Complement of an Idempotent Element

Detailed Solution

Since $e$ is idempotent, we have $$ e^2=e. $$ Let $f=1-e$. We prove that $f$ is also idempotent and that $ef=fe=0$. First, calculate $f^2$: $$ \begin{aligned} f^2&=(1-e)^2\\ &=1-e-e+e^2\\ &=1-2e+e\\ &=1-e\\ &=f. \end{aligned} $$ Here we used $e^2=e$. Thus $f$ is idempotent. Next, using the distributive law, $$ \begin{aligned} ef&=e(1-e)\\ &=e-e^2\\ &=e-e\\ &=0. \end{aligned} $$ Similarly, $$ \begin{aligned} fe&=(1-e)e\\ &=e-e^2\\ &=e-e\\ &=0. \end{aligned} $$ Therefore, $$ \boxed{f^2=f,\qquad ef=fe=0}. $$ These identities require neither commutativity nor any additional assumptions beyond the existence of unity and the idempotency of $e$. Hence option A is correct. Option B is incorrect because the proof of $f^2=f$ uses only distributivity and $e^2=e$, not commutativity. Option C is incorrect. Nontrivial idempotents can also satisfy these identities. For example, in $M_2(\mathbb R)$, take $$ e=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \qquad f=I-e=\begin{pmatrix}0&0\\0&1\end{pmatrix}. $$ Then $e\neq0$, $e\neq I$, and $ef=fe=0$. Option D is incorrect because $f=1-e$ need not be a unit. For example, in $M_2(\mathbb R)$, the matrix $$ f=\begin{pmatrix}0&0\\0&1\end{pmatrix} $$ is idempotent but singular, so it is not invertible. Thus the correct option is $\boxed{\mathrm{A}}$.