# Maximal Ideals

Theorem: Every nonzero ring has a maximal ideal

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

Let $$\mathcal{C}\subset \Sigma$$ be a chain. If $$\mathcal{C}$$ is empty, set $$K =\{0\}$$, otherwise set $$K = \cup_{I\in\mathcal{C}} \{x\in R|x\in I \text{ for some } I\in\mathcal{C}\}$$. It can be checked that $$K\triangleleft R$$. 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 $$\mathcal{C}$$. By Zorn’s Lemma, $$\Sigma$$ has a maximal element $$M$$ which is precisely a maximal ideal of $$R$$.

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

Proof: The nonzero ring $$A/I$$ contains a maximal ideal $$J/I$$ for some $$I\subset J\triangleleft A$$, 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.

Ben Lynn blynn@cs.stanford.edu 💡