Non-commutative ring Local Ring Suppose $R$ is a non-commutative ring with 1. Let $L$ be the unique maximal left ideal of $R$.
Prove that $$R\setminus L=U(R),$$ where $U(R)=\{a\in R\mid ab=1=ba\ \text{for some}\ b\in R\}$ is the set of units of $R$.

I am having some problems proving this, and am even beginning to suspect that it is false. (See Set of non-units in non-commutative ring).
I have proved partial results like $L$ is in fact a two-sided ideal of $R$.
Thanks for any help.
 A: If you proved that $L$ is in fact a two sided ideal, then congratulations, the difficult (or the more ellusive - at least to me) part is over:
Clearly $U(R) \cap L= \emptyset,$ otherwise $L=R$. That is, $U(R) \subseteq R \setminus L$.
Now suppose that $a \in R \setminus L$. Then $Ra$ is a left ideal not contained in the unique left maximal ideal $L$. That is, it is improper, i.e. $Ra=R$. This means that $ba=1$ for some $b \in R$. Now, we again have $b \notin L$, for if $b \in L,$ then $1=ba \in L$ (because $L$ is a right ideal as well) and $L=R,$ contradiction. So we may repeat the argument for $b$ to get a $c \in R$ with $cb=1$. But now we have
$$a=1\cdot a=(cb)a=c(ba)=c \cdot 1=c,$$
so we have $ab=1$ as well, which proves that $a \in U(R)$.
A: It's true.
You can show that the set of nonunits of a ring is closed under addition iff the ring is local. (I think it is already here on math.SE but I couldn't find it.)
You can get at it lots of other ways too.
Here's a direct proof. I guess you agree $U(R)\subseteq R\setminus L$.
If $x$  isn't in $L$, then $Rx=R$. But then $yx=1$ implies $xy$ is a nonzero idempotent of $R$. The only nonzero idempotent of a local ring is $1$, so $x$ is a unit.
