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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.


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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. – Chris Eagle Dec 6 '11 at 23:09
By the way, one of the things you "know" is not true in characteristic $2$. – Chris Eagle Dec 6 '11 at 23:11
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? – Anon Dec 6 '11 at 23:12
It clearly contains $K$. It is clearly contained in $L$. (I didn't mean "strictly intermediate") – Chris Eagle Dec 6 '11 at 23:23
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. – Gerry Myerson Dec 6 '11 at 23:39

Claim: Let $L/K$ be an extension ($ \mbox{char}(K) \neq 2$), with $[L:K] = 4$ and $\mbox{Aut}(L/K) \cong C_2 \times C_2 $. Then $L$ is of the form $ K(\sqrt{a},\sqrt{b}) $ for $a,b \in K$.

Proof: $|\mbox{Aut}(L/K)| = 4 = [L:K] $, so the extension is Galois. The non-trivial subgroups of $C_2 \times C_2$ are three copies of $C_2$, which (by the Galois correspondence) correspond to three subextensions of degree 2. As $\mbox{char}(K) \neq 2$, any degree 2 extension of $K$ is of the form $K(\sqrt{\alpha}) $ for some $\alpha$ in $K$, so we have three subextensions $K(\sqrt{a})/K$, $K(\sqrt{b})/K$ and $K(\sqrt{c})/K$. We must have $K(\sqrt{a},\sqrt{b}) \subseteq L$. We can't have $K(\sqrt{a},\sqrt{b}) = K(\sqrt{a})$, since then we would have that $ K(\sqrt{b}) \subseteq K(\sqrt{a}) $. Similarly, we can't have $K(\sqrt{a},\sqrt{b}) = K(\sqrt{b}) $. If $K(\sqrt{a},\sqrt{b}) = K(\sqrt{c}) $, then $ 2 = [K(\sqrt{a},\sqrt{b}):K] = [K(\sqrt{a},\sqrt{b}):K(\sqrt{b})][K(\sqrt{b}):K] $, and so $K(\sqrt{a},\sqrt{b}) = K(\sqrt{b})$, which we've shown can't be true. So we must have that $ K(\sqrt{a},\sqrt{b}) = L$, as required.

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