle corresponding

to a vertex j/j is even because G is bipartite. It is easy to check that a cycle corresponding to a vertex Xi is even also. Thus G" is bipartite. D Lemma 8.5.3 IfP = {Pi, • • • ,Pr] is a polarization ofG, then by relabeling one may assume that TI € Pi and that ifl 1. It is not hard to see that there is al least one edge

that appears in only one atomic cycle, that we label cr, and is contained in Pr. Label this edge Tr. Let G' be the graph obtained from G by deleting all edges and vertices that do not appear in any atomic cycle except cr. Then G' is planar and bipartite, and is oriented by

Apply the lemma, labeling with TI through T r _i, and we are done.

D

Proposition 8.5.4 If a,...,cr are the atomic cycles, then T C l , . . . , TCr is a regular sequence. Proof. Assume we have a labeling as in the previous lemma. Let > be the lexicographical ordering on Si with

TI > TI > • • • > Tr > • • • > Tq and let in(—) refer to the initial term or ideal with respect to this ordering.

Let mi = in(T Ci ). Note that

so for i •£ j we have gcd(mi,mj) = 1. Hence the height of ( T C l , . . . ,TCr) is r, as required. D

Chapter 8

306

Example 8.5.5 Let G be the following graph and consider the polarization obtained by taking pairs of dashed lines in each atomic cycle.

Cl

I

T4 C4

Thus with the lex ordering Ti > • • • > T4 > T5 > • • • > Ti2 the binomials

TCl , . . . , TC4 form a Grobner basis for the ideal they generate. Fixing more notation

Choose a polarization P = {Pi,...,Pr} of G.

Assume that the edges are labeled such that Ti e Pi. We employ the following notation:

pf

= {TjlTj appears in a} \ Pi

and for any cycle a, with a = 8A:

Pa

=

{Tj\Tj appears in a and Tj e Pi for some Cj C A}.

Let

N = Pc U

1 < i < r and Tj e P,}.

Set 52 = k[Ui, . . . , Ur] and define a homomorphism of fc-algebras by:

0

if Tk 6 Pc

^ if 7 * e 7 Note ker(
i — cii i • • • j cii

(8.2)

Monomial Subrings of Graphs_______________________ 307

such that ii < 12 < • • • < it and ^nTVi

^

0 for .7 = ! , . . . , / - !

7?£ n pc / 0.

To see that this can always be done, choose an atomic cycle w such that

and label that walk w — cr. Let G' be obtained from G by deleting all those edges that appear in no atomic cycle other than cr. Inductively, label the cycles of G' with c\ , . . . , cr-\ . Let >52 be the reverse lexicographic ordering with respect to this order-

ing, with Ui >s2 U2 >s2 • • • >s2 Ur. Now we define a term order on Si as follows: for monomials a and b in Si, say a >- b if (a) >s2 gj be a term order on Si that is a refinement of >-. Remark. For a cycle a, we write Ta = Ta+ - Ta~ where Ta+ = in> Si (Ta). It is easy to see that T Ta+ = iGrobner Bases

In this part we find Grobner bases for (f>(P(G})

and

4>(A(G)} with respect to the term orders introduced in the previous paragraph. We write g ^* h for the reduction of g to h with respect to a set of polynomials and a term order. For any two cycles a and b, we will write T

=J ' '

a

7

1

- y

a

> &

J1

=

f a ,

+-

Note that Ta^ is the S-polynomial of Ta and T& with respect to >sl. If A is a region in the plane, we write A for the closure of A. Proposition^. 5. 6 Let a and b be cycles. Suppose the connected components of A \ B are AI, . . . ,AS and the connected components of B \ A are

B1,...,Bt. ThenTa,b e (Tai,...,Ta,,Tbl,...,Tbt). Proof. The proof is left as an exercise.

D

Note that since the initial terms of the Ta; and the T\,{ are all relatively prime, they form a Grobner basis for the ideal they generate; see Lemma 2.4.14. Thus, it follows from Proposition 8.5.6 that Taj, ^> 0 with respect to T 0 l ,. . . ,T a .,T t l ) . . .,T 6 ( . Lemma 8.5.7 Let A = k[x\, . . . ,xn] and let > be a term order on A. Suppose mg ~~* 0 with respect to /i, . . . , f s , where g, /i, . . . , fs (E A and m is a monomial such that gcd(m,in>(/j)) = 1 for 1 < i < s. Then g ~-» 0 with respect to f i , . . . , f a .

308

Chapter 8

Proof. The proof is left as exercise.

Q

Theorem 8.5.8 ([85]) M = M(G) = {(T c )|c is a cycle } is a Grobner basis for (/>(P(G)) with respect to >s2Proof. For an exponent v of T we will write
") = U^. Let a and b be cycles. Then

Let

By Buchberger's criterion, it suffices to show that t/ 0] f, ~> 0 with respect to {[/c = 0(TC)| c is a cycle }. By Proposition 8.5.6, T ai & ~> 0 with respect to Tai , . . . , T0s , T1^ , . . . , T^t .

Since every Tai and T&; is a binomial or monomial, every step in the reduction T0ib ^> 0 is a binomial or monomial. Further, since Tu >Sj Tv implies [M") >s2 U^ , for any binomial / £ Si we have <£(/) = 0 or ins a (0(/)) = 0(in S l (/). Thus (f>(Ta

6)

=

with respect to Uai , . . . , Ua, , U^ , . . . , Utt • It can be easily seen that Ua^\4>(Ta^) and the quotient is a monomial Uv such that Ui\Vv implies Ct C Ar\B. Thus, by Lemma 8.5.7, Ua,b ~» 0 with respect to {C/c| c is a cycle }. D Note that it follows from this and from Buchberger's criterion that {Tc c is a cycle in G}

is a Grobner basis for P(G) with respect to >gj. In fact, it is actually a universal Grobner basis, even in the case where G is not planar according to Proposition 8.1.10.

Theorem 8.5.9 ([85]) B = B(G) = {(j>(Tc)\c is an atomic cycle } is a Grobner basis for (f>(A(G)) with respect to >s2-

U\

U\ if Ci contains any edges in P°, ' — Uj • • • Uj if Cj doesn't contain any edges in Pc, 1 k

where Cj1 , . . . , Cjk are the atomic cycles that share an edge with Pf, counting multiplicities. Since in> s (UCi) = U\Pi, the result is clear. D

Monomial Subrings of Graphs_______________________309

Applications As applications of the results of the previous paragraph, we show that N is a system of parameters for k[G]. Corollary 8.5.10 height A(G) = r.

Corollary 8.5.11 ([85]) The image of N in Si/A(G) forms a system of parameters. Proof. Both corollaries follow from the fact that 71

''

r

is ( f / i , . . . , L r)-primary, and the fact that TV has the expected number of

elements. From ,_„,

,,_

.,

,_ ._r_ q _ r^

,_„,

one obtains that TV has q — r elements.

D

Proposition 8.5.12 height P(G) = r.

Proof. It follows from Proposition 8.2.12.

D

Corollary 8.5.13 ([85]) The image of N in Si/P(G) = k[G] forms a system of parameters.

Proposition 8.5.14 If a is a cycle and Ta £ (Tc\ c is a cycle, c ^ a), then a has a chord. Proof. If Ta £ (Tc c is a cycle, c 7^ a), then each term of Ta must be divisible by a term of Tc for some c ^ a. D For the purposes of computing Hilbert functions, we clarify that the various rings involved are graded by

deg(ri) = deg(E/i) = 1 and degfo) = 1/2.

Thus the isomorphism

S1/P(G)^k[G}ck[x1,...,xn] is a graded isomorphism of standard fc-algebras. Corollary 8.5.15 The Hilbert series of k[G]

can be written as

where Q(t) is a polynomial with positive integer coefficients

and

310

Chapter 8

Proof. Let ~k\G] = Si/(N + P(G)) ^ S2/(f>(P(G}). Since the edge subring k[G] is Cohen-Macaulay, the image of N in k[G] is a regular sequence. Thus

We have that in >S2 <^(P(G)) = (u[Pl1,..., Ulp"1, other monomials ). Then we have Q(t) = H-j^^(i) with the desired properties.

D

Exercises 8.5.16 Let G be the bipartite graph

X6

X7

and P(G) the toric ideal of k[G] obtained by mapping iy to XiXj . Prove that P(G) is a Cohen-Macaulay ideal of height 5. If we order the if/ variables in reverse lexicographical order: tl4 > £23 > ^12 > ^56 > ^37 > ^26 > ^34 > ^78 > ^15 > ^48 > ^67 > ^58 j

use this order to show that / = (hi, h%, hs, h^, h§) is a complete intersection

inside -P(G), where

Find an appropriate embedding of G in the plane to explain how this order was obtained and why this order is useful to find a maximal regular sequence of binomials inside P(G).

8.5.17 Let G be a graph that can be written as G = GiUG2, where GI and G2 are subgraphs with at most one vertex in common. If GI is bipartite,

then P(G) = (P(Gi),P(G 2 )) and

Monomial Subrings of Graphs_______________________311 8.5.18 Show that the following embedding in the plane of the bipartite graph consisting of two edge disjoint squares that meet at a point has atomic cycles which are not cycles.

8.6

Normality of bipartite graphs

In this section we characterize the torsion freeness of an edge ideal in graph theoretical terms and study the integral closure of the powers of an edge

ideal. Two key notions to our examination are the even cycles and certain class of configurations of the graph closely related to the circuits of the graph.

Torsion freeness of bipartite graphs We shall prove here that edge ideals of bipartite graphs are normally torsion free. In particular, by Proposition 7.3.17 and Theorem 7.2.35, their Rees algebras are normal Cohen-Macaulay domains. The main technical device is the following: Lemma 8.6.1 Let G be a bipartite graph and let I = I(G) be its edge ideal. If Xi is a vertex in G, then x\Is D (I 5+1 :xi) C /s+1 for all s > 0.

Proof. The proof is by induction on s. The case s = 0 is easy to verify. Assume the containment true for all integers less than s. Assume M is a monomial belonging to 2;i/ s n(/ s + 1 :a;i)\/ s + 1 . There are square-free monomials of degree two MI ,..., Ms, NI ,..., Ns+i in / so that M - MI • • • MSM_ and xiM = NI • • • Ns+iL, for some monomial L and for some monomial M divisible by x\. We set M — x\M\. By an induction hypothesis we may assume M^ • • • Mir ^ Njl • • • Njr for all increasing sequences {ik}, {jk} having values in {1,. . . , s} and for all r < s. For convenience, set Ms+i = Ns+z = 1. Given l < / c < s + l w e claim that there are distinct vertices x\,..., x^k of G so that (after re-ordering the TVj's and M^s) Mi = x^x^i+i for i < k - 1, Ni - x 2 i_iX 2 , for i < k, and Mk • • • MSM = x2kNk+i • • • Ns+iH, for some monomial H.

312______________________________________ Chapter 8

The proof of the claim is by induction on k. If k = I , from the equation XiM = Ni---Ns+iH and using that M ^ Is+1 , we obtain Nt = xix2 for some i. We reorder the Ni's such that N\ = 2:1X2, which shows that M = 0:2^2 • • 'Ns+iH. Assume that the claim is valid for k and consider the equality

Mk • • • MS~M = x2kNk+1 • • • NS+1H. Observe that if x2k divides M, then the equation

M

= (x2x3)---(x2k-2X2k-i')Mk---Ms(xiX2k)'Mi = N1---NkMk---MsW1

(8.3)

implies M e /s+1. Hence x2k divides Mt for some k < i < s; as before we re-order the Mi's so that Mk = x2kz for some vertex z. If z 6 \x\ , • • • , x2k}, then either z = x, with i odd and in this case we have an odd cycle, or i is even and we have a relation of the form M,a • • • Mir = Njl • • • Njr : using that G is a bipartite graph together with our previous assumptions we see that none of these cases can occur. Therefore z = x-^k+i satisfies:

and x2k+iMk+i • • • MSM = Nk+i ---Ns+iH. Since M £ Is+1 the last equation shows that x2k+i divides Nt for some i, say Nk+i = x2k+iw for some vertex w. We set w = x2k+2. Notice that by the last argument w £ {xi, . . . x2k+i}, and the induction on k is complete. We now apply the claim for k = s + 1 to obtain M = x2s+2H. To complete the induction on s, if k = s + 1 then an application of Eq. (8.3) above leads to M 6 7S+1, which is a contradiction to the initial choice of M. Q

Lemma 8.6.2 Let G be a graph and U = {xi, . . . , xr} a subset of the vertex set of G. If A is a minimal vertex cover of G, then NG(U) $£ A, where NG(U) is the neighbor set of U .

Proof. Set ./V = NG(XI, . . . , xr). Assume A is a minimal vertex cover of G such that N c A. Take an edge e of G. If x\ G e, then e = {xi,z} with z e N, thus z e A \ {xi} and e n (A \ {xi}) ^ 0. If x\ £ e, then using e n A ^ 0, one has e n (A \ {zi}) ^ 0. Altogether A \ {xi} is a vertex cover of G, which contradicts the minimality of A. Therefore N cannot be contained in A. D Theorem 8.6.3 ([262]) Let G be a graph and let I = I(G) be its edge ideal. The following conditions are equivalent:

(i) G is bipartite. (ii)

/ is normally torsion free.

Monomial Subrings of Graphs_______________________ 313

Proof. Let V = {x\ , . . . , xn} be the vertex set of G and let

R = fc[x] = k[xi , . . . , xn] be a polynomial ring over a field k. We set m = (x). (i) =£• (ii) We will proceed by induction on n. The ideal / is clearly

normally torsion free for n = 1 and n = 2. Assume / is normally torsion free for bipartite graphs with fewer than n vertices. Let P G Ass(R/Is) for some s. First notice that by Lemma 8.6.1

m ^ (Is: M) for every monomial M in the x variables and for all s > 1. As a consequence m is not an associated prime of R/P for all s. Therefore, P is a face ideal properly contained in m, hence Ip can be written (after a permutation of

the variables) as

IP = (xi,...,xk,I(Gi))P, where GI is a bipartite subgraph of G. Using the induction hypothesis we conclude that PRP € Ass(Rp/IP),

which gives P £ Ass(R/I) and the proof is complete. (ii) =>• (i) Assume G is not bipartite and pick an odd cycle with vertices xi, . . . ,xr. Set k = (r + l)/2 and / = xi • • • xr. Note / ^ Ik, because any monomial in Ik has degree at least 2k. Let

N = {xi,...,xr>xr+i,...,xa} be the neighbor set of {xi, ...,xr}. Note that / e I^k\ because according to Lemma 8.6.2 the linear form s = x\ + • • • + xs is a nonzero divisor of R/I and sf £ Ik. Therefore /* ^ I^k\ a contradiction to Proposition 3.3.26. D Let G be a non bipartite graph and / — I(G) its edge ideal. According to [38] the sets Ass(R/In) will stabilize for large n. A recent article [62] gives a constructive method to find this stable set and present an upper

bound on where the stable set will occur. Corollary 8.6.4 ([262]) Let G be a graph and I = I(G) its edge ideal If G is bipartite, then the Rees algebra Tl(I) is a normal domain. Proof. Note that I(G) is a radical ideal which is generically a complete

intersection and use Theorem 3.3.31.

D

314

Chapter 8

Integral closure We discuss here some typical behavior of the integral closure of the powers of an edge ideal. Proposition 8.6.5 // / is the edge ideal of a graph, then I2 is integrally closed.

Proof. Is left as an exercise.

D

M. Hochster pointed out the next example showing that the proposition does not extend to the next power of /. Example 8.6.6 (Hochster's configuration) Let G be the graph

Note that / = (xix2xz}(x5XQXf)

satisfies / & I3 \ I3.

This example leads naturally to the following concept.

Definition 8.6.7 Let G be a graph. A Hochster configuration of order t (E-configuration for short) in G consists of two odd cycles C2r+i and C^s+i satisfying the following conditions:

(i) C2r+i n N(C2s+i) = 0 and t = r + s + 1. (ii) No chord of either C2r+i or C2s+i is an edge of G. Given such a configuration, we consider the product of the monomials corresponding to the two cycles to produce new generators of the integral closure of the appropriate powers of I(G). Proposition 8.6.8 // / is the_ edge ideal of a graph G admitting an H-

configuration of order t, then I1 ^ /*. Proof. The monomial obtained by multiplying all the variables in the two cycles of the configuration is an element of /* \ /*. Indeed, let 6 = zi • • • z2r+i and 7 = wi • • • w2s+i be the monomials corresponding to the two cycles in the configuration. By definition, we have r + s + 1 = t. Therefore dj 6 Il~l \ /*. On the other hand S2 e_/ 2r+1 and 72 € I2s+1, thus ((57)2 e I^+'+i) = I2t. This shows that £7 e /*. D

Monomial Subrings of Graphs_______________________315

Conjecture 8.6.9 ([262]) The edge ideal of a graph G is normal if and only if G admits no H-configurations.

As it will be seen in the next section this conjecture has a positive answer, but first we need to find a good description for the integral closure of the

edge subring of a graph.

Exercises 8.6.10 Let R = k[xi,x2,x3,2/1,2/2,2/3,2/4] and / = /iJ3 + I3Ji, where Ir (resp. Jr) denote the ideal of R generated by the square-free monomials of degree r in the X{ variables (resp. yi variables). If z = xix-zx^yly^yzy't, then z2 6 I4 a n d ^ G J 2 \ / 2 . 8.6.11 Let G be a graph. Explain the differences between a circuit of G and an ^-configuration of G.

8.7

The integral closure of an edge subring

Let G be a graph and let k[G] be the edge subring of G over an arbitrary field k. In general, it is difficult to certify whether k[G] has a given algebraic property or to obtain a measure of its numerical invariants. This arises

because the number q of monomials is usually large. It is similarly difficult to carry out manipulations such as those required in the Noether normalization process. Our goal here is to unfold a construction for the integral closure or normalization k[G] of k[G], for this purpose the underlying graph theoretic aspects are very helpful.

A combinatorial description of normalizations Let us begin by proving the normality of edge subrings associated to graphs which are nearly bipartite.

Proposition 8.7.1 Let G be a connected graph and k a field. If G has at most one odd cycle, then k[G] is normal.

Proof. We may assume that G has a unique odd cycle, for otherwise G is a bipartite graph and by Theorem 8.6.3 k[G] is normal. Pick an edge e of G which is on the unique odd cycle and set H = G\ {e}. Notice that H is a connected bipartite graph because e is not a bridge (it is on a cycle). Hence k[H] is a normal domain of dimension dim k[G] — 1 (see Proposition 8.2.12). Consider the epimorphism

k[H][t] -^ k[G] —>0, induced by (p(t) = /e,

316______________________________________ Chapter 8

where t is a new indeterminate and fe is the monomial corresponding to e. Since fc[.H][i] and k[G] are both Noetherian domains of the same dimension, f is an isomorphism. Therefore k[G] is normal. D

Lemma 8.7.2 Let G be a graph without even cycles. Then the intersection of any two cycles is either a point or the empty set.

Proof. Let Z\ = {ZQ,ZI, . . . ,zn = z0} and Z2 = {WQ,WI, . . . ,wm = w0} be two distinct cycles of G such that Z\ n Z2 ^ 0. We may assume z0 — w0 and u>i $ Zi . Let j = min{? > l\ Wi G Zi}, and write Wj — zs. If 0 < s < n, considering the cycles {w0,Wi,...,Wj

= Zs,Zs-i,...,Z0}

and {WQ,WI,...,WJ = zs,zs+i,. . . ,zn} of lengths s + j and j + n — s, respectively, we obtain that n is even, which is a contradiction. Therefore s — 0 or s = n. To conclude observe that Zi = {WQ,WI,. . . ,WJ-I,ZQ}, hence Z\ n Z2 = {z0}. D Definition 8.7.3 A graph G having just one cycle is said to be unicyclic.

Definition 8.7.4 Let G be a graph without even cycles. A cycle Z of G is said to be a terminal cycle if N(Z) = Z U {v}, for some v ^ Z.

Lemma 8.7.5 Let G be a connected graph without even cycles. Assume that Zi r\N(Z2) = 0, for any two cycles Zi, Z^ of G. If G is not a unicyclic graph, and deg(i>) > 2 for all v g V(G), then G has at least two terminal cycles. Proof. Let Z\, . . . , Zr be the cycles of the graph G. We now construct a new graph T having vertex set

V(G)\\JZi Thus the vertices of T are the cycles of G, together with the vertices of G not lying on any cycle of G. Two points YI, Y% of T are adjacent if and only

if YI n N(Y-2,) =£ 0, here we regard a vertex t; of G as the one point set {v}. Notice that T is a tree and that the terminal cycles of G correspond to the end points of T. Hence G must have at least two terminal cycles. n

Monomial Subrings of Graphs

317

The integral closure in terms of bow ties

Let G be a graph with

vertex set V = {xi, . . . ,xn} and let R = k [ x i , . . . ,xn] be a polynomial ring over a field k. The set { f i , . . . , f q } will denote the set of all monomials XiXj in R such that {xi,Xj} is an edge of G. Note

k[G] = k [ f 1 , . . . , f g ] c R In general it is difficult to predict which elements lie in the integral closure of an affine domain, but in the case of k[G] one has the following general description: k[G] is also a monomial subalgebra generated by monomials /f — —~£i xl

. . . ™Ai x _—Yx/3 n

,

0^ p t |\m i^

,

with the following two properties (see Theorem 7.2.28):

• fm = Y [ f i i ,

m,bi < E N a n d m > 1.

The first condition asserts that / lies in the field of fractions of k[G], the other being the special form the integrality condition takes for such elements. __ To explain the description of k[G] we begin by exhibiting some monomials in the integral closure of k[G]. A bow tie of a graph G is an induced

subgraph w of G consisting of two edge disjoint odd cycles

Z7 =

Zi = {z0,zi,...,zr = z0} and Z2 = {za,zs+i,... ,zt = zs}, joined by a path {zr,..., zs}. In this case one sets Mw = z\ • • • zrzs+i • • • zt. We observe that Z\ and Z<± are allowed to intersect and that only the variables in the cycles occur in Mw, not those in the path itself. If w is a bow tie of a graph G, as above, then Mw is in the integral closure of k[G]. Indeed if gt = Zi-\Zi, then

1 * .n* i odd * 3 eve r
Hence Mw € k[G\.

and

318______________________________________Chapter

8

Lemma 8.7.6 LetG be a connected graph without even cycles, V the vertex set of G, and / i , . . . , fq the monomials defining the edges of G. Assume the conditions:

(a) Z\ r\N(Zz) = 0, for any two cycles Zi,Z^ ofG, and (b) deg(w) > 2 for all v £ V and G has an even number of vertices.

Then Ylv&v v is in k[B], where B = {/i,..., fq} U {Mw w is a bow tie}. Proof. The proof is by induction on the number of cycles of the graph G.

Set z = Ylvev v' ^ G has exactly two cycles, then G is a bow tie and we can write z = f d , where / 6 k[G] and S = MQ is the product of the vertices in the two cycles. We may now assume that G has at least three cycles. By Lemma 8.7.5 there are two terminal cycles

Zi = {XQ,XI,. .. ,xr = x0},

Z2 = {y0,yi,...,ya =y0}.

There is a path {xr,xr+i,... ,xri}, so that r + 1 < ri, deg(:rj) = 2 for all r < j < ri, and deg(a; ri ) > 3. Likewise there is a path {ya,ya+i,... ,ySl}, so that s + 1 < si, deg(j/j) = 2 for all s < j < s\, and deg(ysi) > 3. We may assume r\ and s\ even, for otherwise, if ri is odd, we write

z=

• (xrxr+i) • • • (x r i _ 2 a;r 1 -i) JJ v,

where V\ = V \ {xi, . . . ,x n _i}, and then use induction. It is not hard to show that {xi,...,xri}n{yi,...,ysi-i} = { z i , . . . , i r i _ i } n { t / i , . . . , 2 / S l } = 0.

Set

o~i = (xr+ixr+2) • • • ( x n _ 2 a ; r i _ i ) and
r

Z = OlCf-2 J| J/i JJ Xi JJ V,

where Vi = V \ {x\, . . . , x r i _ i , y i , . . . , j/ S l _i}, hence using induction we get z 6 k[B]. We may now assume that r\ , Si are even and x ri = ysi . Consider the subgraph

H = G \ {xi , . . . , xri _i , j/i , . . . , ysi -i } . If deg(a; ri ) > 4, then applying the induction hypothesis to H we obtain

z e k[B]. Assume deg(x ri ) = 3. Note that H has a terminal cycle •^3 = {ZQ,ZI, . . . ,Zt = ZQ}

Monomial Subrings of Graphs_______________________319

and there is a path {24,24+1, • • • ,zti}> so that t + 1 < ii, deg(zj) = 2 for all t < j < ti, and d e g ( z t l ) > 3, by the previous arguments we are reduced to the case ri,si,ti even and xri = ysi — z^. Observe that this forces the equality V = {xi,...,xri,yi,...,y8l-i,zi,...,ztl-i}.

To complete the proof we use the identity z - (2i2 2 )(232 4 )--77i-iZti)tfi0' 2 to derive z e k[B].

Q

Given a graph G we set AG = {log(/i),..., log(/ g )}, where / i , . . . , /, are the monomials corresponding to the edges of G and log(/j) is the exponent

vector of /j. The affine subsemigroup of W1 generated by AG will be denoted by CG. Thus CG= Na.

The cone generated by CG will be denoted by K+ CG , it is equal to f

r

K+C'G= lZ) a « Note that fe[G] is equal to the affine semigroup ring k[Co]- This cone will be studied closely in Section 8.8. Lemma 8.7.7 Let GI and GI be two finite sets of monomials in disjoint

sets of indeterminates. Then k[G\ U G?\ ~ k[Gi] ®& &[G2] and

U G2] ~ z's i/ze subring generated by the generators of k[Gi] and A;[G2J.

Proof. Set G = GI U G 2 . By Theorem 7.2.28 we have

k[G\ =fc[{a: a |a

On the other hand there is a decomposition

Using this equality the assertion follows readily.

D

Corollary 8.7.8 Let G be a graph and Gi,...,Gr its connected components. Then k[G] is normal if and only if k[Gi] is normal for each i.

320______________________________________Chapter

8

Theorem 8.7.9 ([263]) Let G be a graph and k a field. Then the integral closure k[G] of k[G] is generated as a k-algebra by the set B = { / i , . . . , fq} U {Mw\ w is a bow tie},

where fi,... , f q denote the monomials defining the edges of G.

Proof. Let V = {xi,... ,xn} and E = {ei,... ,eq} be the vertex set and edge set of G, respectively. According to Theorem 7.2.28 one has:

k[G] = k[{xa\ae The proof is by induction on q, the number of edges of G. The case q = 1 is clear. Assume q > 2 and that the result is true for graphs with less than q edges. If ei = {xj,xk}, we set /, = XjXk.

First we remove all the isolated vertices of G, without affecting the subring k[G]. By Lemma 8.7.7 and the induction hypothesis we may assume

that G is a connected graph. We may also assume that deg(u) > 2 for all v € V; because if deg(w) = 1 and fi — vxj, then setting G' = G\ {e^} we have k[G] = k[G'][fi], the latter being a polynomial ring over k[G'}, hence

k[G] = and fi is not part of a bow tie, which by induction yields the required

equality.

__

Let z = xa 6 k[G] be a minimal generator of the integral closure of k[G].

There is a positive integer m so that zm = /r Al • - • /™ A < = A61 • • • /*« , where A< £ 7L and bt € Z+, Vi.

