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I am doing a specific exercise where the quaternion group is realised as a Galois group of some field extension. It goes like this: let $K = \mathbb Q(\sqrt 2,\sqrt 3)$ and $\alpha = (2 +\sqrt 2)(3 +\sqrt 6)$. I want to show that $\alpha$ is not a square in $K$. I tried to solve equations explicitly to derive some contradiction but I didn't succeed. Any suggestions? Thanks in advance.

AYK
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  • That extension has degree 4, so the Galois group cannot be the Quaternion group. – Adam Hughes Nov 21 '14 at 00:25
  • I didn't say the quaternion group is obtained as $G(K/\mathbb Q)$ (this group is just the Klein four-group). But after proving that $\alpha$ is not a square you can adjunct the square root of $\alpha$ to $K$ in order to get a field extension $L/\mathbb Q$ with $G(L/\mathbb Q)$ equal to the quaternion group. – AYK Nov 21 '14 at 00:41
  • Ah, it's good to say your strategy: it was not obvious from your characterization. – Adam Hughes Nov 21 '14 at 00:53

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