CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13

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In characteristic $2$ , distinct quadratic forms can have the same polar form

Statement refuted

Refuted claim: A quadratic form is determined by its polar form in every characteristic.

Facts & Assumptions

Given: On $F_{2}^{2}$ , define $q_{0} (x, y) = 0$ and $q_{1} (x, y) = x^{2}$ .

[L1]

The ring $F_{2}$ is a field of characteristic $2$ (For every prime $p$ , the two operations on $Z / p$ make it a field).

[L2]

A quadratic form has $q (a v) = a^{2} q (v)$ and bilinear polar form $b_{q} (u, v) = q (u + v) - q (u) - q (v)$ (A quadratic form $q$ in arbitrary characteristic and its polar form $b_{q} (u, v) = q (u + v) - q (u) - q (v)$ ).

Counterexample

technique · direct computation

1.1

Both $q_{0}$ and $q_{1}$ have the required degree-two homogeneity. The polar form of $q_{0}$ is zero, while for $u = (x, y)$ and $v = (s, t)$ , $b_{q_{1}} (u, v) = (x + s)^{2} - x^{2} - s^{2} = 2 x s = 0$ by [L1]. Thus both are quadratic forms with zero polar form.

L1L2algebra

1.2

They are distinct because $q_{0} (1, 0) = 0$ while $q_{1} (1, 0) = 1$ .

givenalgebra

2.1

Hence polarization is not injective in characteristic $2$ .

step 1.1step 1.2∎

Depends on

Used by

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Sources

K. Conrad, Bilinear Forms, §7 (standard reference, not scraped)

In characteristic 22, distinct quadratic forms can have the same polar form