The Stacks project

Lemma 15.110.7. Let $R$ be a ring. Let $G$ be a finite group acting on $R$. Let $R^ G \to A$ be a ring map. The map

\[ A \to (A \otimes _{R^ G} R)^ G \]

is an isomorphism if $R^ G \to A$ is flat. In general the map is integral, induces a homeomorphism on spectra, and induces purely inseparable residue field extensions.

Proof. To see the first statement consider the exact sequence $0 \to R^ G \to R \to \bigoplus _{\sigma \in G} R$ where the second map sends $x$ to $(\sigma (x) - x)_{\sigma \in G}$. Tensoring with $A$ the sequence remains exact if $R^ G \to A$ is flat. Thus $A$ is the $G$-invariants in $(A \otimes _{R^ G} R)^ G$.

The second statement follows from Lemma 15.110.6 and Algebra, Lemma 10.46.11. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 15.110: Group actions and integral closure

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Your email address will not be published.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



View Lemma 15.110.7 as pdf