Krull's Theorem

Theorem
Let $R$ be a nonzero ring. Then $R$ has a maximal ideal.

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

We will prove the theorem by applying Zorn's lemma to $P$. First, we check that the conditions for Zorn's lemma are met.

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

Then $I$ is a proper ideal of $R$:


 * $\left\{{I_\alpha}\right\}_{\alpha \in A}$ is non-empty chain, so it contains some ideal $I_\beta$, and $0 \in I_\beta$. Thus $0 \in I$.
 * 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 $-x \in I$.
 * 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 $I_\gamma\subseteq I_\beta$. Without loss of generality, 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 $x + y \in I$.
 * If $x \in I$, then $x \in I_\beta$ for some $\beta\in A$. Since $I_\beta$ is an ideal, $rx \in I_\beta$ for any $r \in R$. Thus $rx \in I$ for any $r \in R$.
 * The ideals $I_\alpha$ are all proper, so none of them contain $1$. Thus $I$ does not contain $1$.

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

Now we can apply Zorn's lemma to $(P, \subseteq)$, and we get some 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 $R$.