Hilbert coefficients and Buchsbaumness of the associated graded ring of filtration
aa r X i v : . [ m a t h . A C ] F e b HILBERT COEFFICIENTS AND BUCHSBAUMNESS OF THE ASSOCIATEDGRADED RING OF FILTRATION
KUMARI SALONI
Abstract.
Let A be a Noetherian local ring with the maximal ideal m and I be an m -primary idealin A . In this paper, we study a boundary condition of an inequality on Hilbert coefficients of an I -admissible filtration I . When A is a Buchsbaum local ring, the above equality forces Buchsbaumnesson the associated graded ring of filtration. Our result provides a positive resolution of a question ofCorso in a general set up of filtration. Introduction
Let ( A, m ) be a Noetherian local ring of dimension d > I an m -primary ideal. For simplicity,we assume that A/ m is infinite. For an I -admissible filtration I , let Q = ( a , . . . , a d ) ⊆ I be a reductionof I i.e. I n +1 = QI n for n ≫
0. We may assume that Q is also a reduction of ideals I and I . Let G ( I ) = ⊕ n ≥ I n /I n +1 be the associated graded ring of I . Let R ( I ) = ⊕ ∞ n =0 I n t n and T = ⊕ ∞ n =0 Q n t n be the Rees algebra of I and Q respectively. We regard them as sub-algebras of the polynomial ring A [ t ]. Let R ≥ m = ⊕ n ≥ m I n /I n +1 be the graded ideal of R ( I ). In particular, we write R + for R ≥ . Theunique homogeneous maximal ideal of R ( I ) and T = R ( Q ) are denoted by N and M respectively, i.e., N = m R + R + and M = m T + T + . We write F = F I ( Q ) = T /IT and F ′ = T /I T. Let ℓ ( M ) denote the length of an A -module M. It is well known that the function ℓ ( A/I n ) is asymp-totically polynomial, i.e., we have a polynomial P ( I , n ) = d X i =0 ( − i e I ( I ) (cid:18) n + d − i − d − i (cid:19) for unique integers e i ( I ) such that P ( I , n ) = ℓ ( A/I n ) for all n ≫
0. The coefficients e i ( I ), 1 ≤ i ≤ d aredependent on each other through several inequalities and the boundary conditions in those inequalitiesare often related with the depth properties of form rings. One of such inequality is the following:(1) e ( I ) − e ( Q ) ≥ e ( I ) − ℓ ( A/I ) + ℓ ( I / ( I + Q ))where Q = ( a , . . . , a d ) ⊂ I is a minimal reduction of I , see [18, Theorem 2.4]. We also prove it inTheorem 13. In the case, when A is Cohen-Macaulay, we have e ( I ) ≥ e ( I ) − ℓ ( A/I ) − ℓ ( I/ ( I + Q ) . In [4] and [7], authors proved that the equality e ( I ) = 2 e ( I ) − ℓ ( A/I ) − ℓ ( I/ ( I + Q ) holds true if andonly if I = QI and Q ∩ I = QI . In this case, the associated graded ring G ( I ) of I is Cohen-Macaulay.Further, Corso [3] conjectured and proved for d = 1 that if the equality in (1) holds for m -adic filtration { m n } and A is a Buchsbaum local ring, then the associated graded ring G ( m ) of m is a Buchsbaumring. Rossi and Valla in [18, Theorem 2.1] generalized Corso’s result for modules. In [15, Theorem1.2], Ozeki extended Corso’s result for arbitrary dimension. Furthermore, Ozeki in [16], discussed theequality in (1) for m -primary ideals in arbitrary dimension. He proved that under the conditions (C1) Date : February 10, 2021.2010
Mathematics Subject Classification.
Primary: 13H10, 13A30, Secondary: 13D45.
Key words and phrases.
Buchsbaum modules, local cohomology, d-sequence, Hilbert coefficients. and (C2) given below, if the equality e ( I ) − e ( Q ) = 2 e ( I ) − ℓ ( A/I ) − ℓ ( I/ ( I + Q ) holds and A is aBuchsbaum local ring then the associated graded ring G ( I ) is Buchsbaum with I ( G ( I )) = I ( A ). In thispaper, we extend Ozeki’s result for an I -admissible filtration I . First consider the following conditions.( C ) The sequence a , . . . , a d is a d-sequence in A .( C ) The above sequence is an unconditioned strong d-sequence.( C ) We have ( a , . . . , ˇ a i , . . . , a d ) : a i ⊆ I .( C ) depth A > I − invariant of an A -module M of dimension s is I ( M ) = s − X i =0 ℓ (H i m ( M ))where H i m ( M ) denote the local cohomology module of M with support in m . We set W := H m ( A ). Theorem 1.
Suppose that the conditions ( C ) and ( C ) hold. Then the following conditions are equiv-alent: (1) e ( I ) − e ( Q ) = 2 e ( I ) − ℓ ( A/I ) + ℓ ( I / ( I + Q )) ; (2) I n +2 ⊆ Q n I + W, ( Q n + W ) ∩ ( I n +1 + W ) = Q n I + W for all n ≥ and ( a , . . . , ˇ a i , . . . , a d ) : a i ⊆ I + Q for ≤ i ≤ d .When this is the case, we have W ⊆ I and the following assertions hold true. (1) for all n ∈ Z , [H M ( G ( I ))] n = W/I ∩ W if n = 2( I n ∩ W ) / ( I n +1 ∩ W ) if n ≥ otherwise (2) For ≤ i ≤ d − , H i M ( G ( I )) = [H i M ( G ( I ))] − i ∼ = H i m ( A ) . (3) The a-invariant a ( G ( I )) ≤ − d. (4) I ( G ( I )) = I ( A ) < ∞ . (5) e ( I ) = e ( Q ) + e ( Q ) + e ( I ) − e ( I ) + ℓ A ( A/I ) if d ≥ . (6) e i ( I ) = e i − ( Q ) + 2 e i − ( Q ) + e i ( Q ) for ≤ i ≤ d if d ≥ . (7) If A is quasi Buchsbaum local ring, then G ( I ) is quasi-Buchsbaum with I ( G ( I )) = I ( A ) . (8) If A is a Buchsbaum local ring, then G ( I ) is a Buchsbaum ring with I ( G ( I )) = I ( A ) . Preliminaries
Sally module.
We define S = S Q ( I ), the Sally module of I with respect to Q , as a cokernel ofthe following map of graded T -modules0 −→ ( I T ) −→ ( R + (+1)) = ⊕ j ≥ I j t j − −→ S −→ . Thus S = ⊕ j ≥ I j +1 /I Q j . Lemma 2.
Suppose that ( C ) , ( C ) and ( C ) are satisfied and the ring T satisfies Serre’s condition ( S ) . Then Ass T ( S ) ⊆ { m T } . Hence, dim T ( S ) = d if S = 0 . The proof is discussed in [6, Lemma 2.3] for I = { I n } and in [17, Lemma 2.2] for I = { I n } . Weprovide an analogous proof for general case for the convenience of reader. ILBERT COEFFICIENTS AND BUCHSBAUMNESS OF THE ASSOCIATED GRADED RING OF FILTRATION 3
Proof.
Let P ∈ Ass T ( S ). Since m l S = 0 for l ≫
0, we have p = m T ⊆ P. Suppose p = P . Note thatdepth T P ( R + (+1)) P ≥ a ∈ p ⊆ P is a non-zero-divisor on R + (+1) = ⊕ j ≥ I j t j − . Applyingdepth Lemma on the exact sequence0 −→ ( I T ) P −→ ( R + (+1)) P −→ S P −→ , of graded T P -modules, we get depth T P ( I T ) P ≥ min { depth T P ( R + (+1)) P , depth T P S P + 1 } = 1 anddepth T P S P = 0 ≥ min = { depth T P ( R + (+1)) P , depth T P ( I T ) P − } = depth T P ( I T ) P −
1. Hencedepth T P ( I T ) P = 1. Now consider the exact sequence0 −→ ( I T ) P −→ T P −→ ( T /I T ) P −→ . Since p = P , htP ≥
2. So dim T P ≥ T p ≥ min { , dim T P } = 2 as T satisfies( S ). Again by depth Lemma, 1 = depth T P ( I T ) P ≥ min { depth T P T P , depth T P ( T /I T ) P + 1 } =depth T P ( T /I T ) P + 1 which implies depth T P ( T /I T ) P = 0 . By Lemma 5,
T /I T is Cohen-Macalaywhich means dim( T /I T ) P = depth T P ( T /I T ) P = 0 . Thus P ∈ min T ( T P /I T P ) = { m T } which is acontradiction. (cid:3) Lemma 3.
