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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/8f/iii/solution

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Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 8F / iii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

Statement (iii) is false over an arbitrary field. Expanding the assumed idempotence gives

(α + β)^{2} = α + β ⟹ α β + β α = 0.

(1)

Multiplying this relation on the left and right by

α

shows

α β = - α β α = β α .

(2)

Thus $2 α β = 0$ , so the conclusion would be true in characteristic different from two. In characteristic two, however, take $α = β = I$ . Then $α$ , $β$ , and $α + β = 0$ are idempotent, but $α β = I \neq = 0$ .

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