How to show that $\sqrt{\{0\}} = \bigcap_{\text{I is prime}}I$ in a commutative ring with unit Here's another question about ring theory. How to show that $\sqrt{\{0\}} = \bigcap_{\text{I is prime}}I$ in a commutative ring with a unit. 
Own attempts
For "$\subseteq$", let $x$ be an arbitrary element in $\sqrt{\{0\}}$, then there exists $n \in \mathbb{N}$ such that $x^n=0$. Now let $I\subset R$ be an arbitrary prime ideal. we have got to show that $x \in I$. If there are no zero devisors, it's obvious because in this case $\sqrt{\{0\}} = \{0\}$, but how is this done without this premise?
To establish "$\supseteq$", I observed that $\sqrt{\{0\}}$ is an ideal, because it's a subgroup concerning "$+$", and for any $x$, $x^n = 0$ implies that $(rx)^n=r^n\cdot 0 = 0$ by commutativity. The nilradical is also prime if there are no zero divisors, which I didn't justify. This would complete the inclusion.
Can you help me with the rest of the proof? 
 A: Hint: Clearly if $x^n=0$, then $x$ is in every prime ideal. Conversely, suppose that $x$ is not nilpotent. Note then that the localization $R_x$ has a maximal ideal, which by the correspondence theorem, corresponds to a prime ideal of $R$ not intersecting $S$, and thus not containing $x$. 
A: Show that if
$$I\le R\;\;\text{ is prime, then }\;\; x^n\in I\implies x\in I$$
For the other direction : suppose $\,x^n\ne 0\;\;\forall\,n\in\Bbb N\;$ , and define
$$X:=\left\{\;J\le R\;;\;\forall\,n\in\Bbb N\,,\,x^n\notin J\;\right\}.\;\text{Observe that}\;\;\{0\}\in K\implies K\neq\emptyset$$
and we can partial order $\,K\,$ by inclusion. Now, if $\,C\subset K\,$ is any totally ordered set (i.e. chain) , then
$$U_C:=\bigcup_{I\in C}I\in K\;,\;\;\text{since the union is an ideal not containing $\,x\,$ (why?)}$$
and clearly $\,U_C\,$ is an upper bound for any element in $\,C\,$ , so by Zorn's Lemma there exists a maximal element $\,M\in K\,$ . Well, now just show $\,M\,$ is a prime ideal in $\,R\,$ and, of course, $\,x\notin M\,$
