Primary Decomposition in a Sequentially Cohen-Macaulay Module

Sara Faridi^{∗}

March 17, 2005

The notion of a sequentially Cohen-Macaulay module was introduced by Stanley [?], following the introduction of a nonpure shellable simplicial complex by Björner and Wachs [BW]. It was known that the Stanley-Reisner ideal of a shellable simplicial complex is Cohen-Macaulay (see [BH]). A shellable simplicial complex is by definition pure (all facets have the same dimension), which is equivalent to its Stanley-Reisner ideal being unmixed. A nonpure shellable simplicial complex, on the other hand, may not be pure, so its Stanley-Reisner ideal may not be unmixed, and hence not Cohen-Macaulay. As it turns out, however, the Stanley-Reisner ideal of a nonpure simplicial complex is “sequentially Cohen-Macaulay” (Definition 1 below).

If the Stanley-Reisner ideal of a simplicial complex is sequentially Cohen-Macaulay, the complex has Cohen-Macaulay pure subcomplexes (see Duval [D] Theorem 3.3, or Stanley [?] Chapter III, Proposition 2.10). In the language of commutative algebra, this is equivalent to all equidimensional components appearing in the primary decomposition of a square-free monomial ideal being Cohen-Macaulay (see [F] for more details).

The purpose of this note is to establish that, more generally, this is what being sequentially Cohen-Macaulay means for any module. Below we use basic facts about primary decomposition of modules to study the structure of the submodules appearing in the (unique) filtration of a sequentially Cohen-Macaulay module. The main result (Theorem 5) states that each submodule appearing in the filtration of a sequentially Cohen-Macaulay module M is the intersection of all primary submodules whose associated primes have a certain height and appear in an irredundant primary decomposition of the 0-submodule of M. Similar results, stated in a different language, appear in [Sc]; the author thanks Jürgen Herzog for pointing this out.

Definition 1 ([St] Chapter III, Definition 2.9). Let M be a finitely generated ℤgraded module over a finitely generated ℕ-graded k-algebra, with R_{0} = k. We say that M is sequentially Cohen-Macaulay if there exists a finite filtration

0 = M_{0} ⊂ M_{1} ⊂ … ⊂ M_{r} = M

of M by graded submodules M_{i} satisfying the following two conditions.

(a) Each quotient M_{i}/M_{i−1} is Cohen-Macaulay; (b) dim (M_{1}/M_{0}) < dim (M_{2}/M_{1}) < … < dim (M_{r}/M_{r−1}), where “dim ” denotes Krull

∗Université du Québec à Montréal, Laboratoire de combinatoire et d’informatique mathématique, Case postale 8888, succursale Centre-Ville, Montréal, QC Canada H3C 3P8. Email: faridi@math.uqam.ca.

This research was supported by a NSERC Postdoctoral Fellowship.

dimension.

Before we begin our study of sequentially Cohen-Macaulay modules, we record two basic lemmas that we shall use later. Throughout the discussions below, we assume that R is a finitely generated algebra over a field, and M is a finite module over R.

Lemma 2. Let Q_{1}, …, Q_{t}, P all be primary submodules of an R-module M, such that Ass(M/Q_{i}) = {q_{i}} and Ass(M/P) = {℘}. If Q_{1} ∩ … ∩ Q_{t} ⊆ P and Q_{i} ⊈ P for some i, then there is a j ≠ i such that q_{j} ⊆ ℘.

Proof. Let x ∈ Q_{i} \ P. For each j ≠ i, pick the positive integer m_{j} such that

q_{j}^{m_{j}}x ⊆ Q_{j}.

So we have that

q_{1}^{m_{1}} … q_{i−1}^{m_{i−1}}q_{i+1}^{m_{i+1}} … q_{t}^{m_{t}}x ⊆ Q_{1} ∩ … ∩ Q_{t} ⊆ P

which implies that, since x ∉ P,

q_{1}^{m_{1}} … q_{i−1}^{m_{i−1}}q_{i+1}^{m_{i+1}} … q_{t}^{m_{t}} ⊆ ℘

and hence for some j ≠ i, q_{j} ⊆ ℘.

