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Prove that if $[L:K] = 4$ and $ \mbox{Aut}(L/K) \cong C_2 \times C_2 $ then $ L$ is of the form $K(\sqrt{a},\sqrt{b}) $.

I know that the extension is Galois, and so I can use the Galois correspondence. $C_2 \times C_2$ has 3 copies of $C_2$ as its non-trivial subgroups. These must correspond to 3 subextensions of $L/K$ of degree 2. I know also that an extension $F/K$ of degree 2 is of the form $K(c) $ for some $c^2$ in $K$. So it seems I have $ K \subseteq K(\sqrt{a}) $, $K(\sqrt{b}), K(\sqrt{c}) \subseteq L$.

From here I'm unsure what to do. It seems strange that I have three of these subextensions... I think I could do it if I had just two of them.

Thanks

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    Suppose $K(\sqrt{a})$ and $K(\sqrt{b})$ are two of the intermediate extensions. Then what is $K(\sqrt{a},\sqrt{b})$? It has to be some intermediate field of the extension, and there aren't very many.2011-12-06
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    By the way, one of the things you "know" is not true in characteristic $2$.2011-12-06
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    Yes, apologies. I'm assuming $ \mathbb{Q} \subset K$. I'm not quite sure what you're suggesting in your first comment. Why does $K(\sqrt{a},\sqrt{b})$ have to be an intermediate field?2011-12-06
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    It clearly contains $K$. It is clearly contained in $L$. (I didn't mean "strictly intermediate")2011-12-06
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    Oh, OK. Well it's either $L$ or one of $K(\sqrt{a})$, $K(\sqrt{b})$ (it can't be the other intermediate field, because this would mean that it had to be a degree 4 extension of $K$). If it's $L$, we're done. If not, then we have (wlog) $\sqrt{b} \in K(\sqrt{a}) $. But then that means $K(\sqrt{b}) \subseteq K(\sqrt{a}) $, which can't be true.2011-12-06
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    Now that you have figured it out, you should post it as an answer. Posting answers to your own questions may seem weird, but it is actually encouraged here. Then you can even accept your answer, and we can all go on to the next question.2011-12-06
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    I've typed one out, but can't post it until 8 hours has passed.2011-12-06

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