PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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The symmetric polynomials form a subring

Statement

For every commutative ring $R$ and every $n \in N$ , the symmetric polynomials $R [x_{1}, \dots, x_{n}]^{{Sym}_{n}}$ form a subring of $R [x_{1}, \dots, x_{n}]$ . For $n = 0$ this subring is $R$ .

Facts & Assumptions

Given: A commutative ring $R$ , a natural number $n$ , and symmetric polynomials $f, g \in R [x_{1}, \dots, x_{n}]$ .

[L1]

A polynomial is symmetric when every variable permutation fixes it (Symmetric polynomials as the invariants of variable permutations).

[L2]

A subset is a subring when it contains $1$ and is closed under addition, additive inverses, and multiplication (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

Proof

technique · direct

1.1

Every variable permutation fixes the constant polynomials $0$ and $1$ , so both are symmetric.

1.2

For every permutation $σ$ , substitution of permuted variables commutes with the ring operations, so $σ (f + g) = σ (f) + σ (g) = f + g$ , $σ (- f) = - σ (f) = - f$ , and $σ (f g) = σ (f) σ (g) = f g$ .

givenL1algebra

2.1

Thus the symmetric polynomials contain $1$ and are closed under addition, additive inverses, and multiplication, so they form a subring. When $n = 0$ every polynomial is a coefficient in $R$ and the assertion gives $R$ itself.

step 1.1step 1.2L2∎

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 15 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

D. Grinberg, An Introduction to Algebraic Combinatorics, Chapter 7, Section 7.1 (standard reference, not scraped)