Each element of a ring is either a unit or a nilpotent element iff the ring has a unique prime ideal

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.

Solution 1:

$F[x]/(x^3)$ 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.

Solution 2:

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. "

Solution 3:

Hint for $\Leftarrow$:

Every ring with identiy has a prime ideal. Let $P$ be a prime ideal of $R$. Then it contains all the nilpotent elements (why?). It does not contain a unit (why?), so it is in fact the set of nilpotent elements of $R$ and hence because of the condition the unique prime ideal.