## The set of nilpotent elements in a commutative ring is an ideal

Let $R$ be a commutative ring. Recall that an element $x \in R$ is called nilpotent if $x^n = 0$ for some nonzero natural number $n$. Prove that the set $\mathfrak{N}(R)$ of all nilpotent elements in $R$ form an ideal. $\mathfrak{N}(R)$ is called the nilradical of $R$. [Hint: Use the Binomial Theorem to prove closure under addition.]

Let $x,y \in \mathfrak{N}(R)$. Then for some nonnegative natural numbers $n$ and $m$, we have $x^n = y^m = 0$.

Consider $(x+y)^{n+m}$. By the Binomial Theorem, we have $(x+y)^{n+m} = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k y^{n+m-k}$. Note that if $k \geq n$, $x^k = 0$, and if $k < n$, then $y^{n+m-k} = 0$. Thus $(x+y)^{n+m} = 0$, and we have $x+y \in \mathfrak{N}(R)$. Moreover, we have $(-x)^n = (-1)^nx^n = 0$, so that $-x \in \mathfrak{N}(R)$. Since $0 = 0^1 \in \mathfrak{N}(R)$, $\mathfrak{N}(R)$ is an additive subgroup of $R$.

Finally, since $R$ is commutative, if $r \in R$ then $(rx)^n = r^nx^n = 0$ and likewise $(xr)^n = 0$. Thus $\mathfrak{N}(R)$ absorbs $R$ on the left and the right, and hence is an ideal.