SEQUENTIALLY COHEN–MACAULAYNESS OF BIGRADED MODULES 3 LEILA OMARMELI AND AHAD RAHIMI 1 0 2 Abstract. Let M be a finitely generated bigraded module over the standard n bigraded polynomial ring S = K[x1,...,xm,y1,...,yn] where K is a field. In a this paper we study the sequentially Cohen–Macaulayness of M with respect to J Q=(y1,...,yn). 9 2 ] C Introduction A Let K be a field and S = K[x ,...,x ,y ,...,y ] be the standard bigraded . 1 m 1 n h K-algebra with degx = (1,0) and degy = (0,1) for all i and j. We set the t i j a bigradedirrelevant idealsP = (x ,...,x )andQ = (y ,...,y ). LetM beafinitely m 1 m 1 n generated bigraded S-module. The largest integer k for which Hk(M) 6= 0, is called [ Q the cohomological dimension of M with respect to Q and denoted by cd(Q,M). A 1 finite filtration D : 0 = D D ··· D = M of bigraded submodules of v 0 1 r 6 M, is called the dimension filtration of M with respect to Q if Di−1 is the largest 4 submodule of D for which cd(Q,D ) < cd(Q,D ) for all i = 1,...,r, see [8]. In i i−1 i 8 Section1, we explicitly describe thestructure ofthesubmodules D thatextends [10, 6 i . Proposition 2.2]. In fact, it is shown that D = N for i = 1,...,r−1 where 1 i Tpj6∈Bi j 0 0 = s N is a reduced primary decomposition of 0 in M with N is p -primary 3 Tj=1 j j j for j = 1,...,s and 1 : v B = {p ∈ Ass(M) : cd(Q,S/p) ≤ cd(Q,D )}. i i i X In [9], we say M is Cohen–Macaulay with respect to Q, if grade(Q,M) = cd(Q,M). r A finite filtration F: 0 = M M ··· M = M of M by bigraded submodules a 0 1 r M, iscalled aCohen–Macaulay filtrationwithrespect toQifeach quotient M /M i i−1 is Cohen–Macaulay with respect to Q and 0 ≤ cd(Q,M /M ) < cd(Q,M /M ) < ··· < cd(Q,M /M ), 1 0 2 1 r r−1 see [8]. If M admits a Cohen–Macaulay filtration with respect to Q, then we say M is sequentially Cohen–Macaulay with respect to Q. Note that if M is sequentially Cohen–Macaulay withrespect toQ, thenthefiltrationF isuniquely determined and it is just the dimension filtration of M with respect to Q, that is, F = D. In Section 2, we give a characterization of sequentially Cohen–Macaulay modules with respect to Q in terms of local cohomology modules which extends [3, Corollary 4.4] and [4, Corollary 3.10]. We apply this result and the description of the submodules M i mentioned earlier, showing that S/I is sequentially Cohen–Macaulay with respect 2000 Mathematics Subject Classification. 16W50, 13C14,13D45, 16W70. 1 to P and Q where I is the Stanley-Reisner ideal that corresponds to the natural triangulation of the projective plane P2. Here S = K[x ,x ,x ,y ,y ,y ], P = 1 2 3 1 2 3 (x ,x ,x ) and Q = (y ,y ,y ). Note that S/I is Cohen–Macaulay of dimension 3, 1 2 3 1 2 3 if charK 6= 2. In the following section, M is a Cohen–Macaulay S-module that is sequentially Cohen–Macaulay with respect to Q. Some properties of M are given in Propo- sitions 3.1 and 3.2. Inspired by Proposition 3.2, we make the following ques- tion: Let M be a finitely generated bigraded Cohen–Macaulay S-module such that Hk(M) 6= 0 for all grade(Q,M) ≤ k ≤ cd(Q,M). Is Hs(M) 6= 0 for all Q P