Question about nilradicals and localization?

Question

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 $f$?

Answer 1

the nilradical is the set of all nilpotents, or the radical of the zero ideal $\sqrt{(0)}$. In general, localizing commutes with taking radicals. In particular, localizing at a particular prime will of course work as well.

To see this, let $S \subset A$ be a multiplicative subset. Consider $\frac{x}{s} \in S^{-1}(\sqrt{(0)})$. then $a^n=0$, so $(\frac{a}{s})^n=0 \in S^{-1}(A)$, so it is also in $\sqrt{S^{-1}(0)}$.

The first answer in the link provides the other direction, but it is again from the definition.

Answer 2

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.