AAPP Atti della Accademia Peloritana dei Pericolanti Classe di Scienze Fisiche, Matematiche e Naturali

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1 DOI: /AAPP.922A3 AAPP Atti della Accademia Peloritana dei Pericolanti Classe di Scienze Fisiche, Matematiche e Naturali ISSN Vol. 92, No. 2, A3 (2014) ON THE SYMMETRIC ALGEBRA OF SYZYGY MODULES OF MONOMIAL IDEALS GAETANA RESTUCCIA a AND PAOLA LEA STAGLIANÒ a ABSTRACT. We consider the symmetric algebra of the first syzygy module of a monomial ideal generated by an s-seuence. We introduce on that algebra an admissible term order which allows us to compute its algebraic invariants. Introduction Let R = K[x 1,...,x n ] be a polynomial ring in n indeterminates over a field K and let M be a finitely generated R-module. The symmetric algebra Sym R (M) is a very important algebra from many points of view. In particular, if M is a linear type ideal I in the sense of Valla (1979), then Sym R (I) = Rees(I) and it is evident its importance in algebraic geometry. The main goal of this paper is to compute the invariants of the symmetric algebra of the first syzygy module of M, Syz 1 (M), where M is a monomial ideal I of R. We look to the computation of invariants of Sym(Syz 1 (I)) via the theory of s-seuences, that has been introduced by Herzog, Restuccia, and Tang (2001) and having a useful role in this direction (La Barbiera and Restuccia 2011; Restuccia 2006; Restuccia, Utano, and Tang 2014). If the R-module Syz 1 (I) is generated by an s-seuence, we give formulas for dimension, depth (with respect to the maximal irrelevant ideal of R), multiplicity and Castelnuovo-Mumford regularity of Sym R (Syz 1 (I)) only in terms of the same invariants of ideals of R. More precisely, in Section 1 we give two sufficient conditions for Syz 1 (I) to be generated by a s-seuence, I a monomial ideal of R. In Section 2, we investigate the module Syz 1 (m), m = (x 1,...,x n ) the maximal irrelevant ideal of R. We prove that Syz 1 (m) is generated by a s-seuence if and only if n = 3. The complete study of the invariants of Sym R (Syz 1 (m)), for n > 3 is given by Restuccia, Utano, and Tang (2014). Also Sym R (Syz n 1 (m)) and Sym R (Syz n (m)) are studied, finding that Syz n 1 (m) and Syz n (m) are generated by an s- seuence, hence we proceed to calculate their invariants.

2 A3-2 G. RESTUCCIA AND P. L. STAGLIANÒ 1. Preliminaries Let R be a Noetherian graded ring and let M a finitely generated R-module. We give a short introduction of the notion of s seuence generating a finitely generated R-module M and we will conclude with the main theorem concerning the computation of the invariants of Sym(M): the Krull dimension of M, dim(m), the multiplicity of M, e(m), the Castelnovo- Mumford regularity of M, reg(m), and the depth of M with respect to the maximal irrelevant ideal of R, depth(m). Let M be a finitely generated module on a Noetherian ring R, with generators f 1, f 2,..., f n. We denote by (a i j ) i=1,...,m the relation matrix, by Sym i (M) the ith symmetric power, and by j=1,...,n Sym R (M) = i 0 Sym(M) i the symmetric algebra of M. Note that where Sym R (M) = R[Y 1,...,Y n ]/J, J = (g 1,...,g m ), and g i = n j=1 a i j Y j. We consider S = R[Y 1,...,Y n ] a graded ring by assigning to each variable Y i the degree 1 and to the elements of R the degree 0. Then J is a graded ideal and the natural epimorphism S Sym R (M) is a homomorphism of graded R algebras. Let < be monomial order on the monomials in Y 1,...,Y n with the order Y 1 < Y 2 <... < Y n. We call such an order admissible. For any polynomial f R[Y 1,...,Y n ], f = α a α Y α, we put in < ( f ) = a α Y α where Y α is the largest monomial in f with a α 0, and we set in < (J) = (in < ( f ) : f J). For i = 1,...,n we set M i = 1 j=1 R f j, and let I i be the colon ideal M i 1 : f i. In other words, I i is the annihilator of the cyclic module M i /M i 1 and so M i /M i 1 = R/Ii. For convenience we also set I 0 = (0). Definition 1.1. The colon ideals I i are called the annihilator ideals of the seuence f 1,..., f n. Notice that (I 1 Y 1,I 2 Y 2,...,I n Y n ) in < (J), and that the two ideals coincide in degree 1. Definition 1.2. The generators f 1,..., f n of M are called s seuence (with respect to an admissible order <), if in < (J) = (I 1 Y 1,I 2 Y 2,...,I n Y n ) If in addition I 1 I 2... I n, then f 1,..., f n is called a strong s seuence. The invariants of the symmetric algebra of a module which is generated by an s-seuence are computed by the corresponding invariants of R. We have:

