Show that an element $x$ belongs to the unique maximal ideal of the commutative ring $R$ if and only if $ 1+xr$ is invertible. Let R be a commutative ring with $1 \neq 0$. Let $M$ be the unique maximal ideal of R. I want to show that $x \in M$ if and only if $1+xr$ is invertible for every $r \in R$. I am a bit stuck on where to begin. Any hints are appreciated.
 A: Suppose that $x\in M, $ for every $r, xr\in M$ since $M$ is an ideal. If $(1+xr)$ is not invertible, it is contained in a maximal ideal which is $M$, you deduce that $1=(1+xr)-xr\in M$. Contradiction.
Suppose that $(1+xr)$  is invertible for every $r$, if $x$ is not in $M$, it is invertible, there exists $u$ such that $ux=1$, $1+(-u)x=0$ is not invertible. Contradiction.
A: It is the special case $\rm\ J = (x)\ $ in $\ (1) \iff (2)\ $ below.
Theorem $\ $ TFAE in ring $\rm\:R\:$ with units $\rm\:U,\:$ ideal $\rm\:J,\:$ and Jacobson radical $\rm\:Jac(R)$
$\rm(1)\quad  J \subseteq Jac(R),\quad $ i.e.  $\rm\:J\:$  lies in every max ideal $\rm\:M\:$ of $\rm\:R$
$\rm(2)\quad  1+J \subseteq U,\quad\ \ $  i.e. $\rm\: 1 + j\:$  is a unit for every $\rm\: j \in J$
$\rm(3)\quad  I\neq 1\ \Rightarrow\  I\,+\,J \neq 1,\quad $  i.e.  proper  ideals survive in $\rm\:R/J$
$\rm(4)\quad\!\! M$ max $\rm\,\Rightarrow  M+J \ne 1,\quad $  i.e. max ideals survive in $\rm\:R/J$
Proof $\: $ (sketch) $\ $  With  $\rm\:i \in I,\ j \in J,\:$ and max ideal $\rm\:M,$
$\rm(1\Rightarrow 2)\quad  j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\:$ unit.
$\rm(2\Rightarrow 3)\quad i+j = 1\ \Rightarrow\ 1-j = i\:$ unit $\rm\:\Rightarrow\:  I = 1$
$\rm(3\Rightarrow 4)\ \ \ $  Let $\rm\:I = M\:$ max.
$\rm(4\Rightarrow 1)\quad  M+J \ne 1 \Rightarrow\ J \subseteq M\:$  by  $\rm\:M\:$ max.
