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.


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 $\mathfrak{N}=0$.

  • $\begingroup$ Do you mind elaborating a bit on the first part of your post? I can certainly see that $\frac{x}{1}=0$ implies $\exists s\in S=R\setminus p$ s.t. $xs=0$. I could derive from $xs=0\in p$ and $s\notin p$, that $x\in p$ (as $p$ is prime). But now I am stuck again. $\endgroup$ – Phil-ZXX Mar 29 '13 at 14:46
  • $\begingroup$ I found a prime ideal $\mathfrak{p}$ containing the annihilator of $x$, i.e. $\operatorname{Ann}(x)=\left\{r\in R \mid xr=0\right\}$. Therefore $xs\neq0$ for every $s\in S=R\setminus \mathfrak{p}$ since $S\cap \operatorname{Ann}(x)=\varnothing$, which by contraposition implies that $\frac{x}{1}\neq 0$. Then you can proceed with your argument, using $s=1$ and $p=\mathfrak{p}$. $\endgroup$ – A.P. Mar 29 '13 at 15:07
  • $\begingroup$ Thanks again, unfortunately I fail to see why $\text{Ann}(x)$ is contained in a prime ideal (or especially in the prime ideal $p$). It's probably a simple fact. :/ $\endgroup$ – Phil-ZXX Mar 29 '13 at 16:07
  • $\begingroup$ Since $\operatorname{Ann}(x)$ is a proper ideal, it is contained in some maximal ideal (see Corollary 1.4 in Atiyah-MacDonald), which is prime. For your argument to work you don't need it to be contained in every prime ideal of $R$, just in one of them, which I called $\mathfrak{p}$. $\endgroup$ – A.P. Mar 29 '13 at 16:11
  • $\begingroup$ Oh, I totally forgot about "Suppose that for every prime ideal $p$ ...", I thought we had a fixed prime ideal. Hence my confusion. Again, thanks a lot! $\endgroup$ – Phil-ZXX Mar 29 '13 at 16:28

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