$\operatorname{rad}(I)=\bigcap_{I\subset P,~P\text{ prime}}P$ $R$ 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?  
 A: In general, if $S$ is a subset of $R$, closed under multiplication, then an ideal which is maximal among those disjoint from $S$ is a prime ideal.
It is easiest to show this if you know about localizations, because this amounts to saying that a maximal ideal of $S^{-1} R$ pulls back to a prime ideal of $R$.  But we can also do the heavy lifting ourselves:
Suppose that $I$ is maximal among ideals disjoint from $S$, and take two elements $x,y\notin I$.  Then the ideal $(I,x)$, which is strictly larger than $I$, must contain an element of $S$.  Similarly, $(I,y)$ also contains an element of $S$.  Use this to show that $(I,xy)$ contains an element of $S$, so $xy\notin I$.
A: 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 $S = \{1 \} \cup \{f^n|n \geq 1\}$.
