FILTER-REGULAR SEQUENCES AND MIXED MULTIPLICITIES1 9 0 NguyenTien Manh and Duong QuocViet 0 2 Department of Mathematics, Hanoi Universityof Education n 136-Xuan Thuy Street, Hanoi, Vietnam a E-mail: [email protected] J 4 2 ABSTRACT: Let S be a finitely generated standard multigraded algebra over an ] Artinian local ring A; M a finitely generated multigraded S-module. This paper C answerstothequestionwhenmixedmultiplicitiesofM arepositiveandcharacterizes A them in terms of lengths of A-modules. As an application, we get interesting results . on mixed multiplicities of ideals, and recover some early results in [Te] and [TV]. h t a m 1. Introduction [ 1 v Throughout this paper, let (A,m) denote an Artinian local ring with maximal ideal 5 m, infinite residue field k = A/m; S = S (d > 0) a finitely gen- 2 n1,...,nd≥0 (n1,...,nd) 8 erated standard d-graded algebra over AL(i.e., S is generated over A by elements of 3 total degree 1); M = M a finitely generated d-graded S-module. . n1,...,nd≥0 (n1,...,nd) 1 Set L 0 a : b∞ = (a : bn); 9 n≥0 0 S△ = S S ; S△ = S ; n≥0 (n,...,n) + n>0 (n,...,n) : v S+ = Ln1+···+nd>0S(n1,...,nd)L; S++ = n1,...,nd>0S(n1,...,nd); Xi Si = LS(0,..., 1 ,...,0) (i = 1,...,d); L r i a M△ = M ; ℓ = dimM△. n≥0|{z}(n,...,n) L Denote by Proj S the set of the homogeneous prime ideals of S which do not contain S . Set ++ Supp M = P Proj S M = 0 . ++ { ∈ | P 6 } By [HHRT, Theorem 4.1] and Remark 3.1, dimSupp M = ℓ 1 and l [M ] ++ − A (n1,...,nd) is a polynomial of degree ℓ 1 for all large n ,...,n (see Remark 3.1, Section 3). 1 d − 1 Mathematics Subject Classification (2000): Primary13H15. Secondary13A02,13E05, 13E10,14C17. Keywordsandphrases:Artinianring,multiplicity,multigradedmodule,filter-regularsequence. 1 2 The terms of total degree ℓ 1 in this polynomial have the form − nk1 nkd e(M;k ,...,k ) 1 ··· d . 1 d k ! k ! k1+···X+kd=ℓ−1 1 ··· d Then e(M;k ,...,k ) are non-negative integers not all zero, called the mixed mul- 1 d tiplicity of type (k ,...,k ) of M [HHRT]. 1 d In the case that (R,n) is a Noetherian local ring with maximal ideal n, J is an n-primary ideal, I ,...,I are ideals of R. Then 1 s JnIn1 Ins T = 1 ··· s Jn+1In1 Ins n,n1M,...,ns≥0 1 ··· s is a finitely generated standard multigraded algebra over Artinian local ring R/J. Mixed multiplicities of T are called mixed multiplicities of ideals J,I ,...,I (see 1 s [Ve] or [HHRT]). The theory mixed multiplicities of n-primary ideals was introduced by Risler and Teissier in 1973 [Te], by Rees in 1984 [Re]. In past years, the positivity and the relationship between mixed multiplicities and Hilbert-Samuel multiplicity of ideals have attracted much attention (see e.g. [Sw], [Ro], [KR1, KR2], [KV], [Vi], [Tr1], [Tr2], [MV], [TV]). We turn now to the positivity of mixed multiplicities of a multigraded module. In the geometric context, this problem was introduced by Kleiman and Thorup in 1996 [KT]. In the algebraic setting, it was investigated by Trung in 2001 [Tr2] in the case of a bigraded ring. Under a totally algebraic point of view, this paper answers to the question when mixed multiplicities of multigraded modules are positive and characterizes these mixed multiplicities in terms of lengths of modules. Our approach is based on the properties of filter-regular sequences. The notion of filter-regular sequences was introduced by Stuckrad and Vogel in their book [SV]. The theory of filter-regular sequences became an important tool to study some classes of singular rings and has been continually developed (see e.g. [Tr1], [BS], [Hy],[Tr2]). As one might expect, we obtain the following result. 