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Definition: An element $a$ in a ring $R$ with identity is called a right(left) divisor if there exist $u \in R$ such that $ua=1$ ($au=1$).

Can we construct a ring $R$ such that there exist an element $a\in R$ , $a$ is a right divisor of $R$ while $a$ is not a left divisor.

My attempt:
$(1)$ Since a is a right divisor, if such a ring exists, R must have an identity $1_R$.
(2) R can not be commutative.

jgon
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J.Guo
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    Two things, firstly, what is a right/left divisor of a ring? Also secondly, please don't put spaces between a word and following punctuation. It makes it substantially harder to read your post, since it's very distracting to see all this punctuation floating around the page. – jgon Feb 19 '19 at 03:00
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    Oh, you mean an element that is left invertible but not right invertible (or vice versa, not sure which one is left invertible and which is right invertible). – jgon Feb 19 '19 at 03:14
  • I put back your work, since your post will be looked on more favorably with that included. – jgon Feb 19 '19 at 03:18
  • Thank you ! I've learned algebra for only a few days with english textbook , so it might be difficult for me to describle the question proper . I hope the post edit now might me more proper . – J.Guo Feb 19 '19 at 03:25
  • Yes, I think this is a good question now. I'm also fairly sure that the answer is yes, but I'm having trouble thinking of a good example. (I mostly work with commutative rings) – jgon Feb 19 '19 at 03:27
  • Found a MSE question that answers your question: https://math.stackexchange.com/q/70777/90543 – jgon Feb 19 '19 at 03:32
  • @jgon I think that question is to prove $ab=1$ while $ba \neq 1$ . But I want to show $bx\neq 1$ for all $x$ . – J.Guo Feb 19 '19 at 03:50
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    Fair point, but the answer addresses your question. Since the example it gives is linear operators on the polynomials, with derivative of integral being the identity, but the integral operator is not surjective, so you can never have integral composed with any operator being the identity. – jgon Feb 19 '19 at 03:52
  • @jgon I see , thank you . – J.Guo Feb 19 '19 at 04:03
  • @J.Guo Also remember to try any of the numerous duplicates linked to the duplicate. Like Is a left invertible element of a ring necessarily right invertible?, which obviously is the same as your question. – rschwieb Feb 19 '19 at 15:14

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