The Stacks project

Lemma 47.16.4. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring with normalized dualizing complex $\omega _ A^\bullet $. Let $E$ be an injective hull of $\kappa $. Then there exists a functorial isomorphism

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

for $N$ running through the finite length $A$-modules.

Proof. By induction on the length of $N$ we see that $R\mathop{\mathrm{Hom}}\nolimits _ A(N, \omega _ A^\bullet )$ is a module of finite length sitting in degree $0$. Thus $R\mathop{\mathrm{Hom}}\nolimits _ A(-, \omega _ A^\bullet )$ induces an anti-equivalence on the category of finite length modules. Since the same is true for $\mathop{\mathrm{Hom}}\nolimits _ A(-, E)$ by Proposition 47.7.8 we see that

\[ N \longmapsto \mathop{\mathrm{Hom}}\nolimits _ A(R\mathop{\mathrm{Hom}}\nolimits _ A(N, \omega _ A^\bullet ), E) \]

is an equivalence as in Lemma 47.16.3. Hence it is isomorphic to the identity functor. Since $\mathop{\mathrm{Hom}}\nolimits _ A(-, E)$ applied twice is the identity (Proposition 47.7.8) we obtain the statement of the lemma. $\square$

Comments (2)

Comment #3231 by Niels on

The proof uses Matlis duality that is stated for complete local Noetherian rings only. Is the reduction to the complete case obvious ?

Comment #3245 by on

Yes, because finite length -modules are the same thing as finite length modules over the completion. However, this type of argument is used a lot and we should explain it somewhere carefully and then refer to it. Maybe we should also restate Matlis duality for arbitrary local rings (where on one side we look at finite modules over the completion).

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