field theory - How to see that $M$ is Galois over $k$ in Lang's proof that solvable extensions are a distinguished class? (Prop. VI.7.1, *Algebra*)

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 $k$?

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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 $k \subset E \subset L$.

Answer

Expanding on the comment above by @renus.

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$L$ 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

We let $M$ be the compositum of all extensions $\sigma L$ for all embeddings $\sigma$ of $L$ over $\color{green}{k}$.

instead as

We let $M$ be the compositum of all extensions $\sigma L$ for all embeddings $\sigma$ of $L$ over $\color{red}{K}$.

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.