Lemma 15.98.5. Let $G$ be a finite group acting on a ring $R$. For any two primes $\mathfrak q, \mathfrak q' \subset R$ lying over the same prime in $R^ G$ there exists a $\sigma \in G$ with $\sigma (\mathfrak q) = \mathfrak q'$.

**Proof.**
The extension $R^ G \subset R$ is integral because every $x \in R$ is a root of the monic polynomial $\prod _{\sigma \in G}(T - \sigma (x))$ in $R^ G[T]$. Thus there are no inclusion relations among the primes lying over a given prime $\mathfrak p$ (Algebra, Lemma 10.35.20). If the lemma is wrong, then we can choose $x \in \mathfrak q'$, $x \not\in \sigma (\mathfrak q)$ for all $\sigma \in G$. See Algebra, Lemma 10.14.2. Then $y = \prod _{\sigma \in G} \sigma (x)$ is in $R^ G$ and in $\mathfrak p = R^ G \cap \mathfrak q'$. On the other hand, $x \not\in \sigma (\mathfrak q)$ for all $\sigma $ means $\sigma (x) \not\in \mathfrak q$ for all $\sigma $. Hence $y \not\in \mathfrak q$ as $\mathfrak q$ is a prime ideal. This is impossible as $y \in \mathfrak p \subset \mathfrak q$.
$\square$

## 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Your email address will not be published.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)