"If $x$ is a non-unit, then $1-ux$ is a unit"

Question

I don't understand these two lines from my book. We are given that $R$ is a local ring.

If every $2$-generator submodule is cyclic and $Ra$, $Rb$ are given, then $Ra+Rb=Rc$, hence $a=xc,b=yc,c=ua+vb$, so $c=uxc+vb$. If $x$ is a non-unit, then $1-ux$ is a unit, so $Ra\subseteq Rc \subseteq Rb$.

First, from the definition of a local ring, if $x$ is a non-unit, then $1-x$ is. But how can we deduce that $1-ux$ is?

Is it a typo? If not, what's the significance of $ux$? Where does it come up? I can only see that rearranging the equations in the text we get $c(1-ux)=vb$.

Answer 1

If $x$ is a non-unit, then so is $ux$ for any $u$. This is true in any commutative ring. Otherwise we'd have $$(ux)^{-1}ux=((ux)^{-1}u)x=1$$ This contradicts the assumption that $x$ is not a unit.

If the ring is noncommutative we can still obtain the result that $ux$ is not a unit from the fact that the nonunits form an ideal in a local ring.

Answer 2

$1-ux$ can not be in the maximal ideal as $x$ is and hence $ux$ is. And since we have a local ring, any element not in the maximal ideal must be a unit.