Lemma 47.15.12. Let $A$ be a Noetherian ring. Let $\omega _ A^\bullet$ be a dualizing complex. Let $\mathfrak m \subset A$ be a maximal ideal and set $\kappa = A/\mathfrak m$. Then $R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa , \omega _ A^\bullet ) \cong \kappa [n]$ for some $n \in \mathbf{Z}$.

Proof. This is true because $R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa , \omega _ A^\bullet )$ is a dualizing complex over $\kappa$ (Lemma 47.15.9), because dualizing complexes over $\kappa$ are unique up to shifts (Lemma 47.15.5), and because $\kappa$ is a dualizing complex over $\kappa$. $\square$

There are also:

• 1 comment(s) on Section 47.15: Dualizing complexes

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).