Suppose that $R$ is a commutative ring with identity $1$

Let $a\in R$ with $ab=0$ for some $b\ne0$.

Under what conditions $a$ must be also nilpotent?

  • $\begingroup$ I don't know of any unusual condition which allows zero divisor $\implies$ nilpotent. Naturally if $a$ is nilpotent, then $a^n=0$ means $a\cdot a^{n-1}=0$, so $a$ is a zero divisor. $\endgroup$ – Ian Coley May 15 '13 at 15:45
  • $\begingroup$ This context sounds too general to provide any more conditions than just the one where $a$ is in every prime ideal. $\endgroup$ – user714630 May 15 '13 at 15:49
  • $\begingroup$ math.stackexchange.com/questions/19132/… see this link , there is a comment in the second answer which deduce that an element is nilpotent from being zero divisor ! but i don't understand why this is true ! $\endgroup$ – Fawzy Hegab May 15 '13 at 15:50
  • 1
    $\begingroup$ @MathsLover I guess the argument in the answer you linked to contains quite a bit of handwaving, at leas tit seems far from immediate. Cf. rather the accepted answer there. $\endgroup$ – Hagen von Eitzen May 15 '13 at 16:08
  • 1
    $\begingroup$ @QiaochuYuan , nope , i meant conditions on the element itself ! $\endgroup$ – Fawzy Hegab May 16 '13 at 10:28

This is technically an answer, though I don't know if it's particularly useful...

Claim: Let $a$ and $b$ be elements of a commutative unital ring such $R$ that $ab = 0$. Then $a$ is nilpotent if and only if every minimal prime containing $b$ also contains $a$.

Proof: If $a$ is nilpotent, then it is contained in all (minimal) prime ideals of $R$. Conversely, suppose that every minimal prime ideal containing $b$ also contains $a$. It is enough to show that $a$ is a member of every minimal prime of $R$ (for then $a$ is a member of every prime of $R$, and is therefore in the nilradical of $R$). So let $P$ be a minimal prime of $R$. If $b \in P$ then by hypothesis $a \in P$ also. On the other hand, if $b \notin P$ then $ab = 0 \in P$ with $P$ prime implies that $a \in P$. QED

In particular, if $b \in R$ is a zero divisor that is not a member of any minimal prime, then every element of $R$ that annihilates $b$ is nilpotent. (For if $ab = 0$, then it's vacuously true that every minimal prime containing $b$ also contains $a$.)

For a specific example of this situation, take the ring $R = k[x,y]/(x^2,xy)$. This has unique minimal prime ideal $(x)$, with $y \notin (x)$. So every element of $R$ that annihilates $y$ (for instance, $x$) must be nilpotent. (In fact, the annihilator of $y$ is quite easily seen to be $(x)$. So it's not a terribly interesting example.)

(You may also wish to see the following question on MathOverflow for vaguely related information: https://mathoverflow.net/questions/20826.)

  • $\begingroup$ Dear Manny, This is a nice answer. Cheers, $\endgroup$ – Matt E May 16 '13 at 1:54
  • $\begingroup$ +1 and kudos for taking up the challenge of finding a condition on the element :) $\endgroup$ – rschwieb May 16 '13 at 10:52

I don't think there is any better answer for this question than the more specific ones given at your other question "Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? ". The most natural conditions are mentioned there, and they are generally conditions on ideals of the ring, not the specific element $a$.

To recap some of the sufficient conditions mentioned there that make zero divisors nilpotent:

  1. $R$ an Artinian local ring

  2. $\{0\}$ a primary ideal.

  • $\begingroup$ @GeorgesElencwajg Very strange: I wonder what I was thinking! Having a prime nilradical obviously only makes "half" the zero divisors nilpotent. I have been aware of that counterexample for quite some time. $\endgroup$ – rschwieb Feb 13 '17 at 12:23
  • 1
    $\begingroup$ Dear rschwieb: indeed I was quite surprised that of all people you would write that! But don't worry it happens to us all and I have often asked myself how I could ever have said or written much worse assertions than yours :-) [I have now removed my previous comment] $\endgroup$ – Georges Elencwajg Feb 13 '17 at 13:59
  • $\begingroup$ @GeorgesElencwajg Thank you for alerting me and giving me the chance to correct the problem. $\endgroup$ – rschwieb Feb 13 '17 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.