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. If $$ N\subset R\ $$ 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.