Nilpotent Ideal

Question

I read the definition of Nilpotent Ideal but having hard time grasping it. I need some simple examples to feel what the definition meant.

Answer 1

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 $$\text{nilrad}(R)=\bigcap_{P\in\text{Spec}(R)}P$$

Answer 2

$\{0\}$ is nilpotent in any ring

$R$ is not a nilpotent ideal in any ring $R$ with identity.

For any ideal $I\lhd R$, $I/I^n$ is a nilpotent ideal of $R/I^n$.

If you take a field and the polynomial ring $S=F[x_1,x_2,\ldots]$ in countably many indeterminates, and let $I$ be the ideal $(x_1, x_2^2, x_3^3, \ldots, x_n^n\ldots)$. In the quotient $R=S/I$, the ideal $J/I$ generated by $x_1, x_2, x_3\ldots$ consists completely of nilpotent elements, but it is not a nilpotent ideal because $(J/I)^k\neq\{0\}$ for any natural number $k$.