# 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 ProofOfAlgebraicIndependenceOfElementarySymmetricPolynomials 2013-03-22 17:38:28 2013-03-22 17:38:28 lalberti (18937) lalberti (18937) 4 lalberti (18937) Proof msc 05E05