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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,

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    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

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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.

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    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$2012-04-29
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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.

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    Yes, it does -- like a rng (that is a ring possibly without unity) without addition. A semigroup is simply a set with an associative operation. The name comes from this example: $(\mathbb N,+)$, which is the additive group of integers "cut in half". I think this statement should be possible to prove without any knowledge of semigroup theory, but this is where it logically belongs. Actually, some similar statements from which this one readily follows are exercises in a classic book on semigroup theory.2012-04-30