If an element has a unique right inverse, is it invertible?

Suppose $u$ is an element of a ring with a right inverse. I'm trying to understand why the following are equivalent.

• $u$ has at least two right inverses
• $u$ is a left zero divisor
• $u$ is not a unit

If $v$ and $w$ are distinct right inverse of $u$, then $u(v-w)=0$, but $v-w\neq 0$, so $u$ is a left zero divisor. It's also clear that if $u$ is a left zero divisor, it cannot be a unit (else I could cancel $u$ from $ub=0$ to see $b=0$).

I'm having a heck of a time seeing why $u$ is not a unit implies $u$ has at least two right inverses. I tried the contrapositive, but saw no good approach. What am I missing?

Solution 1:

If $u$ has only one right inverse $v$, then $u(1-vu)=u-(uv)u=0$ hence $u(1-vu+v)=1$ and by uniqueness $1-vu+v=v$, so $1=vu$ and $v$ is a left inverse. Hence $u$ is a unit.