(8.4)

Observe that if 6, = A; = 0 for some i, then each bowtie of G \ {e^} is a bowtie of G and the induction hypothesis yields z € k[B]. Hence we may further assume that all the monomials fi, • • • , fq must be present in Eq. (8.4). There are two cases to consider. Case (I): Assume that G contains an even cycle Z. We set Vi = log(/j). For convenience we use the edges of G to represent the cycle Z by the sequence {e^ , . . . , 64 }, where two consecutive edges are adjacent. If bir = 0 for some 1 < r < k, then using the relation

= 0, 3 =1

j=l

that comes from the even cycle, we may rewrite Eq. (8.4) as zm = f^

. . ./TO^,

=

fr . . . f b

( where

^ ezv

(8.5)

Monomial Subrings of Graphs_______________________ 321

which by induction yields z € k[B\. We may now assume bir > 0 for all 1 < r < k. We may harmlessly assume that b^ < • • • < 6, fe . Using Eq. (8.4) and Eq. (8.5) we obtain q

k

ma = ^ biVi = 6jj ^ v^ + ^ avi - 26ia ^ v^ + ^ Cji;,, j=i j=i j^ii j even «^*i where the Cj's are non negative integers. The net effect is that now we may

rewrite Eq. (8.4) as zm = Am<51 • • • f™S< = ft1 • • • f^, where fc e Z, d< € Z+, for all i, and djj = (5^ = 0 . Hence by induction z £

Case (II): Assume that all the cycles of G are odd. Let Z\ and Z-^ be two distinct cycles of G, represented by the sequences of edges {e^ , . . • , ejfc } and {e^j, . . . ,64}, respectively. If Zi r\ Z^ ^ 0, then according to Lemma 8.7.2 Zi r\ Zz = {x}, for some x 6 V, say z € supp(/j a ) D supp(/^). Using the identity H

«<: + S

j odd

j even

W

^' =

£

U

*3- + I] ^

j even

j odd

we may proceed as in Case (I). If Z\ D ^2 = 0 and N(Z{) C\ Z% ^ $, say /t = xy, with x 6 supp(/;J and y e supp(/« 1 ). Again using the equality

j odd

j odd

j even

j even

we may proceed as in Case (I) to derive z 6 k[B]. Therefore we are reduced to the case N(Zi) n Z^ = 0, for any two cycles Zi,Z^. If Xj is not in the support of /f 1 • • • / , " , for some j, consider the set D =

Then E i6jD A, = 0 and 6j = 0, for all i e D. Note that, for any z £ D, we may eliminate the monomial /j from Eq. (8.4), hence by induction we get z € k[B}. As a consequence we are reduced to the case supp(z m ) = V, that is, z = x"1 • • • x^n , with di > 1 for all i. By Corollary 7.7.17 we obtain that k[G] is generated as a /c-algebra by monomials of degree at most 2 ( n — |"f ])•

Altogether we get

n < deg(z) < 2 (n -[!]), hence n must be even and z = x\ • • • xn. Using Lemma 8.7.6 we conclude

that z € k[B], as required.

D

322

Chapter 8

There is a version of Theorem 8.7.9, due to Hibi and Ohsugi [164], that allows loops in the graph G. Next we clarify that there are two special types of bowties that are

irrelevant for the description of the integral closure of k[G]. Proposition 8.7.10 Let w be a bowtie consisting of two edge disjoint odd cycles Z\, Z-z that meet at a vertex or are connected by an edge, that is, w has essentially one of the forms:

then Mw is in k[G\.

Proof. Left as an exercise.

D

As an immediate consequence of Theorem 8.7.9 and Proposition 8.7.10 one has the following full characterization of when k[G] is integrally closed.

Corollary 8.7.11 Let G be a connected graph. Then k[G] is normal if and only if for any two edge disjoint odd cycles Zi,Z% either Zi and Z% have a common vertex, or Z\ and Z^ are connected by an edge. Thus the smallest connected graph G such that k[G] is non normal is the Hochster configuration of Example 8.6.6, consisting of a graph G with two disjoint triangles joined by a 2-path. The corollary above prompts the following:

Definition 8.7.12 A graph G is said to satisfy the odd cycle condition if every two vertex disjoint odd cycles of G can be joined by an edge in G.

The odd cycle condition has come up in the literature in connection with the normality of edge subrings [164, 262, 263] and the description of the circuits of a graph [247, 303]; it also occurred before [119]. It could be said that this notion has been rediscovered during the study of monomial subrings of graphs. Corollary 8.7.13 Let G be a connected graph and I = I(G} its edge ideal. Then k[G] is normal if and only if the Rees algebra 11(1} is normal.

Monomial Subrings of Graphs_______________________ 323

Proof. =») Let C(G) be the cone over the graph G; it is obtained by adding a new vertex t to G and joining every vertex of G to t. According to Proposition 8.2.15 there is a isomorphism

1l(I(G)) ~ k[C(G}}. Let Mw be a bowtie of G(G) with edge disjoint cycles Z\ and Z^ joined by the path P, it is enough to verify that Mw e fc[C(G)]. If t £ Zi U Z2 U P, then ID is a bowtie of G and Mw 6 fc[G]. Assume t £ Zi\J Z%, say i € Zi. If Zi n Z2 ^ 0, then M™ € A;[C(G)]. On the other hand if Zrr\Z2 = 0, then Mw € /c[G(G)] because in this case Z\ and Zi are joined by the edge {t, z}, where z is any vertex in Z%. It remains to consider the case t ^ Z\ U Zi and i G P, since G is connected there is a path in G joining Z\ with Zi. Therefore Mw = MWl for some bowtie wi of G and Mw G k[G}. •4=) Conversely if 7Z(/(G)) is normal, then by Proposition 7.4.1 we obtain that k[G] is normal. D

In the corollary above the hypothesis that G is connected is essential to prove that fe[G] normal imply Tl(I) normal; for instance if G consists of two disjoint triangles, then k[G] is a polynomial ring and Tt(I) is non normal.

We give the following counter-example to show that in general it is not true that normality of the Rees algebra implies normality of the monomial subring. Example 8.7.14 Let F = { f 1 : . . . , /6} and / = (F), where /I = XiX-2,/2

= X3X4X5,f3

= XiX3,f^

= X2X4,f5

= X 2 X 5 , / 6 = XiX5.

Then

is clearly not normal but 'R.(I) is normal by the Jacobian criterion.

Proposition 8.7.15 Let G be a graph and let GI, . . . , Gr be the connected components of G. Then Ti(I(G}) is normal if and only if either: (i) G is bipartite, or (ii) exactly one of the components, say GI, is non bipartite and K[G\] is a normal domain.

Proof. One may proceed as in the proof of Corollary 8.7.13.

D

Corollary 8.7.16 ([262]) The edge ideal of a graph G is normal if and only if G admits no H- configurations.

324______________________________________Chapter

8

Remark 8.7.17 (a) Let G be a graph consisting of two disjoint triangles.

Note that G itself is an //-configuration but G is not contained in a bowtie, since the two triangles cannot be joined by a path. (b) If two disjoint odd cycles Zi, Z2 of a graph G are in the same connected component, then the two cycles form an //-configuration if and only if Zi, Z2 cannot be connected by an edge of G and Zi, Z2 have no chords in G Integral closure of Rees algebras Let G be a graph with vertices x i , . . . , x n and / = I(G) its edge ideal. Consider the endomorphism <j>

of the field f c ( x i , . . . :xn,t) defined by x, >->• xrf, t i-> t~2. It induces an isomorphism

R[It] i—> k[C(G)],

Xi*-^rtxi,

XiXjt i-> XjXj,

where /? = f c [ x i , . . . , x n ] is a polynomial ring over a field k and C(G) is the cone over G obtained by adding a new variable t. We thank Wolmer Vasconcelos for showing us how to get the description of the integral closure of R[It] given below. First we describe the integral

closure of k[C(G)}. If Zi = { x i , X 2 , . . . , xr = xi} and Z-^ = {zi,z%,... ,zs = z\] are two edge disjoint odd cycles in C(G). Then one has

Mw = xi • • • xrzi • • • zs for some bow tie w of C(G), this follows readily because C(G) is a connected graph. In particular Mw is in the integral closure of k[C(G)]. Observe that if t occurs in Z\ or Z2, then Mw e k[C(G)] because in this case either Zi and Z2 meet at a point, or Z\ and Z2 are joined by an edge in C(G); see Proposition 8.7.10. Thus in order to compute the integral closure of k[C(G)] the only bow ties that matter are those defining the set: B — {Mw\ w is a bowtie in C(G) such that t $• supp(Mu,)}. Therefore by Theorem 8.7.9 we have proved:

Proposition 8.7.18 k[C(G)] = k[C(G)][B]. To obtain a description of the integral closure of the Rees algebra of / note that using (j> together with the equality r+1 2

s-1 2

Mw = ———————————————

tei

Monomial Subrings of Graphs

325

where xr+i = xi, we see that Mw is mapped back in the Rees algebra in the element T-+8

Mwt

2

r+3

= XiX2 • • - X T Z ] _ Z - i • • • Zst

2

.

If B' is the set of all monomials of this form, then one obtains: Proposition 8.7.19 R[It] = R[It][B'].

Exercises 8.7.20 Prove that k[G] is normal for every connected graph G with six

vertices, where A; is a field. 8.7.21 Let G be a graph on the vertex set V = {xi,... ,xn} and choose

a new set of vertices j / i , . . . ,yn, define S(G), the suspension of G, to be the graph obtained by attaching to each vertex x^ the new vertex j/j and the edge {xi,yi}. If /(G) is a normal ideal, then J(G(G)) and I(S(G)) are normal ideals. 8.7.22 Let G be a graph and let m — (r + s)/2 be the least positive integer such that G has two vertex disjoint odd cycles of lengths r and s. If / = /(G) is the edge ideal of G, prove that /* is integrally closed for i < m.

8.8

The equations of the edge cone

Let G be a graph on the vertex set V = {v\,... ,vn} and R — k[xi,...,xn] a polynomial ring over a field k. The edge cone of G is the cone K_|_.4 C M" spanned by the set A of all vectors 6j + e,j such that Vi is adjacent to Vj, where ej denotes the ith unit vector. In this section we present a combinatorial description of the facets (faces of maximal dimension) of ffi_)_^4 when k[G] has dimension n. We show some estimates for the a-invariant of k[G], for certain G. The problem of finding upper bounds for the a-invariant of more general normal monomial subrings generated by square-free monomials of the same degree was addressed in Chapter 7.

The irreducible representation of the edge cone The computation of the equations defining the facets of the edge cone of a graph G is a crucial step toward the effective determination of the canonical

module and the a-invariant of k[G], whenever k[G] is a normal domain. For simplicity we have focus our attention on connected non bipartite graphs, but the methods can be applied to bipartite graphs as well.

326______________________________________ Chapter 8

Definition 8.8.1 The support of a 6 R" is defined as

supp(a) = {ai| a; ^ 0}. Lemma 8.8.2 Let G be a graph with connected components Gi,...,Gr and M its incidence matrix. If GI is a tree with at least two vertices and (j?2, . . . , G> are unicyclic non bipartite graphs, then ker(M*) = (/3) for some vector j3 whose support is contained in {0, 1, —1}.

Proof. Let V = {i>i, . . . ,vn} be the vertex set of G. If G is a tree, using induction on the number of vertices it is not hard to show that ker(M f ) = (/?), where supp(/3) = {!,-!}. On the other hand if G — Gi for some i > 2, then M* is non singular and ker(M*) = (0). The general case follows readily by writing M* as a "diagonal" matrix

where Mai is the incidence matrix of the graph Gi. Observe the equality

n The facets of the edge cone Let us introduce some more terminology and fix the notation that will be used throughout. To ease the reading we recall some of the relevant notation on polyhedral geometry; see Chapter 7.

We set AG (or simply A if G is understood) equal to the set {cci,... ,aq} of row vectors of the transpose of the incidence matrix MQ of G. The edge cone IR+,4 of G is defined as the cone generated by A, that is: a; e K+ for alH > . J

l~ By Lemma 8.3.2 one has

n - c0(G) = dim k[G] = rank (MG) = dim E+A,

where co(G) is the number of bipartite components of G. If a € K n , a ^ 0, then the set Ha will denote the hyperplane of Rn through the origin with normal vector a, that is,

Ha = {x 6 En (x,a) = 0}. This hyperplane determines two closed half-spaces H+ = {x £ K™ (a;,a) > 0} and

H~ = {x 6 R n | (x,a) < 0}.

A subset F C K™ is a proper face of the edge cone of G if there is a supporting hyperplane Ha such that __

TTT)

A f~\

TT

_]_ CK

— M_j_ .A I 1 ±i-d 7^ W ?

Ttl)

A

/•—

TT —

-1^-)- ^H. C_ .oa ,

Monomial Subrings of Graphs_______________________ 327

Definition 8.8.3 A proper face F of the edge cone is a facet if dim F = dim E+A- 1.

When the edge cone is properly embedded into K" it will turn out that its facets are of two kinds: those defined by independent sets (see definition

below) and those defined by supporting hyperplanes of the form Hei . Lemma 8.8.4 // the incidence matrix of G has maximal rank n, then the

set F = Hei n K+.4 is a proper face of the edge cone. Proof. The proof is left as an exercise.

D

Let A be an independent set of vertices of G. The supporting hyperplane of the edge cone defined by

ViSA

v,eN(A)

will be denoted by HALemma 8.8.5 // the incidence matrix of G has maximal rank n and A is an independent set of vertices of G , then the set F = M^ACiHA is a proper face of the edge cone.

Proof. The proof is left as an exercise.

D

To determine whether HA defines a facet of R_|- A consider the subgraph L — LI U L-2, where LI is the subgraph of G with vertex set and edge set

V(Li) = A U N(A) and E(Li) = {z e E(G)\ z n A ^ 0} respectively, and L2 =<5> is the subgraph of G induced by S = V\V(Li). The vectors in A D HA correspond precisely to the edges of L.

Remark 8.8.6 Let F be a facet of ffi+ A defined by the hyperplane Ha. If the rank of MQ is n, then it is not hard to see that Ha is generated by a linearly independent subset of A. See Corollary 7.2.22. Theorem 8.8.7 ([305]) Let G be a graph with vertex set V = {v\, . . . ,vn} and M its incidence matrix. If rank(M) = n, then F is a facet of R+ A if and only if either

(a) F = Hei n R+.4 for some i, where all the connected components of G \ {vi} are non bipartite graphs, or

(b) F = HA r\E+A for some independent set A C V such that LI is a connected bipartite graph, and the connected components of L% are non bipartite graphs.

328______________________________________ Chapter 8

Proof. Let A = {a\ , . . . , aq} be the set of column vectors of the incidence

matrix of G. Assume that F is a facet of K+yl. By Remark 8.8.6 we may assume that there is a 6 Kn such that

(i) F = ^AnHa,

(a,ai) < 0 Vie {!,..., g}, and

(ii) a i , . . . , a n -i are linearly independent vectors in Ha.

Consider the subgraph D of G whose edges correspond to a\ , . . . , an-i and its vertex set is the union of the vertices in the edges of D. Let F be the transpose of the incidence matrix of D, and D\, . . . ,Dr the connected components of D. Without loss of generality one may assume 0

r=

...

0

o r2 ... o . 0

, rank(F) = n — 1,

0

where I\ is the transpose of the incidence matrix of D{. We set ni and qi equal to the number of vertices and edges of -Dj respectively. Let 0 be the matrix with rows c c i , . . . ,an-i- We consider two cases. Case (I): Assume 0 has a zero column, hence it has exactly one zero column and D has vertex set V(D) = V\ {vi}, for some i. This implies

Since rij < g, for all i, one obtains n; = qi for all i. Note that Co(D) = 0, by

Lemma 8.3.2. Therefore Di,...,Dr are unicyclic non bipartite graphs. As any component of G \ {v^} contains a component of D, we are in case (a). Case (II): Assume that all the columns of 0 are non zero, that is, D has vertex set equal to V. Note that in this case

n=> ^ = 1 + /> / .j

j

. qi •*-

(8.6) * • '

Since rank (T) — n- c0(D) = n - 1, we get that D has exactly one bipartite component, say D\. Hence n\ - 1 < qi and m < qi for all i > 2, which together with Eq. (8.6) yields qi = m — 1 and rij = qi for all i > 2. Altogether DI is a tree and £ > 2 , . . . , Dr are unicyclic non bipartite graphs. By Lemma 8.8.2 there is j3 e Rra such that

ker(r) = (/?) and supp(/3) C {0,1, -1}. We may harmlessly assume a = /?, because ff a = Hp and a G (/3). Set

A = {viE V\ ^ = 1} and B = {^ e V| a» = -1}.

Monomial Subrings of Graphs_______________________ 329

Note that A ^ 0, because D has no isolated vertices and its vertex set is V. We will show that A is an independent set in G and B = N(A), where the neighbor set of A is taken w.r.t. G. On the contrary if {vi,Vj} is an edge of G for some Vi,Vj in A, then O-k = Ci + ej and by (i) we get (a, ctk) < 0, which is impossible because (a,ak) = 2. This proves that A is an independent set.

Next we show N(A) = B. If Vi 6 N(A), then a& = ei + BJ for some Vj in A, using (i) we obtain (a,ak) = a» + 1 < 0 and a, = — 1, hence w, € B. Conversely if Vi G B, since D has no isolated vertices, there i s l < f c < n — 1 so that a/f = ei + BJ, for some j, by (ii) we obtain (a, a^} = — 1 + a, = 0, which shows that Vj e A and Vi £ Af(.A).

Prom the proof of Lemma 8. 8. 2 we obtain V(Di) = A\JN(A). Therefore a i , . . . , a n _ i are in HA and .ff0 = HA- Observe that LI (see notation before Remark 8.8.6) cannot contain odd cycles because A is independent and every edge of L\ contains a vertex in A. Hence LI is bipartite, that LI is connected follows from the fact that D\ is an spanning tree of LI . Note that every connected component of LI is non bipartite because it contains some Di, with i > 2. Conversely assume that F is as in (a). Since the vectors in A fl He.

correspond precisely to the edges of G \ {i>j} one obtains that F is a facet of the edge cone. If F is as in (b), a similar argument shows that F is also a facet. D

Corollary 8.8.8 ([305]) Let G be a connected non bipartite graph with vertex set V = { V I , . . . , V H } . Then a vector x = ( X I , . . . , X H ) 6 R™ is in R_i_»4 if and only if x is a feasible solution of the system of linear inequalities —Xi 5Zu-g.A

Xi

~ SireJvM)

Xi

< 0, i = 1, . . . ,n — ^' for

a

^ independent sets A C V.

Proof. Let

be the irreducible representation of the edge cone as an intersection of closed half-spaces. By Theorem 7.2.18 the set Hbi n R+A is a facet of E+./4, and we may apply Theorem 8.8.7.

D

Proposition 8.8.9 ([305]) Let G be a connected non bipartite graph with n vertices. If k[G] is a normal domain, then a monomial x^1 • • • x^n belongs to k[G] if and only if the following two conditions hold

(i) /? = (/?i , . . . , /?„) is in the edge cone of G, and (ii) J]™=i A *s an el>en integer.

Chapter 8

330

Proof. Set A = AG = {ai, • • • , aq}. Assume /? £ 1+ A and deg(x^) even. We proceed by induction on deg(x/3). Using Corollary 8.3.5 one has the isomorphism hence 2/3 e 1+.4 n Z.4 = N.4 and one may write

2/3 = 2

+

;,

Sj e N and e* € {0, 1},

by induction one may assume ^'=1 Sjaj = 0. Therefore from the equality

above one concludes that the subgraph whose edges are defined by the set {Q!J| Cj = 1} is an edge disjoint union of cycles Zi, . . . , Zr. By induction one may further assume that all the Zj's are odd cycles. Note r > 2, because deg(xb) is even. As G is connected, using Corollary 8.7.11 it follows that x® € k[G]. The converse is clear because k[G] is normal. D Remark 8.8.10 Let us give a simple example to see that the connectivity hypothesis is essential in Proposition 8.8.9. Consider the graph G = with six vertices consisting of two disjoint triangles:

The vector /3 = ( 1 , . . . , 1) belongs to the edge cone because 2/3 6 K14, but x13 is not in k[G]. Let G be a connected non bipartite graph with k[G] normal. The last results can be used to give conditions for a graph to have a perfect matching.

Proposition 8.8.11 Let G be a connected graph with n vertices. If n is even and k[G] is normal of dimension n, then x\- • -xn is in k[G] if and only if \A\ < \N(A)\ for every independent set of vertices A of G. Proof. =>•) Since k[G] is normal:

Hence a = ( 1 , . . . , 1) = ( a i , . . . , a n ) is in M+Aa- Using Corollary 8.8.8 we get that the vector a satisfies the inequalities:

ai=\N(A)\

Monomial Subrings of Graphs_______________________ 331

for every independent set of vertices A of G, as required.

<=) First we use Corollary 8.8.8 to conclude that a is in the edge cone, then apply Proposition 8.8.9 to get a & k[G].

O

Corollary 8.8.12 (Generalized marriage problem) Let G be a graph with an even number of vertices. If G is connected and satisfies the odd cycle condition, then the following are equivalent

(a) G has a perfect matching.

(b) |^4| < |-W(-A)| for all A independent set of vertices of G. Proof. The proof follows using Theorem 6.1.8 and Proposition 8.8.11.

D

Corollary 8.8.13 ([51]) Let G = JCn be the complete graph on n vertices. Then a vector x 6 E™ is in E+,4 if and only if x = (xi, . . . ,£„) satisfies

—Xi

<

0,

i = 1, . . . ,n

In addition if n > 4, these equations define all the facets ofR+A. Example 8.8.14 If G is a triangle with vertices {xi,xi,xz}, then the edge cone of G has three facets defined by Xi < Xi + X3,

Xi < X\ + £3,

X3 < Xi + Xi.

Note that x\ > 0, x% > 0 and £3 > 0 define proper faces of dimension 1.

The a-invariant of some normal subrings Let X be an arbitrary subset of E". The relative interior of X, denoted ri(X), is the interior of X relative to aff(X), the affine hull of X. As usual the interior of X in E™ is denoted by X°. We grade the edge subring k[G] C R of a graph G with the normalized grading

k[G]i = k[G] n Ihi, where i=0

is a polynomial ring over the field k with the standard grading. Recall that k[G] is normal if and only if N^4e = R+As n ZAa, this follows from Corollary 7.2.29. If k[G] is normal, then by Theorem 7.2.34,

the canonical module uik[G] °f k[G\ is given by

w*[G] = (Kl a e NAG n ri(E+ A;)}). This formula will be used throughout to estimate some a-invariants.

(8.7)

332

Chapter 8

Remark 8.8.15 Let G be a connected non bipartite graph and take a vector fi in ri(ffi+.4 G ). If F is a proper face of l+A?, then /3 £ F. This

observation holds for all cones because a proper face is always contained in the boundary [39, Theorems 5.3]. In particular ft > 0 for all i, because Hei n K+ AG is a proper face. Definition 8.8.16 The join GI * G2 of two disjoint graphs G\ and £2

consists of GI U G2 and all the lines joining GI with G 2 .

Proposition 8.8.17 Let GI andG% be two connected graphs onn vertices. If k[Gi\ and fc[G 2 ] are normal, then

a(k[Gi * GI}) — -n. Proof. Let V\ = {vi,... ,vn} and V^ = {vn+i,... ,V2n} be the vertex set of GI and G2 respectively. One may assume n > 2. Set G = GI * G 2 . Note that the vector /3 = ( 1 , . . . , 1) <E K2n satisfies

2_^

:;<-! + Vi€A

Xi

and

x, > 1,

Vi<EN(A)

for all independent sets A of G, because |.A| < n — 1 < |7V(.A)|. Hence by Corollary 8.8.8 and Lemma 8.3.2 we get /? e (t+,4G)0 n R4G. Since fc[G] is normal by Corollary 8.7.11, we apply Eq. (8.7) to get x@ £ Uk[G\- This implies -n < a(k[G\). On the other hand by Corollary 7.7.15, we conclude a(k[G]) = -n. D Proposition 8.8.18 Let G be a connected non bipartite graph on n vertices. If k[G] is normal, then

a(k(G}) - 1 < a(k[C(G)V < - [^1 , I z I where \x] is the least integer greater or equal than x. Proof. Let u>i and w2 be the canonical modules of k[G] and k[C(G)] respectively. Set A = AG and V — V(G). By Lemma 8.3.2 one has )° C En and

r i ( M + ^ c G ) = (1^^ C G ) 0 C

jn+l

As k[G] is normal and according to Eq.(8.7), we may pick j3 & (^+A so that deg(x^) is equal to —a(k[G]). Set 7 = (/3i + l , / 3 2 , . . . ,/? n , 1). To prove the first inequality we will show that 7 is in (E^-Ac(G))° H Ny4c( G ), that is, one must show that 7 is not on any facet of HL±Ac(G)Let -F be a facet of E^Ac(G) and H the hyperplane defining F. By Theorem 8.8.7 there are three cases to consider. If H = Hei for some

Monomial Subrings of Graphs_______________________ 333

1 < i < n, then 7 ^ H ; for otherwise /? would be on the proper face of K+.-4

determined by Hei , which contradicts the choice of/? (see Remark 8.8.15). If H = HSn+l, note (7, e n+ i) = 1 and 7 0 H. Assume that H — HA, for some independent set in C(G). If vn+i s A, then 7 satisfies < -1+

and 7 0 HA- Assume v n +i ^ A, since /? satisfies the inequality

it follows that 7 satisfies

Hence 7 $ jff^. Observe that k[C(G)] is normal by Corollary 8.7.11. Thus using Eq.(8.7) one obtains a;7 G 6J2 and the proof of the first inequality is complete. The second inequality follows from Corollary 7.7.15. D

Proposition 8.8.19 Let G\ and GI be two connected non bipartite graphs.

If k[Gi] and k[G-2\ are normal, then

a(k[Gi]) + a(k[G2}) < a( Proof. Choose vectors a and b in

(M+AarfnNAa!

and ( K ^

G

°n

respectively, such that

deg(x a ) = -a(fc[Gi]) and deg^6) = -a(Jb[G 2 ]). Using arguments similar to those given in the proof of Proposition 8.8.18 it follows that (a, b) is an interior point of M + ^4G 1 *G 2 > which implies the asserted inequality. D

Exercises 8.8.20 Let G = G\ U GI be a graph with two non bipartite connected components GI and GI . Find a relation between the irreducible representation of the cone K_|_ AG and the irreducible representations of the cones R+ AG\ and

Chapter 8

334

8.8.21 Let G be a connected non bipartite graph with n vertices. If k[G] is a normal domain, then x\ • • • xn is in k[G] if and only if ( 1 , . . . , 1) is in the

edge cone of G and n is even. 8.8.22 Prove that the following graph does not have a perfect matching by

exhibiting an independent set A such that |.A| ^ |7V(.A)|.

8.8.23 Prove that the following are the equations of the edge cone of the graph G shown on the left.

<

0

— z4 < 0 -z5 < 0 Z3

<

Zi hZ2 -1-

Z2

<

Zi -h Z3 -1-

Z4 1- Z5

<

Z2 -hz3

Z5

Show that the a-invariant of k[G]

is equal to -4.

z4 -Hzs z4 -Hz5

Chapter 9

Semigroup Rings of Complete Graphs The purpose of this chapter is to study some of the properties of edge subrings of complete graphs, including the bipartite case. Special attention is given to the computation of their Hilbert series, Grobner bases and Noether normalizations.

9.1

Monomial subrings of bipartite graphs

Here we treat two special types of monomial subrings attached to complete bipartite graphs and compute their Hilbert series and a-invariant. First we study the edge subring of a complete bipartite graph, and then we examine the Rees algebra of the edge ideal of a complete bipartite graph.

The monomial subring of a bipartite graph One of the results that simplify the study of an edge subring associated to a complete bipartite graphs is the fact that its toric ideal is a determinantal prime ideal. Let us begin with a classical formula for determinantal ideals. Theorem 9.1.1 Let X = (xij) be an m x n matrix of indeterminates over a field K and I t ( X ) the ideal generated by the t-minors of X. If m < n and 1 < t < m + 1, then

height(/ t (X)) = (m-t+l)(n-t+l). Proof. See [52, Theorem 2.5].

D

336

Chapter 9

Proposition 9.1.2 Let K be afield and K.m>n the complete bipartite graph with vertex set V = {xi, . . . ,xm,yi, . . . , yn}. Then the toric ideal P of -m,n] is generated by the 2 x 2 minors of a generic m x n matrix. Proof. Let

A = K[{Tij 1 < i < TO, 1 < j < n}} be a polynomial ring over a field K in the indeterminates TJJ and consider the graded homomorphism of fi'-algebras

•0: A —> K[lCm,n] = K[xiyj's],

ip(Ti:j) = xtyj.

We claim that the kernel of tp is generated by the 2 x 2 minors of a generic mxn matrix. Indeed let X = (T^). It is clear that the ideal I z ( X ) , generated by the 2 x 2 minors of X, is contained in P = ker(^j). On the other hand, since the graph is bipartite one has

dim K[Km,n] = m + n — I, see Proposition 8.2.12. Therefore height(P) = (mn) - (m + n - 1) = (m - l)(n - 1) = height (J 2 PO), the latter equality by Theorem 9.1.1. Since they are both prime ideals, we have I2(X) = P. Q The Cohen-Macaulayness and Gorensteiness of algebras that include K[£m,n], has been dealt with in great detail in [50]. Proposition 9.1.3 Let K.m,n be the complete bipartite graph and A/P the toric ideal of K [K.m,n]- Ifn>m, then the Hilbert series of A/P is given by m—l /

..