For an I -filtration I , assume that Q ( I and µ = µ ( I \ Q ) . Then there exists an exactsequence (2) T ( − µ φ −→ R ( I ) \ T → S ( − → of graded T -modules.Proof. Let I = Q + ( x , . . . , x µ ). Consider the map T ( − µ φ −→ R ( I ) /T defined as φ ( α , . . . , α µ ) = µ P i =1 α i x i t ∈ R ( I ) /T. Then [Coker φ ] n = I n / ( Q n + Q n − I ) ∼ = I n /Q n − I = S ( − n for n ≥ φ ] = 0. Therefore, Coker φ ∼ = S ( −
1) as graded T -module. (cid:3) The Hilbert coefficients of the Sally module S and the coefficients e i ( I ) of I are related through thefollowing relations: ℓ ( A/I n +1 ) = ℓ ( A/Q n I ) − ℓ ( S n ) and ℓ ( A/Q n I ) = ℓ ( A/Q n ) + ℓ ( Q n /Q n I )for n ≥
0. The next result is a generalization of [16, Proposition 2.4].
Proposition 4.
Let I be an I -admissible filtration. Then (3) e ( I ) ≥ ( e ( Q ) + e ( I ) − ℓ A ( A/I ) + e ( S ) if dim S = d.e ( Q ) + e ( I ) − ℓ A ( A/I ) if dim S ≤ d − . Proof.
We have, for n ≥ Q n /Q n I = A/I ⊗ Q n = ( A/I ⊗ T ) n ∼ = ( T /I T ) n and T /I T is ahomomorphic image of A/I [ X , . . . , X d ]. So for all n ≥ ℓ ( Q n /Q n I ) ≤ ℓ ( A/I ) (cid:18) n + d − d − (cid:19) + lower degree termsThere exists an integer N such that for all n ≥ N , ℓ ( A/Q n ) = d X i =0 ( − i e i ( Q ) (cid:18) n − d − id − i (cid:19) = e ( I ) (cid:18) n + dd (cid:19) − ( e ( I ) + e ( Q )) (cid:18) n + d − d − (cid:19) + d X i =2 ( − i ( e i − ( Q ) + e i ( Q )) (cid:18) n + d − id − i (cid:19) Thus, for all n ≥ N , ℓ ( A/I n +1 ) = ℓ ( A/Q n ) + ℓ ( Q n /Q n I ) − ℓ ( S n ) KUMARI SALONI ≤ e ( I ) (cid:18) n + dd (cid:19) − ( e ( I ) + e ( Q )) (cid:18) n + d − d − (cid:19) + d X i =2 ( − i ( e i − ( Q ) + e i ( Q )) (cid:18) n + d − id − i (cid:19) + { ℓ ( A/I ) (cid:18) n + d − d − (cid:19) + lower degree terms } − ℓ ( S n ) ≤ e ( I ) (cid:18) n + dd (cid:19) − n e ( I ) + e ( Q ) − ℓ ( A/I ) o(cid:18) n + d − d − (cid:19) + lower degree terms+ d X i =2 ( − i ( e i − ( Q ) + e i ( Q )) (cid:18) n + d − id − i (cid:19) − ℓ A ( S n ) . Now the conclusion follows by comparing the coefficients. (cid:3)
We need the following Lemma for Proposition 6.
Lemma 5.
Suppose that conditions ( C ) and ( C ) are satisfied for an I -admissible filtration I . Then F ∼ = ( A/I )[ X , . . . , X d ] and F ′ ∼ = ( A/I )[ X , . . . , X d ] as graded A -algebras. In particular, F and F ′ areCohen-Macaulay rings of dimension d .Proof. See [6, Proposition 2.2] for F = T /IT ∼ = ( A/I )[ X , . . . , X d ]. Then F ′ ∼ = A/I ⊗ T /IT ∼ =( A/I )[ X , . . . , X d ]. (cid:3) Proposition 6.
Let I be an I -admissible filtration. Suppose that the conditions ( C ) and ( C ) aresatisfied. Then, for all n ≥ ,ℓ A ( A/I n +1 ) = e ( I ) (cid:18) n + dd (cid:19) − n e ( I ) + e ( Q ) − ℓ ( A/I ) o(cid:18) n + d − d − (cid:19) + d X i =2 ( − i ( e i − ( Q ) + e i ( Q )) (cid:18) n + d − id − i (cid:19) − ℓ A ( S n )(5) Proof.
Suppose that the the conditions ( C ) and ( C ) hold. Then N = 0 in the proof of Proposition 4and T /I T ∼ = A/I [ X , . . . , X d ] by Lemma 5. Hence equality holds in (4) for all n ≥ . . (cid:3) In (5), we may write ℓ A ( S n ) = e ( S ) (cid:18) n + s − s − (cid:19) − e ( S ) (cid:18) n + s − s − (cid:19) + . . . + ( − s − e s − ( S )for n ≫
0. Then, on comparing the coefficients of both sides, we get the following corollary.
Corollary 7.
Suppose that conditions ( C ) and ( C ) are satisfied and s = dim S . (1) Suppose s = d . Then (a) e ( I ) = e ( I ) + e ( Q ) − ℓ ( A/I ) + e ( S ) and (b) e i ( I ) = e i − ( Q ) + e i ( Q ) + e i − ( S ) for ≤ i ≤ d . (2) Suppose s < d . Then (a) e ( I ) = e ( I ) + e ( Q ) − ℓ ( A/I ) , (b) e i ( I ) = e i − ( Q ) + e i ( Q ) for ≤ i ≤ d − s and (c) e i ( I ) = e i − ( Q ) + e i ( Q ) + ( − d − s e i − d + s − ( S ) for all d − s + 1 ≤ i ≤ d. Generalized depth and generalized Cohen-Macaulayness.
In [8], authors defined the notionof generalized depth. This invariant has been studied by Brodmann [1] and Faltings [5] in their work onthe finiteness properties of local cohomology modules. We recall the definition and some results from[8].
ILBERT COEFFICIENTS AND BUCHSBAUMNESS OF THE ASSOCIATED GRADED RING OF FILTRATION 5
For an A -module M and an ideal J of A , the generalized depth of M with respect to J is g − depth J ( M ) := sup { k ∈ Z | J ⊆ q Ann A H i m ( M ) for all i < k } . Note that, g − depth J M ≥ k if and only if some power of J annihilates H i m ( M ) for 0 ≤ i ≤ k − . Fora non-negatively graded Noetherian ring S = ⊕ S n with S local and a homogeneous ideal J of S , wedefine the generalized depth of a graded S -module L with respect to J as, g − depth J L = g − depth J L N L N where N is the maximal homogeneous ideal of S . Investigating the relationship between the depths of R ( I ) and G ( I ), authors in [8, Proposition 3.2] proved that g − depth R ( I ) + R ( I ) = g − depth G ( I ) + G ( I ) + 1when I = { I n } . We extend the above result for I − admissible filtration in Proposition 9. First we notethe following lemma which is a generalization of [8, Lemma 3.1]. Lemma 8.
Let A be a local ring and I = { I n } be a filtration of A . Let P be a prime ideal of R ( I ) such that R ( I ) + (1) ⊆ P but R ( I ) + * P . Then depth R ( I ) P = depth G ( I ) Q + 1 where Q = P/R ( I ) + (1) .Proof. Choose xt m ∈ R ( I ) + \ P with x ∈ I m . Then ( xt m − ) R ( I ) P ⊆ R ( I ) + (1) R ( I ) P . Indeed,( xt m − ) R ( I ) P = R ( I ) + (1) R P . To see this, let a = yt n ∈ R ( I ) + (1) with y ∈ I n +1 , then yt n = xt m − . yt n +1 xt m ∈ xt m − R ( I ) P . Therefore G ( I ) Q ∼ = R ( I ) P /R ( I ) + (1) R ( I ) P = R ( I ) P / ( xt m − ) R ( I ) P . If xt m − is a zero-divisor in R ( I ) P , then there exists an associated prime P ′ of R ( I ) such that xt m − ∈ P ′ ⊆ P . Suppose P ′ = (0 : R ( I ) f ), then xt m − f = 0= ⇒ xt m f = 0= ⇒ xt m ∈ P ′ ⊆ P , a contradiction.Thus, xt m − is non-zero-divisor in R ( I ) P which gives the desired result. (cid:3) Proposition 9.
Let ( A, m ) be a Noetherian local ring and I = { I n } a filtration of A . Then g − depth G ( I ) + G ( I ) = g − depth R ( I ) + R ( I ) − Proof.
