CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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.

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If $n!$ is invertible, then $p_{1}, \dots, p_{n}$ freely generate the symmetric-polynomial ring

Statement

Let $R$ be a commutative ring in which $n! 1_{R}$ is a unit. Then substitution $P_{k} \mapsto p_{k}$ is an $R$ -algebra isomorphism

$R [P_{1}, \dots, P_{n}] ⟶ R [x_{1}, \dots, x_{n}]^{{Sym}_{n}} .$

If $R$ is a field, the unit hypothesis is equivalent to $char R = 0$ or $char R > n$ .

Facts & Assumptions

Given: A commutative ring $R$ in which $n! 1_{R}$ is invertible.

[L1]

Newton's identities are $k e_{k} = \sum_{i = 1}^{k} (- 1)^{i - 1} e_{k - i} p_{i}$ for $k \geq 1$ (Newton's identities: $k e_{k} = \sum_{i = 1}^{k} (- 1)^{i - 1} e_{k - i} p_{i}$ ).

[L2]

The elementary symmetric polynomials freely generate the symmetric-polynomial ring (Fundamental theorem of symmetric polynomials: unique expression as a polynomial in $e_{1}, \dots, e_{n}$ ).

[L3]

The factorial satisfies $n! = 1 \cdot 2 \dots n$ , with $0! = 1$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

Proof

technique · direct

1.1givenL3algebra

For each $1 \leq k \leq n$ , the element $k 1_{R}$ is a unit: the product of $k 1_{R}$ with the images of all the other factors in $n!$ is the unit $n! 1_{R}$ , and a factor of a unit in a commutative ring is a unit.

2.1step 1.1L1algebra

Using the inverse of $k 1_{R}$ , [L1] recursively expresses $e_{k}$ as a polynomial in $p_{1}, \dots, p_{k}$ . Conversely [L1] expresses $p_{k}$ as $(- 1)^{k - 1} k e_{k}$ plus a polynomial in $e_{1}, \dots, e_{k - 1}$ .

3.1step 2.1L2

These mutually inverse triangular substitutions have unit diagonal coefficients, so they give an isomorphism $R [e_{1}, \dots, e_{n}] ≅ R [p_{1}, \dots, p_{n}]$ . Composing with [L2] proves free generation.

4.1step 1.1algebra∎

In a field, a positive integer image is a unit exactly when it is nonzero. Thus all of $1, \dots, n$ are nonzero exactly in characteristic zero or characteristic greater than $n$ , which is equivalent to the factorial image being nonzero and hence invertible.

Depends on

Used by

Newton's identities through p₄ in three variables Example

Dependency tree · two levels

23 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 3 (standard reference, not scraped)
D. Grinberg, An Introduction to Algebraic Combinatorics, Chapter 7, Section 7.1 (standard reference, not scraped)

If n!n! is invertible, then p1,…,pnp1​,…,pn​ freely generate the symmetric-polynomial ring