CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11

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Over $Z / 2$ , an antisymmetric bilinear form need not be alternating

Statement refuted

The false converse is: every antisymmetric bilinear function is alternating. Over $R = Z / 2$ , define $F : M_{2} (R) \to R$ on columns $x, y \in R^{2}$ by $F ([x ∣ y]) = x_{0} y_{0}$ . Then $F$ is bilinear and antisymmetric but not alternating.

Facts & Assumptions

Given: The field $R = Z / 2$ and the displayed function $F$ .

[L1]

Antisymmetric means that swapping columns negates the value, while alternating means vanishing on equal columns (Column-multilinear, alternating, normalized and antisymmetric functions on square matrices over a commutative ring).

[L2]

$Z / 2$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

[L3]

In $Z / 2$ , addition and multiplication are modulo $2$ , so $- 1 = 1$ (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

Counterexample

technique · direct

1.1

The coordinate product is linear in each column. Moreover $F ([y ∣ x]) = y_{0} x_{0} = x_{0} y_{0} = - F ([x ∣ y])$ because multiplication is commutative and $- 1 = 1$ in $R$ . Thus $F$ is antisymmetric.

L1L2L3algebra

2.1

For $e_{0} = (1, 0)^{T}$ , one has $F ([e_{0} ∣ e_{0}]) = 1$ , so $F$ does not vanish on equal columns and is not alternating.

step 1.1L1algebra∎

Depends on

Used by

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Dependency tree · two levels

16 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

S. New, MATH 146 Linear Algebra 1 Lecture Notes, Theorem 4.19 (standard reference, not scraped)

Over Z/2Z/2, an antisymmetric bilinear form need not be alternating