Informal Seminar on Stanley-Reisner Theory, UMN, Fall 2002 12 December 2002

Shellable complexes are Cohen-Macaulay Speaker: Jeremy Martin

I’m going to prove the following (Theorem 5.1.13 in Bruns and Herzog):

Theorem: Let Δ be a shellable simplicial complex. Then the Stanley-Reisner ring k[Δ] is Cohen-Macaulay (for all fields k).

We need the following preliminary results. Let R be a ℕ^{n}-graded ring and I, J graded ideals of R. (We may as well take R = k[x_{1}, …, x_{n}] and I, J monomial ideals.)

Lemma 1: We have an exact sequence

0 → R/(I ∩ J) ⟶^{α} R/I ⊕ R/J ⟶^{β} R/(I + J) → 0.

where

α(a + (I + J)) = (a + I, a − J),
β(a + I, b + J) = (a + b) + (I + J).

Proof: α obviously has zero kernel, and β is surjective by the Chinese Remainder Theorem. Finally,

im α = I/(I ∩ J) ⊕ J/(I ∩ J)  and
ker β = (I + J)/I ⊕ (I + J)/J

which are equal by basic group theory. □

Lemma 2: If R/I and R/J are Cohen-Macaulay of dimension d and R/(I + J) is Cohen-Macaulay of dimension d − 1, then R/(I ∩ J) is Cohen-Macaulay of dimension d.

Proof: It is easy to check that R/I ⊕ R/J is Cohen-Macaulay of dimension d. Recall that H^{i}(M) is nonzero iff depth M ≤ i ≤ dim M (with equality if M is Cohen-Macaulay). So we want to show that

(1)  H^{i}(R/(I ∩ J) ≠ 0 ⟺ i = d.

Apply the long exact sequence of local cohomology to the short exact sequence of Lemma 1, obtaining (2) … → H^{i−1}(R/I⊕R/J) → H^{i−1}(R/(I+J)) → H^{i}(R/(I∩J) → H^{i}(R/I⊕R/J) → H^{i}(R/(I+J)) → …

If i ≠ d, then the second and fourth terms in (2) are zero, so H^{i}(R/(I ∩ J) = 0. If i = d, then the first and fifth terms are zero and the second and fourth terms are nonzero, so H^{i}(R/(I ∩ J) ≠ 0, giving (1). □

Proof of the theorem: Let Δ be shellable of dimension d − 1 on vertices {v_{1}, …, v_{n}}, with F_{1}, …, F_{m} a shelling order on the facets. That is, for all 1 ≤ j ≤ m + 1, the complex ⟨F_{j+1}⟩ ∩ Δ_{j} is generated by some nonempty set of maximal proper faces of F_{j+1}, where Δ_{j} = ⟨F_{1}, …, F_{j}⟩. For each facet F_{j}, define an ideal

P_{j} = (x_{i} : v_{i} ∉ F_{j});

it is easy to verify that P_{j} is prime and that

I_{Δ} = ⋂_{j=1}^{m} P_{j}.

We will show by induction that for every j, the ring k[Δ_{j}] is Cohen-Macaulay. If j = 1 then Δ_{j} = ⟨F_{1}⟩ is a simplex, so k[Δ_{j}] is the polynomial ring on the variables {x_{i} : v_{i} ∈ F_{1}}.

For j > 1, let I = I_{Δ_{j−1}} and J = P_{j}. Then by Lemma 1 we have a short exact sequence

(3)  0 → k[Δ_{j}] → k[Δ_{j−1}] ⊕ k[⟨F_{j}⟩] → k[⟨F_{j}⟩ ∩ Δ_{j−1}] → 0. 1

By the definition of shellability, ⟨F_{j}⟩ ∩ Δ_{j−1} = ⟨G_{1}, …, G_{l}⟩, where G_{k} = F_{j} \ {x_{h_{k}}} for 1 ≤ k ≤ l. Therefore

k[⟨F_{j}⟩ ∩ Δ_{j−1}] = k[x_{i} : v_{i} ∈ F_{j}] / ∏_{k=1}^{l} x_{h_{k}},

which is Cohen-Macaulay of dimension d − 1. Moreover, k[Δ_{j−1}] and k[⟨F_{j}⟩] are Cohen-Macaulay of dimension d (the former by induction, the latter because it is a polynomial ring). So k[Δ_{j}] is Cohen-Macaulay by Lemma 2. □