Nilradical is an ideal

Statement

The set of nilpotent elements in a commutative unital ring is an ideal (this ideal is termed the nilradical).

Proof

Commutativity is crucial to the proof, as we shall see.

Abelian group structure

It is clear that $0$ is nilpotent, and that if $x$ is nilpotent, so is $-x$. We thus only need to show closure under addition. Suppose $x$ and $y$ are nilpotent with $x^{m}=y^{n}=0$. Then consider:

$(x+y)^{m+n}$

We can expand this by the binomial theorem. We get a sum of monomials. For each monomial, either the power of $x$ is at least $m$ or the power of $y$ is at least $n$. Thus, each of the monomials in the expansion is zero, and so the above expression simplifies to 0.

Commutativity is essentially to rewrite expressions like $xyxy$ as a power of $x$, times a power of $y$.

The ideal property

We need to show that if $x$ is nilpotent, and $a$ is any ring element, then $ax$ is nilpotent. Since there exists a $n$ such that $x^{n}=0$, we have $(ax)^{n}=a^{n}x^{n}=0$. Here, we again use commutativity to move the $a$s past the $x$s.