TITLE: Is the intersection of finite index subfields finite?
QUESTION [5 upvotes]: Suppose that $K$ and $L$ are two fields contained in some larger field, and let $KL$ denote the smallest subfield of the ambient field containing both of them. If $KL$ is a finite extension of both $K$ and $L$, Is it necessarily the case that $K$ is finite over $K\cap L$ (which happens if and only if $L$ is finite over $K\cap L$)?
I see no a priori reason why this should be true (and my initial attempts to liken the situation to intersections of finite index subgroups/vector spaces was unsuccessful), but I don't have enough familiarity with transcendental extensions to build a counterexample.

REPLY [9 votes]: Another example, which you may find more elementary. Let $k$ be any field of characteristic zero, hence infinite. Consider the rational-function field $k(x)$, and the two subfields $k(x^2)$ and $k\bigl((x-1)^2\bigr)$. You see easily that the field degree is $2$ in both cases. But I say that the intersection of these two quadratic underfields is $k$ itself.
Indeed, let $g(x)\in k(x^2)$. Then of course it’s invariant under the automorphism $\sigma$ sending $x$ to $-x$. And if $g\in k\bigl((x-1)^2\bigr)$ as well, then $g$ is also invariant under the automorphism $\tau$ of $k(x)$ that sends $x$ to $2-x$. So $g$ will be invariant under the composition of these two, say $\tau\circ\sigma$, which sends $x$ first to $-x$, and this in turn to $-(2-x)=x-2$. But infinitude of $k$ and characteristic zero implies that $\tau\circ\sigma$ is of infinite order, i.e. nontorsion in the group of all automorphisms of $k(x)$. But no rational function is invariant under this $2$-shift, except for the constant rational functions.