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I have the following problem:

Let $E=\mathbb{Q}[\alpha, \beta]$ where $\alpha^2 \in \mathbb{Q}$, $\beta^2 \in \mathbb{Q}$, and $[E:\mathbb{Q}]=4$ If $\gamma \in E-\mathbb{Q}$ and $\gamma^2 \in \mathbb{Q}$ prove that $\gamma$ is a rational multiple of one of $\alpha, \beta, $or $\alpha \beta$.

I'm thinking a proof my contradiction would work, in combination with an argument concerning the degree of an extension, but it sounds fishy to me.

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For a quick answer you can use the Galois correspondence. Otherwise see that $\{1,\alpha,\beta, \alpha\beta\}$ is a $\mathbb{Q}$-basis in $E$, write $\gamma$ as a linear combination of these four elements and see what it means that $\gamma^2\in\mathbb{Q}$ (some coefficients must be $0$). – user26857 Nov 27 '12 at 22:39
What is the quick answer with the Galois correspondence? – Frank White Nov 27 '12 at 23:16
The Galois group is isomorphic to the Klein group, so there are only three intermediate extensions of degree $2$. – user26857 Nov 27 '12 at 23:35
@YACP How do you know the Galois group is isomorphic to the Klein four group instead of the cyclic group of order four? – Frank White Jan 5 '13 at 3:51
Simply because all automorphisms have order $2$. – user26857 Jan 5 '13 at 9:33
up vote 0 down vote accepted


== It must be $\,[\Bbb Q(\alpha):\Bbb Q]=[\Bbb Q(\beta):\Bbb Q]=[\Bbb Q(\alpha\beta):\Bbb Q]=2\,$

== Since $\gamma\in E\setminus\Bbb Q\,\,\,,\,\,\gamma^2\in Q\,$ , then also $\,[\Bbb Q(\gamma):\Bbb Q]=2\,$

== Make a diagram of the lattice of subfields of $\,E/\Bbb Q\,$ determined by the first hint, and then try to make "fix" $\,\Bbb Q(\gamma)\,$ somewhere in there...

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