We may assume that A is complete, see [8, Remark 2.4]. Consequently, A is a homomorphicimage of a regular local ring and hence R ( I ) and G ( I ) are homomorphic images of regular rings. Nowusing [14, Proposition 2.4], we have that g − depth G ( I ) + G ( I ) = min { depth G ( I ) p + dim G ( I ) / p G ( I ) | p ∈ SpecG( I ) , G( I ) + * p , p is homogeneous } and g − depth R ( I ) + R ( I ) = min { depth R ( I ) p + dim R ( I ) / p R ( I ) | p ∈ SpecR( I ) , R( I ) + * p , p is homogeneous } Suppose g − depth R + R ≥ k . Let q be a prime ideal of G ( I ) not containing G ( I ) + . Then thereexists a prime ideal p in R such that R + (1) ⊆ p and q = p / R + (1). Clearly, R + * p (as G ( I ) + * q ).By Lemma 8, depth R ( I ) p = depth G ( I ) q + 1. Therefore,depth G ( I ) q + dim G ( I ) / q G ( I ) = depth R p − R / p ≥ k − . which implies that g − depth G ( I ) + G ( I ) ≥ g − depth R + R − . For the converse, let g − depth G ( I ) + G ( I ) ≥ k . let p be a homogeneous prime ideal of R suchthat R + * p . Then, there exists a P ∈ Spec( R ), R + * P and ( R + (1) , p ) ⊆ P. Otherwise, for allprime ideal P with ( R + (1) , p ) ⊆ P , we have R + ⊆ P . This means R + ⊆ p ( R + (1) , p ) which implies R l + ⊆ ( R + (1) , p ) for all l ≫
0. Using ( ?? ), ( R l + ) n = ( R + ) n = ( R + (1) , p ) n for all n ≫ . This implies I n t n = I n +1 t n + p n = ⇒ p n = I n t n for all n ≫
0. Thus, for all n ≫
0, ( R + ) n = p n = ⇒R + ⊆ p which is acontradiction. KUMARI SALONI
Now let Q = P/ R + (1). From [8, Remark 2.2] and Lemma 8, we havedepth R p + dim R / p ≥ depth R P + dim R /P ≥ depth G ( I ) Q + 1 + dim G ( I ) /Q ≥ k + 1Hence g − depth R + R ≥ k + 1 which gives g − depth R + R ≥ g − depth G ( I ) + G ( I ) + 1 . (cid:3) Next, we establish a relationship between the generalized depths of Sally module and associatedgraded ring of filtration in Proposition 11. The I -adic case was discussed in [16]. First we need thefollowing lemma. Lemma 10.
Suppose the conditions ( C ) and ( C ) are satisfied. Then, for each ≤ i ≤ d, we have [H i M ( I T )] n = H m ( A ) if i = n = 0H i − m ( A ) if ≤ i ≤ d, − i ≤ n ≤ − otherwisefor all n ∈ Z . Hence I T is a generalized Cohen-Macaulay T -module and dim T I T = d + 1 .Proof. By Lemma 5, F ′ ∼ = ( A/I )[ X , . . . , X d ] is Cohen-Macaulay. So, by the exact sequence,0 → I T → T → F ′ → , we get H i M ( I T ) ∼ = H i M ( T )for 0 ≤ i ≤ d − → H d M ( I T ) → H d M ( T ) → H d M ( F ′ )of graded T -modules. Since [H d M ( T )] n = 0 for all n ≤ − d by [23, Theorem 6.2] and [H d M ( F ′ )] n = 0for all n ≥ − d , we have H d M ( I T ) ∼ = H d M ( T ) as graded T -modules. Now the conclusion follows by[23, Theorem 6.2]. (cid:3) Proposition 11.
Suppose that the conditions ( C ) , ( C ) and ( C ) are satisfied and S = 0 . Let s = g − depth M S. Then g − depth M G = s − if s < d and S is generalized Cohen-Macaulay T -module ifand only if g − depth M G ≥ d − . The later is the case when s = d .Proof. Clearly s ≤ dim T S = d . By Lemma 10, I T is generalized Cohen-Macaulay with dim T I T = d + 1. Consider the exact sequence 0 → I T → R + (1) → S → . If s = d , then it follows that g − depth M R + (1) ≥ d and if s < d , then g − depth M R + (1) = s . Next,we consider the exact sequence 0 → R + → R → A → . Since A is generalized Cohen-Macaulay local ring of dimension d , we get that g − depth M R ≥ d if s = d and g − depth M R = g − depth M R + = s if s < d . By Proposition 9, g − depth M G ( I ) = s − s < d .Further if s = d , the exact sequence 0 → R + (1) → R → G ( I ) → g − depth M G ( I ) ≥ d −
1. By Proposition 9, we have that g − depth M G ( I ) ≥ d − ⇐⇒ g − depth M R ≥ d which implies s = d as otherwise from the discussion above g − depth M R = s < d . Finally, S isgeneralized Cohen-Macaulay if and only if s = d . (cid:3) ILBERT COEFFICIENTS AND BUCHSBAUMNESS OF THE ASSOCIATED GRADED RING OF FILTRATION 7
Reduction steps.
In the theory of Hilbert polynomials, it is very useful to reduce a problem(i) to the case when depth
A > W = H m ( A ), C = A/W and I C denote the filtration { I n C = ( I n + W ) /W } in C which is an IC -admissible filtration. For an element a ∈ A , we write I /a for the filtration { I ′ n = I n + ( a ) / ( a ) } of thering A/ ( a ). If I is an I -admissible filtration, then I / ( a ) is an I/ ( a )-admissible filtration of A/ ( a ). Let A ′ = A/ ( a ) , Q ′ = Q/ ( a ) , I ′ = I/ ( a ) and I ′ = I / ( a ). Lemma 12. (1) If A satisfies ( C ) and ( C ) , then C = A/W satisfies ( C ) and ( C ) . (2) 2 e ( I ) − e ( I )+ e ( Q ) = 2 ℓ A ( A/I )+ ℓ A ( I / ( I + Q )) if and only if e ( I C ) − e ( I C )+ e ( QC ) =2 ℓ A ( C/I C ) + ℓ A ( I C/ ( I C + QC )) and W ⊆ I + Q . (3) Suppose that d ≥ . Let a ∈ Q \ m Q be a superficial element for I = {I n } and Q . Then e ( I ) − e ( I ) + e ( Q ) = 2 ℓ A ( A/I ) + ℓ A ( I /I + Q )) if and only if e ( I ′ ) − e ( I ′ ) + e ( Q ′ ) =2 ℓ A ( A ′ /I ′ ) + ℓ A ( I ′ / ( I ′ + Q ′ )) . The inequality and its boundary conditions
In this section, we prove the inequality e ( I ) − e ( Q ) ≥ e ( I ) − ℓ A ( A/I ) − ℓ A ( I /I + Q ) for an I -admissible filtration. A different proof of the Theorem 13 can be found in [18, Theorem 2.4]. Themain result of this section is Proposition 15 in which we prove some equivalent conditions for equalityto hold. Let L = ⊕ n ≥ I n +2 + I Q n I Q n which is a graded submodule of S . Theorem 13.
