TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Fundamental theorem of symmetric polynomials: unique expression as a polynomial in $e_{1}, \dots, e_{n}$

Statement

For every commutative ring $R$ and every $n \in N$ , substitution $T_{k} \mapsto e_{k}$ is an $R$ -algebra isomorphism

$R [T_{1}, \dots, T_{n}] ⟶ R [x_{1}, \dots, x_{n}]^{{Sym}_{n}} .$

Equivalently, every symmetric polynomial has a unique expression $Q (e_{1}, \dots, e_{n})$ .

Facts & Assumptions

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

[L1]

Every symmetric polynomial is $Q (e_{1}, \dots, e_{n})$ 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: $Q (e_{1}, \dots, e_{n}) = 0$ implies $Q = 0$ (The elementary symmetric polynomials are algebraically independent over the coefficient ring).

Proof

technique · direct

1.1

Substitution $T_{k} \mapsto e_{k}$ defines an $R$ -algebra homomorphism whose image lies in the symmetric-polynomial subring.

givenalgebra

1.2

The map is surjective by [L1].

1.3

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

2.1

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

step 1.1step 1.2step 1.3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 12 results over 7 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

K. Conrad, Symmetric Polynomials, Section 2 (standard reference, not scraped)
D. Grinberg, An Introduction to Algebraic Combinatorics, Theorem 7.1.16 (standard reference, not scraped)

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