CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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

Statement refuted

The false converse is: every antisymmetric bilinear function is alternating. Over $R=\mathbb Z/2$ , define $F:M_2(R)\to R$ on columns $x,y\in R^2$ by $F([x\mid y])=x_0y_0$ . Then $F$ is bilinear and antisymmetric but not alternating.

Facts & Assumptions

Given: The field $R=\mathbb Z/2$ 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]

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

[L3]

In $\mathbb Z/2$ , addition and multiplication are modulo $2$ , so $-1=1$ (For every natural $n$ , $(\mathbb{Z}/n,+)$ 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 $F([y\mid x])=y_0x_0=x_0y_0=-F([x\mid y])$ because multiplication is commutative and $-1=1$ in $R$ . Thus $F$ is antisymmetric.

L1L2L3algebra

2.1

For $e_0=(1,0)^{\mathsf T}$ , one has $F([e_0\mid e_0])=1$ , so $F$ does not vanish on equal columns and is not alternating.

step 1.1L1algebra∎

Depends on

Used by

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Direct dependencies and their dependencies through the next three levels: 60 results over 17 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

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

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