Let ( A, m ) be a Noetherian local ring, I and m -primary ideal and I = { I n } be an I -admissible filtration. Then (6) e ( I ) − e ( Q ) ≥ e ( I ) − ℓ A ( A/I ) − ℓ A ( I /I + Q ) ≥ Proof. If I = Q (= I ), then by Proposition 4, e ( I ) − e ( Q ) ≥ e ( Q ) − ℓ ( A/I ) ≥ e ( Q ) − ℓ ( A/I ) + e ( Q ) − ℓ ( A/I ) as e ( Q ) ≤ ℓ ( A/Q ) . Suppose Q I , µ = µ ( I \ Q ) and I = Q + ( x , . . . , x µ ) as defined in Lemma 2. The canonicalgraded homomorphism F ′ → G = R / R + (1) gives an exact sequence(7) 0 → K (2) → F ′ → G → R / ( R + (1) + T ) → T -modules where K (2) = ker( F ′ → G ). Now consider the graded homomorphism of graded T -modules ( T /I T ( − µ φ −→ R / ( R + (1) + T )defined as φ ( α , . . . , α µ ) = µ X i =1 α i x i t ∈ R ( I ) / ( R + (1) + T )where α i ∈ T /I T for 1 ≤ i ≤ µ . Then [Coker φ ] n = I n /I n +1 + Q n I Q n − /I n +1 + Q n ∼ = I n / ( I n +1 + I Q n − ) = ( S/L ) n − . Thus Coker φ ∼ = S/L ( −
1) and we get the exact sequence(8) 0 → ker φ → F ′ ( − µ φ −→ R ( I ) / ( R + (1) + T ) → S/L ( − → → [ker φ ] → [ F ′ µ ] → [ R / R + (1) + T ] → F ′ to get the following commutative diagram:([ker φ ] ⊗ F ′ )( − / / ψ (cid:15) (cid:15) ([ F ′ µ ] ⊗ F ′ )( − / / ψ (cid:15) (cid:15) ([ R / ( R + (1) + T )] ⊗ F ′ )( − / / ψ (cid:15) (cid:15) / / ker φ f / / [ F ′ ( − µ ] / / Im φ / / KUMARI SALONI of graded T -modules. Since ψ is bijective, we have an exact sequence(9) 0 → K (1) → ([ R ( I ) / ( R + (1) + T )] ⊗ F ′ )( − ψ −−→ Im φ → T -modules where K (1) = ker ψ . Note that K (1) ≤ ([ R / ( R + (1) + T )] ⊗ F ′ )( −
1) and K (2) ≤ F ′ , so dim K ( i ) ≤ d for i = 1 ,
2. Since L ≤ S , dim L ≤ d . We put L = K (3) for convenience ofnotation. From (7), (8) and (9), we have that ℓ ( I n /I n +1 ) = ℓ ( G ( I ) n ) = ℓ ( F ′ n ) + ℓ (([ R / R + (1) + T ] n ) − ℓ ([ K (2) ] n )= ℓ ( F ′ n ) + { [Im φ ] n + ℓ ([ S/L ] n − ) } − ℓ ([ K (2) ] n )= ℓ ( F ′ n ) + { ℓ ( (cid:2) [ R / R + (1) + T ] ⊗ F ′ (cid:3) n − − ℓ ([ K (1) ] n ) } + ℓ ( S n − ) − ℓ ([ K (3) ] n − ) − ℓ ([ K (2) ] n )(10)for all n ≥
0. Since (
A/I )[ X , . . . , X d ] → F ′ = T /I T defined by X i x i t is surjective map and[ R / R + (1) + T ] = I /I + Q , we have that ℓ A ([[ R / R + (1) + T ] ⊗ F ′ ] n − ) ≤ ℓ A ( I /I + Q ) (cid:18) n − d − d − (cid:19) = ℓ A ( I /I + Q ) (cid:16)(cid:18) n + d − d − (cid:19) − (cid:18) n + d − d − (cid:19)(cid:17) and ℓ A (( F ′ ) n ) ≤ ℓ A ( A/I ) (cid:18) n + d − d − (cid:19) Further, we write ℓ A ( G ( I ) n ) = e ( I ) (cid:18) n + d − d − (cid:19) − e ( I ) (cid:18) n + d − d − (cid:19) + . . . + ( − d − e d − ( I ) ℓ A ( S n − ) = e ( S ) (cid:18) n − s − s − (cid:19) − e ( S ) (cid:18) n + s − s − (cid:19) + . . . where s = dim S ≤ d . We get(11) e ( I ) ≤ ( ℓ A ( A/I ) + ℓ A ( I /I + Q ) + e ( S ) if dim S = dℓ A ( A/I ) + ℓ A ( I /I + Q ) if dim S < d.
In (11), e ( S ) ≤ e ( I ) − e ( Q ) − e ( I ) + ℓ A ( A/I ) if dim S = d and e ( I ) − e ( Q ) − e ( I ) + ℓ A ( A/I ) ≥ e ( I ) ≤ ℓ A ( A/I ) + ℓ A ( I /I + Q ) + e ( I ) − e ( Q ) − e ( I )or 2 e ( I ) ≤ ℓ A ( A/I ) + ℓ A ( I /I + Q ) + e ( I ) − e ( Q ) . (12)This completes the proof. (cid:3) Lemma 14.
Suppose that the conditions ( C ) and ( C ) are satisfied and Q I . Then we have, ℓ A ( I n /I n +1 ) = n ℓ A ( A/I ) + ℓ A ( I / ( I + Q )) o(cid:18) n + d − d − (cid:19) − ℓ ( I / ( I + Q )) (cid:18) n + d − d − (cid:19) + ℓ A ( S n − ) − ℓ A ([ K (1) ] n ) − ℓ A ([ K (2) ] n ) − ℓ A ([ K (3) ] n − ) for all n ≥ . Furthermore, e ( I ) − e ( Q ) = 2 e ( I ) − ℓ A ( A/I ) − ℓ A ( I /I + Q ) + X i ∈ Γ e ( K ( i ) ) where Γ = { i | ≤ i ≤ , dim T K ( i ) = d } . Proof is almost same as in [16, Lemma 3.3]. We write it here.
Proof.
By (10), we have that ℓ A ( I n /I n +1 ) = ℓ A ( G ( I ) n ) = ℓ ( F ′ n ) + { ℓ ( (cid:2) [ R / ( R + (1) + T )] ⊗ F ′ (cid:3) n − − ℓ ([ K (1) ] n ) } + ℓ ( S n − ) − ℓ ([ K (3) ] n − ) − ℓ ([ K (2) ] n ) ILBERT COEFFICIENTS AND BUCHSBAUMNESS OF THE ASSOCIATED GRADED RING OF FILTRATION 9 for all n ≥
0. By Lemma 5, F ′ ∼ = ( A/I )[ X , . . . , X d ] and [ R / ( R + (1) + T )] ⊗ F ′ ∼ = ( I / ( I + Q )) ⊗ A/I [ X , . . . , X d ]. Therefore, we have for all n ≥ ,ℓ A ( F ′ n ) = ℓ A ( A/I ) (cid:18) n + d − d − (cid:19) and ℓ A ( (cid:2) [ R / ( R + (1) + T )] ⊗ F ′ (cid:3) n − ) = ℓ A ( I / ( I + Q )) (cid:18) n + d − d − (cid:19) Thus, ℓ A ( I n /I n +1 ) = ℓ A ( A/I ) (cid:18) n + d − d − (cid:19) + ℓ A ( I / ( I + Q )) (cid:18) n + d − d − (cid:19) − ℓ A ([ K (1) ] n )+ ℓ A ( S n − ) − ℓ A ([ K (3) ] n − ) } − ℓ A ([ K (2) ] n )= ℓ A ( A/I ) (cid:18) n + d − d − (cid:19) + ℓ A ( I / ( I + Q )) n(cid:18) n + d − d − (cid:19) − (cid:18) n + d − d − (cid:19)o + ℓ A ( S n − ) − ℓ A ([ K (1) ] n ) − ℓ A ([ K (3) ] n − ) } − ℓ A ([ K (2) ] n )= n ℓ A ( A/I ) + ℓ A ( I / ( I + Q )) o(cid:18) n + d − d − (cid:19) − ℓ A ( I / ( I + Q )) (cid:18) n + d − d − (cid:19) + ℓ A ( S n − ) − ℓ A ([ K (1) ] n ) − ℓ A ([ K (3) ] n − ) } − ℓ A ([ K (2) ] n )for all n ≥
0. On comparing the coefficients, we get that e ( I ) = ℓ A ( A/I ) + ℓ A ( I /I + Q ) + e ( S ) − P i ∈ Γ e ( K ( i ) ) if dim S = dℓ A ( A/I ) + ℓ A ( I /I + Q ) − P i ∈ Γ e ( K ( i ) ) if dim S ≤ d −
1= 2 ℓ A ( A/I ) + ℓ A ( I /I + Q ) + e ( I ) − e ( Q ) − e ( I ) − X i ∈ Γ e ( K ( i ) )since e ( I ) − e ( I ) − e ( Q ) + ℓ ( A/I ) = e ( S ) if dim S = d and e ( I ) − e ( I ) − e ( Q ) + ℓ ( A/I ) =0 if dim S ≤ d − (cid:3) It is easy to observe that Ass T K ( i ) ⊆ { m T } for 1 ≤ i ≤ . . To see this, note that K (1) ≤ [ R / R + (1)+ T ] ⊗F ′ ∼ = ( I /I + Q ) ⊗ ( A/I )[ X , . . . , X d ] which is a maximal Cohen-Macaulay F ′ ∼ = A/I [ X , . . . , X d ]-module, K (2) ≤ F ′ ∼ = A/I [ X , . . . , X d ] and K (3) ≤ S . Therefore, if K ( i ) = 0, then dim K ( i ) = d for 1 ≤ i ≤
3. We will need the following proposition later. Note here that ( C )= ⇒ ( C ) and ( C ) + ( C )= ⇒ T satisfies ( S ). Proposition 15.
