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.