'm -I _

7\n+m — 1

Proof. Let V\ — {x\ , . . . , xm] and Vz = {yi , • • • ,yn} be two disjoint sets of vertices, and R = K[Vi U Vz] a polynomial ring over a field K. Let

be a polynomial ring in the indeterminates TIJ and consider the graded homomorphism of AT-algebras

V>: A —> K[tCmtn} = K[xiyfs\,

^(T^) = Xiyj.

By Proposition 9.1.2 the toric ideal P = ker(V') is equal to I^X], the ideal generated by the 2-minors of the mxn matrix X = (T^ ) . Note that the

Semigroup Rings of Complete Graphs_________________ 337

dimension of K[K.m,n] is equal tom + n — l. The set of the 2 x 2 minors of X form a Grobner basis for I^(X} with respect to the lexicographical ordering induced by TU > • • • > Tin > • • • > Tmi > • • • > Tmn, see [59, 276]. Note that in(P), the initial ideal of P, is generated by the

2-diagonals of X, that is, one has in(P)

= (

By Corollary 2.4.13 one has

thus the proof reduces to compute F(A/'\n.(P),z). Consider the arrays:

and

Let If and Ji be the ideals of A generated by the 2-diagonals of Xt and YI respectively. Note in(P) = /„. For 2 < i < n there are graded exact sequences

0 —> A/Kt[-l] ^4 A/(Tm(l+1),. . .,Tmn, It) —> A / (Tmi, • • • ,Tmn,h-i) —> 0, where

Kt = (Tij | i < m, j < 1) + Jt + (Tm(e+1), . . . , Tmn). As T m i, . . . ,Tmi is a regular sequence on A/Kf, it is not hard to see that one can write PI A IK-

v\

fr(m-l)(n-l+l)

F(A/Kt,z) = (1 _ ^ m+n _! , where and

^ =K^\ l <m^< 3],

338

Chapter 9

that is, the polynomial /i( m _i)( n _^+i) determines the /i-vector of Hence

F(A/In,z)

=

Solving this recursive formula for the Hilbert series of A/In rapidly leads to the required identity.

D

Corollary 9.1.4 Let K,m,n be the complete bipartite graph. If n >m, then the a-invariant and the multiplicity of K[K,min} are given by:

a(K[K,m
and e(K[K.m,n]) =

/ui -i- fi — 2 \

771 — 1

Proof. Use Remark 4.1.11.

D

Another approach to the computation of the a-invariant is through the theory of Segre products, and then appealing directly to the results of [132]. Useful methods to compute the a-invariant of some interesting families of algebras, e.g. algebras with straightening laws or Stanley-Reisner rings, are introduced in [42].

Let us state two general results on determinantal rings generalizing Proposition 9.1.3 and Corollary 9.1.4 respectively

Theorem 9.1.5 ([69]) Let X = (x;./) be anmxn matrix of indeterminates over a field K with m
F(Rr+1,z) = ——————— j-———————

>J=

--r

(d = dimRr+l).

Theorem 9.1.6 ([133, 42]) Let X = (xij) be an m x n matrix of i

terminates over a field K and Ir(X) the ideal of K[xij] generated by the r -minors of the matrix X . If m < n, then

Proposition 9.1.7 Let G be an spanning connected subgraph of K,m^n. If n > m, then the a-invariant of K[G] satisfies

a(K[G}) < a(K[!Cm,n}) = -n.

Semigroup Rings of Complete Graphs_________________ 339

Proof. From Corollary 9.1.4 one has

a(K[K.m,n}} = ~nHence one may proceed as in the proof of Proposition 7.7.14 to rapidly derive the asserted inequality. D An application of Dedekind-Mertens formula ring and / = f ( t ) G R[t] is a polynomial, say

f

= ao +

If R is a commutative

... + amtm,

the content of / is the R-ideal (a0, . . . , o m ). It is denoted by c(f). Given another polynomial g, the Gaussian ideal of / and g is the R-ideal

c(fg). This ideal bears a close relationship to the ideal c(f)c(g), one aspect of which is expressed in the classical lemma of Gauss: If R is a principal ideal domain, then

c(fg) = c(f)c(g). In general, these two ideals are very different but one aspect of their relationship is given by a formula originally due to Dedekind-Mertens (see [91] and [231]): (9.1)

We consider the ideal c(fg) in the case when / and g are generic polynomials. It turns out that some aspects of the theory of Cohen-Macaulay rings

show up very naturally when we closely examine c(fg). One path to our analysis and its applications to Noether normalizations of some semigroup rings, starts by multiplying both sides of (9.1) by c(/) m ; we get )c(g)]m = c(f)c(g)[c(f)c(g}r. (9.2) It will be shown that this last formula is sharp in terms of the exponent m = deg(/). To make this connection, we recall the notion of a reduction of an ideal. Let R be a ring and / an ideal. A reduction of / is an ideal J C / such that, for some non-negative integer r, the equality 7-r+l _

jr

holds. The smallest such integer is the reduction number r j ( I ) of / relative to J, and the reduction number r(I) of / is the smallest reduction number amongst all reductions J of /.

340______________________________________ Chapter 9

Thus (9.2) says that J = c(fg) is a reduction for / = c(/)c(g), and that the reduction number is at most min{deg(/),deg(<7)}. The following result solves the question of Noether normalizations for monomial subrings of complete bipartite graphs.

Theorem 9.1.8 ([91]) Let R = k[xo,. . . ,xm,yo, . . . ,yn] be a polynomial ring over a field k and let

i+j=Q

Then

A = k[h0,hi, . . . ,hm+n] *-> S = k[{xiyj\0 < i < m, 0 < j < n}}

is a Noether normalization of S. Proof. Let R[t] be a polynomial ring in a new variable t and /, g 6 R[t] the generic polynomials

^ and g = g(t) = i=0

j=0

Set / = c(f)c(g) and J = c(fg). Note that J is a reduction o f / by Eq. (9.2) and more precisely JIm = Im+1 . Therefore

R[Jt]

^ R[It]

is an integral extension; indeed R[It] is generated as an .R[J£]-module by a finite set of elements of t-degree at most m. On the other hand, as / and J are generated by homogeneous polynomials of the same degree, one has an integral embedding:

R[Jt]

<8>.R R/m = k[h0, . . .,hm+n] ^ R[It] <8> R/m = k [ x i y j ' s } ,

where m = (XQ . . . , zm,yo, • • • , J/n)- To complete the proof note that the dimension of k[xiyj's] is equal to n + m + 1, by Lemma 8.3.2. n Theorem 9.1.9 ([72]) Let R = k[xo, . . . , xm, yo, • • • , yn] be a polynomial ring over a field k and let f , g & R[t] be the generic polynomials

it1

and g = j=0

with n>m. If I = c(f)c(g)

and J = c(fg),

then r j ( I ) = m.

Semigroup Rings of Complete Graphs_________________ 341

Proof. By Theorem 9.1.8 there is a Noether normalization:

A= k[h0,hi,...,hm+n] <-> S = k[xiyj's], where h

i = X!

i+j=g

Xiy

i'

Set / = (/i, . . . , /s). We note that the ideal / = (xiyj's) is the edge ideal associated to a complete bipartite graph G which is the join of two discrete graphs, one with m + 1 vertices and another with n + 1 vertices. Since k[G] is Cohen-Macaulay (see Corollary 8.7.11), using Proposition 2.2.14 we get

k[G] = Af01 © • • • © Af0k , j3i e Ns . By Corollary 9.1.4 the a-invariant of k[G] is equal to —n — l, we can compute the Hilbert series of k[G] using the decomposition above to conclude that

max |deg(//3i)} = m, where the degree is taken with respect to the normalized grading. On the other hand assume r = rj(I) and JIr = Ir+1, since we already know the inequality r < m. It suffices to prove r > m. Note

where

1= <^ a a = (a\,..., as) € Ns and a = N a, < r If r < m this rapidly leads to a contradiction because there is J C X such that k[G] can be written as

k[G] = 0 Af°. aej

See Exercise 9.1.18.

O

Multiproducts In order to see a different explanation of Eq. (9.2), we extend it to the product of 3 (or more) polynomials, but using the theory of Segre products as a tool. Let X = {x0,. . .,xm}, Y = {y0, . . . ,yn}, and Z = {z0>. . .,zp} be 3 sets of indeterminates. Defining the polynomials m

n it

i

p ti

, g = Y,Vi > j=0

and

h = Y^zktk, k=0

342______________________________________ Chapter 9

one has that J = c(fgh) is a reduction of / = c(f)c(g)c(h) by DedekindMertens formula. If m < n < p, a simple calculation will show <m + n. We now resolve this inequality. Proposition 9.1.10 ([72]) Let X = {XQ, . . . , xm}, Y = {yo, . . . , yn}, and Z = {ZQ, . . . , zp} be sets of distinct indeterminates, let R = k[X, Y, Z] be a polynomial ring over a field k, and let

I = (xiyjZk\Xi £ X, yj G Y, zk e Z). Then I is a normal ideal of R.

Proof. We will show that Iq is complete for all q > 1. Let 7* be the integral closure of Iq and let / € 7* be a monomial. We write 01

Jf —~ xrh

ar bl • • . xT ii • • •Vj, iib"zk! ?Cl • • •zkyCi• i Vji

r

t

Since fw e 7s' for some w > 0 we can write fw _ J

di . . .

— ^qi

X

dx <Jx

M

JW

'

where M is a monomial whose support is contained in Y U Z. Therefore we obtain r

\

a

w 2_] i — /"J di > wq, j=l

i=l

which implies

A similar argument shows s

t

^2bi>q and ]P c, > q. 1=1

Therefore / £ I".

4=1

D

By Theorem 7.2.35, the Rees algebra R[It] is a Cohen-Macaulay ring. Since ^(7) = k [ x i y j z k \ X i e X, yj £Y,zk£ Z] is a normal domain by Proposition 7.4.1, and therefore Cohen-Macaulay by Theorem 7.2.35. We may thus more easily compute the reduction number of the subring .F(7) .

Semigroup Rings of Complete Graphs

343

Theorem 9.1.11 The reduction number rj (I) of the ideal I above is m+n. Proof. Since F(I) is a Cohen-Macaulay algebra, its reduction number can also be obtained from the degrees of the generators of its canonical module. But F ( I ) is a Segre product of standard Cohen-Macaulay algebras and the canonical module is given by an explicit formula from the canonical modules of the factors. Entering the data in [132, Theorem 4.3.1], we get r j ( I ) = m + n. D

Rees algebras of complete bipartite graphs The following result show that Rees algebras of edge ideals associated to complete bipartite graphs have some structure; its proof makes use of general facts about ladder ideals that can be found in the literature. Proposition 9.1.12 LetB/Q be the presentation ofR,(I), the Rees algebra of the edge ideal I = (xiyj\l
2/2

•••

2/n

TII

Ti2

•••

Tin

-/ ml

-/ m2

' ''

J-mn

Xl

Y =

. •Em

and the 2-minors ofY form a Grb'bner basis w.r.t the lex ordering induced by 2/i > • • • > yn > X! > TII > • • • > Tin > x2 > • • • > Tmn. Proof. Consider the presentation of the Rees algebra of /: ^ : B = Klxi's^j's^j's]

—> K(I),

^(Ty) = TxiVj,

Note Iz(Y) C Q = ker(t/)). As they are both prime ideals of height mn — 1 by [67, Section 4] and [228] (cf. [65, Proposition 3.3.1]), one has equality. The assertion that the 2-minors are a Grobner basis of I^(Y} is a general fact of ladder ideals [228]. D Proposition 9.1.13 LetB/Q be the presentation ofTi(I), the Rees algebra

of the edge ideal I = (xiyj\l m, then the Hilbert series of B /Q can be written as

V

/

In particular a(B/Q) = -(n + 1) and e(B/Q) = (m+") - 1.

344______________________________________Chapter

9

Proof. Let

B = K{xi,yj,Ti:j

1 < i < m, 1 < j < n]

be a polynomial ring over a field K and consider the ladders

y-2

2/1

•••

yn

Me =

NI = ^21 -^22

'

TH ^m

J-ml

J-m2

'''

Tf-2

^2n

'''

• • • Tin.

•*• mn

Let It and Ji be the ideals generated by the 2-diagonals of Mf and NI respectively, where a 2-diagonal is the product of the diagonal entries of a 2 x 2 submatrix.

By Proposition 9.1.12 and Corollary 2.4.13 one has that the Hilbert series of B/Q is equal to the Hilbert series of Bjl\. Hence the proof reduces to compute F(B/Ii,z). For 1 < I < m - 1 consider the graded exact sequences

where

KI = (Tij

l + l
Therefore

C

m-1

^

\

''

/

+

As xi,... ,xm is a regular sequence on B/Kf readily obtains

using Proposition 9.1.3 one

E^f": r u' and consequently n

a-*;

m+n-f-l

-

Semigroup Rings of Complete Graphs

345

Using the formula f

x + i\

(x + n + 1

yields the required equality.

D

Theorem 9.1.14 ([306]) Let

S = K[{xiyj,XiX,yjX

1
be the monomial subring of the cone over a complete bipartite graph K.mn. Ifn>m>2, then the canonical module u>$ of S is minimally generated as follows:

(a) If n = m, then

(b) Ifn>m+ 1, then uis is minimally generated by the set of monomials: (i)

xi • • • xmyi • • • ynxb, where n — m + 1 < b < n + m — 1 and n + m + b £ 2N, or

(ii) x"1 • • • x^yi • • -ynxb, where m

2_,ai = n — b + 2, b = 2 , 3 , . . . , n — m + 1, and a,{ > 1 for all i.

Proof. For a monomial M = x^ • • • x^ybl • • -yb^xb, set log(M) — ( a i , . . . , am, bi,..., bn, b). Define

A = {log(M)|M € {xiyj,XiX,yjX\l


As 5 is normal of dimension n + m + 1 (see Theorems.7.9 and Lemma 8.3.2) one can use the Danilov-Stanley formula (see Theorem 7.2.34) to express its canonical module as:

where (KLj_,4) 0 is the interior of the cone generated by A and S has the normalized grading. Let

M = x^ • • •x(^y11 • • -yn"x

346______________________________________Chapter

9

be a minimal generator of the ideal u>s- According to Theorem 8.8.7 a vector J3 e Nm+n+1 is in (R+,4)0 if and only if j3 satisfies the inequalities:

fa, i = 1,... ,m + n

1 < m X—^ ' -* i=l

m-f n ^—"^ ' ^ i=m-\-l

m+n

m

53

i=l

Pm+n+1

\—^ n — 1 + 2_^ f t '

<

Therefore one has:

1

<

QJ,

1 < bj,

m

n

i=l

i=l

l
I <j
(9.4)

h

< -1 + m

n

Using that a E N^4 one also obtains: m

n

(9.5)

Observe that (9.4) and (9.5) yield b > 2. We consider the following cases: I. Assume a^ > 2 for some i. We claim that 6; = 1, for all i. For simplicity assume a\ > 2 and &i > 2. Write:

«=i

i=i

If 1 < i < 2, write M = M'y^x. Since logM = logM' + logyix one has that logM' is in 7LA. It is not hard to see that logM' satisfies all the equations in (9.3), thus logM' e (M_|_y4)°. By the normality of S one concludes

Semigroup Rings of Complete Graphs_________________347

log Af' € N.4, hence M' is in ujs, which contradicts the minimality of M. If

i > 3, consider M = M'xiyi, as before one readily obtains M' £ u>s, which again contradicts the choice of M.

II. Assume n = m. First let us show Oj = 1 and bj = 1, for all i,j. By the previous case and symmetry if a» > 2 for some i (resp. bi > 2), then bi = 1 for all i (resp. a; = 1 for all i). If aj > 2 for some i, say i = 1, then M' G 015, where M = M'x-^x, and this is impossible by the choice of M. Next, using (9.4) and (9.5), it follows that b = 2i for some 1 < i < n — 1, and we are in case (a).

III. Assume n > m+1 > 3. We claim that bi = 1, for all i. If bi > 2, say i = 1, then by case I one has a; = 1 for all i. As before one readily derives a contradiction by writing M = M'y\x. Altogether one can write:

M = xl • • • x^ yi • • • ynx ,

b > 2.

If Oj = 1 for all z, then using (9.4) and (9.5) one has n-m+l
and m + n + b G 2N,

and we are in case (b). Next assume a» > 2 for some i, say i = 1. Combining

(9.4)

and (9.5) one has m

n<^ai + b-2. i=l

If the inequality m

n
m + l
for b = 2, . . . ,n — m + I , and we are in case (b) . Thus any minimal generator of a; 5 must be as in (a) or (b). To complete the argument observe that any of the monomials occurring in (a) or (b) are in LJS and they are not multiple of each other. D Remark 9.1.15 I f n > m = l, then assertion (b) above still holds. To show it replace the very first set of inequalities in (9.3) by /?j > 1, i = 2, . . . , n + m. On the other hand if n — m = 1, then u>s —

Corollary 9.1.16 Let S = Ti(I) be the Rees algebra of the edge ideal I of the complete bipartite graph ICmn. Ifn>m>2 or n > m = 1, then type(5)=

348______________________________________Chapter

9

Proof. I f n = m > 2 o r i f n > m = l the formula follows rapidly from

Theorem 9.1.14(a) and Remark 9.1.15. Assume n > m + 1 > 3. Observe that there are m — 1 generators of us as in Theorem 9.1.14(1), to count the remaining generators of ws as in Theorem 9.1.14(11) note that the sum from b = 2tob = n — m + l o f the m-partitions of n — b + 2 is equal to n-m+l /

E

6=2

,

l n —b + \ ™_ i V

'

t=m+l n—ni /

"

-i

m

Hence Tl

type(S) = (m - 1) + (

\

I

J - 1= (

71

) + (m - 2),

as required.

Exercises 9.1.17 Let .R[t] be a polynomial ring over a ring J?. If / = f ( t ) and
c(fg)c(g)m

= c(S)c(g}m+l

(m = deg(/)).

9.1.18 Let M be a graded free module over a polynomial ring A such that M = Afi © • • • ffi A/ r , where /, is homogeneous for all z. If M is generated by homogeneous elements g ^ , . . . ,gs, then M = Ag^ © • • • ® Agir, for some

9.1.19 Let A = k[xi,...,xm] and R = k [ y i , . . . , y n ] be two polynomial rings over a field k, prove that the Segre product

(Ao ® fc -Ro) © (Ai ®k Ri) ® • • • C A ®k R is isomorphic to S = fc[xj?/j's] C fc[xi's, j/j's]. 9.1.20 Let A and /? be two standard algebras over a field k of dimensions

m and n respectively. If S is the Segre product of A and R, prove the following relation between the multiplicities fd\ = e(A)e(R) f A\ (D\ e(S)

Semigroup Rings of Complete Graphs_________________ 349

9.1.21 Let R = K[xi,...,xn] be a polynomial ring over a field K and m = (xi , . . . , Xn). Prove that the toric ideal of the Rees algebra 7?.(m) is the ideal generated by the 2-minors of the following matrix of indeterminates:

i

!

xi T2

••• xn . . . Tn

9.1.22 Let R = K[x\, . . . ,2:4] be a polynomial ring over a field K and /

the 3rd square-free Veronese subring of R generated by the monomials 3,

X1X2X4,

Prove that the Rees algebra 7^(7) is a complete intersection. 9.1.23 Let R = K[xi,...,xn] be a polynomial ring over a field K and

/ = (xi, . . . ,xn)k. Set s = —\n/k~\ + n, and write n — qk + r\, for some q,r\ £ N such that 0 < k < r\. By a "counting degrees" argument prove: ifn=0, k-l . . . l

k-l 2- g

ri-1 fc-1

X

q+l

X

q+2

X

fc-1 ,- TS n

^

1

9.1.24 Let 7? = AT[a;i, . . . ,xn] be a polynomial ring over a field K and m — (xi, . . . ,xn). If I = mk and J = (a;*, . . . , x*), then J is a reduction of

/ and the reduction number of / relative to J is given by rni

r j ( I ) = — — + n, where \x~\ is the ceiling of x. I K I

9.1.25 Let 5 be a standard algebra over an infinite field K and ...,hd ^ S a Noether normalization of S with hi € Si for alH. If 61 , . . . , bt is a minimal set of homogeneous generators of S and as an A-module, prove (hi,... ,hd)Sr = Sr+i, where r — maxj{deg(6j)} and that r > 0 is the minimum integer where equality occurs. The reduction number of S, denote by r ( S ) , is the minimum r taken over all Noether normalizations. See [298, Chapter 9] for an study of several degrees of complexity associated to a graded module. 9.1.26 Let S be a standard algebra over an infinite field K and a(S) its a-invariant. If S is Cohen-Macaulay prove:

r(S) =o(5)+dim(5).

350______________________________________ Chapter 9

9.1.27 Let (f>: B —> 5 be a graded epimorphism of standard /(T-algebras, where K is an infinite field. If dim(B) = dim(S) and

is a Noether normalization of B with hi £ Si for all z, prove that S is integral over A' = K[(hi), . . . ,
is a Noether normalization of S. 9.1.28 Let (p:B -t S be a graded epimorphism of standard /C-algebras, where K is an infinite field. If dim(B) = dim(S), prove r(B) > r(S).

9.2

Monomial subrings of complete graphs

Lexicographical Grobner bases and Hilbert functions of toric ideals of edge subrings associated to complete graphs will be considered here.

Let us recall some terminology and fix some of the notation that will be used throughout the section.

The letters i,j,k,t,n,r will denote positive integers. We denote by Gan , the graph with vertex set

and edge set

E = {{Ty,Tw}| ( i , j ) , ( k , f ) 6 an, {i,j} n {k,l} = <&,i
Given a set D C Vn, we denote by GD the maximal subgraph of G0n with vertex set D. Let An = k[Vn] be a polynomial ring over a field k and G a graph on

the vertex set Vn. The edge-ideal I(G), of the graph G, is the ideal of An generated by the square free monomials TijTkt so that T^ is adjacent to T^f in G. If all the vertices of a graph G are isolated we set I(G) = (0). The graph G is called Cohen-Macaulay (C-M for short) if I(G) is a C-M

ideal. A subset C C Vn is a minimal vertex cover for G if every edge of G is incident with some vertex in C and there is no proper subset of C with this property.

Semigroup Rings of Complete Graphs_________________ 351

Let R = k[xj_, . . . ,xn] be a polynomial ring over the field k, and K.n the complete graph on the vertex set X = {xi, . . . ,xn}. For i < j we set fij = XiXj. Let be the /c-subring of R spanned by the f i j . Consider the homomorphism

: An = k[{Tij\ l fa. The ideal Pn — ker(^) is the presentation ideal or toric ideal of k[K.n]Proposition 9.2.1 ([304]) Let R = k[xi,...,xn] be a polynomial ring over a field k and Pn the presentation ideal of k[K,n]. Set

B = {TijTkl - TuTjk, TikTji - TuTjk\ I then B is a minimal generating set for Pn and B is a reduced Grobner basis for Pn. Proof. By Corollary 7.1.7 it follows that B is a minimal generating set for Pn. Let /, g e B. Taking into account Theorem 2.4.15, it suffices to verify that S ( f , g), the S-polynomial of / and g, reduces to zero with respect to B. There are two sequences of positive integers i < j < k < I and p < q < r < s so that

f (a) TuTu - TuTjk 1 (b) TjeTik-TieTjk.

7

or, 9

f (c) TpqTrs - TpsTqr \ (d) TprTqs-TpsTqr.

or,

Assume / and g as in (a) and (c) respectively, the others cases can be treated similarly. We may assume that the leading terms of / and g have common factors, for otherwise S ( f , g ) reduces to zero w.r.t {/, g} by Lemma 2.4.14. First we assume Tij — Tpq, in this case i = p and j = q, and the Spolynomial of / and g is equal to S(f, g) = —TrsTpiTqk+TkeTpSTQr. Observe that

off

( Tqr(TtrTps-TrsTpi) i _ J Tps(TqrTsi - TrsTqe) + Trs(-TpiTqs + TpsTqi)

if fc = r, if k - s,

~ } Tkr(TpsTqr - TprTqs) - Tpr(TrsTqk - TqsTkr) [ Tps(-TrsTqk+TskTqr)

if i = r, i f l = s.

U> 91

Hence S(f, g) -*B 0 if {k, 1} n {r, s} ^ 0. We may then assume k and I not in {r, s}. If i > s and r > k then

) — Tps(TkiTqr - TqtTkr) - Tkr(TpsTqi - TptTqs) +Tpi(TrsTqk-TkrTqs}

= 0.

352

Chapter 9

On the other hand if i > s and k > r we have S(f,g) ~ Tps(TkiTqr - TqtTrk) - Trk(TpsTqe - TpiTqs) +Tpl(TrsTqk - TrkTqs) = 0. In both cases S ( f , g ) —>g 0. Assume s > I and k > r. From the equality

S(f,g) ~ Tps(TqrTki - TqiTrk) + Tpt(TqkTrs - TqsTrk) —Trk(TpsTqi — Tp(Tqs) — 0, we derive S(f,g) -^B 0. To complete the argument consider the case s > and r > k, using the equality

S(f, g) + Tpi(TrsTqk - TqsTkr] + Tkr(TptTqs - TpsTqi) -Tps(TkiTqr - TkrTqt) = 0, we obtain S(f,g) ->B 0. Now consider the case Ty = Trs, note that the case Tki = Tpq is symmetric. We have i = r, j = 7 and S(f,g) = TkiTpsTqr - TpqTrtTsk. Since S(f,g) - Tqr(TpsTki - Tp(Tsk) - Tsk(TqrTpi - TpqTri) = 0, we obtain S(f,g) —>B 0. Finally we show the case Tkt = Trs, notice k = r, I = s and •-H/iS)

==

J-ij-'-ps-'-qr ~ -Lpq+ is-ljr-

If i > q then

S(f,g) + Tis(TpqTjr - TprTqj) - Tqj(TqsTpr - TpsTqr) = 0, and S(f,g) -^B 0. If i < q, then S(f,g) + Tis(TpqTjr - TpTTqj] - T^T^T^ - TirTqj) —T QJ-(T u \ ps T-LIT —T-T -Lis-Lpr)\ — — fl ;

Example 9.2.2 If fc[{Ty 1 < i < j ' < 5}] has the lex ordering induced by T45 < T34 < T35 < T23 < T24 < T25 < Tia < T13 < T14 < T15. Then the reduced Grobner basis for the toric ideal P of /c[/Cg] is equal to:

- T45T13, T23Ti5 - T35T12, rrn

rp

rp

rp

J 24-^13 ~~ ^34-^12

Semigroup Rings of Complete Graphs_________________353

Corollary 9.2.3 If Pn is the toric ideal of the subring k[fCn] and n > 4, then Pn is the ideal Iz(Y) generated by the 2-minors of the symmetric array

Y =

Tl2

•

:

:

Tln

Tin

T"23

•• •

T2(7i-2)

T" 2 (n-l)

Tzn

:

•.

:

:

:

' ' '

T(n_^n

T"(n-l)n

•

Tan

and the 2-minors ofY form a Grobner basis for Ii(Y) w.r.t the lex ordering

Lemma 9.2.4 Let n > 4 be a fixed integer, and let r be positive integer so

that 1 < r < n - l . Set D'r = {T^ 6 Vn\r < i < n - 2, r + I < j < n - 1},

Dlr = {TireVn k
If I(Gai) is a Cohen-Macaulay ideal of height

i(i - 3) 2

for all i
forl
N = {T^ 6 Dkr T^ is adjacent to Tkr}.

Since N is equal to

354______________________________________ Chapter 9

it follows that

in — r — 1\ +r k

M =(

J

)

- ~l

and that N is a minimal vertex cover for Gokr • Therefore and

Observe that T^k+l)n is an isolated vertex in &D (fc+1)r ,, hence TV can not be contained in a minimal vertex cover for Go(k+1}r

Consider the exact sequence

0 — >• An/(I(GDkr),N}

^ An/I(GDkr) — > 4 n /(I(GzO,T fcr ) — > 0.

Since the ends of this sequence are both Cohen-Macaulay rings of dimension equal to dimAn/I(GDkr) we derive that An/I(Gokr,} is CM, as required, n Corollary 9.2.5 ([280]) Let

I(GDl(n_l}) = ({Ti(n_1}Tjn l 4. Proof. Take r = n - 1 and fe = 1 in Lemma 9.2.4.

n

Theorem 9.2.6 The edge ideal I (Gan) is Cohen-Macaulay of height equal to n ( n - 3 ) / 2 . Proof. The proof is by induction on n. The assertion is not hard to verify f o r n < 5. Assume the result for all graphs isomorphic to G(Ji for i < n. Set Gr = Vn — i U {-Zlnj^ni • • • jTrn\.

