Definitions: $f$ is separable if every irreducible factor has distinct roots. $E/F$ is a Galois extension if the fixed field of the Galois group Gal$(E/F)$ is $F$
I would like to prove the following statement:
If $f \in F[X]$ is a separable polynomial then the splitting field $E$ is Galois over $F$
I just proved this claim in the case $f$ is irreducible itself. What about if $f$ has more than one irreducible factor?
My attempt: induction on the number of irreducible factors. Let $f=f_1f_2…f_n$ a factorization in irreducible factors $f_i \in F[X]$. Let $E_i$ be the splitting field of $f_i$. The splitting field $E$ of $f$ is the composite $E_1E_2…E_n$. By induction $E_1$ and $E_2…E_n$ are Galois over $F$. Now I have to prove that $E_1E_2…E_n$ is Galois over F. If $E_1 \cap (E_2…E_n)=F$ is done but this is not true in general.
Best Answer
If you want to avoid the primitive element theorem : For each root $\beta$ of $f \in F[x]$ your separable polynomial whose $E$ is the splitting field, if $\beta \not \in F$ then it has a distinct $F$-conjugate $\gamma$, let $\sigma : F(\beta) \to F(\gamma)$ be the natural field homomorphism, it can be extended to an homomorphism $\sigma:E \to \sigma(E) \subset \overline{E}$, since $E/F$ is normal then $\sigma(E) = E$ and hence $\sigma\in Gal(E/F)$ and $\beta \not \in E^{Gal(E/F)}$.
That is to say $E^{Gal(E/F)}=F$ and $E/F$ is Galois.