Proof of the existence of maximal ideals by Zorn's lemma

Proof

We use Zorn's lemma. We write the proof for left ideals. For two-sided ideals, read "two-sided ideal" for "left ideal" throughout; step 6 explains the one extra check.

1. Let $R$ be a ring and let $I$ be a left ideal of $R$ with $I \neq R$.
2. A left ideal $L$ of $R$ equals $R$ if and only if $1 \in L$. Indeed, if $1 \in L$, then $r = r \cdot 1 \in L$ for every $r \in R$.
3. Let $\mathcal{P}$ be the set of left ideals $L$ of $R$ with $I \subseteq L$ and $L \neq R$. Inclusion is a partial order on $\mathcal{P}$. The set $\mathcal{P}$ is nonempty, since $I \in \mathcal{P}$.
4. Let $\mathcal{C} \subseteq \mathcal{P}$ be a chain. If $\mathcal{C}$ is empty, then $I$ is an upper bound of $\mathcal{C}$ in $\mathcal{P}$. So assume that $\mathcal{C}$ is nonempty, and let $U$ be the union of the members of $\mathcal{C}$.
5. The set $U$ is a left ideal. It contains $0$, since some $L \in \mathcal{C}$ exists and $0 \in L$. Let $a, b \in U$, say $a \in L_1$ and $b \in L_2$ with $L_1, L_2 \in \mathcal{C}$. Since $\mathcal{C}$ is a chain, we may assume $L_1 \subseteq L_2$. Then $a - b \in L_2 \subseteq U$. For $r \in R$, $r a \in L_1 \subseteq U$.
6. In the two-sided case, also $a r \in L_1 \subseteq U$ for $a \in L_1$ and $r \in R$. So $U$ is then a two-sided ideal.
7. Every member of $\mathcal{C}$ contains $I$, so $I \subseteq U$. No member contains $1$ by step 2, so $1 \notin U$, and $U \neq R$.
8. So $U \in \mathcal{P}$, and $U$ contains each member of $\mathcal{C}$. Thus every chain in $\mathcal{P}$ has an upper bound in $\mathcal{P}$.
9. By Zorn's lemma, $\mathcal{P}$ has a maximal element $M$. Then $M$ is a left ideal with $I \subseteq M$ and $M \neq R$.
10. Let $L$ be a left ideal of $R$ with $M \subseteq L$ and $L \neq R$. Then $I \subseteq L$, so $L \in \mathcal{P}$. By the maximality of $M$ in $\mathcal{P}$, $L = M$.
11. So every left ideal $L \supseteq M$ satisfies $L = M$ or $L = R$. Hence $M$ is a maximal ideal of $R$ that contains $I$.
12. Now let $R$ be a nonzero ring, so $1 \neq 0$. The set $\set{0}$ is an ideal, and $\set{0} \neq R$ since $1 \notin \set{0}$. Steps 1 to 11 with $I = \set{0}$ give a maximal ideal of $R$.