Suppose that the conditions ( C ) , ( C ) and ( C ) are satisfied. Assume that Q I .Then the following conditions are equivalent: (1) We have e ( I ) − e ( Q ) = 2 e ( I ) − ℓ A ( A/I ) − ℓ A ( I /I + Q ) . (2) There exists exact sequences → ( I / ( I + Q ) ⊗ F ′ )( − → R / ( R + (1) + T ) → S ( − → and → F ′ → G ( I ) → R / ( R + (1) + T ) → of graded T -modules. (3) K ( i ) = 0 for ≤ i ≤ . When this is the case, we have the injective maps H i M ( G ( I )) ֒ → H i M ( R / ( R + (1) + T )) ֒ → H i M ( S )( − of graded T -modules for ≤ i ≤ d − and H iM ( G ( I )) ∼ = H i M ( R / ( R + (1) + T )) ∼ = H i M ( S )( − for ≤ i ≤ d − . Proof. (1) ⇒ (3) Suppose (1) holds and K ( i ) = 0 for some 1 ≤ i ≤
3. Then i ∈ Γ and e ( K ( i ) ) = 0 usingLemma 14 which is a contradiction.(3) ⇒ (2) From the exact sequence (9), we get that Im φ ∼ = (( I /I + Q ) ⊗ F ′ )( − ⇒ (1) Suppose (2) holds. Then, for all n ≥ ℓ A ( I n /I n +1 ) = ℓ A ( G ( I ) n = ℓ A ([ F ′ ] n ) + ℓ A ([ R / R + (1) + T ] n )= ℓ A ([ F ′ ] n ) + ℓ A (cid:0) [( I / ( I + Q )) ⊗ F ′ ] n − (cid:1) + ℓ A ( S n − ) . Now using the same argument as done in Lemma 14, we get the desired equality.Finally, since F ′ ∼ = ( A/I )[ X , . . . , X d ] by Lemma 5 and ( I /I + Q ) ⊗ F ′ is a maximal Cohen-Macaulay module, we get the injective mapsH iM ( G ( I )) ֒ → H iM ( R / R + (1) + T ) ֒ → H iM ( S )( − T -modules for 0 ≤ i ≤ d − iM ( G ( I )) ∼ = H i M ( R / ( R + (1) + T )) ∼ = H i M ( S )( −
1) for0 ≤ i ≤ d − (cid:3) Remark 16.
Suppose e ( I ) − e ( Q ) = 2( e ( Q ) − ℓ ( A/Q )) for an I -admissible filtration I and Q ⊆ I ⊆ I a reduction. Then (1) e ( I ) = e ( Q ) , (2) A is Cohen-Macaulay and (3) I = { I n } . Proof.
Since e ( I ) ≥ e ( Q ) by (6) and e ( Q ) ≤ ℓ ( A/Q ). Therefore, 0 ≤ e ( I ) − e ( Q ) = 2( e ( Q ) − ℓ ( A/Q )) ≤ ⇒ e ( I ) = e ( Q ) and e ( Q ) = ℓ ( A/Q ). Thus A is Cohen-Macaulay, see [2, Corollary4.7.11] and e ( I ) = e ( Q ) = 0 . This gives that I = { I n } by [13, Theorem 3.21]. (cid:3) Corollary 17.
Suppose that the conditions ( C ) , ( C ) and ( C ) are satisfied and the equality e ( I ) − e ( I ) + e ( Q ) = 2 ℓ ( A/I ) + ℓ ( I /I + Q ) holds. Then (1) I n +2 ⊆ Q n I for all n ≥ , (2) Q n ∩ I n +1 = Q n I for all n ≥ , (3) depth G ( I ) > and (4) ( a , a , . . . , ˇ a i , . . . , a d ) : a i ⊆ I + Q for all ≤ i ≤ d. Proof.
Suppose I = Q . Then by Remark 16, A is Cohen-Macaulay and I = { I n = Q n } . Then theconclusion holds. Suppose Q I . (1) By Proposition 15, K (3) = 0 which gives I n +2 ⊆ Q n I for all n ≥ K (2) = 0, so [ K (2) ] n = Q n ∩ I n +1 /Q n I = 0 for all n ≥ . (3) Using the two exact sequences of Proposition 15(2) and depth lemma, it is enough to see that( I /I + Q ) ⊗ F ′ ∼ = ( I /I + Q ) ⊗ A/I [ X , . . . , X d ] and S have positive depths as T -modules.(4) Let x ∈ ( a , a , . . . , ˇ a i , . . . , a d ) : a i . Then xa i = P ≤ j ≤ d,j = i a j x j , x j ∈ A . Here, x, x j ∈ I due tothe condition ( C ). Let g = x ⊗ a i t − X ≤ j ≤ d,j = i x j ⊗ a j t ∈ ( I / ( I + Q )) ⊗ F ′ ILBERT COEFFICIENTS AND BUCHSBAUMNESS OF THE ASSOCIATED GRADED RING OF FILTRATION 11 where x, x j denote the images of x, x j in I / ( I + Q ) and a i t denote the image of a i t in F ′ . Theimage of g under the map 0 → (( I / ( I + Q )) ⊗ F ′ )( − → R / ( R + (1) + T ) is xa i t − X ≤ j ≤ d,j = i x j a j t = 0which implies that g = 0 ∈ ( I / ( I + Q )) ⊗ F ′ ∼ = ( I / ( I + Q )) ⊗ ( A/I )[ a t, . . . , a d t ]. Therefore, x ∈ I + Q. (cid:3) Proof of the main theorem: Part 1
We now prove our main result. First, we establish the equivalence of (1) and 2 of Theorem 1 in thissection. Suppose I n +2 ⊆ Q n I + W for all n ≥
1; ( Q n + W ) ∩ ( I n +1 + W ) = Q n I + W for all n ≥ a , . . . , ˇ a i , . . . , a d ) : A a i ⊆ I + Q for all 1 ≤ i ≤ d ; Claim 1:
The equality e ( I ) − e ( Q ) = 2 e ( I ) − ℓ ( A/I ) − ℓ ( I / ( I + Q )) holds. Proof of Claim 1.
Suppose Q = I . Then I + W = ( I + W ) ∩ ( Q + W ) = QI + W = Q + W and I n +2 ⊆ Q n +2 + W for n ≥ . This implies that I C = { Q n C } . Since W ⊆ ( a , . . . , ˇ a i , . . . , a d ) : a i ⊆ I + Q , it is enough to get the equality e ( I C ) − e ( QC ) = 0 = 2( e ( I C ) − ℓ ( C/QC ))in C using Lemma 12. By [22][Section 4], we have that e ( I ) = e ( Q ) = ℓ ( C/QC ) − ℓ (( a , . . . , a r − ) : a r / (( a , . . . , a r − ) : a r ) ∩ Q )) = ℓ ( C/QC ) as ( a , . . . , a r − ) : a r ⊆ I + Q = Q. Suppose Q I . Since A satisfies ( C ), Q is a standard parameter ideal of A . From [23, Corollary3.9], Q ∩ W = 0. Therefore, W ⊆ ( a , . . . , ˇ a i , . . . , a d ) : a i ⊆ I + Q . We show the required equality inthe ring C and use Lemma 12. By Proposition 15, it is enough to show that K ( i ) = 0 for 1 ≤ i ≤ C , we may assume that the conditions ( C ) , ( C ) and ( C ) are satisfied usingLemma 12 and I n +1 = Q n − I for n ≥ Q n ∩ I n +1 = Q n I for all n ≥
1. It follows easily that K (3) = ⊕ n ≥ ( I n +2 + W ) + ( I Q n + W )( I Q n + W ) = 0 and K (2) = ⊕ n ≥ ( Q n + W ) ∩ ( I n +1 + W ) I Q n + W = 0 . Now we show that K (1) = 0 . From (9), that we have the exact sequence(13) 0 → K (1) → ([ R / ( R + (1) + T )] ⊗ F ′ )( − ψ −−→ Im φ → K (1) = ker ψ . Suppose K (1) = 0. Let n = min { m | [ K (1) ] m = 0 } . Since (cid:2) [ R / ( R + (1) + T )] ⊗ F ′ (cid:3) m − = 0 for all m ≤
