 Solution

The equivalence relation defining the localization at a prime ideal givesThereforeEquivalently, all zero divisors of $R$ must lie in $P$. To prove necessity, if $s\notin P$ and $sr=0$ with $r\ne 0$, then $r/1=0$ in $R_P$, contradicting the hypothesis that the map is injective. Conversely, if every denominator annihilates no nonzero element, the displayed kernel of a ring homomorphism is zero, so the map is injective.

An integral domain automatically satisfies this criterion at every prime ideal, but the criterion itself does not require $R$ to be an integral domain.