User:Calimikester

$ 7.4.37\ $

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

Let $ R\ $ be a commutative local ring with identity. Let $ M\ $ be the unique maximal ideal. Consider $ R-M\ $, the set of all elements in $ R\ $, but not in $ M\ $. Since $ 1\notin M\ $, $ 1 \in R-M\ $. Let $ u\in R-M\ $ and consider $ (u)\ $, the ideal generated by $ u\ $. If $ (u)\ $ is a proper ideal, then $ (u)\ \subset M\ $, but this contradicts the fact that $ u\notin M\ $. This shows that $ (u)\ = R\ $. Thus, $ \exists v\in R $ such that $ uv = 1\ $. This shows that every element in $ R-M\ $ is a unit.

Prove conversely that if $ R\ $ is a commutative ring with $ 1\ $ in which the set of nonunits forms an ideal $ M\ $, then $ R\ $ is a local ring with unique maximal ideal $ M\ $.

Suppose $ R\ $ is a commutative ring with $ 1\ $ in which the set of nonunits forms an ideal $ M\ $. We need to show that $ M\ $ is a unique maximal in $ R\ $. Assume that $ M\ $ is not maximal. Then $ \exists I\in R\ $ that is maximal such that $ M\ \subset \ I \subset \ R\ $ and $ M\ne I\ne \ R $.

We know that $ 1\notin M\ $ (because $ 1\ $ is a unit), and since $ M\ne I\ $, $ \exists u\in I\ $, where $ u\ $ is a unit. Then $ 1 \in I\ $. But that implies $ I= R\ $. And if $ I= R\ $, then $ I\ $ is not a maximal ideal. If $ I\ $ is not maximal, then that implies that $ M\ $ is maximal.

Now, suppose we have an ideal $ N\subseteq R\ $ and $ N\ $ is not a subset of $ M\ $. Then $ \exists u\in N\ $ such that it's a unit. Again, we have $ N= R\ $. This shows that every proper ideal of $ R\ $ is contained in $ M\ $. Thus $ M\ $ is the unique maximal ideal in $ R\ $ and $ R\ $ is local.