The Stacks project

Lemma 15.110.4. Let $R$ be a ring. Let $G$ be a finite group of order $n$ acting on $R$. Let $J \subset R^ G$ be an ideal. Then $R^ G/J \to (R/JR)^ G$ is ring map such that

  1. for $b \in (R/JR)^ G$ there is a monic polynomial $P \in R^ G/J[T]$ whose image in $(R/JR)^ G[T]$ is $(T - b)^ n$,

  2. for $a \in \mathop{\mathrm{Ker}}(R^ G/J \to (R/JR)^ G)$ we have $(T - a)^ n = T^ n$ in $R^ G/J[T]$.

In particular, $R^ G/J \to (R/JR)^ G$ is an integral ring map which induces homeomorphisms on spectra and purely inseparable extensions of residue fields.

Proof. Part (1) follow from Lemma 15.110.1 with $I = JR$. If $a$ is as in part (2), then $a$ is the image of $x \in R^ G \cap JR$. Hence $(T - x)^ n = \prod _{\sigma \in G} (T - \sigma (x))$ is congruent to $T^ n$ modulo $J$ by Lemma 15.110.3. This proves part (2). To see the final statement we may apply 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Your email address will not be published.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 0H35. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 15.110.4 as pdf