Suppose we have the following tower of fields: $\mathbb Q \subset \mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})$. Compute Aut$(\mathbb Q(\sqrt{2})/\mathbb Q)$, Aut$(\mathbb Q(\sqrt[4]{2})/\mathbb Q(\sqrt{2}))$, and Aut$(\mathbb Q(\sqrt[4]{2})/\mathbb Q)$. What do you see?

I do not know how to even do this, please help.

  • $\begingroup$ Do you know what you're being asked to do? Do you know how to specify a homomorphism whose domain is $\mathbb{Q}$ Or $\mathbb{Q}(\sqrt{2})$? $\endgroup$ – Hurkyl May 5 '13 at 1:49

In the first case, you must determine the automorphisms $\Bbb Q(\sqrt{2})\to\Bbb Q(\sqrt{2})$ that act as identity on $\Bbb Q$. In the second, you must determine the automorphisms $\Bbb Q(\sqrt[4]{2})\to\Bbb Q(\sqrt[4]{2})$ that act as identity on $\Bbb Q(\sqrt{2})$.

Suppose that $f$ is an automorphism of $\Bbb Q(\sqrt 2)$ that acts as identity on $\Bbb Q$. Taking an arbitrary element $a+b\sqrt2$ of $\Bbb Q(\sqrt2)$, where $a,b$ are rational, we then have $$f(a+b\sqrt{2})=f(a)+f(b)f(\sqrt2)=a+bf(\sqrt 2).$$ Thus, the members of $\text{Aut}(\Bbb Q(\sqrt2)/\Bbb Q)$ are completely determined by where they send $\sqrt{2}$. What options are there for that, given that $f(2)=2$?

A similar idea will hold for $\text{Aut}(\Bbb Q(\sqrt[4]2)/\Bbb Q(\sqrt2)),$ since our arbitrary elements of $\Bbb Q(\sqrt[4]2)$ will be of the form $$a+b\sqrt[4]2+c\sqrt2+d\sqrt[4]8=a+c\sqrt2+(b+d\sqrt2)\sqrt[4]2,$$ where $a,b,c,d\in\Bbb Q$, and since we'll need $f(\sqrt2)=\sqrt2.$

  • $\begingroup$ for $f(2)=2$ would that be $\{id, \sigma\}$ where $id=\sqrt{2}$ and $\sigma=-\sqrt{2}$? $\endgroup$ – Breezy May 5 '13 at 16:52
  • $\begingroup$ Yes, we need $f(\sqrt2)=\pm\sqrt2$. $\endgroup$ – Cameron Buie May 5 '13 at 18:42

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