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I have a version of natural irrationalities theorem that states:

Let $F/K$ be a extension of fields. Let's denote by $Gal(f/K)$ the Galois group of the extension $K(\alpha_1,\ldots,\alpha_n)$ for $\alpha_i$ all the roots of the separable polynomial $f \in K[X]$. Similarly for $Gal(f/F)$. Then there is a group monomorphism $Gal(f/F) \to Gal(f/K)$.

From this I should deduce the following corollary:

If Gal(f/K) is simple then for each extension F/K, Gal(f/F) = Gal(f/K) or is trivial.

How can I prove that $Gal(f/F)$ needs to be normal in $Gal(f/K)$?

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  • $\begingroup$ I think it'd be a good idea to write down what{s the relation between the (fields...?) $\;F,\,K$ ....And are $\;\alpha_i\;$ all the roots of $\;f\;$ ? Is $\;f\;$ an element of $\;F[x]\;$ ? Lots of info lacking $\endgroup$ – DonAntonio Dec 15 '17 at 23:54
  • $\begingroup$ @DonAntonio all the things you asked are now in the question, do you think is complete now? $\endgroup$ – Javier Dec 15 '17 at 23:56
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This isn't true. Indeed, by the fundamental theorem of Galois theory, if $H\subseteq Gal(f/K)$ is any subgroup, then the fixed field $F=K(\alpha_1,\dots,\alpha_n)^H$ satisfies $Gal(f/F)=H$. So to get the result you ask for, you would need $Gal(f/K)$ to have no nontrivial proper subgroups, not just on nontrivial proper normal subgroups (this is equivalent to $Gal(f/K)$ being abelian as well as simple).

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  • $\begingroup$ would this situation change if i consider $Gal(f/K)$ as the corresponding subgroup of permutations? $\endgroup$ – Javier Dec 16 '17 at 1:07
  • $\begingroup$ No, that's an isomorphic group so nothing changes. $\endgroup$ – Eric Wofsey Dec 16 '17 at 1:11

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