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How would I go about showing that the composition of Galois extensions may not be Galois? I figure I can just provide a counterexample but I'm having a hard time thinking of any.

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    Composition? A compositum of two Galois extensions is Galois. But a Galois extension of a Galois extension need not be Galois. What do you mean by composition? – Angina Seng Apr 19 '19 at 04:56
  • The terminology is tower of Galois extensions the counterexample is $\Bbb{Q}(2^{1/4})$ – reuns Apr 19 '19 at 05:01
  • @reuns could you elaborate on that counterexample please? –  Apr 19 '19 at 05:14
  • Let $N \subset N' \subset S_n$ be finite groups. Suppose that $N' \subset S_n$ is normal and $N \subset N'$ is normal, but $N \subset S_n$ is not normal. You can find several examples here: https://math.stackexchange.com/questions/255274/are-normal-subgroups-transitive . Now take $K = F(x_1, ..., x_n)$. $S_n$ acts on it by permuting the $x_i$, and $K/K^{S_n}$ is Galois with Galois group $S_n$. Now the tower $K^{N'} / K^{N} / K^{S_n}$ works (the consecutive ones are Galois, but $K^{N'} / K^{S_n}$ is not since $N'$ is not normal in $S_n$. – Ronald J. Zallman Apr 20 '19 at 17:51

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