Nontrivial Idempotents in $\mathbb Z_3\times\mathbb Z_3$

Detailed Solution

An element $e$ of a ring is called idempotent if $$ e^2=e. $$ In the product ring $R=\mathbb Z_3\times\mathbb Z_3$, multiplication is defined componentwise: $$ (a,b)(c,d)=(ac,bd), $$ where all operations are performed modulo $3$. For $e=(1,0)$, we obtain $$ e^2=(1,0)(1,0)=(1^2,0^2)=(1,0)=e. $$ Thus $e$ is idempotent. Moreover, $$ e=(1,0)\neq(0,0) $$ and $$ e=(1,0)\neq(1,1)=1_R. $$ Therefore, $e$ is a nontrivial idempotent, proving that option A is correct. We examine the remaining options. Option B is incorrect. Although $$ e(1,1)-e=(1,0)-(1,0)=(0,0), $$ this only shows that $(1,1)$ is the multiplicative identity. It does not establish nilpotency. In fact, since $e^2=e$, we have $$ e^n=e\neq0 $$ for every positive integer $n$. Hence $e$ is not nilpotent. Option C is incorrect. A nonzero element need not be a unit. If $e=(1,0)$ had an inverse $(a,b)$, then $$ (1,0)(a,b)=(1,1). $$ But $$ (1,0)(a,b)=(a,0), $$ whose second component is always $0$. Therefore, no inverse exists, and $e$ is not a unit. Option D is incorrect. Although the only idempotents in $\mathbb Z_3$ are $0$ and $1$, the product ring contains the idempotents $$ (0,0),\quad(1,0),\quad(0,1),\quad(1,1). $$ The two idempotents $(1,0)$ and $(0,1)$ are nontrivial. Thus the correct option is $$ \boxed{\mathrm{A}}. $$