Suppose you have a ring R with 1. If this ring is finite, I think I can prove that any element with a 1-sided inverse also has a two-sided inverse. The question asks for a counter-example when the ring is infinite. I'm having trouble coming up with one.
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See also this answer. – Bill Dubuque Aug 17 '17 at 02:02
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Easy examples of non-commutative rings are matrix rings (i.e. endomorphism rings of some vector space). To get an infinite "matrix" ring, look at an infinite vector space. As an example, take
$$ \bigoplus_{n = 1}^\infty \mathbf{Z}/2\mathbf{Z}. $$
See if you can find an element of
$$ \operatorname{End} \left( \bigoplus_{n = 1}^\infty \mathbf{Z}/2\mathbf{Z} \right) $$
which has a one-sided inverse but not a two-sided one. You can also replace $\mathbf{Z}/2\mathbf{Z}$ by some other ring if you prefer.
Trevor Gunn
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