3 ON THE SYMMETRIC ALGEBRA OF SYZYGY MODULES OF MONOMIAL IDEALS A3-3 Proposition 1.1. Let M be generated by an s-seuence f 1,..., f n with annihilator ideals I 1,...,I n. Then d := dim(sym R (M)) = max e(sym R (M)) = 0 r n, 1 i 1 <...<i r n 0 r n,1 i 1 <...<i r n dim(r/(i i I ir ))=d r In particular, if f 1,..., f n is a strong s seuence, then d = max 0 r n {dim(r/i r ) + r}; e(sym R (M)) = r dim(r/i r )=d r {dim(r/(i i I ir )) + r}; e(r/(i i I ir )). e(r/i r ). Proof. (see Herzog, Restuccia, and Tang 2001, Proposition 2.4) Proposition 1.2. Let R = K[x 1,...,x m ] be a polynomial ring, and let M be a graded R- module. If M is generated by a strong s-seuence and I 1 I n are the annihilator ideals of this seuence, then reg(sym R (M)) max{reg(i i ) : i = 1,...,n}; and depht(sym R (M)) min{depth(r/i i ) + i : i = 0,1,...,n}. Proof. (see Herzog, Restuccia, and Tang 2001, Proposition 2.5) 2. First syzygy module of a monomial ideal Let I be a monomial ideal in R = K[x 1,...,x n ], generated by monomials m 1,...,m s, I = (m 1,...,m s ), and consider the beginning of the Taylor resolution (Eisenbud 1995; Taylor 1966): 2 R s ϕ 2 R s ϕ 1 R R/I 0 We denote by e 1,...,e s the canonical basis of R s. Then ϕ 1 (e i ) = m i for all i and ϕ 2 (e 1 e j ) = lcm(m i,m j ) m i e i lcm(m i,m j ) m j e j for all i < j The Taylor resolution is multigraded with deg(e i ) = deg(m i ), deg(e i e j ) = deg(lcm(m i,m j )). We choose i 1,...,i l {1,...,m} and set s(i 1,...,i l ) = l k=1 lcm(m i1,...,m il ) lcm(m ik,m ik+1 ) e i k e ik+1 (i l+1 = i 1 )

4 A3-4 G. RESTUCCIA AND P. L. STAGLIANÒ Then ϕ 2 (s(i 1,...,i l )) = 0. We call the elements s(i 1,...,i l ) cyclic syzygies. Theorem 2.1. Let I = (m 1,...,m s ) R = K[x 1,...,x n ] be a monomial ideal, let U be a subset of {(i, j) : 1 i < j s} and set F = (i, j) U Se i e j. Then Ker(ϕ 2 F) is generated by cyclic syzygies. Proof. (see Bruns and Herzog 1995, Proposition 5.1) Example 2.1. Let I = (x, y, z, t) K[x, y, z, t] be a monomial ideal. We consider the subsets of cardinality 3 of {1,2,3,4}. We have the following cyclic syzygies s(1,2,3) = ze 1 e 2 + xe 2 e 3 ye 1 e 3 s(1,2,4) = te 1 e 2 + xe 2 e 4 ye 1 e 4 s(1,3,4) = te 1 e 3 + xe 3 e 4 ze 1 e 4 s(2,3,4) = te 2 e 3 + ye 3 e 4 ze 2 e 4 If we consider the subset of cardinality 4, we obtain s(1,2,3,4) = zte 1 e 2 + xte 2 e 3 + xye 3 e 4 yze 1 e 4 = zs(1,2,3) + xs(2,3,4) In fact, for this ideal, any cyclic syzygy has length 3. Same result for I = (x 1,x 2,...,x n ) K[x 1,x 2,...,x n ] Theorem 2.2. Let I = (m 1,...,m t ) be a monomial ideal of K[x 1,x 2,...,x n ]. Let 1 i 1 < i 2 < i 3 t and put: m i1,i 2 i 3 = lcm(m i 1,m i2,m i3 ). lcm(m i2,m i3 ) Suppose that: (a) the second syzygy module of I is generated by trinomials; lcm(mi1,i (b) for i 2 j 2 or i 3 j 3, we have: gcd 2 i 3 ) lcm(m i2,m i3 ), lcm(m j 1, j 2 j 3 ) lcm(m j2,m j3 ) 1 j 1 < j 2 < j 3 t (c) for i 2 = j 2, i 3 = j 3 and i 1 j 1, we have: (1) m i2,i 1 i 3 = m i2, j 1 i 3 = m i2,i 1 j 1 m i3,i 1 i 2 = m i3, j 1 i 2 = m i3,i 1 j 1, m i1, j 1 i 3 = m i1,i 2 i 3 = m i1, j 1 i 2 (2) if = gcd(m i1,i 2 i 3,m j1, j 2 j 3 ), then /m i2,i 1 i 3 and /m i3,i 1 i 2. Then Syz 1 (I) is generated by an s-seuence. Proof. The relation ideal Z of Syz 1 (I) is generated by = 1, 1 i 1 < i 2 < i 3 t, s i1 i 2 i 3 = m i1,i 2 i 3 T i2 i 3 m i2,i 1 i 3 T i1 i 3 + m i3,i 1 i 2 T i1 i 2

