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 closureMathworldPlanetmath K of Q.

Consider the substitution map that associates to values t1,,tnKn the symmetric functions in these variables s1,,sn.


Because K is algebraic closed this map is surjectivePlanetmathPlanetmath. Indeed, fix values vi, then on an algebraic closed field there are roots ti such that


And by developing the right-hand side we get vi=si.

Then we consider the transpositionMathworldPlanetmath morphism of algebras ϕ* :


The capital letters are there to emphasize the Si and Ti are variables and R[S1,,Sn] and R[T1,,Tn] are regarded as function algebras over Kn.

The theorem stating that the symmetric functions are algebraically independentMathworldPlanetmath is no more than saying that this morphism is injectivePlanetmathPlanetmath. As a matter of fact, ϕ*(Si) is the ith symmetric function in the Ti, and ϕ* is clearly a morphism of algebras.
The conclusionMathworldPlanetmath is then straightforward from the surjectivity of ϕ because if fϕ=0 for some f, then by surjectivity of ϕ it means that f was zero in the first place. In other words the kernel of ϕ* 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