Show that if \(R\) is a local ring then \(R\setminus \unitsof{R}\) is a proper ideal that is contained in the Jacobson radical \(J(R)\).
Solution
Let \(\mathfrak m\) be the unique maximal ideal of the local ring \(R\).
1. Every element of \(\mathfrak m\) is a nonunit.
If \(x\in\mathfrak m\) were a unit, then \(1=x^{-1}x\in\mathfrak m\), contradicting the properness of the maximal ideal \(\mathfrak m\). Thus \[ \mathfrak m\subseteq R\setminus \unitsof{R}. \]
2. Every nonunit of \(R\) lies in \(\mathfrak m\).
Let \(x\in R\) be a nonunit. Then the principal ideal \((x)\) is proper, so it is contained in a maximal ideal of \(R\). Since \(R\) is local, its only maximal ideal is \(\mathfrak m\). Hence \(x\in\mathfrak m\), and therefore \[ R\setminus \unitsof{R}\subseteq\mathfrak m. \]
3. The set of nonunits is a proper ideal.
By 1 and 2, \[ R\setminus \unitsof{R}=\mathfrak m. \] Since \(\mathfrak m\) is a proper ideal, so is the set of nonunits.
4. In fact, the set of nonunits equals the Jacobson radical.
By definition, \(J(R)\) is the intersection of all maximal ideals of \(R\). A local ring has the single maximal ideal \(\mathfrak m\), so \[ J(R)=\mathfrak m=R\setminus \unitsof{R}. \] In particular, \(R\setminus \unitsof{R}\subseteq J(R)\).