If $F/E$ is a finite Galois extension, then one can show that $F(X)/E(X)$ is also a finite Galois extension of the same degree (a basis for $F/E$ is also a basis for $F(X)/E(X)$). Since $E[X]$ is a Dedekind domain, what is its integral closure in $F(X)$? Is it $F[X]$?
The reason I'm asking is I'm trying to solve the following problem:
"Let $h$ be monic and irreducible in $E[X]$, and let $h_1, h_2 \in F[X]$ be monic irreducible divisors of $h$ in $F[X]$. Then there exists an automorphism $\theta$ of $F[X]$ which leaves $E[X]$ fixed and sends $h_1$ to $h_2$."
If $F[X]$ is in fact the integral closure of $E[X]$, then I can use the result that $(h_1), (h_2)$ are primes of $F[X]$ which lie over $(h)$, so there is a $\theta \in Gal(F(X)/E(X))$ with $\theta(h_1) = (h_2)$ (which is almost what I want).
Since $E\subset F$ is algebraic, $E[X]\subset F[X]$ is integral (why?). Moreover, $F[X]$ is integrally closed, so the integral closure of $E[X]$ in $F(X)$ is (as you hoped) $F[X]$.