PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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The Vandermonde product transforms by the sign of the root permutation

Statement

Let $f \in F [x]$ be separable of degree $n$ , order its roots as $α_{1}, \dots, α_{n}$ , and put

$δ : = \prod_{1 \leq i < j \leq n} (α_{i} - α_{j}) .$

For $σ$ in the Galois group, let the same symbol denote its induced root permutation. Then $σ (δ) = sgn (σ) δ$ for the Vandermonde product of an ordered root list. Consequently $δ^{2}$ is fixed by the Galois group.

Facts & Assumptions

Given: The faithful root permutation action of A polynomial Galois group acts faithfully on its roots and the Vandermonde product of The Vandermonde polynomial $Δ_{n} = \prod_{i < j} (x_{i} - x_{j})$ .

[F1]

The sign of $σ$ is the integer $sgn (σ) : = (- 1)^{inv (σ)} \in {+ 1, - 1}$ , where $inv (σ)$ counts the pairs $i < j$ with $σ (i) > σ (j)$ (Inversions, inversion number, the sign $sgn (σ) = (- 1)^{inv (σ)}$ , and even and odd permutations).

[L1]

For every natural $n$ , the function $sgn : S_{n} \to {+ 1, - 1}$ is a group homomorphism (The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$ ).

[L2]

Every permutation of a finite set is a product of transpositions, and the identity is represented by the empty product (Every finite permutation is a product of transpositions, so the transpositions generate $S_{n}$ ).

[L3]

If $σ = τ_{1} \dots τ_{r}$ is any factorisation of a finite permutation into transpositions, then $(- 1)^{r} = (- 1)^{inv (σ)}$ (Every transposition factorisation of $σ$ has parity $(- 1)^{inv (σ)}$ ).

Proof

technique · direct

1.1algebra

Swapping two entries of the ordered root list reverses the factor belonging to that pair, while the remaining affected factors exchange in pairs; hence a transposition multiplies $δ$ by $- 1$ .

1.2F1L3

Taking the one-factor factorisation $τ = τ$ in [L3] gives $(- 1)^{inv (τ)} = - 1$ , so $sgn (τ) = - 1$ for every transposition $τ$ by [F1].

2.1step 1.1L2

By [L2] write $σ = τ_{1} \dots τ_{r}$ as a product of transpositions, and apply step 1.1 once for each factor: each application multiplies the current Vandermonde product by $- 1$ , so $σ (δ) = (- 1)^{r} δ$ .

3.1step 2.1step 1.2L1L2

By [L1] and step 1.2, $sgn (σ) = sgn (τ_{1}) \dots sgn (τ_{r}) = (- 1)^{r}$ , so step 2.1 gives $σ (δ) = sgn (σ) δ$ . For $n = 0$ or $n = 1$ the factorisation is empty by [L2], the product defining $δ$ is likewise empty and equals $1$ , and the sign is $1$ , so the identity holds there as $1 = 1$ .

4.1step 3.1algebra∎

Squaring the identity of step 3.1 removes the sign, so $σ (δ^{2}) = δ^{2}$ for every $σ$ . A root equal to zero creates no exception; separability ensures distinct roots and hence $δ \neq 0$ .

Depends on

Used by

For a monic separable polynomial in characteristic not two, the Galois group lies in Aₙ exactly when the discriminant is a square Theorem

Dependency tree · two levels

19 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, v5.10, Proposition 4.1 (standard reference, not scraped)