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$\def\Gal{\operatorname{Gal}}$ I was working on homework, and the problem starts off by saying that I previously showed (I can't find where, though) that with $\def\Q{{\mathbb Q}}\def\Z{{\mathbb Z}}F=\Q$ and $L=\Q(\sqrt{2+\sqrt{2}})$, $G=\Gal(L/\Q)\cong \Z/4\Z$. I can't figure out how this is so.

The min poly for $\sqrt{2+\sqrt{2}}$ over $\Q$ is $x^4-4x^2+2$, whose roots are:

$$ \def\a{{\alpha}}\a_1=\sqrt{2+\sqrt{2}},\,\a_2=-\a_1,\,\a_3=\sqrt{2-\sqrt{2}},\,\a_4=-\a_3 $$

so therefore $\def\s{{\sigma}}\def\t{{\tau}}\s\in G$ can be thought of as a permutation of the $\a_i$. Note that $\a_1\a_3 = \sqrt{2}$. So $G = \{e, \s, \t, \s\t\}$ where:

$$ \s(\a_1)=\a_2,\,\t(\a_1)=\a_3\\ \s(\a_2)=\a_1, \s(\a_3)=\s\left(\frac{\sqrt{2}}{\a_1}\right)=\frac{\sqrt{2}}{\a_2}=\a_4, \s(\a_4)=\a_3\\ \t(\a_3)=\t\left(\frac{\sqrt{2}}{\a_1}\right)=\frac{\sqrt{2}}{\a_3}=\a_1, \t(\a_2)=\t(-\a_1)=\a_4, \t(\a_4)=\a_2,\text{ so}\\ \quad\s\to(12)(34)\\ \quad\t\to(13)(24)\\ \quad\t\s\to(14)(23) $$

In this group, each element is its own inverse, so this should mean that $G \not\cong \Z/4Z$, but $G\cong V$, the Klein four-group.

Where have I gone wrong?

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If the Galois group were $(\mathbb Z/2\mathbb Z)^2$, then $L$ would contain two quadratic subfields. One of them is $\mathbb Q(\sqrt 2)$; what is the other one?

In fact I claim that $L$ contains no other quadratic subfield. A quick way to see this using number theory is that if $L$ contained $\mathbb Q(\sqrt D)$ where $D$ is odd, then $L$ would be ramified above some odd prime. But the discriminant of $L$ divides the discriminant of $x^4-3x^2+2$ which is $32$, hence $disc(L)$ is a power of $2$. So $L$ cannot be ramified above any odd prime. You can definitely come up with an elementary proof, though.

Your mistake is in incorrectly presenting the Galois group. How do you know that that's what the Galois group does to the roots?

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  • $\begingroup$ I thought I properly mapped out each element. I took $\alpha_1\to\alpha_2=-\alpha_1$, then since $\alpha_1\alpha_3=\sqrt{2}$, $\alpha_3=\frac{\sqrt{2}}{\alpha_1}$, (oh wait maybe this is the mistake) $= \frac{\sqrt{2}}{-\alpha_1}=\alpha_4$ $\endgroup$ – Justin Oct 31 '14 at 4:14
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$\def\a{{\alpha}}\def\t{{\tau}}\def\Gal{\operatorname{Gal}}\def\Q{{\mathbb Q}}$Your mistake was the assumption that $\t(\sqrt{2})=\sqrt{2}$; $\t(\sqrt{2})=-\sqrt{2}$:

$$ \a_1 = \sqrt{2+\sqrt{2}}, \a_3 = \sqrt{2-\sqrt{2}}\\ \begin{align} \t(\a_3)&=\t\left(\frac{\sqrt{2}}{\a_1}\right)\\ &=\t\left(\frac{\color{red}{\a_1^2-2}}{\a_1}\right)\\ &=\frac{\a_3^2-2}{\a_3}\\ &=\frac{-\sqrt{2}}{\a_3}\\ &=-\a_1=\a_2 \end{align} $$

And since the $\a_1$ then has order $> 2$, we know that $\t(\a_2)=\a_4, \t(\a_4)=\a_1$. Note that we don't need to worry if $\t(2)=2$ since $\t\in G$ so $\t$ fixes $\Q$

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  • $\begingroup$ I just wanted to add how helpful this question is/was to me, as well as the correction above. It helped me fully understand to express everything in terms of what we "know" for any of the given automorphisms...otherwise, drastic problems arise. Very nice presentation as well as answers/corrections! $\endgroup$ – Procore Apr 30 '15 at 22:49

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