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Let $F_1$, $F_2$ and $K$ be fields of characteristic $0$ such that $F_1 \cap K = F_2 \cap K = M$, the extensions $F_i / (F_i \cap K)$ are Galois, and $[F_1 \cap F_2 : M ]$ is finite. Then is $[F_1 F_2 \cap K : M]$ finite?

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No. First, the extension $\mathbb{C}(z)/\mathbb{C}$ is Galois for any $z$ that is transcendental over $\mathbb{C}$: if $p(z)$ is a nonconstant polynomial, and $\alpha$ is a root and $\beta$ is a nonroot, then the automorphism $\sigma\colon z\mapsto z+\alpha-\beta$ maps $z-\alpha$ to $z-\beta$, so that $p(z)$ has a factor of $z-\alpha$ and no factor of $z-\beta$ but $\sigma p$ has a factor of $z-\beta$, so $p\neq\sigma p$. Thus, no polynomial is fixed by all automorphisms. If $\frac{p(z)}{q(z)}$ is a rational function with $q(z)$ nonconstant, then a similar argument shows that we can find a $\sigma$ that "moves the zeros" of $q$, so that $p(z)/q(z)$ will have a pole at $\alpha$ but no pole at $\beta$, while $\sigma p/\sigma q$ has a pole at $\beta$. Thus, the fixed field of $\mathrm{Aut}(\mathbb{C}(z)/\mathbb{C})$ is $\mathbb{C}$, hence the extension is Galois.

Now, let $x$ and $y$ be transcendental over $\mathbb{C}$. Take $F_1=\mathbb{C}(x)$, $F_2=\mathbb{C}(y)$, $K=\mathbb{C}(xy)$, all subfields of $\mathbb{C}(x,y)$. Then $M=\mathbb{C}$, so $F_i/M$ is Galois; $F_1\cap F_2=\mathbb{C}$, so $[F_1\cap F_2\colon M]=1$. But $F_1F_2\cap K = \mathbb{C}(x,y)\cap \mathbb{C}(xy) = \mathbb{C}(xy)$, and $\mathbb{C}(xy)$ is of infinite degree over $M=\mathbb{C}$

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