I'd appreciate a hint on the following:
Let $E$ and $F$ be fields such that $[E:F]=2$. It is possible that both $E$ and $E$ are isomorphic to $\mathbb{Q}(x)$, the field of rational functions over $Q$?
I haven't been able to come up with an example where it is true, but my bank of examples has been limited to things like quadratic extensions of $\mathbb{Q}$ and $\mathbb{F}_p[x]/(\mbox{degree 2 irreducible polynomial in x})$.
By the primitive element theorem, if $[E:F]=2$ there is some $\alpha \in E$ with $a^2 \in F$ such that $F \cong E(\alpha)$. My intuition is that then we'd have $\mathbb{Q}(x) \cong E \cong F(\alpha) \cong \mathbb{Q}(x)(\beta)$, for $\beta$ such that $\beta^2\in \mathbb{Q}(x)$. But then we'd have something like $2 = [E:F] = [\mathbb{Q}(x)(\beta): \mathbb{Q}(x)]=1$, which is absurd. Is this at least thinking in the right direction? Or is this statement actually true in some cases?