Let $A$ be a ring, let $D$ be the set of the zero-divisors and $S:=A-D$. Show that any minimal prime ideal of $A$ is contained in $D$. I know that this question has already been asked, but (in the one I saw) it was requested a proof not involving localization. Since I took this exercise from Atiyah-Macdonald, I would like to know if my proof (that relies on localization, being it the subject of the chapter) is correct. Thank you
If $p$ is a minimal prime, $T:=A-p$ is a maximal multiplicatively closed set. Being $T$ maximal, $S\subseteq T$: in fact if $x\notin T$ is not a zero-divisor, $T\subset Tx$ is a proper multiplicatively closed set, since $0\notin Tx$. So $p\subseteq D$.