# Why are Dedekind-finite rings called so?

A Dedekind-finite ring is a ring in which $ab=1$ implies $ba=1$.

It seems natural to look for a connection to Dedekind-finite sets, however for such a set any injective endomorphism is surjective, while for a Dedekind-finite ring it goes vice versa, namely, any surjective endomorphism is injective (In other words, such a ring is Hopfian).

So, what is the motivation behind this name (for rings)?

Thanks.

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It would seem to me that you should simply apply your own observation concerning Dedekind-finite sets and their definition to the left/right homotheties involved: given $ab=1$, the right homothety defined by $b$, i.e. $\vartheta_b:R\rightarrow R$, $r\mapsto rb$, viewed as a homomorphism of abelian groups, say, is clearly surjective (one has $\vartheta_b\circ\vartheta_a=\text{id}_R$); iff also $ba=1$, then $\vartheta_a\circ\vartheta_b=\text{id}_R$, making $\vartheta_b$ injective, too (note also that the one-sided multiplicative inverses of an element, when they exist, must coincide due to associativity). The endomorphism rings of finite-dimensional vector spaces over (skew) fields are, of course, standard examples of Dedekind-finite rings, further justifying (possibly) the intuitive feel that such vector spaces (and hence their endomorphism rings) are "small" in a sense. Kind regards, Stephan F. Kroneck.

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Lam has an exercise on this in Lectures on Modules and Rings pp 18:

A module M is called Dedekind finite if $M\cong M\oplus N$ implies $N=0$. $M$ is a Dedekind finite module iff $End(M_R)$ is a Dedekind finite ring. If $M$ is Hopfian, then $End(M_R)$ is Dedekind finite, but not always conversely. The case when $M=R_R$, Dedekind finiteness of $R_R$ turns out to be equivalent to being a Hopfian module, since $R_R$ is projective.

I spent some time looking at rings where $R_R$ was $\textit{coHopfian}$, and found some interesting stuff. For one thing, it's not the same as being a coHopfian object in the category of rings. It took a lot of digging but I finally found an example given by Varadarajan of a left-not-right (module)-coHopfian ring.

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@BillDubuque We are definitely not on those terms yet, but as a compromise, you can find me idling in the crusade of answers chat during the day. – rschwieb Feb 6 '14 at 11:34
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