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Today I was trying to find an infinite ring $R$ whose all nonzero and nonidentity elements were zero divisors and actually found one: $\mathcal R =\text{Fun}(\mathbb N, \mathbb Z/2\mathbb Z)$. Given a function $f \in \mathcal R$ we can define another function $g \in \mathcal R$ by $g(x) = 1 - f(x)$. As everyone can easily check, $$(fg)(x) = f(x) - f(x)^2 = 0.$$

One can of course replace $\mathbb N$ with any infinite set. Now I'm curious if there are more examples of such strange behaving ring, that aren't similar to $\mathcal R$. Is there a notion of such rings in abstract algebra?

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  • $\begingroup$ I'm not sure about having every nonzero and nonidentity element be a zero divisor, but if we take $R$ and $S$ as rings, $R\times S$ has the property that $\forall r \in R$ and $\forall s \in S$, $(r,0)$ and $(0,s)$ are zero divisors, namely $(r,0)\cdot (0,s) = (0,0)$. $\endgroup$ – walkar Jan 28 '15 at 15:30
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    $\begingroup$ Actually your ring is a direct product of countable many copies of $\mathbb Z/2\mathbb Z$. This is an example of Boolean ring. Note that all Boolean rings satisfies your requirement. Maybe this can be helpful. $\endgroup$ – user26857 Jan 28 '15 at 21:41
  • $\begingroup$ One more thing: among the finite commutative rings only $\mathbb Z/2\mathbb Z$ satisfies the requirement. $\endgroup$ – user26857 Jan 28 '15 at 22:01
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A very obvious example is any Boolean ring. Since $x^2=x$ for every element in the ring, $x(1-x)=0$. Thus if $x$ is neither $0$ nor $1$, $x$ must be a zero divisor.

These rings are very easy to come by. You can take any subring of a direct product of any number of copies of the field of two elements.

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  • $\begingroup$ After some debate on M.SE maybe one of its users changed the Wiki page in the sense that zero is acknowledged as being a zero divisor: en.wikipedia.org/wiki/Zero_divisor#Zero_as_a_zero_divisor (no need to mention that this assumption is essential in commutative algebra). $\endgroup$ – user26857 Jan 29 '15 at 7:29
  • $\begingroup$ @user26857 I prefer that convention too, but it does not seem to make any difference to the statement of the question or answer here. $\endgroup$ – rschwieb Jan 29 '15 at 10:55
  • $\begingroup$ What I mean is that nothing here says 0 isn't a zero divisor. $\endgroup$ – rschwieb Jan 29 '15 at 11:02
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In [Malik, Mordeson, Sen, "Fundamentals of Abstract Algebra"] and some other books, they give following definition:

Let $(R,+)$ be a commutative group. Define multiplication on $R$ by $ab=0$ for all $a,b \in R$, where $0$ denotes the identity element of the group $(R,+)$. Then $(R,+,.)$ is a ring called the "zero ring". If $R$ contains more than one element, then $R$ is a commutative ring without $1$ and every nonzero element of $R$ is a zero divisor.

In this sense, $(\mathbb Z(p^{\infty}),+,.) $ seems to satisfy this property by definition.

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  • $\begingroup$ This is almost a perfect answer, but unfortunately I don't want to have $1$ as one of the zero divisors. $\endgroup$ – user207868 Jan 28 '15 at 21:19
  • $\begingroup$ Actually, you will probably not have 1 as one of the zero divisors, unless you get the trivial zero ring {0}, for a better example to illustrate, probably I have seen the ring of {[[n,-n],[n,-n]]} type 2x2 matrices over Z as a subring, or something like that..which also has nonidentity elements and all zero divisors $\endgroup$ – S.B. Jan 28 '15 at 21:38

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