$R$ local ring, $I$ maximal ideal then $x\notin I$ implies $x$ unit

Question

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 $xy-1\in I$

Answer 1

Here's a different idea.

Suppose $x \not \in I$. Then since $R$ is local, $x$ must generate $R$ (unless $x=0$). Hence there is some $z \in R$ such that $xz=1$.

Answer 2

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.

Answer 3

if $x$ is not a unit, then $(x) \neq R$ and since $I$ is the unique maximal Ideal you must have $(x) \subseteq I$, which is impossible.