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Show that if $a$ and $b$ are nilpotent elements of a commutative ring then $a+b$ is also nilpotent.

Let $R$ be a commutative ring, since $a$ and $b$ are nilpotent we know that $a^n = 0$ for some $n \in \mathbb{Z}^+$. Same argument for $b$ with some $m \in \mathbb{Z}^+$.

I wanted to first consider the linear combination of them, $a^n + b^m = 0$ but i'm not sure if this could get me anywhere. I can't really connect multiplication and addition in this way i have it here. And I can't make anything from the statement $a^n = -b^m$ because we aren't given that R is a field. (not sure if i could do anything with that anyways)

furashu
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1 Answers1

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Hint: Try to see what happens when you expand $(a+b)^{n+m}$. Remember that since the ring is commutative- you can use the binomial formula. Show that either the exponent on $a$ is greater than or equal to $n$ (and hence that term is $0$), or the exponent on $b$ is "large", and hence the term is $0$.

voldemort
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  • thank you, i will try this now. is it proper to use "intuition" like this in a proof, i.e., suddenly say "let's consider $(a+b)^{n+m}$ and then end up proving the result from it rather than deriving where you pulled that term from? – furashu Jan 30 '14 at 05:13
  • I think it's standard in a proof to say let's consider this. Usually this intuition comes after a bit of trial and error, or from things you have seen before. Generally you do not need to tell how you got to your idea, as most of the time ideas come in a "non-linear" fashion anyway. – voldemort Jan 30 '14 at 05:16