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 $.
$\square$

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