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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 :)

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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 8 '11 at 4:03

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up vote 15 down vote accepted

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.

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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 9:53
@Groups $I(f) = \int_0^x f(t)dt$ works. –  Bib Apr 26 at 15:47

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 $yx-1$.

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In my example, this is the matrix of integration with respect to a suitable basis. –  lhf Oct 8 '11 at 3:37
A very good and nontrival example! –  Mathemagician1234 Oct 8 '11 at 4:45

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