$\operatorname{rad}(I)=\bigcap_{I\subset P,~P\text{ prime}}P$ $R$ commutative ring with unity. $I$ $R$-ideal. Then $\operatorname{rad}(I)=\bigcap_{I\subset P,~P\text{ prime}}P$. That is, the radical of $I$ is the intersection of all prime ideals containing $I$. 
There is a proof of this in my textbook, but I do not understand a certain piece of it. Here is the entire proof:
Clearly, $\operatorname{rad}(I) \subset \bigcap P$. Conversely, if $f\notin \operatorname{rad}(I)$, then any ideal maximal among those containing $I$ and disjoint from $\{f^n\mid n\geq1\}$ is prime, so $f\notin \bigcap P$. 
I understand the first containment, but I do not know how to show that any ideal maximal among those containing $I$ and disjoint from $\{f^n\mid n\geq1\}$ is prime. 
Could someone please give me a push in the right direction?  

The following theorem, or some equivalent, probably appears in your book.

In a ring $R$, given an ideal $I$ and a multiplicative subset $S$ such that $S \cap I = \emptyset$, there exists a maximal ideal among those containing $I$ and disjoint from $S$; any such ideal is prime.

To prove this, note that such ideals are in one-to-one order-preserving correspondence with proper ideals of $S^{-1}R$ containing $S^{-1}I$. Thus it is enough to take the inverse image of a maximal ideal containing $S^{-1}I$ under the canonical homomorphism $R \to S^{-1}R$. The inverse image of a maximal ideal is always prime.
In the present situation, apply the theorem to $S = \{1 \} \cup \{f^n|n \geq 1\}$.
