= Solution
{c}
\b[No: $R$ cannot contain a nonzero nilpotent element.] We prove the needed principle, <localization detects zero elements>, rather than assume it.
Let $a\ne0$ in $R$. Its <annihilator> $\operatorname{Ann}(a)=\{r:ra=0\}$ is a proper <ideal>, because $1$ does not annihilate $a$. Every proper <ideal> is contained in a <maximal ideal>, by the <Zorn lemma>; choose such a <maximal ideal> $\mathfrak m$ containing $\operatorname{Ann}(a)$. If $a/1=0$ in $R_{\mathfrak m}$, then some $s\notin\mathfrak m$ satisfies $sa=0$. This would put $s$ in $\operatorname{Ann}(a)\subseteq\mathfrak m$, a contradiction. Thus every nonzero element survives in at least one <localization at a prime ideal>.
Now suppose $a^n=0$. For every <prime ideal> $P$, $(a/1)^n=0$ in $R_P$. The assumed absence of nonzero <nilpotent elements> forces $a/1=0$ in every $R_P$. The preceding argument then forces $a=0$. In the terminology of <reduced rings>, we have proved
$$
\boxed{R_P\text{ reduced for every prime }P\ \Longrightarrow\ R\text{ reduced}.}
$$
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