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.
Unique Prime Ideal Implies Every Element Is Nilpotent or a Unit – Ring Theory
ring-theory
Best Answer
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.