TITLE: if $a,b$ are nilpotent elements of a commutative ring $R$, show that $a+b$ is also nilpotent
QUESTION [11 upvotes]: 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.

REPLY [3 votes]: $ (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}$ $\Rightarrow$ $\, i\!+\!j \color{#0a0}{< m\!+\!n})$