I have recently seen the following result in one of my classes,
Theorem. Let $A$ be an integrally closed domain with fraction field $K$, and let $L/K$ be a finite Galois extension with $B$ the integral closure of $A$ in $L$. If $\mathfrak{p} \subset A$ is a maximal ideal and $\mathfrak{q}, \mathfrak{q}' \subset B$, with $\mathfrak{q} \cap A = \mathfrak{p} = \mathfrak{q}' \cap A$, there exists $\sigma \in \operatorname{Gal}(E/K)$ such that $\sigma(\mathfrak{q}) = \mathfrak{q}'$.
As a corollary,
Corollary. Let $A$ be an integrally closed domain with fraction field $K$, and let $L/K$ be a finite Galois extension with $B$ the integral closure of $A$ in $L$. If $\mathfrak{p} \subset A$ is a maximal ideal, there are finitely many primes $\mathfrak{q} \subset B$ such that $\mathfrak{q} \cap A = \mathfrak{p}$.
I have been then asked to proved the latter when $L/K$ is a separable but not necessarily normal extension. My first thought is to consider the normal closure $E$ of $L$, which will be a finite extension over $K$. I will note $B_E$ and $B_L$ for the integral closures of $A$ in $E$ and $L$ respectively.
By the corollary above, there are a finite amount of primes in the integral closure of $A$ in $E$ that contract to $\mathfrak{p}$. It suffices then to show that distinct 'lifts' of $\mathfrak{p}$ to $B_L$ can be 'lifted' once again to primes in $B_E$ to distinct primes that contract to $\mathfrak{p}$, since an infinite amount of lifts in $B_L$ would contradict the fact that there are a finite amount of lifts in $B_E$.
Now, the extension $B_L \subset B_E$ is integral: by definition, an element $b \in B_E$ has an integral equation with coefficients in $A \subset B_L$. Therefore, by the going-up theorem two primes $\mathfrak{q},\mathfrak{q}' \subset B_L$ that contract to $\mathfrak{p}$ can be lifted to primes $\mathfrak{r},\mathfrak{r}'$ such that $\mathfrak{r} \cap B_L = \mathfrak{q}$ and $\mathfrak{r'} \cap B_L = \mathfrak{q}'$ and thus
$$ \mathfrak{r'} \cap A = \mathfrak{r}' \cap B_L \cap A = \mathfrak{q}' \cap A = \mathfrak{q} \cap A = \mathfrak{r} \cap B_L \cap A = \mathfrak{r} \cap A. $$
Since $\mathfrak{q} \neq \mathfrak{q}'$, we know that $\mathfrak{r} \neq \mathfrak{r}'$, and this concludes the proof.
Is the above argument correct?
Thanks in advance.
Your proof is correct. But you have a typo: the prime $\mathfrak r$ is such that $\mathfrak r \cap B_L = \mathfrak q$. And similarly for $\mathfrak r'$.
Another way of proving the result, which uses essentially the same argument, is to show that the contraction map from prime ideals of $B_E$ to prime ideals of $B_L$ is surjective when restricted to the (finite) set of primes of $B_E$ lying above $\mathfrak p$ and correstricted to the set of primes of $B_L$ lying above $\mathfrak p$.
Best, Nicolás.