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Let $R$ be a unital ring. If $a\in R$ is a left-unit then $\exists b\in R\mid ab=1$. If $a$ also happens to be a right-unit then also $\exists c\in R\mid ca=1$. This means we have: $$ab=1\wedge ca=1\Longrightarrow cab=c=b$$ Thus we have a unique inverse for $a$.

I wonder if we can also find a unique inverse when $a$ is only a left-unit. Does anyone know how to prove/disprove this.

gebruiker
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1 Answers1

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Actually, trying to prove uniqueness of left inverses leads to dramatic failure! It's an interesting exercise that if $a$ is a left unit that is not a right unit, then there are infinitely many $b$'s such that $ab=1$.

So the only situations that can arise for a left unit $a$ is that it is a two-sided unit (with a unique inverse that works on either side) or else it is truly one-sided, in which case it has infinitely many distinct left inverses.

Of course, a left-unit need not be a unit. The standard example is to use a vector space of countable dimension and basis $\{b_1,b_2,\ldots\}$ and define linear transformations $f$ and $g$ by $f(b_i)=b_{i+1}$ for all $i$ and $g(b_i)=b_{i-1}$ for all $i\geq 0$, and $g(b_0)=0$. It's clear that $gf=1$ but $fg\neq 1$.

Actually, the choice of $g(b_0)$ can be chosen completely at random from the vectors in the vector space, and each such choice generates a distinct new map $g$. All of them, however, satisfy $gf=1$.

rschwieb
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  • Whether $fg \neq 1$ is possible is not the question here. The question is whether we can construct $h$, different from $g$, such that $hf = 1$. This we can do by defining $h(b_i) = b_{i-1}$ and $h(b_0)$ any nonzero scalar. – spin May 23 '14 at 16:17
  • @spin When I read the question, it says "a unique inverse" not "a unique left inverse." But even then, this solution implies that a left invertible element is either a unit or has infintely many left inverses. I will have to make this more obvious, apparently... – rschwieb May 23 '14 at 18:02
  • Thanks for the answer, however you did raise a new question (for me though)...: You've shown me that the left-inverse need not be unique. In your example you even showed that there can be infinitely many. From the first sentence of your answer it appears that $a$ should have either a unique inverse, (x)or $a$ should have infinitely many left-inverses. Could you elaborate on this, please. (Or would it be more appropriate to ask it as another question?) – gebruiker May 23 '14 at 18:57
  • @rschwieb: sorry, I meant any nonzero vector. I think the answer is better now with the edit. – spin May 23 '14 at 19:10
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    @gebruiker not necessary to ask it again: it appears in one or two places on this site already. I believe this page supplies a proof in the question. – rschwieb May 23 '14 at 19:22