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I couldn't follow the proof of Corollary 1 on p.16 of Algebraic number theory by Lang.

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 $$\psi_1=\psi_2 \circ \sigma.$$

I found the post Proof on p. 16 of Lang's Algebraic Number Theory but I still cannot follow the given proof.

Here is the proof in the post.

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 $\omega \circ \psi_1 = \psi_1 \circ \sigma$.

First of all, I don't know why we can view $\psi_1:B \to \psi_1(B)\cong B/\mathfrak{P}$ as the projection map. The projection map used in the textbook is I though the canonical map $B \to B/\mathfrak{P}$.

Secondly, I know $\omega \in \operatorname{Gal}(B/\mathfrak P | A/\mathfrak p)$ comes from an element $\sigma \in \operatorname{Gal}(L|K)$ byProposition 14. Then isn't it $\omega=\psi_1\circ \sigma$ (if I assume my first question)?

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  • $\begingroup$ @DonAntonio I edited and added the statement. $\endgroup$ – Snow May 5 '17 at 20:57
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The canonical map $p: B \to B/\mathfrak{P}$ and the map $\psi_1: B \to \psi_1(B)$ are related by an isomorphism:

the situation

The isomorphism $\overline{\psi_1}: B/\mathfrak{P} \to \psi_1(B)$ is defined by $\psi_1 = \overline{\psi_1} \circ p$ (*). Given the automorphism $\omega: \psi_1(B) \to \psi_1(B)$ we get an automorphism $\tilde{\omega} : B/\mathfrak{P} \to B/\mathfrak{P}$ satisfying $\overline{\psi_1} \circ \tilde{\omega} = \omega \circ \overline{\psi_1}$. The property $\omega \circ \psi_1 = \psi_2$ implies that $\omega$ is the identity on $\varphi(A) = A/\mathfrak{p}$. The isomorphism $\overline{\psi_1}$ is the identity on $A/\mathfrak{p}$ as well (because $\psi_1$ extends $\varphi$), so $\tilde{\omega}$ is the identity on $A/\mathfrak{p}$. Hence $\tilde{\omega} \in \operatorname{Gal}(B/\mathfrak{P}|A/\mathfrak{p})$; the previous Proposition 14 is applied to $\tilde{\omega}$ to get $\sigma \in G_\mathfrak{P}$ such that $\tilde{\omega} \circ p = p \circ \sigma|_B$ (**). Now $\omega \circ \psi_1 = \psi_1 \circ \sigma|_B$ because both sides equal $\overline{\psi_1} \circ \tilde{\omega} \circ p$ (use the commutativity of the diagram).

(*) In terms of elements: $\overline{\psi_1}(\bar{x}) = \psi_1(x)$.

(**) In terms of elements, this reads $\tilde{\omega}(\overline{x}) = \overline{\sigma(x)}$.

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