grade(P,M) ≤ s ≤ cd(P,M)? Of course the question has positive answer in the case that M has only one(two) non-vanishing local cohomology with respect to Q. The projective plane P2 would also be the case as module with three non-vanishing local cohomology. In Section 3, we call M to be approximately Cohen–Macaulay with respect to Q if M/u (0) is Cohen–Macaulay with respect to Q and grade(Q,M) ≥ cd(Q,M)−1. Q,M Hereu (0)istheunmixedcomponent ofM withrespect toQ. ByProposition4.2, Q,M this definition is equivalent to saying that M is sequentially Cohen–Macaulay with respect to Q and grade(Q,M) ≥ cd(Q,M) − 1. From [8], we obtain some class of approximately Cohen–Macaulay modules with respect to Q. It is then shown that the M¨obius band is approximately Cohen–Macaulay with respect to P and Q, but not approximately Cohen–Macaulay. 1. The dimension filtration with respect to Q Let K be a field and S = K[x ,...,x ,y ,...,y ] the standard bigraded poly- 1 m 1 n nomial ring over K. In other words, degx = (1,0) and degy = (0,1) for all i i j and j. We set the bigraded irrelevant ideals P = (x ,...,x ) and Q = (y ,...,y ), 1 m 1 n and let M be a finitely generated bigraded S-module. Denote by cd(Q,M) the cohomological dimension of M with respect to Q which is the largest integer i for which Hi (M) 6= 0. Notice that 0 ≤ cd(Q,M) ≤ n. Let q ∈ Z. In [9], we say M is Q Cohen–Macaulay with respect to Q if Hi (M) = 0 for all i 6= q. This is equivalent Q to saying that grade(Q,M) = cd(Q,M) = q, see [9, Proposition 1.2]. Remark 1.1. We recall the following facts which will be used in the sequel. By Formula (5) in [9] we always have (1) grade(P,M) ≤ dimM −cd(Q,M), and the equality holds if M is Cohen–Macaulay. If M is Cohen–Macaulay with respect to Q with |K| = ∞, then (2) cd(P,M)+cd(Q,M) = dimM, see, [9, Theorem 3.6]. In view of (1), the converse holds if M is Cohen–Macaulay. The exact sequence 0 → M′ → M → M′′ → 0 of S-modules with M finitely generated yields (3) cd(Q,M) = max{cd(Q,M′),cd(Q,M′′)}, see [2, Proposition 4.4]. 2 If M is a finitely generated S-module, then (4) cd(Q,M) = max{cd(Q,S/p) : p ∈ Ass(M)}, and (5) cd(Q,M) = max{cd(Q,S/p) : p ∈ Supp(M)}, see, [2, Corollary 4.6]. For a finitely generated bigraded S-module M, there is a unique largest bigraded submodule N of M for which cd(Q,N) < cd(Q,M), see [8, Lemma 1.9]. We recall the following definition from [8]. Definition 1.2. Let M be a finitely generated bigraded S-module. We call a filtra- tion D: 0 = D D ··· D = M of bigraded submodules of M the dimension 0 1 r filtration of M with respect to Q if D is the largest submodule of D for which i−1 i cd(Q,D ) < cd(Q,D ) for all i = 1,...,r. i−1 i Remark 1.3. Let D be the dimension filtration of M with respect to Q. For all i, the exact sequence 0 → D → D → D /D → 0 by using (3) yields i−1 i i i−1 cd(Q,D ) = max{cd(Q,D ),cd(Q,D /D )} = cd(Q,D /D ). i i−1 i i−1 i i−1 Thus, cd(Q,D /D ) < cd(Q,D /D ) for all i. i−1 i−2 i i−1 Let D: 0 = D D ··· D = M be the dimension filtration of M with 0 1 r respect to Q. We set Bi = {p ∈ Ass(M) : cd(Q,S/p) ≤ cd(Q,Di)}, Ii = Y p