# The set of all nilpotent element is an ideal of R

Given that R is commutative ring with unity. I want show that set of all nilpotent element 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?

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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 3:03
@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. –  user21436 Apr 16 '12 at 4:09
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 –  user21436 Apr 16 '12 at 4:18
@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. –  user21436 Apr 16 '12 at 4:34
@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 4:35

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 $s\geq t$?

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how can i show $(a-b)^k=0$? Other properties are satisfied from hint bt one this one is left –  Siddhant Trivedi Apr 16 '12 at 4:33
@SiddhantTrivedi: Then don't accept the answer yet! –  Arturo Magidin Apr 16 '12 at 4:34
N=r+s be the largest power then i will get $(a-b)^k=0$ Am i right? –  Siddhant Trivedi Apr 16 '12 at 4:50
With $N=r+s$ you will get $(a-b)^N=0$, yes. In fact, $r+s-1$ will suffice, but $r+s$ will do. –  Arturo Magidin Apr 16 '12 at 4:54
thank u somuch to giving me a such a wonderful guidance and hint to solve my promblem....thanks a lot –  Siddhant Trivedi Apr 16 '12 at 4:59