Here is what I want to prove:
Let $p$ be a prime integer, and consider the ring $R=\mathbb{Z}$. Let $S=\mathbb{Z}\setminus p\mathbb{Z}$. The ring $S^{-1}R$ is called the localization of $\mathbb{Z}$ at $p$. Find all of the ideals of this ring, and describe the maximal ideal (there is only one).
For my proof:
Specifically, we know that $S^{-1}R=\left\{\frac{a}{b}\text{ }|\text{ }a\in R,b\in S\right\}$.
I have a theorem stating the following: If $P$ is a prime ideal in $R$ that does not intersect $S$, and $S$ has no zero divisors, then $\Phi(P)$ is a prime ideal in $S^{-1}R$, where $\Phi:R\to S$ defined by $\Phi(I)=\{\frac{x}{s}|x\in I,s\in S\}$. I think I want to use this here, but I still can't quite picture how these ideals would look in $S^{-1}R$.
It's always true that the prime ideals of $S^{-1}R$ are in bijection with the prime ideals of $R$ that don't intersect $S$. When $R= \mathbb{Z}$ and $S= \mathbb{Z} - p\mathbb{Z}$, the primes that don't intersect $S$ are precisely those contained in the complement of $S$, or $p \mathbb{Z}$. But since $p$ is prime, the only such prime ideal is $p \mathbb{Z}$ itself. As a subset of $S^{-1}R$, this is the set of fractions $x/s$ such that $x \in p \mathbb{Z}$.
Can you describe the rest of the ideals?