Proposition 15

Let A be a noetherian ring satisfying the following two conditions: (i) for every minimal prime ideal $p$ of $A$, the ring $A_p$ is reduced; (ii) for every prime ideal $p$ of A of height $\geq 1$, one has $\mathrm{prof}(A_p) \geq 1$. Then A is reduced.

Let $n$ denote the nilradical of A. For every minimal prime ideal $p$ of A, one has by (i) $n_p = 0$, that is, $p \not\in \mathrm{Supp}(n)$ and $a fortiori$ $p \not\in \mathrm{Ass}_A(n)$. For every $p \in \mathrm{Spec}(A)$ of height $\geq 1$, one has by (ii) $pA_p \not\in \mathrm{Ass}_{A_p}(A_p)$ and $a fortiori$ $pA_p \not\in \mathrm{Ass}_{A_p}(n_p)$, whence $p \not\in \mathrm{Ass}_A(n)$ (IV, § 1, No. 2, prop. 5). Thus the set $\mathrm{Ass}_A(n)$ is empty, which implies that $n$ is zero.