Take the 2-minute tour ×
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.

Please help me to prove that if $R$ is an infinite ring, then $R$ has either an infinite number of zero divisors, or it has no zero divisors.

share|improve this question
4  
Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, some would consider your post rude because it is a command ("Prove"), not a request for help, so please consider rewriting it. –  Zev Chonoles Feb 27 '13 at 9:33
    
Is commutativity assumed? It would be good to add that tag, if so. I'm not sure how Zhen Lin's solution works in the noncommutative case. –  rschwieb Feb 27 '13 at 14:33
    
I don't know ring is commutative or not. There's no information about it in this problem. –  Jane Feb 27 '13 at 17:02
add comment

2 Answers

Hint. Let $a$ be a zero-divisor. Consider $\{ a b : b \in R \} \setminus \{ 0 \}$. This is a non-empty set of zero-divisors in $R$. If it is infinite, then you are done. Otherwise it is finite; what are the implications of this? (Use the pigeonhole principle.)

share|improve this answer
    
so, If I understood it correctly: by pigeonhole principle, there exists an infinite subset X of R and an element x from X such that ax=ay for all y from X. so –  Jane Feb 27 '13 at 10:59
    
Yes, that's correct. –  Zhen Lin Feb 27 '13 at 11:06
    
Therefore we have infinite set y -x and I do not know, what to do with it? –  Jane Feb 27 '13 at 11:15
    
oh, thank you, I get it –  Jane Feb 27 '13 at 11:20
add comment

Hi here is a proof by contradiction:

Assume that there are only finitely many zero divisors in $R$ and let them be $a_1, \ldots a_n$ and denote the set of these $A$. Fix some element $b \in R, b \ne a_i$ for all $i$. Convince yourself that there are infinitely many elements $c \in R, c \ne a_i$ for all $i$ such that $(b-c) \ne a_i$ for all $i$, let $C$ be the set of these. Given a $x \in R$ that is not a zero divisor we must have that $a_i \mapsto x a_i$ is a bijection of $A$ otherwise $x$ would be a zero divisor. But then since there is only finitely many bijections of $A$ since $A$ is finite, and there are infinitely many elements in $C$, we see that we can find an elements $y$ in $C$ that give the same bijection of $A$ as $b$ , but then $(b-y)a_i = b a_i - y a_i = 0$ hence $(b-y)$ is a zero divisor a contradiction.

share|improve this answer
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.