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This question already has an answer here:

It is well-known that if $R$ is an Artin ring,and $ab=1$ in $R$ where $a,b\in R$,then $ba=1a$.(This is not difficult)this is a very hot in Mathematics. If $AB = I$ then $BA = I$

It seems it is not right for arbitrary ring that if $ab=1$,then $ba=1$. Can someone helps to give an example.

Thanks in advance!

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marked as duplicate by B. Mehta, rschwieb abstract-algebra May 17 '18 at 16:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ or this or this or this or any of other similar posts that show up if you just search for "[ring-theory] right left inverse" $\endgroup$ – rschwieb May 17 '18 at 16:05
  • $\begingroup$ @rschwieb thanks $\endgroup$ – Sky May 17 '18 at 23:08
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  1. Staying at linear endomorphisms, this is no longer true for infinite dimensional spaces.
    Let e.g. $e_1,e_2,\dots$ be a basis, and consider $B:=e_k\mapsto e_{k+1}$ and $A:=e_k\mapsto e_{k-1}, \ e_1\mapsto 0$

2.Take any monoid $M$ that satisfies this, then consider its 'group ring' $\Bbb ZM$.

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