Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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 at 3:13
    
ah yes, it is supposed to be nilpotent, sorry, i will edit.. –  terrible at math Feb 26 at 3:13
add comment

1 Answer 1

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

share|improve this answer
    
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 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 at 3:36
1  
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 at 3:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.