On filter - Regular sequences of multi - graded modules

JOURNAL OF SCIENCE OF HNUE Natural Sci., 2008, Vol. 53, N ◦ . 5, pp. 3-8 ON FILTER-REGULAR SEQUENCES OF MULTI-GRADED MODULES Le Van Dinh Hanoi National University of Education Nguyen Tien Manh Hung Vuong University Abstract. Let S be a standard r-graded ring andM be a finitely generated r-graded S-module. In this paper, we prove the existence of filter-regular sequences of M and then use them to study mixed multiplicities of M . 1. Introduction Filter-regular sequences were introduced by Stuckrad and Vogel in their book about Buchsbaum rings [3]. They rapidly became an important tool to study some classes of singular rings. Recently, filter-regular sequences have been used to investi- gate mixed multiplicities and joint reduction numbers in bigraded rings [2,4]. In this paper, we shall exploit filter-regular sequences to give a criterion for the positivity of mixed multiplicities of multi-graded modules. This paper is divided into three sections. Section 2 deals with the existence of filter-regular sequences of multi-graded modules, while Section 3 characterizes the positivity of mixed multiplicities of multi-graded modules in terms of filter-regular sequences. 2. Filter-regular sequences of multi-graded modules We begin by recalling some multi-index notation. Let r be a positive integer. For multi-indexesm = (m1, . . . , mr),n = (n1, . . . , nr) ∈ Nr, we writem > n (m > n) ifmi > ni (mi > ni) for all i = 1, . . . , r. Ifm > n, setm− n = (m1−n1, . . . , mr−nr). Furthermore, if n is a nonnegative integer, we shall use the notation nr = (n, . . . , n). Let S = ⊕ n∈Nr Sn be a standard r-graded ring over a Noetherian local ring A = S0, i.e, S is generated over A by finite number of elements of total degree 1. Denote Si = S(0,..., 1 (i) ,...,0), S + i = SiS = ⊕ ni>0 Sn (i = 1, . . . , r), and S++ = S+1 ∩ . . . ∩ S + r = ⊕ n>0 Sn. Let M = ⊕ n∈Nr Mn be a finitely generated r-graded S-module. Definition 2.1. Let S be a standard r-graded ring and let M be a finitely generated r-graded S-module. A homogeneous element x ∈ S is called a filter-regular element of M if (0M : x)n = 0 for n >> 0. 3 Le Van Dinh and Nguyen Tien Manh A sequence x1, . . . , xt of homogeneous elements in S is called a filter-regular sequence of M if xi is a filter-regular element of M/(x1, . . . , xi−1)M for i = 1, . . . , t. The concept of filter-regular sequences as defined above seems strange in com- pare with that of non-graded cases. But it will be rapidly showed that the two concepts are similar. Proposition 2.1. Let M be a finitely generated r-graded module over a standard r-graded ring S, and let N be an r-graded S-submodule of M . Then there exists some u = (u1, . . . , ur) ∈ Nr such that Nn = Sn−uNu for all n > u. Proof. Since N is finitely generated, we may choose an element u ∈ Nr such that N is generated by ⋃ v6uNv. For n > u, we have Nn = ∑ v6u Sn−vNv = Sn−u ∑ v6u Su−vNv ⊆ Sn−uNu. The converse inclusion is obvious. Proposition 2.2. Let S be a standard r-graded ring and letM be a finitely generated r-graded S-module. The following conditions are equivalent: (i) S(1,...,1) ⊆ √ Ann(M); (ii) Mn = 0 for n >> 0. Proof. (i)⇒(ii): Choose m ∈ N∗ such that Sm(1,...,1) = S(m,...,m) ⊆ Ann(M), or equiv- alently, SmrM = 0. By Proposition 2.1, there exists a u ∈ N r such that Mn = Sn−uMu for all n > u. This implies Mn = Sn−u−mrSmrMu = 0 for all n > u+mr. (ii)⇒(i): We can choosem ∈ N such thatMn = 0 for n >mr. Then SmrMv = 0 for all v ∈ Nr. Thus Sm(1,...,1) = Smr ⊆ Ann(M), and hence, S(1,...,1) ⊆ √ Ann(M). From this proposition it follows that if S(1,...,1) ⊆ √ Ann(M) then every ho- mogeneous element x ∈ S is a filter-regular element of M . Therefore, to prove the existence of filter-regular elements of M , it is enough to consider the case S(1,...,1) * √ Ann(M). 