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Consider the extension $\mathbb{Q} \subset\mathbb{Q} (\sqrt{2}, \sqrt{3})$. How many elements are there in $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})?$ Describe all elements in $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$. Find all subgroups of $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$ and their fixed fields.

I'm using Dummit & Foote Chapter 14 (Galois Theory) if it needs to be referenced. Chapter 13 was a breeze for me so I just feel like I need someone to explain how easy of a connection this is and it'll just click (or maybe it's not and I need to be told that, too). I really just need direction.

Zev Chonoles
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Jared
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3 Answers3

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Hint: An automorphism fixes the integers, indeed the rationals. So it has to carry $\sqrt{2}$ to $\pm\sqrt{2}$, and $\sqrt{3}$ to $\pm\sqrt{3}$. There are not many candidates.

André Nicolas
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  • So the only options are the mapping $id(x)=x$ and $f(x)=-x$? Those are the only two. – Jared May 23 '13 at 03:04
  • @Questioneer Remember, if $\sigma\in\operatorname{Aut}\left(\Bbb{Q}\left(\sqrt{2},\sqrt{3}\right)/\Bbb Q\right)$, then $\sigma$ is determined by how it acts on a basis of $\Bbb{Q}\left(\sqrt{2},\sqrt{3}\right)$ (thinking of $\Bbb{Q}\left(\sqrt{2},\sqrt{3}\right)$ as a $\Bbb Q$ vector space). You might (and probably won't) be able to say $\sigma(x) = x$ or $\sigma(x) = -x$ (meaning $\sigma(x)$ won't be a nice polynomial function of $x$) for an arbitrary $\sigma$. – Stahl May 23 '13 at 03:06
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You know that any automorphism of $\Bbb{Q}\left(\sqrt{2},\sqrt{3}\right)$ fixing $\Bbb{Q}$ has to send $\sqrt{2}$ and $\sqrt{3}$ to another root of their respective minimal polynomials, so that you might want to look at those polynomials/roots and see/count the different (coherent) ways you can permute the roots. Once you do that, you should be able to determine what an arbitrary element does by seeing what it does to the elements of a basis of $\Bbb{Q}\left(\sqrt{2},\sqrt{3}\right)$ (viewed as a $\Bbb Q$ vector space).

Stahl
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Hint:

$\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3}):\mathbb{Q})$ $\cong$ $\mathbb{Z_2} \times\mathbb{Z_2}$

This together with the other hints should give you the answer. You will still need to find the subgroups and their corresponding fixed fields, which is not too difficult.

Islands
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