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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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    Simply because all automorphisms have order $2$.2013-01-05

1 Answers 1

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Hints:

== 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...