= Reducedness is detected by prime localizations
{title2=$A\text{ reduced}\iff A_{\mathfrak p}\text{ reduced for every }\mathfrak p$}
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 $ua^n=0$. Then $(ua)^n=0$, so reducedness of $A$ gives $ua=0$ and hence $a/s=0$.
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