A doubt regarding the proof of "the set of nilpotent elements is an ideal".

Question

On pg.5 of "Commutative Algebra" by Atiyah-Macdonald, Proposition 1.7 states that

The set $A$ of all nilpotent elements in a commutative ring $R$ is an ideal.

Let $x,y\in A$. Clearly, for any $n\in \Bbb{N}$, $x^n\neq 0$ and $y^n\neq 0$. However, why can't $(xy)^n=x^ny^n=0$? Nowhere is it stated that $R$ is an integral domain. And if $xy\notin A$, then $A$ is not an ideal.

Where am I going wrong?

Thanks in advance!

Answer 1

If $x \in A$ then it means that $x^n=0$ for some $n$. You are trying to prove that the set of elements which are NOT nilpotent are an ideal, which of course is not true.

P.S. Here are two hints for the proof:

If $x,y \in A$ then $x^n=0$ and $y^m=0$ for some $m,n$. Then $(x+y)^{m+n}=0$.

Also, if $x \in A, r \in R$ then $x^n=0$ hence $(rx)^n=0$.