We claim that Gcr is Cohen-Macaulay with

for 1 < r < n — 1. The proof of the claim is by induction on r. If r — 1 then /(Gci) = I(G
Semigroup Rings of Complete Graphs_________________ 355

for i < r. Let

Ur - {Tij e Cr\Tij is adjacent to Trn},

and let D\r be as in Lemma 9.2.4. Observe that .

Set Lr = (I(GCr},Ur} and Jr = (/(G Cr ),T rn ). Since

(/(GcJ.E/r) = (I(GDlr),Ur) and (/(G c J,T rn ) = (/(£<;,._! ),T rn ), an application of Lemma 9.2.4 yields that Lr is Cohen-Macaulay and

> — 1

(r-2)

Hence

On the other hand by the second induction hypothesis ht ( J r ) = ht (L r ) and

Jr is CM. Altogether we have that A/Jr and A/LT are Cohen-Macaulay rings with the same dimension. Since Lr is also equal to (I(Gcr._1), Ur) we have

Therefore Ur can not be a subset of a minimal cover for Gcr_1, and thus

From the exact sequence

0 — »• (An/Lr)[-l]

^ An/I(GcJ —> An/Jr — >. 0.

we obtain I(Gcr) Cohen-Macaulay. In particular for r = n — 1 we conclude that /(Go- n ) is Cohen-Macaulay of height equal to n(n — 3)/2. D Proposition 9.2.7 ief F n (z) = hn(z)(l — z)~l be the Hilbert series of the

ring An/I(Gan). Then the polynomial hn(z) satisfies the difference equation hn(z) = 2/i n _ 1 (z) + (z- l)hn^(z) + (n- 2)z2 for n > 5 with boundary conditions hs(z) = 1 and h^(z) = 1 + 2z + z2.

356

Chapter 9

Proof. Set A = An and Cr = Vn-^ U {Tln,T2n,.. .,Trn}. Let Dlr and Ur be as in Lemma 9.2.4 and Theorem 9.2.6 respectively. Consider the exact sequences

o,

0 — > (A/Ln-i) [-1] ^" A/I(GCn_,) and

0 — »• (A/i r ) [-1] ^

0,

where

and Using these sequences recursively, for r — n - 2 , . . . , 3, 2 yields

Fn(z) = z [F(A/Ln--i,z) + ••• + F(A/L2,z)} + F(A/J2,z), where Ji = (T<2n, . . . ,T( n _ 1 ) n ,/(G cr , v _ 1 )). On the other hand, by induction on r, we readily obtain ifr 6 { n - l , n - 2 } , if2
(l-z)n Therefore ra-l

n-l

hn(z) = z

- k)z lk=n-2

and n-2

n-2 (k

hn-l(z) = Z

_

.fc=n-3

subtracting the last two equations we obtain

hn(z) = 2hn-!(z) + (z- l)hn^(Z) + (n- 2)z 2 . The boundary conditions follow readily since I(Gaa) = (0).

d

Corollary 9.2.8 ([262]) The ring k[!Cn] is a Cohen-Macaulay ring. Proof. Let An/Pn be the presentation of k[fCn}. Note that in(P), the initial ideal of P, is equal to I(G0n). By Theorem 9.2.4 we obtain that A/\n(P) is Cohen-Macaulay, and this implies (see [133, 208]) that A/P is Cohen-Macaulay as well. D

Semigroup Rings of Complete Graphs_________________ 357 Remark 9.2.9 Let

Tl(I) =k[{xl,...,xn,Txixj

l
be the Rees algebra of the edge-ideal / = /(/C n ) and

fc[/C n+ i] = k({xiXj\ 1 < i < j < n + I}}. Then by Proposition 8.2.15 there is an isomorphism

induced by

ip(xi) = XiXn+i and Therefore by Lemma 9.2.8 we obtain that the Rees algebra of the edgeideal of a complete graph is Cohen-Macaulay. In fact such algebras are

even normal by Proposition 7.4.5. Theorem 9.2.10 ([304]) Let K.n be the complete graph on n vertices and An/Pn be the presentation of k[K,n}. If n > 3, then the Hilbert series Fn(z) of An/Pn satisfies:

Proof. Let Fn(z) = F(An/Pn,z). Since in(P n ), the initial ideal of Pn, equals I(Gan), by Corollary 2.4.13 the Hilbert series of An/I(Gan ) is equal to Fn(z). Let (1 - z)nFn(z) = hn(z) = a0n + alnz +••• + ag(n]nze(n\

and let

00

Gmx = n=3

be the generating function of the sequence (a mn ) n >3. By Proposition 9.2.7 we have

{

2a m (n-l) + «(m-l)(n-2) ~ «m(n-2)

if n > 5, HI £ {0, 2},

+ (n-2)

if n > 5, m = 2.

Recall that 00

G0(x) = ^xn, h3(z) = l, and h4(z) = I + 2z n=3

358______________________________________Chapter

9

Therefore for m > 3 we obtain oo

TJ=5

00

00

ra=5

00

ra=5

n=5

2

2

">Td r (7 Ajj^ni \CT"> J -I* "^ ^-* m, — i1 ('r'l V*^/ — r (7 "^ T7i (V) V /'

— —

Jj

and consequently

for m > 3. By a similar argument we derive

Gi(a;) =

£ 4 (2 - z) n _ ^

^ n(n - 3) , . ,5 4 ., J = 2^ -^— x ' and G 2(x) = x 4 /(l - x) 5 .

X)

Hence a\n = n(n - 3)/2

n=3

*

for n > 3. To complete the proof observe that

ra=2m

x

2m

for m > 2. Hence amra = (2^) for TO > 2 and n > 3.

D

Remark 9.2.11 The Hilbert function of k[ICn] can be computed using the fact that the Ehrhart polynomial of the second hypersimplex of order n is equal to the Hilbert function of k[Kn]. See Chapter 7 and [79, 277]. Corollary 9.2.12 The multiplicity e(k[K.n]), of the ring k[ICn], is equal to 2"-1 -n.

Exercises 9.2.13 Let k be a field and A = k({Ti:j\ I < i < j < 5}]. Consider the permutation of the variables: Tl2

TIS

Ti4

Ti5

T23

T24

T^

Ts4

TSS

T45

T45

Ts4

TSS

T23

T24

T25

3\2

T\3

Ti4

TIB

Prove that the toric ideal P of /c[/Cs] is not invariant under the automorphism of A induced by
Semigroup Rings of Complete Graphs

359

9.2.14 If Q is the toric ideal of the Rees algebra ??.(/) of the monomial ideal / = (xiXj\l < i < j < n — 1), then Q is the ideal Ii(Y) generated by the 2-minors of the symmetric array •

Tl2

Til

•

Tl3

T23

•

'

'

'

Tl(n-l)

2(n-i)

sO-l)

'••

Kl

£2

£3

'••

Y

=

Tl3

•••

Tzz

7\(n_2)

Tl(n-l)

••• T2(n_2) •••

'

T2(n_i)

T3(n_2)

•

TZ(n_i)

'

Xl Xz X3

'

(n_2)(n-l)

Xn-2

.

•

Xn-1

Xn-l

•

and the 2-minors of Y form a Grobner basis for Iz(Y} w.r.t the lex ordering Ti2 < • • • < T^n_i) < xi < T23 < • • • < T 2 ( n _i) < z 2 < • • • < x n _i.

ffini Set B = AT[Ty 1 < i < j < n] and x{ - Tm for 1 < i < n - 1. There is a commutative diagram

B

-^

^

a

^

, '

K[ICn} ot(xi)

=

XiXn, 1 < i < n

Since a is a isomorphism one has Q = ker(V') = ker(
9.3

Noether normalizations of edge subrings

In this section two explicit Noether normalization of the edge subring of a complete graph are presented, it will turn out that those normalizations

can serve as a model to deal with some specific graphs. The results of this section are due to A. Alcantar [2, 3]. There are few classes of graded algebras where homogeneous Noether normalizations are known, such classes include quotients of monomials ideals [298, Proposition 2.3.3], edge subrings of complete bipartite graphs [72] and edge subrings of bipartite planar graphs [85]. Systems of parameters of edge subrings

Let G be a graph on the

vertex set V — {xi,... ,xn} and R = K[XI, ... ,xn] a polynomial ring over a field K. One can express the toric ideal of the edge subring K[G] as the kernel of the epimorphism of K-algebras:

tp: A = K[{Tij i < j and x^ is adjacent to Xj}] -)• K[G]

360

Chapter 9

induced by ^(T,j) = XiXj, where {T,j} is a new set of variables in one to one correspondence with the edges of G. We shall always assume that A and R have the usual grading and that K[G\ has the normalized grading

thus ip is a graded map. In the sequel we denote ker(i/>) by P(G). An standard Noether normalization of A/P(G) is an integral extension

where hi, . . . ,hj are homogeneous polynomials of degree one and d is the

Krull dimension of .RT[G]. Note that (*) is a Noether normalization if and only if

where 971 is the maximal ideal of A generated by all the variables T^ . See Chapter 2. Edge subrings of complete graphs The complete graph with vertices x\ , . . . , xn, denoted by /C n , has every pair of its n vertices adjacent. Let

K[fCn] = K[xiXj

I < i < j < n}} C R

be the edge subring of R associated to K,n. Thus the toric ideal P(K,n) of K[ICn] is the kernel of the homomorphism of .^-algebras

?/>: An = K{{T{j 1 1 < i < j < n}} ->. K[Kn] induced by Tij ->• x^Xj.

Notation For use below we set Pn = P(K,n)- If j < i, we set TJJ := T^. The homogeneous maximal ideal of An will be denoted by 9Jtn . Lemma 9.3.1 ([3]) Let Qn = (Pn,Jn) C An be the ideal of An generated by Pn and Jn = (hi, . . . , hn), where

Tki. k
Let P be an associated prime ideal of Qn. Assume Qp is an ySlp-primary ideal of Ap for 4 < p < n. If P contains a variable Trs, then P — 9Jtra.

Proof. Let P be an associated prime ideal of Qn so that Trs £ P, to simplify notation assume r = n — 1 and s = n. Since 2^.( n _ 1 )Tj n — TkjT(n-i)n € Pn,

Semigroup Rings of Complete Graphs

361

we obtain Tfc(ra_1)T.,-n 6 P for j ^ k. If T^-i) ^ P for some k, then

Tj,, 6 P for all j ^ k, j < n - 1. As hn e P we have

hence Qn-i C P. Using the hypothesis yields Tk(n-i) 6 P, a contradiction. Therefore Tk(n_l} e P for all k ^ n - I. Let Q = (Qn, Ti(n-l) i • • • > 2"(n-2)(ra-l) > ^(n-l)ra)-

Because Q C P, from the hypothesis we derive that P contains all variables not in {T fc(n _ 1 }^ n _ 1 . Therefore P = 9Jln. D Theorem 9.3.2 ([3]) //A' zs either a field of characteristic zero or K is a field of characteristic p > 2 and 2, . . . , n — 1 are relatively prime to p, then the ideal Qn is an 9)ln -primary ideal of An. Proof. We proceed by induction on n. It is not hard to verify that

for 3 < i < 6 (see arguments below). Assume n > 7 and the result true

for m < n. Let P be an associated prime ideal of Qn. By Lemma 9.3.1 it suffices to prove that P contains one of the T^-'s variables. Set

Consider the equality n-2 i=3

and define the following permutations of {1,..., n}

1

2

n-1 1

/ 1 2 ~\n-3n-l

n-3 2

n-2 f f 2

2 n-2

n

n n-1

n-2 n

_ f I 2 n-3 n-2 ^lnn-2n-l 2

t 7 3

n- 1 1

n-1 1

n n-2

n n-3

362

Chapter 9

Observe that a permutation a of {1,..., n] induces an automorphism

a': An —> An such that cr'(Tjj) = T^J^Q). Since

a'(Pn) C Pn and cr'(J n ) C J»

we have a'(Qn) C Qn. Applying a\ to Eq. (9.6) yields <7<(T 2n L n ) 6 Qn. Note that

Tk(k+i)(Tik ~'

if i = 1, n > 5, k -n-

1

Tk(k+i) Tk(k+i) (Tik ~ . Tk(k+1)(Tlk-T1(k+1})

ifi

If Tk(k+i) £ P for all 2 < k < n - 1, then TH - TIJ G P for i ^ j. Using the identity

= nj=3

we obtain Ti2 e P.

D

Corollary 9.3.3 ([3]) //-ftT w afield of characteristic zero, then

K[hl,...,hn]^An/Pn~K{!Cn} is a Noether normalization of K[K,n]. Proof. Note that the monomial subring A"[/Cn] is integral over the subring K[hi,...,hn] by Theorem 9.3.2. D Corollary 9.3.4 If K is a field of characteristic zero, then An/Pn is a finitely generated free module over K[hi, . . . , hn]. Proof.

Since K[K,n] is a Cohen-Macaulay graded algebra of dimension

n, by Proposition 6.3.1 we obtain that An/Pn is a finitely generated free module over K[hi, . . . , hn]. Note that {hi, . . . , hn} is a regular system of parameters for An/Pn. D

Semigroup Rings of Complete Graphs_________________ 363

Another Noether normalization Let us present here a simpler Noether normalization of the monomial subring of the complete graph. Since any variable Ty occurs only in hi and hj, one has:

2/i = hi + • • • + hn, where h = Setting

{

—h + hi + hi+i hi+i

for i = 1,2, for 3 < i < n — 1,

h — hi

for i = n,

it is not hard to see that f = {51, . . . ,gn} is a Grobner basis for (hi, . . . , /i n ) w.r.t. the lex ordering Ti2 > • • • > T\n > T^ > • • • > 7( n _i) n . Thus one may try to eliminate some of the first variables from F with the intention of

producing some "triangulated" Noether normalization whose first elements are variables. The next result proves that one may replace g\ by Ti2 and still get a normalization.

Proposition 9.3.5 Let K be a field of characteristic p. If 2, . . . , n — 1 are relatively prime to p, then K[Tl2, 52, . . . ,Sn] -+ An/Pn ~ K[)Cn]

is a Noether normalization of K[K-n}. Proof. From the equalities

h = (hi-h) + h2 + --- + hn

(9.7)

and

Khi + ••• + Khn = Kgi + --- + Kgn, one has

As K[K,n] is Cohen-Macaulay (see Proposition 7.4.5) by Theorem 9.3.2 one concludes that gi, . . . ,gn is a regular system of parameters for An/Pn. In particular 52, • • • ,9n is a regular sequence on An/Pn. To finish the proof it suffices to show that T\2 is a regular element on

S = An/(Pn,g2,...,gn). Let Q be an associated prime of S and assume Ti2 is in Q. If /ij € Q, then hs, . . . , /i n , /i are in Q, which together with Eq. (9.7) gives h% 6 Q. Hence gi € Q; a contradiction to the fact that g\ is regular on S. Therefore h\ cannot be in Q and consequently there is k > 3 so that Tik ^ Q-

364______________________________________Chapter

9

If {1, A;} n {2, j} = 0 and j ^ 2, then TuT2j- - T12Tkj is in Pn and thus TZJ is in Q for all j ^ {2, k}. Note T^k is not in Q; for otherwise hz,... ,hn and h — hi are in Q and from Eq. (9.7) /i € Q, and hence hi & Q which is impossible. By considering T^Ty — T^jTik, one obtains Ty G Q for all ij so that i,j are not in {2, k} and z / j. It follows readily that all the variables Tij not occurring in hk belong to Q. Using Eq. (9.7) and noticing that T^ occurs exactly twice in hi + • • • + hn, one gets 2(h-hk) 6 Q, but since hk is also in Q one derives h, h2 6 Q and consequently g\ is in Q, which is again a contradiction. Altogether T12 is not in the union of the associated primes of S and is therefore a regular element on 5. D Examples and one open question The minimal vertex covers of the complement G of a graph G seem to be related to the problem of finding a Noether normalization of the edge subring K[G\.

Example 9.3.6 Let G be the following graph and G its complement.

G Using Macaulay one has P(G) = (91,92,93,94), where 9l = £l3*25 - ^15*23,

52 =

93 — £12*35 ~ ^15^23)

94 =

By Corollary 8.7.11 it follows that the edge subring K[G] is a normal CohenMacaulay domain. To find a Noether normalization for K[G] over a field K of characteristic zero consider the sequence:

unfortunately ht(P(G),$i, . . . ,#5) = 7, but we can modify the coefficients of the variables in the support of Oi to obtain the required height 8, more precisely a Noether normalization for K [G] is:

. ..,h5}^ K(G] ~ K [ t t j ] / P ( G ) ,

Semigroup Rings of Complete Graphs_________________ 365

where hi = ti2 + il3 + ti5,

h% = ti2 + t-22 + ^25,

hz = tis + t23 + 3^34 + £35,

/l4 = 3^34 + £45,

hs, = iis + £25 + £35 + £45Note that the coefficients of the variables occurring^in 6i were modified by first finding a minimal vertex cover C = {^4} for G and then changing t^j by itij, if Xj is in C1. Remark 9.3.7 To verify the assertion that ht(P(G), hi, . . . , /i s ) = 8 in the Example 9.3.6, note that hi, . . . , h5 specializes nicely, that is,

hi , • • • , /ij_i , "j+i i • • • , "<5 is a system of parameters of K[G \ {x,}] for each 1 < i < 5, where h'j is obtained making tik = 0 in the sequence hi, . . . , h$. Hence we may proceed essentially as in the proof of Proposition 9.3.5. Take an associated prime Q of (P(G),hi, . . . ,h5). If ii 2 6 Q, then *is*23 G <9- Note one may assume £ 2 3 € <3; otherwise i15 and i13 are in Q and from: li, . . . , /16, «12, *13, *16) = P(G\{li}), /12, ^3, /14, /l^, *12, tis, *15) C Q,

one concludes ht(Q) = 8. Using:

again one concludes ht(<5) = 8. Similar arguments show that if £13 or ti& are in Q, then ht(Q) = 8. Hence one may assume £12,^13,^15 not in Q. From the equalities:

93-91-92 + (tl5 ~ ^35)^1 + *13/l5 = <15(tl2 + 2iis + £15 + £34 -

one readily derives that Q is the irrelevant maximal ideal of K[tij]. All the normalizations of monomial subrings of graphs that we have found, through the procedure of Example 9.3.6, suggest the following: Question 9.3.8 Let G be a graph with n vertices and A/P(G) the presentation of K[G] over a field K of characteristic zero. Assume K[G] is a normal subring of dimension n such that deg(u) > 2 and dim K[G \ {v}} = n — 1 for any vertex v of G. Is there a Noether normalization of the form:

K[hi, . ..,hn}^ A/P(G), hj = ^ ^ijTij for all 1 < j < n, £ij 6 K 1

Chapter 9

366

The next instance of a graph is rather more complicated than the graph given in Example 9.3.6, but exhibits the same sensitivity to modifying the coefficients linked to subsets of minimal covers of the complement of the graph. Example 9.3.9 Let G be the following graph on 10 vertices and G its

complement.

htP(G)=5. _ Minimal cover for G: C = {x3,X4,X5,XT,Xs,Xg,XiQ}.

By the description of the integral closure of K[G] given in Corollary 8.7.11 it follows that K[G] is a normal domain. It is not hard to verify that a Noether normalization for K[G] over a field K of characteristic zero is:

K[hi,..., /no] -> K[G\ ~ where hi = *1,2 + *1,5 + tl,6 + * 10, /l3 = 2i 2 ,3 + 3*3,4 + 3*3,7 - 3*3,8,

/*5 = *1,5 + 4*4,5 + 5^5,9

5*5,10,

L

__

J.

I

O^-

_L 4

_L

O^

"2 — '1,2 r ^£2,3 T 13,6 "t" ^'2,7, /l4 = 3*3,4 + 4*4,5 + 4*4,8 •H- 4*4,9 ^6 = ^1,6 "+• ^2,6

h7 = 2*2,7 + 3*3,7,

h& = 3*3,8 + 4*4,8

hg = 4*4,9 + 5*5,9,

^10 = ^1,10 + 5*5,10-

The coefficients of the variables occurring in hi were chosen using a minimal vertex cover C for G and then modifying the coefficient of *y, if Xj is in C.

Remark 9.3.10 The monomial subring of Example 9.3.9 was studied by Hibi and Ohsugi [165] to show a normal subring K[G] of a graph G such that none of the initial ideals of P(G) is squarefree.

Chapter 10

Monomial Curves In this chapter we present some material on toric ideals of monomial curves and symmetric semigroups. The emphasis will be on space curves and monomial curves in the affine space of dimension four whose toric ideals are generated by critical binomials. We should point out that Chapter 10 and Chapter 11 overlap at certain points. A few almost identical results will be proven in both chapters with slightly different proofs.

10.1

Defining equations of monomial curves

Let k be a field and

the semigroup generated by a finite sequence di , . . . , dn of relatively prime positive integers. A monomial curve T in the affine space 77 over k is given parametrically by r- - tdi iLl

—— L

,

that is, we have Let R = k[xi, . . , ,xn] and k[t] be two polynomial rings over k graded by

deg(£i) = di for all i and deg(i) = 1 respectively. Let <j> be the graded homomorphism of fc-algebras

The image of (f>, denoted by k[S] or k[T], is the semigroup ring associated with 5 and the homogeneous ideal P = ker() is the presentation ideal or toric ideal of k[S].

367

368_____________________________________ Chapter 10

By Proposition 7.1.2 the toric ideal P is generated by a finite set of binomials. Hence the exact sequence 0 — > K = ker(^) —> 1" -^ Z —> 0, ^(ei) = d* is related to the presentation ideal P of k[S] in the following manner: if g = xa - x13 is a binomial, then g e P if and only if a - /? e ker(^). Given a binomial g = :ra — xb , we set g = a — &. Given a subset / C R we denote its zero set in A£ by V ( I ) , and given a subset X C A£ we denote its vanishing ideal in R by I ( X ) . If hi, . . . , hr G Z", we denote by

(hi,. . . ,hr) the subgroup of Z" generated by these elements. Let G C Z n be a subgroup. We associate to G an equivalence relation on the monomials in R = k[x\, . . . ,xn], by xa ~G x® if a — j3 6 G. Note that this relation is compatible with the product, i.e., xa ~G x@ implies x~<xa ~G x^x13 for all 7 e Nn . Definition 10.1.1 A nonzero polynomial

in R is simple (with respect to ~G) if all its monomials, i.e., those xa with nonzero coefficient A a , are equivalent under ~G. Given any polynomial / 6 R\ {0}, we can group together its monomials by equivalence classes under ~G, thereby obtaining a decomposition / = hi + • • • + hm

with the property that each summand hi is simple, and that no monomial in hi is equivalent with a monomial in hj if j ^ i. Such a decomposition of / as a sum of maximal simple subpolynomials is unique up to order. We will refer to the summands hi as the simple components of / (with respect

to ~G). Lemma 10.1.2 Let gi, . . . ,gr be binomials in R and G C Z n the subgroup generated by 'gi, . . . ,~gr. If f 6 (g\, . . . , g r ) , then every simple component of f with respect to ~Q also belongs to (g\ , . . . , gr) . Proof. First, each generator gi = xai — z/3i is simple, as its two constituting monomials xa\x@i are equivalent under ~G by construction. Moreover, xjgi remains simple for any 7 e N", since the relation ~ G on monomials

Monomial Curves________________________________369

is compatible with the product. Let now / be any element in the ideal ( < 7 i > - • • i <7r)- Then, / is a linear combination of polynomials of the form x^gi, with i G {1,..., r} and 7 € N n , which are simple. Hence, every simple component of / is also a linear combination of some x1 $i and therefore belongs to (gi,...,gr). D Theorem 10.1.3 ([104]) Let gi,...,gr be a system of binomials in the toric ideal P and I = (g\,..., gr). If char(/c) = p ^ 0 (resp. char(fc) = 0), then rad(J) — P if and only if

(a) pmker(VO C (gi,...,gr) for some m <E N (resp. ker(^) = ( < ? i , . . .,
(b) V ( g i , . . . , g r , X i ) = {0}, for alii. Proof. =>•) (a) First we consider the case char(fc) = p ^ 0. By hypothesis,

there exists m > 0 such that fp™ e I for all / e P. Let 7 e ker(VO and write 7 = a - /? with a,/3 G N" and 5 = a;" - x$. Thus 5 belongs to P and 5 = a — /? = 7. Set G ={?!,..., ?r)CZ".

We know that gp™ e / and gp™ = xapm - x@pm. Since the simple components of gp belong to I and hence to P (see Lemma 10.1.2), and since xapm does not belong to / because it does not belong to P, it follows that

gpm is simple, i.e., pma ~ G pm/3. Hence pma - pmf3 e G. This shows pmker(V>) C G. Next we treat the case char(/c) = 0; this case was shown in [100] using simple components. Let G = (51,... ,"gr) and 0 ^ r\ 6 ker(V'). Note that one can write rj = a — 0, where a and /3 are in N™ and have disjoint support, thus / = xa — x13 £ P and fq is in / for some prime q 2> 0. Hence

where 7, — ^ are in G and c, € fc \ {0}, for all i. Set

h = 2_,ci(xJi ~ z<5i )i

where Z = {i | qa — 7^ e I/}.

Observe that h equals the sum of all the monomials CiX6i occurring in fq such that qa — $, £ G, because qa — 7, G G iff gen — Si £ G. Expanding / 9 one has:

9-1

If qa-q/3 <£ G, then = 3**

370_____________________________________ Chapter 10

for some J contained in {1, . . . ,q — 1}. As q is prime, q must divide (?) for 0 < z < q. Thus making Xi = 1 yields Ik — (s
Since gcd(g, r) = 1, we get a - /? £ G. (b) Let a € F(#i, . . . , g r , X i ) , since F(P) = V(I) and zfj' - z^ are in P for all j ^ i, it follows that a = 0.

•<=) First we assume char(fc) = p ^ 0. Let f = x° — xe be a binomial in P and write ^ = xai — x!3i . By (a) there are integers s» such that

one may assume s, > 0 by replacing xai/x/3i by its inverse if necessary. Hence writing z^/a^ = ((xa
and using the binomial theorem it follows that M fpm is in /, for some monomial M. Since P is generated by a finite set of binomials, there is a monomial a;7 such that z7 fp™ 6 /, for all / G P. Let PI be a minimal prime of /, we will show that PI contains P. As / is graded observe that PI is also graded (see Lemma 1.2.3). If none of the variables in the support of x7 are in PI , then P C PI. Otherwise, after a permutation of the variables and the binomials, one has the inclusions:

I C (xi,...,xm,G') C PI C ( x i , . . . , x n ) ,

where G' = {
variables occurring in any binomial in G1 is disjoint from x\, . . . ,xm and

the binomials gq+i, . . . ,gr belong to the ideal (x\, . . . ,xm). Set aj = 0 for 1 < i < m and a, = 1 for i > m. Since

one derives that G' must be the empty set. If m < n, then (GJ) is again in V(<7i, . . - ,gr,xi), which yield m = n and thus P C PI- Hence P C rad(7) and rad(/) = P. Note that this part of the proof works if char(A:) = 0 by making pm = 1 . D Remark 10.1.4 The condition (a) above is equivalent to require that the group ker(^)/G is a finite p-group, and this condition can be readily validated by noticing that ker(^)/G is a finite p-group if and only if there is an isomorphism

Zn/G ~ Z x H, where H is a finite p-group.

