In a ring $R$ with a $1_R$ with $u,v, u+v$ are units, show that $u^{-1} + v^{-1}$ is also a unit.

Question

In a non-commutative ring $R$ with a $1_R$ with $u,v, u+v$ are units, show that $u^{-1} + v^{-1}$ is also a unit.

The question seems relatively simple, but I'm having a tricky time explicitly finding the formula.

I have tried :

$(u^{-1} + v^{-1})(u-v)=1-u^{-1}v+v^{-1}u-1=-u^{-1}v+v^{-1}u$

Hints appreciated.

Answer 1

$$ (u^{-1}+v^{-1})v(u+v)^{-1}u = (u^{-1}v + 1)(u+v)^{-1}u = u^{-1}(v+u)(u+v)^{-1}u = 1. $$ Now, try the other direction.

Answer 2

$(u^{-1}+v^{-1}) u v=v+u$ implies that $(u^{-1}+v^{-1}) (u v (v+u)^{-1})=1$. But only in the commutative case.

Answer 3

Lemma: In a ring with $1$, if $a$ and $c$ are units, and if $abc=1$, then $b$ is a unit. In fact $b^{-1}=ca$.

Proof: $b(ca)=a^{-1}abca=a^{-1}1a=1$. Showing that $(ca)b = 1$ is similar.

For your situation, look at $u(u^{-1}+v^{-1})v(u+v)^{-1}$

Then use the Lemma to conclude that $u^{-1}+v^{-1}$ is invertible.