Alphabeta Math
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 square of the Vandermonde polynomial is symmetric

Statement

For every commutative ring R and every n, the polynomial

Δn2=1i<jn(xixj)2

is symmetric. This includes characteristic two and the empty products for n=0,1.

Facts & Assumptions

Given: A commutative ring R and variables x1,,xn.

[L1]

The Vandermonde polynomial is Δn=i<j(xixj) (The Vandermonde polynomial Δn=i<j(xixj)).

[L2]

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

Proof

technique · direct
1.1

A variable permutation bijects the unordered pairs {i,j} with themselves. For each pair, it sends (xixj)2 to either (xσ(i)xσ(j))2 or the same factor with its two terms reversed.

L1L3
2.1

Reversing a difference has no effect after squaring, since (uv)2=(vu)2 in every commutative ring, including characteristic two. Hence the permutation merely reorders the factors of Δn2.

step 1.1algebra
3.1

Every variable permutation fixes Δn2, so it is symmetric by [L2]. For n=0,1, the product is 1 and the same conclusion holds.

step 2.1L1L2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 14 results over 10 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