Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
See also this thread. and this one. – Gone Apr 29 '12 at 20:41

2 Answers

up vote 2 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.