# 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.

*blynn@cs.stanford.edu*💡