Field not algebraic over an intersection but algebraic over each original field

Question

What is an example of a field $K$ which is of degree $2$ over two distinct subfields $E$ and $F$ respectively, but not algebraic over $E \cap F$?

Answer 1

Let $X$ be an indeterminate and consider $E=\mathbb C(X^2),F=\mathbb C(X^2+X),K=\mathbb C(X)$. Clearly $K$ is quadratic over both $E$ and $F$. We claim $E\cap F=\mathbb C$.

Suppose first $p(X^2)=q(X^2+X)$ for some rational functions $p,q$ over $\mathbb C$. Consider them as equal to a function $f(X)$ over $\mathbb C$ in a variable $X$. This equality tells that if $z$ is a zero (resp. a pole) of $f$, then $-z$ and $-1-z$ are also zeros (resp. poles) of $f$. Unless $f$ is constant, it has either a zero or a pole, and by repeatedly applying $z\mapsto -z,z\mapsto -1-z$ we find it has infinitely many of these. But this is impossible for rational function unless it's identically zero. It follows that $f$ is constant, so $p(X^2)=q(X^2+X)\in\mathbb C$.