Given that R is commutative ring with unity, I want show that set of all nilpotent elements is an ideal of R.

I know how to show ideal if set is given but here set is not given to me. Can anyone help me?

  • 1
    $\begingroup$ The set is given. It is $\{x\in R \mid x^n=0 \text{ for some } n\}$. What you need to show is that if $x^n=0$ and $y^m=0$, then $(x+y)^k=0$ for some $k$. The other properties of that set being an ideal should be straight forward. $\endgroup$
    – Aaron
    Apr 16, 2012 at 3:03
  • $\begingroup$ @JyrkiLahtonen I thought quaternions gave a counter example but, well I could be wrong. Something is wrong with me: I keep saying it all wrongly. $\endgroup$
    – user21436
    Apr 16, 2012 at 4:09
  • $\begingroup$ And, I agree with ring of matrices...I do know of examples there...And, it is easy to get an example there without knowing all the nilpotent elements... @JyrkiLahtonen $\endgroup$
    – user21436
    Apr 16, 2012 at 4:18
  • $\begingroup$ @JyrkiLahtonen I agree we [I and the user Matt] concluded this wrongly--We have a CA group here where we discussed and solved exercises--Sadly, this error has crept in. Thanks for helping me realize. Now, I realize we are wrong. Thank you once again. $\endgroup$
    – user21436
    Apr 16, 2012 at 4:34
  • 4
    $\begingroup$ @Kannappan Sampath: Dear Kannappan, One of the key properties of the quaternions is that they are a division algebra. In particular, as Jyrki notes, they contain no non-zero nilpotent elements. Thus the set of nilpotents in the quaternions is an ideal, the zero ideal. Regards, $\endgroup$
    – Matt E
    Apr 16, 2012 at 4:35

1 Answer 1


The set is given: it is $$\{a \in R\mid a\text{ is nilpotent}\}.$$ Remember that $a\in R$ is nilpotent if and only if there exists $n\gt 0$ such that $a^n=0$.


  1. Is $0$ in the set?

  2. If $a$ and $b$ are in the set, can you guarantee that a large enough power of $(a-b)$ is equal to $0$? Think binomial expansion.

  3. If $a$ is in the set and $r\in R$, is $ra$ in the set? Here, too, commutativity will be key.

Added. For $2$: if $a^k=0$, then $a^r=0$ for all $r\geq k$; if $b^t=0$, then $b^s=0$ for all $s\geq t$. Can you find an $N$ such that each term in the binomial expansion of $(a-b)^N$ will have either $a^r$ with $r\geq k$ or $b^s$ with $s\geq t$?

  • $\begingroup$ how can i show $(a-b)^k=0$? Other properties are satisfied from hint bt one this one is left $\endgroup$ Apr 16, 2012 at 4:33
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    $\begingroup$ N=r+s be the largest power then i will get $(a-b)^k=0$ Am i right? $\endgroup$ Apr 16, 2012 at 4:50
  • 2
    $\begingroup$ With $N=r+s$ you will get $(a-b)^N=0$, yes. In fact, $r+s-1$ will suffice, but $r+s$ will do. $\endgroup$ Apr 16, 2012 at 4:54
  • $\begingroup$ thank u somuch to giving me a such a wonderful guidance and hint to solve my promblem....thanks a lot $\endgroup$ Apr 16, 2012 at 4:59

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