Show that \(\bigcup_{\mathfrak{m} \in \operatorname{MaxSpec}(R)} \mathfrak{m} = R \setminus R^\times\).
Solution
1. Every element of a maximal ideal is a nonunit.
Let \(\mathfrak m\) be maximal and let \(x\in\mathfrak m\). If \(x\) were a unit, then \[ 1=x^{-1}x\in\mathfrak m, \] contradicting that \(\mathfrak m\) is proper. Hence \[ \bigcup_{\mathfrak m\in\operatorname{MaxSpec}(R)}\mathfrak m \subseteq R\setminus R^\times. \]
2. Every nonunit belongs to a maximal ideal.
Let \(x\in R\) be a nonunit. Then \((x)\ne R\), so \((x)\) is a proper ideal. Every proper ideal of a ring with identity is contained in a maximal ideal (Zorn’s lemma), so \((x)\) is contained in some maximal ideal \(\mathfrak m\). Thus \(x\in\mathfrak m\), and therefore \[ R\setminus R^\times \subseteq \bigcup_{\mathfrak m\in\operatorname{MaxSpec}(R)}\mathfrak m. \]
Combining the two inclusions gives \[ \boxed{\bigcup_{\mathfrak m\in\operatorname{MaxSpec}(R)}\mathfrak m=R\setminus R^\times.} \]