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.