TITLE: Each element of a ring is either a unit or a nilpotent element iff the ring has a unique prime ideal
QUESTION [13 upvotes]: 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.

REPLY [8 votes]: http://am-solutions.wikispaces.com/Solutions+to+Chapter+1
"Let $A$ be a ring, $R$ its nilradical. Show that the following are equivalent:
1) $A$ has exactly one prime ideal;
2) every element of $A$ is either a unit or nilpotent;
3) $A/R$ is a field.
Proof. 1) ⇒ 2). Observe that $R$, which is the intersection of the prime ideals, is equal to the given prime ideal; and that $A$ is a local ring. Thus $A−R=A^∗$ and by definition $R$ consists of all nilpotent elements.
2) ⇒ 3). The quotient map $A→A/R$ is surjective. Since ring homomorphisms map units to units, $x∈A/R$ is either $0$ or a unit.
3) ⇒ 1). All prime ideals contain $R$, and $R$ is a maximal ideal: hence there is one prime ideal. "