Nilradical ($\sqrt{(0)}$)

The nilradical of a commutative ring is its ideal of nilpotent elements, equivalently the radical of an ideal applied to its zero ideal.

For any multiplicative subset $S$ of a commutative ring $R$, fractions with nilpotent numerators are nilpotent in $S^{-1}R$. Conversely, if $(a/s{)}^n=0$, the vanishing criterion in a module localization gives $t\in S$ with $t{a}^n=0$. Then $(ta{)}^n=t^{n-1}t{a}^n=0$ and $a/s=ta/(ts)$ has a nilpotent numerator. This proves the formula for the nilradical without any finiteness assumption.

In $R=k[t_1,t_2,\dots ]/(t_1^2,t_2^2,\dots )$, the nilradical is generated by all ${\bar{t}}_i$. Every element uses finitely many variables and is nilpotent, while each square-free product ${\bar{t}}_1\cdots {\bar{t}}_j$ is nonzero. Thus this nilradical is not a nilpotent ideal. A Noetherian ring cannot have this behavior: finitely many nilpotent generators give a nilpotent ideal.

For every commutative ring $R$,Every nilpotent element lies in every prime ideal. If $a$ is not nilpotent, an ideal maximal among those disjoint from $\{1,a,a^2,\dots \}$ is a prime ideal avoiding $a$, by the Zorn lemma and the maximality argument.