GORENSTEIN HILBERT COEFFICIENTS Sabine El Khoury Department of Mathematics, American University of Beirut, Beirut, Lebanon se24@aubedulb Hema Srinivasan Department of Mathematics, University of Missouri, Columbia, Missouri, USA hema@mathmissouriedu Abstract We prove upper and lower bounds for all the coefficients in the Hilbert Polynomial of a graded Gorenstein algebra S R/I with a quasi-pure resolution over R The bounds are in terms of the minimal and the maximal shifts in the resolution of R These bounds are analogous to the bounds for the multiplicity found in [8] and are stronger than the bounds for the Cohen Macaulay algebras found in [6] 1 Introduction Let S S i be a standard graded k-algebra of dimension d, finitely generated in degree one HS, i dim k S i is the Hilbert function of S It is well known that HS, i, for i >> 0, is a polynomial P S x, called the the Hilbert polynomial of S P S x has degree d 1 If we write, d 1 x + d 1 i P S x 1 i e i x e 0 d 1! xd 1 + + 1 d 1 e d 1 Then the coefficients e i are called the Hilbert coefficients of S The first one, e 0 called the multiplicity is the most studied and is denoted by e If we write S R/I, where R is the polynomial ring in n variables and I is a homogeneous ideal of R, then all these coefficients can be computed from the shifts in the minimal homogenous R- resolution F of S given as follows: 0 M s jm s R j βsj δs M i jm i R j βij δ i M 1 jm 1 R j β1j δ1 R R/I 0 Let h height of I, so that h s In 1995, Herzog and Srinivasan [5] proved that if this resolution is quasi-pure, ie if m i M i 1, then s m i s! es 1 s M i, if h s s!
2 GORENSTEIN HILBERT COEFFICIENTS h and es M i if h < s h! Further, Herzog, Huneke and Srinivasan conjectured this to hold for all homogeneous algebras S which came to be known as the multiplicity conjecture When S is Gorenstein, Srinivasan established stronger bounds for the multiplicity Theorem [Srinivasan [8]] If S is a homogeneous Gorenstein algebra with quasi-pure resolution of length s 2k or 2k + 1, then m 1 m k M k+1 M s s! es M 1 M k m k+1 m s s! In this paper, we establish bounds for all the remaining Hilbert coefficients of Gorenstein Algebras with quasi-pure resolutions analogous to the above bounds for the multiplicity We prove in 42 Theorem 42 If S is a homogeneous Gorenstein Algebra with quasi-pure resolution of length s 2k or 2k + 1 Then, for 0 l n s, f l m 1 m k M k+1 M s m 1m k M k+1 M s s+l! e l S f l M 1 M k m k+1 m s M 1M k m k+1 m s s+l! l with f l a 1, a s a it i t + t 1 and f 0 1 1 i 1 i l s t1 Boij and Söderberg [1] conjectured that Betti sequences of all graded algebras can be written uniquely as sums of positive rational multiples of betti sequences of pure algebras which in turn implied the multiplicity conjecture In 2008, these conjectures were proved by Eisenbud and Schreyer [3] for C-M