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.


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.

  • $\begingroup$ I guess what is throwing me off is. why is when $i \leq n, b^{m+n-i}=0$ and for $i \ge n, a^i =0$ $\endgroup$ – oliverjones Feb 18 '15 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 '15 at 2:16
  • $\begingroup$ yes thank you! much clearer! $\endgroup$ – oliverjones Feb 18 '15 at 2:19

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.


$ (a\!+\!b)^{m+n}\! = \sum c_{ij}a^i b^j$ with $\ i\!+\!j\, =\, m\!+\!n\,\Rightarrow\ \begin{align}\overbrace{i\ge\, n^{\phantom{I}}\!\!\!}^{\Large a^i\,=\ 0}\\ {\rm or}\,\ \ \underbrace{j\ge m}_{\Large b^j\,=\ 0}\end{align}\,\ $ (else $\ \begin{align}i<m\\ j<n\,\end{align}$ $\,\Rightarrow\,$ $\, i\!+\!j < m\!+\!n\,\Rightarrow\!\Leftarrow)$

  • $\begingroup$ I always present it in the above symmetric form, which I think makes the argument clearer. $\endgroup$ – Bill Dubuque Feb 18 '15 at 3:16

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