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.

Is there a ring $~R~$ with non-trivial multiplication (i.e. $~\exists a,b\in R ~~~ ab\neq 0$) such that each non-zero element of $~R~$ is a zero-divisor?

share|improve this question
3  
Presumably, then, you allow rings without identity, since the identity cannot be a zero divisor. –  Thomas Andrews May 20 '13 at 14:00
    
@ThomasAndrews I use the following definition of a ring: (R,+) is an abelian group; multiplication is associative and distributive. Multiplication identity isn't required. –  Igor May 20 '13 at 14:03

3 Answers 3

up vote 5 down vote accepted

The simplest example might be $R=2\mathbb Z/8\mathbb Z$. Then $2\cdot 2\neq 0$ but $a\cdot 4=0$ for all $a\in R$.

share|improve this answer

To get a general class of examples, you can take the nilradical of any commutative ring (ie, the set of nilpotent elements). This forms an ideal consisting solely of nilpotent elements (and hence of zero-divisors), but will usually have non-trivial multiplication.

share|improve this answer

A (trivialish) example is $\{0, 1\}$ with $1 \cdot 1 = 0$.

share|improve this answer
1  
This doesn't satisfy the requirement of having non-trivial multiplication. –  Zev Chonoles May 21 '13 at 0:09

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.