modules in characteristic zero and extended to non C-M modules by Boij and Söderberg [1] Using these results, Herzog and Zheng [6] showed that if S is Cohen-Macaulay of codimension s, then all Hilbert coefficients satisfy m 1 m 2 m s s + i! with h i d 1, d s h i m 1, m s e i S M 1M 2 M s h i M 1, M s s + i! 1 j 1 j i s k1 i d jk j k + k 1 and h 0 d 1, d s 1 Our results extend those of Srinivasan [8] as well as the above result [6] to all coefficients of Gorenstein algebras with quasi-pure resolutions In section 3 we give an explicit formula of the Hilbert coefficients as a function of the shifts of the minimal resolution of a Gorenstein algebra These expressions depend on whether the projective dimension is even or odd In section 4, we establish the stronger bounds for the higher Hilbert coefficients when the algbera has a quasi-pure resolution
GORENSTEIN HILBERT COEFFICIENTS 3 2 Preliminaries and Notations Let R K[x 1, x n ] be the poynomial ring in n variables, I be a homogeneous ideal contained in x 1, x 2,, x n and S R/I Let F be the minimal homogeneous resolution of S over R given by: 0 b s b i R d sj δs R d ij δ i R d 1j δ 1 R R/I 0 Definition 21 A resolution is called quasi-pure if d ij d i 1,l for all j and l, that is, if m i M i 1 for all i Suppose S is Gorenstein Then by duality of the resolution, the resolution of S can be written as follows If I is of height 2k + 1 then 0 R c R c a 1j and if I is of height 2k then 0 R c b k R c a 1j b k R a kj b k /2r k R c a kj R c a kj R a 1j R 1 b k /2r k R a kj R a 1j R 2 Remark 22 1 The minimal shifts in the resolution are: m i min j a ij 1 i k c max j a s i,j k + 1 i < s c i s The maximal shifts in the resolution are: M i max j a ij 1 i k c min j a s i,j k + 1 i < s c i s
4 GORENSTEIN HILBERT COEFFICIENTS and M s m s c 2 Let α ij a ij c a ij for i k with p i min j α ij m i M s i and P i max j α ij M i m s i Definition 23 Given α 1, α 2,, α k a sequence of real numbers, we denote the following Vandermonde determinants by 1 1 1 α 1 α 2 α k α1 V t V t α 1, α 2,, α k 2 α2 2 α 2 k α1 k 2 α2 k 2 α k 2 k α1 k 1+t α2 k 1+t +t k 1 j<i k α i α j β 1 +β 2 +β k t α β 1 1 α β 2 2 α β k Remark 24 V t α 1, α 2,, α k 0 if the sequence is in ascending order n p As a convention, for any non-negative integers n, p, we set the binomial coefficient 0 if n < p The following binomial identities are essential to our theorems In [8], Srinivasan showed Lemma 25 For all k 0, c, a 1 [ n 2 ] n t 1 c a n a n 1 t a t c a t c 2ac n 2t 1 t [ n 2 ] [ n t n 2 ] n t 1 c a n +a n 1 t a t c a t c n 2t + 1 t a t c a t c n 2t t t 1 The proof goes along the same lines as in [[8], lemmas 2-3] t1 k 3 Hilbert Coefficients of Gorenstein Algebras Let R K[x 1, x n ] and I a graded ideal Let F be the minimal resolution of S R/I, 0 b s b i R d sj δs R d ij δ i R d 1j δ 1 R R/I 0
