= Solution
Statement (iii) is \b[false over an arbitrary field]. Expanding the assumed idempotence gives
$$
(\alpha+\beta)^2=\alpha+\beta
\quad\Longrightarrow\quad
\alpha\beta+\beta\alpha=0.
$$
Multiplying this relation on the left and right by $\alpha$ shows
$$
\alpha\beta=-\alpha\beta\alpha=\beta\alpha.
$$
\b[Thus $2\alpha\beta=0$, 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 $\alpha\beta=I\ne0$.]
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