Problem is this:
Suppose for any element $r$ in a ring $R$ with unity $1$, $r$ or $1-r$ is invertible. Then show that non-invertible elements form an ideal of $R$.
It is crystal clear that $-r$ is non-invertible if $r$ is. So I tried to show the sum of two non-invertible elements is non-invertible, and the product of non-invertible elements with any ring element is non-invertible but I'm stuck.
Any help will be appreciated.
Best Answer
Assuming the statement of the problem is that exactly one of $r$ or $1-r$ is invertible for any $r\in R$:
Let $J$ denote the set of noninvertible elements in $R$. Suppose $r,s\in J$, and that $r+s$ is invertible. Observe that $r$ noninvertible implies that $-r$ is noninvertible, so $1-(-r)=1+r$ is invertible. Then we have that:
$s=(r+s)-r\implies 1-s=1-(r+s)+r\implies (1-s)(r+s)^{-1}=(1+r)(r+s)^{-1}-1$.
But note that $(1-s)(r+s)^{-1}$ is invertible, so it must be that $(1+r)(r+s)^{-1}-1$ is invertible, but $(1+r)(r+s)^{-1}$ is invertible, so it must be that $(1+r)(r+s)^{-1}-1=-(1-(1+r)(r+s)^{-1}$ is noninvertible. Therefore, we have a contradiction, so $r+s$ is noninvertible.
(**)Hopefully, it is clear that the product of two invertible elements is again invertible. Now, let $r\in J$ and $q\in R$, and suppose $qr$ is invertible. We split the proof into two cases: First, suppose that $q$ is invertible. Note that $qr$ invertible implies that there exists $t\in R$ such that $tqr=1=qrt$. But then $t$ is invertible, so $qr=t^{-1}qrt$ and using $q$ invertible implies $r=q^{-1}t^{-1}qrt$, but then $r$ is the product of invertible elements and hence is invertible, a contradiction.
For the other case, suppose $q$ is noninvertible. Then $1-q$ must be invertible, so by the argument of the previous case, $(1-q)r$ must be noninvertible, and hence $1-(r-qr)$ is invertible. But then rearranging yields $-r+(1+qr)$ is invertible, but $-r$ is noninvertible, and $-qr$ is invertible so $1+qr$ is noninvertible, so $1-(r-qr)$ is the sum of noninvertible elements, and by what we have above, must be noninvertible, a contradiction.
Therefore, $J$ is a (left) ideal in $R$. To show it is also a right ideal, the proof is similar.
EDIT: At the request of @Junkhook9000 in the comments below, here is a proof that $J$ is an ideal which only requires the weaker property that at least one of $r$ and $1-r$ is invertible (as opposed to having exactly one of them invertible). First, prove (as above, starred) that the product of an invertible element with a noninvertible element is noninvertible. Now, suppose that $r,s\in J$ but that $r+s$ is invertible. Then there exists some element $q\in R$ with $q(r+s)=1$ and rearranging yields $r=q^{-1}(1-qs)$. But $1-qs$ is invertible since $qs$ is the product of an invertible element with a noninverible element and hence $qs$ is noninvertible. Therefore $r$ is the product of two invertible elements and is therefore invertible, a contradiction. So $J$ is closed under addition. Now suppose $rs$ is invertible. Then $(rs)r$ is noninvertible since $r\in J$, so $1-rsr$ is invertible. However, $1-rsr=(rs)^{-1}(rs-r)=-(rs)^{-1}(1-s)r$, which is noninvertible since $rs$ is invertible and since $1-s$ is invertible because $s$ is not, but the product of an invertible element (in this case, $-(rs)^{-1}(1-s)$) times a noninvertible element ($r$) is noninvertible. Therefore $rs$ cannot be invertible, so $rs\in J$.