Advances in Mathematics 152, 7277 (2000) doi:10.1006aima.1998.1888, available online at http:www.idealibrary.com on
Ideals with Stable Betti Numbers Annetta Aramova and Jurgen Herzog FB6 Mathematik und Informatik, Universitat GHS Essen, Postfach 103764, 45117 Essen, Germany E-mail: [A.Aramova,juergen.herzog]uni-essen.de
and Takayuki Hibi Department of Mathematics, Graduate School of Science, Osaka University, Toyonaka, Osaka 560-0043, Japan E-mail: hibimath.sci.osaka-u.ac.jp Received November 17, 1998; accepted December 4, 1998
Componentwise linear ideals were introduced earlier to generalize the result that the StanleyReisner ideal I 2 of a simplicial complex 2 has a linear resolution if and only if its Alexander dual 2* is CohenMacaulay. It turns out that I 2 is componentwise linear if and only if 2* is sequentially CohenMacaulay. In this paper we discuss Betti number properties of componentwise linear ideals. Let I be a graded ideal of a polynomial ring S and Gin(I ) the generic initial ideal of I with respect to the reverse lexicographic term order on S. Our main result is that I and Gin(I ) admit the same graded Betti numbers if and only if I is componentwise linear. For the proof of this fact, we describe some properties of the Betti diagram of a generic initial ideal. Combinatorial implications for shifted complexes will also be discussed. 2000 Academic Press
1. GINS AND COMPONENTWISE LINEAR IDEALS Let K be a field of characteristic 0 and S=K[x 1 , ..., x n ] the polynomial ring over K with each deg x i =1. We work with the reverse lexicographic term order on S induced by x 1 > } } } >x n . For a graded ideal I of S, let Gin(I) denote the generic initial ideal of I with respect to this term order, see, e.g., [8, 10]. If I is a graded ideal of S, then we write I ( j ) for the ideal generated by all homogeneous polynomials of degree j belonging to I. Moreover, we write I d for the ideal generated by all homogeneous polynomials of I whose degree is greater than or equal to d. We say that a graded ideal I/S 72 0001-870800 35.00 Copyright 2000 by Academic Press All rights of reproduction in any form reserved.
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is componentwise linear if I ( j ) has a linear resolution for all j. We refer the reader to [11, 13] for detailed information on componentwise linear ideals. Let [ ; i, i+ j (I )] i, j denote the graded Betti numbers of a graded ideal I. It is known that ; i, i+ j (I ); i, i+ j (Gin(I)) for all i and j [10]. The main result of the present paper is the following: Theorem 1.1. A graded ideal I of S satisfies ; i, i+ j (I )=; i, i+ j (Gin(I )) for all i and j if and only if I is componentwise linear. In order to prove Theorem 1.1, the following result concerning the homological data of generic initial ideals will be required. Theorem 1.2.
Let I be a graded ideal generated in degree d. Then
(a)
If ; i, i+ j (Gin(I )){0, then ; i $, i $+ j (Gin(I )){0 for all i $
(b)
If ; 0, j (Gin(I)){0, then ; 0, j $(Gin(I )){0 for all d j $< j.
Proof. Since the generic initial ideal is strongly stable, statement (a) follows from the EliahouKervaire resolution [9] of stable ideals. Let g 1 , ..., g m be the generators of I of degree d. Suppose that ; 0, j&1(Gin(I)) =0. Then consider the ideal I j&2 . Since Gin(I j&2 )=Gin(I) j&2 , we may assume that ; 0, d+1(Gin(I))=0. We have to show that Gin(I) is generated in degree d. It follows from ; 0, d+1(Gin(I))=0 that all S-polynomials of degree d+1 reduce to zero with respect to [ g 1 , ..., g m ]. By virtue of Remark 1.3 below, this implies that [ g 1 , ..., g m ] is a Grobner basis of I, equivalently, Gin(I) is generated in degree d. In fact, since (in(g 1 ), ..., in(g m )) is a strongly stable ideal, its first syzygy module is generated in degree d+1. Remark 1.3. Recall the following fact from Grobner basis theory which is a stronger version of the well known Buchberger criterion [8, Sect. 15.4]: A set of generators G=[ g 1 , ..., g m ] of I is a Grobner basis of I if and only if each S-polynomial which comes from a minimal set of homogeneous generators of the first syzygy module of the ideal (in( g 1 ), ..., in( g m )) reduces to zero with respect to G. Before starting our proof of Theorem 1.1, we state the following: Lemma 1.4. Let I and J be graded ideals of S generated in degree d with the same graded Betti numbers. Then I d+1 and J d+1 have the same graded Betti numbers. Proof.
