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.

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