# Algebraic independence of elementary symmetric polynomials

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

$$e_1=x_1+x_2+\cdots+x_n\\ e_2=\sum_{i Let $$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.

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.