# 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.

## Author: mathtuition88

http://mathtuition88.com

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