A valuation ring of a field $F$ is a subring $\mathcal O\subsetneq F$ such that either $z\in\mathcal O$ or $z^{-1}\in\mathcal O$ for all $z\in F$. Show that any valuation ring of $\mathbb Q$ is $\mathbb Z_{(p)}$ for some prime $p$, where $\mathbb Z_{(p)}=\{a/b\in\mathbb Q\mid a,b\in\mathbb Z,p\nmid b\}$.
My attempt:
Suppose that $\mathfrak m$ is the unique maximal ideal of $\mathcal O$. Since $\mathfrak m$ is a prime ideal, $\mathfrak m\cap\mathbb Z$ is a prime ideal in $\mathbb Z$, so that $\mathfrak m\cap\mathbb Z=p\mathbb Z$ for some prime number $p$. Then $p\mathcal O=(\mathfrak m\cap\mathbb Z)\mathcal O\subseteq\mathfrak m$. If $a/b\in\mathcal O$ with relatively prime integers $a$ and $b$, then $pa/b\in p\mathcal O\subseteq\mathfrak m$, and $b/(pa)\notin\mathcal O$. Assume that $p$ divides $b$, so that $\gcd(a,p)=1$. It follows that $a\notin\mathfrak m$ and $a^{-1}\in\mathcal O$. Then $b/(pa)=(b/p)a^{-1}\in\mathcal O$, a contradiction. Hence $\mathcal O\subseteq\mathbb Z_{(p)}$. For the reverse inclusion, if $a/b\in\mathbb Z_{(p)}$ with relatively prime integers $a$ and $b$, then $b/a\notin p\mathbb Z_{(p)}$. On the other hand, $\mathfrak m\subseteq\mathfrak m\mathbb Z_{(p)}\subseteq p\mathbb Z_{(p)}$, so $b/a\notin\mathfrak m$. If $b/a\in\mathcal O$, then it is a unit in $\mathcal O$ and $a/b\in\mathcal O$; otherwise $a/b\in\mathcal O$ by the definition. Therefore $\mathcal O=\mathbb Z_{(p)}$.
EDIT: If $\mathfrak m\cap\mathbb Z$ is the zero ideal, then $z\notin\mathfrak m$ and $z^{-1}\in\mathcal O$ for all $z\in\mathbb Z^*$, contradicting $\mathcal O\subsetneq\mathbb Q$.
Is this valid? Besides, I wonder if there is some more general conclusions to prove this; e.g., if a valuation ring is a maximal subring.
If $\mathfrak m\cap\mathbb Z$ is the zero ideal. Because $\mathcal O$ is a subring of $ \mathbb Q $ we have $1 \in \mathcal O $ which again implies that $ \forall a \in \mathbb Z^*$ one will have $a \in \mathcal O $. However, because $\mathfrak m\cap\mathbb Z= (0)$ we will have every none nul intergers invertible in $\mathcal O $ from where we have $a, a^{-1} \in \mathcal O $ for all $a \in \mathbb Z^*$. Thus $\mathbb Q \subseteq \mathcal O $ because for $a/b \in \mathbb Q $ we have $ (a/b)= a b^{-1} \in \mathcal O$. which contradicts $\mathcal O\subsetneq\mathbb Q$.