3 Main theorem (Theorem 3.4). Let S be a finitely generated standard d-graded algebra over an Artinian local ring A and M a finitely generated d-graded S-module such that S is not contained in √Ann M. Set ℓ = dimM△. Then the follow- (1,1,...,1) S ing statements hold. (i) e(M;k ,...,k ) = 0 if and only if there exists an S -filter-regular sequence 1 d ++ 6 x ,...,x with respect to M consisting of k elements of S ,...,k elements 1 ℓ−1 1 1 d M △ of S and dimM = 1, where M = . d (x ,...,x )M 1 ℓ−1 (ii) Suppose that e(M;k ,...,k ) = 0 and x ,...,x is an S -filter-regular se- 1 d 1 ℓ−1 ++ 6 quence with respect to M consisting of k elements of S ,...,k elements of 1 1 d M S . Set M = and r = r(S△;M△) the reduction number of S△ d (x ,...,x )M + + 1 ℓ−1 △ M with respect to M△. Then e(M;k ,...,k ) = l n for all n r. 1 d A(cid:20)(0M△ : S+△∞)n(cid:21) ≥ As interesting consequences of main result, we get results on mixed multiplicities of ideals (Proposition 4.4 and Theorem 4.5), and recover the results of Risler and Teissier [Te](see Remark 4.8), Trung and Verma [TV](see Remark 4.6). 2. On Filter-Regular Sequences Recently, filter-regular sequences have been used to investigate mixed multiplici- ties andjoint reduction numbers in bigradedrings (see e.g. [Hy], [Tr2]). This section gives some properties of filter-regular sequences in multigraded modules. Definition 2.1. Let S = S be a finitely generated standard d- n1,...,nd≥0 (n1,...,nd) graded algebra over an ArtiLnian local ring A; M = M a finitely n1,...,nd≥0 (n1,...,nd) generated d-graded S-module such that S is nLot contained in √Ann M. A (1,1,...,1) S homogeneous element x S is called an S -filter-regular element with respect to ++ ∈ M if x / P for any P Ass M and P does not contain S . That means S ++ ∈ ∈ x / P. ∈ S++*P,P∈AssSM Let x ,...,x be homogeneous elemSents in S. We call that x ,...,x is an S -filter- 1 t 1 t ++ regular sequence with respect to M if x is an S -filter-regular element with respect i ++ M to for all i = 1,...,t. (x ,...,x )M 1 i−1 4 Now, we briefly give some comments on filter-regular sequences of a finitely gen- erated multigraded module. Note: Since Ass [M/(0 : S∞ )] = P Ass M S * P , it follows that S M ++ { ∈ S | ++ } a homogeneous element x S is an S -filter-regular element if and only if x is ++ ∈ non-zero-divisor in M/(0 : S∞ ). Moreover, we have the following notes. M ++ (i) It is easy to see that a homogeneous element x S is an S -filter-regular ++ ∈ element with respect to M if and only if 0 : x 0 : S∞ . M ⊆ M ++ (ii) If S √Ann M then 0 : S∞ = M. Hence for any homogeneous ele- (1,1,...,1) ⊆ S M ++ ment x S, we always have 0 : x 0 : S∞ . This only obstructs and does ∈ M ⊆ M ++ not