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.

Every symmetric polynomial over a commutative ring is a polynomial in the elementary symmetric polynomials

Statement

For every commutative ring $R$ and every $n \in N$ , each symmetric polynomial $f \in R [x_{1}, \dots, x_{n}]$ has the form

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

for some $Q \in R [T_{1}, \dots, T_{n}]$ .

Facts & Assumptions

Given: A commutative ring $R$ , a natural number $n$ , and a symmetric polynomial $f$ .

[L1]

The leading multidegree of a nonzero symmetric polynomial is weakly decreasing (The leading multidegree of a symmetric polynomial is weakly decreasing).

[L2]

Over a commutative ring with $1 \neq 0$ , the leading multidegree of $e_{1}^{b_{1}} \dots e_{n}^{b_{n}}$ is the cumulative-sum tuple of $b$ , 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).

[L3]

The symmetric polynomials form a subring (The symmetric polynomials form a subring).

Proof

technique · direct

1.1givenL3

The assertion is immediate for $f = 0$ and for $n = 0$ , when the symmetric-polynomial ring is $R$ . Assume now that $n > 0$ and $f \neq 0$ ; then some coefficient of $f$ is nonzero, so $1 \neq 0$ in $R$ and [L2] applies.

1.2givenL1

Let $a = (a_{1}, \dots, a_{n})$ be the leading multidegree of $f$ and let $c$ be its leading coefficient. By [L1], set $b_{i} = a_{i} - a_{i + 1}$ for $i < n$ and $b_{n} = a_{n}$ , all natural numbers.

2.1step 1.2L2L3algebra

By [L2], the polynomial $c e_{1}^{b_{1}} \dots e_{n}^{b_{n}}$ has the same leading term as $f$ , so their difference $f_{1}$ is either zero or has strictly smaller leading multidegree. It remains symmetric by [L3].

3.1step 2.1

Repeat step 2.1 while the remainder is nonzero. The process terminates because all exponent tuples encountered are bounded coordinatewise by the finite support box of the original polynomial and strictly decrease lexicographically at each subtraction.

4.1step 3.1algebra∎

Summing the finitely many subtracted monomials in the $e_{i}$ gives a polynomial $Q$ with $f = Q (e_{1}, \dots, e_{n})$ .

Depends on

Used by

Dependency tree · two levels

7 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)
D. Grinberg, An Introduction to Algebraic Combinatorics, Theorem 7.1.16 (standard reference, not scraped)