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 $$(ax)^n = a^nx^n = 0$$. Here, we again use commutativity to move the $$a$$s past the $$x$$s.