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For a ring $R$, I must show that if R has a left identity that is not a right identity, then there is more than one left identity.

Could I do this by contradiction, using 2x2 matrices and show that a given set has two left identities that are not right identities?

user1729
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tdashrom
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    A left identity matrix is usually a right identity matrix as well, so it's not what I would consider a natural first idea. – Arthur Mar 29 '18 at 13:58
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    No, you cannot proceed as you suggest: You must show that all rings with a left identity that is not a right identity has more than one left identity, not that the property is true for some ring. – Travis Willse Mar 29 '18 at 14:01
  • @fredgoodman Sorry, it looks like I did confuse the task here and there (finding infinitely many left identities as opposed to left inverses). To resolve it more quickly, it would have been more constructive to actually mention this obvious possibility, rather than being opaquely saying merely (in effect) "I think you're wrong." Of course I've reopened, now that I finally noticed my cognitive error. – rschwieb Mar 29 '18 at 14:55

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Let $u$ be a left identity that is not a right identity. It suffices to find an element $v \ne 0$ such that $v y = 0$ for all $y \in R$, for then $u + v$ is a left identity. Since $u$ is not a right identity, there exists an element $x \in R$ such that $v = x u -x \ne 0$. Check that $v y = 0$ for all $y \in R$. Thus $v$ has the desired properties.

fred goodman
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