p∈Bi and A = {p ∈ Ass(M) : p ∈ V(I )} for i = 1,...,r. i i Lemma 1.4. Let the notation be as above. Then the following statements hold (a) B = A for i = 1,...,r. i i (b) Supp(D ) ⊆ V(I ) for i = 1,...,r. i i Proof. Fortheproof(a),theinclusionB ⊆ A fori = 1,...,r isobvious. Letp ∈ A . i i i Then p ∈ Ass(M) with I ⊆ p. Hence q ⊆ p for some q ∈ Ass(M) with cd(Q,S/q) ≤ i cd(Q,D ). The canonical epimorphism S/q → S/p yields cd(Q,S/p) ≤ cd(Q,S/q) i by using (3). It follows that p ∈ B and hence A ⊆ B . i i i For the proof (b), we let p ∈ Supp(D ). Here we distinguish two cases: If p ∈ i Min(Supp(D )), then p ∈ Ass(D ) and cd(Q,S/p) ≤ cd(Q,D ) by (5). It follows i i i that p ∈ B = A by part(a) and hence p ∈ V(I ). If p 6∈ Min(Supp(D )). Then i i i i there exists q ∈ Min(Supp(D )) such that q ⊆ p. It follows from the first case that i q ∈ V(I ) and hence p ∈ V(I ), as desired. i i In the following we describe the structure of the submodules D in the dimension i filtration of F with respect to Q which extends [10, Proposition 2.2]. 3 Proposition 1.5. Let D be the dimension filtration of M with respect to Q. Then Di = HI0i(M) = \ Nj pj6∈Bi for i = 1,...,r −1 where 0 = s N is a reduced primary decomposition of 0 in Tj=1 j M with N is p -primary for j = 1,...,s. j j Proof. In order to prove the first equality, we have V(Ann(D )) = Supp(D ) ⊆ V(I ) i i i for i = 1,...,r −1 by Lemma 1.4(b). Since I is finitely generated, it follows that i Iki ⊆ Ann(D ) for some k and hence IkiD = 0 for some k . Thus D = H0(D ) ⊆ i i i i i i i Ii i H0(M) for i = 1,...,r − 1. Now we prove the equality by decreasing induction Ii on i. For i = r − 1, we assume that D H0 (M) ⊆ D = M. It follows r−1 Ir−1 r from the definition dimension filtration that cd(Q,H0 (M)) = cd(Q,M). Note Ir−1 that AssH0(M) = A = B for i = 1,...,r−1 by [6, Proposition 3.13](c). Thus by Ii i i using (4) we have cd(Q,H0 (M)) = max{cd(Q,S/p) : p ∈ AssH0 (M)} Ir−1 Ir−1 = max{cd(Q,S/p) : p ∈ Ass(M), cd(Q,S/p) ≤ cd(Q,D )}. r−1 Hence cd(Q,M) = cd(Q,H0 (M)) ≤ cd(Q,D ), a contradiction. Thus D = Ir−1 r−1 r−1 H0 (M). Now let 1 < i < r − 1, and assume that D = H0(M). We show Ir−1 i Ii D = H0 (M). Assume D H0 (M). Since H0 (M) ⊆ H0(M) = D , it i−1 Ii−1 i−1 Ii−1 Ii−1 Ii i follows from the definition dimension filtration that cd(Q,H0 (M)) ≥ cd(Q,D ). Ii−1 i There exists p ∈ Ass(M) such that cd(Q,H0 (M)) = cd(Q,S/p) ≤ cd(Q,D ). Ii−1 i−1 Hence cd(Q,D ) ≥ cd(Q,D ), a contradiction. Therefore D = H0 (M). The i−1 i i−1 Ii−1 (cid:3) second equality follows from Lemma 1.4(a) and [6, Proposition 3.13](a). Remark 1.6. Let D be the dimension filtration of M with respect to Q with cd(Q,M) = q. We call the submodule Dr−1 = \ Nj = \ Nj, pj6∈Br−1 cd(Q,S/pj)=q the unmixed component of M with respect to Q and denote it by u (0). Notice Q,M that u (0) = u (0) introduced by Schenzel in [10]. If M is relatively unmixed m,M M with respect to Q, that is, cd(Q,M) = cd(Q,S/p) for all p ∈ Ass(M), then by Proposition 1.5 we have s Di = \ Nj = \Nj = 0 for all i < r. pj6∈Bi j=1 Corollary 1.7. Let D be the dimension filtration of M with respect to Q. Then for i = 1,...,r we have (a) Ass(D ) = B , i i (b) Ass(M/D ) = Ass(M)−B . i i 4 Proof. Parts (a) and (b) follow from Proposition 1.5, Lemma 1.4(a) and the facts that AssH0(M) = A and AssM/H0(M) = Ass(M) − A , see [6, Proposition Ii i Ii (cid:3) i 3.13](c). 