# 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?

• $R$ is just a ring? – Rafael Holanda Sep 5 '16 at 9:07
• It has multiplicative identity. – tony Sep 5 '16 at 9:09
• Since $uvu=u$, you can write $u(1-vu)=0$ and $(uv-1)u=0$. How you have showed that $u\neq 0$, if the ring was an integral domain would be proved. – Rafael Holanda Sep 5 '16 at 9:17
• there is nothing here about the ring being an integral domain. – tony Sep 5 '16 at 9:18
• are you asking me how i proved u not equal to 0? – tony Sep 5 '16 at 9:19

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 $uv=1$.
• And then, of course, you could symmetrically show $vu=1$, if the OP was not already convinced that $u$ wasn't single-sided-invertible. – rschwieb Sep 5 '16 at 18:25
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, $$v = u^{-1}$$.
• Why is $uvu = u(vu)^{-1}$? – epimorphic Apr 17 at 3:22