carry usefulness. That is why in Definition 2.1, one has to exclude the case that S √Ann M. (1,1,...,1) S ⊆ (iii) An S -filter-regular sequence x ,...,x with respect to M is a maximal S - ++ 1 t ++ filter-regular sequence if S * Ann [M/(x ,...,x )M] (1,1,...,1) S 1 t−1 p and S Ann [M/(x ,...,x )M]. (1,1,...,1) S 1 t ⊆ p Let m = (m ,...,m ),n = (n ,...,n ) Zd. We recall that m n if m n 1 d 1 d i i ∈ ≤ ≤ for all i = 1,...,d. Lemma 2.2. Let N be an d-graded S-submodule of M. Then there exist posi- tive integers u ,...,u such that N = S N for all n 1 d (n1,...,nd) (n1−u1,...,nd−ud) (u1,...,ud) 1 ≥ u ,...,n u . 1 d d ≥ Proof. Since M is a finitely generated S-module and S is a Noetherian ring, N is also a finitely generated S-module. It implies that there exist positive integers u ,...,u such that N is generated by N . Set u = (u ,...,u ), 1 d v1≤u1,...,vd≤ud (v1,...,vd) 1 d v = (v ,...,v ), n =(n ,...,n ). We hSave 1 d 1 d Nn = Sn−vNv Sn−uNu ⊇ Xv≤u for all n u. Since S is a finitely generated standard d-graded ring, SmSn = Sm+n ≥ for all m,n. It follows that Sn−vNv = Sn−u(Su−vNv) Sn−uNu for all v u. ⊆ ≤ Hence Nn Sn−uNu for all n u. So Nn = Sn−uNu for all n u. (cid:4) ⊆ ≥ ≥ 5 Remark 2.3. Since M is Noetherian, there exists a positive integer n such that 0 : S∞ = 0 : Sn . By Lemma 2.2, there exist positive integers u ,...,u such M ++ M ++ 1 d that [0 : Sn ] = S [0 : Sn ] M ++ (n1,...,nd) (n1−u1,...,nd−ud) M ++ (u1,...,ud) for all n u ,...,n u . This fact follows that [0 : Sn ] = 0 for all 1 ≥ 1 d ≥ d M ++ (n1,...,nd) n n+u ,...,n n+u . Hence (0 : S∞ ) = 0 for all large n ,...,n . 1 ≥ 1 d ≥ d M ++ (n1,...,nd) 1 d The following characteristic of filter-regular sequences will often be used. Proposition 2.4. Let x S be a homogeneous element. Then x is an S - ++ ∈ filter-regular element with respect to M if and only if (0 : x) = 0 for all M (n1,...,nd) large n ,...,n . 1 d Proof. Suppose that x is an S -filter-regular element with respect to M. Then ++ 0 : x 0 : S∞ by Note (i). By Remark 2.3, M ⊆ M ++ (0 : x) (0 : S∞ ) = 0 M (n1,...,nd) ⊆ M ++ (n1,...,nd) for all large n ,...,n . Now assume that (0 : x) = 0 for all large n ,...,n . 1 d M (n1,...,nd) 1 d It is enough to prove that (0 : x) 0 : S∞ for all v ,...,v . Indeed, if M (v1,...,vd) ⊆ M ++ 1 d a (0 : x) then xS a = 0 for every n ,...,n . Hence ∈ M (v1,...,vd) (n1,...,nd) 1 d S a (0 : x) = 0 (n1,...,nd) ⊆ M (n1+v1,...,nd+vd) for largen ,...,n . It implies that a 0 : Sn for alllarge n.Hence a 0 : S∞ . 1 d ∈ M ++ ∈ M ++ So (0 : x) 0 : S∞ for all v ,...,v . Hence x is an S -filter-regular M (v1,...,vd) ⊆ M ++ 1 d ++ element with respect to M. (cid:4) Remark 2.5. Suppose that x S is an S -filter-regular element with ∈ (a1,...,ad) ++ respect to M. Consider λ : M xM ,y xy. x (n1−a1,...,nd−ad) −→ (n1−a1,...,nd−ad) 7→ It is clear that λ is surjective and Kerλ = (0 : x) . Since x is an x x M (n1−a1,...,nd−ad) S -filter-regular element with respect to M, Kerλ = 0 for all large n ,...,n by ++ x 1 d Proposition 2.4. Therefore, xM = M and (n1−a1,...,nd−ad) ∼ (n1−a1,...,nd−ad) l [(M/xM) ] = l [M ] l [xM ] A (n1,...,nd) A (n1,...,nd) − A (n1−a1,...,nd−ad) = l [M ] l [M ] A (n1,...,nd) − A (n1−a1,...,nd−ad) for all large n ,...,n . 