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Let $\mathbb{Q}(i)$ be given, where $i^2=-1$. I want to know:

Whether there is a Galois extension $K/\mathbb{Q}(i)$ such that $\mathrm{Gal}(K/\mathbb{Q}(i))\cong D_8$, where $D_8$ is the dihedral group of size 8?

I know the Fundamental Theorem of Galois theory, and how to calculate Galois groups.

I tried several guesses, such as $\mathbb{Q}(2^{1/4},i)$, $\mathbb{Q}(2^{1/8},i)$, but none worked.

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  • $\begingroup$ Take a dihedral extension of the rationals not containing i and lift it. $\endgroup$ May 17, 2017 at 14:47
  • $\begingroup$ But @franzlemmermeyer, all the examples of extensions with group $D_8$ that we see in basic Galois Theory classes are $\Bbb Q(m^{1/4},i)$ over $\Bbb Q$. $\endgroup$
    – Lubin
    May 17, 2017 at 15:14
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    $\begingroup$ I'm trying to prove $\mathbb{Q}(\sqrt{3+\sqrt{2}}, \sqrt{3-\sqrt{2}})/\mathbb{Q}$ has Galois group $D_8$ $\endgroup$ May 17, 2017 at 15:17
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    $\begingroup$ Nevermind, I found one. $\mathbb{Q}(\sqrt{3+5\sqrt{2}}, \sqrt{3-5\sqrt{2}})/\mathbb{Q}$, the splitting field of $x^4-6x^2-41$. $\endgroup$ May 17, 2017 at 16:34
  • $\begingroup$ um... what's wrong with $\Bbb Q(2^{1/4}, i)$ ?? $\endgroup$
    – mercio
    May 18, 2017 at 9:30

1 Answer 1

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You only need to know whether one exists, not what it is, so no need to guess at what an actual extension would be.

Instead, you could think only about the Galois groups. You can use the Fundamental Theorem of Galois Theory to show that such an extension exists if and only if there exists a group $G$ such that $G\cong D_8$ and $\operatorname{Gal}(K/\mathbb{Q}(i))$ is a normal subgroup of $G$. If you show that, and find such a group (up to isomorphism), then you're done.

Why a normal subgroup in particular? Well, it has to be a Galois extension, not just any extension.

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