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.