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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.

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    $\begingroup$ 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. $\endgroup$ Feb 24, 2019 at 18:00
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    $\begingroup$ 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 $\alpha$. $\endgroup$
    – Mathmo123
    Feb 25, 2019 at 15:20
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    $\begingroup$ @Mathmo123 but... In Weintraub’s ‘Galois theory’ primitive element theorem is a conseguence of the statement which I want to prove. $\endgroup$
    – ictibones
    Feb 26, 2019 at 21:54
  • $\begingroup$ @LeonardoVannini Interesting... one shouldn't need any Galois theory to prove the primitive element theorem (see this proof for example). $\endgroup$
    – Mathmo123
    Feb 27, 2019 at 14:45

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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.

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  • $\begingroup$ Are you assuming $E/F$ separable? Why we can assume that? $\endgroup$
    – ictibones
    Feb 26, 2019 at 23:39
  • $\begingroup$ @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 $\endgroup$
    – reuns
    Feb 26, 2019 at 23:53

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