4 On filter-regular sequences of multi-graded modules Proposition 2.3. Let S be a standard r-graded ring and letM be a finitely generated r-graded S-module. If m is a positive integer, then (i) 0M : S m ++ = 0M : Smr ; (ii) (0M : S m ++)n = 0 for n >> 0. Proof. (i) As S++ = S(1,...,1)S, we have Sm++ = S m (1,...,1)S = SmrS = ⊕ n>mr Sn. This gives 0M : S m ++ = ⋂ n>mr (0M : Sn) = 0M : Smr . (ii) By Proposition 2.1, there exists u ∈ Nr such that (0M : S m ++)n = Sn−u(0M : S m ++)u for n > u. This and (i) give (0M : S m ++)n = Sn−u(0M : Smr)u ⊆ Sn−u(0M : Smr). The last module equals to zero for n > u+mr, so the claim follows. The following proposition shows that the concept of filter-regular sequences of M as defined is nothing other than that of filter-regular sequences of M with respect to S++ if we forget the grading of M. Proposition 2.4. Let S be a standard r-graded ring and letM be a finitely generated r-graded S-module. If x ∈ S is a homogeneous element, then x is a filter-regular element of M if and only if 0M : x ⊆ 0M : S ∞ ++. Proof. It follows from Proposition 2.3 that (0M : S ∞ ++)n = 0 for n >> 0. Therefore, if 0M : x ⊆ 0M : S ∞ ++, it is clear that x is a filter-regular element of M . Conversely, let a ∈ (0M : x)n for some n ∈ Nr. Then Smrax = 0 for all m ∈ N, and hence aSmr ⊆ (0M : x)n+mr = 0 for m >> 0. Now using Proposition 2.3 we obtain a ∈ 0M : S ∞ ++. We are now ready to prove the existence of filter-regular sequences of multi- graded modules. Theorem 2.1. Let S be a standard r-graded ring, and let M be a finitely generated r-graded S-module such that S(1,...,1) * √ Ann(M). If the residue field k of A = S0 is infinite, then for i = 1, . . . , r there exists x ∈ Si which is a filter-regular element of M. 5 Le Van Dinh and Nguyen Tien Manh Proof. It follows from Proposition 2.2 and Proposition 2.3 that M∗ = M/0M : S∞++ 6= 0. Thus, Ass(M ∗) is a non-empty finite set. We have Ass(M∗) = {P ∈ Ass(M) | P + S++} = {P ∈ Ass(M) | P + S1 · · ·Sr}, since S++ = S1 · · ·SrS. This implies Si * ⋃ P∈Ass(M∗) P for each i ∈ {1, . . . , r}. Because the residue field k is infinite, there is an x ∈ Si which is a nonzero divisor on M∗. By Proposition 2.4, x is a filter-regular element of M. From Proposition 2.4, it is easy to see that if x, y are filter-regular elements of M then xy is also a filter-regular element of M . Indeed, we have 0M : xy = (0M : x) : y ⊆ (0M : S ∞ ++) : y = 0M : S ∞ ++. Theorem 2.1 now implies the existence of filter-regular element of arbitrary degree. Nevertheless, we are frequently interested in filter-regular sequences in ⋃r i=1 Si. One can ask about the lengths of maximal filter-regular sequences of M in ⋃r i=1 Si. This topic will be discussed in another paper. 