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.

The elementary symmetric polynomials are algebraically independent over the coefficient ring

Statement

The elementary symmetric polynomials $e_{1}, \dots, e_{n}$ are algebraically independent over $R$ : if $Q \in R [T_{1}, \dots, T_{n}]$ satisfies $Q (e_{1}, \dots, e_{n}) = 0$ , then $Q = 0$ .

Facts & Assumptions

Given: A commutative ring $R$ and a polynomial $Q \in R [T_{1}, \dots, T_{n}]$ .

[L1]

Over a commutative ring with $1 \neq 0$ , distinct exponent tuples $b$ give the monomials $e_{1}^{b_{1}} \dots e_{n}^{b_{n}}$ distinct leading multidegrees, each with leading coefficient $1$ (The leading multidegree of $e_{1}^{b_{1}} \dots e_{n}^{b_{n}}$ is $(b_{1} + \dots + b_{n}, b_{2} + \dots + b_{n}, \dots, b_{n})$ with coefficient one).

[L2]

A polynomial in an iterated polynomial ring has finite support and is zero exactly when every coefficient is zero (Polynomial rings in finitely many commuting indeterminates by iteration).

Proof

technique · contradiction

1.1assume-contra

Suppose, for contradiction, that $Q \neq 0$ but $Q (e_{1}, \dots, e_{n}) = 0$ .

2.1step 1.1givenL1L2choose

Since $Q \neq 0$ , some coefficient $c_{b}$ is nonzero by [L2], so $1 \neq 0$ in $R$ and [L1] applies. Among the finitely many monomials $c_{b} T_{1}^{b_{1}} \dots T_{n}^{b_{n}}$ of $Q$ with $c_{b} \neq 0$ , choose one whose substituted leading multidegree is greatest.

3.1step 2.1L1algebra

By [L1], no other substituted monomial has that leading multidegree, and the chosen substituted monomial has leading coefficient $c_{b} \neq 0$ . Hence this term cannot cancel in $Q (e_{1}, \dots, e_{n})$ , even if $R$ has zero divisors.

4.1step 1.1step 3.1L2discharge-contradiction∎

This contradicts $Q (e_{1}, \dots, e_{n}) = 0$ , whose every coefficient is zero by [L2]. Therefore $Q = 0$ .

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

K. Conrad, Symmetric Polynomials, Section 2 (standard reference, not scraped)