Non-algebraic subfield intersection

Question

Construct an example in which a field $F$ is of degree 2 over two distinct subfields $A$ and $B$, but so that $F$ is not algebraic over $A\cap B$.

I'm having trouble thinking of an explicit example here.

Answer 1

Let $F=\mathbb{Q}(\pi,i)$, with $A=\mathbb{Q}(\pi)$ and $B=\mathbb{Q}(i\pi)$. Then $F$ is not algebraic over $A\cap B=\mathbb{Q}$.

Answer 2

Here’s another example, purely geometric. Consider $\Bbb Q$ and the field $\Bbb Q(t)$ of rational functions in one variable. Your $F$ will be $\Bbb Q(t)$. It has two obvious automorphisms of order two, namely $\sigma(t)=-t$ and $\tau(t)=2-t$. The fixed field of $\sigma$ is “quadratic beneath” F, in fact this fixed field is $\Bbb Q(t^2)$, which will be your $A$.

Similarly, you see that the fixed field of $\tau$ is $\Bbb Q((t-1)^2)$, because $\tau(t-1)=1-t=-(t-1)$. So your $B$ will be $\Bbb Q((t-1)^2)$. Now I claim that $A\cap B=\Bbb Q$. Indeed, something fixed by both $\sigma$ and $\tau$ will be fixed by $\sigma\circ\tau$, which sends $t$ first to $2-t$ and then to $2+t$. But this is translation by $2$ units, and the only rational functions that are fixed under a translation are the constants.