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Let $R$ denote a commutative ring and $Nil(R)$ is the ideal consisting of all nilpotent elements in $R$. I am attempting to prove that $Nil(R)$ is closed under addition.

My work so far is summarized in the following. I know that $a,b\in{Nil(R)}$ implies there exists integers $k_{1},k_{2}$ such that $a^{k_{1}}=b^{k_{2}}=0_{R}$ where $0_{R}$ is the additive identity of $R$. To assert that $a+b\in{Nil(R)}$ is to assert the existence of an integer $k_{3}$ satisfying $(a+b)^{k_{3}}=0_{R}$, for explicit rings such as the real or complex numbers this is very straight forward and one can merely use exponent properties but I am admittedly uncertain how to go about showing the Nilradical is additively closed for an arbitrary Ring where the elements may not be numbers but objects. Any hints or helpful observations showing how to prove this would be much appreciated.

user26857
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user328158
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1 Answers1

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Observe that you can use Newton's Binomial Theorem because of commutativity, so take

$$(a+b)^{k_1+k_2}=\sum_{i=0}^{k_1+k_2}\binom{k_1+k_2}i a^ib^{k_1+k_2-i}$$

Now check what happens with each summand above (by the way, can you see how is it possible to take an even lower exponent in the above?)

DonAntonio
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