Algebraic independence of elementary symmetric polynomials

Question

I am following the book Lectures on Algebra by Abhyankar, p. 638.

Let $k$ be a field, $x_1,x_2,\ldots,x_n$ be independent variables, and define

$$e_1=x_1+x_2+\cdots+x_n\\ e_2=\sum_{i<j} x_ix_j\\ \cdots\\ e_n=x_1x_2\cdots x_n.$$ Let $L=k(x_1,\ldots,x_n)$, the field of rational functions in $x_1,\ldots,x_n$ over $k$.

Let $K:=k(e_1,e_2,\ldots,e_n)$. Now the author says the following:

$L/K$ is a splitting field (extension) of the separable polynomial $(y-x_1)(y-x_2)\cdots (y-x_n)\in K[y]$. Consequently, $(e_1,e_2,\ldots, e_n)$ is a transcendence basis of $L/K$.

Question: I think, he wants to say that it is a transcendence basis of $L/k$, which by definition according to Wikipedia says that:

$e_1,e_2,\ldots,e_n$ are algebraically independent over $k$; and $L$ is algebraic extension of $k(e_1,e_2,\ldots,e_n)$.

So, is it correct that $(e_1,\ldots,e_n)$ is transcendence basis of $L/k$? My main question is the following:

$L/K$ is a splitting field (extension) of the separable polynomial $(y-x_1)(y-x_2)\cdots (y-x_n)\in K[y]$ (OKAY). Consequently [HOW?], $(e_1,e_2,\ldots, e_n)$ is a transcendence basis of $L/k$.

I do not get how the author concludes about the set $(e_1,e_2,\ldots, e_n)$ as transcendence basis of $L/k$ without any computation/justification? I tried to understand from the splitting field of separable polynomial, but I could not touch it properly.

Answer 1

You are correct: The elementary symmetric polynomials $e_1,\ldots,e_n$ are elements of $K$, so they are certainly not algebraically independent over $K$, and hence they cannot form a transcendence basis of anything over $K$.

The first quote shows that $L$ is algebraic over $K=k(e_1,\ldots,e_n)$ because it is the splitting field of a certain polynomial in $K[y]$. So to show that $(e_1,\ldots,e_n)$ is a transcendence basis of $L/k$ it remains to show that $(e_1,\ldots,e_n)$ is algebraically independent over $k$.

Suppose toward a contradiction that $(e_1,\ldots,e_n)$ is algebraically dependent over $k$. Then $L/k$ has a transcendence basis of fewer than $n$ elements, and so the transcendence degree of $L/k$ is less than $n$. But of course $(x_1,\ldots,x_n)$ is also a transcendence basis of $L/k$, and so the transcendence degree of $L/k$ equals $n$; a contradiction.