## Characterization of commutative rings with a unique prime ideal

Let $R$ be a commutative ring with $1 \neq 0$. Prove that the following are equivalent.

1. $R$ has exactly one prime ideal.
2. Every element of $R$ is either nilpotent or a unit.
3. $R/\mathfrak{N}(R)$ is a field.

$(1) \Rightarrow (2)$ Suppose $R$ has exactly one prime ideal. Since every maximal ideal of $R$ is prime, $R$ is local. By §7.4 #26, $\mathfrak{N}(R)$ is the unique maximal ideal. By §7.4 #37, every element of $R \setminus \mathfrak{N}(R)$ is a unit, and (by definition) the remaining elements are nilpotent.

$(2) \Rightarrow (3)$ Suppose $x + \mathfrak{N}(R)$ is nonzero; then $x \notin \mathfrak{N}(R)$. Since $x$ is not nilpotent in $R$, $x$ is a unit. Then $x^{-1}$ exists in $R$, and we have $(x + \mathfrak{N}(R))(x^{-1} + \mathfrak{N}(R)) = 1$. Thus every nonzero element of $R/\mathfrak{N}(R)$ is a unit. Since this ring is commutative, $R/\mathfrak{N}(R)$ is a field.

$(3) \Rightarrow (1)$ Suppose $R/\mathfrak{N}(R)$ is a field. By §7.4 #26, $\mathfrak{N}(R)$ is contained in every prime ideal of $R$. By the Lattice Isomorphism Theorem for rings, the only possible proper prime ideal is $\mathfrak{N}(R)$; moreover, $\mathfrak{N}(R)$ is prime since it is maximal in $R$. Thus $\mathfrak{N}(R)$ is the unique prime ideal of $R$.