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

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.

• I guess what is throwing me off is. why is when $i \leq n, b^{m+n-i}=0$ and for $i \ge n, a^i =0$ Feb 18, 2015 at 2:15
• @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? Feb 18, 2015 at 2:16
• yes thank you! much clearer! Feb 18, 2015 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}\! =\! \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 $$\Rightarrow$$ $$\, i\!+\!j \color{#0a0}{< m\!+\!n})$$

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