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Let $N/K$ be a finite Galois extension such that $G = Gal(N/K)$ is an abelian group, and let $M$ be an intermediate field of $N/K$. Show that $M/K$ is normal and that $Gal(M/K)$ is abelian.

Here is what I have: Let $\alpha \in N$ have the minimal polynomial $f(x)$ over $K$, and the minimal polynomial $g(x)$ over $M$. Then $g(x)$ divides $f(x)$. Now as $N/K$ is normal and has one root $\alpha \in N$ we know that $f(x)$ splits into linear factors over $N$ and that these factors are all distinct. Therefore $g(x)$ splits into distinct factors in $N$, meaning $N/M$ is normal and separable.

Also, if $\alpha \in M$ then $f(x)$ can't have a repeated root in $M$ since it doesn't in $N$, meaning $M/K$ is separable.

To prove that $M/K$ is normal we need to show that $M$ contains all of the roots of $f(x)$, but I'm not sure where to go from here.

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  • $\begingroup$ Do you have the fundamental theorem of Galois theory at your disposal? $\endgroup$ Mar 18, 2016 at 1:52
  • $\begingroup$ @AlexWertheim Yes but I'm not sure how to use it here $\endgroup$
    – user323857
    Mar 18, 2016 at 1:54

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Hint: it is always the case that $N/M$ is Galois, as you observe. The fundamental theorem of Galois theory says that the extension $M/K$ is Galois if and only if $\mathrm{Gal}(N/M)$ is a normal subgroup of $\mathrm{Gal}(N/K)$. Furthermore, it tells you that $\mathrm{Gal}(N/K)/\mathrm{Gal}(N/M) \cong \mathrm{Gal}(M/K)$. Can you see how to use the information you're given now?

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  • $\begingroup$ I'm still confused as to what the next step should be. Am I now trying to show that $Gal(N/M)$ is normal? $\endgroup$
    – user323857
    Mar 18, 2016 at 2:16
  • $\begingroup$ Yes. $\mathrm{Gal}(N/M)$ is a subgroup of the abelian group $\mathrm{Gal}(N/K)$. What is true of any subgroup of an abelian group? $\endgroup$ Mar 18, 2016 at 2:32

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