2. Sequentially Cohen–Macaulay with respect to Q We recall the following definition from [8]. Definition 2.1. Let M be a finitely generated bigraded S-module. We call a finite filtration F: 0 = M M ··· M = M of M by bigraded submodules M a 0 1 r Cohen–Macaulay filtration with respect to Q if (a) Each quotient M /M is Cohen–Macaulay with respect to Q; i i−1 (b) 0 ≤ cd(Q,M /M ) < cd(Q,M /M ) < ··· < cd(Q,M /M ). 1 0 2 1 r r−1 We call M to be sequentially Cohen–Macaulay with respect to Q if M admits a Cohen–Macaulay filtration with respect to Q. Note that if M is sequentially Cohen–Macaulay with respect to Q, then the fil- tration F in the definition above is uniquely determined and it is just the dimension filtration of M with respect to Q defined in Definition 1.2, see [8, Proposision 1.12]. We have the following characterization of sequentially Cohen–Macaulay modules with respect to Q in terms of local cohomology modules which extends [3, Corollary 4.4] and [4, Corollary 3.10]. Proposition 2.2. Let F: 0 = M M ··· M = M be the dimension 0 1 r filtration of M with respect to Q. Then the following statements are equivalent: (a) M is sequentially Cohen–Macaulay with respect to Q; (b) Hk(M/M ) = 0 for i = 1,...,r and k < cd(Q,M ); Q i−1 i (c) grade(Q,M/M ) = cd(Q,M ) for i = 1,...,r. i−1 i Proof. (a) ⇒ (b): We proceed by decreasing induction on i. As M /M is Cohen– i i−1 Macaulay with respect to Q for all i. Thus for i = r we have Hk(M/M ) = 0 Q r−1 for k < cd(Q,M). Now let 1 < i < r, and assume that Hk(M/M ) = 0 for Q i−1 k < cd(Q,M ). The exact sequence i 0 → M /M → M/M → M/M → 0, i−1 i−2 i−2 i−1 induces the following long exact sequence (6) ··· → Hk(M /M ) → Hk(M/M ) → Hk(M/M ) → ··· . Q i−1 i−2 Q i−2 Q i−1 As M /M is Cohen–Macaulay with respect to Q, we have Hk(M /M ) = 0 i−1 i−2 Q i−1 i−2 for k < cd(Q,M ). By Remark 1.3, we have cd(Q,M ) = cd(Q,M /M ) < i−1 i−1 i−1 i−2 cd(Q,M ).Hencebyusing(6)andtheinductionhypothesis, wehaveHk(M/M ) = i Q i−2 0 for k < cd(Q,M ), as desired. i−1 (b) ⇒ (a): By Remark 1.3 we have cd(Q,M /M ) < cd(Q,M /M ) for all i. i i−1 i+1 i Thus it suffices to show that M /M is Cohen–Macaulay with respect to Q for all i i−1 i. We prove this statement by decreasing induction on i. In condition (b), we first assume i = r. It follows that M/M is Cohen–Macaulay with respect to Q. Now r−1 5 let 1 < i < r, and assume that M /M is Cohen–Macaulay with respect to Q. The i i−1 exact sequence 0 → M /M → M/M → M/M → 0, i i−1 i−1 i induces the following long exact sequence (7) ··· → Hk−1(M /M ) → Hk−1(M/M ) → Hk−1(M/M ) → ··· . Q i i−1 Q i−1 Q i Suppose k < cd(Q,M ). Induction hypothesis and our assumption say that i−1 Hk−1(M /M ) = Hk−1(M/M ) = 0. Hence Hk−1(M/M ) = 0 by (7). We Q i i−1 Q i Q i−1 have Hk(M/M ) = 0 for k < cd(Q,M ) because of our assumption again. Thus