The exact sequence 0 Ä I d+1 Ä I Ä K(&d ) ;0, d Ä 0
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induces the long exact sequence } } } Ä Tor i+1(K, I d+1 ) (i+1)+ ( j&1) Ä Tor i+1(K, I ) (i+1)+( j&1) ;
0, d Ä Tor i+1(K, K ) (i+1)+ j&(d+1) Ä Tor i (K, I d+1 ) i+ j Ä Tor i (K, I ) i+ j
;
0, d Ä Tor i (K, K) i+ j&d Ä } } } .
It then follows that ; i, i+ j (I d+1 )=; i, i+ j (I ) for all i and for all j{d, d+1. Also, ; i, i+ j (I d+1 )=0 if jd. Now, if j=d+1, then the above long exact sequence becomes 0, d 0 Ä Tor i+1(K, I ) i+1+d Ä Tor i+1(K, K ) ;i+1
Ä Tor i (K, I d+1 ) i+d+1 Ä Tor i (K, I ) i+d+1 Ä 0. n ) ; 0, d (I)&; i+1, i+1+d (I). Hence, ; i, i+d+1(I d+1 )=; i, i+d+1(I )+( i+1 The same formulae are valid for ; i, i+ j (J ). This completes the proof.
K
We are now in the position to give a proof of Theorem 1.1. Proof of Theorem 1.1. First, suppose that I is componentwise linear. The following formula for the graded Betti numbers of a componentwise linear ideal I is known [11]: ; i, i+ j (I )=; i (I ( j ) )&; i (mI ( j&1) ). Here m is the irrelevant maximal ideal (x 1 , ..., x n ) of S. Since a strongly stable ideal is componentwise linear and since Gin(I ) is strongly stable, the same formula is valid for Gin(I ). Therefore, it suffices to prove that ; i (I ( j ) )=; i (Gin(I ) ( j ) ) and ; i (mI ( j&1) )=; i (m Gin(I ) ( j&1) ). Since I ( j ) has a linear resolution, it follows from the BayerStillman theorem [3], which guarantees that the regularity of an ideal and that of its generic initial ideal coincide, that Gin(I ( j ) )=Gin(I ) ( j ) . Since I ( j ) and Gin(I ( j ) ) have the same Hilbert function, and since the Betti numbers of a module with linear resolution are determined by its Hilbert function, the first equality follows. To prove the second one, we note that mI ( j&1) has again a linear resolution and that, by the same reason as before, m Gin(I) ( j&1) =Gin(mI ( j&1) ). Second, suppose that I and Gin(I ) have the same graded Betti numbers. Let max(I ) (resp. min(I)) denote the maximal (resp. minimal) degree of a homogeneous generator of I. To show that I is componentwise linear, we work with induction on r=max(I )&min(I ). Set d=min(I ). Let r=0. Since I and Gin(I ) have the same graded Betti numbers, it follows that Gin(I ) is generated in degree d. Since Gin(I ) is a strongly stable ideal, we have that Gin(I ) has a linear resolution, hence I has a linear resolution.
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Now, suppose that r>0. Since Gin(I d+1 )=Gin(I ) d+1 , our induction hypothesis and Lemma 1.4 imply that I d+1 is componentwise linear. Thus, it suffices to prove that I ( d ) has a linear resolution. Suppose this is not the case. Then, by the BayerStillman theorem, Gin(I ( d ) ) has regularity >d. Moreover, since Gin(I ( d ) ) is strongly stable, its regularity equals max(Gin(I ( d ) )). It follows from Theorem 1.2 that Gin(I ( d ) ) has a generator of degree d+1. Now, ; 0, d+1(I)=dim I d+1 &dim(mI ( d ) ) d+1 =dim I d+1 &dim(I ( d ) ) d+1 , and ; 0, d+1(Gin(I ))=dim Gin(I ) d+1 &dim(m Gin(I ) ( d ) ) d+1 =dim Gin(I ) d+1 &dim(m Gin(I ( d ) )) d+1 >dim Gin(I ) d+1 &dim Gin(I ( d ) ) d+1 , because (m Gin(I ( d ) )) d+1 is properly contained in Gin(I ( d ) ) d+1 . Hence ; 0, d+1(Gin(I ))>; 0, d+1(I ), a contradiction. This completes our proof.
