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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?

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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
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 $a\in R$.

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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.

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

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

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