= Solution
{c}
The equivalence relation defining the <localization at a prime ideal> gives
$$
\ker(R\to R_P)=\{r\in R:sr=0\text{ for some }s\notin P\}.
$$
Therefore
$$
\boxed{R\longrightarrow R_P\text{ is injective }
\iff\text{ every }s\in R\setminus P\text{ annihilates no nonzero element of }R.}
$$
Equivalently, all <zero divisors> of $R$ must lie in $P$. To prove necessity, if $s\notin P$ and $sr=0$ with $r\ne0$, 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>.
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