Let $R$ be a ring. Prove that each element of $R$ is either a unit or a nilpotent element iff the ring $R$ has a unique prime ideal.
Help me some hints.
Ring Theory – Units and Nilpotent Elements with Unique Prime Ideal
abstract-algebraring-theory
Related Solutions
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 $\mathfrak{N}=0$.
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.
Best Answer
$F[x]/(x^3)$ consists of units and nilpotent elements, but has four ideals, so this suggests you meant something more like unique prime ideal.
This is indeed true for commutative rings. The hypothesis that nonunits are nilpotent means that the nilradical is a maximal ideal. But considering that all prime ideals contain the nilradical, the nilradical is precisely the one prime ideal in the ring.
Conversely, if you assume the ring has one prime ideal, then there is clearly only one maximal ideal, and everything inside it is a nonunit, hence nilpotent.
The statement is false for noncommutative rings. $M_2(R)$ has exactly one prime ideal: $\{0\}$. Needless to say there are non-nilpotent non-units in this ring (for example $\begin{bmatrix}1&0\\0&0\end{bmatrix}$.)
It might be interesting though to follow up and see if any of the new one-sided prime ideal definitions makes this work in noncommutative rings.