# extending automorphisms in a tower of fields

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?

• 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 $K$. – Brett Frankel Apr 5 at 16:43
• @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. – 0k33 Apr 5 at 16:54
• 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. – Alexandros Apr 7 at 20:49
• @Alexandros great, thank you! – 0k33 Apr 8 at 18:45