I am trying to answer a question that has already been posted in here (About Nilradical and Localization). I did not have much success with the first two answers, and the other two mention sheafs, which is something I am not acquainted with.
Basically, I want to show that $\mathcal{N}(R_P) = 0$, for all prime ideals $P$ of a ring $R$, then $\mathcal{N}(R) = 0$, where here $\mathcal{N}$ is the nilradical.
I am aware that localization commutes with taking radicals and intersections, but trying to use the fact that $\mathcal{N}(R) = \cap \{ \text{all prime ideals of } R \} $ and $\mathcal{N}(R) = \sqrt{(0)}$ hasn't led me far.
All I've got so far that makes sense is that $0 = \mathcal{N}(R_P) = (\mathcal{N}(R))_P$, for all prime ideals $P$ of $R$.
Any hints would be very much appreciated. Or perhaps hints that could help me understand the hints given in the link above!
Suppose $a\in\mathcal N(R)$ and consider the ideal $\operatorname{ann}(a)=\{r\in R\mid ra=0\}$. Notice that $$a=0\iff 1\in\operatorname{ann}(a)\iff\operatorname{ann}(a)=R.$$ If $\operatorname{ann}(a)$ is a proper ideal, then it is contained in some maximal ideal $M$. But notice that in the localization $R_M$ we have $a/1\in\mathcal N(R_M)=0$; use this to conclude there is some $s\notin M$ such that $sa=0$. Now use the fact that $\operatorname{ann}(a)\subseteq M$ to obtain a contradiction.