GORENSTEIN HILBERT COEFFICIENTS 5 Theorem 31 Peskine-Szpiro Suppose S is C-M then these shifts d ij are known to satisfy [[9], 1] s b i 1 k 0 1 i d k ij 0 1 k < s 1 s s!e k s These equations can be thought of as defining the multiplicity, e e 0 S In fact, the higher Hilbert Coefficients can also be expressed in terms of the shifts in the resolution[4] We include a simple proof for the sake of completeness Theorem 32 with ν l r 1 s s + l!e l 1 ξ 1 <ξ 2 <<ξ l r s+l 1 l 1 l r ν l r r0 s b i 1 i ξ 1 ξ 2 ξ l r and ν 0 1 Proof We know that the Hilbert function of R/I is s 1i b i td ij 1 t n Qt 1 t d where d dim R/I n s and Qi 1 e i! i We get s b i 1 i t d ij Qt1 t s 3 We denote these two quantities by S R/I t We differentiate both sides l + s times, and evaluate them at t 1 We first start by the right hand side S s+l s + l R/I t 1s s!q l t + 1 tp t l where P t is a polynomial in t Evaluating at t 1: S s+l s + l R/I 1 1s s!q l 1 + 0 l 1 s s + l!e l d s+r ij On the other hand, S R/I t s b i 1 i t d ij So
6 GORENSTEIN HILBERT COEFFICIENTS with ν l r with ν s+l r Note that s R/I 1 b i 1 i S l 1 ξ 1 <ξ 2 < <ξ l r <l 1 S s+l b i l 1 dij l l! s 1 i d ij r s b i 1 i r0 r1 ξ 1 ξ 2 ξ l r and ν 0 1 l 1 l r ν l r d r ij s R/I 1 b i s+l 1 i 1 s+l r ν s+l r d r ij s+l r1 1 s+l r ν s+l r r1 1 ξ 1 <ξ 2 <<ξ s+l r s+l 1 s b i 1 i d r ij 0 when r < s and S s+l and hence the result s b i 1 i ξ 1 ξ 2 ξ s+l r and ν 0 1 s+l R/I 1 1 s+l r ν s+l r rs l 1 l r ν l r r0 s b i 1 i s b i 1 i d s+r ij d r ij d r ij Let I be Gorenstein The minimal free resolution of I is written as in 1 and 2 depending on whether the projective dimension of I is even or odd In [8], Srinivasan gave a more simplified expression for the multiplicity in both cases She proved Theorem 33 Srinivasan Let I be Gorenstein of grade s 2k +1 and the minimal graded resolution of S R/I be as in 1 Then, k b i 1 i a t ijc a ij t c 2a ij 0 if 1 t < k 1 k 2k + 1!eS if t k c if t 0
GORENSTEIN HILBERT COEFFICIENTS 7 Theorem 34 Srinivasan Let I be Gorenstein of grade s 2k and the minimal graded resolution of S R/I be as in 2 Then, k b i 1 i a t ijc a ij t 0 if 1 t < k 1 k 2k! es 2 if t k 1 if t 0 We extend these results to all coefficients and we show Theorem 35 Let I be Gorenstein of grade s 2k + 1 and the minimal resolution of S R/I be as in 1 Then 1 k s + l!e l is equal to [ r 2 ] k + r t 1 l r ν l r 1 t k + t 0 r l c r 2t k b i 1 i a k+t ij c a ij k+t c 2a ij Proof Following the result of Theorem 32, it suffices to show that equals that k b i [ r 2 ] k + r t 1 k+i 1 t k + t s b i 1 i d s+r ij s b i 1 i d s+r ij c r 2t a k+t ij c a ij k+t c 2a ij We have c s+r + k c s+r i,j b j a s+r ij 1 i + k b j 1 i [c a ij s+r a s+r ij ] 1 2k+1 i c a ij s+r c s+r i,j c s+r i,j [ s+r 2 ] 1 i 1 i k+[ r+1 2 ] 1 t s + r 1 t t 1 t 2k + r t t a t ijc a ij t c 2a ij c s+r 1 2t by lemma 25 a t ijc a ij t c 2a ij c 2k+r 2t
