# Show that if $R$ is a ring with a single ideal prime minimal $\neq(0)$ then $R$ has non-trivial nilpotent elements

Show that if $R$ is a ring with a single ideal prime minimal $\neq(0)$ then $R$ has non-trivial nilpotent elements, that is, elements to $a\neq 0$ for those who $a^n=0$, for a certain power $n> 0$.

I've thought about this problem and I really do not know where to start, any suggestions? I have tried to reason for the absurd and suppose that $R$ does not have nilpotent elements and try to build two minimal prime ideals but I do not know how to do this, could someone help me? Thank you very much.

• I would try to show that the elements of that ideal are nilpotent. – Torsten Schoeneberg Sep 7 '18 at 0:27
• The word order is a bit confusing though. I assume the ideal is minimal among those which are prime, and further it is unique with that property, and non-zero? – Torsten Schoeneberg Sep 7 '18 at 0:29

## 1 Answer

Recall that the radical $\sqrt{I}$ of an ideal $I\subseteq R$ is defined as $$\sqrt{I} := \{x\in R\mid x^n\in I\textrm{ for some }n\geq 1\}.$$ $\sqrt{I}$ is also an ideal of $R.$ Notice that in particular, $\sqrt{(0)}$ is precisely the nilpotent elements of $R.$ What you want to prove is a corollary of the following general lemma:

Lemma: Let $R$ be a commutative ring. Then $$\sqrt{(0)} = \underset{\mathfrak{p}\textrm{ prime}}{\bigcap_{\mathfrak{p}\subseteq R}}\mathfrak{p}.$$ Proof: As any ideal contains $0,$ it follows that for any nilpotent $x\in R$ with $x^n = 0,$ and any prime ideal $\mathfrak{p}\subseteq R,$ we have $$0=x^n\in\mathfrak{p}\implies x\in\mathfrak{p}.$$ Thus, $\sqrt{(0)}\subseteq\bigcap_{\mathfrak{p}}\mathfrak{p}.$

Conversely, suppose $x\in R$ is not nilpotent. We will prove that there exists a prime ideal $\mathfrak{p}_x$ not containing $x.$ Indeed, consider the localization $R[x^{-1}]$ of $R.$ There is a bijection $$\{\mathfrak{p}\subseteq R[x^{-1}]\textrm{ prime}\}\leftrightarrow\{\mathfrak{q}\subseteq R\textrm{ prime such that }x\not\in\mathfrak{q}\}.$$ (Prove this is you're not familiar with the statement!) As $x$ is assumed to not be nilpotent, $R[x^{-1}]$ is not the zero ring, and as such, contains a prime ideal. Via the bijection above, we obtain a prime ideal $\mathfrak{q}$ not containing $x,$ as desired.