2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ $R$ 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
8
$\begingroup$

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 $uv=1$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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
-1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.