8 GORENSTEIN HILBERT COEFFICIENTS By theorem 33 the only remaining terms in the sum are t 0 and t k, so s b i 1 i d s+r ij i,j i,j 1 i k+[ r+1 2 ] tk [ r 2 ] 1 t 2k + r t t k + r t 1 i 1 t+k k + t Example 36 1 k s + 1!e 1 [ ν 1 + k+1 k 1 k s + 2!e 2 [ ν 2 ν 1 k+1 k a t ijc a ij t c 2a ij c 2k+r 2t a k+t ij c a ij k+t c 2a ij c r 2t k b i c] 1 i a k ijc a ij k c 2a ij b i c + k+2 ] k c 2 1 i a k ijc a ij k c 2a ij k k b i 1 i a k+1 ij c a ij k+1 c 2a ij We now consider the case when s is even Theorem 37 Let I be Gorenstein of grade s 2k and the minimal resolution of S R/I be as in 1 Then 1 k s + l!e l is equal to [ r 2 ] t 0 0 r l [ k + r t 1 l r+t ν l r + k + t k + r t 1 k + t 1 ] c r 2t In the summation j runs from 1 to b i if i < k and from 1 to b k /2 if i k Proof We proceed the same way as the proof of theorem 35 We have s b i 1 i d s+r ij c s+r + k c s+r + i,j c 2k+r + i,j b j a s+r ij 1 i + 1 2k i c a ij s+r 1 i [ c a ij s+r + a s+r ij 1 i k+[ r 2 ] k+[ r 2 ] + t1 ] 1 t 2k + r t t 1 t 2k + r 1 t t 1 k b i 1 i a k+t ij c a ij k+t a t ijc a ij t c 2k+r 2t a t ijc a ij t c 2k+r 2t by lemma 25
GORENSTEIN HILBERT COEFFICIENTS 9 By theorem 34 the only remaining terms in the sum are t 0 and t k, we s b i obtain that 1 i d s+r ij [ r 2 ] 1 i k + r t 1 t+k a t+k ij c a ij t+k c r 2t t + k i,j [ r 2 ] + 1 t+k k + r 1 t t + k 1 a t+k ij c a ij t+k c r 2t Example 38 1 k s + 1!e 1 [ ν k 1 k + k 1 k 1 + k+1 k + k ] k 1 c 1 k s + 2!e 2 [ ν k 2 k + k 1 k 1 k+1 ν1 k k b i [ k + 1 1 i a k ijc a ij k + k + 1 + k k 1 k k k b i 1 i a k ijc a ij k c + k+2 k + k+1 k 1 ] c 2 ] k b i 1 i a k+1 ij c a ij k+1 4 Bounds for the coefficients with quasi-pure resolutions Definition 41 For any ordered s-tuple of positive integers, and f 0 1 f l y 1,, y s In this section, we prove 1 i 1 i l s l y it i t + t 1, 1 l s t1 Theorem 42 If S is a homogeneous Gorenstein Algebra with quasi-pure resolution of length s 2k or 2k + 1 Then f l m 1 m k M k+1 M s m 1m k M k+1 M s s+l! e l S f l M 1 M k m k+1 m s M 1M k m k+1 m s s+l! Remark 43 1 M n m n c 2 These bounds are strictly stronger than the bounds in the conjecture found by Herzog and Zheng in [6]
10 GORENSTEIN HILBERT COEFFICIENTS Let S R/I Then S has a R free resolution of length s 2k + 1 or s 2k The first half of the resolution is given below and b k /2r k b k R c a kj R a kj b k /2r k R a 1j R, s 2k + 1 4 R a kj R a 1j R, s 2k 5 Thus, we let r i b i, i k and r k b k if s is odd and r k b k 2 if s is even Without loss of generality we may take a i1 a i2 a ibi for all i If s 2k, we pick a kj so that c a krk a krk The symmetry of the resolution and the exactness criterion forces b k to be even Srinivasan showed in [[8], 5], that Lemma 44 If S is Gorenstein with a quasi-pure resolution then c 2a ij for all i, j Proof Since quasi-purity means the a ij increase