Is the intersection of finite index subfields finite?

Question

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.

Answer 1

No. To get some intuition for this, restrict to the case that $K$ and $L$ are fields containing a field $k$ and contained in its algebraic closure $\bar{k}$, and let $G = \text{Gal}(\bar{k}/k)$. Then, via the infinite Galois correspondence,

• $K$ and $L$ are the fixed fields of two closed subgroups $H_1, H_2$ of $G$,
• $KL$ is the fixed field of $H_3 = H_1 \cap H_2$, and
• $K \cap L$ is the fixed field of the closure $H_4$ of the subgroup generated by $H_1$ and $H_2$.

By hypothesis, $H_3$ is a finite index subgroup of $H_1$ and of $H_2$, and the question is whether it follows that $H_1$ or $H_2$ are finite index subgroups of $H_4$. Of course the answer to this group-theoretic question is no: the problem is that $H_1$ and $H_2$ can fail very badly to commute. For example, they can both be finite, but $H_4$ might nevertheless be infinite.

It remains to check that this situation can actually happen in a Galois group. Explicitly, let $k = \mathbb{Q}$, let $H_1$ be the subgroup of $G = \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ generated by complex conjugation, and let $H_2$ be the subgroup of $G$ generated by a nontrivial conjugate of complex conjugation. The fixed field $K$ is the real algebraic numbers, which have index $2$ in $\overline{\mathbb{Q}}$, and the fixed field $L$ is something else, so $KL = \overline{\mathbb{Q}}$ and $K$ and $L$ both have index $2$ in it.

I claim that $K \cap L$ necessarily has infinite index in $K$ and $L$, or equivalently in $KL = \overline{\mathbb{Q}}$. The reason is that by the Artin-Schreier theorem, if $K \cap L$ has finite index in $\overline{\mathbb{Q}}$ then it must have index $2$, hence must be equal to both $K$ and $L$, and by hypothesis $K \neq L$.

Answer 2

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.