Let p be a prime number, Consider the subring of $\mathbb{Q} $
$$ \mathbb Z_{(p)} = \{ a/b: gcd(a, b) =1 \text{ and } p\text{ does not divide } b\} $$
I need to show that there is only one maximal ideal that it is $(p) $ and that the set of units is $(p) $
So my idea is to show that all proper ideals are contained in $(p) $ but I don't know how to begin with this, so I would appreciate a hint.
I will be working under the assumption that you're meant to show that the set of units is $\mathbb{Z}_{(p)} \setminus (p)$.
Here's an outline of an approach to prove this: First prove that $x$ is a unit iff $x \not\in (p)$. Then, you can proceed as follows: Let $I$ be any ideal. If $I \subseteq (p)$, we're done. Otherwise, there is some $x \in I \setminus (p)$. By the above, $x$ is a unit - what does that mean for $I$?
Filling in the details: Let $x$ be a unit. Then $x = a/b$ for some $a,b \in \mathbb{Z}$ with $\operatorname{gcd}(a,b) = 1$ and $p \nmid b$, and (since inverses are unique) $b/a \in \mathbb{Z}_{(p)}$. But this implies $p \nmid a$ as well. If $a/b \in (p)$, this implies $a/b = a'/b' \cdot p/1$; since $p \nmid b,b'$, we have $p \mid a$, therefore $a/b \not\in (p)$.
Conversely, suppose $a/b \in (p)$. Then $p \mid a$, so $b/a \not\in \mathbb{Z}_{(p)}$, and $a/b$ is not a unit.
$(p)$ is clearly an ideal of $\mathbb{Z}_{(p)}$.
To prove that $(p)$ is maximal, let $I$ be an ideal of $\mathbb{Z}_{(p)}$. If $I \subseteq (p)$, we are done. Otherwise, there is some $x \in I \setminus (p)$, and by the above, $x$ is a unit. But since $I$ is closed under multiplication with ring elements, $x \cdot x^{-1} = 1 \in I$, so $I = \mathbb{Z}_{(p)}$.
