Maximal Ideals

Theorem: Every nonzero ring has a maximal ideal

Proof: Let \(R\) be a nonzero ring and put \(\Sigma=\{I|I\triangleleft R, I\ne R\}\). Then \(\Sigma\) is a poset with respect to \(\subset\). Also \(\Sigma\) is nonempty since it contains the zero ideal.

Let \(\mathcal{C}\subset \Sigma\) be a chain. If \(\mathcal{C}\) is empty, set \(K =\{0\}\), otherwise set \(K = \cup_{I\in\mathcal{C}} \{x\in R|x\in I \text{ for some } I\in\mathcal{C}\}\). It can be checked that \(K\triangleleft R\). Furthermore, \(K \ne R\) otherwise one of the ideals of the chain would contain \(1\) and thus all of \(R\), a contradiction since \(\Sigma\) does not contain \(R\). Thus \(K\in\Sigma\) and \(K\) is an upper bound for \(\mathcal{C}\). By Zorn’s Lemma, \(\Sigma\) has a maximal element \(M\) which is precisely a maximal ideal of \(R\).

Corollary: Let \(I\triangleleft R, I\ne R\). Then there exists a maximal ideal \(R\) containing \(I\).

Proof: The nonzero ring \(A/I\) contains a maximal ideal \(J/I\) for some \(I\subset J\triangleleft A\), and this \(J\) must be maximal in \(A\).

Corollary: Every nonunit of a nonzero ring \(R\) is contained in some maximal ideal.

Proof: Apply the previous corollary to the prinipal ideal generated by the nonunits.

Remark: If it is known that \(R\) is Noetherian, that is, \(R\) satisfies the ascending chain condition (a.c.c.): for all sequences of ideals \(I_1\subset I_2\subset...\) we have \(I_n = I_{n+1} = ...\) for some \(n\ge 1\), then we may avoid using Zorn’s Lemma. We take \(I_1 = \{0\}\) (or any other ideal) and if \(I_1\) is not maximal, we set \(I_2\) to some ideal strictly containing \(I_1\). Continuing in this fashion gives us an ascending chain of ideals, and by the a.c.c. we know we must eventually reach a maximal ideal.

Ben Lynn 💡