Lemma 15.115.6. Let $A \to B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. Let $K_1/K$ be a normal extension. Say $G = \text{Aut}(K_1/K)$. Then $G$ acts on the rings $K_1$, $L_1$, $A_1$ and $B_1$ of Remark 15.114.1 and acts transitively on the set of maximal ideals of $B_1$.

**Proof.**
Everything is clear apart from the last assertion. If there are two or more orbits of the action, then we can find an element $b \in B_1$ which vanishes at all the maximal ideals of one orbit and has residue $1$ at all the maximal ideals in another orbit. Then $b' = \prod _{\sigma \in G} \sigma (b)$ is a $G$-invariant element of $B_1 \subset L_1 = (L \otimes _ K K_1)_{red}$ which is in some maximal ideals of $B_1$ but not in all maximal ideals of $B_1$. Lifting it to an element of $L \otimes _ K K_1$ and raising to a high power we obtain a $G$-invariant element $b''$ of $L \otimes _ K K_1$ mapping to $(b')^ N$ for some $N > 0$; in fact, we only need to do this in case the characteristic is $p > 0$ and in this case raising to a suitably large $p$-power $q$ defines a canonical map $(L \otimes _ K K_1)_{red} \to L \otimes _ K K_1$. Since $K = (K_1)^ G$ we conclude that $b'' \in L$. Since $b''$ maps to an element of $B_1$ we see that $b'' \in B$ (as $B$ is normal). Then on the one hand it must be true that $b'' \in \mathfrak m_ B$ as $b'$ is in some maximal ideal of $B_1$ and on the other hand it must be true that $b'' \not\in \mathfrak m_ B$ as $b'$ is not in all maximal ideals of $B_1$. This contradiction finishes the proof of the lemma.
$\square$

## 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Your email address will not be published.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)

There are also: