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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?

egreg
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Fawzy Hegab
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  • 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. – Ian Coley May 15 '13 at 15:45
  • This context sounds too general to provide any more conditions than just the one where $a$ is in every prime ideal. – Karl Kroningfeld May 15 '13 at 15:49
  • http://math.stackexchange.com/questions/19132/characterizing-units-in-polynomial-rings-rx 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 ! – Fawzy Hegab May 15 '13 at 15:50
  • One can answer a different question: in what rings *every* zero divisor is also nilpotent? The present question, I'm afraid, has only the obvious answer "$a$ is nilpotent when it is nilpotent". – egreg May 15 '13 at 15:52
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    @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. – Hagen von Eitzen May 15 '13 at 16:08
  • @egreg The user already asked "When are all zero divisors nilpotent?" and it was was well-answered here: http://math.stackexchange.com/questions/378206/under-what-conditions-does-a-ring-r-have-the-property-that-every-zero-divisor-is – rschwieb May 15 '13 at 16:24
  • Conditions on what? $R$? – Qiaochu Yuan May 15 '13 at 21:03
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    @QiaochuYuan , nope , i meant conditions on the element itself ! – Fawzy Hegab May 16 '13 at 10:28
  • @HagenvonEitzen , didn't understand your point ! – Fawzy Hegab May 16 '13 at 10:28

2 Answers2

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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.)

Manny Reyes
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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.

rschwieb
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  • @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](http://math.stackexchange.com/a/728535/29335). – rschwieb Feb 13 '17 at 12:23
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    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] – Georges Elencwajg Feb 13 '17 at 13:59
  • @GeorgesElencwajg Thank you for alerting me and giving me the chance to correct the problem. – rschwieb Feb 13 '17 at 14:46