Firstly, the nilradical is contained in every prime ideal. Indeed, if ${\displaystyle r\in {\mathfrak {N}}_{R},}$ one has ${\displaystyle r^{n}=0}$ for some positive integer ${\displaystyle n.}$ Since every ideal contains 0 and every prime ideal that contains a product, here ${\displaystyle r^{n}=0,}$ contains one of its factors, one deduces that every prime ideal contains ${\displaystyle r.}$

Conversely, let ${\displaystyle f\notin {\mathfrak {N}}_{R};}$ we have to prove that there is a prime ideal that does not contains ${\displaystyle f.}$ Consider the set ${\displaystyle \Sigma }$ of all ideals that do not contain any power of ${\displaystyle f.}$ One has ${\displaystyle (0)\in \Sigma ,}$ by definition of the nilradical. For every chain ${\displaystyle J_{1}\subseteq J_{2}\subseteq \dots }$ of ideals in ${\displaystyle \Sigma ,}$ the union ${\textstyle J=\bigcup _{i\geq 1}J_{i}}$ is an ideal that belongs to ${\displaystyle \Sigma ,}$ since otherwise it would contain a power of ${\displaystyle f,}$ that must belong to some ${\displaystyle J_{i},}$ contradicting the definition of ${\displaystyle J_{i}.}$

So, ${\displaystyle \Sigma }$ is a partially ordered by set inclusion such that every chain has a least upper bound. Thus, Zorn's lemma applies, and there exists a maximal element ${\displaystyle {\mathfrak {m}}\in \Sigma }$. We have to prove that ${\displaystyle {\mathfrak {m}}}$ is a prime ideal. If it were not prime there would be two elements ${\displaystyle g\in R}$ and ${\displaystyle h\in R}$ such that ${\displaystyle g\notin {\mathfrak {m}},}$ ${\displaystyle h\notin {\mathfrak {m}},}$ and ${\displaystyle gh\in {\mathfrak {m}}}$. By maximality of ${\displaystyle {\mathfrak {m}},}$ one has ${\displaystyle {\mathfrak {m}}+(g)\notin \Sigma }$ and ${\displaystyle {\mathfrak {m}}+(h)\notin \Sigma .}$ So there exist positive integers ${\displaystyle r}$ and ${\displaystyle s}$ such that ${\displaystyle f^{r}\in {\mathfrak {m}}+(g)}$ and ${\displaystyle f^{s}\in {\mathfrak {m}}+(h).}$ It follows that ${\displaystyle f^{r}f^{s}=f^{r+s}\in {\mathfrak {m}}+(gh)={\mathfrak {m}},}$ contadicting the fact that ${\displaystyle {\mathfrak {m}}}$ is in ${\displaystyle \Sigma }$. This finishes the proof, since we have proved the existence of a prime ideal that does not contain ${\displaystyle f.}$