5 ON THE SYMMETRIC ALGEBRA OF SYZYGY MODULES OF MONOMIAL IDEALS A3-5 with T i1 i 2 < T i1 i 3 < T i2 i 3. We consider two trinomials s i1 i 2 i 3 and s j1 j 2 j 3, then in < (s i1 i 2 i 3 ) = m i1,i 2 i 3 T i2 i 3 and in < (s j1 j 2 j 3 ) = m j1, j 2 j 3 T j2 j 3. The S polynomial is (b): m j1, j S(s i1 i 2 i 3,s j1 j 2 j 3 ) = 2 j 3 T j2 j 3 gcd(m i1,i 2 i 3 T i2 i 3,m j1, j 2 j 3 T j2 j 3 ) s m i1,i i 1 i 2 i 3 2 i 3 T i2 i 3 gcd(m i1,i 2 i 3 T i2 i 3,m j1, j 2 j 3 T j2 j 3 ) s j 1 j 2 j 3 (i) If j 2 i 2 or j 3 i 3 and j 1 i 1 then gcd(m i1,i 2 i 3 T i2 i 3,m j1, j 2 j 3 T j2 j 3 ) = gcd(m i1,i 2 i 3,m j1, j 2 j 3 ) = 1 by hypothesis. Then the S polynomial S(s i1 i 2 i 3,s j1 j 2 j 3 ) reduces to zero. (ii) If j 1 = i 1, j 2 i 2 or j 3 i 3, then s j1 j 2 j 3 = s i1 j 2 j 3 and gcd(m i1,i 2 i 3 T i2 i 3,m i1 j 2 j 3 T j2 j 3 ) = gcd(m i1,i 2 i 3,m i1, j 2 j 3 ) = 1 by hypothesis. Then the S polynomial S(s i1 i 2 i 3,s i1 j 2 j 3 ) reduces to zero. (c): If j 2 = i 2 and j 3 = i 3, i 1 j 1, we can suppose i 1 < j 1. Consider the set {i 1, j 1,i 2,i 3 }, where i 1 < j 1 < i 2 < i 3, then we have: gcd(m i1,i 2 i 3 T i2 i 3,m j1,i 2 i 3 T i2 i 3 ) = gcd(m i1,i 2 i 3,m j1,i 2 i 3 )T i2 T i3 Put gcd(m i1,i 2 i 3,m j1,i 2 i 3 ) =, then the S polynomial is S(s i1 i 2 i 3,s j1 i 2 i 3 ) = m j 1,i 2 i 3 s i1 i 2 i 3 m i 1,i 2 i 3 s j1 i 2 i 3 = = m j 1,i 2 i 3 m i1,i mi2,i 1 i 3 T i1 i 3 + m i3,i 1 i 2 T i1 i 2 2 i 3 mi2, j 1 i 3 T j1 i 3 + m i3, j 1 i 2 T j1 i 2. In the following gcd(m i1,m i2 ) = [m i1,m i2 ], for i 1 i 2. By m i2,i 1 i 3 = m i2, j 1 i 3 and by m i3,i 1 i 2 = m i3, j 1 i 2, we can write: S(s i1 i 2 i 3,s j1 i 2 i 3 ) = m i2, j 1 i 3 ( m j 1,i 2 i 3 T i1 i 3 + m i 1,i 2 i 3 T j1 i 3 ) + m i3,i 1 i 2 ( m j 1,i 2 i 3 T i1 i 2 m i 1,i 2 i 3 T j1 i 2 ) = = m i2,i 1 i 3 ( m i 3,i 1 j 1 T i1 j 1 m j 1,i 2 i 3 +m i3,i 1 i 2 ( m j 1,i 2 i 3 = m i2,i 1 i 3 ( m i 1,i 2 i 3 T i1 i 3 + m i 1,i 2 i 3 T i1 i 2 m i 1,i 2 i 3 T j1 i 2 ) = T j1 i 3 m j 1,i 2 i 3 m i3,i T j1 i 3 ) m 1 j 1 i2,i 1 i 3 T i1 j 1 + T i1 i 3 + m i 3,i 1 j 1 T i1 j 1 )+ m i 3,i 1 i 2 ( m i2,i 1 i 3 T i1 j 1 + m j1,i 2 i 3 T i1 i 2 m i1,i 2 i 3 T j1 i 2 ).

6 A3-6 G. RESTUCCIA AND P. L. STAGLIANÒ By m i3,i 1 j 1 = m i3,i 1 i 2, m i1, j 1 i 3 = m i1,i 2 i 3, m i1, j 1 i 2 = m i1,i 2 i 3, m j1,i 2 i 3 = m j1,i 1 i 2, we can write: S(s i1 i 2 i 3,s j1 i 2 i 3 ) = m i 2,i 1 i 3 (m i1, j 1 i 3 T j1 i 3 m j1,i 1 i 3 T i1 i 3 + m i3,i 1 j 1 T i1 j 1 )+ m i 3,i 1 i 2 (m i1, j 1 i 2 T j1 i 2 m j1,i 1 i 2 T i1 i 2 + m i2,i 1 j 1 T i1 j 1 ) = = m i 2,i 1 i 3 s i1 j 1 i 3 m i 3,i 1 i 2 s i1 j 1 i 2 and S(s i1 i 2 i 3,s j1 i 2 i 3 ) reduces to zero by s i1 j 1 i 3,s i1 j 1 i 2 Example 2.2. Let I k be the suare-free monomial ideal of K[x 1,...,x n ] generated by all suare-free monomials of degree k. For k = n 1, I k satisfies all conditions of Theorem 2.2, then Syz 1 (I n 1 ) is generated by an s-seuence. The following lemmas contain some eualities between monomials of K[x 1,...,x n ], in view of giving other sufficient conditions in order that the S-polynomials of the generators of the relation ideal of the symmetric algebra of the first syzygy module of I reduce to zero. Lemma 2.1. Let m 1,m 2,m 3 monomials of K[x 1,...,x n ]. Then lcm(m i1,m i2,m i3 ) lcm(m i1,m i3 ) = m i1 lcm(gcd(m i1,m i2 ),gcd(m i1,m i3 )) Proof. Put m 1 = x i 1 1 x i x i n, m 2 = x j 1 1 x j xj n n, m 3 = x k 1 1 xk xk n n Moreover = lcm(m 1,m 2,m 3 ) lcm(m 2,m 3 ) = xmax(i 1, j 1,k 1 ) 1...x max(i n, j n,k n ) n x max( j 1,k 1 ) 1...x max( j n,k n ) n = x max(i 1, j 1,k 1 ) max( j 1,k 1 ) 1...x max(i n, j n,k n ) max( j n,k n ) n m 1 lcm(gcd(m 1,m 2 ),gcd(m 1,m 3 )) = x i x i n lcm(gcd(x i 1 1,x j 1 1 ),...,gcd(xi n,x j n n ),gcd(x i 1 1,x k 1 1 ),...,gcd(xi n,xn k,n )) = x i 1 max(min(i 1, j 1 ),min(i 1,k 1 )) 1...x i n max(min(i n, j n ),min(i n,k n )) n Suppose i t < j t < k t, t, 1 t n, it results: max(i t, j t,k t ) max( j t,k t ) = k t k t = 0 = and i t max(min(i t, j t ),min(i t,k t )) = i t max(i t,k t ) = i t i t = 0. Hence the euality holds. For the other cases, it is easy to verify the assertion.

