Prove $uvu=u$, $v$ unique implies $u$ invertible in a ring $R$ Suppose $R$ is a ring and $u,v \in R$ are so, such that $uvu=u$ and $v$ is unique. How do I show that $u$ is invertible? R is assumed to have multiplicative identity.
So far I have been able to prove that $u,v \neq 0$. I have also proven that $u$ cannot be a non-invertible element with a one-sided inverse.
Can I get a few more hints?
Given that $uvu=u$ for some unique $v$, note that $u^2vu=u^2$. Consider the quantity : $u(v+uv-1)u$. Note that $u(v+uv-1)u = uvu + u^2vu - u^2 = u + u^2 - u^2  = u$. But then, $v$ was the unique element with this property, hence $v=v+uv-1$ hence $uv=1$.
In fact, there is another generalization form of this lemma. For an element $u$ in the ring $R$, the sufficient and necessary condition for its invertibility is either (1) $uvu=u$, $vu^2v=1$ or (2) $uvu=u$ and such $v$ is unique.

The necessity is easy to verify and part 2 of the sufficiency has been given. Therefore, I would like to provide the proof of part 1. Since the condition (1) implies that the semigroup $(R,\cdot)$ contains an identity element and $vu^2v=(vu)(uv)=1$, we have $uv=(vu)^{-1}$. Note that $u = uvu = u(vu)^{-1} = u^2v$. Then we have $vu=vu^2v=1$, which implies that $u$ is invertible and furthermore, $v = u^{-1}$.