Sequentially Cohen--Macaulayness of bigraded modules

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by  Leila Omarmeli

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.
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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