Nilpotent elements in commutative rings

Question

Let $A$ be a commutative ring, $a, a+b \in A$ are nilpotent. Does this imply that $b$ is nilpotent?

Answer 1

Yes b is also a nilpotent element in R. If n is the nilpotence degree of a and m is the nilpotence degree of a+b, considering b=(a+b-a)^nm and using binomial theorem, you can easily obtain b is nilpotent. Here with no loss of generalization, take m>n or m>= n+1 and after expanding the above expression and playing m and n, the result will follow.

Answer 2

Yes. The set of all nilpotent elements in $A$, denoted by $\operatorname{Nil}(A)$, is an ideal, in particular it's an additive group. Therefore $(a + b) - a = b \in \operatorname{Nil}(A)$.

The important part you need to show that $\operatorname{Nil}(A)$ is an ideal is the following: say $a, b \in \operatorname{Nil}(A)$, then there exist $m, n \in \mathbb N$ such that $a^m = 0$ and $a^n = 0$. Now observe that, by the Binomial Theorem, $$(a + b)^{m + n} = \sum_{k = 0}^{m + n} \binom{m + n}{k} a^{m + n - k}b^k = 0.$$ Why is that? Well, if $k > n$, then $b^k = 0$ and if $k \leq n$, then $n - k \geq 0$ which means $m + n - k \geq m$ which means $a^{m + n - k} = 0$.