Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16
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Fundamental theorem of symmetric polynomials: unique expression as a polynomial in e1,…,en

Statement

For every commutative ring R and every n∈N, substitution Tk↦ek is an R-algebra isomorphism

R[T1,…,Tn]⟶R[x1,…,xn]Sym⁡n.

Equivalently, every symmetric polynomial has a unique expression Q(e1,…,en).

Facts & Assumptions

Given: A commutative ring R and a natural number n.

[L1]

Every symmetric polynomial is Q(e1,…,en) for some polynomial Q (Every symmetric polynomial over a commutative ring is a polynomial in the elementary symmetric polynomials).

[L2]

The elementary symmetric polynomials are algebraically independent: Q(e1,…,en)=0 implies Q=0 (The elementary symmetric polynomials are algebraically independent over the coefficient ring).

Proof

technique · direct
1.1givenalgebra

Substitution Tk↦ek defines an R-algebra homomorphism whose image lies in the symmetric-polynomial subring.

1.2L1

The map is surjective by [L1].

1.3L2

Its kernel is zero by [L2], so it is injective.

2.1step 1.1step 1.2step 1.3∎

The substitution map is therefore an isomorphism. Surjectivity gives existence of an expression, and injectivity gives its uniqueness.

Depends on

Used by

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources