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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?

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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? – Qiaochu Yuan Jun 9 '12 at 23:46
Thanks @QiaochuYuan, I was able to solve the first from your hint. I'll give the second a shot. – Hana Bailey Jun 10 '12 at 0:00
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 – Yuki Jun 10 '12 at 1:24
Is is true in a monoid? – user23211 Jun 10 '12 at 8:54

1 Answer 1

up vote 8 down vote accepted

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.

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