TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

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 $Q (e_{1}, \dots, e_{n})$ .

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

[L1]

[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).

technique · direct

1.1givenalgebra

Substitution $T_{k} \mapsto e_{k}$ 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