The Stacks project

Lemma 47.16.1. Let $(A, \mathfrak m, \kappa ) \to (B, \mathfrak m', \kappa ')$ be a finite local map of Noetherian local rings. Let $\omega _ A^\bullet $ be a normalized dualizing complex. Then $\omega _ B^\bullet = R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$ is a normalized dualizing complex for $B$.

Proof. By Lemma 47.15.8 the complex $\omega _ B^\bullet $ is dualizing for $B$. We have

\[ R\mathop{\mathrm{Hom}}\nolimits _ B(\kappa ', \omega _ B^\bullet ) = R\mathop{\mathrm{Hom}}\nolimits _ B(\kappa ', R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )) = R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa ', \omega _ A^\bullet ) \]

by Lemma 47.13.1. Since $\kappa '$ is isomorphic to a finite direct sum of copies of $\kappa $ as an $A$-module and since $\omega _ A^\bullet $ is normalized, we see that this complex only has cohomology placed in degree $0$. Thus $\omega _ B^\bullet $ is a normalized dualizing complex as well. $\square$

Comments (0)

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Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AX1. Beware of the difference between the letter 'O' and the digit '0'.