Monomial Curves________________________________ 371 Example 10.1.5 Consider the monomial curve T = {(t,t)\t€ k} C A2,, where A: is a field of characteristic p > 0. If g = xp — yp G k[x, y], then

g=(x-y? and T = V(g). Note that g does not generate the subgroup ker(^) = ((1, —1)) of Z 2 . Proposition 10.1.6 If gi, . . . , gr is a set of binomials generating the toric ideal P of k[S], then #1, . • . ,
where 7; — Si are in L and a G k, for all i. Set

X = {i\ a - 7; e L} = {i a - Si e L} and J = { i | / 3 - 7 i e L } = {i|/?-( Since X l i r f z u " Cj(x 7i — x"5^ = 0, one has

Using x7i - x5i = (xe - x6i) - (xe - x 7 '), with 9 = a or 9 — /3 one may rewrite / as

where a — ji, j3 — 5j 6 L \ {0}, for all i and j. One may assume /? ^ 7; and a 7^ 6j for all i, j; otherwise one has a — /3 e L. Hence s

m

s

m

^ d = 1, ]T 6 j = -1 and ]T Ci2;7i = - ^ e^^', consequently x7i = a;53' for some i and j, and this shows rj = a — /? G I/. D

372_____________________________________Chapter

10

Lemma 10.1.7 Let

ip:Zn —>Z be the homomorphism induced by ij)(ei) = d{, where d\,... ,dn is a sequence of relatively prime integers. If L is a submodule o/ker(7/>), then L = ker(^) if and only if Zn/L is a free 1,-module of rank 1. Proof. Notice that ker('0)/L is torsion free if and only if Z™/I/ is torsion free. Therefore one has the equality ker(-0) = L if and only if Z™/L is a free Z-module of rank 1. d

Proposition 10.1.8 Let d\,... ,dn be a sequence of relatively prime integers and ip: Z™ —> 1 the linear map induced by ip(ei) = d{, then the set

1 < i < j
"'•

is a generating set for ker(^). Here (di,dj) = g c d ( d i , d j ) and Cj is the ith unit vector. Proof. By Lemma 10.1.7 it suffices to prove that Zn/(L) is free of rank equal to 1. Let A be the matrix with rows

re,; -

and let /» be the ideal generated by the i-rowed minors of A. Let us show that In-i = Z. For a fix k consider the (n — 1) x n submatrix B of A with rows _ (di,dk) % (di,dk) i ^ k. It is not hard to see that the integers

n

/,

dk

,^

occur as n — 1 minors of -B. If a prime p divides ak, then it divides dk, therefore In-i = Z. Since /j C Ii-i all the invariant factors of A must be equal to 1 and the proof is complete. D

Corollary 10.1.9 Let I be the ideal generated by the set of binomials dj/(di,di) di/(di,dj) £/' V " - Xj M ' "

,, , . • ^ \ (I < I < J < n).

Then rad (/) = P, where P is the toric ideal of k[S].

Monomial Curves

373

Corollary 10.1.10 7/7 = ({x^ - xf

1 < i < j < n}), then rad(7) = P.

Proof. Using the method of proof of Proposition 10.1.8 it follows that ker(f/>) is generated by

thus rad(/) = P by Theorem 10.1.3.

D

Remark 10.1.11 Let gi,...,gr be a finite set of binomials in the toric ideal P and / the ideal generated by gi , . . . , gr . In general the algebraic equality

rad(J) = P and the geometric equality

V(P) = are not equivalent (see Exercise 10.1.20), but they are related as follows: . if k is any field and rad(7) = P, then V(I] = V(P), • on the other hand if k is algebraically closed and V(I) = V(P), then by the Nullstellensatz rad(7) = P.

Now, we want to investigate what it means geometrically for a set of binomials gi, . . . , gr in P to satisfy only condition (a) in Theorem 10.1.3.

The next result is a refinement of [100, Lemma 3] in positive characteristic. Proposition 10.1.12 ([104]) Let G — (pi, . . . , where gi , . . . , gr is a set of binomials in P . If k is a field of char(A;) = p ^ 0 (resp. any field k) and pmker(?/>) C G (resp. ker(-0) — G), then

V(9l,...,gr)cTUV(x1---xn). Proof. Assume char(/c) ^ 0; for the other case see [100]. Let a = (ai, . . . ,an) be in V(gi, . . . ,
gi(a) = aai - aft = 0 Vi, where gi = xai — x^i . Hence aai — a!3i . Since all the entries of a are distinct from zero, this can also be expressed as aai~®i — 1. Now, if g - xa - x0 e P, then a - (3 e ker(^) and pm(a - /?) <E G. Thus one can write pm(a — /?) = J^j ^i(ai — A)> for some l{ € Z. Hence

anda p "* a = apm/3. Therefore apr"a -apm/3 = (a a -a / 3 ) p m = 0, which implies g(a) = aa — a® = 0. As P is generated by binomials it follows that a is in

V(P) = T, as claimed.

D

374_____________________________________Chapter

10

Monomial curves are affine toric varieties An interesting property of monomial curves is that they are affine varieties defined by binomials. In

Section 11.2 we give a criterion to decide when a set defined parametrically by monomials has this property. Lemma 10.1.13 Let k be afield and a, b G fc\{0}. If n, mare two prime integers and an — bm , then the equation zm = a has a solution in k. Proof. Since 1 = ns + qm, one has

a = (a n ) s a« m = (6 m ) s a ?m = (bsaq)m = zm, as required.

D

Proposition 10.1.14 Let I be the ideal of R generated by the set

[x^ -xf

1 < i < j < nj .

Then F = V(I), where V(I) is the zero set of I.

Proof. It is enough to prove the containment V(I) C T, because clearly one has F C V ( I ) . Let a = (ai, . . . ,a n ) be an element in V(I). One may assume o» ^ 0 for all f, otherwise a = 0. Assume n = 2. As a^ = a2l by Lemma 10.1.13 there is t £ k such that ai = tdl . Note that u = a2/td2 satisfies udl = 1 and consequently gcd(o(w),d 2 ) = 1. Thus there is an integer s so that sd2 = 1 mod o(u), setting tQ = tus readily gives aj = tfr, for i = 1,2. Hence a 6 F. Set h = gcd(di, . . . ,d n -i) and for 1 < i < n — 1 write di = d(h, where i, . . . , d'n_1) = 1. Let /' be the ideal generated by the binomials:

By induction assume

y(/')cr'= Using the equalities:

dj/(di,dj)

xi

- Xj

di/(di,dj)'

d;./(d;,d;.)' - Xjd;/(dj,d;.)' ,

= xi

.

.

1< i <j
one concludes that a' = (a\, . . . , a n _i) is in F'. Hence there t e k such that

Cj = td>', for all i < n. Set b = a^ and z — b/tdn. From the identities

Monomial Curves________________________________ 375

one readily derives zd''/^di'd^ = 1, thus o(z) divides d't for i < n, that is, o(z) = I and tdn — a£. By Lemma 10.1.13 there is ti 6 k satisfying

t = t%. Set v = an/tdn , since vh = 1 the order of w divides h and this implies gcd(o(v),dn) = 1. Hence sdn = 1 mod o(w), for some integer s. To complete the proof set to = tivs and note

tiv~ = t"v = an

and

tdi = tdivsd, = td'iv,di = a.

(l
where the last equality uses that o(v) divides h and that h divides di for i < n. D

For an alternative shorter argument to prove Proposition 10.1.14 one can use the next elementary fact.

Proposition 10.1.15 Let a\, . . . ,an be a sequence of non zero elements of a field k. If d\,..., dn are relatively prime integers such that a"1 = adi ftor all i,j, then there is t € k with a^ = tdi for all i. Proof. One can write 1 = c\d\ + • • • + cndn, for some Cj € Z. From the equalities (af ' ' ' fln") ' = al1 ' ' ' ' af

* ' ' ' an" * —ailCl ' ' ' aT ' '"ain " = a »)

one derives that t — aj 1 • • • a£™ is the required element.

D

Remark 10.1.16 Assume k algebraically closed, then Proposition 10.1.14 is valid without assuming g c d ( d j , . . . , dn) = 1. More generally, if X is the curve in A£ given parametrically by x•*>l- -—— Jf-(t] ZW!

where fi(t) 6 k[t] \ k for i = 1,... ,n, then by the extension theorem it follows that V(P) = X, where P is the toric ideal of k [ f i ( t ) , . . . , f n ( t ) ] . beautiful exposition of the extension theorem.

See [74, Chapter 3] for a

Proposition 10.1.17 Let F = {(tdl, • • • , tdn)\ t € k} be a monomial curve in the affine space A£. If di > d% >j_- • > dn and k is the field of complex numbers, then the protective closure F of F is equal to p _ jr^i td2Udl~d2

£dnudi-dn

udi~\

g pn

u

^ g ^|

H Q 1)

376_____________________________________ Chapter 10

Proof. Let F' be the right hand side of Eq. (10.1). By Proposition 2.4.29 one has F = V(I(<^(F))), where ip is the map

given by (p(a) = [(a, 1)]. Let a = [ a i , . . . , a n , & ] e T,

then /(ai, . . . , a n , b) = 0 for any polynomial / in k[x\, . . . ,xn, w] such that f(tdl, . . . ,tdn , 1) = 0 for all t € k and / homogeneous. One may assume b ^ 0, otherwise if b = 0, then f = xdl - xdiwdl~di is in /(^(F)) for z > 2, hence making w = 0 and t = ai J / d l we get a € F'. Since b ^ 0 we readily get f(ai/b, . . . ,an/b) = 0 for every binomial / in the toric ideal P

of k[td\...,td"}, hence (ai/b,...,an/b) € V(P}. As F = V(P), there is ti 6 fc such that 0^/6 = if for all i. Making t = bl/dlti and u = bl/dl one concludes a 6 F'.

Conversely let

be an element of F' and / a homogeneous polynomial of degree r such that / vanishes on ip(F). From the equality

i, ...,xn,w)= wrf(xi/w, . . .,xn/w, 1)

one readily derives that f ( a ) = 0, that is, a G Y.

Exercises 10.1.18 Let di, . . . ,dn be a sequence of relatively prime integers and ^ : Z n —> Z

the linear map induced by ^( e j) = ^»- If G1 is a subgroup of ker(i/>), then

Torsion(Zn/G) C ker(^)/G, with equality if ker(-0)/G( is a finite group. 10.1.19 Let D = (dy) be an m x n matrix with integer entries and

the linear map induced by V'( e i)

=

-^i> where D; is the ith column of D.

If G is a subgroup of ker('0), prove that G = ker(^) if and only if Z"/G is torsion free of rank equal to rank(£>).

Monomial Curves________________________________377

10.1.20 Let P be the toric ideal of the subring k[t6,t8,t9} and / = (92,93), where g% = x\ — x\, g% = x\ — x\. Then

(a) P — (§2,xl - £1X3) for any field k,

(b) Z 3 / < 3 2 , ? 3 ) - 2 x Z 3 ,

(c) rad(/) = P if char(fc) = 3, and rad(J) / P if char(fc) = 2,

(d) r = {(£V8,;9)|i e k} = {(o,o,o),(i, i, i)} = v(i), if k = z 2 . 10.1.21 Let P be the toric ideal of 12[t6,t8,ts] and let 6

3i

»3 — ^2 ~

A

3-

Then

(a) P = rad (91,53) = rad (91,9215s)

an

d S? is a binomial for i = 1,2,3.

2

(b) I(X) = (x1+x3,x2+x3,x3+x 3), where X = {(t6,ts, td)\ t € Z2}. 10.1.22 Let d i , . . . ,dn be a sequence of nonzero integers and if> the linear map from TLn to Z induced by T/>(CJ) = dj. Prove that the set

I
10.1.23 An affine variety X in fc™ is said to be a set-theoretic complete intersection if its vanishing ideal I ( X ) is a set-theoretic complete intersection. If k is algebraically closed prove that X is a set-theoretic complete intersection if and only if X can be defined by n — dim(.X') polynomials in

n variables with coefficients in k.

10.2

Symmetric semigroups

Let S ^ (0) be a semigroup of N, that is, S is a subset of N which is closed under addition and 0 € S. Note that there exists a sequence di < • • • < dn of positive integers such that

If g c d ( d i , . . . , dn) = 1, it is said that 5 is a numerical semigroup. Lemma 10.2.1 // S C N is a numerical semigroup, then

for some c £ S.

378_____________________________________ Chapter 10

Proof. Let di < • • • < dn be a sequence of positive integers generating S and such that gcd(di, . . . ,dn) = 1. To determine c write

and set

c= (\ri\di + • • • + \rn\dn)dn. It is not hard to see that c satisfies the required property.

D

Definition 10.2.2 Let 5 be a numerical semigroup and c the greatest integer not in S, this number is called the Frobenius number of S. Definition 10.2.3 Let S be a numerical semigroup and c its Frobenius number, the semigroup S is said to be symmetric if c — z G S for all z £ S.

Proposition 10.2.4 Let S be a numerical semigroup o/N and let c be its Frobenius number. Then S is symmetric if and only if

(c+l)/2=|N\5|. Proof. Consider the map


Because of the decomposition [0,c] = ( N \ 5 ) U ( S n [ 0 ) c ) ) )

one obtains that S is symmetric if and only i f c + l = 2|N\5|.

Definition 10.2.5 A discrete valuation of a field K is a mapping

v : K —> Z U {00} with the following properties. For all a, b e K: (a) v(ab) = v(a) + v ( b ) ,

(b) v(a + b) > mm{v(a),v(b)}, (c) v(a) = oo if and only if a = 0.

The valuation ring of v is the set of all a g K with v(a) > 0.

D

Monomial Curves

379

Example 10.2.6 Let K — k ( t ) be the field of rational functions in one variable over a field k. For f ( t ) € k[t] write f ( t ) = tmg(t), where m 6 N and g(t) is a polynomial with 5(0) ^ 0. Define v(f) = TO, and extend v to a discrete valuation of K by setting

The valuation v is the t-adic valuation of K. Proposition 10.2.7 Let (A,m,k) be Cohen-Macaulay local ring with total ring of fractions Q and x 6 m a nonzero divisor. If dim (A) = I , then there is an isomorphism of k-vector spaces

Soc(A/xA) -trr 1 /^,

where m"1 = {z £ Q\ zm C A}. Proof. The "multiplication by x" map

(f-.m~l I A -» Soc(A/xA), (f(z + A) = xz + xA yields the required isomorphism.

D

One of the motivations to study symmetric semigroups comes from the following result. Here we adapt a proof of E. Kunz to make it work for the graded case.

Theorem 10.2.8 ([203]) Let A - k[tdl , . . . , tdn] be the monomial subring

of k[t] generated by tdl , . . . ,tdn over the field k and S = v(A), where v is the t-adic valuation on K = k ( t ) . If S C N is a numerical semigroup, then S is symmetric if and only if A is a Gorenstein ring. Proof. Assume S is symmetric. Set m = A+ and

m""1 - {x <E K\xm C A}.

Let h e m"1 \ A. If v(h) e S, then p = v(h) > 1. Hence h e k[t] and we can write h = h\ + h^ where hi £ A and v(hz) ^ S. We claim that v(h^) = c, where c is the greatest integer not in S. If i>(/i2) < c > then since

S is symmetric 0 < p = c — v(li2) 6 S. Choose g G A with v(g) = p, then

c = v(h2) +v(g) = v ( h 2 g ) , therefore h^g ^ A, contradicting h^ 6 m"1. Thus v(h2~) > c. To complete the proof of the claim notice that v (h^) < c, for otherwise v(hz) 6 S.

Next we show that m"1 = A + Af for any / e m"1 \ A. We may assume v(f) $ S, otherwise write / = /i + /2 with f i £ A and /2 ^ S and notice

380_____________________________________ Chapter 10

A + Af = A + A/2- Hence v(f) = c and / is a polynomial in t. We may assume, by dividing / by an appropriate constant, that f ( t ) — tcg(t) where c, which gives x — / £ A and x G A + A f . Therefore the length of m"1 /A as an A-module is 1 and according to Proposition 10.2.7 this implies that A is Gorenstein. Note that the proof shows m"1 C k[t]. Conversely assume A is a Gorenstein ring. Let c be the greatest integer not in S and

{tj +tc~1A\Q<j
B

-

C

= {tc+1+i + tc+lA\0
First note dim& A/tc+lA = c + 1, because the set B U C is a fc-basis of A/tc+lA. Define (p: B -+ C

by


We now show that
socle of A/tc+1A lives only in degree 2c + 1, there is s € S1 \ {0} such that

hence s + i is not in S and thus i < s + i < S and c — (i + s) ^ S, which contradicts the choice of i. Hence c — i must be in 5 and (p is onto. Since (p is clearly injective we obtain 2

= |[o,c]ns|

and S is symmetric by Proposition 10.2.4.

D

A nice generalization of Theorem 10.2.8 due to R. Froberg will be presented below; the idea is to find an appropriate expression for the CohenMacaulay type of A in terms of the semigroup using an Artinian reduction of A. Proposition 10.2.9 ([117]) Let

T(S) = {x 6 N \ 5 x + s e 5, Vs e 5, s > 0}, where S is a numerical semigroup of N with Frobenius number c. Then S is symmetric if and only i f T ( S ) = {c}.

Monomial Curves________________________________381

Proof. =£•) Let x G T(S) and assume x < c, then 0 < c — x £ 5 and x + (c — x) = c is in 5, which is impossible, hence x = c.

<=) Let z ^ S, z > 0, one must show c — z & S. If c — z is not in 5, choose the least positive integer z such that c — z is not in 5. Since c — z ^ c,

by definition of T(S) there is s 6 5, s > 0 with (c - z) + s not in 5, hence z — s > 0, which contradicts the choice of z.

D

Theorem 10.2.10 ([117]) Let A = k[S] be the semigroup ring of a numerical semigroup S over a field k and

T(S) = {x 6 N \ S | x + s <E S, Vs <E 5, s > 0}. Then \T(S)\ is equal to the Cohen-Macaulay type A.

Proof. Let d\ < • • • < dn be a minimal generating set of 5 with di,... ,dn relatively prime and s 6 5, s > 0. Consider the map

if. T(S) -> Soc(A/tsA), (p(x) = tx+s + tsA. It is enough to verify that the image of T(S) is a /c-vector space basis of Soc(A/tsA), because if is clearly injective. Note that Soc(A/tsA) has a fc-basis consisting of monomials of the form tw + tsA, with w - s ^ S and tdi (tw + tsA) — tsA for all i, write di + w = s + Sj, where s, is in 5.

We claim w - s 6 T(S). It follows that (w-s) + s' is in S for all s' <E 5, s' > 0. Observe that w — s > 0; otherwise from w — s = si — d\ = 82 — d% we derive si = 0 and d% = s% +di, hence d% = [idi, which is impossible because di and d% are part of a minimal generating set of S. Hence w — s € T(S) and
Arithmetical symmetric semigroups Let 5 be a numerical semigroup and c its Frobenius number, that is, c is the greatest integer not in 5. The problem of Frobenius consists in determining the integer c; in non-mathematical terms this problem can be explained as follows. Given a sufficient supply of coins of various denominations find the largest amount that cannot be formed with these coins. The problem of computing and estimating the Frobenius number has been examined by several authors [117, 248, 254].

Lemma 10.2.11 Let u,v € N+ and assume gcd(u,v) = 1, then

c = uv — u — v

is the largest integer not in U = uN + vN.

382_____________________________________Chapter

10

Proof. It is easy to see that c is not in U. Indeed if c e U, then uv — u — v — riu + r^v for some ri,rz in N, that is,

u(v — n — 1) = v(r-2 + 1)

and since gcd(u, v) = 1 one derives v — r\ — 1 = qv > v, which is impossible. Next we show c + i e U for i 6 N + . One has 1 = n\u + (j,2v, hence c + i = (v — 1 + ifii)u + (-1 + ifj-2)v, and by the division algorithm one may write c + i = b\u + b%v, where 0 < b-2 < u. From the identity

c — bin — b-^v = (—bi — l)u + (u — 1 — b-2)v = — i ^ U, one concludes &i > 0 and consequently c + i G U.

d

Lemma 10.2.12 Let n > 3 and v > 1 be two integers and S the semigroup

S = diN+--- + dn^, where di — di + (i — l)v and di = r + k(n — 1) for some k,r 6 N. // gcd(di, v) — 1 and 2 < r < n, then the greatest integer not in S is equal to (k + v)di — v.

Proof. Set c = (A; + v)d\ — v, we will show c + i G 5 for i > 1. One may assume 1 < i < di. Notice

c + i - (k + l)di + X + i, where A = vdi — d\ —v. Consider U = d\N + vK By Lemma 10.2.11 A is the greatest integer not in U. Hence A + i = ad\ + bv, for some a, b in N. One has vdi + i = (a + l)di + (b + l)v. Hence b < (n — l)(k + a + l) and we can write b = j(k + a+ 1) +1, for some \<(.
Monomial Curves

383

for some ai in N. Setting

note that one can rewrite c as: n

c = (k + l)di + A = ^ aidi = sdi + pv, for some p e N. Since X £ U, one derives s < fc and c < sdn < kdn, which is a contradiction since c = kdn + (r - l)v. D Theorem 10.2.13 ([3, 107,

195]) Letr,k,n,v e N and let

di = r + k(n — 1) and di = d\ + (i — l)v be an arithmetical sequence. / / 2 < r < n ^ 2 and gcd(di,v) = I , then r = 2 if and only if the semigroup

is symmetric. Proof. =>) Let d\ = 2 + (n - l)fc. Set

c = (k + v)di - v — kdn + v and U — diN + vN. Assume z £ S. First consider the case z € U, write z — rdi + sv, r, s € N.

By the Euclidean algorithm and Lemma 10.2.12 one can write z = adn + bv, where a, b are integers so that 0 < a < k and 1 < b < d\. Writing a = k — i, where 0 < i < k, we claim that 1 < b < (n — l)i + 1. If b > (n — l)i + 1, then z = x\d\ + x<2.v, with x\ = k + v — i and x-^ = b — (n — l)i — 2. Since 0 < x-2 < (n — l)zi it is not hard to show that z G S, which is

impossible. Hence 1 < b < (n- l)i + 1. Set y\ = i and y2 = (n—l)i-b+l. Since 0 < 7/2 < (n — l)yi from the equality c — z = y\di + y^b, one gets c - z € S. Now consider the case z <£ U. One can write z = ad\ + bv, with 1 < 6 < d\ and a < 0. Set w\ = k + 1 and w? = di — b — 1. Using c — z = widi — (1 + a)di + wzv and 0 < w^ < (n — l)u>i, gives c — z 6 S. Therefore S is symmetric. •4=) Assume 5 is symmetric and keep the same notations as above. Let 2 < r < n and d\ — r + k(n - 1). As v £ S one has c - v € 5 and

c —v = where 6j € N. Set a = ST=i ^*

an<

^ n°tice that

c — v = (k + l)di + A — v = adi + qv, for some q G N. Therefore a < k, otherwise one obtains A € U. From the inequality c — v < adn < kdn, it follows that r = 2. D

384_____________________________________ Chapter 10

Exercises 10.2.14 Let S ^ (0) be a semigroup of N. Prove that the following are equivalent:

(a) S is a numerical semigroup.

(b) c + N C 5 for some c 6 5.

(c) | N \ 5 |
5 = 4M + 5N + 6N is symmetric. 10.2.17 Let S be the numerical semigroup

S = 5N + 6N + 7N + 8N and k a field. Prove that k[S] is a Gorenstein ring which is not a complete intersection.

10.2.18 Let S C N be a numerical semigroup with Frobenius number c and k[S] its semigroup ring over a field k. If

A = k[S] C k[t] has the induced grading, show that the degree of the Hilbert series of A/tc+lAis 2 c + l . 10.2.19 Let k be a field and let

where xt has degree d» 6 N+ . If gcd(di , d 2 ) = 1 and c + 1 = a\di + a2d2, show that the Hilbert series of B is

where c is the Frobenius number of the proper semigroup S =

Note 5 ~ A/t c+1 A, where A = k [ t d l , t d 2 ] .

Monomial Curves

10.3

385

Ideals generated by critical binomials

Let us begin by fixing some notation to be used throughout this section. We will always assume that

5 = M! + • • • + mn is a numerical semigroup generated by a sequence d\ , . . . , dn of relatively prime positive integers and that F is the corresponding monomial curve. The kernel of the homomorphism of fc-algebras <j>: R = k[xi , . . . , xn] —> k[S] = k[tdl , . . . , t d » ] , induced by <j>(xi) = tdi , is the toric ideal of k[S] and will be denoted by P, where R is a polynomial ring in n variables over a field k. Definition 10.3.1 A binomial

is called a critical binomial with respect to Xi if m, is the least positive integer such that rriidi € y^ dj N. Definition 10.3.2 A set {/i, . . . ,/„} is called a full set of critical binomials if fi is a critical binomial with respect to Xj for all i. Given a full set F of critical binomials, a main problem is to determine when the ideal generated by F is equal to P. The interest in studying ideals generated by critical binomials comes from a result of J. Herzog [146] showing that the ideal P of a monomial space curve is generated by a full set of critical binomials (see Theorem 10.3.10). This is not longer the case for toric ideals of monomial curves in higher dimensions: Example 10.3.3 ([99]) Let m be a positive integer and let ( d i , d z , d3,d*t) = (m 2 , 2m2 — 1,3m2 + m,4m 2 + m — 1), then v(P) > TO, where v(P) is the minimum number of generators of P.

If 5 is symmetric and generated non redundantly by four integer, then P is minimally generated by 3 or 5 elements [34]; thus even in this case P is not generated by critical binomials. For simplicity of notation /j will stand for a critical binomial with respect to Xi, for i = 1, . . . ,n. Note that we do not assume ±/i, . . . , ±/ra distinct.

386_____________________________________ Chapter 10

Lemma 10.3.4 ([146]) Let f — ™ Q i _ ^2 a3 f _ bz _ Ji 63 r _ c _ ci C2 jl — .t,^ ^2 Xg , J 2 — -t-2 J^i J/3 , J3 — ^33 J^ ^2

be a full set of critical binomials. If bi > 0, c$ > 0 /or a// i, i/ien Oj > 0 /or all i and a\ = bi + c\ , 62 = a 2 + c2 , c3 = a3 + 63 . Proof. If 02 = 0, then 03 > 03. Using the equality f

\

ras~C3 f

/I ~+~ ^3

_ ~0l

/3 — x i

~.Cl

C 2X_ a s - C 3

— -Z-i -^2

3

we derive a contradiction, hence a^ > 0, by a similar argument 03 > 0.

Define a = (-ai,a 2 ,a 3 ), /? = (61, -62,63), 7 = (ci,c 2 ,-c 3 ).

and (i>i,w 2 ,u 3 ) = a + & + 7. Next we show Vi > 0 for all i. Note a\ > 61, otherwise using xJ 1-ai X3 3 /i + /2 one derives a contradiction. Hence using x33 f i + x^~ : /2, one derives 03 + 63 > 03 and 1)3 > 0. Similarly one shows vi,v% > 0. Since vidi + v^d-i + ^3^3 = 0 we have v i = 0 for alH. D Proposition 10.3.5 Lei

6e a /M|/ sei of critical binomials and I — (/i,/2,/s)- If 6j > Q,CJ > 0 /or all i, then {/1,/2,/s} w a, Grobner basis for I with respect to the revlex ordering xi > xi > £3 . Proof. Use Lemma 10.3.4 and Buchberger's criterion Theorem 2.4.15.

d

Proposition 10.3.6 Let f,

— T a i — 7- a 2 T a 3

Jl — X±

f n — T62 _

X^ X^ , /2 — -Z-2

T-'l^^S

f

_

•''I •2'3 , 73 —

_™ x C iX C2

i

2

'

, XT C 3

3

be a full set of critical binomials. If bi > 0,Ci > 0 for all i, then the ideal (/1,/2,/s) is equal to the toric ideal P of k[S].

Proof. Let / be a binomial in P and assume we order the monomials with the revlex ordering x\ > x% > £3. As /i,/2,/s are binomials, by the division algorithm Proposition 2.4.6 we can write / = hifi + /i2/2 + /i3/3 + r, where r = x^x^x^ - x^xfxf

6 P,

ri, si < GI, f2, «2 < 62 and such that none of the monomials occurring in r is divisible by the leading term of /3. To agree with Proposition 10.3.5 one

must show r = 0.

Monomial Curves________________________________387

Assume r ^ 0. As x^3 or x%3 is a factor of r, one may assume that 83 = 0. If si < TI or s2 < r-2 we get a contradiction with the minimality of 62 or ai respectively, hence s\ > TI and s2 > r 2 . From the equality r = x^x^h,

where h = Xg 3 — xsll~r'ia;^2"7"2, one concludes h € P and rs > 03. Using the identity h ^.,r -cs r _ _rs-C3 ci c2 _ s i — n s 2 -r- 2 /t — 0,0 3 ^ 3 — j/3 x-^ x 2 — Xi 0^9

and the minimality of &2 and 01 allow us to conclude si — ri > ci and «2 - r-2 > c 2 , a contradiction because x\lxs^ is not divisible by x^x^Q Remark 10.3.7 The proof of Proposition 10.3.6 does not use the hypothesis that the integers di,d^,ds are relatively prime. Proposition 10.3.8 // ai

fi —— T Jl — •''l

— X T° 2

2

—— — fo — j2i

C3

—— T jfr, 3 — ^3

— T C l T C2

-^i J'2

is a /u/Z sei of critical binomials, then the ideal (/i,/s) is equal to the toric ideal P ofk[S}. Proof. Let / be a binomial in P and assume we order the monomials with the revlex ordering xs > x\ > x 2 . As /i,/s are binomials, by the division algorithm we can write

f = h1f1+h3f3+r,

where r = x^x^x^3 —x^x^x^3 € P, such that TI,SI < 01 and rs,S3 < €3. Assume r ^ 0. As x 2 2 or x%* is a factor of r, one may assume that s2 = 0. Note si > ri and s3 > r 3 . From the equality r = xllxr33h, where h = Xg 2 ~ xsl1~Tlx'33~r3, one concludes h € P and r% > a 2 . Therefore the identity /i _|_ ~''2-a2 r _ T Qi™»-2-02 _ si-ri s 3 -r 3 ft -f- Z 2 /i — Z j X 2 — Xj^ X3 ,

rapidly leads to contradiction with the minimality of cs because one has the inequality QI > si — TI .

