Unique prime ideal implies every element is nilpotent or a unit.

Let $R$ be a commutative ring with only one prime ideal. I want to show that every element of $R$ is either a unit or nilpotent, or equivalently, that the nilradical is the unique maximal/prime ideal. Is there a way to prove this without using the fact that the nilradical is in fact EQUAL to the intersection of all prime ideals, instead of just contained in that intersection? I ask because all the solutions I've seen posted to this Dummit and Foote problem used that fact, which had not been proved yet in the book.

• Just for reference: Exercise 10 from Introduction to Commutative Algebra by Atiyah & Macdonald also gives an approach. – user188634 May 19 '16 at 18:59

I would try this approach. It uses, however, localizations, which is (if I remember correctly) the only tool needed for proving that the nilradical is the intersection of all prime ideals.

Let $a \in R$ be some non-invertible element. Thus, it is contained in the only maximal (= the only prime) ideal of $R$. Consider the localization $S^{-1}R$ of $R$, where $S=\{a^k \; | \; k \in \mathbb{N}\}$. Now the "correspondence theorem for localizations" says that prime ideals of $S^{-1}R$ are in one-to -one correspondence with prime ideals of $R$ not intersecting $S$. But since $a$ is a member of the only prime ideal of $R$, it follows that $S^{-1}R$ must be the zero ring (it is an unital ring with no prime ideals, hence no maximal ideals).

Thus, we have $(1/1)=(0/1)$ in $S^{-1}R$, which by definition means that $a^n=a^n(1-0)=0$ in $R$ for some $n \in \mathbb{N}$.

Another approach (more elementary one): I have a feeling this is just a rephrasing of the proof above, however, it may be more transparent.

Let $a$ be a non-invertible element which is not nilpotent. Then $a$ is contained in some maximal ideal $M$, which is prime.

Consider the family of ideals

$$\mathcal{M}=\{I \;|\; a^k \notin I\; \forall k\}$$

Ordered by inclusion. Since $a$ is not nilpotent, $0 \in \mathcal{M}$, hence the collection is nonempty. It is clearly closed under taking unions of chains of ideals. Hence it contains some maximal element $P$. The claim is that $P$ is prime ideal. Consider two elements $x,y \in R \setminus P$. Then we have $xR+P, yR+P \supsetneq P$. Since $P$ was maximal in $\mathcal{M}$, it follows that

$$a^m=xr+p, a^n=ys+q$$

for some $n,m \in \mathbb{N},\;\; r,s \in R, \;\; p,q \in P$. Then $$a^{n+m}=xyrs+xrq+ysp+pq,$$ where $xrq+ysp+pq \in P$. It follows that $xyrs \notin P$, since $a^{n+m}\notin P$. Thus, $xy \notin P$.

We have proved that $P$ is a prime ideal not containing $a$. In particular, $M$ and $P$ are two distinct prime ideals in $R$. Thus, assuming $R$ has only one prime ideal, all non-invertible elements must be nilpotent.