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if $a,b$ are nilpotent elements of a commutative ring $R$, show $a+b$ is also nilpotent

So then $a^n=0, b^m = 0, n,m \in \mathbb{Z}^+$ I know this is solvable using the binomial theorem but I would much rather solve it another way if possible. The answer using the binomial theorem isn't making the most sense at the present moment.

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3 Answers 3

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You can use the binomial theorem without working out the binomial coefficients if it makes it easier for you. Take $(a+b)^{m+n} = \sum_{i=0}^{m+n}c_i a^{i}b^{m+n-i}$ where $c_i$ is what the coefficients should be. Note for each $i \le n$, $b^{m+n-i} =0$ and for $i \ge n$, $a^i = 0$. Therefore each element of this sum is $0$.

Note however that this is false in a general ring, i.e. the sum of nilpotent elements needs not be nilpotent in general. Take as your ring $M_{2\times 2}(F)$ with $a = \left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right]$ and $b = \left[\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right]$. Then $a^2 = b^2 = \left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right]$, but $(a+b)^2 = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]$ which is certainly not nilpotent.

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  • $\begingroup$ I guess what is throwing me off is. why is when $i \leq n, b^{m+n-i}=0$ and for $i \ge n, a^i =0$ $\endgroup$ Feb 18, 2015 at 2:15
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    $\begingroup$ @oliverjones When $i\le n$, $m+n-i \ge m$, right? So then $b^k = 0$ for $k\ge m$ as $b^m = 0.$ Then in other cases, i.e. when $i\ge n$, $a^i = 0$ as $a^n = 0$. Does that make it clearer? $\endgroup$
    – walkar
    Feb 18, 2015 at 2:16
  • $\begingroup$ yes thank you! much clearer! $\endgroup$ Feb 18, 2015 at 2:19
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I'm not sure how to avoid the binomial theorem, but here is a way to prove the proposition.

$$(a+b)^{n+m}=\sum_{k=0}^{n+m}{n+m\choose k} a^kb^{n+m-k}=\sum_{k=0}^{n-1}{m+n\choose k}a^kb^{m+n-k}.$$

The last equality follows because $a^k=0$ for $n\le k\le m+n$. So, $k<n$ for the terms that remain, which implies $m+n-k>m$. So, $b^{m+n-k}=0$ and these terms vanish also.

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$ (a\!+\!b)^{m+n}\! =\! \overbrace{\sum\, c_{ij}\,a^i b^j}^{\textstyle i\!+\!j \color{#0a0}{= m\!+\!n}}\!\Rightarrow\ \begin{align}\overbrace{i\ge\, n^{\phantom{I}}\!\!\!}^{\Large a^i\,=\ 0}\\[.1em] \color{#c00}{{\rm or}}\,\ \ \underbrace{j\ge m}_{\Large b^j\,=\ 0}\end{align}\,\,$ $\ (\color{#c00}{\rm else}\, $ $ \begin{align}i<m\\ j<n\,\end{align}$ $\Rightarrow$ $\, i\!+\!j \color{#0a0}{< m\!+\!n})$

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    $\begingroup$ I always present it in the above symmetric form, which I think makes the argument clearer. $\endgroup$ Feb 18, 2015 at 3:16

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