7 ON THE SYMMETRIC ALGEBRA OF SYZYGY MODULES OF MONOMIAL IDEALS A3-7 Lemma 2.2. Let I = (m 1,...,m t ) K[x 1,x 2,...,x n ] be a monomial ideal. Let 1 i 1 < i 2 < i 3 t and 1 j 1 < j 2 < j 3 t. Put: Suppose: Then m i1,i 2 i 3 = lcm(m i 1,m i2,m i3 ). lcm(m i2,m i3 ) lcm(m i1,m i2 ) = lcm(m i1,m j2 ), lcm(m i1,m i3 ) = lcm(m i1,m j3 ), lcm(m i2,m i3 ) = lcm(m j2,m j3 ). m i1,i 2 i 3 = m i1, j 2 j 3. Proof. Using the definition, we have: m i1,i 2 i 3 = lcm(m i 1,m i2,m i3 ) lcm(m i2,m i3 ) = lcm(lcm(m i 1,m j2 ),lcm(m i1,m j3 )) lcm(m j2,m j3 ) = lcm(lcm(m i 1,m i2 ),lcm(m i1,m i3 )) lcm(m i2,m i3 ) = lcm(m i 1,m j2,m j3 ) lcm(m j2,m j3 ) = = m i1, j 2 j 3 Lemma 2.3. Let I = (m 1,...,m s ) K[x 1,x 2,...,x n ] be a monomial ideal. Let 1 i < j < k s and 1 l < h < g s. Put: m i m i, jk = lcm(gcd(m i,m j ),gcd(m i,m k )). Suppose that: and Then m i, jk = m i,lh. gcd(m i,m j ) = gcd(m i,m l ) gcd(m i,m k ) = gcd(m i,m h ). Proof. Using the definition of m i, jk, we have: m i, jk = m i lcm(gcd(m i,m j ),gcd(m i,m k )) = m i lcm(gcd(m i,m l ),gcd(m i,m h )) = m i,lh Theorem 2.3. Let I = (m 1,...,m s ) K[x 1,x 2,...,x n ] be a monomial ideal. Put: m i m i, jk = lcm(gcd(m i,m j ),gcd(m i,m k )). Suppose that:

8 A3-8 G. RESTUCCIA AND P. L. STAGLIANÒ (a) the second syzygy module, Syz 2 (I), is generated by trinomials: {s i jk = m i, jk g jk m j,ik g ik + m k,i j g i j,1 i < j < k s} where g i j is the free basis of 2 R s ; (b) for j h or k g, gcd(m i, jk,m l,hg ) = 1, 1 i < j < k s, 1 l < h < g s; (c) for h = j, g = k, i l (1) gcd(m i,m j ) = gcd(m i,m l ) = gcd(m j,m l ); gcd(m k,m l ) = gcd(m i,m k ) = gcd(m j,m k ); (2) if = gcd(m i, jk,m l, jk ), then /m j,ik and /m k,i j. Then Syz 1 (I) is generated by an s-seuence Proof. It is sufficient to prove that the relation ideal of Syz 1 (I), Z = (s i jk = m i, jk T jk m j,ik T ik + m k,i j T i j,1 i < j < k s) K[x 1,...,x n ][T i j, 1 i < j < j n] has a linear Gröbner basis on the variables T i j. We fix a term order compatible with the following order on the variables: T 12 < T 13 <... < T s 1,s and x i < T i j, i, j. Then in < (s i jk ) = m i, jk T jk. Let s i jk and s lhg where 1 i < l < h < g n be two second syzygies, in < (s lhg ) = m l,hg T hg. We consider the S polynomial S(s i jk,s lhg ) = m l, jk n s i jk m i, jk n s l jk. (b) If j h or k g, i l, the variables T jk and T hg are different and gcd(m i, jk T jk,m l,hg T hg ) = gcd(m i, jk,m l,hg ) = 1 by hypothesis. Then the S polynomial S(s i jk,s lhg ) reduces to zero. If h j or g k, i = l, then s lhg = s ihg, in < (s ihg ) = m i,hg T hg, gcd(m i,hg T hg,m i, jk T jk ) = gcd(m i,hg,m i, jk ) = 1 by hypothesis and the S polynomial S = (s i jk,s ih j ) reduces to zero. (c) Let i l,h = j, g = k, then T hg = T jk and gcd(m i, jk T jk,m l, jk T jk ) = gcd(m i, jk,m l, jk )T jk Put = gcd(m i, jk,m l, jk ), where is a monomial, then the S polynomial is S(s i jk,s l jk ) = m l, jk s i jk m i, jk s l jk = = m l, jk (m i, jkt jk m j,ik T ik + m k,i j T i j ) m i, jk (m l, jkt jk m j,lk T lk + m k,l j T l j ) =