Lemma 3. Let M be an R-module and N be a submodule of M. Then for every ℘ ∈ Ass(M/N), if ℘ ⊉ Ann(N), then ℘ ∈ Ass(M).

Proof. Since ℘ ∈ Ass(M/N), there exists x ∈ M \ N such that ℘ = Ann(x); in other words

℘x ⊆ N.

Suppose Ann(N) ⊈ ℘, and let y ∈ Ann(N) \ ℘. Now y℘x = 0, and so ℘ ⊆ Ann(yx) in M.

On the other hand, if z ∈ Ann(yx), then zyx = 0 ⊆ N and so zy ∈ ℘. But y ∉ ℘, so z ∈ ℘. Therefore ℘ ∈ Ass(M).

Suppose M is a sequentially Cohen-Macaulay module with filtration as in Definition 1. We adopt the following notation. For a given integer j, we let

Ass(M)_{j} = {℘ ∈ Ass(M) | height ℘ = j}.

Suppose that all the j where Ass(M)_{j} ≠ ∅ form the sequence of integers

0 ≤ h_{1} < … < h_{c} ≤ dim R

so that

Ass(M) = ⋃_{1≤j≤c} Ass(M)_{h_{j}}.

We can now make the following observations.

Proposition 4. For all i = 0, …, r − 1, we have

1. Ass(M_{i+1}/M_{i}) ∩ Ass(M) ≠ ∅; 2. Ass(M)_{h_{r−i}} ⊆ Ass(M_{i+1}/M_{i}) and c = r;

3. If ℘ ∈ Ass(M_{i+1}), then height ℘ ≥ h_{r−i};

4. If ℘ ∈ Ass(M_{i+1}/M_{i}), then Ann(M_{i}) ⊈ ℘;

5. Ass(M_{i+1}/M_{i}) ⊆ Ass(M);

6. Ass(M_{i+1}/M_{i}) = Ass(M)_{h_{r−i}};

7. Ass(M/M_{i}) = Ass(M)_{≤h_{r−i}};

8. Ass(M_{i+1}) = Ass(M)_{≥h_{r−i}}.

Proof. 1. We use induction on the length r of the filtration of M. The case r = 1 is clear, as we have a filtration 0 ⊂ M, and the assertion follows. Now suppose the statement holds for sequentially Cohen-Macaulay modules with filtrations of length less than r. Notice that M_{r−1} that appears in the filtration of M in Definition 1 is also sequentially Cohen-Macaulay, and so by the induction hypothesis, we have

Ass(M_{i+1}/M_{i}) ∩ Ass(M_{r−1}) ≠ ∅ for i = 0, …, r − 2

and since Ass(M_{r−1}) ⊆ Ass(M) it follows that

Ass(M_{i+1}/M_{i}) ∩ Ass(M) ≠ ∅ for i = 0, …, r − 2.

It remains to show that Ass(M/M_{r−1}) ∩ Ass(M) ≠ ∅. For each i, M_{i−1} ⊂ M_{i}, so we have ([B] Chapter IV)

Ass(M_{1}) ⊆ Ass(M_{2}) ⊆ Ass(M_{1}) ∪ Ass(M_{2}/M_{1})  (1)

The inclusion M_{2} ⊆ M_{3} along with the inclusions in (1) imply that

Ass(M_{2}) ⊆ Ass(M_{3}) ⊆ Ass(M_{2})∪Ass(M_{3}/M_{2}) ⊆ Ass(M_{1})∪Ass(M_{2}/M_{1})∪Ass(M_{3}/M_{2}).

