Why does a minimal prime ideal consist of zerodivisors? Let $A$ be a commutative ring. Suppose $P \subset A$ is a minimal prime ideal. Then it is a theorem that $P$ consists of zero-divisors.
This can be proved using localization, when $A$ is noetherian: $A_P$ is local artinian, so every element of $PA_P$ is nilpotent.  Hence every element of $P$ is a zero-divisor. (As Matt E has observed, when $A$ is nonnoetherian, one can still use a similar argument: $PA_P$ is the only prime in $A_P$, hence is the radical of $A_P$ by elementary commutative algebra.)
Can this be proved without using localization?
 A: This is a comment on the proof sketch in the question: the localization $A_P$ is local of dimension zero (its unique maximal ideal is also a minimal prime ideal, and hence is the unique prime ideal of $A_P$), but need not be Artinian, as far as I can tell.  E.g. if $A =\mathbb C[[x^{1/2},x^{1/3},\dots]]/(x)$, then $A$ is local with a unique prime ideal (namely the ideal generated by all the $x^{1/n}$), but is not Artinian (equivalently, not Noetherian), since if $(a_i)$ is any strictly descending sequence of rational numbers in the interval $(0,1)$, then the ideals $x^{a_i}A$ form a strictly descending sequence of ideals in $A$.
(Hopefully I'm not blundering here; if I am, someone please let me know!)
(Also, I should add that it is still the case that since $PA_P$ is the unique minimal prime of $A_P$, every element of $PA_P$ is nilpotent, and hence every element of $P$ is a zero divisor, so my comment is very nitpicky: it is just about the use of the terminology Artinian.)
A: In the early days of commutative algebra people did not use the language of localization, so probably such proof exists. In particular, I have been told Kaplansky's early book (around 1950?) might be such reference, but I don't have it right now.  
In any case, Exercise 2.3 of Eisenbud tells you how to cheat and localize without admitting it. I might be wrong here, but what you want to prove is about some multiplicative properties of elements in $R$, so perhaps any proof you can find is just localization in disguise. 
A: Denote set complements in $\rm A $ by $\rm\,\bar T = A - T.\, $ Consider the monoid $\rm\,S\,$ generated by $\rm\,\bar P\,$ and $\rm\,\bar Z,\ $ for $\rm\,Z = $ all zero-divisors in $\rm A $ (including $0).\,$ $\rm\,0\not\in S\ $ (else $\rm\, 0 = ab,$ $\rm\ a\in \bar P,$ $\rm\ b\in \bar Z\ $ $\rm \Rightarrow b\in Z),\,$ so we can enlarge $\,0\,$ to an ideal $\rm\,Q\,$ maximally disjoint from $\rm\,S.\, $ Since $\rm\,S\,$ is a monoid, $\rm\,Q\,$ is prime.  $\rm\, S\,\supset\, \bar P \cup \bar Z\ \Rightarrow\ Q \subset \bar S \subset P\cap Z,\, $ so by minimality of $\rm\,P\,$ we infer $\rm\, P = Q \subset Z.\quad$ QED
