Nilradical equals intersection of prime ideals

Nilradical equals intersection of prime ideals: Let $R$ be a commutative ring. Then

\mathrm{Nil}(R)=\bigcap_{\mathfrak p\in \mathrm{Spec}(R)} \mathfrak p,

i.e., an element is nilpotent if and only if it lies in every prime ideal. Equivalently, $\mathrm{Nil}(R)=\sqrt{(0)}$ .

Remarks

The nilradical is the ideal of all nilpotent elements; this proposition expresses it as an intersection of prime ideals. Equivalently, it is the radical of $(0)$ , so $R/\mathrm{Nil}(R)$ is reduced.