Q i−2 i−1 Hk(M /M ) = 0 for k < cd(Q,M ) by (6). Therefore M /M is Cohen– Q i−1 i−2 i−1 i−1 i−2 Macaulay with respect to Q, as desired. (b) ⇒ (c): We set cd(Q,M ) = cd(Q,M /M ) = q for i = 1,...,r. Our i i i−1 i assumption says that grade(Q,M/M ) ≥ q for i = 1,...,r. We only need to i−1 i know Hqi(M/M ) 6= 0. Consider the long exact sequence Q i−1 (8) ··· → Hqi−1(M/M ) → Hqi(M /M ) → Hqi(M/M ) → ··· . Q i Q i i−1 Q i−1 Since q − 1 < q < q , it follows from our assumption that Hqi−1(M/M ) = 0. i i i+1 Q i Thus if Hqi(M/M ) = 0, then by (8) we have Hqi(M /M ) = 0, a contradiction. Q i−1 Q i i−1 The implication (c) ⇒ (b) is obvious. (cid:3) As an application of Proposition 1.5 and Proposition 2.2 we have Example 2.3. Let I be the Stanley-Reisner ideal that corresponds to the natural triangulation of the projective plane P2. Then I = (x x x ,x x y ,x x y ,x y y ,x y y ,x x y ,x y y ,x y y ,x y y ,x y y ). 1 2 3 1 2 1 1 3 2 1 1 3 1 2 3 2 3 3 2 1 2 2 2 3 3 1 2 3 1 3 We set R = S/I where S = K[x ,x ,x ,y ,y ,y ], P = (x ,x ,x ) and Q = 1 2 3 1 2 3 1 2 3 (y ,y ,y ). One has that if charK 6= 2, then R is Cohen–Macaulay of dimen- 1 2 3 sion 3. The ideal I has the minimal primary decomposition I = 10 p where p = Ti=1 i 1 (x ,y ,y ),p = (x ,y ,y ),p = (x ,y ,y ),p = (x ,y ,y ),p = (x ,y ,y ),p = 3 1 3 2 1 1 3 3 2 1 2 4 3 1 2 5 1 2 3 6 (x ,y ,y ),p = (x ,x ,y ),p = (x ,x ,y ),p = (x ,x ,y ),p = (x ,x ,x ). As 2 2 3 7 2 3 3 8 1 2 1 9 1 3 2 10 1 2 3 P = p ∈ Ass(R), we have grade(P,R) = 0. Using (4) we have cd(P,R) = 2. Since 10 R is Cohen–Macaulay, it follows from (1) that grade(Q,R) = 1 and cd(Q,R) = 3. We first show that R is sequentially Cohen–Macaulay with respect to P. By Propo- sition 1.5, R has the dimension filtration 0 = R R R R = R, 0 1 2 3 with respect to P where 9 6 R1 = \pi/I and R2 = \pi/I. i=1 i=1 We set I = 9 p and I = 6 p . In view of Proposition 2.2, we need to show 1 Ti=1 i 2 Ti=1 i grade(P,R /R ) = grade(P,S/I ) = cd(P,R ) = 1 3 1 1 2 and grade(P,R /R ) = grade(P,S/I ) = cd(P,R) = 2. 3 2 2 6 Since P 6⊆ p for i = 1,...,9, it follows that grade(P,S/I ) ≥ 1. On the other i 1 hand, grade(P,S/I ) ≤ dimS/I −cd(Q,S/I ) = 3−2 = 1. Thus the first equality 1 1 1 holds. In order to show the second equality, we note that S/I has dimension 3 2 and by using CoCoA depth 2. Thus Formula (1) can not be used to compute grade(P,S/I ). We set q = p ∩p = (x x ,y ,y ), q = p ∩p = (x x ,y ,y ) and 2 1 1 2 1 3 1 3 2 3 4 2 3 1 2 q = p ∩p = (x x ,y ,y ). Consider the exact sequence 3 5 6 1 2 2 3 0 → S/q ∩q → S/q ⊕S/q → S/(q +q ) → 0. 1 2 1 2 1 2 Since grade(P,S/q ⊕ S/q ) = 2 and grade(P,S/(q + q )) = 1, it follows that 1 2 1 2 grade(P,S/(q ∩q )) ≥ 2. As cd(P,S/(q ∩q )) = 2, we have grade(P,S/(q ∩q )) = 1 2 1 2 1 2 2. Consider the exact sequence (9) 0 → S/I → S/q ∩q ⊕S/q → S/(q +q )∩(q +q ) → 0. 2 1 2 3 1 3 2 3 The exact sequence 0 → S/(q +q )∩(q +q ) → S/(q +q )⊕S/(q +q ) → S/(q +q +q ) → 0 1 3 2 3 1 3 2 3 1 2 3 yieldsthatgrade(P,S/(q +q )∩(q +q )) ≥ 1.Henceby(9)wehavegrade(P,S/I ) ≥ 1 3 2 3 2 2. As cd(P,S/I ) = 2, we conclude that grade(P,S/I ) = 2, as desired. 