9 ON THE SYMMETRIC ALGEBRA OF SYZYGY MODULES OF MONOMIAL IDEALS A3-9 = m l, jk m i, jkt jk m l, jk m j,ikt ik + m l, jk m k,i jt i j m i, jk m l, jkt jk + m i, jk m j,lkt lk + m i, jk m k,l jt l j = m l, jk m j,ikt ik + m l, jk m k,i jt i j + m i, jk m j,lkt lk m i, jk m k,l jt l j. Consider the set {i,l, j,k}. We can suppose i < l and we have to consider only the case i < l < j < k. In the following gcd(m i,m j ) = [m i,m j ]. By [m j,m k ] = [m l,m k ], we have m k,i j = m k,l j and by [m i,m j ] = [m j,m l ], we have m j,ik = m j,lk (Lemma 2.3). By [m k,m j ] = [m l,m k ], we have m k,il = m k,i j (Lemma 2.3). Now we are exactly in the case (c) of the Theorem 2.2, by Lemma 2.1 then the assertion holds. Remark 2.1. One deduces easily that the conditions (c) in the last theorem imply the conditions (c) reuired in Theorem 2.2, but their formulation is more explicit and we do not know if there is an euivalence. Example 2.3. The seuence of monomials xy,xz,xu,xv in K[x,y,z,u,v] satisfies all the conditions of Theorem 2.2 and Theorem 2.3, as we can easy verify. If I = (xy,xz,xu,xv), Syz 2 (I) is generated by the cyclic syzygies: ye 2 e 3 ze 1 e 3 + ue 1 e 2, ye 3 e 4 ue 1 e 4 + ve 1 e 3, ye 2 e 4 ze 1 e 4 ve 1 e 2, ze 3 e 4 ue 2 e 4 + ve 2 e 3 For the 2-syzygy generated by four elements of the basis, we have: uve 1 e 2 + yve 2 e 3 + yze 3 e 4 + zue 4 e 1 = u(ye 2 e 4 ze 1 e 4 ve 1 e 2 ) y(ze 3 e 4 ue 2 e 4 + ve 2 e 3 ), then Syz 2 (I) is generated by trinomials. In the following example, the second module of syzygies is generated by an s-seuence, but the seuence does not satisfy the conditions of Theorem 2.2 and Theorem 2.3, showing the sufficiency of the conditions. Example 2.4. Let I = (x 1 x 2 x 4,x 1 x 3 x 4,x 2 x 3 x 4,x 1 x 2 x 5,x 1 x 3 x 5,x 2 x 3 x 5 ) be a monomial ideal of the polynomial ring R = K[x 1,x 2,x 3,x 4,x 5 ]. Let 0 S 2 ( 5) S 7 ( 4) 6 i=1 R ie i = S 6 ( 3) I 0 be the minimal graded free resolution. Syz 1 (I) = x 4 e 6 x 5 e 3,x 4 e 5 x 5 e 2,x 4 e 4 x 5 e 1,x 1 e 6 x 2 e 5,x 2 e 5 x 3 e 4,x 1 e 3 x 2 e 2,x 2 e 2 x 3 e 1

10 A3-10 G. RESTUCCIA AND P. L. STAGLIANÒ J = (x 4 T 6 x 5 T 3,x 4 T 5 x 5 T 2,x 4 T 4 x 5 T 1,x 1 T 6 x 2 T 5,x 2 T 5 x 3 T 4,x 1 T 3 x 2 T 2,x 2 T 2 x 3 T 1 ) Syz 2 (I) = x 1 g 7 x 2 g 6 x 4 g 4 + x 5 g 2,x 2 g 6 x 3 g 5 x 4 g 3 + x 5 g 1 Z = (x 1 Y 7 x 2 Y 6 x 4 Y 4 + x 5 Y 2,x 2 Y 6 x 3 Y 5 x 4 Y 3 + x 5 Y 1 ) J does not admit a linear Gröbner basis for any term order in R[T 1,...,T 6 ] with x i < T j, for all i, j, and T 1 < T 2 <... < T 6. In fact, if we compute the Gröbner basis G for J, we obtain: G = {T 2 x 2 T 1 x 3,T 3 x 1 T 1 x 3,T 5 x 2 T 4 x 3,T 6 x 1 T 4 x 3,T 4 x 4 T 1 x 5, T 5 x 4 T 2 x 5,T 6 x 4 T 3 x 5,T 3 T 5 x 5 T 2 T 6 x 5,T 3 T 4 x 5 T 1 T 6 x 5, T 2 T 4 x 5 T 1 T 5 x 5,T 2 T 4 x 3 T 1 T 5 x 3,T 3 T 4 x 3 T 1 T 6 x 3, T 1 T 3 T 5 x 3 + T 1 T 2 T 6 x 3 } Nevertheless, Z admits a linear Gröbner basis in the variables Y i s: G (1) = {Y 6 x 2 Y 5 x 3 Y 3 x 4 +Y 1 x 5,Y 7 x 1 Y 5 x 3 Y 3 x 4 Y 4 x 4 +Y 1 x 5 +Y 2 x 5 } so J is generated by a (not-strong) s seuence: in < (Z) = ((x 2 )Y 6,(x 1 )Y 7 ) with I 1 = I 2 = I 3 = I 4 = I 5 = (0), I 6 = (x 2 ), I 7 = (x 1 ) 3. Syzygy modules of the maximal ideal Let K be a field, R = K[x 1,x 2,...,x n ] a polynomial ring and m = (x 1,x 2,...,x n ) the maximal homogeneous ideal. We want to study the symmetric algebra of the first syzygy module of m, Syz 1 (m) and of the (n 1)-th syzygy module of m, Syz n 1 (m). Consider the minimal free resolution of K = R/(x 1,x 2,...,x n ) given by the K.(x 1,x 2,...,x n ), the Koszul complex of S with respect to the set of elements {x 1,x 2,...,x n } = {x} K(x) := 0 K n K n 1... K 2 K 1 K 0 m 0 We are going to study Syz 1 (m). By the morphism... K 2 Syz 1 we have the following presentation of the symmetric algebra Sym R (Syz 1 (m)) = Sym R (K 2 /Z) where Z is the relation ideal of Syz 1 (m). Since Syz 1 (m) = x i e j x j e i,1 i < j n and Syz 1 (m) K 1, if we put σ i j = x i e j x j e i, we

