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Let $ K/ L /F $ be fields. If $K / L$ is Galois and $ L / F $ is Galois, then $ K / F$ is Galois. We mentioned this very quickly in today's class without justifying. But I have trouble seeing this. Any help is appreciated.

Our definition for $ K / F$ being Galois is the $F$-automorphism group $G = G( K / F) : = \{ \sigma \in aut (G) : \sigma_F = id_F \}$ fixes only $F$, i.e. $ \{ x \in K : \sigma (x) = x \text { for all } \sigma \in G \} = F$.

But we also showed that $ K /F$ is Galois if and only if $ K/F$ is normal and separable.

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    $\begingroup$ It would probably help any answerers to know what exactly what definition you have of being Galois (and, probably, if you are only considering finite extensions) $\endgroup$
    – Mariano Suárez-Álvarez
    May 1, 2014 at 0:59
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    $\begingroup$ Galoisness is not transitive. $\endgroup$
    – blue
    May 1, 2014 at 1:01
  • $\begingroup$ Not true, as @Matt points out in his response. It’s also an error that’s all too often made. $\endgroup$
    – Lubin
    May 1, 2014 at 1:08
  • $\begingroup$ Thank you for all of you who answers! $\endgroup$
    – user112564
    May 1, 2014 at 1:10
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    $\begingroup$ Galoisity? ${}{}{}$ $\endgroup$
    – Mariano Suárez-Álvarez
    May 1, 2014 at 1:11

2 Answers 2

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This is not true. E.g. $\mathbb Q(\sqrt{1+2i})$ is Galois over $\mathbb Q(i)$, which is Galois over $\mathbb Q$. But $\mathbb Q(\sqrt{1+2i})$ is not Galois over $\mathbb Q$.

Another example, maybe easier to check: $\mathbb Q(\sqrt{1+\sqrt{2}}) \supset \mathbb Q(\sqrt{2})\supset \mathbb Q$.

The Galois closure will of the first field over $\mathbb Q$ will contain $\sqrt{1-\sqrt{2}}$ as well; but this is not a real number, whereas $\mathbb Q(\sqrt{1+\sqrt{2}})$ consists entirely of real numbers.

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  • $\begingroup$ Thanks! But how can I see that $ \mathbb Q ( \sqrt { 1 + 2i } )$ is not Galois over $\mathbb Q $? $\endgroup$
    – user112564
    May 1, 2014 at 1:17
  • $\begingroup$ @user112564: Dear user, Its Galois closure will also contain $\mathbb Q(\sqrt{1-2i})$, and hence also $\sqrt{5}$. If it were already Galois, then (for degree reasons) it would have to equal $\mathbb Q(i,\sqrt{5})$. But you can check (with a little effort) that no element in this field squares to give $1 + 2i$. (There are other ways to see this that use a little more number theory. And there are surely other examples that are easier to check; this is just the first one that came to mind. Actually, I added another one.) Regards, $\endgroup$
    – Matt E
    May 1, 2014 at 1:18
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    $\begingroup$ Perhaps an easier example is $K=\Bbb Q(\root4\of 2)$, $L=\Bbb Q(\sqrt2)$, $F=\Bbb Q$. The splitting field of $f(x)=x^4-2$ is $\Bbb Q(\root4\of 2,i)$, after all. :) $\endgroup$
    – Ted Shifrin
    May 1, 2014 at 1:23
  • $\begingroup$ @TedShifrin: Dear Ted, Indeed. I added another more easily checked example too. (I guess it's easier to tell real from imaginary than to look at the splitting behaviour of primes above $5$!) Cheers, $\endgroup$
    – Matt E
    May 1, 2014 at 1:24
  • $\begingroup$ I got it! Thank you so much! Also thanks to Ted Shifrin. $\endgroup$
    – user112564
    May 1, 2014 at 1:28
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In the book Abstract Algebra, third edition, by David S. Dummit and Richard M. Foote, they mention the example $\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}(\sqrt[4]{2})$ as a pair of Galois extensions which are not transitive:

The field $\mathbb{Q}(\sqrt[4]{2})$ is not Galois over $\mathbb{Q}$ since any automorphism is determined by where it sends $\sqrt[4]{2}$ and of the four possibilities $\{\pm\sqrt[4]{2},\pm i\sqrt[4]{2}\}$, only two are elements of the field (the two real roots).

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