Suppose \(\mfm \in \mspec R\) is a proper maximal ideal. Show that under either of the following two conditions, \(R\) is local:
\(R\sm \mfm \subseteq \unitsof{R}\), so every element of \(R\sm \mfm\) is a unit, or
\(1 + \mfm \subseteq \unitsof{R}\)
Solution
Let \(R\) be a commutative ring with identity and \(N \da R\sm \unitsof{R}\) its set of nonunits.
1. If \(R\sm \mfm \subseteq \unitsof{R}\), then \(\mfm\) is the only maximal ideal of \(R\).
The proper ideal \(\mfm\) contains no unit, so \(\mfm\subseteq N\), and the hypothesis gives \(N\subseteq\mfm\); hence \(\mfm=N\). A proper ideal \(I\) contains no unit, so \(I\subseteq N=\mfm\).
2. If \(1 + \mfm \subseteq \unitsof{R}\), then \(R\sm \mfm \subseteq \unitsof{R}\), so \(R\) is local by step
1.
Let \(r\in R\sm \mfm\). By maximality, \(\mfm+\gens{r} = R\), so \(rt + m = 1\) for some \(t\in R\) and \(m\in \mfm\). Then \(rt = 1-m \in 1 + \mfm \subseteq \unitsof{R}\), so \(r\) is a unit.