Proof

Let ${\displaystyle r\in {\mathfrak {N}}_{R}}$  and ${\displaystyle {\mathfrak {p}}}$  be a prime ideal, then ${\displaystyle r^{n}=0}$  for some ${\displaystyle n\in \mathbb {Z} _{>0}}$ . Thus

${\displaystyle r^{n}=r\cdot r^{n-1}=0\in {\mathfrak {p}}}$ ,

since ${\displaystyle {\mathfrak {p}}}$  is an ideal, which implies ${\displaystyle r\in {\mathfrak {p}}}$  or ${\displaystyle r^{n-1}\in {\mathfrak {p}}}$ . In the second case, suppose ${\displaystyle r^{m}\in {\mathfrak {p}}}$  for some ${\displaystyle m\leq n-1}$ , then ${\displaystyle r^{m}=r\cdot r^{m-1}\in {\mathfrak {p}}}$  thus ${\displaystyle r\in {\mathfrak {p}}}$  or ${\displaystyle r^{m-1}\in {\mathfrak {p}}}$  and, by induction on ${\displaystyle m\geq 1}$ , we conclude ${\displaystyle r^{m}\in {\mathfrak {p}},\,\forall m:0<m\leq n-1}$ , in particular ${\displaystyle r\in {\mathfrak {p}}}$ . Therefore ${\displaystyle r}$  is contained in any prime ideal and ${\displaystyle {\mathfrak {N}}_{R}\subseteq \bigcap _{{\mathfrak {p}}\varsubsetneq R{\text{ prime}}}{\mathfrak {p}}}$ .

Conversely, we suppose ${\displaystyle f\notin {\mathfrak {N}}_{R}}$  and consider the set

${\displaystyle \Sigma :=\lbrace J\subseteq R\mid J{\text{ is an ideal and }}f^{m}\notin J{\text{ for all }}m\in \mathbb {Z} _{>0}\rbrace }$ 

which is non-empty, indeed ${\displaystyle (0)\in \Sigma }$ . ${\displaystyle \Sigma }$  is partially ordered by ${\displaystyle \subseteq }$  and any chain ${\displaystyle J_{1}\subseteq J_{2}\subseteq \dots }$  has an upper bound given by ${\displaystyle J=\bigcup _{i\geq 1}J_{i}\in \Sigma }$ , indeed: ${\displaystyle J}$  is an ideal^{[Note 1]} and if ${\displaystyle f^{m}\in J}$  for some ${\displaystyle m}$  then ${\displaystyle f^{m}\in J_{l}}$  for some ${\displaystyle l}$ , which is impossible since ${\displaystyle J_{l}\in \Sigma }$ ; thus any chain in ${\displaystyle \Sigma \neq \emptyset }$  has an upper bound and we can apply Zorn's lemma: there exists a maximal element ${\displaystyle {\mathfrak {m}}\in \Sigma }$ . We need to prove that ${\displaystyle {\mathfrak {m}}}$  is a prime ideal: let ${\displaystyle g,h\notin {\mathfrak {m}},gh\in {\mathfrak {m}}}$ , then ${\displaystyle {\mathfrak {m}}\varsubsetneq {\mathfrak {m}}+(g),{\mathfrak {m}}+(h)\notin \Sigma }$  since ${\displaystyle {\mathfrak {m}}}$  is maximal in ${\displaystyle \Sigma }$ , which is to say, there exist ${\displaystyle r,s\in \mathbb {Z} _{>0}}$  such that ${\displaystyle f^{r}\in {\mathfrak {m}}+(g),}$  ${\displaystyle f^{s}\in {\mathfrak {m}}+(h)}$ , but then ${\displaystyle f^{r}f^{s}=f^{r+s}\in {\mathfrak {m}}+(gh)={\mathfrak {m}}\in \Sigma }$ , which is absurd. Therefore if ${\displaystyle f\notin {\mathfrak {N}}_{R}}$ , ${\displaystyle f}$  is not contained in some prime ideal or equivalently ${\displaystyle R\setminus {\mathfrak {N}}_{R}\subseteq \bigcup _{{\mathfrak {p}}\varsubsetneq R{\text{ prime}}}(R\setminus {\mathfrak {p}})}$  and finally ${\displaystyle {\mathfrak {N}}_{R}\supseteq \bigcap _{{\mathfrak {p}}\varsubsetneq R{\text{ prime}}}{\mathfrak {p}}}$ .