Let $p \in \mathbb{Z}$ be prime. Let $\mathbb{Z}_{(p)}$ be the localization of $\mathbb{Z}$ with respect to $\mathbb{Z}\setminus (p)$. Prove that $\frac{p}{1}$ is the only prime element of $\mathbb{Z}_{(p)}$ up to associates.
First, we need to show that $\frac{p}{1}$ is prime. We note that $\frac{p}{1} \neq \frac{0}{1}$ as this would imply that $pt=0$ for some $t\not\in (p)$. But $t \neq 0$ since $0 \not \in \mathbb{Z}\setminus (p)$, so that $p =0$, a contradiction. Also, $\frac{p}{1}$ is not a unit since $p \not\in \mathbb{Z}\setminus (p)$. We assume that $\frac{p}{1}$ divides $\frac{a}{b}\cdot \frac{c}{d}$ where $a,c \in \mathbb{Z}$ and $c,d \in (p)$. We want to show $\frac{p}{1}$ divides $\frac{a}{b}$ or $\frac{c}{d}$.
How do I proceed?
To show $p$ is prime in $\mathbb Z_{(p)}$: assuming $p$ divides the product $\frac{a}{b} \frac{c}{d}$, this means that there exists an $\frac{e}{f} \in \mathbb Z_{(p)}$ such that $\frac{pe}{f} = \frac{ac}{bd}$. This tells you that $pebd = acf$, so $p$ divides $acf$ in $\mathbb Z$. Since $p$ is prime in $\mathbb Z$, this tells you that $p$ divides $a$, $c$, or $f$ in $\mathbb Z$. But $f$ is not divisible by $p$, so $p$ must divide $a$ or $c$ in $\mathbb Z$.
Let's say that $p$ divides $a$ in $\mathbb Z$. So $pk = a$ for an integer $k$. Then $p \frac{k}{b} = \frac{a}{b}$, so $p$ divides $\frac{a}{b}$ in $\mathbb Z_{(p)}$. Therefore, $p$ is prime in $\mathbb Z_{(p)}$.
To show that $p$ is the only prime up to units, consider a general nonzero element $\frac{a}{b}$ in $\mathbb Z_{(p)}$. Write $a$ and $b$ as products of primes in $\mathbb Z$. All the primes in $\mathbb Z$ except for $p$ are units in $\mathbb Z_{(p)}$, which tells you that up to a unit, every element of $\mathbb Z_{(p)}$ is of the form $p^k$ for $k \geq 0$, and this isn't prime unless $k =1$.