User:Kcbetancourt/AlgebraHW5

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 $$ M\subseteq (x) $$. But again, since $$ M $$ is the maximal ideal, this means that $$ (x) = R $$. Since $$ 1 \in R, \exists y \in R : xy = 1 $$. This implies that $$ x\ $$ is a unit. Therefore, every element of $$ R-M\ $$ is a unit.

<= Let R be a commutative ring with 1 and the set of all non-units be M. We want to show that M is a Unique Maximal in R. Assume that M is not maximal. Then $$ \exists \ $$ an ideal, I, in R such that M $$ \subset \ $$ I $$ \subset \ $$ R. And M $$ \ne \ $$ I $$ \ne \ $$ R. We know that 1 does not exist in M, because 1 is a unit. So if M is not equal to I, then $$ \exists \ $$ u $$ \in \ $$ I, where u is a unit. Then 1 $$ \in \ $$ I. But that implies I=R. But if I=R, then I is not a maximal idea. If I is not maximal, then that implies that M is maximal. Now assume that M is not unique. Then $$ \exists \ $$ M' such that M' is a maximal ideal in R. Then M' $$ \subset \ $$ R and M $$ \ne \ $$ R. So 1 $$ \notin \ $$ M'. Thus M' is the set of all non-units. That implies that M'=M. Therefore M is a unique maximal ideal in R.