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 \( \math