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 $uv=1$.