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Let $R$ be a ring with identity. Prove that if $1-ab$ is invertible for some $a,b \in R$, then $1-ba$ is also invertible.

Ok, si if $R$ is a ring with unity, then we have $R$ with $1 \ne 0$ We have to show that $\forall a,b \in R$, $(1-ab)$ is invertible, so then $(1-ba)$ is invertible as well.

My thought process if to maybe show that if $(1-ab)x=1, x(1-ab)=1$ for some $x \in R$. Then we would have to show that $(1-ba)y=1,$ so then $y(1-ba)=1$ for some $y \in R$????

marwalix
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cele
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    See also this question. – Bill Dubuque Apr 22 '15 at 18:36
  • I asked to reopen because the question @Bill Dubuque refers to is in the ring of matrices while this one considers a general ring with unity. – marwalix Apr 22 '15 at 18:38
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    @marwalix My answer there works generally, and the 2nd linked dupe is an exact dupe, i.e. the ring case. So it is surely a dupe (I inadvertantly posted the wrong target in the link, hence the comment). There are also other dupes iirc. – Bill Dubuque Apr 22 '15 at 18:39
  • Sorry @BillDubuque I had not seen your answer. Had stopped at the one accepted. It is definitely a duplicate. – marwalix Apr 22 '15 at 18:44
  • I did not realize this. I will look this over and comment if I have any questions – cele Apr 22 '15 at 18:45
  • @cele If the other answers do not suffice then you can ask questions on those answers, or, if that is not fruitful, then you can post a new (non-duplicate) question on the specific points that are not clear. – Bill Dubuque Apr 22 '15 at 18:47

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