Krull's Theorem

Theorem
Let $R$ be a nonzero ring.

Then $R$ has 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 $\left\{{I_\alpha}\right\}_{\alpha \in A}$.

Property 1: $0 \in I$
Since $\left\{{I_\alpha}\right\}_{\alpha \in A}$ is non-empty chain, it must contain some ideal $I_\beta$

Since $I_\beta$ is an ideal, $0 \in I_\beta$.

Thus $0 \in I$.

Property 2: $x \in I \implies -x \in I$
If $x \in I$, then $x \in I_\beta$ for some $\beta\in A$.

Since $I_\beta$ is an ideal, $-x \in I_\beta$.

Thus $-x \in I$.

Property 3: $x, y \in I \implies x + y \in I$
If $x,y \in I$, then $x \in I_\beta$ for some $\beta\in A$, and $y \in I_\gamma$ for some $\gamma \in A$.

Since $\left\{{I_\alpha}\right\}_{\alpha \in A}$ is totally ordered, $I_\beta \subseteq I_\gamma$ or $I_\gamma\subseteq I_\beta$.

, we can assume $I_\beta \subseteq I_\gamma$, which gives us $x,y \in I_\gamma$.

Since $I_\gamma$ is an ideal, $x + y \in I_\gamma$.

Thus $x + y \in I$.

Property 4: $x \in I \land r \in R \implies rx, xr \in I$
If $x \in I$, then $x \in I_\beta$ for some $\beta\in A$.

If $r \in R$, then since $I_\beta$ is an ideal, $rx, xr \in I_\beta$.

Thus $rx, xr \in I$.

Property 5: $I \subsetneq R$
The ideals $I_\alpha$ are all proper, so none of them contain $1$.

Thus $I$ does not contain $1$, which means $I \subsetneq R$.

Since $I$ satisfies these 5 properties, it 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 $R$.