Nội dung

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 and M 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 investigate 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-indexes m = (m1 , . . . , mr ), n = (n1 , . . . , nr ) ∈ Nr , we write m > n (m > n) if mi > ni (mi > ni ) for all i = 1, . . . , r. If m > n, set m − n = (m1 −n1 , . . . , mr −nr ). Furthermore, if n is a nonnegative integer, we shall use the notation nr = (n, . . . , n). L 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 finiteLnumber of elements of total degree 1. Denote Si = S(0,..., 1 ,...,0) , Si+ = Si S = ni >0 Sn (i = 1, . . . , r), and S++ = (i) L L S1+ ∩ . . . ∩ Sr+ = 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 compare 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−u Nu for all n > u. r Proof. Since N is S finitely generated, we may choose an element u ∈ N such that N is generated by v6u Nv . For n > u, we have Nn = X Sn−v Nv = Sn−u v6u X Su−v Nv ⊆ Sn−u Nu . v6u The converse inclusion is obvious. Proposition 2.2. Let S be a standard r-graded ring and let M be a finitely generated r -graded S -module. The following conditions are equivalent: p (i) S(1,...,1) ⊆ Ann(M); (ii) Mn = 0 for n >> 0. m Proof. (i)⇒(ii): Choose m ∈ N∗ such that S(1,...,1) = S(m,...,m) ⊆ Ann(M), or equivalently, Smr M = 0. By Proposition 2.1, there exists a u ∈ Nr such that Mn = Sn−u Mu for all n > u. This implies Mn = Sn−u−mr Smr Mu = 0 for all n > u+mr . (ii)⇒(i): We can choose m ∈ N such that Mn = 0 for n >mr . Then p Smr Mv = 0 r m for all v ∈ N . Thus S(1,...,1) = Smr ⊆ Ann(M), and hence, S(1,...,1) ⊆ Ann(M). p From this proposition it follows that if S(1,...,1) ⊆ Ann(M) then every homogeneous element x ∈ S is a filter-regular element of M . Therefore, to prove the existence p 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 let M be a finitely generated r -graded S -module. If m is a positive integer, then m (i) 0M : S++ = 0 M : Sm r ; m (ii) (0M : S++ )n = 0 for n >> 0. Proof. (i) As S++ = S(1,...,1) S , we have m m S++ = S(1,...,1) S = Sm r S = M Sn . n>mr This gives m 0M : S++ = \ (0M : Sn ) = 0M : Smr . n>mr (ii) By Proposition 2.1, there exists u ∈ Nr such that m m (0M : S++ )n = Sn−u (0M : S++ )u for n > u. This and (i) give m (0M : S++ )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 let M 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++ . ∞ )n = 0 for n >> 0. Therefore, Proof. It follows from Proposition 2.3 that (0M : S++ ∞ 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 Smr ax = 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 multigraded modules. Theorem 2.1. Let S be a standard r -graded ring, and let M be a finitely generated p 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 }, S since S++ = S1 · · · Sr S. 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 Sr degree. Nevertheless, we are frequently interested in filter-regular sequences in S i=1 Si . One can ask about the lengths of maximal filter-regular sequences of M in ri=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 nk11 · · · nkr 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 of M 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 ∈ S Then i be a filter-regular element of M . Assume that P (M, n) 6= 0. (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 x 0 −−−→ M/0M : 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 ki−1 ki −1 ki+1 coefficient of nk11 · · · ni−1 ni ni+1 · · · nkr r in P (M/xM, n) is exactly the coefficient of that monomial in e(M; k1 , . . . , kr ) k1 e(M; k1 , . . . , kr ) k1 ki−1 ki+1 n1 · · · nkr r − n1 · · · ni−1 (ni − 1)ki ni+1 · · · nkr r . k1 ! · · · kr ! k1 ! · · · kr ! 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 k , . . . , k r be nonnegative integers such that k1 +· · ·+kr = d (d = (1) (1) (r) (r) dim Supp++ M). Let x1 , . . . , xk1 , . . . , x1 , . . . , xkr be a filter-regular sequence of M (i) (i) such that x1 , . . . , xki ∈ Si (i = 1, . . . , r). Then e(M; k1 , . . . , kr ) > 0 if and only if 1 dim Supp++ M = 0, (1) (r) (r) where M = M/(x(1) 1 , . . . , xk1 , . . . , x1 , . . . , xkr )M. (1) (1) (r) (r) Proof. Put Q = (x1 , . . . , xk1 , . . . , x1 , . . . , xkr ). 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 dim Supp++ 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 dim Supp++ M = 0. 7 Le Van Dinh and Nguyen Tien Manh Corollary 3.1. Let q = dim Supp ++ S and let k1 , . . . , kr be nonnegative (1) (1) (r) (r) that x1 , . . . , xk1 , . . . , x1 , . . . , xkr is a that k1 + · · · + kr = q. Assume (i) (i) sequence of S with x1 , . . . , xki ∈ Si (i = 1, . . . , r). Denote (1) (1) (r) integers such filter-regular (r) Q = (x1 , . . . , xk1 , . . . , x1 , . . . , xkr ). If e(M; k1 , . . . , kr ) > 0 and the conditions ∞ ∞ 6 dim S/Q : S++ −t dim S/Si+1 + · · · + Si+t + Q : S++ 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 = Si+ 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. It is easy to see from the hypothesis that 6 dim S ∗ − t dim S ∗ /Si∗+ + · · · + Si∗+ t 1 for all 1 6 i1 < · · · < it 6 r , where Si∗+ = Si+ S ∗ (i = 1, . . . , r). Now using [1, Theo. 4.3] we obtain e(S; k1, . . . , kr ) = e(S; 0, . . . , 0) = e(S ∗ ; 0, . . . , 0) = e(S ∗ ). REFERENCES [1] M. Herrmann, E. Hyry and J. Ribbe, and Z. Tang, 1997. Reduction numbers and multiplicities of multigraded structures. J. Algebra, 197, pp. 311-341. [2] E. Hyry, 1999. The diagonal subring and the Cohen-Macaulay property of a multigraded ring. Trans. Amer. Math. Soc., 351, pp. 2213-2232. [3] J. Stuckrad and W. Vogel, 1986. Buchsbaum rings and applications. VEB Deutscher Verlag der Wisssenschaften, Berlin. [4] N.V. Trung, 2001. Positivity of Mixed multiplicities. Math. Ann., 319, N◦ . 1, pp. 33-63. 8