0

Today i was working on a problem on a ring with goes as follows. Let A be a ring, prove that any element of A has either $0,1$ or an infinity of left inverses.

I think i found a proof of this statement, but the official solution looks nothing like mine. It is said that i should show that if any element has two distinct left inverses then there's an infinity of them (which i did) by considering the kernel of $f:x\rightarrow ax$. My try goes as follow:

let $x,y$ be such inverses then for all $p,q\in\mathbb Z$ satisfying $p+q=1$, $px+qy$ is another left inverse, because : $(px+qy)a=p+q=1$. And for all $p\in\mathbb Z$ we have :

$px-(p-1)y=(p-1)x-(p-2)y$ that implies $x=y$. Hence there are infinitely many distinct elements of the form $px+qy$. Is my proof flawed ?

Shaun
  • 44,997
1 2 3
  • 199
  • 9
  • The sequence $0,1,0,1,0,1\cdots$ only takes two values, even though any two consecutive terms are distinct. – tkf Nov 22 '20 at 18:45
  • You proved that if $x\ne y$ are two distinct left inverses of (something, let's call it $u$), then, for all $p\in\Bbb Z$, $px+(1-p)y$ and $(p-1)x+(2-p)y$ are two distinct left inverses of $u$. You did not prove that, for all two distinct $p_1,p_2\in \Bbb Z$, $p_1x+(1-p_1)y$ and $p_2x+(1-p_2)y$ are distinct. –  Nov 22 '20 at 18:47
  • @JoséCarlosSantos I think you misread the question. The OP is already aware of a proof of the theorem. They are asking why an alternative set of left inverses are not in general distinct. They should be able to construct a counterexample from the comments, but if not, I don't see why someone should not spell it out as an answer at some point. – tkf Nov 22 '20 at 19:35
  • Why do you claim that the OP is already aware of a proof of the theorem? – José Carlos Santos Nov 22 '20 at 19:47
  • @JoséCarlosSantos "but the official solution looks nothing like mine." – tkf Nov 22 '20 at 19:56
  • @tkf Right. I missed that. Anyway, this is a site for questions and answers. Since the OP now knows that the question has already been posted here, they can always post their solution as an answer. – José Carlos Santos Nov 22 '20 at 19:59
  • @JoséCarlosSantos No they correctly suspected that their solution was wrong. The error in their logic was pointed out to them. A complete answer though would be a counterexample, where their set of inverses are not infinite. In any case there is no reason for this question to be closed. – tkf Nov 22 '20 at 20:04

0 Answers0