If we continue this process inductively, at the i-th stage we have

Ass(M_{i}) ⊆ Ass(M_{i−1}) ∪ Ass(M_{i}/M_{i−1})
⊆ Ass(M_{1}) ∪ Ass(M_{2}/M_{1}) ∪ Ass(M_{3}/M_{2}) ∪ … ∪ Ass(M_{i}/M_{i−1})

and finally, when i = r it gives

Ass(M) ⊆ Ass(M_{1}) ∪ Ass(M_{2}/M_{1}) ∪ Ass(M_{3}/M_{2}) ∪ … ∪ Ass(M/M_{r−1}).  (2)

Because of Condition (b) in Definition 1, and the fact that each M_{i+1}/M_{i} is Cohen-Macaulay (and hence all its associated primes have the same height; see [BH] Chapter 2), if for every i we pick ℘_{i} ∈ Ass(M_{i+1}/M_{i}), then

h_{c} ≥ height ℘_{0} > height ℘_{1} > … > height ℘_{r−1}.

where the left-hand-side inequality comes from the fact that Ass(M_{1}) ⊆ Ass(M). By our induction hypothesis, Ass(M) intersects Ass(M_{i+1}/M_{i}) for all i ≤ r − 2, and so because of (2) we conclude that

height ℘_{i} = h_{c−i}, and Ass(M)_{h_{c−i}} ⊆ Ass(M_{i+1}/M_{i}) for 0 ≤ i ≤ r − 2.

And now Ass(M)_{h_{0}} has no choice but to be included in Ass(M/M_{r−1}), which settles our claim. It also follows that c = r.

2. See the proof for part 1.

3. We use induction. The case i = 0 is clear, since for every ℘ ∈ Ass(M_{1}) = Ass(M_{1}/M_{0}) we know from part 2 that height ℘ = h_{r}. Suppose the statement holds for all indices up to i − 1. Consider the inclusion

Ass(M_{i}) ⊆ Ass(M_{i+1}) ⊆ Ass(M_{i}) ∪ Ass(M_{i+1}/M_{i}).

From part 2 and the induction hypothesis it follows that if ℘ ∈ Ass(M_{i+1}) then height ℘ ≥ h_{r−i}.

4. Suppose Ann(M_{i}) ⊆ ℘. Since \sqrt{Ann(M_{i})} = ⋂_{℘′∈Ass(M_{i})} ℘′, we have

⋂_{℘′∈Ass(M_{i})} ℘′ ⊆ ℘

so there is a ℘′ ∈ Ass(M_{i}) such that ℘′ ⊆ ℘. But by part 2 and part 3 above

height ℘′ ≥ h_{r−i+1} and height ℘ = h_{r−i}

which is a contradiction.

5. From part 4 and Lemma 3, it follows that

Ass(M_{i+1}/M_{i}) ⊆ Ass(M_{i+1}) ⊆ Ass(M).

6. This follows from parts 2 and 5, and the fact that M_{i+1}/M_{i} is Cohen-Macaulay, and hence all associated primes have the same height.

7. We show this by induction on e = r − i. The case e = 1 (or i = r − 1) is clear, because

by part 6

Ass(M/M_{r−1}) = Ass(M)_{h_{1}} = Ass(M)_{≤h_{1}}.

Now suppose the equation holds for all integers up to e−1 (namely i = r −e+1), and we would like to prove the statement for e (or i = r − e). Since M_{i+1}/M_{i} ⊆ M/M_{i}, we have

Ass(M_{i+1}/M_{i}) ⊆ Ass(M/M_{i}) ⊆ Ass(M_{i+1}/M_{i}) ∪ Ass(M/M_{i+1})  (3)

By the induction hypothesis and part 6 we know that

Ass(M/M_{i+1}) = Ass(M)_{≤h_{r−i−1}} and Ass(M_{i+1}/M_{i}) = Ass(M)_{h_{r−i}},

which put together with (3) implies that

Ass(M)_{h_{r−i}} ⊆ Ass(M/M_{i}) ⊆ Ass(M)_{≤h_{r−i}}

We still have to show that Ass(M/M_{i}) ⊇ Ass(M)_{≤h_{r−i−1}}. Let

℘ ∈ Ass(M)_{≤h_{r−i−1}} = Ass(M/M_{i+1}) = Ass((M/M_{i})/(M_{i+1}/M_{i})).

