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This question already has an answer here:

Let $A$ be a ring and $a\in A$ an element that has a right-inverse but does not have a left-inverse. Show that $a$ has infinitely many right-inverses.

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marked as duplicate by Alexander Gruber, rschwieb, Seirios, Davide Giraudo, Emily Mar 13 '13 at 18:35

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.

  • $\begingroup$ The title should be more helpful $\endgroup$ – NECing Mar 13 '13 at 18:08
  • $\begingroup$ Please formulate this question more clearly. $\endgroup$ – Alexander Gruber Mar 13 '13 at 18:08
  • $\begingroup$ Hmm, I duped it as the wrong question. I'm pretty sure this has already been asked though. $\endgroup$ – rschwieb Mar 13 '13 at 18:10
  • $\begingroup$ The question is almost certainly "If $a$ is left invertible but not right invertible, show $a$ has infinitely many distinct left inverses", a classic result. $\endgroup$ – rschwieb Mar 13 '13 at 18:11
  • $\begingroup$ So I understand this question was closed because of the unbelievable language it was "written" in? Because it is not the duplicate of Kaplansky's theorem, at least not directly. $\endgroup$ – DonAntonio Mar 14 '13 at 0:15
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Hint: Let $b$ be a right-inverse of $a$. For any $i \geq 0$, we define $b_i = (1-ba)a^i + b$. Show that if $a$ doesn't have a left-inverse, the $b_i$ are pairwise distinct right-inverses of $a$.

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