1 d 6 Next, we show that the existence of filter-regular sequences is universal. Proposition 2.6. Assume that S is not contained in √Ann M. Then for (1,1,...,1) S each i = 1,...,d, there exists an S -filter-regular element x S with respect to ++ i ∈ M. Proof. Set M∗ = M/(0 : S∞ ). Note that M ++ Ass M∗ = P Ass M S S = S * P , S S 1 d (1,1,...,1) { ∈ | ··· } S * P Ass M∗ for all i = 1,...,d. Since k is an infinite field and Ass M∗ is i S S ∈ finite. Hence for each i = 1,...,d, there exists x S P. On the other ∈ i \ P∈AssSM∗ hand S * √Ann M, x is an S -filter-regular eleSment with respect to M.(cid:4) (1,1,...,1) S ++ Suppose that S is not contained in √Ann M. Then S M = 0 for all (1,1,...,1) S (u,...,u) 6 u > 0. Hence for any u > 0, there exist m ,...,m such that S M = 0. 1 d (u,...,u) (m1,...,md) 6 Since S M M , M = 0. So for any u > 0, (u,...,u) (m1,...,md) ⊆ (m1+u,...,md+u) (m1+u,...,md+u) 6 there exist n ,...,n u such that M = 0. Now, assume that for each 1 d ≥ (n1,...,nd) 6 u > 0, there exist n ,...,n u such that M = 0. We will show that 1 d ≥ (n1,...,nd) 6 S * √Ann M. Indeed, if S √Ann M then S M = 0 for some (1,1,...,1) S (1,1,...,1) S (n,...,n) ⊆ n. By Lemma 2.2, there exist positive integers u ,...,u such that 1 d M = S M (n1,...,nd) (n1−u1,...,nd−ud) (u1,...,ud) for all n u ,...,n u . Hence 1 1 d d ≥ ≥ M = S S M = 0 (n1,...,nd) (n1−n−u1,...,nd−n−ud) (n,...,n) (u1,...,ud) for all n n + u ,...,n n + u . This contracdicts the hypothesis. So 1 1 d d ≥ ≥ S * √Ann M. From these facts, we get the following proposition. (1,1,...,1) S Proposition 2.7. Let S be a finitely generated standard d-graded algebra over an Artinian local ring A; M a finitely generated d-graded S-module. Then the following conditions are equivalent: (i) S is contained in √Ann M. (1,1,...,1) S (ii) M = 0 for all large n ,...,n . (n1,...,nd) 1 d LetS = S beafinitelygeneratedstandardgradedalgebraoveranArtinian n≥0 n local ring ALand J a homogeneous ideal of S is generated by elements of degree 1. Let M = M be a finitely generated graded S-module. Set S = S . n≥0 n + n>0 n We call J La reduction of S with respect to M if (JM) = M for all largLe n. The + n n least integer n such that (JM) = M is called the reduction number of S with n+1 n+1 + 7 respect to J and M [NR]. We denote this integer by r (S ;M). A reduction J of S J + + with respect to M is called a minimal reduction if it does not properly contain any other reduction of S with respect to M. The reduction number of S with respect + + to M is defined by r(S ;M) = min r (S ;M) J is a minimal reduction of S with respect to M . + J + + { | } The following lemma will determine the relationship between filter-regular se- quences and minimal reductions in graded rings and modules. Lemma 2.8 (see [Tr1]). Let S = S be a finitely generated standard graded n≥0 n algebra over an Artinian local ring LA; M = M a finitely generated graded S- n≥0 n module such that S is not contained in √ALnn M. Set ℓ = dimM. Assume that J 1 S is a minimal reduction of S with respect to M. Then J is generated by an S -filter- + + regular sequence with respect