The Stacks project

Lemma 47.16.7. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring with normalized dualizing complex $\omega _ A^\bullet $. Let $M$ be a finite $A$-module. The following are equivalent

  1. $M$ is Cohen-Macaulay,

  2. $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(M, \omega _ A^\bullet )$ is nonzero for at most one $i$,

  3. $\mathop{\mathrm{Ext}}\nolimits ^{-i}_ A(M, \omega _ A^\bullet )$ is zero for $i \not= \dim (\text{Supp}(M))$.

Denote $CM_ d$ the category of finite Cohen-Macaulay $A$-modules of depth $d$. Then $M \mapsto \mathop{\mathrm{Ext}}\nolimits ^{-d}_ A(M, \omega _ A^\bullet )$ defines an anti-auto-equivalence of $CM_ d$.

Proof. We will use the results of Lemma 47.16.5 without further mention. Fix a finite module $M$. If $M$ is Cohen-Macaulay, then only $\mathop{\mathrm{Ext}}\nolimits ^{-d}_ A(M, \omega _ A^\bullet )$ can be nonzero, hence (1) $\Rightarrow $ (3). The implication (3) $\Rightarrow $ (2) is immediate. Assume (2) and let $N = \mathop{\mathrm{Ext}}\nolimits ^{-\delta }_ A(M, \omega _ A^\bullet )$ be the nonzero $\mathop{\mathrm{Ext}}\nolimits $ where $\delta = \text{depth}(M)$. Then, since

\[ M[0] = R\mathop{\mathrm{Hom}}\nolimits _ A(R\mathop{\mathrm{Hom}}\nolimits _ A(M, \omega _ A^\bullet ), \omega _ A^\bullet ) = R\mathop{\mathrm{Hom}}\nolimits _ A(N[\delta ], \omega _ A^\bullet ) \]

(Lemma 47.15.3) we conclude that $M = \mathop{\mathrm{Ext}}\nolimits _ A^{-\delta }(N, \omega _ A^\bullet )$. Thus $\delta \geq \dim (\text{Supp}(M))$. However, since we also know that $\delta \leq \dim (\text{Supp}(M))$ (Algebra, Lemma 10.72.3) we conclude that $M$ is Cohen-Macaulay.

To prove the final statement, it suffices to show that $N = \mathop{\mathrm{Ext}}\nolimits ^{-d}_ A(M, \omega _ A^\bullet )$ is in $CM_ d$ for $M$ in $CM_ d$. Above we have seen that $M[0] = R\mathop{\mathrm{Hom}}\nolimits _ A(N[d], \omega _ A^\bullet )$ and this proves the desired result by the equivalence of (1) and (3). $\square$

Comments (5)

Comment #3585 by Kestutis Cesnavicius on

In part (3) of the statement, I think the exponent should be .

Comment #8491 by Haohao Liu on

Strictly speaking, part (2) should be rephrased as "for at most one " to include the trivial case .

Comment #8492 by Haohao Liu on

Another issue is that, should the zero module be called Cohen-Macaulay?

Comment #9103 by on

Yes, the zero module is Cohen-Macaulay. Thanks for pointing out the issue with the statement. I fixed the statement as you suggested here.

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