Under what conditions does a ring $R$ have the property that every zero divisor is a nilpotent element ?

If we have a ring $R$, we know that every nilpotent element is either zero or a zero divisor.

If $A$ is the set of all nilpotents except the zero, and $B$ is the set of all nonzero zero divisors, then under what conditions on $R$ do we have $A = B$ ?

  • $\begingroup$ For example in $\,\Bbb Z_{p^n}\;,\;\;n>1\,$ , every non-zero zero divisor is of the form $\,mp^k \pmod{p^n}\;,\;\;k<n\;,\;\,1\le m\le p\,$ , so it is also a nilpotent element...and in this kind of rings it is an if and only if condition. $\endgroup$ – DonAntonio May 1 '13 at 13:11
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    $\begingroup$ A quick note: If this holds, then the characteristic of the ring is either a prime or $0$. $\endgroup$ – Tobias Kildetoft May 1 '13 at 13:13
  • $\begingroup$ @DonAntonio no, only for integral domains (which clearly satisfy the above conditions). $\endgroup$ – Tobias Kildetoft May 1 '13 at 13:16
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    $\begingroup$ I mean that if every zero-divisor is nilpotent, the the characteristic is either a prime or $0$. If $R$ is an integral domain, then clearly all the zero-divisors are nilpotent. $\endgroup$ – Tobias Kildetoft May 1 '13 at 13:19
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    $\begingroup$ Actually, my previous comment was not quite correct. The characteristic could also be a power of a prime, as demonstrated by the very first example given by @DonAntonio. $\endgroup$ – Tobias Kildetoft May 1 '13 at 13:27

Most generally speaking, a nontrivial idempotent is a zero divisor that is never nilpotent. From this we deduce that such a ring cannot have any nontrivial idempotents.

The exclusion of nontrivial idempotents is really strong. In particular, it implies that none of the nontrivial right or left ideals are summands. (Commutative domains are a case of this, of course, but in that case the zero divisors and nilpotent elements are all 0 :) )

Here is a crisp partial result. In the case when $R$ is right Artinian, all elements are either units or zero divisors. A right Artinian ring without nontrivial idempotents is local. Since the maximal ideal is nilpotent, it's clear that all the zero divisors are nilpotent. In summary, this says that among right Artinian rings, the local ones are exactly the ones with the "ZD implies nilpotent" property.

I've seen Jacoson call commutative rings without nontrivial idempotents connected rings, but I have to say that I don't have many exotic examples of them in mind. I'm guessing that being connected probably doesn't characterize the "ZD implies nilpotent" property.

There are a lot of examples of reduced rings which aren't domains which are non-examples for your property. For example, $F\times F$ for a field $F$ has no nilpotent elements but obviously has zero divisors. This ring is already pretty 'nice' (commutative Artinian!) and you can even make it finite by using a finite field.

D. Lazard pointed out to me in a conversation that $\{0\}$ is a primary ideal iff $Nil(R)$ is prime and there is no other associated prime. This is very cool too :)

  • $\begingroup$ i think that the answer of my question is not easy as i thought ! i really didn't study artinian rings or idempotents ! i just have started rings from some weeks ! just know the basics and this question arised to my head while i was proving a problem ! $\endgroup$ – Fawzy Hegab May 3 '13 at 19:17
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    $\begingroup$ @MathsLover Fun question :) $\endgroup$ – rschwieb May 3 '13 at 19:26
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    $\begingroup$ @MathsLover Also, I added a line about an additional observation I heard from an acquaintance. $\endgroup$ – rschwieb May 3 '13 at 19:32
  • $\begingroup$ thanx , i think that i will come back to this question when i study these concepts in the future ! $\endgroup$ – Fawzy Hegab May 3 '13 at 19:36

If $A$ is a commutative ring with this property, then the zero-divisors form an ideal of $A$ (since the nilradical is an ideal), in fact a prime ideal, and thus it's the unique minimal prime ideal of $A$. Conversely, if $A$ has a unique minimal prime ideal containing all the zero-divisors, then this has to be the nilradical, so every zero-divisor is nilpotent.

People seem to be calling these rings 'primary rings,' since this condition characterizes the rings you get after quotienting an arbitrary ring by a primary ideal. I don't know how widely used this term is.

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    $\begingroup$ The term is widely used among people working with non-commutative rings. $\endgroup$ – Mariano Suárez-Álvarez May 1 '13 at 14:54
  • $\begingroup$ Hi: sorry, I'm not very familiar with ideals consisting of zero divisors, and I only get eight hits when I google "prime ideal of zero divisors." Can you explain briefly what it is? $\endgroup$ – rschwieb May 1 '13 at 20:07
  • $\begingroup$ @MarianoSuárez-Alvarez Is it? I know Jacobson used it to mean rings for which R/rad(R) is simple Artinian. Are there other good versions to be aware of? $\endgroup$ – rschwieb May 1 '13 at 20:08
  • $\begingroup$ The same people that talk about primitive rings (which are rings obtained by quotienting something by a primitive ideal) talk about primary rings. I can't think of references right now but the book by McConnel and Robson should use the term, I guesa. $\endgroup$ – Mariano Suárez-Álvarez May 1 '13 at 20:12
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    $\begingroup$ @PaulVanKoughnett, the zero divisors rarely form a proper ideal. Consider the ring $\mathbb R\times\mathbb R$. $\endgroup$ – Mariano Suárez-Álvarez May 2 '13 at 0:29

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