Lemma 49.2.12. Let $A \to B$ be a quasi-finite homomorphism of Noetherian rings. Let $\omega _{B/A}^\bullet \in D(B)$ be the algebraic relative dualizing complex discussed in Dualizing Complexes, Section 47.25. Then there is a (nonunique) isomorphism $\omega _{B/A} = H^0(\omega _{B/A}^\bullet )$.

**Proof.**
Choose a factorization $A \to B' \to B$ where $A \to B'$ is finite and $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(B)$ is an open immersion. Then $\omega _{B/A}^\bullet = \omega _{B'/A}^\bullet \otimes _ B^\mathbf {L} B'$ by Dualizing Complexes, Lemmas 47.24.7 and 47.24.9 and the definition of $\omega _{B/A}^\bullet $. Hence it suffices to show there is an isomorphism when $A \to B$ is finite. In this case we can use Dualizing Complexes, Lemma 47.24.8 to see that $\omega _{B/A}^\bullet = R\mathop{\mathrm{Hom}}\nolimits (B, A)$ and hence $H^0(\omega ^\bullet _{B/A}) = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ as desired.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)