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. $\square$

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