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Let $A$ be a unitary ring. The question is simply: can the product of two non invertible elements in $A$ be invertible?

I proved that the answer is negative if $A$ does not have zero divisors, because if you have $a,b$ non invertible elements in $A$ and $abx=xab=1$, then $a(bx)=1$ so $a$ is right invertible and if we put $d=bxa \implies db=b(xab)=b1=b \implies d=1$ (because obviously $b$ is not $0$), so $a$ is left invertible, as desired. If $A$ does have zero divisors, I don't see how to adapt the proof...

Thank you!

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Yes. Consider the ring of bounded linear operators on $\ell^2$. Let $$ a = (x_1,x_2,\dots) \mapsto (0,x_1,x_2,\dots), \\ b = (x_1,x_2,\dots) \mapsto (x_2,x_3,\dots) .$$ Then $ba = I$, but neither $a$ nor $b$ are invertible.

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    $\begingroup$ Don't you means "Yes"??? $\endgroup$ – Gina Jan 5 '14 at 1:25
  • $\begingroup$ Fixed it, thanks. There is a theorem that says if $a$ in a unitary ring has a left inverse, but no right inverse, then it has infinitely many left inverses. I don't remember the proof, and I remember it being a hard problem. So this does mean that my example cannot be made much simpler. $\endgroup$ – Stephen Montgomery-Smith Jan 5 '14 at 1:34
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    $\begingroup$ Here is a proof that the ring has to be infinite: math.stackexchange.com/questions/138541/… $\endgroup$ – Stephen Montgomery-Smith Jan 5 '14 at 1:37
  • $\begingroup$ Thanks, I try to search for the theorem you mentioned. The closest I got is Kaplansky-Jacobson theorem, which say that if it have more than 1 right inverse then it have infinitely many. But I don't think that's what you're talking about. Do you know the name of the theorem? $\endgroup$ – Gina Jan 5 '14 at 1:43
  • $\begingroup$ Has anyone found the name of the theorem? $\endgroup$ – Léo Léopold Hertz 준영 Jan 16 '14 at 11:16
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Here is a more algebraic example. Let $R$ be the ring of all $\infty\times\infty$ matrices with real entries such that in each column there is only a finite number of non-zero entries. Let $A = \left(\begin{array}{CCCcC} 0 & 1 & 0 & 0& \cdots \\ 0 & 0 & 2& 0& \cdots \\ 0 & 0 & 0 & 3 & \cdots \\ \vdots & \vdots & \vdots &\end{array}\right)$ and $B = \left(\begin{array}{CCCcC} 0 & 0 & 0 & 0& \cdots \\ 1 & 0 & 0 & 0& \cdots \\ 0 & \frac 12 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots &\end{array}\right)$. Then it is clear that $AB = I$ therefore $AB$ is invertible but neither $A$ nor $B$ are not invertible. (since for example the first row of $B$ is zero this mean that for any matrix $C \in R$ the $(1,1)$ entries of $BC$ is zero so this is impossible to have $BC = I$.)

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