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The proposition is as follows :

Let $A$ be integrally closed in its quotient field $K$, and $B$ be integral closure of $A$ in finite Galois extension $L$ of $B$, with group $G$. Let $\mathfrak p $ be maximal ideal of $A$, $\mathfrak q$ be maximal ideal of B lying above $\mathfrak p $. Then $B/\mathfrak{q} $ is normal over $A/\mathfrak p$, and mapping $\sigma \mapsto \bar{\sigma} $ is homomorphism of $G_{\mathfrak q}$ (decomposition group of $\mathfrak q $) onto the Galois group of $B/\mathfrak q $ over $A/\mathfrak p $.

I have no problem with the part proving $\bar{B}/\bar{A} $ is normal extension. As the automorphism group is referred to as 'Galois group of $\bar{B}/\bar{A} $', I assume the proposition means that $\bar{B}/\bar{A} $ is normal and separable extension, but it seems that the separability part of the proof doesn't appear in the text.

I find it hard to prove the separability myself and actually it seems that it is not generally separable.

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The separability does not always hold. It is true, however, if the residue field $A/\mathfrak{p}$ is finite.

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