## 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.