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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asked Oct 8, 2011 at 3:25
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5$\begingroup$ 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. $\endgroup$– Ross MillikanOct 8, 2011 at 4:03
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$\begingroup$ 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. $\endgroup$– JoeApr 12 at 18:50
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3 Answers
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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.
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1$\begingroup$ 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 $I(0)=0$? $\endgroup$– GroupsDec 25, 2014 at 9:53
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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 $yx-1$.
answered Oct 8, 2011 at 3:30
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14$\begingroup$ In my example, this is the matrix of integration with respect to a suitable basis. $\endgroup$– lhfOct 8, 2011 at 3:37
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$\begingroup$ A very good and nontrival example! $\endgroup$ Oct 8, 2011 at 4:45
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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.
