TITLE: Question on finite field extensions
QUESTION [5 upvotes]: If M/K is a finite field extension and $L \subset M$, must $L/(K \cap L)$ be finite?
I have failed to prove this is true but am not sure where to look for counterexamples - starting with M = $\mathbb{C}$ or $\mathbb{C}(X)$ doesn't seem fruitful as there aren't many subfields extending finitely to it.

REPLY [2 votes]: I absolutely endorse user8268's nice answer. Proffering the following alternative. It may be easier to believe intuitively, but I couldn't find an easier rigorous proof for the key claim :-(

Let $x$ and $y$ be two algebraically independent transcendentals over $\Bbb{Q}$. When $M=\Bbb{Q}(x,y)$, $K=\Bbb{Q}(x,y^2)$ and $L=\Bbb{Q}(x+y)$, we have $[M:K]=2$, $K\cap L=\Bbb{Q}$ and $[L:K\cap L]=\infty$.

Consider the following two automorphisms of $M$, determined by how they map the generating elements. First $\sigma: x\mapsto x, y\mapsto -y$ and then $\tau: x\mapsto x+1, y\mapsto y-1$.
Of these $\sigma$ has order two, and we can immediately identify its fixed field as $K=\Bbb{Q}(x,y^2)$. We also immediately see that $\tau$ is an $L$-automorphism of $M$. It remains to prove the following
Claim. $L\cap K=\Bbb{Q}$.
Proof. Any element of $L\cap K$ is a fixed point of both $\sigma$ and $\tau$. Consider the commutator $\delta:=\sigma\tau\sigma^{-1}\tau^{-1}$. We see that
$\delta(x)=x$ and $\delta(y)=y+2$. So if $F=\Bbb{Q}(x)$, then $\delta$ is an $F$-automorphism of $M$. Reusing a relevant part of an old answer of mine we see that the fixed points of $\delta$ are exactly the elements of $F$. In that old answer I looked at the fixed points of translation by one (instead of two), but that is irrelevant.
On the other hand, the automorphism $\rho:=\tau\sigma\tau\sigma^{-1}$ maps the generators as follows $\rho(y)=y$, $\rho(x)=x+2$. Repeating the earlier argument once more shows that the fixed points of $\rho$ are the elements of $\Bbb{Q}(y)$.
Hence any element of $L\cap K$ must be an element of both $\Bbb{Q}(x)$ as well as $\Bbb{Q}(y)$. The claim follows from this.