Alphabeta Math
CorollaryStatement: 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 complete homogeneous symmetric polynomials h1,,hn freely generate the symmetric-polynomial ring

Statement

Substitution Ukhk is an R-algebra isomorphism

R[U1,,Un]R[x1,,xn]Symn.

Thus h1,,hn freely generate the symmetric-polynomial ring.

Facts & Assumptions

Given: A commutative ring R and a natural number n.

[L1]

The identity E(t)H(t)=1 gives hke1hk1++(1)kek=0 for 1kn (The generating-series identity E(t)H(t)=1).

[L2]

Substitution Tkek is an isomorphism from a polynomial ring onto the symmetric-polynomial ring (Fundamental theorem of symmetric polynomials: unique expression as a polynomial in e1,,en).

Proof

technique · direct
1.1

The recurrence in [L1] expresses hk as (1)k1ek plus a polynomial in e1,,ek1, and also expresses ek as (1)k1hk plus a polynomial in h1,,hk1.

L1algebra
2.1

Recursion on k therefore gives mutually inverse triangular substitutions between R[e1,,en] and R[h1,,hn]; every diagonal coefficient is 1 or 1, hence a unit in R.

step 1.1algebra
3.1

Composing either triangular isomorphism with [L2] shows that Ukhk is an R-algebra isomorphism onto the symmetric-polynomial ring.

step 2.1L2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 12 results over 6 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