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