with i, we just need to check c 2a krk If s 2k + 1 then c a krk m k+1 M k a krk So c 2a krk If s 2k, then c 2a krk by choice and hence the result To be able to prove theorem 42 we need to consider two different determinants depending on whether s is even or odd Suppose s is odd Let b i b k α 1j c 2a 1j α ij c 2a ij α kj c 2a kj b i b k α1jc 2 2a 1j αijc 2 2a ij αkjc 2 2a kj M t 1j c 2a 1j α k+t 1j c 2a 1j b i where α ij a ij c a ij Then, M t Now consider b i ij c 2a ij α k+t ij c 2a ij 1 j i b i b k b k kj c 2a kj α k+t kj c 2a kj α iji c 2a iji V t α 1j1, α kjk l k + r t 1 l r ν l r 1 t c r 2t M t k + t r0
GORENSTEIN HILBERT COEFFICIENTS 11 It is equal to 1 j i b i α iji c 2a iji V α 1j1 α kjk l k + r t 1 l r ν l r 1 t k + t r0 c r 2t We thank László Székely for his help with the following lemma βi t Lemma 45 We have for all c, a i > 0 and α i a i c a i k + r t 1 t c r 2t α β i i a β i i k + t c a i β k+i c β 2k+1 βi t β 1 + +β 2k+1 r α β i ij i Proof β 1 + +β 2k+1 r a β i i c a i β k+i c β 2k+1 is the coefficient of x r in 1 1 cx 1 1 1 a i x 1 c a i x 1 1 cx γ 1,,γ k+1 the coefficient of x r is in this last expression is: k 2β i+γ i β i +γ k+1 r k β i+γ i +γ k+1 r c r 2t 1 t t 0 c r 2t 1 t t 0 γi γi β i β i 1 1 cx a i c a i x 2 cx ai c a i x 2 γ i cx γ k+1 c γ i β i 1 β i a i c a i β i c γ k+1 c γ i β i 1 β i a i c a i β i c γ k+1 k+1 γ ir t β 1 +β k t β 1 +β k t a i c a i β i a i c a i β i k+1 γ ir t γi β i γi β i
12 GORENSTEIN HILBERT COEFFICIENTS It remains to show that, k+1 γ ir t γi β i r t + k This can be proved by t + k induction on n, t, k Alternatively, consider the negative binomial theorem 1 x βi+1 n + βi x n β n i x β i 1 x βi+1 n + βi x n+β i γi x γ i, β n i β γ i i γi so that is the coefficient of x r t γ k+1 in γ 1 +γ k+1 r t β i x β i 1 x β i+1 x t 1 x t+k This in turn, is equal to the coefficient of x r 2t γ k+1 in 1 So, 1 x t+k γi r t r γk+1 2t + t + k 1 t + k 1 γ 1 +γ k+1 r t β i γ k+1 0 r t γ k+1 0 r t + k t + k r t + k 1 γk+1 t + k 1 Now, when s is even we consider the following determinant: N t r 1 r 1 r 1 r 1 α 1j α 2 1j 1j α k+t 1j r i r i α ij αij 2 r i r i ij α k+t ij r k α kj r k αkj 2 r k kj r k α k+t kj
GORENSTEIN HILBERT COEFFICIENTS 13 with α ij a ij c a ij, r i b i for i 1, k 1 and r k b k /2 Then, N t α iji V t α 1j1,, α kjk 1 j i b i Now consider [ ] k + r t k + r t 1 1 l r ν l r 1 t + c r 2t N t k + t k + t 1 0 r l It is equal to α iji V α 1j1 α kjk j i 0 r l [ ] k + r t k + r t 1 1 l r ν l r 1 t + c r 2t k + t k + t 1 The following is the version of lemma 45 for the case where s is even Lemma 46 For all c, a i > 0 and α i a i c a i, we have [ ] k + r t k + r t 1 1 t + c r 2t α β i i k + t k + t 1 β 1 +β sr βi t βi t α β i ij i k 1 a β i i c a i β k+i c β 2k a β k k + c a k β k Proof The proof goes along the same lines as the odd case k 1 a β i i c a i β k+i c β 2k a β k k + c a k β k is the coefficient of x r in β 1 +β sr k 1 1 1 cx 1 1 cx 1 1 1 a i x 1 c a i x 1 1 a k x + 1 1 c a k x 1 1 1 a i x 1 c a i x [1 c a kx + 1 a k x] 1 1 cx 1 1 [2 cx] 1 a i x 1 c a i x 1 1 1 a i x 1 c a i x [ 1 + 1 ] 1 cx
