Local Ring Equivalent Conditions

If $R$ is a commutative ring with 1 then the following conditions are equivalent.
(i) $R$ is a local ring, that is, a commutative ring with 1 which has a unique maximal ideal.
(ii) All nonunits of $R$ are contained in some ideal $M\neq R$ .
(iii) The nonunits of $R$ form an ideal.

Proof
(H pg 147)

(i) $\implies$ (ii): If $a\in R$ is a nonunit, then $(a)\neq R$ since $1\notin (a)$ . Therefore $(a)$ (and hence $a$ ) is contained in the unique maximal ideal $M$ of $R$ , since $M$ must contain every ideal of $R$ (except $R$ itself).

(ii) $\implies$ (iii): Let $S$ be the set of all nonunits of $R$ . We have $S\subseteq M\neq R$ . Let $x\in M$ . Since $M\neq R$ , $x$ cannot be a unit. So $x\in S$ . Thus $M\subseteq S$ . Hence $S=M$ , which is an ideal.

(iii) $\implies$ (i): Assume $S$ , the set of nonunits, form an ideal. Let $I\neq R$ be a maximal ideal. Let $a\in I$ , then $a$ cannot be a unit so $a\in S$ . Thus $I\subseteq S\neq R$ . By maximality $S=I$ and this shows $S$ is the unique maximal ideal.