Splitting field of separable polynomial is Galois extension

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.

• Incidentally, that is not the standard definition of a separable polynomial (the standard definition is just that all the roots of $f$ in a splitting field are distinct). The weaker definition you're using is sufficient for this result, though. – Eric Wofsey Feb 24 at 18:00
• You could cheat, and note that since $E/F$ is finite and separable, then by the primitive element theorem, $E = F(\alpha)$ for some $\alpha \in E$. Then apply your theorem to the minimal polynomial of $\alpha$. – Mathmo123 Feb 25 at 15:20
• @Mathmo123 but... In Weintraub’s ‘Galois theory’ primitive element theorem is a conseguence of the statement which I want to prove. – Leonardo Vannini Feb 26 at 21:54
• @LeonardoVannini Interesting... one shouldn't need any Galois theory to prove the primitive element theorem (see this proof for example). – Mathmo123 Feb 27 at 14:45

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.
• Are you assuming $E/F$ separable? Why we can assume that? – Leonardo Vannini Feb 26 at 23:39
• @LeonardoVannini In characteristic $0$ that's direct by looking at $\gcd(m,m')$, in characteristic $p$ to show $f$ is separable implies so is the minimal polynomial of $\beta$ is much less obvious, I would need symmetric polynomials plus several lemmas. Now here all you have to do is replacing $\beta$ by the roots of $f$, see my edit – reuns Feb 26 at 23:53