Assuming choice, commutative ring has a maximal ideal

Let $𝑅$ be a commutative ring and $𝐼 ◃ 𝑅$ be a proper ideal. Then there exists a maximal ideal $𝑀 ◃ 𝑅$ containing $𝐼$ , invoking Zorn's lemma. #m/thm/ring

Proof

Define

I = {𝐽 ⊊ 𝑅 : 𝐼 ⊴ 𝐽 ◃ 𝑅}

as a Poset ordered by inclusion. Let $C \subseteq I$ be a chain, and define $𝑈 C = ⋃ C$ . Then for any $𝑥 \in 𝑈 C$ , there exists a $𝐶 \in C$ such that $𝑥 \in 𝐶$ , and therefore $𝑅 𝑥 \subseteq 𝐶 \subseteq 𝑈 C$ , so $𝑈 C$ is an ideal containing $𝐼$ , and therefore an upper bound of $C$ in $I$ . By Zorn's lemma, $I$ has a maximal element.

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