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

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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

1.1

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.

givenL3

1.2

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.

givenL1

2.1

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].

step 1.2L2L3algebra

3.1

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.

step 2.1

4.1

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

step 3.1algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 14 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)