1
$\begingroup$

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,

  • 0
    It is impossible in Dedekind-finite rings, as there $xyz=1\implies y(zx)=1$. And finite matrices over a field are Dedekind-finite.2012-04-29
  • 2
    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.2012-04-29
  • 0
    @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!2012-04-29
  • 0
    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.2012-04-29
  • 1
    ... 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.2012-04-29
  • 0
    Start with something simpler: try to find a ring $R$ with $x,z\in R$ such that $xz=1$ but $zx\neq1$. Then $y:=zx$ is a nontrivial idempotent and hence cannot have a left or right inverse.2012-05-02

2 Answers 2