Let $R$ be a conmutative local ring, $I$ its maximal ideal. I want to prove that $x\notin I$ implies $x$ unit.
So far I have: Let $x\notin I$, I consider $x+I\in A/I$, which is a field (because $I$ is maximal). So there exists $y+I\in A/I$ such that $(x+I)(y+I)=1+I\Longrightarrow xy+I=1+I$. How do I deduce that $y$ is the inverse of $x$ (which I think it is true). All I have is $xy-1\in I$
Best Answer
Suppose $x$ is not a unit. Then by Zorns lemma there exists a maximal ideal in $R$ containing $x$. But $R$ is a local ring, so has a unique maximal ideal, which is a contradiction.