For every prime ideal $p$ the local ring $R_p$ has no nilpotent elements, then $R$ has no nilpotent elements

Question:

Let $R$ be a ring. Suppose that for every prime ideal $p \lhd R$ the local ring $R_p$ ($=(R\setminus p)^{-1}R$) has no non-zero nilpotent elements. Prove that $R$ has no non-zero nilpotent elements in this case. (Hint: look at the nilradical of $R_p$?)

I tried the following:

Assume for contradiction $0\ne x\in R$ is a non-zero nilpotent element of $R$, so $x^n=0$ for some $n>1$. Now $\frac{x}{s}$ is in $R_p$ for any $s\in R\setminus p$. And $(\frac{x}{s})^n=\frac{x^n}{s^n}=\frac{0}{s^n}=\frac{0}{1}=0$. So $R_p$ has non-zero nilpotent elements, a contradiction. Therefore $R$ cannot have non-zero nilpotent elements.

My question: May I assume that $\frac{x}{s}$ is non-zero? If so, I seem to have found a proof which works, but doesn't use the nilradical. So what would a proof using the nilradical look like?

In case it helps, I am going through Chapter 3 of Atiyah, MacDonald.

Solution 1:

Recall that $\frac{x}{1}=0$ implies that there is an $s\in S$ such that $xs=0$ (for example look at page 37 of Atiyah-MacDonald). Since $\operatorname{Ann(x)}$ is an ideal and $x$ is nilpotent, then there is a prime ideal $\mathfrak{p}$ such that $x^{n-1}\in\operatorname{Ann}(x)\subseteq\mathfrak{p}$ and $x\in\mathfrak{p}$. Thus by contraposition $\frac{x}{1}\neq 0$ in $R_{\mathfrak{p}}$.

Your proof is essentially a full write-down of the nilradical proof:

By Corollary 3.12 if $\mathfrak{N}$ is the nilradical of $R$, then $\mathfrak{N}_\mathfrak{p}$ is the nilradical of $R_\mathfrak{p}$. Since being the $0$ module is a local property, $\mathfrak{N}_\mathfrak{p}=0$ for every prime ideal $\mathfrak{p}\subset R$ implies that $\mathfrak{N}=0$.