CI

Definition 10.3.9 The support of a binomial f = xa — x@, denoted by

supp(/), is denned as supp(x a )Usupp(a; / 3 ), where supp(x a ) = {xi\ai > 0}. Theorem 10.3.10 ([146]) // ai a2 3 b2 bl f, — Jl — rXl — T X2 r° £3 , fo J2 — —ra 2 — rX j

is a full set of critical binomials, then the ideal (/i,/ 2 ,/s) is equal to the toric ideal P of k[S}.

388_____________________________________Chapter

10

Proof. By Proposition 10.3.6 one may assume supp(/j) 7^ {x\,x^,xz} for

at least two fa. Hence one may assume a 3 = 0 and the cases to consider are the following: (i) b± = 0, (ii) 63 = 0, (iii) c\ = 0 and (iv) c2 = 0. (i) Note 63 > cs, if 63 =03, then /2 is a critical binomial w.r.t £3 and by Proposition 10.3.8 one concludes that P is generated by {/i,/2}- If &s > £3, from the identity t i ^bs-ca f _ b2 /2 ~r X 3 /3 — -12

^.bs-cax ciX c2 ^3 l 2 >

we get C2 = 0. Thus

/2 = /2 + Z3 3 ~ C3 /3 = 42 - x^'x? and f

_

£,

i

™Cl-Ol /

/3 — 73 T 2
_

C3

_ C l - O lX 02

7l — X 3 "~ • l i

2 •

Applying Lemma 10.3.4 to /i, /2, /3 yield ci = a\. Hence /3 is critical w.r.t xi, using Proposition 10.3.8 one concludes that P is generated by {/2,/s}(ii) As a2 > 62 and 61 > a1} from /i - xf~b2 /2 = x^1 - z"2"62^1 we get A = -/2 and P = (/i, /3) by Proposition 10.3.8. One may now assume 61 > 0 and 63 > 0, (iii) using x%2~ 2 / 2 + fs we readily see that this case cannot occur, (iv) using x^~ 2 / 2 + /i we derive a

contradiction. This proof is due to A. Alcantar [3].

d

Proposition 10.3.11 ([146]) If P is the toric ideal of k[tdl, td\ t d s ] over

a field k and d\, d 2 , dy, are relatively prime, then P is a set theoretic complete intersection.

Critical binomials in four variables Let / be an ideal generated by a full set of critical binomials / i , . . . , /„, one fi for each variable £;. It would be desirable to find effective means for

determining when / is equal to the toric ideal P. Here we address the case n = 4 and char(fc) = 0 by presenting the elements of an algorithm, having / i , . . . , /4 as input, that decides when / = P occurs. In the remaining of this Section we study ideals generated by critical binomials in four variables. Our aim is to determine when the ideal generated by a full set of critical binomials is prime and in this case determine its type. If the presentation ideal P of a monomial curve in A^ is generated by a full set of critical binomials, produced by any critical-binomial-algorithm,

we predict how these generators should look like (see Theorem 10.3.15). Some of the results presented here can be shown using different methods introduced by W. Gastinger, E. Kunz and R. Waldi [124, 125], and by H. Bresinsky [37]. A complete description of almost complete intersection monomial curves in A^ was given in [124].

Monomial Curves_________________

__

389

Lemma 10.3.12 Let a

f. — T i ji — -i-i

_ ~ai Ta-k Ta,e

bi bj bk &j AJ. u,( ,tj e_— J-i •L'J -i>k

_

bt r _ -LI , j j — -ijCj

ck ct Ci &i -IK -LI

be a sequence of critical binomials. One has

(i) // dj > 0, at > 0, bi > 0, then a, > bi, bt > a^. (ii) // flj > 0, at > 0,bi > 0 and ak = bk = 0, then a,j + bj > Cj. Proof, (i) If hi > aj, the following equality implies that jt is not critical ^bi-cii bk bj f , _ f bt _ 6 ; -a; aj+bj ak+bk ae x x •^i ^k ^j JiJt ~ ^l •''i j k •''t '

Hence aj > 6j. By symmetry b( > ai . (ii) The minimality of Cj together with the identity xbj ji - xf-bi f t = x^-bixbee - xa^ ' xatl yields the inequality a

i + bj > Cj.

D

Lemma 10.3.13 Let f\, . . . , /4 be the full set of critical binomials: f — ™°i _ ™ Q 2 ™ a 3 ™ a 4 J I — •*•! ^2 3 4 ' f _ «,e4 ei e 2 £3

Ji — -^4

f _ 62 _ J>i bs J/ b^ J 2 — J'2 ^1 "^3 4 '

r _ c3 _ J 3 — ^3

x

ci c2 c4 i J'2 4 '

•t'l ^1 ^3 •

// ttj > 0, bi > 0 for all i, then I = (fi,/2,h, A) is not a prime ideal.

Proof. Assume I is prime. As in Lemma 10.3.12 we have a\ > &i and 62 > 0"2- From the identity ™&3 ,,64 f i ~oi-6i j? _ i a,2( x ai-bi X 6 2 -a 2 •Z-S -C4 /I T 2-1 72 — • 2 ^ l 2 "~

3

4

I ~

2

l

we derive gi £ / and there is a monomial M so that , , ~oi -61 62-02 _

Xl

X2

f (\\ ) W

MTCI rC2 r°4 J W J / i J-2 X 4

Unr I

('m9 s \

>

~ \ (ii) M^ 1 ^ 2 ^ 3 .

(W Z)

'

Therefore c4 = 0 or e3 = 0. First we consider the case c4 = 0. If e\ > 0, then a\ > ei and 64 > 04. From the identity ~ez 63 f , ai -ei r _ _ 0 4 / ria i - e i C64-04 _ •C2 X3 /I T •cl 74 — ^4 V' l • 4

7.a2 + e 2 2:

2

o3+e3N _ a4 _ -^3 / ~~ ^4 52

we obtain g2 € / and there is a monomial M so that (i) (ii)

Mxl x2 x4 Mxeixe2xea

— MX^XZ

or,

(in^ (,-LU.O;

If (i) holds, then c2 = 0 and a\—e\ > c\, since /i is critical and /s = Xg 3 — xj1 we have a contradiction. If (ii) holds then, 62 = 63 = 0 and ai — ei > ei, which is impossible because /i is critical and /4 = x^4 — xe^. Hence e\ = 0.

390_____________________________________Chapter

10

Next we show e2 = 0. Assume 62 > 0, then &2 > £2 and 64 > b^. From the identity T x

ei

l

e3 -r f -4- T 6 2- e 2 f< — T.^f'i. 62 " 62 C^ 64 " 64 _ Tbi+ei bs+e3\ _ X™6 4 X 3 /2n + X 2 /4 — X 4 (,X2 " 4 •''I •''S / ~ 4 93

we obtain 53 € J and there is a monomial M so that ™6 2 -62,,64-64 _

tLr)

J/ A

——

Cl C2 C4 l X
(i} J( W

\

I

A//> Mx

/• • \

T I ^ - P l P o f Q

I 11 1

A/7 'T*

V . V /

T

*>»

1 2 3

^

— — •——

Cl C2 A//> JW X X r 71 f

1 2 fr>

/I// /r*

Pi

* CY>

2

3

O Ior

i

f i n 41

IJ-\J.TTj v

•*

'

If (i) occurs, then c\ = 0 and b2 -e2 > c 2 , which is a contradiction since /2 is critical and /3 = Xg 3 - x 2 2 . If (ii) occurs, then e\ = e3 = 0 and &2 - e2 > e 2 , which is also a contradiction since /2 is critical and /4 = x^4 - x 2 2 . Hence e2 = 0. We claim that c\ = 0. If c\ > 0, then ai > c\ and c3 > a 3 . Using the equality ™ C 2,-C4 f _|_

a i — c i c __

a,3(

ai—ci

eg — as _

02+62

04 + 64 -j _

03

we see that 54 G / and there is a monomial M so that MTCI XT C 2XT C 4 JW:r l 2 4 7 i *1 - f i P o f 1 ^

f (\\ ~ai-ei,_C3-a3 _ J W U/ o — S /.. \

it/1

J.

O

i

Mil

A/7 T i

v^-*y

T

• '-*- *^i "^9

^ /y« 'iJ

3

— — ——

—

7WVClX T C2 Mx l 2 TI /r Pi /I// /-f>^3

3

ornr

;

/ i1n- L U^\ .tJy V

'

If (i) holds, then C2 = 0 and a\ > c\, which is impossible since /j is critical and /3 = Xg 3 - xj 1 . If (ii) holds, then c3 > e 3 , which is impossible because /3 is critical and /4 = x^4 — x^3. Therefore GI = 0. Altogether we have

ci = 04 = 0 and e\ — 62 = 0. We claim that 62 > c 2 , otherwise if c2 > 62, then from the equality X2

/2 + /3 — Xg — X j X 2

Xg £4

and using the minimality of c3 we obtain a contradiction. To complete the proof for the case c4 = 0 notice that the inequality 62 > c2 also yields a contradiction since /3 = Xg 3 — x22 and /2 is critical. It remains to consider the case e3 = 0. We may assume c\ > 0. We claim that ci = 0. If ci > 0, then by (10.5) either (i) or (ii) hold. Since c4 > 0 (ii) holds, as a result e2 = 0 and a\ > e\, which is a contradiction since /i is critical and /4 = x44 — x^ 1 . Hence c\ — 0. Next we show e\ = 0. Assume et > 0, by (10.3) we have that (i) or (ii) hold. If (i) holds, then c2 = 0 and 64 > 04, which is a contradiction since /4 is critical and /3 = Xg 3 — xc£. If (ii) holds, then 62 = 0 and a\ > ei, which is a contradiction since /i is critical and /4 = x 4 4 — x^ 1 . Hence e\ = 0. Altogether we have c\ = 0 and e\ = e3 = 0. We claim that 62 > e 2 , if on the contrary e2 > &2 from the identity 1 62—62 j

i

r __

64 _

61 62 — b2 b3 b4

we contradict the minimality of 64. To complete the proof notice that 62 > 62 also leads to a contradiction since /2 is critical and /4 = x^4 — x 2 2 .

Therefore / is not a prime ideal.

D

Monomial Curves

391

Proposition 10.3.14 Let I — (/i, /2,/3,/4) be an ideal generated by a full set of critical binomial such that f\ = x^1 — xfx^sx^, Oi > 0 for all i and fk = xk ~ x/ f°r some k. If I is prime, then we can re-order the variables so that /2, /s, /4 can be written in one of the following forms

(a) /2 = 42 - xl*, h = -h and /4 = <4 - x e j x f , where a > 0. fUN

f

_

~,b2 _

\y) /2 — •I' 2

™C 3

X

f

_

C3

3 J /3 — -<<3

64

i

_

64

•t4 i /4 — ^4

62

^2 •

Proof. As Oi > 0 for all i we may assume ji = x22 — z 3 3 . Write J3 ~~ *^3

— ^i *^2 *^4

and

J 4 ~~~ *^4

— *^i *£o *^3 '

If GI > 0, then Lemma 10.3.12 and Lemma 10.3.13 yield c2 = 0, ai > ci, GS > as and 04 > 0. Using the equality T X

C4

f,

-I- T a i ~ C l fn

4 j l T J/i

— T° 3 ( ' T a l ~ c l T C 3 - a 3

^ 3 — J'S ^i

^3

~0,2

0,4+C4\ _

^2 ^.4

^0-s

j — Xg (/

we obtain g £ / and o^1"^1^3"'13 = Mx^x^Xg 3 for some monomial M. Therefore e2 = 0, and by Lemma 10.3.12 and Lemma 10.3.13 we obtain ei > 0 and 63 > 0. From the equation 3

ei+ci\ _ —

i, — — /4

we obtain g & I and this is a contradiction since Xg3"61^4"04 is not a multiple of any of the monomials that occur in /i , . . . , /4. Therefore ci = 0. Assume 04 > 0, note that Lemma 10.3.12 yields C2 = 0. If 63 > 0, then ei = e2 = 0 and it follows (see argument below) that /4 = — /3, as required. On the other hand if 63 = 0, we have 62 > 0 and from the equation ei

03\ _

ft — /4 —

—

we obtain 5 £ / and there is a monomial M so that x®l~eixe44~a4 = Hence e\ = 0 and we are in case (b). We may now assume c\ = c4 — 0. By minimality c^ > 62 and since /3 + x^ 2 " 2 /2 = a^3 — o;^2" 2 x 3 3 we obtain the identity c^,d^ = (02 — ^2)^2 + ^3^3, which implies fz = —ji-

Write

/4 = x^4 — x^x^x^3 , we claim that e\ = 0. If e\ > 0, then a\ > e.\ and 64 > 04. Therefore from the equality ^ez e3 f , ™ o i - e i 7- _ _ a 4 / a i - e i J 64-04 2 •"'S J l ' -"-l J 4 — ^4 V-^l '4

X

J/

02 + 62 ~a3+e3 \ _ 04 2 •''S / — •''4 a

we have 5 6 /. There is a monomial M so that z"1"6^4"04 it follows that 62 — 63 = 0, which is a contradiction since /i is critical. Therefore e\ = 0 and the proof is complete. Cl Next we present a classification theorem.

392_____________________________________ Chapter 10

Theorem 10.3.15 ([5]) Let /; be a critical binomial with respect to the variable Xi for all i and I — (/i , /2, /s; /4) • If I is a prime ideal of height 3, then, up to permutation of the variables, /i,/2,/3,/4 can be written in one of the following forms f A\ A

f

1 J

f

— ™0l

~,a

04

f

_

t>2

„,&!

Jl — ^1 ~ £33 £4 ; /2 — X2 — r ei e 2 e 3 _ 64

J 4 — •*';[ ^2

3

64

f

_

™C

C

C

— Xl Z 4 , /3 — £33 — X2 2 X4 4 ,

4 '

where ai > 0, 6j > 0, c^ > 0, e^ > 0 for all i. (f>\ 13

I- ;

f

_ ™°i

f

_ r-,0-1 _ X™02

/: _ ™fl2 _ ~(>3 /• _ -63

f

_

J1

_

T,02™03

/•

/I —xi

~

Jl — x l

™<M

/I — #1 f

X™,a2X

1

3

a 3X a 4

r

_

4 i J 2 — X2

2 > /2 — X 2

T,"2

™02

~ ^2 i /2 — ^2

— _0l _

_

b2

63 f

f

_

b3

~ X3 i /3 — X 3 ~ 62

i)l™63

Jl — J-i ^2 ^3 ! J 2 — J-2 f. —T e 4 _ r e i T e 2 T es J 4 — ^4 **^ i "^2 3 '

•*!

X

/•

/2, /4 — £4

X 2 ^3 •

ai

^ _

64

C4

f

C4

ei e 2 «,e 3

-i-i i J 4 — ^4

•''S i y3 — -^3 ~,b3

f , —
_ _f

~~ X3 i /3 —

_

X

_

4 i /4 — ^4 C3 _

™Ci

3 J /3 — ^S

•*•!

X

-f-j ^2 ^3 • „,«!

—xi •

C2

2 )

where ai > 0, 6j > 0, Cj > 0 and e» > 0 for all i. (T\\ V1^/

f Jl f, J4

— _o _ _ — •''l •''2 X 3 •L4 ' J 2 — ^ 2 e — T e4 _ A T 2™e3 — J<4 2 •''S >

_

_ 3 ' ^3 — ^3

1

•''i •i2 >

where 03 > 0, 04 > 0 and &i > 0, Cj > 0 for all i. CP\

l^J

f

— T01

Jl — xl

XT°

4

f

— T^2

4 ) /2 — -^

/Y-'l^'s

f

_

r,.c3

^CiX

-^l -C3 > /3 — X 3 ~ •''l 1

C2

f

_

_6 4

2 i /4 — -(/4

li

62

63

2 •''S >

where 6» > 0, Cj > 0 /or aH i ana 63 > 0.

Proof. We set Sj = supp(/j). Assume that |5j| = 4 for some f, say |5i| = 4. Let /i = xl1 - x^xfxl\ ^ > 0 for all i. By Lemma 10.3.13 we have |S;| < 3 for i = 2,3,4. Assume that |5j| = 3 for i > 2. If xi ^ (52 U 5s U 54), then (C) occurs, hence we may further assume x\ € 5, for some i > 2, say zx e 52- We claim that x\ e SB U 54. If x\ $ S3 U 54, then (after permutation of variables) we have f

— ~ai _ J~a2 a3 a* l — •'' ' """

f

—

b2 _

bi

63 r _ —

c3

where a^ > 0, 6j > 0, Cj > 0, e$ > 0 for all i. Using the identity ™62 f

I ™C3-e3 f

^2 /3 + X3

_

C4/

C3-e3

/4 — ^4 1^3

e4-C4

^4

™C2 + e 2 \ _

~ 2-2

™C4rt

J ~ •C4 5

we obtain g G / and this is a contradiction since Xg 3 " 63 ^^ 4 " 04 is not a multiple of any of the monomials that occur in the fa's. Hence x\ € S3 1)84. Assume that x\ £ 83 \ 83 D ^4 (the case x\ £ 84 \ Ss fl 84 is similar). Notice that 52 = £3 , otherwise after permutation of variables we have

and, as before, this leads to a contradiction. Thus 52 = 53 and we are in case (D). Now we may also assume x\ G 83 fl 64, by similar arguments it

Monomial Curves________________________________ 393

is not hard to see that case (A) must occur. This completes the proof for

the case \Si\ = 4 and \Si\ - 3 for i = 2,3,4. Notice that if |5i| = 4 and l^l = 2, then by Proposition 10.3.14 we are in case (B) or (Bl). The remaining cases are: (i) |5»| = 3 for all i, (ii) |5j| = 2 for all i and (iii) \Si =3 for some i, \Sj\ = 2 for some j and \St < 3 for all i. Assume (ii). If Si — Sj, then fi = — fj and it follows that we are in case

(B). Hence we may assume 5» ^ Sj for i ^ j. Up to re-ordering of variables there are two possibilities: („,} f — x~ai _ X™a 2 r _ X 62 _ X 63 f _ ™C 3 _ ci j- _ x e4 I",) 7l — i 2 •> J 2 — 2 3 i /3 — 2-3 2-1 ; /4 — £ ( a\ f _ x~ai ™a2 f _ ,_i>2 (P) II — l ~ X2 i J2 — X2 ~

ej, k '

x

63 /• _ c3 c4 jr _ £4 ™ei 3 i /3 — ^3 ~~ £4 i /4 — £4 —x\ •

X

Assume (f)) and consider the equality h = x^-°'(/i + ^2 2 ~' 2 / 2 ) + /4 = xt - ^i1'*1^2-"2^3-

Applying Lemma 10.3.12 to /s and h yields a\ = e\ and a^ = 62- Because of the equality /i +/2 +/4 = x1^4 — Xg3 it also follows 64 = 04 and 63 — c^, that is, we are in case (B2). The same argument can be used in (a) to obtain (Bl), more precisely, if (a) holds we can use h — x°11~ai f i + /3 = Xg 3 - xj1"0^2 and /2 to show that c\ — a\ , and fi+fs= x^3 — x%2 and /2 to show that GS = 63 and 02 = 62 • Assume (i). There are essentially the following six possibilities: (r,\ f, _ ™°i I, a^ ^1 — • t i a

™.a2 03 /• _ b2 _ hi bs r _ c 3 •i2 ^3 > ^ 2 — X 2 •''I •''S > J3 — -^s a

Tl-i^ f, _ ™ i _ ™ 2 ™ a s U

f

— ^.bz

C r \ f, _ ^.a-i

ai,

f

_

by

02

a3

JT JT

_

62 _

02

a

/• _ ~bi

V 7 Jl — X j

^2 ^3 i 11 — J-2

ti

1,^7 J 1 — •''l f,

_„,
Jl — X j fp^ f c

— ™ai

a-i

2

3 ' J2 — "^2

J/2 Xg , _/2 —

bi

b

f

6l

63

f

hi

b*

f

_ -C

J-j J^g 3 , J3 — Xg 3

ci c4 r _ c 4 -^i J'4 > J4 — ^^ a

C4 f

_ ™6 4 _

Cl

C4

JT

ci

c

r _

x-^ x^ , _/4 — x^

ei 63 x^ X j . e _,e

X 2 2x^ 3 .

b3

•''S

_

— _

™C 3

c _

-64

e

61

62

ez

e

\ ) Jl — x^

X2 ^3 3 , ^ 2 — X 2

X j x^ , js — Xg 3

x-^ x^ 4 , J4 — x^ 4

X 2 Xg 3 .

mr r _ „,«! l j /I — -Z-i

o2 03 <• _ b2 _ hi 64 / _ ™C 3 ^2 -""S ' /2 — -^2 •''I ^4 ) /3 — •E3

C2_c 4 r _ e4 X2 £4 , /4 — X4

ei e 3 X x Xg .

Observe that (c) and (d) correspond to (C) and (D) respectively. It is not hard to check that in the other cases, using f i and /2, / is not prime.

Assume (iii). By symmetry we may assume /i = a;"1 — x^Xg 3 , a$ > 0 for all f and j = 2 or j = 4. First we consider the case j = 4, that is, we can write /4 = x^4 — xekk for some k. Set /2 = x^2 — x^x^x^4 and /s = Xg3 — x^x^x^ 4 . If 61 > 0, then a\ > hi, b% > 02 and from the identity 9,637.''4

3

j- _i_ , _ a i — &i f _ 0 2 / 0 1 - 6 1 &2-a 2 ™a3 + 63«6 4 \ _ a2 4 •> 1 ' 1 ^2 — X2 ^x^ X2 — Xg X^ ) — X2 y

we obtain g 6 J and x"1"6^2"02 = Mx^x^x^ 4 for some monomial M. Hence ai > ci, 62 > C2 and 04 = 0, it follows readily that ci > 0 and C2 > 0. A similar argument shows 64 = 0 and we are in case (C). The case GI > 0

Chapter 10

394

is symmetric since we can interchange x% and £3. We may now assume hi = c\ = 0. If 64 = 04 = 0, then /3 = -/2 and (B) occurs. If &4 > 0 and C4 > 0, then k = 2,3 lead to case (B) and k = 1 lead to case (E). If 64 > 0 and 04 = 0, then k = 2 or k = 3 and (B) or (Bl) occur. On the other hand if 64 = 0 and £4 > 0, then c% = 0 and k = 1 cannot occur, observe that k = 2 and A; = 3 lead to cases (Bl) and (B) respectively. To complete the proof of

(iii) it remains to consider the case j = 2. Assume j = 2, in this case we can write /2 = x22 — xkk for some k and /4 = x|4 — x\lx^xe^ , note that k ^ 1.

First assume ci > 0, using f\ and /s we obtain x^~Clxc^~a:i = Mx^x^x^3 , hence 62 = 0, ai > e\ and GS > 63. Notice that e\ > 0 and 63 > 0, otherwise we are in case j = 4. Therefore 04 = 0, k = 4 and we are in (E). We may now assume c\ = 0, notice k ^ 1. If k = 3, then (B) or (Bl) occur. If k = 4 and e\ = €3 = 0, then (B) occurs. If fc = 4, ei = 0 and 63 > 0, then 62 = 02 = 0 and (B) occurs. We are going to show, by contradiction, that the case k = 4, c\ = 0 and e\ > 0 cannot occur. If k = 4, ci = 0 and e\ > 0, then using /2 and /4 gives 62 = 0 and we can write f

_

~ a l _ J,,,a

J l — •''i

'2

where a» > 0 for all i and f — /4 —

e

> 0. Observe that 64 > 64 and consider * f,

4-

fn

— Tb2

/4 ~r /2 — ^2

e ™ e l™ X 3™,('4 X

~ xl

3

4

e4

r

fc

T

•* •

It follows, using /4 and /i, that 62 > ^2 and ai > ei From the identity ™Q3 /:' i ™62-a2 r _ _ei /-oi-ei X 62-02 _ X7,O3+e3X 7 ,f 4-e4 \ _ -.ei _ p r J — ^l 3 t -") 3 /4 i •C2 /I — -^l I2-! 2 3

X

we obtain g £ 7 and there is a monomial M so that ai_e

62_a2

_ ~

i (ii)

Mxc2ax?

or,

If (i) occurs, then 04 = 0 and 62 > C2 > 0, which contradicts the minimality of 6 2 - On the other hand if (ii) occurs, then e3 = 0 and ai > e\ > 0, which contradicts the minimality of a\ . Therefore in both cases we obtain a contradiction and the proof is complete. D Remark 10.3.16 Here we keep the assumptions and notation of the theorem above, (i) In (E) a4 > 64 and e2 = 0 is not possible. If this is the case we have fo — , J3 —

f, — , /4 —

-63 L 3 '

This gives the polynomial x^~ /4 + f i = x^1 - x^'^xf & I from which since e3 > c3. Thinking about it as a diophani - i 2 3 23 C3 e 1 tine equation, > ci since c2 > 0. Thus - xx ~ ~ ' e / and

Monomial Curves

395

a\ — ci > 0 since c 0, a contradiction to «i being minimal. Therefore in

(E) we have b^ > 0, Cj > 0 for all i, e3 > 0 and 62 > 0. In (E) if 04 > 64, then I = (/i,/2,/3,/4), where /{ = A +a^~ e '/4 - z?1 -a#z£X 4 ~ e 4 . that is > / has a generating set as in (D). In (E) if 04 = 64, then / = (/{, AJ A > A0> where f{ = A + /4 = ^'i1 — £22;C33 an<^ -^ has a generating set as in (C). (ii) Assume (D), then it follows from the proof of Proposition 10.3.23

that / has a generating set as in (C). (iii) If (B2) holds, then / = (A, A, fa, A), where /£ = /3 + /4 = xb3s -z?1 and / has a generating set as in (Bl). If (Bl) holds, then I — (f1> /2, /g, /4), where /3 = A + /s = — AJ after permutation of xi and 24 we obtain that / has a generating set as in (B). From this remark we obtain the following

Corollary 10.3.17 Let I be an ideal generated by four critical binomials, one for each variable. If I is a prime ideal of height 3, then there are critical binomials A j A ) / 3 , / 4 with respect to X i , X 2 , x ^ , X 4 generating I so that, up to permutation of the variables, /i,/2,/3,/4 can be written in one of the following forms.

where at > 0, 6» > 0, a > 0, et > 0 for all i.

where 0$ > 0,6j > 0, c, > 0 and &i > 0 /or a/? z. Let A > A, A > /4 be a full set of critical binomials having one of the forms (A)-(E) as in the theorem above, and / = (A, • • • , A)- The rest of the section is devoted to find the arithmetical conditions, on the initial data, for / to be prime. Proposition 10.3.18 ([37, 124]) Let I = (A, A, A, A) be an ideal generated by a full set of critical binomials:

so that a,i > 0, bi > 0, Cj > 0, e, > 0 /or all i. Then I is prime if and only if (a) a\ = bi + ei, 62 = c2 + e 2 , c3 = 03 + e 3 , 64 = a4 + 64 + c 4 , and (b) d4 =

396_____________________________________ Chapter 10

Proof. Assume / is prime. By Lemma 10.3.12 we have <2i > BI,

63 > 62,

63 > 63,

64 > 04,

64 > 64,

64 > 04.

From the equalities ~e 2 e 3 j? ™ « i - e i j? 2-2 Z3 /I ~ ^1 /4 -ei 63 j,,62-62 f •^l 2-3 /2 — ^2 /4 ~ei

^1

Z

62 J?

™C3-e3 f

2 /3 ~ X3

/4

_ — _ — __

—

a^r e 2 03+63 #4 ( — %2 % ™&4 f x™ei + 6i e 3 "^4 V l ^3 C4/

61

^4 V"21!