0, we have [ K (1) ] m = 0 for m ≤ . For m = 1, noticethat (cid:2) [ R / ( R + (1) + T )] ⊗ F ′ (cid:3) ∼ = I / ( I + Q ) ⊗ A/I ∼ = I / ( I + Q )as I I ⊆ I . Let g = P ≤ i ≤ p z i ⊗ r i ∈ [ K (1) ] where z i denote the image of z i ∈ I in I / ( I + Q )and r i denote the image of r i in A/I . Then ψ ( g ) = P ≤ i ≤ p z i r i t = 0 where z i r i t denote the image inIm φ ⊆ [ R / ( R + (1) + T )] = I / ( I + Q ). which implies P ≤ i ≤ p z i r i ∈ I + Q i.e. g = P ≤ i ≤ p z i ⊗ r i = P ≤ i ≤ p z i r i ⊗ ¯1 = 0 in (cid:2) [ R / ( R + (1) + T )] ⊗ F ′ (cid:3) . This gives that [ K (1) ] = 0. Therefore, n ≥ . Let 0 = g ∈ [ K (1) ] n = [ker ψ ] n . PutΓ = { ( α , . . . , α d ) ∈ Z d | α ≥ , α i ≥ ≤ i ≤ d and d X i =1 α i = n − } Γ = { (0 , β , . . . , β d ) ∈ Z d | β i ≥ ≤ i ≤ d and d X i =1 β i = n − } . Then Γ S Γ = { ( α , . . . , α d ) ∈ Z d | α i ≥ ≤ i ≤ d and P di =1 α i = n − } . We may write g = X α ∈ Γ S Γ x α ⊗ ( a t ) α ( a t ) α . . . ( a d t ) α d = X α ∈ Γ x α ⊗ ( a t ) α ( a t ) α . . . ( a d t ) α d + X β ∈ Γ x β ⊗ ( a t ) β . . . ( a d t ) β d where x α and x β denote the images of x α , x β ∈ I in I / ( I + Q ), and ( a t ) α ( a t ) α . . . ( a d t ) α d and( a t ) β . . . ( a d t ) β d denote the images of corresponding elements in F ′ n − . Then ψ ( g ) = X α ∈ Γ x α a α a α . . . a α d d t n + X β ∈ Γ x β a β . . . a β d d t n = 0 ∈ [ R / R + (1) + T ] n (14) ⇒ X α ∈ Γ x α a α a α . . . a α d d + X β ∈ Γ x β a β . . . a β d d ∈ I n +1 + Q n . (15)We have for all n ≥ I n +1 + Q n = Q n − I + Q n ⊆ a Q n − I + ( a , . . . , a d ) n − I + a Q n − +( a , . . . , a d ) n ⊆ a Q n − I + a Q n − + ( a , . . . , a d ) n − and P β ∈ Γ x β a β . . . a β d d ∈ ( a , . . . , a d ) n − . So,there exists q ∈ Q n − I and q ′ ∈ Q n − such that P α ∈ Γ x α a α a α . . . a α d d + a q + a q ′ ∈ ( a , . . . , a d ) n − ⇒ a . { P α ∈ Γ x α a α − a α . . . a α d d + q + q ′ } ∈ ( a , . . . , a d ) n − ⇒ P α ∈ Γ x α a α − a α . . . a α d d + q + q ′ ∈ ( a , . . . , a d ) n − ( I + Q ) ⊆ I n + Q n − since ( a , . . . , a d ) n − : a ⊆ ( a , . . . , a d ) n − . (( a , . . . , a d ) : a ) by [6, Lemma 3.5] and ( a , . . . , a d ) : a ⊆ I + Q is given. Hence, we get X α ∈ Γ x α a α − a α . . . a α d d ∈ I n + Q n − ⇒ X α ∈ Γ x α ⊗ ( a t ) α − ( a t ) α . . . ( a d t ) α d ∈ [ K (1) ] n − = 0which implies that g = X β ∈ Γ x β ⊗ ( a t ) β . . . ( a d t ) β d . Therefore g = 0 since, by symmetry of the elements a i t , the similar steps can be repeated for any a i ,1 ≤ i ≤ d. This is a contradiction. Hence K (1) = 0 . This completes the proof. (cid:3)
Next, we prove (1) ⇒ (2). Suppose the equality e ( I ) − e ( Q ) = 2 e ( I ) − ℓ A ( A/I ) − ℓ A ( I / ( I + Q ))holds. Then e ( I C ) − e ( QC ) = 2 e ( I C ) − ℓ A ( C/I C ) − ℓ A ( I C/ ( I C + QC )) and W ⊆ I + Q byLemma 12. Since ( C ), ( C ) and ( C ) are satisfied for C , by Corollary 17 we get ( Q n + W ) ∩ ( I n +1 + W ) = Q n I + W for n ≥ a , . . . , ˇ a i , . . . , a d ) : A a i ⊆ I + Q for all 1 ≤ i ≤ d ;. It remains to show that I n +2 ⊆ Q n I + W for n ≥ W ⊆ I . Theorem 18.
Suppose that the conditions ( C ) and ( C ) are satisfied and the equality e ( I ) − e ( Q ) =2 e ( I ) − ℓ A ( A/I ) − ℓ A ( I / ( I + Q )) holds. Then I n +2 ⊆ Q n I + W for n ≥ and W ⊆ I . ILBERT COEFFICIENTS AND BUCHSBAUMNESS OF THE ASSOCIATED GRADED RING OF FILTRATION 13
Proof.
We first show that W ⊆ I . Let w = i + q ∈ W ⊆ I + Q where i ∈ I and q ∈ Q . Then q = w − i ∈ ( Q + W ) ∩ ( I + W ) = QI + W by Corollary 17. So, q ∈ ( QI + W ) ∩ Q = QI since Q ∩ W = 0 which implies w = i + q ∈ I . Hence W ⊆ I .For the rest of the proof, we may pass to the ring C and assume that the conditions ( C ), ( C ) and( C ) are satisfied. We show, by induction on d , that I n +2 = Q n I for all n ≥
1. Suppose the equality e ( I ) − e ( Q ) = 2 e ( I ) − ℓ A ( A/I ) − ℓ A ( I / ( I + Q )) holds. Then by Corollary 17, depth G ( I ) > I n +2 ⊆ Q n I for all n ≥ . We may assume that f = a t ∈ ( R is a regular element of G .Suppose d = 1. Since I n +2 ⊆ Q n I ⊆ ( a n ) and f n is a regular element, we get that I n +2 = I n +2 ∩ ( a n ) = a n I = Q n I for all n ≥ d ≥ . Let A ′ = A/ ( a ) , Q ′ = Q/ ( a ) , I ′ = I/ ( a ) and I ′ = I / ( a ) = { I n A ′ } be the filtration in A ′ . Then the conditions ( C ) and ( C ) are satisfied for A ′ and e ( I ′ ) − e ( Q ′ ) = 2 e ( I ′ ) − ℓ A ( A ′ /I ′ ) − ℓ A ( I ′ / ( I ′ + Q ′ )) by Lemma 12. Further, ( C ), ( C ) and ( C ) are satisfied for C ′ = A ′ / H m ( A ′ ). Let I C ′′ = { I ′ n + H m ( A ′ ) / H m ( A ′ ) } be the I ′ C ′ -admissible filtration in C ′ . By induction hypothesis, we getthat for all n ≥ I n +2 A ′ ⊆ Q n I A ′ + H m ( A ′ ) ⇒ I n +2 ⊆ Q n I + ( a ) + H m ( A ′ ) ⊆ Q n I + (( a ) : Q )since a , . . . , a d is a d-sequence and H m ( A ′ ) = (( a ) : Q ). Now let x ∈ I n +2 , write x = y + z where y ∈ Q n I and z ∈ (( a ) : Q ). Then z = x − y ∈ (( a ) : Q ) ∩ I n +2 ⊆ (( a ) : Q ) ∩ Q n I ⊆ (( a ) : Q ) ∩ Q as I n +2 ⊆ Q n I by Corollary 17. Since a , . . . , a d is a d-sequence, This implies z ∈ ( a ) ∩ I n +2 = a I n +1 as a t is a regular element for G ( I ). Thus x = y + z ∈ Q n I . This completes the proof. (cid:3) Proof of the main theorem: Part 2
In this section, we complete the proof of Theorem 1 by establishing the assertions (1)-(8).