3. On the positivity of mixed multiplicities of multi-graded modules In this section, all the assumptions and notation of the previous section will be kept, except that the ring A = S0 will be assumed to be local Artinian. Then it was proved in [1, Theo. 4.1] that the Hilbert function of M H(M,n) = `A(Mn) is a polynomial of degree d = dim Supp++M for n >> 0, where Supp++M = {P a homogeneous prime ideal of S, P + S++ | MP 6= 0}. Denote this polynomial by P (M,n). If k1, . . . , kr are nonnegative integers, write e(M ; k1, . . . , kr)/k1! · · ·kr! for the coefficient of monomial n k1 1 · · ·n kr r in P (M,n). In the case that k1+ · · ·+kr = d, e(M ; k1, . . . , kr) is a nonnegative integer and is called mixed multiplicity ofM of type (k1, . . . , kr). This section characterizes the positivity of mixed multiplicities of M in terms of filter-regular sequences. Lemma 3.1. Let x ∈ Si be a filter-regular element ofM . Assume that P (M,n) 6= 0. Then (i) degP (M/xM,n) 6 degP (M,n)− 1; (ii) if k1 + · · ·+ kr = d then e(M ; k1, . . . , kr) = e(M/xM ; k1, . . . , ki−1, ki − 1, ki+1, . . . , kr). 6 On filter-regular sequences of multi-graded modules Proof. Since (0M : x)n = 0 for n >> 0, we have P (M/0M : x,n) = P (M,n). From the exact sequence 0 −−−→ M/0M : x x −−−→ M −−−→ M/xM −−−→ 0 we obtain P (M/xM,n) = P (M,n)− P (M/0M : x, n1, . . . , ni−1, ni − 1, ni+1, . . . , nr) = P (M,n)− P (M,n1, . . . , ni−1, ni − 1, ni+1, . . . , nr). This implies degP (M/xM,n) 6 degP (M,n)− 1. Moreover, if k1+ · · ·+ kr = d, the coefficient of nk11 · · ·n ki−1 i−1 n ki−1 i n ki+1 i+1 · · ·n kr r in P (M/xM,n) is exactly the coefficient of that monomial in e(M ; k1, . . . , kr) k1! · · · kr! nk11 · · ·n kr r − e(M ; k1, . . . , kr) k1! · · ·kr! nk11 · · ·n ki−1 i−1 (ni − 1) kin ki+1 i+1 · · ·n kr r . And that coefficient obviously is e(M ; k1, . . . , kr)/k1! · · ·ki−1!(ki − 1)!ki+1! · · · kr!. Thus, e(M/xM ; k1, . . . , ki−1, ki − 1, ki+1, . . . , kr) = e(M ; k1, . . . , kr). The lemma has so been proved. The following result gives a criterion for the positivity of the mixed multiplicity e(M ; k1, . . . , kr). Theorem 3.1. Let k1, . . . , kr be nonnegative integers such that k1+· · ·+kr = d (d = dim Supp++M). Let x (1) 1 , . . . , x (1) k1 , . . . , x (r) 1 , . . . , x (r) kr be a filter-regular sequence of M such that x (i) 1 , . . . , x (i) ki ∈ Si (i = 1, . . . , r). Then e(M ; k1, . . . , kr) > 0 if and only if dimSupp++M = 0, where M =M/(x (1) 1 , . . . , x (1) k1 , . . . , x (r) 1 , . . . , x (r) kr )M. Proof. Put Q = (x (1) 1 , . . . , x (1) k1 , . . . , x (r) 1 , . . . , x (r) kr ). Applying Lemma 3.1 successively we obtain degP (M/QM,n) 6 degP (M,n)− d = 0, e(M ; k1, . . . , kr) = e(M/QM ; 0, . . . , 0). If dimSupp++M = 0 then degP (M,n) = 0, so that e(M ; k1, . . . , kr) = e(M ; 0, . . . , 0) > 0. Conversely, if e(M ; k1, . . . , kr) = e(M ; 0, . . . , 0) > 0 then degP (M,n) > 0. Combining this with the inequality in the beginning of the proof we obtain degP (M,n) = 0, which implies dimSupp++M = 0. 7 Le Van Dinh and Nguyen Tien Manh Corollary 3.1. Let q = dim Supp++S and let k1, . . . , kr be nonnegative integers such that k1 + · · ·+ kr = q. Assume that x (1) 1 , . . . , x (1) k1 , . . . , x (r) 1 , . . . , x (r) kr is a filter-regular sequence of S with x (i) 1 , . . . , x (i) ki ∈ Si (i = 1, . . . , r). Denote Q = (x (1) 1 , . . . , x (1) k1 , . . . , x (r) 1 , . . . , x (r) kr ). If e(M ; k1, . . . , kr) > 0 and the conditions dimS/S+i1 + · · ·+ S + it +Q : S∞++ 6 dimS/Q : S ∞ ++ − t are satisfied for all 1 6 i1 < · · · < it 6 r, then e(S; k1, . . . , kr) = e(S/Q : S ∞ ++), where e(S/Q : S∞++) is the Hilbert-Samuel multiplicity of the ring S/Q : S ∞ ++. Proof. Denote S = S/Q, S + i = S + i S (i = 1, . . . , r), S++ = S++S, and S∗ = S/0 : S ∞ ++ = S/Q : S ∞ ++. It follows from the Proposition 2.3 that (0 : S ∞ ++)n = 0 for n >> 0, hence P (S,n) = P (S∗,n). Since e(S; k1, . . . , kr) > 0, we have degP (S∗,n) = degP (S,n) = 0 by Theorem 3.1. 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