I'm trying to figure out why an element $u$ in some ring is invertible with inverse $z$ if any only if

  • $uzu=u$ and $zu^2z=1$


  • $uzu=u$ and $z$ is the unique element meeting this condition.

Clearly, both conditions follow if $u$ is a unit with inverse $z$. However, I can't see why either condition implies that $z=u^{-1}$.

I haven't been able to make any decent progress on my own, so does anyone have hints or suggestions on where to go? Thanks.

Edit: From Qiaochu's hint, $zu$ and $uz$ are idempotent. So $(zu)^2=zu$. But $zu$ has right inverse $uz$, so $(zu)^2(uz)=(zu)(uz)\implies zu=1$. The analogous argument for $uz$ shows $uz=1$, so $z=u^{-1}$.

Does anyone have an idea for the second?

  • 6
    $\begingroup$ For the first: the condition $uzu = u$ implies that both $zu$ and $uz$ are idempotent. The condition $zuuz = 1$ implies that $zu$ has a right inverse (and $uz$ has a left inverse). Can you do anything with this? For the second: I don't have a complete solution but have you tried proving the contrapositive? $\endgroup$ – Qiaochu Yuan Jun 9 '12 at 23:46
  • $\begingroup$ Thanks @QiaochuYuan, I was able to solve the first from your hint. I'll give the second a shot. $\endgroup$ – Hana Bailey Jun 10 '12 at 0:00
  • 6
    $\begingroup$ Note that, for any $k$, if $uk=0$, then $uk+uz=uz\Rightarrow u(k+z)=uz$. Then, we have $u(k+z)u=uzu=u$, so, since $z$ is the unique element with this property, we have $k+z=z$, so, $k=0$. From this you can conclude! =p $\endgroup$ – Yuki Jun 10 '12 at 1:24
  • $\begingroup$ Is is true in a monoid? $\endgroup$ – user23211 Jun 10 '12 at 8:54

For the sake of having an answer:

The strategy is this: since we have $u=uzu$, we also have $0=u(zu-1)=(uz-1)u$. If it can be shown that $u$ is "regular" (in the sense that it is not a nonzero zero-divisor), then we have $zu-1=uz-1=0$, establishing the result.

As per Yuki's comment above, if $u\alpha=0$, then $u(z+\alpha)u=uzu=u$. By uniqueness of $z$, we have $z+\alpha=z$, and so $\alpha=0$. A symmetric argument establishes that if $\alpha u=0$, then $\alpha=0$. Thus, $u$ is regular.


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.