Sided inverses in a non-commutative ring

Question

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,

Answer 1

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.

Answer 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.