 Reducedness is detected by prime localizations

If a nonzero nilpotent element $a$ existed in $A$, its proper annihilator would lie in a maximal ideal $\mathfrak{m}$. No element outside $\mathfrak{m}$ annihilates $a$, so $a/1$ is nonzero in the localization at a prime ideal $A_{\mathfrak{m}}$. Conversely, if $(a/s{)}^n=0$ in a localization, some $u$ in the multiplicative subset satisfies $u{a}^n=0$. Then $(ua{)}^n=0$, so reducedness of $A$ gives $ua=0$ and hence $a/s=0$.