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 $k \subset E \subset L$.
Best Answer
Expanding on the comment above by @renus.
$L$ is taken to be a solvable Galois extension of $K$ containing $EK$, so in particular, $L/K$ is separable. Since $K/k$ is taken to be a finite Galois extension, $K/k$ is also separable. Thus, $L/k$ is separable. The compositum $M$ of all the embeddings of $L$ into an algebraic closure of $k$ is a normal and separable extension of $k$, and thus Galois (see the discussion on page 242 of Lang's Algebra (third edition) just after the proof of Theorem 4.5).
My mistake was in misreading the line
instead as
It is also worthwhile to note that $L/k$ is a finite extension: since solvable extensions are a priori finite extensions as per Lang's definition, $L/K$ is finite, and since $K/k$ is taken to be finite, by the Tower Law we have that $L/k$ is finite. Thus, $M$ is a compositum of only finitely many fields, and is thus a finite extension of $k$. This is an assurance that there is no contradiction when we show later on in the proof that $M/k$ is solvable.