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?

  • $\begingroup$ $R$ is just a ring? $\endgroup$ – Rafael Holanda Sep 5 '16 at 9:07
  • $\begingroup$ It has multiplicative identity. $\endgroup$ – tony Sep 5 '16 at 9:09
  • $\begingroup$ 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. $\endgroup$ – Rafael Holanda Sep 5 '16 at 9:17
  • $\begingroup$ there is nothing here about the ring being an integral domain. $\endgroup$ – tony Sep 5 '16 at 9:18
  • $\begingroup$ are you asking me how i proved u not equal to 0? $\endgroup$ – 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 $uv=1$.

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    $\begingroup$ Thanks. How did you come to this solution? $\endgroup$ – tony Sep 5 '16 at 9:23
  • $\begingroup$ It just flashed through, and seemed correct. So it is. $\endgroup$ – Teresa Lisbon Sep 5 '16 at 9:29
  • $\begingroup$ And then, of course, you could symmetrically show $vu=1$, if the OP was not already convinced that $u$ wasn't single-sided-invertible. $\endgroup$ – rschwieb Sep 5 '16 at 18:25
  • $\begingroup$ Yeah, I did it. Thanks. $\endgroup$ – tony Sep 6 '16 at 5:49

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, $v = u^{-1}$.

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  • $\begingroup$ Why is $uvu = u(vu)^{-1}$? $\endgroup$ – epimorphic Apr 17 at 3:22

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