Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm interested in knowing/collecting some properties of epimorphisms of rings (with identity) that are true over commutative rings but are false in the non-commutative case.

Example: I learned from MO that if $R \hookrightarrow S$ is an epimorphism of commutative rings then $S/R$ is a torsion left $R$-module. But there are counter-examples to this property if $R$ is permitted to be non-comutative.

share|improve this question
    
I believe that the property that you mention must be non-commutative, or why consider it as lef $R$-module? –  Murphy Mar 14 '13 at 1:00
    
I consider it as left module, because basically all my modules are left modules (unless I really need a right module). –  tj_ Mar 14 '13 at 6:10
add comment

2 Answers 2

up vote 4 down vote accepted

A commutative finitely generated ring is Hopfian (proved by Malcev) i.e. every surjective endomorphism is an automorphism. That is not true for non-commutative finitely generated rings.

share|improve this answer
    
Mark, thanks for the answer. Do you have a counter-example for the non-commutative case ? –  tj_ Mar 14 '13 at 6:12
    
@Moderators: Why has this answer (along with my question) been downvoted ? I've seen the discussion on downvotes on meta, and perhaps it's possible to find out some details about the downvoter by help of this new sample. –  tj_ Mar 14 '13 at 6:18
    
@TJ: Take a non-Hopfian finitely generated group (en.wikipedia.org/wiki/Hopfian_group) $G$, then its group ring $\mathbb{Z}G$ is not Hopfian and finitely generated. –  Mark Sapir Mar 14 '13 at 7:56
    
@TJ: I am not a moderator, but people downvote if they think they do not like a question or an answer. It is not surprising that among 10000 people reading this site there are a few people who do not like any particular question. It is actually quite surprising that not every question/answer is downvoted multiple times. –  Mark Sapir Mar 14 '13 at 8:14
    
Mark, this is true for any commutative Noetherian ring and does not require residual finiteness. See for instance my answer to mathoverflow.net/questions/71185/… –  Benjamin Steinberg Mar 15 '13 at 1:03
show 3 more comments

For Artinian commutative rings $R$ all ring epimorphisms $R\to S$ are surjective.

In general, this is false if $R$ is non-commutative. For, Isbell has constructed an epimorphism $R \to S$ where $A$ is finite (hence Artinian) and $S$ infinite. I'll have a look if I can find Isbell's example later on.

share|improve this answer
2  
The example that Manny Reyes gave for a different purpose in the thread mathoverflow.net/questions/121468/… is also an example of this. –  Jeremy Rickard Mar 14 '13 at 11:44
1  
Jeremy, thanks, that's exactly the example (Isbell: Epimorphisms and Dominions IV, J. London Math. Soc. (1969) s2-1 (1): 265-273. Cf. page 268, after 2.5). –  Ralph Mar 14 '13 at 19:55
add comment

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.