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I'm trying to understand the proof of Corollary $1$ on p. $16$. It says

Corollary $1$. Let $A$ be a ring integrally closed in its quotient field $K$. Let $L$ be a finite Galois extension of $K$, and $B$ the integral closure of $A$ in $L$. Let $\mathfrak{p}$ be a maximal ideal of $A$. Let $\phi:A \rightarrow A/\mathfrak{p}$ be the canonical homomorphism, and let $\psi_1, \psi_2$ be two homomorphisms of $B$ extending $\phi$ in a given algebraic closure of $A/\mathfrak{p}$. Then there exists an automorphism $\sigma$ of $L$ over $K$ such that $$\psi_1=\psi_2 \circ \sigma.$$

He shows that

Without loss of generality, we may therefore assume that $\psi_1,\psi_2$ have the same kernel $\mathfrak{P}$.

which I'm happy with. He then says

Hence there exists an automorphism $\omega$ of $\psi_1(B)$ onto $\psi_2(B)$ such that $\omega \circ \psi_1 =\psi_2$.

I think we get this automorphism by observing that $$\psi_1(B)\cong B/\mathfrak{P} \cong \psi_2(B)$$ and then using the induced isomorphisms $\overline{\psi_1}$, $\overline{\psi_2}$ to get $$\omega=\overline{\psi_2}\overline\psi_1^{-1}.$$ The he writes

There exists an element $\sigma$ of $G_{\mathfrak{P}}$ such that $\omega \circ \psi_1=\psi_1 \circ \sigma$, by the preceding proposition.

which I don't follow. The preceding proposition is

Proposition 14. Let $A$ be integrally closed in its quotient field $K$, and let $B$ be its integral closure in a finite Galois extension $L$ of $K$, with group $G$. Let $\mathfrak{p}$ be a maximal ideal of $A$, and $\mathfrak{P}$ a maximal ideal of $B$ lying above $\mathfrak{p}$. Then $B/\mathfrak{P}$ is a normal extension of $A/\mathfrak{p}$, and the map $\sigma \mapsto \overline \sigma$ induces a homomorphism of $G_\mathfrak{P}$ onto the Galois group of $B/\mathfrak{P}$ over $A/\mathfrak{p}$.

$G_\mathfrak{P}=\{\tau \in \text{Gal}(L/K|\tau \mathfrak{P}=\mathfrak{P}\}$

So what I think I need to show is that there is an element $\overline{\sigma} \in\text{Gal}((B/\mathfrak{P})/(A/\mathfrak{p}))$ which needs to be induced by $\sigma$. However, the image of $\psi_1 \circ \sigma$ is $\psi_1(B)$, whereas the image of $\omega \circ \psi_1$ is $\psi_2(B)$, so I don't see how these maps can be equal.

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By proposition 14, $B/\mathfrak P$ is a normal extension of $A/\mathfrak p$, so the images of $\psi_1$ and $\psi_2$ inside the algebraic closure of $A/\mathfrak p$ are actually the same: $\psi_1(B) = \psi_2(B)$. That's why $\omega: \psi_1(B) \to \psi_2(B)$ is called an automorphism rather than an isomorphism. So $\omega$ is an element of $\operatorname{Gal}(B/\mathfrak P | A/\mathfrak p)$, hence it comes from an element $\sigma \in \operatorname{Gal}(L|K)$ with $\sigma \mathfrak P = \mathfrak P$. Viewing $\psi_1: B \to \psi_1(B) \cong B/\mathfrak P$ as the projection map inducing the surjection $\operatorname{Gal}(L|K)_{\mathfrak P} \to \operatorname{Gal}(B/\mathfrak P | A/\mathfrak p)$, this means $\omega \circ \psi_1 = \psi_1 \circ \sigma$.

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