# Nilpotent Elements of Commutative Ring form Ideal

## Theorem

Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$ and whose unity is $1_R$.

The subset of nilpotent elements of $R$ form an ideal of $R$.

## Proof

Let $N$ be the subset of nilpotent elements.

Because $0_R$ is nilpotent, $0_R \in N$ and so $N$ is non-empty.

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

We have:

 $\, \displaystyle \exists n \in \Z_{>0}: \,$ $\displaystyle x^n$ $=$ $\displaystyle 0_R$ Definition of Nilpotent Ring Element $\displaystyle \leadsto \ \$ $\displaystyle a^n \circ x^n$ $=$ $\displaystyle 0_R$ Definition of Ring Zero $\displaystyle \leadsto \ \$ $\displaystyle \paren {a \circ x}^n$ $=$ $\displaystyle 0_R$ Power of Product of Commutative Elements in Semigroup $\displaystyle \leadsto \ \$ $\displaystyle a x$ $\in$ $\displaystyle N$ Definition of $N$

Let $x, y \in N$.

Let $x^n = 0$ and $y^m = 0$.

By the Binomial Theorem, $\paren {x - y}^p = 0$ for $p \ge n + m - 1$.

Thus $x - y \in N$.

Thus from Test for Ideal, $N$ is an ideal.

$\blacksquare$