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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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