Nilradical of a ring

Definition: Let \( A \) be a commutative ring. We define the nilradical of \( A \) to be
\[ Nil(A) := \{ x \in A ~|~ x^n = 0 \text{ for some } n \geq 1 \} \]

Then we get a nice equivalence

\[
  Nil(A) = \bigcap_{\mathfrak{p} \in Spec(A)} \mathfrak{p}
\]

Let \( x \in A \) be nilpotent, i.e. \( x^n = 0 \) for some \( n \in \mathbb{N} \) then \( x^n = 0 \in \mathfrak{p} \) for any prime ideal \( \mathfrak{p} \). By induction, this shows that \( x \in \mathfrak{p} \).

Now suppose that \( x \notin A \), then \( x^n \neq 0 \) for any \( n \in \mathbb{N} \). In particular, the following collection of ideals is non-empty,
\[
   S = \{ \mathfrak{A} ~|~ \mathfrak{A} \text{ ideal of } A \text{ and } x^n \notin \mathfrak{A} \text{ for all } n \}
\]

because the zero ideal \( (0) \) is in \( S \). If we have a chain of ideals
\[
\mathfrak{A}_1 \subset \mathfrak{A}_2 \subset \cdots
\]
then we can take the union of all terms in the chain, call it, \( \mathfrak{B} \). Then, it turns out that \( \mathfrak{B} \) is an ideal. Hence we can use the Zorn's lemma to find the maximal element, say, \( \mathfrak{p} \). We claim that \( \mathfrak{p} \) is a prime ideal.

Suppose for contradiction that there exists \( a,b \notin \mathfrak{p} \) such that \( ab \in \mathfrak{p} \). We have a proper inclusion
\[
\mathfrak{p} \subset \mathfrak{p} + (a) \text{ and } \mathfrak{p} \subset \mathfrak{p} + (b)
\]
by maximality, there exists \( n, m \) such that \( x^n \in  \mathfrak{p} + (a) \) and  \( x^m \in  \mathfrak{p} + (b) \). As a consequence,
\[
x^{n+m} \in (\mathfrak{p} + (a)) ( \mathfrak{p} + (b)) \subset \mathfrak{p}
\]
which is a contradiction. This shows that \( \mathfrak{p} \) is a prime ideal. As \( \mathfrak{p} \) is an element in \( S \), it follows that \( x \notin \mathfrak{p} \), hence
\[
x \notin \bigcap_{\mathfrak{p} \in Spec(A)} \mathfrak{p}
\]
This completes the proof.