TITLE: If $z$ is the unique element such that $uzu=u$, why is $z=u^{-1}$?
QUESTION [7 upvotes]: 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 $zu^2z=1$

OR

$uzu=u$ 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?

REPLY [8 votes]: 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.