= Solution
{c}
\b[Yes: $R$ can have nonzero zero divisors.] Let $R=k\times k$, a <direct product of rings>, where $k$ is any <field>. The nonzero <idempotents> $e_1=(1,0)$ and $e_2=(0,1)$ satisfy $e_1e_2=0$, so both are <zero divisors>.
There are exactly two <prime ideals>:
$$
P_1=0\times k,\qquad P_2=k\times0.
$$
Indeed, any <prime ideal> must contain $e_1$ or $e_2$, since their product is zero. If it contains $e_1$, it contains $k\times0$ and corresponds to a <prime ideal> of the <quotient ring> $k$, whose only <prime ideal> is zero; hence it equals $P_2$. The other case gives $P_1$.
For the <localization at a prime ideal> $P_1$, every denominator $(u,v)\notin P_1$ has $u\ne0$, and
$$
R_{P_1}\longrightarrow k,\qquad
\frac{(a,b)}{(u,v)}\longmapsto\frac a u
$$
is an isomorphism. It is surjective using constant first-coordinate fractions. It is injective because a numerator with first coordinate zero is annihilated by $e_1\notin P_1$, so its fraction is zero. Interchanging the coordinates proves $R_{P_2}\cong k$.
Thus the <localization at a prime ideal> is a <field> in every case, although $R$ has <zero divisors>. The obstruction to using the previous part's argument is that the two nonzero factors of a zero product can survive at different <prime ideals>.
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