Nilradical is an ideal

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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 as past the xs.