Equivalence of Definitions of Nilradical of Ring

Theorem
Let $A$ be a commutative ring.

Proof
By Nilpotent Element is Contained in Prime Ideals, $\Nil A$ is contained in the intersection of all prime ideals.

It remains to prove the other inclusion.

Let $f \in A$ be not nilpotent.

Let $S$ be the set of ideals of $A$ that are disjoint from $\set {f^n: n \in \N}$.

By Zorn's Lemma, $S$ has a maximal element $P$.

In particular, $f \notin P$.

We want to show that $P$ is prime.

Let $a, b \in A$ with $a, b \notin P$.

Then the sums of ideals $\ideal a + P$ and $\ideal b + P$ contain $P$ strictly.

By the maximality of $P$, there exist $n, m \in \N$ with $f^n \in \ideal a + P$ and $f^m \in \ideal b + P$.

Then:
 * $f^{m + n} \in \ideal {a b} + P$

Thus:
 * $\ideal {a b} + P \notin S$

In particular:
 * $a b \notin P$

Thus $P$ is prime, and so:
 * $\displaystyle f \notin \bigcap_{\mathfrak p \mathop \in \Spec A} \mathfrak p$