TITLE: Question about nilradicals and localization?
QUESTION [6 upvotes]: Suppose we have a commutative ring $A$ with identity. Is it true that if $\mathfrak{p}$ is a prime ideal of $A$, then we have
$$ \mathcal{N}( A_{\mathfrak{p}}) \simeq  \mathcal{N}(A)_{\mathfrak{p}} \quad ? $$
Where $\mathcal{N}$ denotes taking the nilradical. Furthermore, is it still true that if we replace localization at $\mathfrak{p}$ with localization at the multiplicative set generated by some non-nilpotent element $f$?

REPLY [11 votes]: Besides referring to more general results, you can also prove the statement directly.
Suppose $a/s$ is nilpotent in $A_{\mathfrak{p}}$. This means $a^n/s^n=0$, for some $n>1$, so $ta^n=0$ for some $t\in A\setminus\mathfrak{p}$. Hence also $(ta)^n=0$ and therefore
$$
\frac{a}{s}=\frac{ta}{ts}\in\mathcal{N}(A)_{\mathfrak{p}}
$$
The converse inclusion is obvious.