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If $R$ is a nonzero ring with identity then I have seen the group of units denoted by $R^{\times}$ or possibly $R^*$ in some texts. In a classical ring there is a trichotomy which declares each element in $R$ is either zero, a unit, or a zero-divisor. Naturally, for a classical ring, the set of zero-divisors and $\{0\}$ is the complement of $R$ by the group of units; $\text{set of zero divisors in $R$} \cup \{0 \} = R - R^{ \times}$.

My question is simply this:

Is there a common notation for the set of zero-divisors in a ring ?

Thanks in advance for your insight.

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  • $\begingroup$ What are some nontrivial examples of classical rings? It seems weird that $\mathbb Z$ and $k[x]$ are not classical. $\endgroup$ – Hoot Jul 8 '16 at 18:32
  • $\begingroup$ @Hoot one sided artinian rings (or even rings satisfying DCC on chains of right ideals generated by the powers of an element), Von Neumann regular rings. $\endgroup$ – rschwieb Jul 8 '16 at 21:25
  • $\begingroup$ For example, any linear associative algebra forms a classical ring. $\endgroup$ – James S. Cook Jul 9 '16 at 13:11
  • $\begingroup$ (my algebra is over the real numbers in the comment above) $\endgroup$ – James S. Cook Jul 9 '16 at 14:59
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The one I see the most is like $Z(R)$, sometimes writing $Z^*(R)$ for the ones that aren't zero.

This collides with some other usages though, notably the center of the ring and the singular ideal of a ring.

I think I've seen it as $ZD(R)$ too, somewhere.

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    $\begingroup$ Atiyah & MacDonald use $D$ for the set of zero divisors. This is also not the best notation since it could be confused with divisors. $\endgroup$ – Hamed Jul 8 '16 at 17:35
  • $\begingroup$ Thanks, this is helpful. I was pondering the use of $Z$, or $Z(R)$, but I also dislike it for the same reasons. At the moment, I think I'm going to use $\mathbf{zd}(R)$. $\endgroup$ – James S. Cook Jul 9 '16 at 13:09
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There is a common notation for the set of zero divisors as the union of (radicals of) annihilators, i.e., $$"Ann_D(R)"=\text{the set of zero divisors }=\bigcup_{0\ne x\in R} \sqrt{\text{Ann}(x)}=\bigcup_{0\ne x\in R}\text{Ann}(x).$$

Edit: Still there remains the question for a short notation of the set of zero divisors.

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    $\begingroup$ Maybe a dumb question: why do you need to take radicals? $\endgroup$ – Hoot Jul 8 '16 at 18:27
  • $\begingroup$ It is not a dump question - see here. $\endgroup$ – Dietrich Burde Jul 8 '16 at 18:41
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    $\begingroup$ I have to admit that I've never seen this "notation". $\endgroup$ – user26857 Jul 8 '16 at 19:29
  • $\begingroup$ @user26857 It is related to annihilators, which is not an uncommon notation. Or did you mean the first notation ? There you are right. $\endgroup$ – Dietrich Burde Jul 8 '16 at 19:30
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    $\begingroup$ Yes, I mean $Ann_D(R)$. Writing the set of zero-divisors like the union of annihilators is not a notation, I suppose. $\endgroup$ – user26857 Jul 8 '16 at 19:32

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