Attached is prop 14 on page 15 of Lang's 'algebraic number theory' 2ed. Note, $G_\mathfrak{B} \subset G$ is the subgroup fixing $\mathfrak{B}$
Can someone perhaps add a sentence or two to explain that $\bar{B}$ is normal over $\bar{A}$.
As I see it, he's taken a separable element and shown that all its conjugates lie in $\bar{B}$. However, for a general element $\bar{y} \in \bar{B}$, we might only say that some power $\bar{x} := \bar{y}^{p^\mu}$ is separable. The minimal polynomial for $\bar{y}$ over $\bar{A}$ will look like $$ f(X) = h(X^{p^\mu}) = (X^{p^\mu}-\bar{x})(X^{p^\mu}-\bar{x}_2)\dots(X^{p^\mu}-\bar{x}_n). $$ So, if $(X^{p^\mu}-\bar{x})$ splits into linear factors in $\bar{B}[X]$ why should the others as well?
Wlog assume $K$ is complete $\mathfrak{p}$-adically, thus $(A,\mathfrak{p})$ and $(B,\mathfrak{P})$ are complete DVRs.
Let $P \in A[X]$ be irreducible mod $\mathfrak{p}$, and assume that it has a root in $B/\mathfrak{P}$. Then in $B/\mathfrak{P}$, we can write $P=LQ$ with $L,Q$ coprime polynomials, and $L$ is a power of a linear factor. By Hensel the decomposition lifts to $B$, ie $P=LQ$ with $L,Q \in B[X]$ and $L$ is, mod $\mathfrak{p}$, a power of a linear factor. In particular, one of the irreducible divisors of $P$ is, mod $\mathfrak{P}$, a power of a linear factor.
Now, $P \in A[X]$ is irreducible, so the Galois group acts transitively on the set of its irreducible factors, so all of its irreducible factors are, mod $\mathfrak{P}$, powers of linear factors. Thus $P$ is split in $B/\mathfrak{P}$.