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CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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Ordinary trace on a p-singular unipotent element is not a Brauer-character value

Statement refuted

For every element of a modular representation, the ordinary matrix trace is the Brauer-character value.

Facts & Assumptions

Given: The cyclic group Cp=u over a field k of characteristic p, with u acting on k2 by (1101).

[L1]

The false statement above is the claim to be refuted (FALSE: a Brauer character is defined on all elements by the usual trace).

[F1]

Brauer characters are defined only on p-regular elements (Brauer character of a finite-dimensional kG-module).

Counterexample

technique · direct
1.1

The element u has order p, so it is p-singular. Its action matrix still has ordinary trace 2.

givenalgebra
2.1

By [F1], the Brauer character of this module is not defined at u, because uCp0. So the trace from step 1.1 cannot be a Brauer-character value.

F1step 1.1
3.1

This directly refutes the statement [L1].

L1step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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