PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck 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

$Δ_{n}^{2} = \prod_{1 \leq i < j \leq n} (x_{i} - x_{j})^{2}$

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

Facts & Assumptions

Given: A commutative ring $R$ and variables $x_{1}, \dots, x_{n}$ .

[L1]

The Vandermonde polynomial is $Δ_{n} = \prod_{i < j} (x_{i} - x_{j})$ (The Vandermonde polynomial $Δ_{n} = \prod_{i < j} (x_{i} - x_{j})$ ).

[L2]

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

[L3]

A permutation is a bijection of the index set (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

Proof

technique · direct

1.1L1L3

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

2.1step 1.1algebra

Reversing a difference has no effect after squaring, since $(u - v)^{2} = (v - u)^{2}$ in every commutative ring, including characteristic two. Hence the permutation merely reorders the factors of $Δ_{n}^{2}$ .

3.1step 2.1L1L2∎

Every variable permutation fixes $Δ_{n}^{2}$ , so it is symmetric by [L2]. For $n = 0, 1$ , the product is $1$ and the same conclusion holds.

Depends on

Used by

The discriminant of a monic polynomial as the coefficient expression of Δₙ² Definition

Dependency tree · two levels

8 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

J. S. Milne, Fields and Galois Theory, Proposition 4.35 through Example 4.37 (standard reference, not scraped)