14 GORENSTEIN HILBERT COEFFICIENTS By an argument similar to the proof of lemma 45 applied to the two sums separately, we see that this is the coefficient of x r in k 1 + r t 1 t k 1 + t [ k + r t 1 t k + t c r 2t βi t k + r t 1 + k + t 1 α β i i + ] c r 2t 1 t k + r t k + t Now in both cases whether s is even or odd we have Lemma 47 1 l r ν l r 0 r l 1 i 1 i l s 1 l r ν l r 0 r l β 1 +β sr l d itjit i t + t 1 t1 β 1 + +β 2k+1 r βi t α β i i a β i ij i c a iji β k+i c β 2k+1 c r 2t and βi t k 1 a β i ij i c a iji β k+i c β 2k a β k kj k + c a kjk β k are both equal to Proof For any given r tuples 1 α 1 α r s with 0 r l we will have 1 β 1 β l r s such that {α 1, α r } {β 1, β l r } is equal to {i 1, i l } l In the product d itjit i t + t 1, the coefficient of 1 i 1 i l s t1 1 α 1 α r s α β i i d αtj αt is 1 l r β 1 β l r 1 l r ν l r, since i t + t 1 is strictly r 1 β 1 <<β l r s+l 1 r increasing until i l +l 1 Further, if s is odd d ij a ij, i k and d ij c a 2k+1 i,j, i > k and if s is even d ij a ij, i < k ; d ij c a 2k i,j, i > k and d kj a kj or c a kj Hence, 1 i 1 i l s 1 l r ν l r 0 r l 1 l r ν l r 0 r l l d itjit i t + t 1 t1 β 1 +β sr β 1 + +β 2k+1 r k 1 a β i ij i c a iji β k+i c β 2k+1, s 2k + 1 a β i ij i c a iji β k+i c β 2k a β k kj k + c a kjk β k, s 2k This completes the proof
Remark 48 1 i 1 i l s GORENSTEIN HILBERT COEFFICIENTS 15 l d itjit i t + t 1 f l a 1j1,, c, when s 2k + 1 is t1 odd However, when s 2k, d kj can equal either a kj or c a kj and hence l d itjit i t +t 1 f l a 1j1, a kjk, c + f l a 1j1,, c a kjk, c 1 i 1 i l s t1 Thus, when s 2k + 1, 1 l r ν l r 0 r l r0 β 1 +β sr β 1 + +β 2k+1 r a β i ij i c a iji β k+i c β 2k+1 f l a 1j1,, c and when s 2k, l k 1 1 l r ν l r a β i ij i c a iji β k+i c β 2k a β k kj k + c a kjk β k f l a 1j1, a kjk, c + f l a 1j1,, c a kjk, c Remark 49 Suppose R/I has a quasi-pure resolution Since the Hilbert function of R/I is unaltered by any cancellations, we may assume d ij > d i 1j for all i Since d ij i for all i, we get d ij i d i 1 j i 1 and hence d pj p d qj q for l all p q Now assume that not all factors in d itjit i t + t 1 are positive and let p be the smallest integer with d ipjip i p + p 1 < 0 Then p > 1 and d ip 1 j ip 1 i p 1 + p 2 > 0 It follows that t1 d ip 1 j ip 1 i p 1 + p 2 d ipj ip i p + p 1 2 or equivalently i p i p 1 d ipj ip d ip 1 j ip 1 + 1 d ip 1 j ip 1 i p 1 d ipjip i p + 1 > d ipjip i p l which is a contradiction, and we get that d itjit i t + t 1 0 Thus, f l d 1j1,, d sjs 0, for all s-tuples d 1j1,, d sjs, provided the resolution is quasi-pure t1 Proof of theorem 42 We have α ij a ij c a ij and α ij α ij a ij a ij c a ij a ij By the quasi-purity of the resolution of S and lemma 44, c 2a ij for all i, j So a i1 a i2 a ibi implies that α i1 α i2 α ibi We also have for all 1 i k, p i min j α ij m i M s i and P i max j α ij M i m s i We treat the even and odd cases separately Case 1 s 2k + 1 is odd The resolution starts as in 4 and by lemmas 45, 47 and
