If the set of all nonunits in a commutative ring is an ideal, then it forms the unique maximal ideal

A commutative ring R is called local if it has a unique maximal ideal. Prove that if R is a local ring with a maximal ideal M then every element of R \setminus M is a unit. Prove conversely that if R is a commutative ring with 1 \neq 0 in which the set of nonunits forms an ideal M, then R is a local ring with unique maximal ideal M.

Let u \in R \setminus M. Consider the ideal (u); if proper, this ideal must be contained in some maximal ideal, and M is the only maximal ideal. Thus (u) \subseteq M, a contradiction since u \notin M. Thus (u) = R, and for some v \in R we have uv = vu = 1. Thus u is a unit.

Suppose now that the set M of nonunits in R form an ideal. First we show that M is maximal. If M \subsetneq I for some ideal I, then I contains a unit, so that I = R. Now we show that M is the unique maximal ideal. Suppose that we have an ideal N \subseteq R such that N \not\subseteq M. Then N contains some element x not in M, which is a unit. Thus N = R. In particular, every proper ideal of R is contained in M. Thus M is the unique proper ideal of R, and R is local.

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