to M consisting of ℓ homogeneous elements of degree 1. Denote by r(S ;M) the reduction number of S with respect to M. The fol- + + lowing lemma makes up an important role in establishing the relationship between mixed multiplicities of multigraded modules and the length of modules. Lemma 2.9. Let S = S be a finitely generated standard graded algebra over n≥0 n an Artinian local ring AL; M = M a finitely generated graded S-module such n≥0 n that S is not contained in √AnLn M. Assume that dimM = 1. Set r = r(S ;M) 1 S + M and M∗ = . Then 0 : S∞ M + (i) l (M∗) = l (M∗) for all n r. A n A r ≥ (ii) l (M ) = l (M∗) for all large n. A n A r Proof. Since M l (M∗) = l n = l (M ) l [(0 : S∞) ] A n A(cid:20)(0 : S∞) (cid:21) A n − A M + n M + n and (0 : S∞) = 0 for all large n by Remark 2.3, l (M∗) = l (M ) for all large n. M + n A n A n The proof of (i): Since dimM = 1, by Lemma 2.8, there exists an S -filter- + regular element x S with respect to M such that xS is a minimal reduction of 1 ∈ S with respect to M and r = r (S ;M). It is clear that M∗ = xn−rM∗ for all + xS + n r n r. Therefore, ≥ l (M∗) = l (xn−rM∗) A n A r 8 for all n r. Since x is an S -filter-regular element, x is non-zero-divisor in M∗. It + ≥ follows that xn−rM∗ = M∗. Hence r ∼ r l (M∗) = l (xn−rM∗) = l (M∗) A n A r A r for all n r. ≥ The proof of (ii): Since l (M∗) = l (M ) for all large n and by (i), we immedi- A n A n ately obtain l (M ) = l (M∗) for all large n. (cid:4) A n A r 3. Mixed Multiplicities of Multigraded Modules Basing on the properties of filter-regular sequences (Section 2), this section an- swers to the question when mixed multiplicities of multigraded modules are positive and characterizes them in terms of lengths of modules. Recall that a polynomial P(n ,...,n ) is called the Hilbert polynomial of the 1 d function l [M ] if P(n ,...,n ) = l [M ] for all large n ,...,n . A (n1,...,nd) 1 d A (n1,...,nd) 1 d Remark 3.1. Set ℓ = dimM△. Denote by P(n ,...,n ) the Hilbert polynomial of 1 d l [M ]. A (n1,...,nd) Assume that ℓ > 0. By [HHRT, Theorem 4.1], degP(n ,...,n ) = dimSupp M 1 d ++ and all coefficients of monomials of highest degree in P(n ,...,n ) are non-negative 1 d integers not all zero (see [Ba]). So degP(n ,...,n ) = degP(n,...,n). Since 1 d P(n,...,n) = l [M ] = l (M△) A (n,...,n) A n for all large n, degP(n,...,n) = dimM△ 1 = ℓ 1. − − Hence degP(n ,...,n ) = dimSupp M = ℓ 1. 1 d ++ − Remark 3.2. dimM△ = ℓ > 0 if and only if S * √Ann M. Indeed, (1,1,...,1) S dimM△ = 0 is equivalent to [M△] = 0 for all large n. The latter means the same n as M = 0 for all large n, but it is easily seen to be equivalent to M = 0 (n,...,n) (n1,...,nd) for all large n ,...,n . Hence dimM△ = 0 if and only if S √Ann M by 1 d (1,1,...,1) S ⊆ Proposition 2.7. Denote by L(M) the set of the lengths of maximal S -filter-regular sequences ++ d in S with respect to M. Then we have the following proposition. j=1 j S 9 Proposition 3.3. Let S be a finitely generated standard d-graded algebra over an Artinian local ring A and M a finitely generated d-graded S-module such that S is not contained in √Ann M. Set ℓ = dimM△. Assume that (1,1,...,1) S e(M;k ,...,k ) = 0, 1 d 6 where k ,...,k are non-negative integers such that k + +k = ℓ 1. Then the 1 d 1 d ··· − following statements hold. (i) If k > 0 and x S is an S -filter-regular element with respect to M then i i ++ ∈ e(M;k ,...,k ) = e(M/xM;k ,...,k 1,...,k ) 1 d 1 i d − and dim(M/xM)△ = ℓ 1. − (ii) There exists an S -filter-regular sequence in d S with respect to M con- ++ j=1 j sisting of k elements of S ,...,k elements ofSS . 