Z

, Q I - C I 64-04*, + #1 X4 J , X 6 2 -e 2 e 4 -6 4 \ ~T~ 2 4 /

C2 + C2 _ ,

2

_C3-63

+ ^3

64 — C 4 \

^4

J

_ — _ — _

—

™O4 „ X 4
C4

4 53

we obtain 51 , 52, 5s 6 /• Let M be the set of monomials that occur in /j for i = 1,2,3,4. Notice that any monomial p that occur in gi can be written as p = pip^, where pi is a monomial and p% is in M. As a consequence «i = &i + ei , 62 = c2 + e2 , c3 = a3 + e3 . Set a = (-ai,0,a 3 ,a 4 ),

/3 = (&i, -63,0,64),

7 = (0,c 2 ,-c 3 ,c 4 ),

6 = (ei,e 2 ,e 3 , -e 4 ),

and consider the exact sequence. 0 —> K = ker(7/j) —^ Z 4 A Z —> 0, V(ej) = d{.

Since w = a + /? + 7 + (5e K we obtain e4 = 04 + 64 + 64. By Theorem 10.1.3

K is generated by a,^,j,6. Hence Iz(A) = Z, where yl is the matrix with rows a,/3,7, <5 and /s(A) is the ideal generated by the minors of order 3 of A. The equality —6 = a + /3 + 7 shows /s(A) = (^41,^43,^43,^44), where A^j are the cofactors of A corresponding to the last row. Since the column vectors C = (^41,^43,^43,^44) and 77 = (di,d2, 1^3,^4) are in the kernel of A we have ker(A) = (JL = r]TL. Therefore d^ = 016303 — 036163 and the first part of the proof is complete. Conversely assume (a) and (b) hold. Observe that £ 6 ker(^l) = rjl and write £ = 077, a € Z. Using (b) gives |a = 1, that is, Z4/(a,/3,j,5) is a torsion free Z-module. This readily implies ker(i/;) = (a,/?, 7, (5). An application of Theorem 10.1.3 gives rad (/) = P. We claim that /i , /2, /s, /4 form a Grobner basis with respect to the reverse lexicographical ordering xi > x-2 > X3 > £4, it is enough to show that the 5-polynomial of /j and fj reduces to zero. As we may assume that the leading terms of fi and /_/ have common factors the claim follows from the identities S(fl, A) = ~X?X?f3

- <4+C 4 /2,

5(/2, / 4 ) = -^44/l - ^4+64 /3

From the claim we derive that £4 is a regular element on R/I, i.e., fi// is a one dimensional Cohen-Macaulay ring. Altogether P is the only associated

Monomial Curves________________________________397

prime of /. Consider the cofactor J^ of the Jacobian matrix J of /, it is given by 03-1 3-1

64-1

04 + 64-1 X4

Since f £ P, Theorem 3.5.10 shows rad (/) = /. Therefore I = P.

D

In the following corollary we keep the same assumptions and notation as in the proposition above. Corollary 10.3.19 // / is prime, then R/I is a Cohen-Macaulay ring of type 3. Proof.

Since / is prime a\ > e\ > l,b^ > &i > l,cs > e3 > 1. Set

S = R/I and A = S/x^S. Notice A ~ Ri/I\, where R\ = k [ x i , X 2 , x 3 ] and J.\ — V 1 "i "^fy

t Xv ; X -> Xn Xo }• oGTi

V = Socle(A) = (J i: m)/Ji. It is readily seen that the socle of A is equal to V

—— (J.'^.X-i

Xf)

X o

J'^-'l

Xf)

Xo

I'^'l

Xiy

X>y

} I J.]_ ,

Therefore dimk(V) = 3 and the type of 5 is 3.

D

Proposition 10.3.20 Let f, — T ai —Xr a 2 7X,a 3X a4 f _ X 62 Jl — xl 2 3 4 > J2 — 2

63 / • _ _ / • t e4 3 i /3 — /2 , /4 — #4

X

e2 e3 ^2 ^3

&e a sei o/ critical binomials with respect to xi,x2,x3,X4 respectively. Let I

=

( / i i / 2 , / 4 ) - T"ften I is a prime ideal if and only if d3 = 016264.

Proof. The proof is the same as that of Proposition 10.3.18. It is enough to make two observations: (i) /i, /2, /4 form a Grobner bases with the rev. lex. ordering x\ > o;4 > x3 > x%, hence x2 is regular on R/I. (ii) The polynomial is an element in the Jacobian ideal of / which is regular on R/I.

D

Proposition 10.3.21 Let f _ ™ f i _ ia 2 a 3 f _ a 62 _ 61 63 f _ Jj™C3 _ ci L c2 j _ i e 4 j l — •''i • 2 ^3 ' ^ 2 — '2 •''I •''S > J3 ~ 3 -^i • 2 > J4 ~ • 4

ei e 2 e 3 *'\ J'2 •''S

6e a set of critical binomials with respect to x\,X2,x3, £4 respectively. Let I = (A, /2, /s, /<0- // «i > 0, bi > 0, Ci > 0 and e» > 0 /or «Z? i, then I is a prime ideal if and only if di = 64(6223 — 6322).

398

Chapter 10

Proof. We proceed as in the proof of Proposition 10.3.18. It is enough to make three observations: (i) By Remark 10.3.7 /i = (/1,/2,/s) is a prime ideal of height two and /4 is regular on R/I\. Hence R/I is Cohen-Macaulay. (ii) The polynomial

is an element in the Jacobian ideal of / which is regular on R/I. (iii) By the previous lemma a\ = b\ + GI , 62 = «2 + c-2 and 03 — 03 + 63. n

Let I be a prime ideal as in the proposition above, then we can also determine its type. We have

Corollary 10.3.22 / / / is prime, then R/I is a Cohen-Macaulay ring of type 2.

Proof. Set 5 = R/I, A = S/xiS, and /i = (x^x^x^x^Xg 3 ). Since

Socle(A) ~ (/^x^"1^3"1^4""1,^2"1^3^1^4^1)/7! for e4 > 1 5 is a Cohen-Macaulay algebra of type 2.

D

Proposition 10.3.23 Let f _ ~ai ™a 2 03 04 f _ 62 ~,bi b3 f _ c3 c\ c2 f _ e4 e2 e3 Jl — xl ~ X2 X3 X4 ' /2 —X2 ~ xl X3 i / 3 —X3 ~ xl X2 > /4 —X4 ~ X2 X3

be a set of critical binomials w.r.txi,x<2,X3,X4 respectively. Ifa^ >0,<24 >0 and 6j > 0,Ci > 0 for all i, then I — (/i,/2,/3,/4) is « prime ideal if and only if di = 64(^3 — bzc^) and 0,4 = qe^ for some q G N. Proof. Assume / is prime. Consider the identity -r-C2 f, X

_1_ ,.01-Cl f-

1 n +xl

— -OS^-Ol-Cl

/3 — 2-3 ^i

C3-03 _

2^3

02+C2X 0 4 \ _ X^^s

^2

4 J ~ 3 5-

Since I is prime 3 6 / and there is a monomial M so that ai-d^cs-os _ J ''I

W

JKia

-l *3

°r'

(ii)

Since (ii) can not occur we have 01 > b\ + GI, hence Lemma 10.3.12 yields 01 = bi + GI. Let a( be the least positive integers so that a(di G c^N + dsN and let /" = x"1 — x^Xg 3 . By Lemma 10.3.4 a( - bi + ci, 62 = a2 + c 2 , 03 = 03+ 63, in particular a( = a\. Notice o2 > 02 and o3 > 03 (see argument below). From Q a f ' l _ f, — ,,,02 Q 3 / a 4 _ —

2- 2 03-03 x

Monomial Curves________________________________ 399

we obtain 04 > 64. By the division algorithm we can write 04 = qe^ + r, with 0 < r < 64 and q > 1. The equality /
\

f _ ~oi -.02+962 X 03 + 963 X r _ f , j; I V^ o 2 + (z-l)e 2 a 3 + («-l)e3 o 4 -ie 4 7l — •El ~ ^2 3 4 — /I i" 74 I / y X2 X3 X4 I \i=l /

shows /' = (/{, /2, /s, /i) = /. We now prove r = 0. Since fl

_

Jl

fH

_ ~a2

Qj

-.02 + 962

J l — •i2 ^3

X

2

03 + 963

• fc 3

X

r

4

we conclude =

(a2 - a2 - ge 2 )d 2 + (03 - a3 -

=

(62 - c2 - o2 - qe2)d2 + (c3 - 03 - 63 -

To prove r = 0 it is enough to show that the coefficients of d^ and d3 are non-negative. From the equality _ 0 l -Cl JT

^1

,

C 2 f' _

03 + 963 /

/3 + ^2 /I — ^3

03-03-763

Qi-Cl

x

02+C2+962

~~ X2

l

1^3

r\

X

4)

it follows that there is a monomial M so that X-l

-o -

I (i)

X<3

—— S

/..\

\. \(n)/

MX x A/x 2 x3 n *-

pi

or

pq

'

As (ii) can not occur we obtain a'3 — 0,3 — qe3 > 0. Similarly using /{ and /2 yields a2 - o2 - 0. As in the proof of Theorem 10.3.18 we readily obtain di = 64(0203 — 6302). For the converse note that by Proposition 10.3.21 and Lemma 10.3.4 /' is a prime ideal. n Corollary 10.3.24 Let f

— ™Ql _

Jl — •"'i

r-.au

f

—

62 _

-^4 i J2 — -*"2

61

f>3

f

— ™C 3 _

-^i ^3 i J3 — •i<3

Ci

C2 r

_

64 _

•*-! J'2 ' J 4 — J/ 4

62—63

-t/2 ^s

6e a set of critical binomials so that /, is critical w.r.t Xj /or aH i. Ifbi > 0, Cj > 0 /or all i and e3 > 0, then I = (/i,/ 2 ,/3,/4) is a prime ideal if and only if di — e 4 (6 2 c 3 - 6 3 c 2 ) and 04 — qe^ for some q € N. Proof. Let /{ = fi+x^^f^

= x^ -xfxfxf~e\

As / = (/{,/ 2 l / 3 ) / 4 )

the proof readily follows from Propositions 10.3.21 and 10.3.23.

10.4

D

An algorithm for critical binomials

The critical binomials in the next example were computed using the algorithm below. As an application of the results in Section 3 we have

400___________________________________ Chapter 10

Example (a) If di -7,d2- 8, d3 - 9, d4 = 17, then fl=X\- £3X4, h=X\- XiX3, f

3

-X

3

-

X\Xl, /4 = £4 - X2X3

is a set of critical binomials generating a height 3 prime ideal of type 2.

(b) If di = 204, d2 = 855, d3 = 1216, d4 = 1260, then f - -47 J l —xl

J

4 3 2 f _ 8 '2 J/ 3' ( '4i J 2 — -kg

15 3 f 6 1 4 ' J3 — X 3

19 4 f _ 5_ 13 3 -^i J '2' .M — X 4 ^i ••'3

is a set of critical binomials generating a height 3 prime ideal of type 3. (c) If di = 9, d-2 = 12, d3 = 18, d4 = 19, then

is a complete intersection prime ideal.

Let us present an algorithm to compute a full set f i , . • . ,fn °f critical binomials, one /, for each variable Xj. Let S = diN + • • • + dnN C N and P the toric ideal of k[S], where gcd(di, . . . , dn) — 1. For a = ( a i , . . . , a n ) e Nn we set deg(a) = E"=iQ!i- Let a,^ 6 N n . The graded reverse lexicographical ordering in N" is defined by a < /? if deg(a) < deg(/3), or deg(a) = deg(/?) and the first nonzero entry of /? — a is negative. Since xt' — xd^ G P for j ^ i, an upper bound for m, is given by d = min{c?i ,... ,di, . . . , dn}. For each i search for a 6 N"'"1 so that

a-(di,...,di,...,dn) = 0 m o d d j ,

(10.6)

where • denotes the usual inner product, all this from degree zero up to degree d. For the a's satisfying (10.6) compute

ma = a- (di,...,di,...,dn)/di and take mi = mao = min{mQ | a satisfies (10.6)}.

If we search the a's in increasing order with respect to < we can take Q.Q minimum with respect to this ordering. To generate the a's in increasing order, from (0, . . . ,0) up to a certain degree g, can be accomplished with the following: Algorithm 10.4.1 for ii = 0 to g do for ii = 0 to i\ do

for in_i = 0 to i n _ 2 do a = i\ - ii,ii - i3, •

Monomial Curves________________________________ 401

Exercises 10.4.2 Let P be the toric ideal of k[td\td2, td3}, where k is a field and di ,d2,ds are relatively prime positive integers. Prove that the following are equivalent: (a) P is a complete intersection.

(b) P is Gorenstein. (c) 5 = diN + d 2 N + d 3 N is symmetric. 10.4.3 If /i, /2, /s is a full set of critical binomials, prove the equality (/i,/2,/3)=ker(
where tp: Z 3 —> Z is the Z-linearmap induced by ^(e^) = di. Recall / = a—/?

if f = xa -x13.

10.4.4 Let /i = Z?1 -4 2 ^ 3 ,/ 2 = x%> - x1? xb33 , f3 = -^xf + x°33 be a full set of critical binomials. If 6j > 0, Cj > 0 for all i, prove the equalities 10.4.5 Let <7i, . . . ,sy be a set of binomials in I? = fc[a;i, . . . ,xn] such that ^(0) = 0 for all i. If V(gi, . . . ,gr,%i) = {0} for all z, prove:

(a) U[=1 supp(g,) = {xi , . . . , xn} (b) (U i6J supp( 5i )) n (L%j supp( 5i )) ^ 0, V J C {1, . . . , r}.

Chapter 11

Affine Toric Varieties and Toric Ideals We will study general toric sets and toric ideals over arbitrary fields. The aim is to examine when a given set of binomials define a toric ideal up to radical and to give a criterion to characterize when a given toric set is an affine toric variety. Here we generalize in a natural way some of the results of Chapter 10.

11.1

Systems of binomials in toric ideals

Let k be any field and D a fixed m x n matrix with non negative integer entries dij and with nonzero columns. The toric set F determined by the matrix D is the set in the affine space A£ given parametrically by

Thus one has p _ / (fdn l 1 —| ( l

,dmil

m

fdln i ' ' ' ' <-l

j.dmn \ l m

J

Let

R= k[xi,...,xn]

and B = k [ t i , . . . ,tm]

be two polynomial rings over k, graded by deg(xj)

=

dn + • • • + dmi for alH, and

deg(ij)

=

1 for all j,

respectively. Let 0 be the graded homomorphism of fc-algebras:

403

K 1c

404_____________________________________Chapter

11

where Di = (dn,..., dmi) is the transpose of the ith column of D and tDi = tdn

tdmi

We denote the image of 0 by k[T]. The kernel of , denoted by P, is the toric ideal of k[T]. Closely related to the map (j> is the homomorphism

determined by the matrix D in the standard bases of Z n and Z m . Indeed, we have (xa) = t^ for all a € N n . As a consequence, a binomial g = xa - x® belongs to P — ker(0) if and only if g = a - f3 belongs to ker(^).

To prove the following two results we will use the notion of simple component of a polynomial with respect to a subgroup of Zn; see Section 10.1. Proposition 11.1.1 If gi, . . . ,gr is a set of binomials generating the toric ideal P of k[T], then
if a - /3 e G. Let 7 e ker(^) and write 7 = a - /?, where a,/3 £ N n . Proving that 7 £ G amounts to prove that xa ~G x@ , i.e., that the binomial g = xa — x13 is simple. Now, g belongs to the toric ideal P= (g\, . . . , g r ) , hence the simple component h of g containing xa also belongs to P, by Lemma 10.1.2. Since xa £ P, it follows that h = g, hence that g is simple, as required. D Definition 11.1.2 Let / be an ideal of a ring R and / G R, the saturation

of / with respect to / is oo

(I:/°°) = \J(I:f) »=i

={reR\rf<€l,

for some i > 1} .

If .R is a polynomial ring the saturation can be computed using Grobner bases and the equality

where t is a new variable. See Chapter 2. For toric ideals the condition (a) of Proposition 2.4.24 will be replaced by a condition which is much easier to verify.

Affine Toric Varieties and Toric Ideals_________________ 405 Proposition 11.1.3 ([104]) Let gi, . . . , gr a set of binomials in the toric ideal P of k[T] and I = (gi, . . . ,gr). //char(fc) = p ^ 0 (resp. char(&) = 0), then the following two conditions are equivalent:

(ai) P = rad(J:^°°), where z = xv • • • xn and (I\z°°) = \Ji>l(I:zi). (a2) p"ker(^) C (gi,...,gr) for some u e N (resp. ker(V>) = (?i, - . . ,• (a 2 ): First we consider the case char(fc) = p 7^ 0. By hypothesis, there exists u > 0 such that /p" 6 (/: z°°) for all f & P.

Let 7 e ker(^). Write 7 = a - J3 with a,/3 € N" and g - xa - x0 , thus

g 6 P and g — a — /3 = j. Set

We know that z'V" € I for some 5 € N™ and

Since the simple components of xsgpU belong to / and hence to P (see

Lemma 10.1.2), and since xsxapU does not belong to / because it does not belong to P, it follows that xsgpU is simple, i.e.,

5 + pui ~ G PUP + o. Hence pua - pu/3 e G. This shows pukei(^) C G. Now we consider the case char(fc) = 0. Let G = ( < 7 i , - - . , 5 r ) and let 0 •£ TJ 6 ker('0) . Note rj = a — /3, where a and /? are in N" , thus / = xa — x@ is in P and x'1 fq is in / for some prime q 3> 0 and for some 7 € N n . Expanding fq one has: f

7 " = = X" /«

"^

"0^

Let /i be the simple component of xjf

i=l

*

^

with respect to G containing

If qa - q/3 £ G, then x9/3+7 does not occur in h. Hence

h= for some J contained in { 1 , . . . , q — I } . Note that h 6 / by Lemma 10.1.2. As q is prime, q must divide (?) for 0 < i < q. Thus making Xi = 1 yields 1& = (sg)lfc, s 6 Z, but since char(fc) = 0 one has 1 = qs, which is

406_____________________________________ Chapter 11

impossible. Hence q(a — /?) G G. Pick a prime r ^ q such that x 7 / r £ /, by the previous argument r(a - /?) e G. Since gcd(
= (a2): It suffices to prove

because the other contention is clear. First we assume char(/c) = p ^ 0. Let

/ = xc-xe be a binomial in P and write

gi - xai - x0i .

By (&%) there are integers s, such that cp"

where we may assume Sj > 0 by replacing xai /x13' by its inverse if necessary. Hence writing

and using the binomial theorem it follows that z 7 / p is in /, for some monomial x 7 . Therefore / e xa,d(I:z°°). In the case of characteristic zero, the same arguments as above will work, after replacing pu by 1 throughout.

D

Theorem 11.1.4 ([104]) Let T be a toric set and g\,... ,gr a set of bino-

mials in the toric ideal P of k[T]. Set I = (gi,..., g r ) . If char(/c) — p ^ 0 (resp. char(/e) = 0), then rad(/) — P if and only if

(a) p"ker(?/0 C(gi,...,gr) for some u 6 N (resp. ker(VO = (gi,... , g r } ) (b) rad(/,o;j) = rad(P, xi) for all i.

Proof. It is a consequence of Proposition 2.4.24 and Proposition 11.1.3. D

Corollary 11.1.5 ([104]) Let gi,. • • ,gr be a set of binomials in the ideal P of the monomial curve F. Set I = (gi,.., ,gr). //char(fc) = p ^ 0 (resp. char(/c) = 0), then rad(7) = P if and only if (a) pmker(-0) C (gi,...,gr) for some m € N (resp. ker(^) = (?i, • • • .5r)) (b)

Afflne Toric Varieties and Toric Ideals_________________ 407

Proof. =>•) By Theorem 11.1.4 one derives condition (a) and

rad(/, Xi) — rad(P,Xi) for all i. On the other hand

rad(P,z;) = (xi,...,xn),

because the height of P is n - I and Xi is regular on R/P. Therefore

•4=) By Proposition 11.1.3 one has P — rad(/: z°°), where z = x\---xn.

Hence there is monomial x5 and an integer N such that xsPN C I. In order to prove that rad(I) = P, we will show that every prime ideal Q containing / also contains P. If Q contains no variable, then the inclusions xs PN C I C Q imply that P C Q, as desired. If Q contains at least one variable, then we claim that Q contains all the variables, implying

Q = (Xl,...,xn) DP. This is the point in the proof where we need a special argument, compensating for the fact that the condition (b) above is weaker than that in Theorem 11.1.4. Indeed, up to a renumbering of the variables X i , . . . ,xn, one may assume that {xi, ...,xg} is the list of all the variables contained in the prime ideal Q, for some integer 1 < s < n. Let a = (a$) be the point in affine space given by a\ — • • • = as = 0, a s+ i = • • • = an = 1.

We claim that a 6 V ( I ) . Indeed, let g = xa - x13 be any binomial belonging to /. Since g belongs to Q, it easily follows that xa € ( x i , . . . , x s ) i f and only if x@ € (xi, . . . ,xs). As a consequence, aa = a^, this common value being 0 if xa G (xi, . . . ,xs), or 1 otherwise. In either case, we have g(a) — 0. It follows that a e V(7), as claimed. But a\ = 0, hence a belongs to V(I,xi), which is {0} by condition (b). Thus a = 0, which by definition of a implies that s — n. Hence, Q contains all the variables and consequently contains P, as desired. D Remark 11.1.6 In the proof of Corollary 11.1.5 the underlying hypothesis gcd(dt , . . . , dn) = 1 is unnecessary.

408_____________________________________Chapter

11

Corollary 11.1.7 Let gi,...,gr be a set of binomials in a toric ideal P.

If k is a field of char(fc) = p ^ 0 (resp. char(fc) = 0) and the containment p m ker(?/>) C G = (ffi, • • • , 0 (resp. ker(f/>) = G), then

V(g1,...,gr)cV(P)(jV(x1---xn). Proof. By Proposition 11.1.3, there is a monomial xs and an integer TV

such that x5PN C I. It follows that

V(I) C V(PN) U V ( x 5 ) C V(P) U V(Xl • • • z n ), as required.

D

Exercises 11.1.8 Let F be a toric set and P the toric ideal of k[T]. If k is an infinite field, prove that _ P = J(r) and V(P) = r,

where F is the Zariski closure F. 11.1.9 Let P be the ideal of a toric set F. If / is an ideal of R and \/7 = P, prove that ^fIR = PR1,

where R' = k f e f 1 , . . . , x^1] is the ring of Laurent polynomials. Hint Use that R' is the localization of R at the multiplicative set of monomials. 11.1.10 Let / be an ideal generated by a set of binomials in the ideal P of the toric set F. Prove that the following conditions are equivalent: ( ai ) rad(J • R1) = P • R', where R1 = k[xf\ .. .,x^]

(ag) xsPN C / for some monomial xs 6 R and some integer N > 1 11.1.11 Let D be an m x n integral matrix and tf>: Zn ->• Z m the homomorphism determined by D. If G is a subgroup of ker(^), prove that

T(Z"/G) = T(ker()/G), where T denotes torsion. 11.1.12 Let F be an arbitrary toric set and P the toric ideal of fc[F]. If gi,..., gr is a set of binomials in P such that

V(gi,...,gr,Xi) = {0}

for all i, prove that dim k\T] = d\m(R/P) = 1. Hint Use the proof of Corollary 11.1.5 to show rad (P,Xi) = (xi,... ,xn).

Affine Toric Varieties and Toric Ideals_________________409

11.1.13 Let D be an m x n integral matrix with nonzero columns and ifj-.Zn —> Z m the homomorphism determined by D. If G is a subgroup of ker(^) and p > 2 a prime number, prove that the following are equivalent:

(ai) p"ker(t/j) c G for some u £ N. (a,-2) ker('0)/G is a finite p-group or ker(ib)/G = (0).

(3,3) Z n /G ~ Z s x H, where H is a finite p-group or H = (0), and s is equal to rank(D). 11.1.14 (Rational quartic curve in P3) Let A; be a field and P the toric ideal of fc[F], where F is the toric set defined by the matrix: D =

/ 4 3 1 0 (0 1 3 4

Prove rad (x^x^— xiX4, x\—x^x\, x\—x\xz) = P and note x\x\ — x\x± £ P.

11.2

Affine toric varieties

Our aim here is to use linear algebra to characterize when a toric set F is an affine toric variety in terms of the existence of certain roots in the base field

and a vanishing condition. We will make use of the fact that any integral matrix is equivalent to a diagonal matrix which is in Smith normal form. First we fix some notation. Let F be a toric set defined by an m x n matrix D = (<%). According to [229, Theorem II.9], there are invertible integral matrices U = (1%) and Q = (g^) of orders m and n respectively such that

where s is the rank of D and AI, . . . , A s are the invariant factors of D, that is, AJ divides A,+i and Aj > 0 for all i.

Notation For use below set

U'1 = (/ y ) and Q-1 = (bij). In the sequel 6; will denote the ith unit vector in Z n . Proposition 11.2.1 // if}\ Z n —> Zm is the homomorphism determined by D in the standard bases, then

ker(^) = Zqs+i ® • • • 0 Zqn,

where Qi corresponds to the ith column of Q.

410_____________________________________ Chapter 11

Proof. Let x £ Z™ and make the change of variables y = Q~lx. As

it follows that Dx = 0 if and only if Ly = 0. Set y = (y\ , . . . , yn).

First note g» £ ker(^) for i > s + 1, because LQ~lqi = Le^ where a is the ith unit vector in Z n . On the other hand if x is in ker(t/>), then Ajy, = 0 for i = 1, . . . , s. Thus

n

x = Qy= ]P 2M»i=s+l

To complete the proof observe that the columns of Q are a basis for Z n . D Proposition 11.2.2 Let V>:Z"^Zm

be the linear map determined by D and

j=l

where qi correspond to the ith column of Q. If e\,..., en is the standard basis of Zn, then

is a generating set for ker (?/>). Proof. Set Q-1 = (6^)-

Note

i-l l

for all j, because QQ~ — I. Hence one can write

i=l

i=s+l

and using Proposition 11.2.1 we obtain Vj — 6j e ker(;0) for j = I , . . . ,n.

On the other hand from the equality above:

\z=s+l

/

z=s+l

j=l

n i=s+l

for all k > s + I. As usual 6ik = 1 if i = k and 60, — 0 otherwise.

D

Affine Toric Varieties and Toric Ideals_________________411

Theorem 11.2.3 ([242]) Let k be a field and P the toric ideal of k[T], where F C fc" is the toric set defined by the matrix D. Then F = V(P) if and only if the following two conditions are satisfied:

(a) // (a,) e V(P) and a* ^ 0 Mi, then a\li • • • a^' has a \i-root in k for i = l,...,s.

(b) V(P,Xi) c r / o r i = l , . . . , n . Proof. 4=) One invariably has F C V(P). To prove the other containment

take a point a = ( a i , . . . , a n ) in V(P), by condition (b) one may assume a; ^ 0 for all z. Thus using (a) there are £' 1 ; ..., t'g in A; such that («;)Ai=af"..-a^=a«

(i = l , . . . ) S ) .

(11.1)

For convenience of notation we extend the definition of ij by putting tj = 1

for i = s + 1 , . . . , m and t' = ( i j , . . . , t'm). Set *J- = (*i) U l '-"(C) U m i

(j = l , . . . , m ) .

(11.2)

We claim that td" = t*lk • • • td^k = ak for k = 1 , . . . , n. Setting U~l = (faj) and comparing columns in the equality U~1L = DQ one has: n

^fi = ^2qjidj j—i

(i = l , 2 , . . . ) S ) )

(11.3)

where fa — ( f a , . . . , fmi) and dj = ( d i j , . . . , dmj) denote the ith and jth columns of U~l and D respectively. Next we compare columns in

to get:

s

where Q~l = (6y). Using UU~l = I and Eq.( 11.2) we rapidly conclude: tfk I

— +' — If,.

(I- — — 1 1 , . . . , Ttlj. m\ \n>

(~\1 1 ^ 1 . 0Z,\)

From Proposition 11.2.2 we derive DVJ = Dej = dj for j = 1,..., n, where /

E

l

8 ^——"^

\ \^———^

1

1)^-0^= Ii /_^ N Ql£bl<>..... / Qni^ti J j Z_^ II

(?'=:l?t>-;^')>

(11*^)

Hence D(VJ)+ = D(CJ + ( v j ) - ) , that is, x^+ - xe'+(v^~ belongs to the toric ideal P. Using that a e V(P) yields a (t< ^+ = a^ + ( ^')-, thus avi=aei=a.

(j-= ! , . . . , „ ) .