K
The following result supports a conjecture of Buchsbaum and Eisenbud [5]. Corollary 1.5. Let I be a componentwise linear ideal of projective p ) for all i0. dimension p. Then ; i (I )( i+1 Proof. In fact, the conjecture is known to be true for monomial ideals, see, e.g., [12]. Theorem 1.1 guarantees that the Betti numbers of a componentwise linear ideal are attained by a monomial ideal. K We conclude this section by noting that Theorem 1.1 and Theorem 1.2(b) are not valid in positive characteristic. Indeed, if characteristic p>0, then I=(x p, y p ) provides a counterexample.
3. SHIFTED COMPLEXES Let, as before, S=K[x 1 , ..., x n ] be the polynomial ring over a field K. Let 2 be a simplicial complex on the vertex set [n]=[1, ..., n] and I 2 the StanleyReisner ideal of 2. Let 2* be the Alexander dual 2*=[F/[n] : [n]"F Â 2]
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of 2, and let 2 s denote the algebraic shifting (cf. [4]) of 2 introduced by G. Kalai [14]. It is shown in [11] (and [13]) that I 2 is componentwise linear if and only if 2 is sequentially CohenMacaulay. We now present the exterior algebra version of Theorem 1.1 whose proof is completely analogue. We refer the reader to [2] for information on Grobner basis theory in exterior algebras. Theorem 2.1. Let J be a graded ideal of the exterior algebra. Then J and Gin(J) have the same graded Betti numbers if and only if J is componentwise linear. Let J 2 be the StanleyReisner ideal of 2 in the exterior algebra E over K. Then the following facts are known: (a) I 2 is componentwise linear if and only if J 2 is componentwise linear [11]; i+ j&1 (b) ; i, i+ j (EJ 2 )= ik=0 ( k+ j&1 ) ; k, k+ j (SI 2 ) for all i and all j [1], in particular, the graded Betti numbers of J 2 are determined by those of I 2 , and vice versa. Corollary 2.2. Let I 2 be the StanleyReisner ideal of a simplicial complex 2. Then I 2 is componentwise linear if and only if I 2 and I 2 s have the same graded Betti numbers. Proof. Recall that J 2 s =Gin(J 2 ). Hence, Theorem 2.1 together with the above observations (a) and (b) complete the proof. K For a componentwise linear ideal I 2 , the h-triangle of 2* and the graded Betti numbers of I 2 are related as follows [11, 13]: For all j, we have : ; i, i+ j (I 2 ) t i = : h i, n& j&1(2*)(t+1) i. i0
i0
Combining Corollary 2.2 with the above formula and noting that I 2* is componentwise linear if and only if 2 is sequentially CohenMacaulay over K, it follows that the h-triangles of 2 and 2 s coincide if 2 is sequentially CohenMacaulay over K (since (2*) s =(2 s )*). This is one direction of [6, Theorem 5.1]. The combinatorial version of Theorem 1.2 is the following: Theorem 2.3. Let 2 be a pure simplicial complex of dimension d. Then 2 is CohenMacaulay if and only if 2 s has no facet of dimension d&1. Proof. Note that 2 is pure of dimension d if and only if J 2* in the exterior algebra is generated in degree n&d&1. Now, the analogue of Theorem 1.2 for the exterior algebra yields that Gin(J 2* ) is generated in
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degree n&d&1 if and only if Gin(J 2* ) has no generator of degree n&d. Since Gin(J 2* )=J (2*) s and since (2*) s =(2 s )*, we have the desired conclusion by using the fact [14] that a simplicial complex is CohenMacaulay if and only if its shifted complex is pure. K
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