# Left/right inverses in non-commutative ring

Let the setting be a non-commutative unital ring $A$ and $a,b \in A$. Assume $a$ is invertible, meaning it has either a left inverse, a right inverse or both.

Does it then hold that $ab$ is invertible if and only if $b$ is?

I believe not: If $a$ has a right inverse $aa^{-1} = 1$ and $b$ has a left inverse: $b^{-1}b=1$ then I can't see how to invert $ab$. Am I missing something?

• The statement is correct if invertible is interpreted as 'having a right- and a leftinverse'. If it is interpreted differently (I think you better do not) then I don't think so. – drhab Feb 15 '14 at 10:10
• @drhab But if the setting is not commutative one cannot assume an invertible element has both left and right inverse. Or am I missing something here? – newb Feb 15 '14 at 10:28
• Are you assuming that every element of $A$ has either a left or a right inverse, but some may have left, others right? Or as the question seems, are you only asking whether assuming $ab$ has a one-sided inverse (left or right) then also $b$ has a one sided inverse (also left or right)? – coffeemath Feb 15 '14 at 10:47
• Normally this is practicized as definition: an element in a ring (commutative or not) is invertible if it has a right- and a leftinverse. If that is the case then it can be shown that these inverses are the same. Have a look at: en.wikipedia.org/wiki/Unit_(ring_theory) – drhab Feb 15 '14 at 10:48
• @drhab The question of post maybe better worded: is it true that $ab$ has a left or right inverse iff $b$ has a left or right inverse? [The OP newb should clear this up, in my opinion.] – coffeemath Feb 15 '14 at 11:02

Any standard example of elements such that $ba=1$ and $ab\neq 0,1$ is a counterexample for you.
The first says that $a,b$ are both one-sided invertible. Then you can deduce that $(ab)^2=ab$, so $ab$ is a nonzero idempotent. A nontrivial idempotent in a ring with identity (meaning one other than $0,1$) is never left or right invertible because it is both a left and right zero divisor: $(1-ab)ab=ab(1-ab)=0$.
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 $ab\neq 0,1$.
• In the first line, did you mean $ab = 1$? – newb Feb 24 '14 at 14:32
• And $ba \neq 0,1$? – newb Feb 24 '14 at 14:33
• Dear @newb : No, I didn't intend that, because I'm giving an example where $ab$ is not invertible as a counterexample to the posted claim. However, thanks to your comment I can see that the $ba$'s in the second paragraph ought to be $ab$'s... thanks for spotting this typo. I usually type this example so that $ab=1$ and I managed to mix that with this. – rschwieb Feb 24 '14 at 14:54