Proposition 2

— Let A be a commutative ring. The set of nilpotent elements of A is an ideal of A equal to the intersection of the set of all prime ideals of A.

Let x be a nilpotent element of A and p a prime ideal of A. The residue class of $x \mod p$ is a nilpotent element of the integral domain $A/p$, hence is zero; thus we have $x \in p$.

Let x be a non-nilpotent element of A. By Prop. 1 the principal ideal $(1 - xT)$ of $A[T]$ is distinct from $A[T]$. By Krull's theorem (I, p. 104) there exists a maximal ideal m of $A[T]$ containing $1 - xT$. Then m is a prime ideal of $A[T]$, hence $p = A \cap m$ is a prime ideal of A. We have $1 \notin m$ and $1 - Tx \in m$, hence $Tx \notin m$ and a fortiori $x \notin p$.

We have thus shown that the set $\mathfrak{n}$ of nilpotent elements of A is the intersection of the set of all prime ideals of A; since every intersection of ideals is an ideal, $\mathfrak{n}$ is an ideal.