Remark 19. (1)
When the equality e ( I ) − e ( Q ) = 2 e ( I ) − ℓ ( A/I ) − ℓ ( I / ( I + Q )) holds,then H M (ker φ ) = ker φ = H M ( G ( I )) and H i M ( G ( I )) ∼ = H i M ( G ( I C )) for i ≥ where φ is thecanonical homomorphism of graded rings G ( I ) φ −→ G ( I C ) → . (2) Let I ⊆ Q and e ( I ) − e ( Q ) = 2 e ( I ) − ℓ ( A/I ) − ℓ ( I / ( I + Q )) . Then A is Cohen-Macaulay ring and I n +1 = Q n I for n ≥ . Thus S = (0) and G ( I ) is Cohen-Macaulay with a ( G ( I )) ≤ − d. Proof. (1) It is easy to see that [ker φ ] n ∼ = ( I n ∩ W ) / ( I n +1 ∩ W ) which is (0) for n ≫
0. ThereforeH M (ker φ ) = ker φ and H i M (ker φ ) = 0 for i ≥
1. Since depth G ( I C ) > M (ker φ ) = H M ( G ( I )) and H i M ( G ( I )) ∼ = H i M ( G ( I C )) for i ≥ . (2) Since I ⊆ Q and e ( I ) − e ( Q ) ≥ e ( I ) − ℓ ( A/I ) from Proposition 4. Hence,0 = e ( I ) − e ( Q ) − e ( I ) + 2 ℓ ( A/I ) + ℓ ( I /Q )= e ( I ) − e ( Q ) − e ( I ) + ℓ ( A/I ) + ℓ ( A/Q ) ≥ ℓ ( A/Q ) − e ( I ) ≥ . Thus, ℓ A ( A/Q ) = e ( Q ) which implies that A is Cohen-Macaulay, see [2, Section 4.7] and e ( I ) − e ( Q ) = e ( I ) − ℓ ( A/I ). Using Corollary 7, S = (0) as dim S = d . Recall that if A is Cohen-Macaulay, then either S = (0) or dim S = d , see [24, Proposition 2.1]. Consequently, I n +1 = Q n I for all n ≥ G ( I ) is Cohen-Macaulay by Vallabrega-Valla criteria. Since G ( I ) is Cohen-Macaulay, we have a d ( G ( I ) + d ≤ (cid:3) Theorem 20.
Suppose that the conditions ( C ) and ( C ) are satisfied and e ( I ) − e ( Q ) = 2 e ( I ) − ℓ A ( A/I ) − ℓ A ( I / ( I + Q )) holds. Then G ( I ) is generalized Cohen-Macaulay and S is a generalizedCohen-Macaulay T -module.Proof. By Remark 19, we may assume that I * Q , S = 0 andH i M ( G ( I )) ∼ = H i M ( G ( I C ))for i ≥
1. Thus we may pass to the ring C and assume that ( C ) , ( C ) and ( C ) are satisfied. ByProposition 15, we have thatH i M ( G ( I )) ֒ → H i M ( R / ( R + (1) + T )) ֒ → H iM ( S )( − ≤ i ≤ d −
1. Then g − depth M S ≤ g − depth M G ( I ). Therefore g − depth M S = g − depth M G ( I ) = d by Proposition 11. (cid:3) Proposition 21.
Suppose that the conditions ( C ) and ( C ) are satisfied and e ( I ) − e ( Q ) = 2 e ( I ) − ℓ ( A/I ) − ℓ ( I / ( I + Q )) . Then the following hold true. (1) For all n ∈ Z , [H M ( G ( I ))] n ∼ = W/I ∩ W if n = 2( I n ∩ W ) / ( I n +1 ∩ W ) if n ≥ otherwise (2) We have H i M ( G ( I )) = [H i M ( G ( I ))] − i ∼ = H i m ( A ) for ≤ i ≤ d − . (3) We have a ( G ( I )) ≤ − d. Proof.
In view of Remark 19, we may assume that I * Q , S = (0) and [H M ( G ( I ))] n = ( I n ∩ W ) / ( I n +1 ∩ W ) = 0 for n = 0 , W ⊆ I from Theorem 18. Further, [H M ( G ( I ))] = W/I ∩ W and[H M ( G ( I ))] n = I n ∩ W/I n +1 ∩ W for n ≥
3. This proves (1).Now by passing to the ring C , we may assume that ( C ) , ( C ) and ( C ) are satisfied, depth G ( I ) > f = a t is a non-zero-divisor on G ( I ). The exact sequence 0 → G ( I )( − f −→ G ( I ) → G ( I ) /f G ( I ) → → G ( I ) /f G ( I ) → H M ( G ( I ))( − f −→ H M ( G ( I )) → H M ( G ( I ) /f G ( I )) → . . .. . . → H i − M ( G ( I ) /f G ( I )) → H i M ( G ( I ))( − f −→ H i M ( G ( I )) → H i M ( G ( I ) /f G ( I )) → . . .. . . H d − M ( G ( I ) /f G ( I )) → H d M ( G ( I ))( − f −→ H d M ( G ( I )) → . (16)We apply induction on d to prove the assertions (2) and (3). Suppose d = 1. Then I = a I byTheorem 18. We have G ( I ) /f G ( I ) = A/I ⊕ I / ( I + Q ) ⊕ I /QI and [H M ( G ( I ))] n = 0 for all n ≫ a ( G ( I )) ≤ d ≥ d −
1. Let A ′ = A/ ( a ) , I ′ = I/ ( a ), Q ′ = Q/ ( a ) and I ′ = I /a . Note that G ( I ′ ) ∼ = G ( I ) /f G ( I ) . Then by Lemma 12, e ( I ′ ) − e ( Q ′ ) = 2 e ( I ′ ) − ℓ A ( A ′ /I ′ ) − ℓ A ( I ′ / ( I ′ + Q ′ ))and the conditions ( C ) and ( C ) hold for A ′ . By induction hypothesis, we have that a d − ( G ( I ′ )) ≤ − ( d −
1) = 3 − d and for all 1 ≤ i ≤ d − , (17) H i M ( G ( I ′ )) = [H i M ( G ( I ′ ))] − i ∼ = H i m ( A ′ ) . ILBERT COEFFICIENTS AND BUCHSBAUMNESS OF THE ASSOCIATED GRADED RING OF FILTRATION 15
First let us observe that (16) gives the exact sequence[H d − M G ( I ′ )] a ( G ( I ))+1 → [H d M ( G ( I )( − a ( G ( I ))+1 → d − M G ( I ′ )] a ( G ( I ))+1 = 0. Therefore a ( G ( I )) + 1 ≤ a ( G ( I ′ )) ≤ − d i.e. a ( G ( I )) ≤ − d. Now the condition ( C ) implies that Q = ( a , . . . , a d ) is a standard system of parameters of A . Then( a , . . . , a d ) is a standard system of parameter of A ′ . Since I n +2 = Q n I for all n ≥ I ′ n ∩ H m ( A ′ ) ⊆ Q ′ ∩ H m ( A ′ ) = 0 for all n ≥ M ( G ( I ′ )) = [H M ( G ( I ′ ))] = H m ( A ′ ) . Since H i − M ( G ( I ′ )) = [H i − M ( G ( I ′ ))] − i for all 1 ≤ i ≤ d − → [H i M ( G ( I ))] n − → [H i M ( G ( I ))] n for all n ≥ − i and 1 ≤ i ≤ d −
1. This gives [H i M ( G ( I ))] n = 0 for all n ≥ − i and 1 ≤ i ≤ d − . From (16) and (17), we also get[H i M ( G ( I ))] n − → [H i M ( G ( I ))] n → n ≤ − i and 1 ≤ i ≤ d −
2. By Theorem 20, G ( I ) is generalized Cohen-Macaulay which impliesthat [H i M ( G ( I ))] n = 0 for n ≪ ≤ i ≤ d −
2. Therefore we have that [H i M ( G ( I ))] n = (0) for all1 ≤ i ≤ d − n ≤ − i . Thus H i M ( G ( I )) = [H i M ( G ( I ))] − i for all 1 ≤ i ≤ d −
2. Next, we show that [H d − M ( G ( I ))] n = (0) for all n ≤ − d . Considering themonomorphisms 0 → [H d − M ( G ( I ))] n − → [H d − M ( G ( I ))] n for all n ≤ − d , it is enough to show that [H d − M ( G ( I ))] − d = (0) . By Proposition 15, we have theinjections(19) [H d − M ( G ( I ))] − d ֒ → [H d − M ( R / R + (1) + T )] − d ֒ → [H d − M ( S )] − d . We show that [H d − M ( S )] − d = (0). Claim 2 :
For 1 ≤ i ≤ d − , we have [H i M ( R + (1))] n = (0) for all n ≤ − i and n < Proof of Claim 2.
