I was reading in Lang's Algebraic Number Theory (Second Edition page 15-16) and the following proposition occured.

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 $\mathfrak{B}$ a maximal ideal of $B$ lying above $\mathfrak{p}$. Then $B/\mathfrak{B}$ is a normal extension of $A/\mathfrak{p}$, and the map $\sigma \mapsto \overline{\sigma}$ induces a homomorphism of $G_\mathfrak{B}$ onto the Galois group of $B/\mathfrak{B}$ over $A/\mathfrak{p}$.

I worked through the proof but now I'm trying to understand the following corollary to this proposition.

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 $\phi : A \to A/\mathfrak{p}$ 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 $$\psi_1 = \psi_2 \circ \sigma.$$

Proof. The kernels of $\psi_1$, $\psi_2$ are prime ideals of B which are conjugate by Proposition 11 (proposition 11 is exactly this). Hence there exists an element $\tau$ of the Galois group $G$ such that $\psi_1$, $\psi_2 \circ \tau$ have the same kernel. Without loss of generality, we may therefore assume that $\psi_1$, $\psi_2$ have the same kernel $\mathfrak{B}$.

Up to this point, I understand the proof.

Hence there exists an automorphism $\omega$ of $\psi_1(B)$ onto $\psi_2(B)$ such that $\omega \circ \psi_1 = \psi_2$. Is this the case because $\psi_1(B) \cong B/\mathfrak{B} \cong \psi_1(B)$?

There exists an element $\sigma$ of $G_\mathfrak{B}$ such that $\omega \circ \psi_1 = \psi_1 \circ \sigma$, by the preceding proposition. Why does such a $\sigma$ exist? My rough idea is that $\omega$ corresponds to a $\sigma$ that does actually the same but 'before' $\psi_1$. Can someone work out the details of the proof? I have a rough idea of what's happening but the proof didn't convince me.

Then, there's another corollary.

Corollary 2. Let the assumptions be as in Corollary 1, and assume that $\mathfrak{B}$ is the only prime of $B$ lying above $\mathfrak{p}$. Let $f(X)$ be a polynomial in $A[X]$ with leading coefficient 1. Assume that f is irreducible in $K[X]$, and has a root $\alpha$ in $B$. then the reduced polynomial $\overline{f}$ is a power of an irreducible polynomial in $\overline{A}[X]$.

Proof. By Corollary 1, we know that any two roots of $\overline{f}$ are conjugate under some isomorphism of $\overline{B}$ over $\overline{A}$ (How does this follow from the previous corollary?), and hence that $\overline{f}$ cannot split into relative prime polynomials. Therefore, $\overline{f}$ is a power of an irreducible polynomial.


The first part of your question has already been answered here: Proof on p. $16 \;$ of Lang's Algebraic Number Theory

As for the second corollary:

If $\alpha$ is a root of $f(x)$ in $B$, then since $L/K$ is Galois and therefore normal, $f(x)$ has all its roots in $L$ and clearly (since $f$ is monic) in $B$.

Moreover, we have an automorphism $\sigma$ of $L$ which takes $\alpha$ onto $\beta$ from basic field theory. Since any automorphism of $L$ takes $\mathfrak{P}$ onto a prime ideal lying above $p$, it has to fix $\mathfrak{P}$ and therefore it comes from the decomposition group of $\mathfrak{P}$.

This is why $\bar{\sigma}$ induces an automorphism of $B/\mathfrak{P}$ which takes $\bar{\alpha}$ to $\bar{\beta}$.


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