Total ring of fractions ($Q(A)=S^{-1}A$)

Here $S$ consists of the non-zero-divisors of a commutative ring $A$. The localization map $A\to Q(A)$ is injective: a fraction $a/1$ is zero only if some $s\in S$ annihilates $a$. For a reduced ring with finitely many minimal prime ideals, its zero divisors form the union of these primes. Indeed, if $ab=0$ and $a$ avoids every minimal prime, reduction to each domain quotient forces $b$ into their intersection, which is zero. Conversely, for $a$ in a minimal prime ${\mathfrak{p}}_i$, choose $b$ in every other minimal prime but outside ${\mathfrak{p}}_i$, using their incomparability and multiplying suitable elements. Then $b\ne 0$ and $ab=0$.