11 ON THE SYMMETRIC ALGEBRA OF SYZYGY MODULES OF MONOMIAL IDEALS A3-11 obtain Sym R (Syz 1 (m)) = R[Y i j,1 i < j n]/z where Z is the relation ideal. Y i j σ i j Z = (x i Y jk x j Y ik + x k Y i j,1 i < j < k n) Theorem 3.1. The first syzygy module, Syz 1 (m), of the maximal ideal (x 1,x 2,...,x n ) of the polynomial ring K[x 1,x 2,...,x n ] is generated by an s-seuence if and only if n = 3. Proof. For n = 3, the relation ideal Z of Sym R (Syz 1 (m)) is generated by the trinomial Z = (x 1 Y 23 x 2 Y 13 + x 3 Y 12 ) and in < (Z) = (x 1 Y 23 ) for the order on the variables Y 23 > Y 13 > Y 12, hence the assertion hold. Suppose n > 3, than the first syzygy module of the ideal m cannot be generated by an s-seuence and then the assertion holds. Let σ i jk = x i Y jk x j Y ik + x k Y i j, 1 i < j < k n, and σ ghl = x g Y hl x h Y gl + x l Y gh, 1 g < h < l n two elements of the set σ. For the admissible order on the variables Y i j, in < (σ i jk ) = x i Y jk and in < (σ ghl ) = x g Y hl and each S polynomial must to be zero by definition of s-seuence. (i) If g i, h j or l k, gcd(x i Y jk,x g Y hl ) = gcd(x i,x g ) = 1 and then since the leading terms of σ i jk and σ ghl are coprime, the S polynomial S(σ i jk,σ ghl ) reduces to zero. If g = i, h j or l k, in < (σ i jk ) = x i Y hl, in < (σ ghl ) = x i Y jk, gcd(x i Y jk,x i Y hl ) = x i S(σ i jk,σ ghl ) = Y jk (x i Y jk x j Y ik + x k Y i j ) Y hl (x i Y hl x h Y il + x l Y ih ) = = x j Y ik Y hl + x k Y i j Y hl + x h Y il Y jk x l Y ih Y jk It is easy to verify that no term of this sum divides the initial term of all the elements of the set {σ i jk,1 i < j < k n}, for no admissible term order on the monomials in the variables Y i j. Consider the monomial x j Y ik Y hl. For i = g, h j, l k, we have: i = g < j < k, i = g < h < l and in correspondence we have the monomials x i Y jk, x i Y hl, initial terms of the trinomials x i Y jk x j Y ik + x k Y i j, x i Y hl x h Y il + x l Y ih and these monomials do not divide x j Y ik Y hl. Since they not divide none of the remaing monomials, x k Y i j Y hl, x h Y il Y jk, x l Y ih Y jk, the assertion follows. Then S(σ i jk,σ gkl ) cannot be reduced by the set {σ i jk, i < j < k n}.

12 A3-12 G. RESTUCCIA AND P. L. STAGLIANÒ (ii) If i g, h = j, l = k, in < (σ i jk ) = x i Y jk, in < (σ ghl ) = x g Y jk, gcd(x i Y jk,x g Y jk ) = Y jk S(σ i jk,σ g jk ) = x gy jk (x i Y jk x j Y ik + x k Y i j ) x iy jk (x g Y jk x j Y gk + x k Y g j ) = Y jk Y jk = x g x j Y ik + x g x k Y i j + x i x j Y gk x i x k Y g j. Consider the set {i,g, j,k}. We can suppose i < g and we have to consider only the case i < g < j < k. We can write S(σ i jk,σ g jk ) = x j ( x g Y ik + x i Y gk + x k Y ig ) x k (x j Y ig x g Y i j + x i Y g j ) = = x j (x i Y gk x g Y ik + x k Y ig x k (x i Y g j x g Y i j + x j Y i j ) = x j σ igk x k σ ig j and S(σ i jk,σ g jk ) reduces to zero by {σ igk,σ ig j }. Corollary 3.1. For n > 3, the set σ = {x i Y jk x j Y ik + x k Y i j,1 i < j < k n} is not a Gröbner basis of Z, for any term order on the monomials Y i j and such that x i < Y 12 < Y 13 <... < Y n 1,n Theorem 3.2. Let I i j, 1 i < j n, be the annihilator ideals of the seuence {σ i j,1 i < j n} generating Syz 1 (m). Then I i j = (x 1,...,x i 1 ), 1 i < j n, i = 2,...,n, and I 1 j = 0, for all j n. Proof. Put I i j = (σ 12,σ 13,...,σ i, j 1 ) : (σ i j ). Then I 12 = (0) : (σ 12 ) = (0) I 13 = (0,σ 12 ) : (σ 13 ) = (0) In fact σ 12 = x 1 e 2 x 2 e 1, σ 13 = x 1 e 3 x 3 e 1. Let λ I 13 such that λσ 13 = µσ 12, λ,µ S, that is Then hence λ = µ = 0 and I 13 = (0) λ(x 1 e 3 x 3 e 1 ) = µ(x 1 e 2 x 2 e 1 ) λx 1 e 3 λx 3 e 1 = µx 1 e 2 µx 2 e 1 λx 1 = 0 µx 1 = 0 λx 3 µx 2 = 0 I 14 = (σ 12,σ 13 ) : (σ 14 )