If ℘ ⊇ Ann(M_{i+1}/M_{i}), then (by part 6)

℘ ⊇ ⋂_{q∈Ass(M)_{h_{r−i}}} q ⟹ ℘ ⊇ q for some q ∈ Ass(M)_{h_{r−i}}

which is a contradiction, as height ℘ ≤ h_{r−i−1} < height q. It follows from Lemma 3 that ℘ ∈ Ass(M/M_{i}).

8. The argument is based on induction, and exactly the same as the one in part 4, using more information; from

Ass(M_{i}) ⊆ Ass(M_{i+1}) ⊆ Ass(M_{i}) ∪ Ass(M_{i+1}/M_{i}),

the induction hypothesis, and part 6 we deduce that

Ass(M)_{≥h_{r−i+1}} ⊆ Ass(M_{i+1}) ⊆ Ass(M)_{≥h_{r−i+1}} ∪ Ass(M)_{h_{r−i}},

which put together with part 4, along with Lemma 3 produces the equality.

Now suppose that as a submodule of M, M_{0} = 0 has an irredundant primary decomposition of the form:

M_{0} = 0 = ⋂_{1≤j≤r} Q_{1}^{h_{j}} ∩ … ∩ Q_{s_{j}}^{h_{j}}  (4)

where for a fixed j ≤ r and e ≤ s_{j}, Q_{e}^{h_{j}} is a primary submodule of M with

Ass(M/Q_{e}^{h_{j}}) = {℘_{e}^{h_{j}}} and Ass(M)_{h_{j}} = {℘_{1}^{h_{j}}, …, ℘_{s_{j}}^{h_{j}}}.

Theorem 5. Let M be a sequentially Cohen-Macaulay module with filtration as in Definition 1, and suppose that M_{0} = 0 has a primary decomposition as in (4). Then for each i = 0, …, r − 1, M_{i} has the following primary decomposition

M_{i} = ⋂_{1≤j≤r−i} Q_{1}^{h_{j}} ∩ … ∩ Q_{s_{j}}^{h_{j}}.  (5)

Proof. We prove this by induction on r (length of the filtration). The case r = 1 is clear, as the filtration is of the form 0 = M_{0} ⊂ M. Now consider M with filtration

0 = M_{0} ⊂ M_{1} ⊂ … ⊂ M_{r} = M.

Since M_{r−1} is a sequentially Cohen-Macaulay module of length r − 1, it satisfies the statement of the theorem. We first show that M_{r−1} has a primary decomposition as described in (5). From part 7 of Proposition 4 it follows that

Ass(M/M_{r−1}) = Ass(M)_{h_{1}}

and so for some ℘_{e}^{h_{1}}-primary submodules P_{e}^{h_{1}} of M (1 ≤ e ≤ s_{j}), we have

M_{r−1} = P_{1}^{h_{1}} ∩ … ∩ P_{s_{1}}^{h_{1}}.  (6)

We would like to show that Q_{e}^{h_{1}} = P_{e}^{h_{1}} for e = 1, …, s_{1}.

Fix e = 1 and assume Q_{1}^{h_{1}} ⊄ P_{1}^{h_{1}}. From the inclusion M_{0} ⊂ P_{1}^{h_{1}} and Lemma 2 it follows that for some e and j (with e ≠ 1 if j = 1), we have ℘_{e}^{h_{j}} ⊆ ℘_{1}^{h_{1}}. Because of the difference in heights of these ideals the only conclusion is ℘_{e}^{h_{j}} = ℘_{1}^{h_{1}}, which is not possible. With a similar argument we deduce that Q_{e}^{h_{1}} ⊂ P_{e}^{h_{1}}, for e = 1, …, s_{1}.

Now fix j ∈ {1, …, r} and e ∈ {1, …, s_{j}}. If M_{r−1} = Q_{e}^{h_{j}} we are done. Otherwise, note that for every j and ℘_{e}^{h_{j}}-primary submodule Q_{e}^{h_{j}} of M,

Q_{e}^{h_{j}} ∩ M_{r−1}

is a ℘_{e}^{h_{j}}-primary submodule of M_{r−1} (as ∅ ≠ Ass(M_{r−1}/(Q_{e}^{h_{j}} ∩ M_{r−1})) = Ass((M_{r−1} + Q_{e}^{h_{j}})/Q_{e}^{h_{j}}) ⊆ Ass(M/Q_{e}^{h_{j}}) = {℘_{e}^{h_{j}}}). So M_{0} = 0 as a submodule of M_{r−1} has a primary decomposition

M_{0} ∩ M_{r−1} = 0 = ⋂_{1≤j≤r}(Q_{1}^{h_{j}} ∩ M_{r−1}) ∩ … ∩ (Q_{s_{j}}^{h_{j}} ∩ M_{r−1}).

