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Let $R$ be a commutative ring. Show that for any $a,b \in R$ nilpotent that $a+b$ is also nilpotent in $R$.

We know $a^n = 0$ for some n and $b^m = 0 $ for some m, so consider $(a+b)^{m+n} = (a+b)(a+b)...(a+b)$ $n+m$ times.

I don't see how $R$ being commutative helps me here.

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marked as duplicate by Najib Idrissi, Lord_Farin, Milo Brandt, PVAL, Jonas Meyer Feb 22 '15 at 21:07

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.

You seem to be confusing idempotent with nilpotent. Can you clarify which is correct? Idempotent is $a^2 = a$, nilpotent is $a^n = 0$ for some $n$. – RghtHndSd Feb 26 '14 at 3:13
ah yes, it is supposed to be nilpotent, sorry, i will edit.. – terrible at math Feb 26 '14 at 3:13
up vote 2 down vote accepted

Suppose $a^m = 0$ and $b^n = 0$. We now seek a positive integer $k$ such that $(a+b)^k = 0$. Ignoring coefficients, recall that the polynomial expansion of $(a+b)^k$ begins with $a^k b^0$, after which the power of $a$ decreases by $1$ for each successive term while the power of $b$ increases by $1$ for each successive term.

Our strategy, then, will be to choose $k$ so large that the terms before $b$ is raised to the $n$ are all $0$ because $a$ is raised to a high power, after which $b$ will ensure the remaining terms are all $0$.

Expanding: $$(a+b)^k = c_0 a^k + c_{1} a^{k-1} b + c_{2} a^{k-2} b^2 + \cdots + c_{n-1} a^{k-n+1} b^{n-1} + c_{n} a^{k - n} b^{n} + \cdots + c_{k} b^k$$

where the $c_i$ are coefficients found in, e.g., Pascal's Triangle. Then every term from the one starting with $c_{n}$ on will have $b^n = 0$ as a factor and hence be $0$. For the earlier terms, the smallest power of $a$ is $k-n+1$, so that it suffices to ensure $k-n+1 \geq m$, i.e., $k \geq m + n - 1$. Then every term is of the form $c_i a^{k-i} b^i = c_i \cdot 0 = 0$, so that we have a sum of all zeros, which yields $0$. Thus, $a+b$ is nilpotent as desired, where we obtain the sharpest general bound by setting $k = m + n - 1$. QED

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Woah, this is a bit over my head :) looking up the binomial formula, I see I can write what I first wanted to consider as $(a+b)^{n+m} = \sum\limits_{k=0}^{n+m} \binom{n+m}{k} a^{(n+m)-k}b^k$ trying to keep this as easy as possible for myself -- is there anything I can do with this alone? – terrible at math Feb 26 '14 at 3:29
@terribleatmath The binomial formula gives exactly the coefficients $c_i$ described in my response above (and also found in Pascal's triangle). Try computing $(a+b)^{k}$ for small values of $k$ and read through my answer slowly... – Benjamin Dickman Feb 26 '14 at 3:36
oh ok, for some reason I didn't realize that for all terms greater than that such k we also have the term is zero, that makes a lot more sense. – terrible at math Feb 26 '14 at 3:44

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