Nilpotent Elements of Commutative Ring form Ideal

Theorem
Let $A$ be a commutative ring.

The subset of nilpotent elements of $A$ is an ideal.

Proof
Let $N$ be the subset of nilpotent elements.

Since $0$ is nilpotent, $N$ is nonempty.

Let $x\in N$ and $a\in A$.

Let $n\in\N$ be such that $x^n=0$.

Because $A$ is commutative, $(ax)^n=0$.

Thus $ax\in N$.

Let $x,y\in N$, and let $x^n=0$ and $y^m=0$.

By the Binomial Theorem, $(x-y)^p=0$ for $p\geq n+m-1$.

Thus $x-y\in N$.

Thus $N$ is an ideal.

Also see

 * Definition:Nilradical