maximal ideals are coprime

Let \(R\) be a ring and \(\mathfrak{m},\mathfrak{m}'\) different maximal ideals. Then they are coprime

By assumption, there exists an \(r \in \mathfrak{m} \setminus \mathfrak{m}'\). Therefore, \(\mathfrak{m}' \subsetneq (r, \mathfrak{m}')\) is also an ideal. By assumption of maximality, \((r, \mathfrak{m}') = R\). Therefore

\begin{align*} \mathfrak{m} + \mathfrak{m}' = R \end{align*}

(see: generated ideal and sum)