Say we have the following tower of fields: $\mathbb{Q} \subset \mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\sqrt[4]{5})$. For ease, we can write $\mathbb{Q} = F, \mathbb{Q}(\sqrt{5}) = E, \mathbb{Q}(\sqrt[4]{5}) = K$.

I know that $Aut_F(E) = \{id, \tau: \sqrt{5} \mapsto -\sqrt{5} \}$.

I am told that $\tau$ as defined is not an F-automorphism of $K$. Can someone help me figure out why? I am wondering what $\tau$ would have to do to $\sqrt[4]{5}$ in order to be an $F$-automorphism of $K$, and why it doesn't do what it's supposed to do. I don't think the following computation is correct (although it gives the right result), but I'm not sure how to fix it: $$\tau(\sqrt[4]{5}) = \tau(\sqrt{5}^{1/2}) = \tau(\sqrt{5})^{1/2} = -\sqrt{5}^{1/2}$$ where the second-to-last equality follows because $\tau$ is an automorphism. This does not describe a $K$-automorphism of $F$ because $-\sqrt{5}^{1/2} \not\in K$.

I know that $E$ is a Galois extension of $F$, and $K$ is not since $i \not\in K$. Could this be relevant, ie. is there a stronger result hiding here?

  • $\begingroup$ It's not clear to me what you are asking. In particular, you appear to answer the question of why $\tau$ is not an automorphism of $K$. $\endgroup$ – Brett Frankel Apr 5 at 16:43
  • $\begingroup$ @BrettFrankel haha, it's not clear to me either :) I wanted to either check that my work is correct (or receive a suggestion for why it isn't), or potentially see a stronger/more general result of this kind for other cases in which $E$ is a Galois extension of $F$, but $K$ is not. $\endgroup$ – 0k33 Apr 5 at 16:54
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    $\begingroup$ Your work appears correct to me, as any automorphism of K:F which sends $\sqrt-5$ to $-\sqrt-5$, would have to send $\sqrt[4]{5}$ to $i\sqrt[4]{5}$, and that is not possible in a non-imaginary field. $\endgroup$ – Alexandros Apr 7 at 20:49
  • $\begingroup$ @Alexandros great, thank you! $\endgroup$ – 0k33 Apr 8 at 18:45

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