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The set of all nilpotent element is an ideal of R

Given a commutative ring $R$ and two nilpotent elements $r$, $s$ there exists an $n \in \mathbb{N}$ such that

$$ (r+s)^n = 0.$$

I want to prove that in order to show that the nilpotent elements of a commutative ring are closed under addition (to show that the nilpotent elements form an ideal within a commutative ring). Is that a good way to do it? I'm a little stuck.

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marked as duplicate by Dylan Moreland, Rudy the Reindeer, Qiaochu Yuan Apr 16 '12 at 20:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Yes, that will do it. Hint: binomial theorem. – Robert Israel Apr 16 '12 at 20:33
up vote 2 down vote accepted

By the binomial theorem, $$(r+s)^n = \sum_{k=0}^n \binom{n}{k} r^k s^{n-k}.$$ If $r^a = s^b = 0$ choose $n = a+b$. Then for each $k=0,1,\ldots,n$ either $k \geq a$ or $n-k \geq b$ so $r^k = 0$ or $s^{n-k} = 0$.

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