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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 $zu^2z=1$

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?

  • 0
    Is is true in a monoid?2012-06-10

1 Answers 1

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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.