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

Statement (iii) is false over an arbitrary field. Expanding the assumed idempotence givesMultiplying this relation on the left and right by $\alpha$ showsThus $2\alpha \beta =0$, so the conclusion would be true in characteristic different from two. In characteristic two, however, take $\alpha =\beta =I$. Then $\alpha$, $\beta$, and $\alpha +\beta =0$ are idempotent, but $\alpha \beta =I\ne 0$.