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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?

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @QiaochuYuan , nope , i meant conditions on the element itself ! $\endgroup$ – Fawzy Hegab May 16 '13 at 10:28
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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.)

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ @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
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    $\begingroup$ 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

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