The Stacks project

Lemma 48.25.1. Let $X$ be a locally Noetherian scheme over the field $k$. Let $k'/k$ be a finitely generated field extension. Let $x \in X$ be a point, and let $x' \in X_{k'}$ be a point lying over $x$. Then we have

\[ \mathcal{O}_{X, x}\text{ is Gorenstein} \Leftrightarrow \mathcal{O}_{X_{k'}, x'}\text{ is Gorenstein} \]

If $X$ is locally of finite type over $k$, the same holds for any field extension $k'/k$.

Proof. In both cases the ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{X_{k'}, x'}$ is a faithfully flat local homomorphism of Noetherian local rings. Thus if $\mathcal{O}_{X_{k'}, x'}$ is Gorenstein, then so is $\mathcal{O}_{X, x}$ by Dualizing Complexes, Lemma 47.21.8. To go up, we use Dualizing Complexes, Lemma 47.21.8 as well. Thus we have to show that

\[ \mathcal{O}_{X_{k'}, x'}/\mathfrak m_ x \mathcal{O}_{X_{k'}, x'} = \kappa (x) \otimes _ k k' \]

is Gorenstein. Note that in the first case $k \to k'$ is finitely generated and in the second case $k \to \kappa (x)$ is finitely generated. Hence this follows as property (A) holds for Gorenstein, see Dualizing Complexes, Lemma 47.23.1. $\square$


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Your email address will not be published.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 0C03. Beware of the difference between the letter 'O' and the digit '0'.