Example of ring satisfying nilpotent condition

Question

Let $R$ be a commutative ring with identity, let $N = \{x|x \in R, x^k=0$ for some natural number k$\}$, and let $L = \{x|x \in R, 1+xr$ invertible for all $ r \in R\}$. I am trying to find an example of a ring R such that $N$ is strictly smaller than $L$.

It is not hard to show that $N \subseteq L$ and so I have reduced this to finding an element in R that is in $L$ but not in $N$. It seems like $R=$ the real numbers might work, since for example $a \in R$ is in $L$ since $(1+ar)(\frac{1}{1+ar})=1$ but $a^k \neq 0$ for any natural number k if $a \neq 0$.

I'd appreciate any help with this since I'm not sure whether or not I'm on the right track here and most of the other rings such as $Z/nZ$ I've considered don't seem to have this property. Thanks!

Answer 1

Any local domain should work. If $R$ is a local domain with unique maximal ideal $\mathfrak{m}$, then $L=\mathfrak{m}$. But as a domain, $\{0\}$ is a prime ideal, so $N=\{0\}$.

For some concrete examples, you could look at valuation rings, which are always local domains, and DVRs, which are always local (principal ideal) domains.

Answer 2

The ideal $L$ is the Jacobson radical, which is also the intersection of all maximal ideals.

To wit, let $x\in L$ and let $M$ be a maximal ideal of $R$. If $x\notin M$, then $1=xr+y$, for some $y\in M$ and $r\in R$; then $y=1+x(-r)$ is invertible: contradiction. Suppose conversely that $x$ belongs to every maximal ideal and let $r\in R$. If $y=1+xr$ is not invertible, then it belongs to a maximal ideal $M$: then $1=y-xr\in M$: contradiction.

The ideal $N$ is the intersection of all prime ideals. Indeed, a nilpotent element belongs to every prime ideal (easy); conversely, if $x$ is not nilpotent, then there is a prime ideal not containing $x$ (see https://math.stackexchange.com/a/2192902/62967).

You must consider a non Artinian ring, in order to find a ring such that $N\ne L$, because if a commutative ring is Artinian, then every prime ideal is maximal: indeed, if $P$ is a prime ideal, then $R/P$ is an Artinian domain, hence a field, so $P$ is maximal. Thus finite rings like $\mathbb{Z}/n\mathbb{Z}$ cannot be an example where $N\ne L$, nor can be a field, which is obviously Artinian.

Suppose instead $R$ has a nonmaximal prime ideal $P$. Let $M$ be a maximal ideal properly containing $P$. Then, in the localization $R_M$ of $R$ at $M$, there is a single maximal ideal, namely $M_M$, but there is also a proper prime ideal, namely $P_M$. Thus in $R_M$ the intersection of all prime ideals is different from the maximal ideal.