Commutative Algebra

Theorem: Every nonzero ring has a maximal ideal

Proof: Let $R$ be a nonzero ring and put $\Sigma =\{I\mid I◃R,I\ne R\}$. Then $\Sigma$ is a poset with respect to $\subset$. Also $\Sigma$ is nonempty since it contains the zero ideal.

Let $𝒞\subset \Sigma$ be a chain. If $𝒞$ is empty, set $K=\{0\}$, otherwise set $K={\cup }_{I\in 𝒞}\{x\in R\mid x\in I\text{for some}I\in 𝒞\}$. It can be checked that $K◃R$. Furthermore, $K\ne R$ otherwise one of the ideals of the chain would contain $1$ and thus all of $R$, a contradiction since $\Sigma$ does not contain $R$. Thus $K\in \Sigma$ and $K$ is an upper bound for $𝒞$. By Zorn's Lemma, $\Sigma$ has a maximal element $M$ which is precisely a maximal ideal of $R$.

Corollary: Let $I◃R,I\ne R$. Then there exists a maximal ideal $R$ containing $I$.

Proof: The nonzero ring $A/I$ contains a maximal ideal $J/I$ for some $I\subset J◃A$, and this $J$ must be maximal in $A$.

Corollary: Every nonunit of a nonzero ring $R$ is contained in some maximal ideal.

Proof: Apply the previous corollary to the prinipal ideal generated by the nonunits.

Remark: If it is known that $R$ is Noetherian, that is, $R$ satisfies the ascending chain condition (a.c.c.): for all sequences of ideals $I_1\subset I_2\subset ...$ we have $I_n=I_{n+1}=...$ for some $n\ge 1$, then we may avoid using Zorn's Lemma. We take $I_1=\{0\}$ (or any other ideal) and if $I_1$ is not maximal, we set $I_2$ to some ideal strictly containing $I_1$. Continuing in this fashion gives us an ascending chain of ideals, and by the a.c.c. we know we must eventually reach a maximal ideal.