Nilpotent elements of Z_9

Detailed Solution

An element $x$ is nilpotent if $x^k\equiv0\pmod9$ for some positive integer $k$. Checking each residue: $3^2=9\equiv0$, so $3$ is nilpotent; $6^2=36\equiv0\pmod9$ (since $36=4\times9$), so $6$ is nilpotent; and $0$ is trivially nilpotent. Every other nonzero residue ($1,2,4,5,7,8$) is coprime to $9$, hence a unit in $\mathbb Z_9$, and a unit can never be nilpotent (if $x^k=0$ and $x$ were invertible, multiplying by $x^{-k}$ would give $1=0$, absurd). So the complete set of nilpotents is $\{0,3,6\}$. Thus the correct option is $\boxed{\mathrm{C}}$.