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.
 A: 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 $\mathfrak{N}=0$.
A: Let $x\in R$ such that $x^n=0$ for some $n$ and $U=Ann(x)=\{u \in R;~~ux=0\}$.
If we prove that $U=R$ then, in particular, $1\in U$ and therefore $x=0$.
Suppose that $U\subsetneq R$. Then there exists a maximal ideal $m\supset U$. Considere the localization $R_m$ and observe that
$$\left(\dfrac{x}{1}\right)^n = \dfrac{x^n}{1} = \dfrac{0}{1}.$$
So
$$\dfrac{x}{1} = \dfrac{0}{1}$$
and then $\exists u\in R\setminus m$ such that $ux=0$. That is a contradiction because $u\in R\setminus m$ implies $u\notin U$.
Therefore $U=R$ as we want to demonstrate.
