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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    $\begingroup$ 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? $\endgroup$ Apr 19, 2019 at 4:56
  • $\begingroup$ The terminology is tower of Galois extensions the counterexample is $\Bbb{Q}(2^{1/4})$ $\endgroup$
    – reuns
    Apr 19, 2019 at 5:01
  • $\begingroup$ @reuns could you elaborate on that counterexample please? $\endgroup$
    – user555558
    Apr 19, 2019 at 5:14
  • $\begingroup$ 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: math.stackexchange.com/questions/255274/… . 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 $S_n$. $\endgroup$ Apr 20, 2019 at 17:51