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.
Here is a counterexample:
Take $M=\mathbb C$, $K=\mathbb R$, and $L=\bigcup_{n=1}^{\infty} L_n$, where $L_n=\mathbb Q(a_n)$ and finally $a_n = \exp(2\pi i/2^n)\, 2^{1/2^n}$ (we have $a_n^2= a_{n-1}$, so $L_{n-1}\subset L_n$). Then $L\cap(K=\mathbb R)=\mathbb Q(2^{1/2})$, as follows say from a bit of Galois theory (see below) and so $[L:L\cap K]=\infty$.
To see that $L_n\cap \mathbb R=\mathbb Q(2^{1/2})$ for every $n$ (and so $L\cap \mathbb R=\mathbb Q(2^{1/2})$), consider the Galois extension $F=\mathbb Q\bigl(\xi_n:=\exp(2\pi i/2^n), b_n:=2^{1/2^n}\bigr)$ of $\mathbb Q$. The Galois group $G$ is given by $\xi_n\mapsto\xi_n^\alpha$, $b_n\mapsto\xi_n^\beta b_n$, $\alpha\in (\mathbb Z/2^n\mathbb Z)^*$, $\beta\in\mathbb Z/2^n\mathbb Z$. The real numbers in $F$ are those that are fixed by the element $\alpha=-1$, $\beta=0$ (the complex conjugation) and (trivially) by the identity. $L_n$ is the fixed field of the subgroup that preserves $a_n$, which is given by the equation $\alpha +\beta =1$. These two subgroups of $G$ generate all the elements of $G$ with $2|\beta$, and the fixed field of this subgroup (of index $2$ in $G$) is $\mathbb Q(2^{1/2})$.
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.
${\bf Correction:}$
The answer below is wrong, let's see what may be the problem.
Assume we are inside a Galois extension $F\supset F^G$ (possibly infinite), $K= F^{G_1}$, $M = F^{G_2}$, and $L=F^{G_3}$, where $G_1, G_3 \subset G_2$ are subgroups of $G$. We know that $[M\colon K] = [F^{G_2}\colon F^{G_1}] =[G_1\colon G_2]$ is finite. However, $[L\colon K\cap L] = [F^{G_3}\colon F^{G_1}\cap F^{G_3}] =[(G_1, G_3) \colon G_3]$.
Now, we do have $[G_1 G_3 \colon G_3] = [G_1\colon G_1 \cap G_3]$, but $(G_1, G_3)$ might be strictly larger than the set $G_1 G_3$. Now we see why we may be in trouble.
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${\bf Wrong answer below}$
The answer is Yes in characteristic $0$.
We may assume $M\supset K$ is finite Galois (otherwise take the normal closure of $M/K$). Let $G= \operatorname{Gal}(M/K)$, a finite group. We have $K = M^G$, the set of fixed points of $G$.
Now, every $l \in L$ is a root of a polynomial $$\prod_{g \in G} (X - g l)$$ with coefficients in $M^G\cap L = K \cap L$ ( that is wrong, $L$ may not be invariant under $G$ ! ). Therefore the degree of $l$ over $K\cap L$ is at most $|G|$.
Now consider $l_0\in L$ the element in $L$ of maximal degree over $L'\colon = K\cap L$. Let $l$ any other element. We have $$L'(l_0, l) = L'(l_1)$$ since the extension is finite separable, so simple. Since $l_0$ is of maximal degree, we have $L' (l_1) = L'(l_0)$. Conclusion: $L = L'(l_0)$, finite, of degree at most $|G|$.
