A commutative ring \(R\) is called a local ring if it has a unique maximal ideal. Prove that if \(R\) is a local ring with maximal ideal \(M\) then an element \(a \in R\) is a unit if and only if \(a \notin M\). Conversely, prove that if \(R\) is a commutative ring in which the set of non-units forms an ideal, then \(R\) is a local ring.
Solution
Rings are understood to have an identity, and maximal ideals are proper.
1. Every nonunit belongs to a maximal ideal, and no proper idealcontains a unit.
If an ideal contains a unit \(u\), it contains \(u^{-1}u=1\) and therefore the whole ring. If \(a\) is a nonunit, its principal ideal \((a)\) is proper, since \(1=ra\) would make \(a\) invertible.
Consider the partially ordered set of proper ideals containing \((a)\). It is nonempty. The union of any nonempty chain is an ideal containing \((a)\), and is proper: if it contained \(1\), one ideal in the chain would already contain \(1\). The empty chain has \((a)\) as an upper bound. Zorn’s lemma gives a maximal member, which is a maximal ideal of \(R\) containing \(a\).
2. If \(R\) has a unique maximal ideal \(M\), its units are exactlythe elements outside \(M\).
No unit lies in \(M\) because \(M\) is proper. Conversely every nonunit lies in some maximal ideal by step 1, and that ideal must be \(M\). Thus \(a\) is a unit if and only if \(a\notin M\).
3. If the set \(I\) of nonunits is an ideal, it is the uniquemaximal ideal of \(R\).
1. Thus \(I\)is maximal.
Every maximal ideal \(M\) is proper and so contains no units. Consequently \(M\subseteq I\). Since \(M\) is maximal and \(I\) is proper, \(M=I\). This proves uniqueness and hence locality.