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Let $f=\sum_{n=0}^{\infty}a_nx^n$. If $f$ and $a_0$ is nilpotent how I can prove that $f-a_0$ is nilpotent? Or if $f^n=0$ and $a_0^n=0$ how can I prove that $(f-a_0)^n=0$, where $n\in \mathbb N$.

user26857
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someone
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    It shouldn't be true in general that if $f^n = 0$ and $a_0^n = 0$ then $(f - a_0)^n = 0$. What will be true is that $(f - a_0)^{2n} = 0$. – Qiaochu Yuan Mar 13 '13 at 05:15

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In general, if $\alpha^n = 0$ and $\beta^m = 0$ then $(\alpha + \beta)^{n + m - 1} = 0$. Prove this by looking at the binomial expansion, in each term either the exponent of $\alpha$ is $\geq n$ or the exponent of $\beta$ is $\geq m$.

Then for your question note that if $a_0$ is nilpotent then also $-a_0$ is nilpotent. Take $\alpha = f$ and $\beta = -a_0$.

Jim
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    +1: you are one of the very rare mathematicians who rise $\alpha +\beta$ to the power $n+m-1$ instead of the most common $n+m$. Of course it doesn't really matter, but your exponent has a nice minimalist appeal to it :-) – Georges Elencwajg Mar 13 '13 at 11:46