I've asked myself the following question : does there exist a non-commutative ring $R$ with unity $1$ and elements $x,y,z \in R$ such that $xyz = 1$ but $y$ has no left nor right inverses?

(Perhaps I don't need the whole ring structure to ask myself this question but only the multiplicative structure in the ring...)

I have tried several examples (matrix rings, common function spaces examples) but everytime I build $x,y,z$ such that $xyz = 1$ I always end up having a left or a right inverse, because to keep the product to be $1$, I want to "keep all the information from $y$", hence it has an inverse for some weird reason (this kind of problem happened to me over function spaces). Over matrix rings I had the non-zero determinant problem.

Thanks in advance,

  • $\begingroup$ It is impossible in Dedekind-finite rings, as there $xyz=1\implies y(zx)=1$. And finite matrices over a field are Dedekind-finite. $\endgroup$
    – user23211
    Apr 29, 2012 at 6:31
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    $\begingroup$ My instinct would be to look for an counterexample in some kind of "free algebra" generated by noncommuting $x$, $y$, and $z$ (with coefficients in your favorite commutative ring, I guess), and then see if modding out by the single relation $xyz-1$ gives you a ring with the desired property. $\endgroup$ Apr 29, 2012 at 6:40
  • $\begingroup$ @leslie : So perhaps that with $R$ a commutative ring with $1$ and the free algebra generated by non-commuting $x$,$y$ and $z$, one can quotient this by the principal ideal generated by $xyz-1$ and obtain a ring where $\overline x \overline y \overline z = \overline 1$ and $y$ has no left or right inverse by assumption. That's cool =D Thanks for the idea! It seems like a nice trick to generate counter examples, to go through the free algebra and quotient by relations. If you transform your comment in an answer I would check it! $\endgroup$ Apr 29, 2012 at 7:23
  • $\begingroup$ I'd check one of the others, if only because it requires work to prove that $\overline{y}$ has no left or right inverse in the quotient. (The work is simplified by the freeness of the algebra, and the fact that elements in the algebra can be represented uniquely as "noncommutative polynomials" in $x,y,z$, but it's still work). Generally, the "free" idea is very useful for counterexamples because it gives you a specific thing to try (often, if there is an example, the quotient of the free thing must to be an example). But it isn't always the easiest example to prove statements about. $\endgroup$ Apr 29, 2012 at 18:52
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    $\begingroup$ ... for example if you were looking for a ring with three elements $x,y,z$ satisfying some more complicated set of relations, the "quotient a free $R$-algebra by relations" idea still gives you a starting point. But if there were a lot of relations, or complicated relations, the simplest Way to prove that that quotient had the properties you want would probably be to exhibit a concrete $R$ algebra with actual elements $x_0,y_0,z_0$ having the desired properties. So you don't get much out of the idea in that situation. $\endgroup$ Apr 29, 2012 at 18:56

2 Answers 2


Let $R$ be the ring of $\mathbb R$-endomorphisms of the vector space $V$ equal to the direct sum of $\aleph_0$ copies of $\mathbb R.$ Let $e_1,e_2,e_3,\ldots$ be a basis of $V.$ Let $\gamma\in R$ be given by $$e_1\mapsto e_3$$ $$e_2\mapsto e_4$$ $$\vdots$$

$\beta\in R$ by $$e_1\mapsto 0$$ $$e_2\mapsto 0$$ $$e_3\mapsto e_2$$ $$e_4\mapsto e_3$$ $$\vdots$$

And $\alpha\in R$ by $$e_1\mapsto 0$$ $$e_2\mapsto e_1$$ $$e_3\mapsto e_2$$ $$\vdots$$

Now $$\alpha\beta\gamma=1.$$

However $\beta$ is neither surjective nor injective and therefore is neither a left unit nor a right unit.

  • $\begingroup$ Hm. I tried this ring but didn't find an example... I guess I didn't try hard enough! Thanks. My problem when I thought of this example was that I made the middle guy surjective ; I had let $(e_1, e_2, \dots) \mapsto ( e_2, e_3 \dots)$. Doing the backshift operator by removing a component is a good idea. $\endgroup$ Apr 29, 2012 at 7:15
  • $\begingroup$ It is a slightly modified standard example of a ring that is not Dedekind-finite. The standard example is that $\alpha(\beta\gamma)=1$ but $(\beta\gamma)\alpha\neq 1$ $\endgroup$
    – user23211
    Apr 29, 2012 at 7:26

Let $V=\oplus_{i\in \mathbb{N}}\mathbb{Q}$. Let $R=\operatorname{Hom}\,_{\mathbb{Q}}(V,V)$. Then $R$ is an example.

Consider the matrix $$ A=\left(\begin{array}{ccc} 1&0&0&\ldots\\ 0&0&0&\ldots\\ 0&0&1&\ldots\\ \vdots & \vdots & \vdots & \ddots \end{array}\right) $$

Namely, the $2i-1$th row of $A$ is $e_{2i-1}$ but $2i$th row of $A$ is zero.

Then $A$ is the require $y$.

OKay, though I have not said the structure of $R$.

The ring $R$ is a ring of matrices but having infinite dimension and the columns of an element of $R$ have only finitely many non-zero entries while the rows are arbitary.

An interesting fact about $R$ is: there is only one nontrivial two-sided ideal $I$ which is generated by $E_{11}$, the matrix whose $(1,1)$-entry is 1 and other entries are zero.

  • $\begingroup$ Not giving the $x$ and the $z$ makes your counter example quite incomplete... add them if you want me to upvote your answer. The matrix part is quite useless though ; you only need to know what basis vector maps to what basis vector; you're adding a ton of zeros for no reason. =P $\endgroup$ Apr 29, 2012 at 7:16
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    $\begingroup$ @PatrickDaSilva I believe every endomorphism with rank $\aleph_0$ which is neither surjective nor injective will be an example. I have to go to the doctor now so while I wait for my turn I'll try to prove that. $\endgroup$
    – user23211
    Apr 29, 2012 at 7:32
  • $\begingroup$ @ymar : I don't think the zero map will find $x$ and $z$ such that $x0z$ will be the identity map. But some endomorphisms may be this lucky, I agree. $\endgroup$ Apr 29, 2012 at 7:37
  • $\begingroup$ @PatrickDaSilva $0$ doesn't have rank $\aleph_0$! $\endgroup$
    – user23211
    Apr 29, 2012 at 7:38
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    $\begingroup$ @PatrickDaSilva OK, I won't post the proof then. It would require some effort because I would have to remember a certain proof in semigroup theory to be able to convince you. But it's true. $\endgroup$
    – user23211
    Apr 29, 2012 at 9:25

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