2 2 Next, we show that R is sequentially Cohen–Macaulay with respect to Q. By Proposition 1.5, R has the dimension filtration 0 = R R R R = R with 0 1 2 3 respect to Q where R = 10 p /I and R = p /I. We set J = 10 p . In view of 1 Ti=7 i 2 10 Ti=7 i Proposition 2.2, we need to show grade(Q,R /R ) = grade(Q,S/p ) = cd(Q,R) = 3 3 2 10 and grade(Q,R /R ) = grade(Q,S/J) = cd(Q,R ) = 2. 3 1 2 The first statement is clear. In order to prove the second equality, consider the exact sequence (10) 0 → S/J → S/∩9 p ⊕S/p → S/∩9 (p +p ) → 0. i=7 i 10 i=7 i 10 An exact sequence argument shows that grade(Q,S/∩9 p ) = grade(Q,S/∩9 (p +p )) = 2. i=7 i i=7 i 10 Thus it follows from (10) that grade(Q,S/J) ≥ 2. On the other hand, grade(Q,S/J) ≤ dimS/J −cd(P,S/J) = 3−1 = 2. Therefore, grade(Q,S/J) = 2, as desired. 3. Cohen–Macaulay modules that are sequentially Cohen–Macaulay with respect to Q For a Cohen–Macaulay filtration F with respect to Q we recall the following fact from [8, Corollary 1.8] (11) grade(Q,M ) = grade(Q,M) for i = 1,...,r. i The following equalities can be seen from a Cohen–Macaulay filtration with respect to Q provided that the module itself is Cohen–Macaulay. 7 Proposition 3.1. Let M be a finitely generated bigraded Cohen–Macaulay S-module with |K| = ∞. Suppose M is sequentially Cohen–Macaulay with respect to Q with the Cohen–Macaulay filtration 0 = M M ··· M = M with respect to Q. 0 1 r Then (a) cd(P,M ) = cd(P,M) for i = 1,...,r. i (b) grade(Q,M )+cd(P,M ) = dimM for i = 1,...,r. i i i Proof. In order to prove (a), since M is Cohen–Macaulay with respect to Q, it fol- 1 lows from (2) that cd(P,M )+cd(Q,M ) = dimM . By (11) we have cd(Q,M ) = 1 1 1 1 grade(Q,M ) = grade(Q,M). Since M is Cohen–Macaulay, it follows from [8, 1 Lemma 1.11] that dimM = dimM and cd(P,M) = dimM − grade(Q,M) by 1 (1). Thus we conclude that cd(P,M ) = cd(P,M). As by (3) we have cd(P,M ) ≤ 1 i−1 cd(P,M ) for all i, the first equality follows for all i. i For the proof (b), by [8, Lemma 1.11] we have dimM = dimM for i = 1,...,r. i (cid:3) Thus the second equalities follow from (1), (11) and part (a). Proposition 3.2. Let the assumptions and the notation be as in Proposition 3.1. Then the following statements are equivalent: (a) cd(P,M)+cd(Q,M) = dimM +r −1; (b) Hs(M) 6= 0 for all grade(Q,M) ≤ s ≤ cd(Q,M). Q Proof. We first assume that r = 1. As M is Cohen–Macaulay, by (2) we have cd(P,M)+cd(Q,M) = dimM if and only if M is Cohen–Macaulay with respect to Q. Thus the proposition holds in this case. Now let r ≥ 2. By using (1) we have cd(P,M)+cd(Q,M) = dimM+r−1 if and only if cd(Q,M)−grade(Q,M) = r−1. This is equivalent to saying that cd(Q,M ) = cd(Q,M ) + 1 for i = 1,...,r − 1 i+1 i by (11). By [8, Proposition, 1.7] this is equivalent to saying that Hs(M) 6= 0 for all Q grade(Q,M) ≤ s ≤ cd(Q,M). (cid:3) The following example shows that the condition that ”M is Cohen–Macaulay” is required for Proposition 3.2. Example 3.3. We set K[x] = K[x ,...,x ] and K[y] = K[y ,...,y ]. Let L be a 1 m 1 n non-zerofinitely generatedgradedK[x]-moduleofdepth0anddimension 1, andN a