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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.

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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 $K \neq L$.

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  • $\begingroup$ When you say a conjugate of complex conjugation, you mean something of the form $ghg^{-1}$ where $g,h\in \operatorname{Gal})(\overline{\mathbb Q}/\mathbb Q)$ and $h$ is the field automorphism sending swapping the two roots of $x^2+1$? I just want to be clear, since conjugation is being used in two distinct ways in one sentence. $\endgroup$ – Aaron May 10 '15 at 3:59
  • $\begingroup$ @Aaron: yes, that's what I mean. $\endgroup$ – Qiaochu Yuan May 10 '15 at 3:59
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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.

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    $\begingroup$ Thanks. Even though I didn't mention it in the question, looking at the intersection of function fields is what led me to this line of questioning. This example has the benefit that the fields in question aren't even algebraic over the intersection field. $\endgroup$ – Aaron May 10 '15 at 5:22
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    $\begingroup$ Yes, when I was in grad school, I had just such a question. I don’t think I came across this argument till much later, though. $\endgroup$ – Lubin May 10 '15 at 19:21

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