I am studying from Lang's Algebra, and in Chapter VI Galois Theory, $\S$7 Solvable and Radical Extensions, he states and proves the following proposition (on pages 291-292, third edition):
Proposition 7.1. Solvable extensions form a distinguished class of extensions.
The proof goes as follows:
First, we show that if $E/k$ is solvable and $F/k$ is any extension such that $E$ and $F$ are both subfields of some algebraically closed field, then $EF/F$ is solvable.
Second, we observe that if $E/k$ is solvable, then $E/F$ and $F/k$ are also solvable for any intermediate field $F$.
Lastly, assume that $E \supset F \supset k$ is an extension of fields such that $E/F$ and $F/k$ are solvable. We need to show that $E/k$ is solvable. Start by letting $K$ be a finite Galois extension of $k$ containing $F$. By the first part, $EK/K$ is solvable, so let $L$ be a solvable Galois extension of $K$ containing $EK$.
Now, Lang writes:
If $\sigma$ is any embedding of $L$ over $k$ in a given algebraic closure, then $\sigma K = K$ and hence $\sigma L$ is a solvable extension of $K$. We let $M$ be the compositum of all extensions $\sigma L$ for all embeddings $\sigma$ of $L$ over $k$. Then $M$ is Galois over $k$, and is therefore Galois over $K$.
I don't understand why $M$ is Galois over $k$. We chose $L$ to be a solvable Galois extension of $K$, hence all the conjugates of $L$ in an algebraic closure are also solvable Galois extensions of $K$; hence, the compositum $M$ is a Galois extension of $K$. How does Lang say that $M$ is Galois over $k$?
For the sake of completeness, the rest of the proof runs as follows: The Galois group of $M$ over $K$ is a subgroup of the product $$ \prod_\sigma G(\sigma L/K) $$ and hence it is solvable. The map $G(M/k) \to G(K/k)$ given by restriction is a surjective homomorphism with kernel $G(M/K)$. Hence, the Galois group of $M/k$ has a solvable normal subgroup $G(M/K)$ whose factor group $G(K/k)$ is solvable. Thus, $G(M/k)$ is itself solvable. Since $E \subset M$, we are done.
Also, here is Lang's definition of a solvable extension:
A finite extension $E/K$ (which we shall assume separable for convenience) is said to be solvable if the Galois group of the smallest Galois extension $K$ of $k$ containing $E$ is a solvable group. This is equivalent to saying that there exists a solvable Galois extension $L$ of $k$ such that $k \subset E \subset L$.