$K⊂L⊂M$ and $L/K$ and $M/K$ is galois. If $Gal(L/K)$ is isomorphic to $Gal(M/K)$ as group, is it true that $L＝M$? Let $K,L,M$ be fields.
Suppose $K⊂L⊂M$ and $L/K$ and $M/K$ is galois.
If $Gal(L/K)$ is isomorphic to  $Gal(M/K)$ as group, is it true that $L＝M$?
I think this is obviously true in the case $M/K$ is finite because isomorphism of galois group implies $L$ and $M$ has the same extension degree over $K$.
But what about $M/K$ is infinite case ? Are there any counterexamples?
 A: There are lots of counterexamples in the infinite case. Take, for example, $S$ and $T$ to be two infinite sets of prime numbers with $S \subset T$, and let $L = \mathbb{Q}(\sqrt{p} : p \in S)$ and $M = \mathbb{Q}(\sqrt{p} : p \in T)$. Clearly $\operatorname{Gal}(M/\mathbb{Q}) \cong \operatorname{Gal}(L/\mathbb{Q})$, and $L$ and $M$ aren't isomorphic since some rational primes are squares in $M$ but not in $L$.
A: Let $k$ be a field of characteristic different from $2$ and let $K=k(x_0,x_1,x_2,\dots)$ be a field of rational functions over $k$ in infinitely many variables.  Let $L=K(\sqrt{x_1},\sqrt{x_2},\dots)$ and let $M=L(\sqrt{x_0})$.  Then $Gal(M/K)$ and $Gal(L/K)$ are isomorphic, since both are an infinite product of cyclic groups of order $2$ (each factor corresponding to flipping the sign of one of the square roots).
(In fact, every profinite group is the Galois group of some field extension, so giving an example amounts to just finding any profinite group $G$ with a nontrivial closed normal subgroup $H$ such that $G/H\cong G$ (take $M/K$ with Galois group $G$ and let $L$ be the fixed field of $H$.  The example above is just an explicit implementation of this for $G=(\mathbb{Z}/2\mathbb{Z})^\mathbb{N}$.)
