A non-Noetherian nilradical need not be nilpotent

Let $\mathfrak{n}=({\bar{x}}_1,{\bar{x}}_2,\dots )$. Each generator ${\bar{x}}_i$ is nilpotent, and every element of $\mathfrak{n}$ involves only finitely many generators, so in this commutative ring it is a finite sum of commuting nilpotents and hence nilpotent. Therefore $\mathfrak{n}\subseteq \mathrm{Nil}(A)$. Conversely, the quotient $A\mathrm{/}\mathfrak{n}\cong k$ is reduced, so every nilpotent element of $A$ lies in $\mathfrak{n}$. Thus $\mathrm{Nil}(A)=\mathfrak{n}$.

For every integer $N\ge 1$, the element ${\bar{x}}_{N+1}^N$ lies in ${\mathfrak{n}}^N$. It is nonzero in $A$ because only the $(N+1)$-st power of $x_{N+1}$ is killed by the defining relations. Hence ${\mathfrak{n}}^N\mathrm{\ne }0$ for every $N$, so $\mathfrak{n}$ is not nilpotent.

This shows that the Noetherian hypothesis in the nilradical-nilpotence theorem is essential.