Let RR be a . Write R×R^\times for its group of units and set

m:=RR×\mathfrak m := R\setminus R^\times

for the set of nonunits. Then the following are equivalent:

  1. RR is a .
  2. The set m\mathfrak m is an ideal of RR.

When these conditions hold, m\mathfrak m is the unique maximal ideal of RR.

Remarks

For RpR_{\mathfrak p}, the , the unique maximal ideal is pRp\mathfrak pR_{\mathfrak p}, and the associated is Rp/pRpR_{\mathfrak p}/\mathfrak pR_{\mathfrak p}.

This units-versus-maximal-ideal dichotomy is used in .

Examples
  1. Z(p)\mathbb Z_{(p)}. In R=Z(p)R=\mathbb Z_{(p)}, a fraction ab\frac{a}{b}, where pbp\nmid b, is a unit if and only if pap\nmid a. Thus the maximal ideal is pZ(p)p\mathbb Z_{(p)}.
  1. k[x](x)k[x]_{(x)}. In R=k[x](x)R=k[x]_{(x)}, the units are exactly the fractions fg\frac{f}{g} with f(0)0f(0)\neq 0. Hence the maximal ideal is generated by xx.
  1. A field. In a field kk, the only nonunit is 00, so m=(0)\mathfrak m=(0).