(11.7)

412

Chapter 11

Therefore putting altogether

tdk

*=4

t^'i=lXjbikfi

ll

( a ?i) 6 ^ . . . (a'-) 6 -* = 08i*i*+-+9.*.* ^6 a"* 1=7 ak

=l

= ( t f l ) X l b l k • • • (tf>)x*b*k

:

=5 ( t ' ) X l b l k . . . (t1 )x'b'h

for & = 1, . . . , n. Thus a e F, as required.

=>) It is clear that (b) holds because V(P,Xi) C V(P). To prove (a) take (aj) in V(P) with aj / 0 for all z, then by definition of F there are £1 , . . . , tm in k such that QJ = tdj for j = 1, . . . ,n. Therefore by Eq. (11.3) one has:

Thus ( t f i ) X i = af ; • • • a« ni , as required.

D

Corollary 11.2.4 If k is an algebraically closed field, then

Proof. Let a = (aj) 6 V(P) such that a, ^ 0 for all i. Since fc is algebraically closed condition (a) above holds. Therefore one may proceed as in the first part of the proof of Theorem 11.2.3 to get a 6 F. n Next we present another consequence that can be used to prove that monomial curves over arbitrary fields are affine toric varieties and generalizes Proposition 10.1.14.

Corollary 11.2.5 If the columns of D generate Zm as It-module, then the equality F = V(P) holds if and only if V(P, Xi) C F for all i. Proof. Since

Zd1 + --- + Zdn = 1m, one has Aj — 1 for all i, thus condition (a) holds. Therefore F is an affine toric variety if and only if (b) holds. D Corollary 11.2.6 If k is algebraically closed, then T = V(P) if and only i f V ( P , X i ) C T for alii.

Proof. If k is algebraically closed, then (a) is satisfied, thus F is a toric variety if and only if V(P,Xi) C F for all i. n

As a more concrete application we now show that Veronese toric sets are affine toric varieties. Corollary 11.2.7 Let d be a positive integer and

If k is an algebraically closed field and D the matrix whose columns are the vectors in A, then the toric set F determined by D is an affine toric variety.

Afflne Toric Varieties and Toric Ideals_________________413

Proof. Let

B={ta a £A} = { f i , . . . , f m , f m + i , . . . , f , } , where s=

(d + m-l \\ m — i1

One can order the /; such that fi = tf for i — 1 , . . . , m and |supp(/j)| > 2 for i > m, where supp(£ a ) = {ij| aj > 0}.

Fix an integer 1 < i < s, it suffices to prove

V(P,Xi)cT, where P is the toric ideal associated with D. We use induction on m. Take a e V(P, Xi). If i > m and 4>(xi) — fi — ^ ''' ^"m > note that the binomial

belongs to P, hence a G V(P,Xj]

for some 1 < j' < m. Therefore one may

harmlessly assume 1 < i < m and ^(xj) = tf, for simplicity of notation we assume i = 1. Observe that for every fj = t[l • • -V^ with r\ > 0 one has a;- = 0; indeed since x'j — x\l • • • xr^ belongs to P and a € V(P,x\) one

has aj = 0. Let D1 be the submatrix of D obtained by removing the first row and all the columns with nonzero first entry (from top to bottom), and P1 the toric ideal of D'. The vector a' = (a^i £ supp(/j)) is in V(P'), because P' C P. Since V(P') C T' U V(x2 • • • £ « ) , where T' is the toric set associated with D', by induction one readily obtain a G F. D In the light of Exercise 11.2.10, a natural question is whether F can be a variety but not a toric variety, to clarify consider:

Example 11.2.8 Let k = Z 3 and D = (2,4), then F = {(0,0)} U {(1,1)} = V(n - x2,x% - x2). On the other hand P = (xl - x\} and (1,2) e ]/(P). Thus F ^ T/(P)

Exercises 11.2.9 Let D be an m x n integral matrix such that there are invertible integral matrices U and Q such that UDQ = d i a g ( A i , . . . , A s , 0 . . . , 0). If Qi (resp. Qz) denote the submatrix of Q (resp. Q"1) obtained by fixing the first s columns (resp. s rows) of Q (resp. Q~l), prove the equality

414_____________________________________Chapter

11

11.2.10 Let F be a toric set and P the toric ideal of k[T]. If k is an infinite field and F is a variety defined by an ideal J, then

J c P and r = V(J) = V(P). 11.2.11 Let k be an algebraically closed field and F the toric set associated with the matrix: 4 3 1 0 0 1 3 4

Prove that F is an affine toric variety. 11.2.12 Prove that Corollary 11.2.4 and Corollary 11.2.6 are valid assuming condition (a) of Theorem 11.2.3, instead of assuming k algebraically closed. 11.2.13 (Segre variety) Let m,n be two positive integers. If

F = { ( t i U j ) e kmn ti e k, Uj e k} C Af" is the toric set associated with the incidence matrix I? of a complete bipartite graph /C min , prove that F is an affine toric variety over any field k. Hint Use that D is totally unimodular (i.e., every minor is 0 or ±1).

11.3

Curves in positive characteristic

Let / be an ideal of a ring R the arithmetical rank of /, denoted by

r = ara(7),

is the least positive integer r such that there are / i , . . . , fr with raxl(/1,...)/r)=rad(J). By Krull's principal ideal theorem one has ara(7) > ht(7), note that equality occurs if 7 is a set-theoretic complete intersection. Let k be a field and P the associated toric ideal of a monomial curve F in the affine space A£. In characteristic zero is an open problem whether P is a set theoretic complete intersections, several authors have contributed toward the solution of this problem, see [99, 215, 224, 282], the case n = 3 is treated in [35]. If char(/c) > 0, then by [73, Theorem 2] and [73, Remark 1] it follows that P is a set-theoretic complete intersection; a result of T. T. Moh [224] shows that P is indeed generated up to radical by n — 1 binomials. Our aim is to give a different constructive proof of Moh's result due to Alcantar, Reyes and Zarate [3, 4] by identifying special generating sets for the solutions of linear diophantine equations.

Affine Toric Varieties and Toric Ideals

415

Lemma 11.3.1 Let D = (di,... ,dn) be a 1 x n matrix with d, 6 Z \ {0}, then there is a unimodular matrix Q = ( 9 2 , . . . ,qn,qi) with entries in 7L whose first n — 1 columns are of the form:

in \ i > 2 and qu > 0,

o

V o such that

(a) DQ •= (0, . . . , 0 , A ) , and

(b) ker(D) = 9 2 Z + • • • + g n Z ; where A is equal to gcd(di,... ,dn). Proof. The idea is to use an algorithm similar to the one used to diagonalize a matrix with entries in an Euclidean domain. To "diagonalize" D we use elementary operations on the columns of the following matrix:

D I where / is the identity matrix of order n. Note that using elementary operations one can obtain the following reductions: \ • • • • •

0 0

921

biz

922

^22

0

0

i

n

n

n

0 0 0 1

0

0

921

0 0

931 932 933 0

n

r\

922

\

"n

• o

&13 ^23 &33 0

n

0 0 0

1 n

•• •• •• ••

0 0 0 0 1

,

where r 2 = gcd(di,d2) and r% — gcd(r2,ds). In the first reduction we have used elementary operations on the first two columns only, those operations come from the Euclidean algorithm used to compute the greatest common

Chapter 11

416

divisor of di and d?. Similarly the second reduction can be obtained by performing elementary operations on the second and third columns only, where

those operations come from the Euclidean algorithm to compute gcd(r2, (is). Thus by induction we rapidly get the required matrix Q.

To finish the proof use Proposition 11.2.1 to conclude that g 2 , . . . , g n O

generate ker(D) as Z-module.

Proposition 11.3.2 Let D = (di,..., dn) be a matrix with d» € Z \ {0}.

Then there are fa, • • • , fn € Z" such that fu (a) fi =

0

V 0 ] (b) for each fi with i > 2, it holds fu > 0 and fij < 0 for 2 < j < i — I , (c) ker(D) is generated by f a , . . . ,fn as Z-module. Proof. Let q2,...,qn be as in Lemma 11.3.1 and let 2 < k < n. It will be shown recursively that there are fa, • • • , fk satisfying conditions (a),(b), and

We proceed by induction on k. For k = 2, take fa = q^. Assume 2 < k < n and that we have constructed fa, • • • , fk satisfying (a), (b), and such that lq-2 + •

1qk = Ifa

+ Zfk.

(*)

Next define fk+i,k+i = qk+i,k+i > 0 and fk+i,k = qk+i,k ~ >^kfk,k, with Xk large enough to have fk+i,k < 0. In general for 2 < i < k and haven chosen A j + i , . . . , A f c , define choosing A; > 0 to have fk+i,i < 0. Finally define

note that we are not requiring fk+i,i < 0. If we set i,i

fk+l =

fk+l,k+l

0

\

Affine Toric Varieties and Toric Ideals

417

clearly / 2 , - - - , f k + i satisfy the hypotheses (a), (b) of the proposition. On the other hand /fc+l = Qk+l — Afc/fc — • • • — A 2 / 2 ,

using Eq. (*) this implies:

fk+i e Z
in addition qk+i 6 Z/ 2 + • • - + Iifk+i, therefore again by Eq. (*) one obtains Zf2+... + 1fk+1 = Zq2+---+Zqk+l.

D

Lemma 11.3.3 Let D — (d\,... ,dn) be a matrix with di G Z \ {0} and /2, • • • > fn in Zn such that

a

i=

o (b) f a ^ 0 for all i,

(c) ker(L>) is generated by f ? , . . . , fn as Z -module.

Then fa divides d\ for alii > 2. Proof. Since d\&i — d{&\ belong to ker(D) for i > 2, one obtains

Clearly one has d\ = ^nfnn if i = n and p,n = 0 if i < n. By reverse induction follows that fj,n = fj,n-i = • • • = fj,i+1 = 0 and d\ = fiifu. D Theorem 11.3.4 ([224]) Let D = (dj, . . . , dn) be an integral matrix with positive entries and P the toric ideal of k[tdl ,... ,tdn]. If char(fc) = p > 0, then P = rad(#i, . . . ,gn-i) for some binomials gi, . . . ,gn-i-

Proof. Let /2, . . . , fn be as in Proposition 11.3.2. First we prove recursively that for 2 < k < n there are w%, . . . , Wk € Z ra and r^ € N so that

(i)

with Wij < 0 for j = 1,... ,i — l y wu > 0,

418

Chapter 11

(ii) prk (Z/ 2 + • • • + Z/fc) C Zw2 + • • • + 1wk.

To prove this we proceed by induction on k, for k = 2 it is enough to take Wt = /2 and f2 = 0. Let 2 < k < n and assume there are 102, • • • , wk € Z n satisfying (i) and rk € N such that Prfe (Z/2 + • • • + Z/ fc ) C Zw 2 + • • • + Ziut.

(**)

By Lemma 11.3.3, we have /u|di for i — 2, . . . , n. Then di = fk+i,k+i^ for some £, let p afc + 1 be the greatest power of p dividing I , that is, we can write di = fk+i,k+iPai>+1sk+i,

(1)

where p does not divide Sk+i, hence using that the image p of p in Z S(b+1 is non zero, one obtains that p has finite order. Therefore there is 0k+i G N+ such that p0k+1 = Imodsfc+i, hence p^+i" = Imods^+i, for all u € N, consequently there is ck+i € N (depending on u) such that / f c + l U = Sfc + 1 c f c + 1 +l,

(2)

choose u large enough to guarantee pak+l fk+i,i - dk+ick+\ < 0 and define the entries of wk+i as:

,i =Pak+1fk+i,i ~ ,i =pa^fk+lii i = 2 wk+i,k+i =Pak+lfk+i,k+i ij = 0

for j > k + I

Observe that wk+i,i < 0 for i = l , . . . , f c and 10^+1^+1 > 0. Note the equality

wk+i =pak+1fk+i Therefore

Dwk+i =pa^Dfk+i + (dlCk+l)Dek+l - (dk+lck+l)Dei = 0, as a result wk+i 6 ker(D), hence +i

=Pak+1fk+i,k+i

= Pak+1 fk+i,k+i

setting qk+i = ak+i + @k+\u, one concludes Wk+i,k+i=Pqk+lfk+i,k+i-

(3)

Afflne Toric Varieties and Toric Ideals

419

Now, since only the first k + 1 entries of Wk+i and fk+i can possible be different from zero (all the remaining entries are zero), we get from Eq. (3) that only the first k entries of Wk+i — (pqh+l)fk+i can possible be different from zero (the rest are zero), hence using that ker(D) is generated by / 2 , . . . , / n , one derives:

wk+1 -(p^+OA+i ez/2 + "- + z/fc, which together with Eq.(**) gives prk(wk+i ~ ( p q k + 1 } f k + i ) 6 Z(w2, . . . ,wk), and consequently

setting r/t+i = qk+i + Tk, one concludes

. . . , fk+i) C Z,(w2, . . .,wk

Altogether this terminates the induction argument and ends the first step of the proof. Making k = n, we obtain that there are wz, . . . ,wn such that the first i — 1 entries of Wi are negative or zero, the ith entry of Wi is positive, and for j > i the entries of 10, are zero. Furthermore there is pTn such that

Set Qi = Wi+i, this means that g± is a binomial such that ~gi = Wi+\. Let

I = ( g i , . . . ,0 n _i). Take a <E V ( I , X i ) , with a = (ai,. . . ,a n ), and suppose that a 7^ 0 and let Oj be the first entry of a equal to zero (such entry exists

because at least a; — 0). If j ^ 1, one would have —

Wa

93-1. =Wj=Xj"

Wi\

W; ,-_1

-X,'1 ---Xj^

and aj = 0, then some ar = 0 for r < j, which is not possible, thus j must be equal to 1. Let us see that a = (0, . . . , 0) by induction on the coordinates of a, assume ai = • • • = a/t = 0 for 1 < k < n. One has:

and since a e y(7, Xj) C V(7) and #/t £ J, one obtains a^+i — 0. Therefore a = (0, . . . , 0). As a result V(7, a;,) = 0 for i = 1, . . . , n, this fact together with p r -ker(D) cZ(w2,...,wn) prove that the conditions (a) and (b) of Theorem 10.1.3 are satisfied, and consequently P = rad(
Appendix A

Graph Diagrams For convenience we will display some Cohen-Macaulay graphs and unmixed graphs with small number of vertices.

A.I

Cohen-Macaulay graphs

The complete list of Cohen-Macaulay connected graphs with at most six vertices is:

421

422

Appendix A

Graph Diagrams

423

Appendix A

424

A.2

Unmixed graphs

Next we display the set of all unmixed non Cohen-Macaulay connected graphs with at most six vertices.

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Notation (/: J°°), saturation of /, 53 (Ni:RN2), ideal quotient, 4 A(P), Ehrhart ring, 232 A[F], subring generated by F, 23 Ap, normalized Ehrhart ring, 226 G(G), cone over a graph G, 293 F(M,t), Hilbert series, 97 GI * (?2, join of two graphs, 332 H(M,i), Hilbert function, 14 H(y,c), hyperplane, 210 H+(y,c), halfspace, 210 /("), symbolic power, 76 /A, Stanley-Reisner ideal, 142 Ic, ideal of covers, 196 K, a field, 2 K*(x), Koszul complex, 29 NG(U), neighbor set, 163 R(—a), shift in the graduation, 59 R^, kth Veronese subring, 203 Rf, localization at a prime, 3 S ( f , g ) , the ^-polynomial, 49 5~ 1 (M), localization, 3 V(7), prime ideals over 7, 2 Ass(7), associated primes, 5 Assfl(M), associated primes, 5 A", ^-skeleton of A, 145 A/, Stanley-Reisner complex, 142 Spec(7?), spectrum of a ring, 2 Supp(M), support of a module, 5 T(G), minimal covers, 166 \\x\\, Euclidean norm, 212 a+, positive part of a vector, 203 ao(G), covering number, 165 ht (7), height of an ideal, 4

ann/((M), annihilator of M, 4 /?i(G), independence number, 165 IR(M), length of a module, 9 gcd, greatest common divisor, 218 {, ), usual inner product, 210 \a], ceiling of a, 234 1cm, least common multiple, 49 lk(F), link of a face, 143 log(F), log set of F, 236 log(x a ), log of a monomial, 236 A£, affine space, 45 N+, positive integers, 31 P£, projective space, 56 v ( M ) , number of generators, 8 us, canonical module, 106 7 or 70, integral closure, 69 \/7 or rad (7), radical of 7, 4 |a;|, diamond norm, 217 e(M), multiplicity of M, 101 e(q,M), relative multiplicity, 85 ei, unit vector, 156 /(A), the /-vector of A, 150 / ->F 0, non negative vector, 218 xc, monomial, 201 Cn, cycle, 163 447

448

/C n , complete graph, 163 K-m,n, complete bipartite, 182 71(1), Rees algebra, 66 Z(M), zero divisors, 6 X(A), Euler characteristic, 141 R+ = {x e «\x > 0}, 211 IR+.4, cone generated by A, 211 Soc(M), socle of M, 21 Sym^(M), symmetric algebra, 65 adj(.ff), adjoint of a matrix, 8 char(/?), characteristic of R, 2 conv(A), convex hull of A, 210 deg(w), degree of a vertex, 162 depth(M), depth of a module, 18 dim(M), Krull dimension, 4 gr / (fl), associated algebra, 73 iff, if and only if, 2 in(J), initial ideal, 48 n(X), relative interior, 215 s.o.p, system of parameters, 16 star(cr), star of a face, 148 supp(/3), support of a vector, 203 supp(x a ), support of xa, 131 trdeg, transcendence degree, 34 type(M), type of a module, 21 w.r.t, with respect to, 14 Hq(A;A), reduced homology, 139

Notation

Index a-invariant, 99 adjacency matrix, 190 admissible grading, 220 affine fc-algebra, 31 combination, 210 space, 45, 210 generated by a set, 210 algebra, 23 finite, 23 finitely generated, 23 of finite type, 23 almost integral, 74 annihilator, 4 arithmetical rank, 414 Artin-Rees lemma, 85 Artinian reduction, 123 ring, 10 ascending chain condition, 2 associated graded algebra, 73 matrix of a monomial subring, 202 prime of a module, 5 of an ideal, 5 atomic cycles, 303 Betti numbers, 61

initial, 188 initial virtual, 188 binomial, 202 expansion, 100, 119 ideal, 202 449

of a closed walk, 282 bipartite graph, 163 boundary complex, 156 bow tie, 317 Buchberger algorithm, 50 criterion, 51

canonical module, 106 catenary ring, 21

Cayley-Hamilton, 22 ceiling of a vector, 234 characteristic of a ring, 2 Chinese remainder theorem, 10

chord of a cycle, 194 chordal graph, 194

circuit, 297 of a graph, 300 CoCoA see computer algebra, 94 codimension of a module, 4 of an ideal, 4 Cohen-Macaulay graph, 168 ideal, 20 module, 18 ring, 20 complement of a graph, 175 complementary complex, 172 complete bipartite graph, 182 graph, 163 ideal, 69 intersection, 20

Index

450

generically, 68

set theoretic, 20 composition series, 9 computer algebra systems

CoCoA, 94 Macaulay, 53 Normaliz, 236 PORTA, 211 cone, 210

discrete graph, 162 valuation, 378

distance, 163 division algorithm, 49 dominant set, 175

duality in monomial subrings, 268 in Stanley-Reisner ideals, 196

finitely generated, 211

of a complex, 140 over a graph, 293 conormal module, 67

edge, 161 cone, 326

critical graph, 168

content of a polynomial, 339

generator, 282

convex combination, 154, 210 cone, 210 hull, 154, 210

graph, 188 ideal, 168 generalized, 256 subring, 282

polyhedron, 210

set, 210 cover complexity, 166 critical binomials, 385

full set, 385 cutpoint, 175 cycle, 163 basis, 291 space, 291 cyclic polytope, 158 d-tree, 194 Danilov-Stanley formula, 221

Dedekind-Mertens formula, 339 degree of a module, 101 of a vertex, 162 Dehn-Sommerville equations, 157 depth lemma, 18 of a module, 18 Dickson's lemma, 48 dimension

of a set, 210

Ehrhart polynomial, 232 normalized, 227 ring, 232

normalized, 226 elementary integral vector, 297 vector, 296 zonotope, 302 elimination order, 52

embedded prime, 10 end vertex, 162 equations of a cone, 216 Euler characteristic, 141 reduced, 141 formula, 156 even cycle, 163 exact sequence, 5 short, 5 extended Rees algebra, 72 extremal

Cohen-Macaulay ring, 117 Gorenstein ring, 115

of a simplicial complex, 138

theorem, 16

f-vector

Index of a complex, 150

of a polytope, 155 face, 138, 211 ideal, 130 ring, 143 facets

of a cone, 325 of a simplicial complex, 144 faithfully flat, 27 Farkas's lemma, 218 field of fractions, 8 filtration of a ring, 81 finite homomorphism, 23

length, 9 flag complexes, 170 flat homomorphism, 27 forest, 163 forms, 13

Frobenius number, 378 Gauss lemma, 339 geometrically linked, 30 going down, 26

up, 26 Gorenstein graded ideal, 112 ideal, 22 ring, 22 Grobner basis, 49 lexicographical, 276 of toric ideals, 203 reduced, 49 graded algebra, 16 ideal, 13 map, 13 module, 13 ring, 12 graph, 161 connected, 162 connected components, 162 Graver basis, 284

451

H-configuration, 314 h-vector

of a C-M complex, 152 of a complex, 152 of a polytope, 156 of a standard algebra, 108

height of an ideal, 4 Herzog-Kiihl formulas, 63 Hilbert basis theorem, 2 function, 14, 97 polynomial, 101 of a face ring, 150 over an Artinian ring, 15

series, 97 of a face ring, 150 Hilbert-Burch theorem, 62 Hochster configuration, 314 homogeneous element, 13 ideal, 13 resolution, 61 ring, 14 subring, 224

homogenization of an ideal, 54 homomorphism integral, 23

of algebras, 23 of rings, 2

ideal of a subset, 45 of linear type, 67 of relations, 35 quotient, 4 improper face

of a polyhedral set, 211 incidence matrix, 294 Smith form, 295

independence number, 165 independent set of edges, 163 of vertices, 165 induced subgraph, 195

452

initial

degree, 110 ideal, 48 integral closure, 23 commute with localizations, 70, 71 of a domain, 26 of an edge subring, 320 of ideals, 69 of monomial ideals, 234 of monomial subrings, 218 integral extension, 23 integrally closed ideal, 69 ring, 26 irreducible representation, 214 of a cone, 215 submodule, 6 irredundant decomposition of modules, 7 irrelevant maximal ideal, 35

isolated prime, 10 vertex, 162 Jacobian criterion, 91 ideal, 92 matrix, 91 Jacobson radical, 95 join of graphs, 332 of ideals, 245 of simplicial complexes, 140 Konig Theorem, 165 Koszul complex, 28 Krull dimension, 4 intersection theorem, 83 principal ideal theorem, 19

Kruskal-Katona criterion, 151

Index

lattice, 135 distributive, 135 of monomial ideals, 135 Laurent polynomials, 80 leading coefficient, 48 monomial, 48 term, 48 length of a module, 9 level algebra, 274 lex order, 31, 48 linear resolution, 63 variety, 210 link of a face, 143 linkage class, 30 even, 30 of ideals, 30 local ring, 2 localization, 3 at a prime, 3 locally a complete intersection, 78 log set, 236 Macaulay see computer algebra, 53 symbol, 119 theorem, 119 marriage problem, 165 a generalization, 330, 331 minimal primes of a ring, 2 of a module, 9 resolution, 61 minimum number of gens, 58 module of finite length, 9 of fractions, 3 monomial curve, 367 ideal, 130 of minimal covers, 196

453

Index of mixed products, 248 order, 48 ring, 130 subring, 201 subring of a graph, 282 walk, 283 multiplicity of a face ring, 154 of a graded module, 15, 101 over a local ring, 85

Nakayama's lemma general version, 8 graded version, 14 neighbor set of a subset, 163 of a vertex, 182 neighbourly polytope, 158 nilpotent element, 4 nilradical, 4 Noether normalization homogeneous, 39 lemma, 34 of a monomial ring, 39 of an edge subring, 309, 340, 362 Noetherian module, 1 ring, 1 normal ideal, 69 polytope, 228 ring, 26 normality descent, 80, 240 Normaliz, 236 normalization of a ring, 26 of an edge subring, 320 of monomial subrings, 218 normalized degree, 203 grading, 265 volume, 228

normally torsion free, 77 edge ideal, 312 versus normal, 79, 238 Nullstellensatz, 46 numerical semigroup, 377 odd cycle condition, 322 order complex, 170

partition, 150 Pascal identity, 15 path, 162 perfect matching, 165 polarization, 131 of a bipartite planar graph, 304 of an even closed walk, 304 polyhedral cone, 210 set, 210 face, 211 facet, 214 polynomial function, 15 quasi-homogeneous, 13 polytopal subring, 228 polytope, 154 lattice, 228

PORTA, 211 poset, 169 positively graded algebra, 35 presentation ideal, 35, 66, 202 of a graded algebra, 35 primary ideal, 6 extension of, 27 submodule, 6 primary decomposition irredundant, 7 of a graded module, 14 of a module, 7 of an ideal, 7 of monomial ideals, 132

Index

454

prime avoidance

general version, 10 graded version, 41 primitive binomial, 284 projective

closure, 57 of a monomial curve, 375 dimension, 61 space, 56 proper face of a polyhedral set, 211 of a polytope, 155 pure resolution, 62 pure simplicial complex, 145

quasi-regular sequence, 83 radical ideal, 76 radical of an ideal, 4 rank of a module, 10 reduced ring, 4 reduced simplicial homology, 139 reduction number

of an algebra, 349 of an ideal, 339 reduction of a polynomial, 49 Rees algebra, 66 of a filtration, 81 regular element, 6 regular ring, 19 is Cohen-Macaulay, 89 regular sequence, 16 Reisner Theorem, 143 relative interior, 215 remainder, 49 residue field, 2 revlex order, 48 ring extension, 23 S-polynomial, 49 Samuel function, 85 saturated graph, 177 saturation of an ideal, 53 Segre

product, 105 variety, 414 semigroup, 377 ring, 201, 367 Serre's normality criterion, 26 set of fc-products, 256 Shellable complex, 146 shelling, 146 shift in the graduation, 59 simple components, 368 simple module, 9 simplex in affine space, 155 lattice, 228 of a simplicial complex, 138 oriented, 139 unimodular, 228 simplicial complex, 138 Alexander dual, 198 Cohen-Macaulay, 143 geometric realization, 156 simplicial polytope, 155 simplicial sphere, 156 Gorenstein property, 157 skeleton of a complex, 145 sliding depth, 29 socle of a graded algebra, 111 of a module, 21 spanning subgraph, 162 spectrum of a ring, 2 square, 163 square-free monomial, 130 standard algebra, 35 grading, 13 Stanley-Reisner complex, 142 ideal, 142 ring, 142 star, 187 star of a face, 148 strongly Cohen-Macaulay, 29

Index

subgraph, 162 subring of ^-products, 203 support of a binomial, 387 of a module, 5 of a monomial, 131, 299 of a vector, 296 supporting hyperplane, 211 suspension of a graph, 325 symbolic power, 76 Rees algebra, 238 symmetric algebra, 65, 67 semigroup, 378 system of parameters, 19 for modules, 16 homogeneous, 36 syzygetic ideal, 67 syzygy module, 51, 61 computation of, 52 tensor algebra, 65

term order, 48 terminal cycle, 316 terms, 48 toric

ideal, 202, 404 of an edge subring, 282 set, 403 variety, 208 total ring of fractions, 8 transcendence basis, 34 degree, 34 tree, 163 triangle, 163 triangulated graph, 194 twists of a graded module, 61 type, 266 Cohen-Macaulay, 112 of a module, 21 unicyclic graph, 316

455 unimodular covering, 230 unimodular matrix, 295 totally, 414

unmixed graph, 168 unmixed ideal, 20 upper bound conjecture, 159

usual grading, 13 valuation ring, 378 variety

affine, 45 coordinate ring, 45 dimension of, 45 irreducible, 45 irreducible components, 46

projective, 56 toric, 208 Veronese subring, 203 square-free, 203 Veronese variety, 412 vertex, 212 vertex cover, 165 minimal, 165 vertex covering number, 165 vertex critical graph, 168 walk, 162 closed, 162 Zariski closure, 207 topology, 45 of the prime spectrum, 2 zero divisor, 6 zero set of an ideal, 45

колхоз 4/25/06