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 ${\displaystyle {\mathfrak {N))_{R}\supseteq \bigcap _((\mathfrak {p))\varsubsetneq R{\text{ prime))}{\mathfrak {p))}$.