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The following proposition comes from Lang's Algebraic Number Theory on page $12$:

Let $A$ be a ring integrally closed in its field of fractions $K$. Let $L$ be a finite Galois extension of $K$ with $G := \operatorname{Gal}(L/K)$. Let $\mathfrak{p}$ be a maximal ideal of $A$ and let $\mathfrak{P}, \mathfrak{O}$ be prime ideals of the integral closure $B$ of $A$ in $L$ which lie above $\mathfrak{p}$. Then there exists a $\sigma \in G$ such that $\sigma \mathfrak{P} = \mathfrak{O}$.

I understand this but I need some clarification on a corollary given on the next page. Namely

Let $A$ be a ring integrally closed in its field of fractions $K$. Let $E$ be a finite separable extension of $K$ and $B$ the integral closure of $A$ in $E$. Let $\mathfrak{p}$ be a maximal ideal of $A$. Then there exist only finitely many prime ideals of $B$ lying above $\mathfrak{p}$.

The proof is as follows:

Let $L$ be the smallest Galois extension of $K$ containing $E$, let $\mathfrak{O}_1, \mathfrak{O}_2$ be distinct prime ideals of $B$ lying above $\mathfrak{p}$ and let $\mathfrak{P}_1, \mathfrak{P}_2$ be prime ideals of the integral closure of $A$ in $L$ lying above $\mathfrak{O}_1$ and $\mathfrak{O}_2$ respectively. Then $\mathfrak{P}_1 \neq \mathfrak{P}_2$. This argument reduces our assertion to the case where $E$ is Galois over $K$, and then it becomes an immediate consequence of the proposition.

I understand that if $E$ is Galois over $K$, the proposition makes the assertion obvious; since we can always find a $\sigma \in \operatorname{Gal}(E/K)$ such that for some pair of prime ideals $\mathfrak{P}$ and $\mathfrak{O}$, we have $\sigma \mathfrak{P} = \mathfrak{O}$, and since $\operatorname{Gal}(E/K)$ is finite, there can only be finitely many such $\sigma$. I just don't understand how Lang concludes that $\mathfrak{P}_1 \neq \mathfrak{P}_2$ in the corollary and so I don't understand why we can always reduce to the case where $E$ is Galois over $K$.

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If $\mathfrak{P}_1=\mathfrak{P}_2$, then $\mathfrak{O}_1 = B \cap \mathfrak{P}_1 = B \cap \mathfrak{P}_2 = \mathfrak{O}_2$, but $\mathfrak{O}_1$ and $\mathfrak{O}_2$ are assumed to be distinct.

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