proof of algebraic independence of elementary symmetric polynomials

Geometric proof, works when R is a division ring.

Consider the quotient field Q of R and then the algebraic closure $K$ of $Q$ .

Consider the substitution map that associates to values $t_{1},\ldots,t_{n}\in K^{n}$ the symmetric functions in these variables $s_{1},\ldots,s_{n}$ .

\begin{array}[]{lccr}\phi:&K^{n}&\to&K^{n}\\ &(t_{i})&\mapsto&(s_{i})\end{array}

Because $K$ is algebraic closed this map is surjective. Indeed, fix values $v_{i}$ , then on an algebraic closed field there are roots $t_{i}$ such that

X^{n}+\sum_{i}v_{i}X^{i}=\Pi_{i}(X+t_{i})

And by developing the right-hand side we get $v_{i}=s_{i}$ .

Then we consider the transposition morphism of algebras $\phi^{*}$ :

\begin{array}[]{lccr}\phi^{*}:&R[S_{1},\ldots,S_{n}]&\to&R[T_{1},\ldots,T_{n}]% \\ &f&\mapsto&f\circ\phi\end{array}

The capital letters are there to emphasize the $S_{i}$ and $T_{i}$ are variables and $R[S_{1},\ldots,S_{n}]$ and $R[T_{1},\ldots,T_{n}]$ are regarded as function algebras over $K^{n}$ .

The theorem stating that the symmetric functions are algebraically independent is no more than saying that this morphism is injective. As a matter of fact, $\phi^{*}(S_{i})$ is the $i^{th}$ symmetric function in the $T_{i}$ , and $\phi^{*}$ is clearly a morphism of algebras.
The conclusion is then straightforward from the surjectivity of $\phi$ because if $f\circ\phi=0$ for some $f$ , then by surjectivity of $\phi$ it means that $f$ was zero in the first place. In other words the kernel of $\phi^{*}$ is reduced to 0.

Title	proof of algebraic independence of elementary symmetric polynomials
Canonical name	ProofOfAlgebraicIndependenceOfElementarySymmetricPolynomials
Date of creation	2013-03-22 17:38:28
Last modified on	2013-03-22 17:38:28
Owner	lalberti (18937)
Last modified by	lalberti (18937)
Numerical id	4
Author	lalberti (18937)
Entry type	Proof
Classification	msc 05E05