Krull's Theorem

Theorem
Let $R$ be a nontrivial ring.

Then $R$ has a maximal ideal.

Outline of Proof
We use Zorn's Lemma to construct a maximal ideal.

Proof
Let $\left({P, \subseteq}\right)$ be the ordered set consisting of all proper ideals of $R$, ordered by inclusion.

The theorem is proved by applying Zorn's Lemma to $P$.

First, we check that the conditions for Zorn's Lemma are met: $P$ must be non-empty, and every non-empty chain in $P$ must have an upper bound.

$P$ is non-empty
Since $R$ is nonzero, $0$ is a proper ideal of $R$, and thus an element of $P$.

Every non-empty chain in $P$ has an upper bound in $P$
Let $\left\{{I_\alpha}\right\}_{\alpha \in A}$ be a non-empty chain of ideals in $P$.

Let $\displaystyle I = \bigcup_{\alpha \in A} I_\alpha$.

We will show that $I$ is an upper bound in $P$ for the chain $\left\{{I_\alpha}\right\}_{\alpha \in A}$.

$I$ is a proper ideal of $R$
By By Union of Chain of Proper Ideals is Proper Ideal, $I$ is a proper ideal of $R$.

$I$ is an upper bound for the chain $\left\{{I_\alpha}\right\}$
Since $I$ is a proper ideal of $R$, it is an element of our ordered set $P$.

$I$ is the union of the $I_\alpha$, so $I_\alpha \subseteq I$ for all $\alpha \in A$.

This means that $I$ is an upper bound in $P$ for the chain $\left\{{I_\alpha}\right\}_{\alpha \in A}$.

Applying Zorn's Lemma
We have shown that the conditions for Zorn's Lemma are met:
 * $(1): \quad P$ is non-empty
 * $(2): \quad$ every non-empty chain in $P$ has an upper bound.

Applying Zorn's Lemma to $\left({P, \subseteq}\right)$ gives us a maximal element $M$.

This $M$ is a proper ideal of $R$ which is not contained in any other proper ideal.

So by definition, $M$ is a maximal ideal of $R$.