## Existence of maximal ideals

Let $R$ be a commutative ring with 1. Then there exists a maximal ideal in $R,$ i.e. a proper ideal $I$ such that the only ideal that contains $I$ is $R$ itself.

We did not discuss this in the class, and the main reason is that it requires a form of the Axiom of Choice. We will use its form, known as Zorn’s Lemma

Every partially ordered set in which every chain (i.e. totally ordered subset) has an upper bound contains at least one maximal element.

We take the set $\mathcal{I}$ of all proper ideals of $R$ as the set ordered by inclusion, i.e. $K\leq J$ iff $K\subseteq J,$ for $K,J\in\mathcal{I}.$
Any chain is then a set of ideals $\mathcal{J}$ such that for any $K\neq J\in \mathcal{J}$ either $K\subset J$ or $J\subset K.$ We need to check that $\mathcal{J}$ has an upper bound, i.e. $U\in\mathcal{I}$ such that $K\subseteq U$ for any $K\in\mathcal{J}.$ Set $U=\cup_{K\in\mathcal{J}} K.$
We need to check that $U\in\mathcal{J},$ i.e. that it is a proper ideal. As $1\not\in U,$ we only need to check that it is an ideal. It suffices, and is easy, to see that

• for any $x,y\in U$ there exists $K\in\mathcal{J}$ such that $x,y\in K,$ and thus $x+y\in K\subseteq U.$ (Indeed, there exists $K',K''\in\mathcal{J}$ such that $x\in K',$ $y\in K'',$ and w.l.o.g $K'\subseteq K''=K,$ as $\mathcal{J}$ is a chain)
• for any $x\in U$ and $r\in R$ here exists $K\in\mathcal{J}$ such that $rx\in K,$ and thus $rx\in U.$ (Indeed, there exists $K\in\mathcal{J}$ such that $x\in K,$ and $rx\in K,$ as $K$ is an ideal.)

Thus Zorn’s Lemma is applicable to $\mathcal{J}$, so it has maximal element, that is obviously a maximal proper ideal in $R.$ QED.

As a corollary, we see that every proper ideal $K\subseteq R$ is contained in a maximal proper ideal $I.$ Indeed, $R/K$ contains a maximal proper ideal $F,$ and $F=J/I$ for a maximal ideal $J\supset I.$
In particular, any non-invertible element $x\in R$ is contained in such an ideal, by applying the latter construction to $K=(x).$