13 ON THE SYMMETRIC ALGEBRA OF SYZYGY MODULES OF MONOMIAL IDEALS A3-13 From the seuence 1 < 2 < 3 < 4, we have the forms where σ 14 appears: f 124 = x 1 σ 24 x 2 σ 14 + x 4 σ 12, in < ( f 124 ) = x 1 σ 24 f 134 = x 1 σ 34 x 3 σ 14 + x 4 σ 13, in < ( f 134 ) = x 1 σ 34 Conseuently I 14 = 0. On the other hand, by direct calculations: Let λ I 14 such that λσ 14 = µσ 12 + νσ 13, λ,µ,ν S, Then λ(x 1 e 4 x 4 e 1 ) = µ(x 1 e 2 x 2 e 1 ) + ν(x 1 e 3 x 3 e 1 ) λx 1 e 4 λx 4 e 1 = µx 1 e 2 µx 2 e 1 + νx 1 e 3 νx 3 e 1 λx 1 = 0 νx 1 = 0 λx 4 µx 2 = 0 µx 1 νx 3 = 0 hence λ = µ = ν = 0 and I 14 = (0) Likewise for the other ideals until I 1n. The first non zero annihilator ideal is I 23. In fact, for I 23 we have: I 23 = (σ 12,σ 13,...,σ 1n ) : (σ 23 ) = (x 1 ) From the seuence 1 < 2 < 3, we have the forms in which σ 23 appears: f 123 = x 1 σ 23 x 2 σ 13 + x 3 σ 12, in < ( f 123 ) = x 1 σ 23. I 24 = (σ 12,σ 13,...,σ 1n,σ 23 ) : (σ 24 ) = (x 1 ). In fact, from the seuence 1 < 2 < 3 < 4, we have the forms in which σ 24 is the leading term: f 124 = x 1 σ 24 x 2 σ 14 + x 4 σ 12, in < ( f 124 ) = x 1 σ 24 And so on. In general, we have: I 2 j = (x 1 ) for j > 2 For the ideal I 34, from the seuence 1 < 2 < 3 < 4, we have the forms in which σ 34 is the leading term: f 134 = x 1 σ 34 x 3 σ 14 + x 4 σ 13, in < ( f 134 ) = x 1 σ 34 then I 34 = (x 1,x 2 ) and in general f 234 = x 2 σ 34 x 3 σ 24 + x 4 σ 23, in < ( f 234 ) = x 2 σ 34 I 3 j = (x 1,x 2 ) for j > 3 By induction I n 1,n = (x 1,x 2,...,x n 2 )

14 A3-14 G. RESTUCCIA AND P. L. STAGLIANÒ Therefore I i j = (x 1,...,x i 1 ), for 1 < i < j n and I 1 j = (0) j n Proposition 3.1. Let Syz 1 (m) the first syzygy module of m = (x 1,x 2,x 3 ) R = K[x 1,x 2,x 3 ]. Then we have: (i) dim(sym R (Syz 1 (m))) = 4 (ii) e(sym R (Syz 1 (m))) = 4 (iii) depth(sym R (Syz 1 (m))) = 4 (iv) reg(sym R (Syz 1 (m))) = 1 Proof. Since Syz 1 (m) is generated by the s seuence s 12, s 13, s 23, Sym R (Syz 1 (m)) = R[Y i j,1 i < j 3]/Z, being Z = (x 1 Y 23 x 2 Y 13 + x 3 Y 12 ). Then we have: in < (Z) = (x 1 T 23 ) (i) (ii) dim(sym R (Syz 1 (m)) = dim(r[y i j,1 i < j 3]/in < (Z)) = = dim(r/i 23 ) + 2 = dim(r/(x 1 )) + 2 = 4. e(sym R (Syz 1 (m)) = e(r[y i j,1 i < j 3]/in < (Z)) = e(r/i 23 ) = e(k[x 2,x 3 ]) = 4 (iii) depth(sym R (Syz 1 (m)) depth(r[y i j,1 i < j 3]/in < (Z)) = = depth(r/i 23 ) + 2 = 4 (iv) It follows the euality, since depth(a) dim(a) for any finitely generated K algebra. reg(sym R (Syz 1 (m)) = reg(r[y i j,1 i < j 3]/Z) reg(r[y i j,1 i < j 3]/in < (Z)) reg(r/i 23 ) = 1 We recall here a Groebner basis for the relation ideal of Syz 1 (m), not linear in the variables Y i j, obtained by Herzog, Tang, and Zarzuela (2003) for a suitable order on the variables and for the lexicographic order on the monomial in the variables Y i j. Put P 2 (Y ) = {Y i j Y kl Y jk Y il +Y ik Y jl,1 i < j < k < l n}, then, we have:

15 ON THE SYMMETRIC ALGEBRA OF SYZYGY MODULES OF MONOMIAL IDEALS A3-15 Theorem 3.3. The set G = {x i Y jk x j Y ik +x k Y i j,1 i < j < k n} {x r P 2 (Y ),1 r n} is a Groebner basis of Z with respect to any term order with Y n 1,n <... < Y 12 < x 1 < x 2 <... < x n. Proof. (see Herzog, Tang, and Zarzuela 2003, Lemma 3.1) Now we present a variation of the Groebner basis that appears in the previous Theorem, obtained by varying the order on the variables and the admissible order. The obtained Groebner basis is coherent with our definition of s-seuence. Theorem 3.4. The set G = {x i Y jk x j Y ik +x k Y i j,1 i < j < k n} {x r P 2 (Y ),1 r n} is a Groebner basis of Z with respect to x n >... > x 1 > Y 12 <... < Y n 1,n Proof. By symmetry, G is also a Groebner basis with respect to any term order with x n >... > x 1 > Y 12 <... < Y n 1,n. Let u in < (Z). Then u in < (I), and so u = vw where w = in(g) for some g G. If g belongs to {x i Y jk x j Y ik + x k Y i j,1 i < j < k n}, then there is nothing to prove. Otherwise g P(Y ), and w is a monomial only in the Y i j. On the other hand, u = in(h) for some h Z, and hence the leading term of h is divisible by some x i. Therefore v is divisible by some x i, and so u = v (x i w) for some monomial v. Since x i w = in(x i g), and x i g {x r P 2 (Y ),1 r n}, the conclusion follows. Remark 3.1. We note that the Groebner basis given by Theorem 3.3 is linear in the variables x 1,x 2,...,x n. This fact will be crucial for proving that the Jacobian dual of Syz 1 (m) is generated by the s seuence x 1,x 2,...,x n, where x 1,x 2,...,x n are the images of x 1,x 2,...,x n in Sym(Syz 1 (m)) Now, we study two syzygy modules of m whose symmetric algebra has a simple presentation, as shown by Restuccia, Utano, and Tang (2014). Theorem 3.5. Let Syz n 1 (m) be the (n 1) th syzygy module of m, then Syz n 1 (m) is generated by a s-seuence. Proof. By the complex K.(x), Syz n 1 (m) has the presentation 0 K n K n 1 Syz n 1 (m) 0 where Since Sym R (Syz n 1 (m)) = Sym R (K n 1 )/Z, n 1 Sym R (K n 1 ) = Sym R ( P n ) n 1 n (n 1) P n = P n = P n = Pe 1 Pe 2... Pe n Sym R (K n 1 ) = Sym R (R n ) = R[Y 1,...,Y n ],

