Let $F$ be a field and $E/F$ a galois extension. Let $f\in F[x]$ be an irreducible polynomial. I'm trying to prove that all the irreducible factors of $f(x)$ over $E[x]$ have the same degree.
If $\sigma \in Gal(E/F)$ and if $e(x)\in E[x]$ it's an irreducible then $\sigma(e(x))$ it's another factor of $f(x)$. And they have the same degree. It's enough to prove that all the irreducible factors of $f(x)$ over $E$ are of this form. Given two irreducible factors $ e(x),e'(x)$ of $f(x)$ over $E[x]$. Let $\alpha$ be a root of $e(x)$ and let $\alpha'$ be a root of $e'(x)$. I proved that there exist $\sigma \in Gal(E/F)$ such that $\sigma(\alpha)=\alpha'$. I think that this implies $\sigma(e(x))=e'(x)$ because they share a root. But I don't know exactly the reason of this. Can someone explain it to me )=?
Maybe the new question is this:
Let $E/F$ be a Galois extension and let $f(x)\in F[x]$ be an irreducible polynomial. Let $e(x),e'(x)$ two factors of $f(x)$ over $E[x]$ that share a root $\alpha \in \overline{F}$ then $e(x)=e'(x)$.
First, let me point out that it doesn't make sense to say that there exists $\sigma \in G(E/F)$ such that $\sigma \alpha = \alpha'$, because the root $\alpha$ lives in $\overline{E}$, not necessarily in $E$.
Here is how I suggest you prove this: write $f(x) = \prod_{i=1}^n e_i(x)$ for the factorization of $f$ in $E[x]$. Let $G$ act on the set $\{e_i\}$. Let $S\subseteq \{e_i\}$ be an orbit for this action. Let $g(x) = \prod_{s \in S} s$. Show that $g$ is invariant under the action of $G(E/F)$, hence $g(x) \in F[x]$. Show that $f(x)/g(x) \in F[x]$, then use the fact that $f$ is irreducible to conclude that $g=f$, so that $G$ acts transitively on $\{e_i\}$.