i was looking a proof of the following:

Let $k\subset F, F\subset K$ be Galois field extensions, then $Gal (K/F)$ is a normal subgroup of $Gal(K/k)$.

I understand the proof but it start using the fact that $k\subset K$ is Galois and i can't see it.

Is it trivial?

Thanks in advance.

  • $\begingroup$ Have you covered equivalent conditions of being Galois? Which ones are you familiar with? What's your main definition? $\endgroup$
    – anon
    Nov 28, 2013 at 0:57
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    $\begingroup$ Finite normal and separable extension. splitting field of a separable polynomial. The fixed field of $Gal(K/k)$ is $k$. $\endgroup$ Nov 28, 2013 at 1:02
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    $\begingroup$ Actually being Galois is not transitive. (That is, $L/F$ and $F/k$ being Galois does not imply $L/k$ is Galois.) Presumably, since the source speaks of ${\rm Gal}(K/k)$, it is assumed transitive. $\endgroup$
    – anon
    Nov 28, 2013 at 1:07
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    $\begingroup$ I think that this works perfectly if the statement say: Let $k\subset F \subset K$ fields extensions with $k\subset F$ and $k\subset K$ Galois. Because in this case is known that $F\subset K$ is Galois by the second definition i post above. $\endgroup$ Nov 28, 2013 at 1:11
  • $\begingroup$ Absolutely. ${}$ $\endgroup$
    – anon
    Nov 28, 2013 at 1:13

1 Answer 1


The relation "$L$ is Galois over $M$" is not transitive. For example if $n$ is a squarefree integer then the extensions $\Bbb Q(\sqrt[4]{n})/{\Bbb Q}(\sqrt{n})$ and ${\Bbb Q}(\sqrt{n})/\Bbb Q$ are both quadratic hence automatically Galois and yet the extension $\Bbb Q(\sqrt[4]{n})/\Bbb Q$ is not Galois (there are nonreal conjugates of $\sqrt[4]{n}$).

If $K/k$ is Galois and $K/F/k$ an intermediate field, then automatically $K/F$ is Galois (if $K$ is the splitting field of some family of $k[x]$ polynomials, then it is also the splitting field of some $F[x]$ polynomials, since $k\subset F$). However $F/k$ is not automatically Galois. Hence the proof would make sense if it began assuming $K/k$ and $F/k$ were Galois with $K/F/k$ intermediate, as you note.

On these hypotheses, all $G(K/k)$-actions fix $F$ setwise (since $F$ is Galois over $k$), so there is a map given by restriction $G(K/k)\to G(F/k):\sigma\mapsto\sigma|_F$ with kernel $G(K/F)$, those automorphisms which fix $F$ pointwise. Hence $G(K/F)\triangleleft G(K/k)$. (I assume this is the proof you're reading.)


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