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Is it true that if $ab=u$ where $u\in U(R)$ is a unit of the noncommutative ring $R$, then $a,b\in U(R)$? If $R$ is commutative, then this can be seen by

$$a(bu^{-1})=uu^{-1}=1=(bu^{-1})a,$$

but when $R$ is not commutative I am not seeing an easy identity like the above. Is there a counterexample?

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  • $\begingroup$ Consider the left ideal $aR$ and compare it to the left ideal generated by $u$. $\endgroup$ – Hans Engler Dec 4 '14 at 17:06
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No. There exists a ring with $a,b$ such that $ab=1$ and $ba\neq 1$. Even though $1$ is a unit, neither $a$ nor $b$ are units.

Looks like I had trouble finding this before when I was writing this solution:

Examples of such rings with $a$ and $b$ like that are scattered throughout the site, but a little hard to search for. I found one here at this related question, although you don't really need to say "bounded linear operators on $\ell^2$," you can just say "linear transformations from $V\to V$ where $V$ is a vector space with countably infinite dimension." For a fixed basis, the "right shift" $a$ and the "left shift" $b$ on the basis elements create linear transformations such that $ba=1$ and $ab\neq 0,1$.

That is, given a basis $\{v_i\mid i\in \Bbb N\}$, $a$ maps $v_i\mapsto v_{i+1}$, and the other map $b$ maps $v_i\mapsto v_{i-1}$ when $i\geq 1$, and it just maps $v_0$ to zero.

A possible fix

But if both $ab$ and $ba$ are units, then $a$ and $b$ are units. Suppose $abx=1$ and $yba=1$. Then $a^{-1}=bx=yb$. Finally, $a^{-1}ab=b$ is a unit because it is a product of two units.

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  • $\begingroup$ Which ring is that? $\endgroup$ – Raclette Dec 4 '14 at 17:08
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    $\begingroup$ @Raclette I'm searching the site for links to the example right now. It's the ring of linear transformations of an infinite dimensional vector space. It appears in at least a half-dozen solutions. $\endgroup$ – rschwieb Dec 4 '14 at 17:09
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    $\begingroup$ @Raclette More details now provided $\endgroup$ – rschwieb Dec 4 '14 at 17:51
  • $\begingroup$ I don't follow your proof in the positive direction. How do you know that $bx=yb$? I'm pretty sure it is possible to have an element $a$ such that $ax=1=ya$ yet $a$ is not a unit (and does not have a unique left or right inverse). $\endgroup$ – Mario Carneiro Dec 5 '14 at 12:58
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    $\begingroup$ @MarioCarneiro Just notice that $y=y(ax)=(ya)x=x$. That is, having an inverse on both sides implies there is a unique two sided inverse. $\endgroup$ – rschwieb Dec 5 '14 at 17:42
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Following up on @rschwieb's answer:

$R = L(\ell^2)$ (ring of linear transformations), $a$ = left shift $(x_1,x_2, \dots) \mapsto (x_2,x_3, \dots)$, $b$ = right shift $(x_1,x_2, \dots) \mapsto (0, x_1,x_2, \dots)$.

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  • $\begingroup$ It does not really seem very nice to preempt another user like this, but I guess it's not a big deal and that's all I'm going to say. You might want to remove your original comment under the OP, though, since it looks like maybe you were hinting at a positive answer in it, and it apparently has some typos. $\endgroup$ – rschwieb Dec 4 '14 at 17:55
  • $\begingroup$ @rschwieb No need to direct your ire at the answerer for my actions - he had the answer before you could find your example so I credited him for it. Now that you've improved your answer with more details, I can move the checkmark. $\endgroup$ – Mario Carneiro Dec 5 '14 at 12:56
  • $\begingroup$ @MarioCarneiro no ire here, and the check doesn't really matter to me. Just a little common sense social advice that can be heeded or ignored. $\endgroup$ – rschwieb Dec 5 '14 at 17:46
  • $\begingroup$ @all - this is a fast-moving discussion forum about science. Topics belong to all, one should not expect to lay claim to a piece of turf, and the rewards for contributors are in the solution of the problem that has been raised and can therefore be shared - at least in my view. In my post, I attributed prior work and then followed up, in line with common scientific practice. I hope that we can now put this matter to rest. $\endgroup$ – Hans Engler Dec 5 '14 at 18:28

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