non-zero finitely generated graded K[y]-module of depth 0 and dimension 1. We set M = L⊗ N andconsideritasS-module. OnehasdepthM = 0anddimM = 2and K hence M is not Cohen–Macaulay. On the other hand, grade(Q,M) = depthN = 0 and cd(Q,M) = dimN = 1 = dimL = cd(P,M). Hence M is sequentially Cohen– Macaulay with respect to Q which satisfies condition (b) in Proposition 3.2, while the equality (a) does not hold. The following question is inspired by Proposition 3.2. Question 3.4. Let M be a finitely generated bigraded Cohen–Macaulay S-module such that Hk(M) 6= 0 for all grade(Q,M) ≤ k ≤ cd(Q,M). Is Hs(M) 6= 0 for all Q P grade(P,M) ≤ s ≤ cd(P,M)? Remark 3.5. Of course the question has positive answer in the following cases, namely, if M has only one(two) non-vanishing local cohomology with respect to Q. 8 This immediately follows by Formula (1). The projective plane P2 given in Example 2.3 is also the case as module with three non-vanishing local cohomology. 4. Approximately Cohen–Macaulay with respect to Q Definition 4.1. Let M be a finitely generated bigraded S-module. We call M to be approximately Cohen–Macaulay with respect to Q if M/u (0) is Cohen–Macaulay Q,M with respect to Q and grade(Q,M) ≥ cd(Q,M) −1. Here u (0) is the unmixed Q,M component of M with respect to Q introduced in Remark 1.6. An approximately Cohen–Macaulay module introduced by Schenzel [10], is in our terminologyanapproximatelyCohen–Macaulaymodulewithrespecttothemaximal ideal m = P +Q. Proposition 4.2. Let M be a finitely generated bigraded S-module with |K| = ∞. Then the following statements are equivalent: (a) M is approximately Cohen–Macaulay with respect to Q; (b) M is sequentially Cohen–Macaulay with respect to Q and grade(Q,M) ≥ cd(Q,M)−1. Proof. We may assume that M is not Cohen–Macaulay with respect to Q. (a) ⇒ (b): Let F: 0 = M M ··· M = M be the dimension filtration 0 1 r of M with respect to Q, as it always exists. Observe that cd(Q,M ) = q − 1 r−1 where q = cd(Q,M). In fact, our assumption and [7, Proposition 1.7] imply that q −1 = grade(Q,M) ≤ cd(Q,S/p) for all p ∈ Ass(M). Thus by (4) we have q−1 ≤ max{cd(Q,S/p) : p ∈ Ass(M )} = cd(Q,M ) < cd(Q,M ) = q. r−1 r−1 r Therefore, cd(Q,M ) = q−1. ItfollowsfromProposition1.5thatM = u (0) r−1 r−1 Q,M and M = 0 for i = 1,...,r −2 and hence F: 0 u (0) M is the dimension i Q,M filtration of M with respect to Q. Since by our assumption, M/u (0) is Cohen– Q,M Macaulay with respect to Q, it follows from Proposition 2.2 that M is sequentially Cohen–Macaulay with respect to Q. (b) ⇒ (a): Let F be the dimension filtration of M with respect to Q. As grade(Q,M) = cd(Q,M) − 1, by the same argument as above we have M = r−1 u (0) with cd(Q,M ) = q − 1 and M = 0 for i = 1,...,r − 2. Thus by our Q,M r−1 i assumption F: 0 u (0) M is a Cohen–Macaulay filtration with respect to Q,M Q. This implies that M/u (0) is Cohen–Macaulay with respect to Q, as desired. Q,M (cid:3) In view of Corollary 1.7 and Proposition 4.2 we have Corollary 4.3. If M is approximately Cohen–Macaulay with respect to Q, then (a) Ass(u (0)) = {p ∈ Ass(M) : cd(Q,S/p) = cd(Q,M)−1}, Q,M (b) Ass(M/u (0)) = {p ∈ Ass(M) : cd(Q,S/p) = cd(Q,M)}. Q,M Remark 4.4. Let M be a finitely generated bigraded Cohen–Macaulay S-module such that cd(Q,M) = grade(Q,M)+1. Then by (1) we have cd(P,M)+cd(Q,M) = dimM +1. 