Consider the ring $R=\mathbb Z_{36}$, and let $I$ be a nilpotent ideal.
If $a\in I$, then $a^k=0$, so $36\mid a^k$. Since $36=2^2\cdot3^2$ we have $2,3\mid36$. So $$I=\\{0,6,12,18,24,30\\}=(6)$$ is a the largest nilpotent ideal. $I$ is sometimes written as $$\text{nilrad}(R)=\sqrt{(0)}=\\{a\in R\mid a^k=0\text{ for some }k\in\mathbb N\\}=(6)$$ Other examples of nilpotent ideals are $(0), (12), (18)\subset (6)$.
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A useful theorem:
> Let $R$ be a commutative ring and let $\text{Spec}(R)$ denote the set of prime ideals in $R$, then $$\text{nilrad}(R)=\bigcap_{P\in\text{Spec}(R)}P$$