Proposition 12

Let $p$ be a minimal prime ideal of a ring $A$. For all $x \in p$, there exists $s \in A - p$ and an integer $k > 0$ such that $sx^k = 0$.

The set $S$ of elements of the form $sx^k$ ($k$ an integer $> 0$, $s \in A - p$) is a multiplicative subset of $A$. If $0 \notin S$, there would exist a prime ideal $p'$ not meeting $S$ (no. 5, Corollary 2 to Proposition 11). Then $p' = p$ and $p' \# p$ since $x \notin p'$, contrary to the hypothesis that $p$ is minimal.