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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?

user26857
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Siddhant Trivedi
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    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. – Aaron Apr 16 '12 at 03:03
  • @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. –  Apr 16 '12 at 04:09
  • 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 –  Apr 16 '12 at 04:18
  • @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. –  Apr 16 '12 at 04:34
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    @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, – Matt E Apr 16 '12 at 04:35
  • @Jyrki Please ping me if I make a mistake. I learn from them. The last cyclicc groups incident was rather unfortunate and not intended to offend you. Once again, Apologies on that. –  Apr 16 '12 at 04:36
  • @MattE Thank you for enlightening me. Looking forward to such helps on this site, when you drop in. Thank you. –  Apr 16 '12 at 04:40
  • @KannappanSampath: The fourth word in the question (which at this point in time was not edited) is "commutative". So the quaternions are not a counter-example on two counts. BTW, if non-commutative rings had been intended, the word "ideal" would have been ambiguous (left/right/two-sided?). – Marc van Leeuwen Apr 16 '12 at 09:25
  • @MarcvanLeeuwen I brought the idea of dropping the commutativity hypothesis suggesting that Quaterninons gave a counter example. But, I was wrong. Of course, we have that the result holds for commutative rings, no doubt. And, yes, by an Ideal, I have always meant two-sided ones. –  Apr 16 '12 at 11:00

1 Answers1

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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 $a^n=0$.

Hints.

  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 $s\geq t$?

Arturo Magidin
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