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 $$k$$?

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 $$k \subset E \subset L$$.

• If $L/k$ is a finite extension then $L = k(a),a \in L^n$ and $M=\prod_{\sigma \in Aut(\overline{L}/k)} \sigma(L)=L( \{ \sigma(a),\sigma \in Aut(\overline{L}/k)\})$ is a normal extension of $k$, if also $L/k$ is separable then $M/k$ is Galois, it is the splitting field of $\prod_{j=1}^n \text{minpoly}_k(a_j)$. Also why don't you show the solvable extensions are the radical extensions ? – reuns Jun 6 at 23:35
• @reuns Ah, thank you for your comment, I see it now. In my head, I was thinking that we are taking the embeddings of $L$ over $K$, not $k$. Lang proves that solvable extensions are radical extensions in the next proposition. – Brahadeesh Jun 7 at 4:15

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

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

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.