# A ring element with a left inverse but no right inverse?

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

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