Degree of a field over intersection

Question

Let $F \subset L \subset M$ be a tower of field extensions and let $K \subset M$ be some other field. Is it true in general that $[F: F \cap K] \leq [L: L \cap K]$? It looks like it to me, but I can't quite find a proof or counter-example.

Most of the results I've seen talk about identities and inequalities regarding the composite, but not many about intersections. Is there anything in general we can say about degrees of intersections? Is there any relation, for example, between $[L:F]$ and $[L \cap K: F \cap K]$?

I tried messing around with the tower law, but with no avail...

Thanks in advance!

Answer 1

This turns out to be false. Perhaps the easiest type of counterexample is given by this answer to a related question, which I'll adapt for this specific question.

Let $f(x)$ be an irreducible cubic polynomial over $\Bbb Q$ whose roots are $\alpha_1,\alpha_2,\alpha_3$, so that $[\Bbb Q(\alpha_j):\Bbb Q]=3$ for each $j$. If we assume that the discriminant of $f(x)$ is not a square, then the splitting field $L=\Bbb Q(\alpha_1,\alpha_2,\alpha_3)$ satisfies $[L:\Bbb Q] = 6$.

Now take $F=\Bbb Q(\alpha_1)$ and $K=\Bbb Q(\alpha_2)$, so that $[F:\Bbb Q] = [K:\Bbb Q] = 3$. Since $K\subset L$, we have $[L:L\cap K] = [L:K] = [L:\Bbb Q]/[K:\Bbb Q] = 6/3 = 2$; but $[F:F\cap K] = [F:\Bbb Q] = 3$ which is larger.

Answer 2

As Greg Martin explained, the claim is false. I proffer the following, possibly more striking, counterexample.

Let $F=\Bbb{Q}(x)$ be the field of rational functions in the indeterminate $x$, and $L=M=\Bbb{Q}(x,y)$ the field of rational functions in two algeraically indepdendent indeterminates $x$ and $y$. Let $\sigma$ be the order two automorphism of $L$ that interchanges $x$ and $y$, and let $K$ be the fixed field of $\sigma$. That is, $K=\Bbb{Q}(x+y,xy)$. Then $K\subseteq L$ and, by basic Galois theory, $[L:K]=|\langle\sigma\rangle|=2$, so $[L:L\cap K]=2$.

But $K\cap F=\Bbb{Q}$, so $[F:F\cap K]=\infty$.