16 GORENSTEIN HILBERT COEFFICIENTS remark 48 we have that α iji c 2a iji V α 1j1, α kjk 1 j i b i l k + r t 1 l r ν l r 1 t k + t r0 1 j i b i c r 2t βi t α iji c 2a iji V α 1j1 α kjk f l a 1j1,, c α β i ij i By remark 49, f l d 1j1, d sjs 0 Further, c 2a iji by lemma 44 and V α 1j1,, α kjk 0, for α 1j1 α kjk So p i c 2a iji V α 1j1,, α kjk f l m 1, m k, M k+1, M s j i j i α iji c 2a iji V α 1j1 α kjk f l a 1j1,, c which is the same as j i f l m 1, m k, M k+1, M s where Q j i P i c 2a iji V α 1j1,, α kjk f l M 1, M k, m k+1, m s p i detq α iji c 2a iji V α 1j1 α kjk f l a 1j1,, c c 2a 1j α 1j c 2a 1j 1j c 2a 1j f l M 1, M k, m k+1, m s b i b i c 2a ij α ij c 2a ij b i ij c 2a ij b k b k b k P i detq c 2a kj α kj c 2a kj kj c 2a kj
GORENSTEIN HILBERT COEFFICIENTS 17 detq > 0 since at least one of the V α 1j1,, α kjk > 0 and c 2a iji > 0 Replacing the last column by the alternating sums of columns in Q and using theorems 33 and 37, we get c detq det cdetl L 0 So f l m 1, m k, M k+1, M s j i p i cdetl α iji c 2a iji V α 1j1 α kjk f l a 1j1,, c f l M 1, M k, m k+1, m s P i cdetl On the other hand, we start with M t again Replacing the last column of M t by alternating sums of the columns and using theorem 35 we get M t L 0 k b i 1 i a k+t ij c a ij k+t c 2a ij So j i result α iji c 2a iji V α 1j1 α kjk f l d 1j1,, d sjs s+l!e l detl and hence the Case 2 s 2k is even Now the resolution starts as in 5 and by lemmas 46, 47 and remark 48 we obtain α iji V α 1j1 α kjk j i [ ] k + r t k + r t 1 1 l r ν l r 1 t + c r 2t k + t k + t 1 0 r l 1 j i b i βi t α β i ij i α iji V α 1j1,, α kjk f l a 1j1, a kjk,, c + f l a 1j1, c a kjk,, c
18 GORENSTEIN HILBERT COEFFICIENTS Since f l d 1j1, d sjs 0 and V α 1j1,, α kjk 0, we have p i V α 1j1,, α kjk f l m 1, m k, M k+1, M s + f l m 1, m k 1, M k, M s j i and 1 j i b i j i But, α iji V α 1j1,, α kjk f l a 1j1, a kjk,, c + f l a 1j1, c a kjk,, c P i V α 1j1,, α kjk f l M 1, M k, m k+1, m s +f l M 1, M k 1, m k, m s f l m 1, m k, M k+1, M s f l m 1, m k 1, M k, M s f l M 1, M k 1, m k, m s f l M 1, M k, m k+1, m s Therefore, p i V α 1j1,, α kjk 2f l m 1, m k, M k+1, M s j i 1 j i b i j i which is the same as α iji V α 1j1,, α kjk f l a 1j1, a kjk,, c + f l a 1j1, c a kjk,, c P i V α 1j1,, α kjk 2f l M 1, M k, m k+1, m s 2f l m 1, m k, M k+1, M s 1 j i b i where p i detq α iji V α 1j1,, α kjk f l a 1j1, a kjk,, c + f l a 1j1, c a kjk,, c Q 2f l M 1, M k, m k+1, m s r 1 r i r k r i r k α 1j α ij r 1 r 1 1j r i ij r k α kj kj P i detq
