User:Kcbetancourt

7.4.37 A commutative ring $$ R \ $$ is called a local ring if it has a maximal ideal. Prove that if $$ R \ $$ is a local ring with maximal ideal $$M \ $$, then every element of $$R-M \ $$ is a unit.

Since $$M \ $$ is a maximal ideal, it is prime. Let $$x,y\in R-M $$. If we had $$xy\in M \ $$, then since $$M \ $$ is prime, we have $$x\in M \ $$ or $$y \in M \ $$. Hence, $$R-M \ $$ is closed under multiplication.

Let $$ x\in R-M $$. Consider the ideal $$ (x)\ $$ generated by $$ x\ $$. Then, since $$ M\ $$ is the maximal ideal, $$ (x)\subseteq M $$. But $$ x\notin M $$ so we must have