The exact sequence 0 → R + (1) → R → G ( I ) → T -modules induces theexact sequence H i − M ( G ( I )) → H i M ( R + (1)) → H i M ( R )of local cohomology modules. Since H i − M ( G ( I )) = [H i − M ( G ( I ))] − i , we have the injective maps[H i M ( R + (1))] n ֒ → [H i M ( R )] n for n = 3 − i. On the other hand, the exact sequence 0 → R + ( − → R → A → i M ( R )] n ∼ = [H i M ( R + )] n − for all n <
0. Thus,(20) [H i M ( R + (1))] n ֒ → [H i M ( R )] n ∼ = [H i M ( R + )] n − for all n ≤ − i and n < . Since G ( I ) is generalized Cohen-Macaulay by Theorem 20, R is generalizedCohen-Macaulay by [21, Proposition 3.5] which implies that H i M ( R ) is finitely graded for 1 ≤ i ≤ d − i M ( R + (1))] n = 0 for all n ≤ − i and n < (cid:3) Now consider the exact sequence(21) 0 → I T → R + (1) → S → d − M ( R + (1))] − d → [H d − M ( S )] − d → [H d M ( I T )] − d . Since [H d M ( I T )] − d = (0) by Lemma 10 and [H d − M ( R + (1))] − d = (0) by Claim 2, we get [H d − M ( S )] − d =(0) and so [H d − M ( G ( I ))] − d = 0 by (19).It remains to show that [H i M ( G ( I ))] − i ∼ = H i M ( A ) for 1 ≤ i ≤ d − . The exact sequence 0 → A a −→ A → A ′ → m ( A ′ ) ∼ = H m ( A ) since a H m ( A ) = (0) . Therefore, using (18),we get that H M ( G ( I )) = [H M ( G ( I ))] ∼ = [H M ( G ( I ′ ))] ∼ = H m ( A ′ ) ∼ = H m ( A )Suppose 2 ≤ i ≤ d − . From Proposition 21, we have[H i M ( G ( I ))] − i ֒ → [H i M ( R / ( R + (1) + T ))] − i ֒ → [H i M ( S )] − i and from (21), we have [H i M ( R + (1))] − i → [H i M ( S )] − i → [H i +1 M ( I T )] − i . Since [H i M ( R + (1))] − i = 0 by Claim 2 and [H i +1 M ( I T )] − i ∼ = H i m ( A ) by Lemma 10, we get thatH i M ( G ( I )) = [H i M ( G ( I ))] − i ֒ → H i m ( A )for 2 ≤ i ≤ d − . Then I ( A ) ≤ I ( G ( I )) = d − X i =1 (cid:18) d − i (cid:19) ℓ (H i M ( G ( I ))) ≤ d − X i =1 (cid:18) d − i (cid:19) ℓ A (H i m ( A )) = I ( A )see [23, Corollary 5.2] for the first inequality. Then H i M ( G ( I )) = [H i M G ( I ))] − i = H i m ( A ) for 1 ≤ i ≤ d − . This completes the proof. (cid:3)
As a consequence, we get the following interesting corollary on the structure of Sally module offiltration when the equality holds in Theorem 13.
Corollary 22.
Suppose that S = (0) and the conditions ( C ) , ( C ) and ( C ) are satisfied. Suppose theequality e ( I ) − e ( Q ) = 2 e ( I ) − ℓ ( A/I ) − ℓ ( I / ( I + Q )) holds. Then (1) e ( S ) = e ( I ) − ℓ ( A/I ) − ℓ ( I / ( I + Q )) , (2) e ( S ) = e ( I ) − e ( I ) + ℓ ( A/I ) , (3) e i ( S ) = e i − ( Q ) + e i ( Q ) for ≤ i ≤ d − , (4) H i M ( S ) = [H i M ( S )] − i ∼ = H i m ( A ) for ≤ i ≤ d − and (5) a ( S ) ≤ − d. Proof.
We may assume that Q I . Otherwise, by Remark 16, I = { I n } and S = (0). By Lemma 14and Proposition 15, we have ℓ ( S n − ) = ℓ ( I n /I n +1 ) − { ℓ ( A/I ) + ℓ ( I /I + Q ) } (cid:18) n + d − d − (cid:19) + ℓ ( I / ( I + Q )) (cid:18) n + d − d − (cid:19) for all n ≥ . However using the Hilbert-polynomial of S , we have ℓ ( S n − ) = d − X i =0 ( − i e i ( S ) (cid:18) n + d − − id − − i (cid:19) = e ( S ) (cid:18) n + d − d − (cid:19) + d − X i =1 ( − i ( e i ( S ) + e i − ( S )) (cid:18) n + d − − id − − i (cid:19) for all n ≫
0. On comparing the coefficients, we get e ( S ) = e ( I ) − ( ℓ ( A/I ) + ℓ ( I /I + Q )) ,e ( S ) + e ( S ) = e ( I ) − ℓ ( I /I + Q ) and e i ( S ) + e i − ( S ) = e i ( I ) for 2 ≤ i ≤ d − . Now, (1) and (2) follows easily and by Corollary 7, e i ( S ) = e i ( I ) − e i − ( S ) = { e i − ( Q ) + e ( Q ) } − e i − ( S ) = e i − ( Q ) + e i ( Q ) ILBERT COEFFICIENTS AND BUCHSBAUMNESS OF THE ASSOCIATED GRADED RING OF FILTRATION 17 for 2 ≤ i ≤ d − i M ( S )( − ∼ = H i M ( G ( I )) ∼ = [H i M ( G ( I ))] − i ∼ =H i m ( A ) for 0 ≤ i ≤ d −
2. This implies thatH i M ( S ) ∼ = [H i M ( S )] − i ∼ = H i m ( A )for 0 ≤ i ≤ d − → H d − M ( R / ( R + (1) + T )) → H d − M ( S )( − → H d M ( I / ( I + Q ) ⊗ F ′ )( −
1) and(22) 0 → H d − M ( G ( I )) → H d − M ( R / ( R + (1) + T )) → H d M ( F ′ )(23)induced from the two exact sequences in Proposition 15(2). Since [H d − M G ( I )] n = (0) for n = 3 − d byProposition 21 and [H d M ( F ′ )] n = (0) for n > − d , we get [H d − M ( R / ( R + (1) + T ))] n = (0) for n > − d .Further, since [H d M (( I / ( I + Q )) ⊗ F ′ )( − n = (0) for n > − d , we get [ h d − M ( S )] n = (0) for all n ≥ − d . Now consider the exact sequence 0 → I T → R + (1) → S → T -modules and theinduced exact sequence of local cohomology modulesH d − M ( R + (1)) → H d − M ( S ) → H d M ( I T ) . Since [H d M ( I T )] n = (0) for n ≤ − d by Lemma 10 and [H d − M ( R + (1))] n = (0) for n ≤ − d < d − M ( S )] n = (0) for n ≤ − d and H d − M ( S ) = [H d − M ( S ))] − d ∼ = [H d − M ( G I ))] − d ∼ = H d − m ( A )using the exact sequences (22) and (23) along with Proposition 21(2). This completes the proof of (4) . For the last part, we consider the two exact sequences of Proposition 21(2). Then H d M ( R / ( R + (1) + T ))] n = 0 for n > − d as a ( G ( I )) ≤ − d . This implies that a ( S ) ≤ − d. (cid:3) We now complete the proof of second part of our main result. We extend the method of [16] forparts (5) and (6). For the last part, we recall some results from [19].
Proof of Theorem 1.
The assertions (1), (2), (3) and (4) follow from Proposition 21. We now proveassertions (5) and (6). If I ⊆ Q , then A is Cohen-Macaulay and I n +1 = Q n I for n ≥ e ( Q ) = 0 and e ( I ) = 2 e ( I ) − ℓ A ( A/I ) − ℓ A ( I /Q ) = e ( I ) − ℓ A ( A/I ) whichimplies e i ( I ) = 0 for 2 ≤ i ≤ d , see [7, Corollary 2.4]. Since e i ( Q ) = 0 for 1 ≤ i ≤ d , the assertions (5)and (6) hold.Suppose I * Q. Then e ( I C ) − e ( QC ) = 2 e ( I C ) − ℓ ( C/I C ) − ℓ ( I C/I C + QC ) by Lemma 2.We also have that e i ( I C ) = e i ( I ), e i ( QC ) = e i ( Q ) for 1 ≤ i ≤ d − e d ( I ) = e d ( I C ) + ( − d ℓ A ( W ), e d ( Q ) = e d ( QC ) + ( − d ℓ A ( W ) and ℓ A ( A/I ) = ℓ A ( C/I C ) as W ⊆ I by Theorem 18. Therefore, wemay pass to the ring C and assume that ( C ), ( C ) and ( C ) are satisfied.By Corollary 7 and Corollary 22, e ( I ) = e ( Q ) + e ( Q ) + e ( S ) = e ( Q ) + e ( Q ) + e ( I ) − e ( I ) + ℓ A ( A/I ) and e i ( I ) = e i − ( Q ) + e i ( Q ) + e i − ( S ) = e i − ( Q ) + 2 e i − ( Q ) + e i ( Q )for 3 ≤ i ≤ d. To prove part (7) (part (8) respectively) of Theorem 1, suppose A is quasi-Buchsbaum (Buchs-baum respectively). Then G ( I ) is quasi-Buchsbaum (Buchsbaum respectively) by [19][Theorem 1.1]([19][Theorem 1.2] respectively). This completes the proof of Theorem 1. (cid:3) References [1] M. Brodmann,
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Department of Mathematics, Indian Institute of Technology Patna, Bihta, Patna 801106, India
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