16 A3-16 G. RESTUCCIA AND P. L. STAGLIANÒ a polynomial ring in n variables on R, being K n = n P n, the module K n is generated by the uniue generator, t = e 1 e 2... e n. Under the Koszul maps: we map t n i=1 0 K n K n 1, ( 1) i+1 x i e 1 e 2... e i... e n Concerning the generators of Syz n 1 (m), they are the image of the generators of K n 1 = n 1 P n, under the map n 1 (e 1 e 2... e i... e n ) = j=1 ( 1) j+1 x j e 1... e i e n (1) Then Syz n 1 (m) has n 1 generators given by the elements of the right-hand member of (1). So, Sym R (Syz n 1 (m)) = R[Y 1,...,Y n ]/Z and Z = (t), where t = n i=1 ( 1)i+1 x i Y i, the uniue linear form in the variables Y i and in < (t) = ( 1) n+1 x n Y n, hence in < (Z) = (( 1) n+1 x n )Y n = ((xn )Y n ) and it follows for the annihilator ideal. I n = (x n ) Proposition 3.2. Let Syz n 1 (m) the (n 1)-th syzygy module of m. Then we have: (i) dim(sym R (Syz n 1 (m))) = 2n 1 (ii) e(sym R (Syz n 1 (m))) = 0 (iii) depth(sym R (Syz n 1 (m))) = 2n 1 (iv) reg(sym R (Syz n 1 (m))) = 1 Proof. By the structure of in < (Z) = (x n Y n ) and I n = (x n ), we can compute the invariants. We have Sym R (Syz n 1 (m)) = R[Y 1,...,Y n ]/Z It is easy to study when the n th syzygy module of the ideal m, Syz n (m), is generated by s-seuences. We recall that Syz n (m) = K n = n P n = n n P n = R. Therefore for the symmetric algebra of Syz n (m), we have Sym R (Syz n (m)) = Sym R (R) = Sym R (Re), where e is the uniue free generator. Hence Sym R (R) = R[Y ], a polynomial ring in a new variable Y on R, Sym R (R) = K[x 1,x 2,...,x n,y ] a polynomial ring in n + 1 variables on K. The relation ideal of Sym R (R) is Z = (0) and then the syzygy module Syz n (m) is generated by the s seuence e. We have only one annihilator ideal I 1 = (0) = 0 : e = (0).

17 ON THE SYMMETRIC ALGEBRA OF SYZYGY MODULES OF MONOMIAL IDEALS A3-17 The result shows the claim: Proposition 3.3. Let Syz n (m) the n-th syzygy module of m. Then one has: (i) dim(sym R (Syz n (m))) = n + 1 (ii) e(sym R (Syz n (m))) = 0 (iii) depth(sym R (Syz n (m))) = n + 1 (iv) reg(sym R (Syz n (m))) = 0 Acknowledgements The authors are grateful to the referees for useful remarks and for having pointed out some statements. This work has been supported by National Group for Algebraic and Geometric Structure and their Applications (GNSAGA-INDAM). References Bruns, W. and Herzog, J. (1995). On multigraded resolutions. Mathematical Proceedings of the Cambridge Philosophical Society 118(2), DOI: /S X. Eisenbud, D. (1995). Commutative Algebra with a View Toward Algebraic Geometry. Vol Graduate Texts in Mathematics. New York: Springer. URL: algebra/book/ Herzog, J., Restuccia, G., and Tang, Z. (2001). s-seuences and symmetric algebras. Manuscripta Mathematica 104(4), DOI: /s Herzog, J., Tang, Z., and Zarzuela, S. (2003). Symmetric and Rees algebras of Koszul cycles and their Gröbner bases. Manuscripta Mathematica 112(4), DOI: /s La Barbiera, M. and Restuccia, G. (2011). Mixed product ideals generated by s-seuences. Algebra Collouium 18(4), DOI: /S Restuccia, G. (2006). Symmetric algebras of finitely generated graded modules and s-seuences. Rend. Sem. Mat. Univ. Pol. Torino 64(4), Restuccia, G., Utano, R., and Tang, Z. (2014). On the symmetric algebras of the first syzygy module of a graded maximal ideal. Submitted. Taylor, D. K. (1966). Ideals generated by monomials in an R-seuence. PhD thesis. University of Chicago, Department of Mathematics. Valla, G. (1979). On the symmetric and Rees algebras of an ideal. Manuscripta Mathematica 30(3), DOI: /BF

18 A3-18 G. RESTUCCIA AND P. L. STAGLIANÒ a Dipartimento di Matematica e Informatica Università degli Studi di Messina Via Ferdinando Stagno d Alcontres 31, Messina, Italy To whom correspondence should be addressed grest@unime.it Communicated 30 May 2013; published online 13 December 2014 This article is an open access article licensed under a Creative Commons Attribution 3.0 Unported License 2014 by the Author(s) licensee Accademia Peloritana dei Pericolanti (Messina, Italy)