GORENSTEIN HILBERT COEFFICIENTS 19 detq > 0 since at least on of the V α 1j1,, α kjk > 0 Replacing the last column by the alternating sums of columns and using theorems 34 and 37, we get Then 2f l m 1, m k, M k+1, M s 1 j i b i 1 detq det L 0 p i detl detl α iji V α 1j1,, α kjk f l a 1j1, a kjk,, c + f l a 1j1, c a kjk,, c 2f l M 1, M k, m k+1, m s P i detl On the other hand, we start with N t again Replacing the last column of N t by alternating sums of the columns and using theorem 37, we get: N t So s + l!e l detl L 0 k b i 1 i a k+t ij c a ij k+t α iji V α 1j1,, α kjk f l a 1j1, a kjk,, c + f l a 1j1, c a kjk,, c 1 j i b i We get, 2f l m 1, m k, M k+1, M s p i s + l!e l 2f l M 1, M k, m k+1, m s P i 2f l m 1, m k, M k+1, M s m 1 m k M k M k+1 M 2k 1, s + l!e l 2f l M 1, M k, m k+1, m s M 1 M k m k m k+1 m 2k 1 f l m 1, m k, M k+1, M s m 1 m k M k+1 M 2k 1 2M k, s + l!e l
20 GORENSTEIN HILBERT COEFFICIENTS f l M 1, M k, m k+1, m s M 1 M k m k+1 m 2k 1 2m k Now M k c a k1 and m k a k1 Clearly c 2a k1 2m k Also 2c a k1 c + c 2a k1 c So M 2k c 2M k and 2m k c m 2k and hence f l m 1, m k, M k+1, M s m 1 m k M k+1 M 2k 1 M 2k, s + l!e l This completes the proof f l M 1, M k, m k+1, m s M 1 M k m k+1 m 2k 1 m 2k Let R k[x 1,, x n ] and S R/I where I is a homogeneous ideal of height s Corollary 410 Let S R/I be as above Suppose the betti diagram of S is symmetric, that is β ij β s i,c j, 0 i s and β sj 0, j c Then the Hilbert coefficients will satisfy the same bounds as in theorem 42 Corollary 411 Let S R/I be as above Suppose the betti diagram of S is a positive rational linear combinations of quasi-pure Gorenstein symmetric betti diagrams Then the Hilbert Coefficients of S satisfy the same bounds as in theorem 42 References [1] M Boij and J Söderberg Graded Betti Numbers of Cohen-Macaulay Modules and the Multiplicity conjecture, math AC/0611081, 2 78, 2008, 1, 85-106 [2] W Bruns and J Herzog Cohen-Macaulay rings, Cambridge University Press, 1993, 147-205 [3] D Eisenbud and F Schreyer Betti numbers of graded modules and cohomology of vector bundles, J Amer Math Soc, 22, 2009, 859-888 [4] S El Khoury Upper Bounds for the Hilbert Coefficients of almost Cohen-Macaulay Algebras, Int J Algebra, Vol 5, 2011, no 14, 679-690 [5] J Herzog and H Srinivasan Bounds for multiplicities, Transactions of the AMS 350 nbr 7, 1998, 2879-2902 [6] J Herzog and X Zheng Bounds for Hilbert Coefficients, Proceedings of the AMS, 37 nbr 2, 2009, 487-494 [7] C Huneke and MMIller A note on the multiplicity of Cohen-Macaulay algebras with pure resolutions, CanadJ Math, 37, 1985, 1149-1162 [8] H Srinivasan A note on the Multiplicities of Gorenstein Algebras, J Algebra 208, 1998, 425-443 [9] C Peskine and L Szpiro Syzygies and multiplicities, CR Acad Sci Paris Ser A 278, 1974, 1421-1424