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Can I have a hint on how to construct a ring $A$ such that there are $a, b \in A$ for which $ab = 1$ but $ba \neq 1$, please? It seems that square matrices over a field are out of question because of the determinants, and that implies that no faithful finite-dimensional representation must exist, and my imagination seems to have given up on me :)

Aleksei Averchenko
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    The two hints you have been given have a common thread: you need to lose information in one direction and cannot recover it in the other. – Ross Millikan Oct 08 '11 at 04:03
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    For future readers: a ring in which $xy=1\implies yx=1$ is called Dedekind-finite. As the answers to this question show, not all rings are Dedekind-finite, but there are several important classes of rings which are always Dedekind-finite, e.g. commutative rings, finite rings, domains, etc. – Joe Apr 12 '23 at 18:50

3 Answers3

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Take the ring of linear operators on the space of polynomials. Then consider (formal) integration and differentiation. Integration is injective but not surjective. Differentiation is surjective but not injective.

lhf
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    I am slightly confusing. Let $D$ denotes differentiation, and $I$ integration. I want to clarify whether $(D\circ I)(p(x))$ is not necessarily $p(x)$ or $I\circ D(p(x))$ is not necessarily $p(x)$. Should we take $I(0)=0$? – Groups Dec 25 '14 at 09:53
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    @Groups $I(f) = \int_0^x f(t)dt$ works. – Bib Apr 26 '15 at 15:47
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Consider the ring of infinite matrices which have finitely many non-zero elements both in each row and in each column and the matrix $$a=\begin{pmatrix}0&0&0&\cdots\\1&0&0&\cdots\\0&1&0&\cdots\\\ddots&\ddots&\ddots&\ddots\end{pmatrix}.$$

A canonical example is the quotient $A$ of the free algebra $k\langle x,y\rangle$ by the two-sided ideal generated by $yx-1$.

Mariano Suárez-Álvarez
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    In my example, this is the matrix of integration with respect to a suitable basis. – lhf Oct 08 '11 at 03:37
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Let $G$ be the group of sequences in $\mathbb R$ with pointwise addition, and let $A$ be its endomorphism ring. The right shift operator has a left inverse, but not a right inverse.

Joe
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