From Proposition 4 part 8 it follows that

Ass(M_{r−1}) = Ass(M)_{≥h_{2}}

so the components Q_{t}^{h_{1}} ∩ M_{r−1} are redundant for t = 1, …, s_{1}, so for each such t we have

⋂_{Q_{e}^{h_{j}} ≠Q_{t}^{h_{1}}}(Q_{1}^{h_{j}} ∩ M_{r−1}) ⊆ Q_{t}^{h_{1}} ∩ M_{r−1}.

If Q_{e}^{h_{j}} ∩ M_{r−1} ⊈ Q_{t}^{h_{1}} ∩ M_{r−1} for some e and j (with Q_{e}^{h_{j}} ≠ Q_{t}^{h_{1}}), then by Lemma 2 for some such e and j we have ℘_{e}^{h_{j}} ⊆ ℘_{t}^{h_{1}}, which is a contradiction (because of the difference of heights).

Therefore, for each t (1 ≤ t ≤ s_{1}), there exists indices e and j (with Q_{e}^{h_{j}} ≠ Q_{t}^{h_{1}}) such that

Q_{e}^{h_{j}} ∩ M_{r−1} ⊆ Q_{t}^{h_{1}} ∩ M_{r−1}.

It follows now, from the primary decomposition of M_{r−1} in (6) that for a fixed t

P_{1}^{h_{1}} ∩ … ∩ P_{s_{1}}^{h_{1}} ∩ Q_{e}^{h_{j}} ⊆ Q_{t}^{h_{1}}.

Assume P_{t}^{h_{1}} ⊈ Q_{t}^{h_{1}}. Applying Lemma 2 again, we deduce that

℘_{e}^{h_{j}} ⊆ ℘_{t}^{h_{1}}, or there is t′ ≠ t such that ℘_{t′}^{h_{1}} ⊆ ℘_{t}^{h_{1}}.

Neither of these is possible, so P_{t}^{h_{1}} ⊆ Q_{t}^{h_{1}} for all t.

We have therefore proved that

M_{r−1} = Q_{1}^{h_{1}} ∩ … ∩ Q_{s_{1}}^{h_{1}}.

By induction hypothesis, for each i ≤ r −2, M_{i} has the following primary decomposition

M_{i} = ⋂_{2≤j≤r−i}(Q_{1}^{h_{j}} ∩ M_{r−1}) ∩ … ∩ (Q_{s_{j}}^{h_{j}} ∩ M_{r−1}) = ⋂_{1≤j≤r−i} Q_{1}^{h_{j}} ∩ … ∩ Q_{s_{j}}^{h_{j}}

which proves the theorem.

References

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[BH] Bruns, W., Herzog, J. Cohen-Macaulay rings, vol. 39, Cambridge studies in advanced mathematics, revised edition, 1998.

[BW] Björner, A., Wachs, M.L. Shellable nonpure complexes and posets, I. Trans. Amer. Math. Soc. 348 (1996), no. 4, 1299–1327.

[D] Duval, A.M. Algebraic shifting and sequentially Cohen-Macaulay simplicial complexes, Electron. J. Combin. 3 (1996), no. 1, Research Paper 21.

[F] Faridi, S. Simplicial trees are sequentially Cohen-Macaulay, J. Pure and Applied Algebra, Volume 190, Issues 1-3, Pages 121-136 (June 2004).

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[St] Stanley, R.P. Combinatorics and commutative algebra, Second edition, Progress in Mathematics 41, Birkhuser Boston, Inc., Boston, MA, 1996.