9 In particular, if M is approximately Cohen–Macaulay with respect to Q that is not Cohen–Macaulay with respect to Q, then the above equality holds. The following corollaries give some characterizations of approximately Cohen– Macaulay with respect to Q that are originally obtained from [8]. Corollary 4.5. If cd(Q,M) ≤ 1, then M is approximately Cohen–Macaulay with respect to Q by [8, Example 1.6] and Proposition 4.2. Corollary 4.6. Let R = S/fS where f ∈ S be a bihomogeneous element of degree (a,b). Then the following statements are equivalent: (a) R is approximately Cohen–Macaulay with respect to Q; (b) f = h h where degh = (a,0) with a ≥ 0 and degh = (0,b) with b ≥ 0. 1 2 1 2 Proof. The assertion follows from Proposition 4.2 and [8, Theorem 4.4]. (cid:3) Corollary 4.7. Let L and N be two non-zero finitely generated graded modules over K[x] and K[y], respectively. We set M = L⊗ N. Then the following statements K are equivalent: (a) M is approximately Cohen–Macaulay S-module with respect to Q; (b) N is approximately Cohen–Macaulay K[y]-module. Proof. The assertion follows from Proposition 4.2, [8, Theorem 3.2] and the facts that grade(Q,M) = depthN and cd(Q,M) = dimN. (cid:3) Corollary 4.8. Let M be a finitelygenerated bigraded S-module with grade(Q,M) > 0 and |K| = ∞. If M is approximately Cohen–Macaulay with respect to Q, then there exists a bihomogeneous M-regular element y ∈ Q of degree (0,1) such that M/yM is approximately Cohen–Macaulay with respect to Q. Proof. The assertion follows from Proposition 4.2 and [8, Proposition 2.2]. (cid:3) In the following example, we discuss the approximately Cohen–Macaulayness of the ring R with respect to P, Q and P +Q = m. Example 4.9. ConsidertheM¨obiusbandR = S/I whereS = K[x ,x ,x ,y ,y ,y ] 1 2 3 1 2 3 and I = (x x ,x y ,x y ,x x y ,x y y ,y y y ). The ideal I has the minimal pri- 1 3 1 2 2 3 2 3 1 3 1 2 1 2 3 mary decomposition I = 6 p where p = (x ,x ,y ),p = (x ,x ,y ),p = Ti=1 i 1 2 3 2 2 1 2 2 3 (x ,x ,y ),p = (x ,x ,y ),p = (x ,y ,y ),p = (x ,y ,y ). The ring R has di- 1 2 1 4 1 3 3 5 1 1 3 6 3 2 3 mension3andbyusingCoCoAdepth2. WefirstobservethatRisnotapproximately ∼ Cohen–Macaulay. Indeed, u (0) = p modulo I and hence R/u (0) = R R TdimS/p=3 R is not Cohen–Macaulay. Next, we show that R is approximately Cohen–Macaulay with respect to Q. By using (4) we have cd(Q,R) = cd(P,R) = 2. Since Q 6⊆ p for all p ∈ Ass(R), it follows that grade(Q,R) ≥ 1. On the other hand, by (1) we have grade(Q,R) ≤ dimR−cd(P,R) = 3−2 = 1, and therefore grade(Q,R) = 1. Thus, we need to show that R/u (0) is Cohen–Macaulay with respect to Q where Q,R u (0) = 4 p modulo I. As cd(Q,R/u (0)) = cd(Q,S/ 4 p ) = 2 by (4), Q,R Ti=1 i Q,R Ti=1 i it suffices to show that grade(Q,S/ 4 p ) = 2. Consider the exact sequence Ti=1 i 4 (12) 0 → S/\pi → S/(p2 ∩p3)⊕S/(p1 ∩p4) → S/(p2 ∩p3 +p1 ∩p4) → 0. i=1 10