1 1 d d (iii) maxL(M) = ℓ. Proof. Denote by P(n ,...,n ) the polynomial of l [M ]. Since 1 d A (n1,...,nd) S * Ann M, (1,1,...,1) S p by Remark 3.1 and Remark 3.2 we have degP = ℓ 1 0. − ≥ The proof of (i): By Remark 2.5, l [(M/xM) ] = l [M ] l [M ] A (n1,...,nd) A (n1,...,nd) − A (n1,...,ni−1,...,nd) for all large n ,...,n . Denote by Q(n ,...,n ) the polynomial of 1 d 1 d l [(M/xM) ]. A (n1,...,nd) From the above fact, we have Q(n ,...,n ) = P(n ,...,n ,...,n ) P(n ,...,n 1,...,n ). 1 d 1 i d 1 i d − − Since e(M;k ,...,k ) = 0 and k > 0, we get degQ = degP 1 and 1 d i 6 − e(M;k ,...,k ) = e(M/xM;k ,...,k 1,...,k ). 1 d 1 i d − By Remark 3.1, degQ = dim(M/xM)△ 1. Therefore − dim(M/xM)△ = degQ+1. 10 Since degQ = degP 1 and degP = ℓ 1, degQ = ℓ 2. Hence − − − dim(M/xM)△ = degQ+1 = (ℓ 2)+1 = ℓ 1. − − The proof of (ii): Since S * √Ann M, ℓ 1 by Remark 3.2. The proof (1,1,...,1) S ≥ is by induction on ℓ. For ℓ = 1, the result is trivial. Assume that ℓ > 1. Since k + +k = ℓ 1 > 0, there exists k > 0. By Proposition 2.6, there exists an 1 d j ··· − S -filter-regular element x S with respect to M. By (i), ++ 1 j ∈ e(M;k ,...,k ) = e(M/x M;k ,...,k 1,...,k ). 1 d 1 1 j d − Since k + + (k 1) + + k = ℓ 2, by the inductive assumption, there 1 j d ··· − ··· − d exists an S -filter-regular sequence x ,...,x S with respect to M/x M ++ 2 ℓ−1 ∈ j=1 j 1 consisting of k elements of S ,...,(k 1) elemenSts of S , ...,k elements of S . 1 1 j j d d − Hence d x ,...,x S 1 ℓ−1 j ∈ j[=1 is an S -filter-regular sequence with respect to M consisting of k elements of ++ 1 S ,...,k elements of S ,...,k elements of S . 1 j j d d The proof of (iii): We first show that maxL(M) ℓ. By (ii), there exists an ≥ d S -filter-regular sequence x ,...,x S with respect to M consisting of ++ 1 ℓ−1 ∈ j=1 j k elements of S ,...,k elements of S . BSy (i), 1 1 d d e(M/(x ,...,x )M;0,...,0) = e(M;k ,...,k ) = 0. 1 ℓ−1 1 d 6 Since e(M/(x ,...,x )M;0,...,0) = 0, it follows that 1 ℓ−1 6 dim[M/(x ,...,x )M]△ > 0. 1 ℓ−1 Hence by Remark 3.2, S * Ann [M/(x ,...,x )M]. (1,1,...,1) S 1 ℓ−1 p d By Proposition 2.6, there exists an S -filter-regular element x S with ++ ℓ ∈ j=1 j respect to M/(x1,...,xℓ−1)M. Hence x1,...,xℓ is an S++-filter-regulaSr sequence in d S with respect to M. Since the length of this sequence is ℓ, maxL(M) ℓ. j=1 j ≥ SNow, we need to show that maxL(M) ℓ. Note that ℓ = degP + 1. We will ≤ prove that maxL(M) degP + 1 by induction on degP. For each j = 1,...,d, ≤ suppose that x S is an S -filter-regular element with respect to M. Denote by 1 j ++ ∈ Q (n ,...,n ) the polynomial of l [(M/x M) ]. By Remark 2.5, 1 1 d A 1 (n1,...,nd) l [(M/x M) ] = l [M ] l [M ] A 1 (n1,...,nd) A (n1,...,nj,...,nd) − A (n1,...,nj−1,...,nd)