Commutative ring: sum of nilpotent elements

Question

I know that in a commutative ring, if $a$ and $b$ are nilpotent, then $a + b$ is nilpotent, and this can be proved using the binomial theorem. I cannot figure out, however, where the assumption of commutativity is used, because once this is lifted, the result is no longer necessarily true. Is commutativity required to invoke the binomial theorem? Is there another step where it's used?

Answer 1

The binomial theorem requires commutativity indeed (or, at least, that $a$ and $b$ commute). For instance, in a commutative ring you have$$(a+b)^2=a^2+2ab+b^2,$$but, in general, all that you can say is that$$(a+b)^2=a^2+ab+ba+b^2.$$Now, suppose that $a^2=b^2=0$. The fact that that $(a+b)^3=0$ (in a commutative ring) follows from the fact that$$(a+b)^3=a^3+3a^2b+3ab^2+b^3.$$But you would not be able to deduce that $(a+b)^3=0$ from$$(a+b)^3=a^3+a^2b+aba+ab^2+ba^2+bab+b^2a+b^3.$$