# Dimension of compositum of field extensions

Suppose we have fields $L$, $M$ and $N$ all infinite algebraic Galois extensions of a field $k$ such that $L \cap M$, $L \cap N$ and $N \cap M$ are finite dimensional extensions of $k$. Then is $L \cap MN$ a finite dimensional extension of $k$?

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Take $k=\mathbb{Q}$, take $L$ to be the splitting field of the set $\{x^{2^n}-2\mid n\text{ a positive integer}\}$, $M$ to be the splitting field of the set $\{x^{2^n} - 3 \mid n\text{ a positive integer}\}$, and $N$ to be the splitting field of the set $\{x^{2^n} - 6\mid n\text{ a positive integer}\}$.
Then $L\cap M = L\cap N = N\cap M = \mathbb{Q}$ (any proper subextension of $N/\mathbb{Q}$ contains $\sqrt{2}$, hence $L$ does not contain $\sqrt{3}$; but by a symmetric argument, every proper subextension of $M/\mathbb{Q}$ contains $\sqrt{3}$ and hence $M$ does not contain $\sqrt{2}$; thus $L\cap M$ cannot be a proper extension of $\mathbb{Q}$; similar arguments for the other two intersections).
However, $MN$ contains all $2^n$th roots of both $3$ and $6$, hence contains all $2^n$th roots of $2$, hence contains $L$, so $L\cap MN = L$ is infinite dimensional over $\mathbb{Q}$. Similar arguments show $M\cap LN=M$ and $N\cap LM=N$.
Alternatively, take the set of primes and partition it into two infinite sets, $P_1$ and $P_2$; then take $L = \mathbb{Q}(\{\sqrt{p}\mid p\in P_1\})$, $M=\mathbb{Q}(\{\sqrt{p}\mid p\in P_2\})$. Then write $P_1=\{p_1\lt p_2\lt p_3\lt\cdots\}$, $P_2 = \{q_1\lt q_2\lt q_3\lt\cdots\}$, and define $T = \{p_iq_i\mid i=1,2,3,\ldots\}$ and let $N=\mathbb{Q}(\{\sqrt{t}\mid t\in T\})$; again, $L\cap M=L \cap N=M\cap N = \mathbb{Q}$, but $L\subseteq MN$, $M\subseteq LN$, and $N\subseteq LM$.