Show that if a ring $R$ satisfies that for every $x\in R$, $x$ or $1-x$ is invertible then $R$ has a unique max ideal

Question

I tried to prove that if $R$ is like that the group of all not invertible in $R$ is ideal and from there to prove the last but that didnt work for me.

Answer 1

Showing that $m$ is indeed an ideal: Let $a\in m$ and $r\in R$, suppose $ra$ becomes a unit then we find some $b\in R$ with $b(ra)=1$ but then $(br)a=1$ and thus $a$ is a unit, which contradicts the definition of $m$ hence $ra\in m$

Now let $x,y\in m$ suppose $x+y$ becomes a unit, then $1=b(x+y)=bx+by$ and hence $bx=1-by$. We already now, that $by$ is not a unit by above argumentation. By assumption $1-by$ must me a unit and therefore $bx$ is a unit, making $x\in m$ a unit. Which is a contradiction.