Subrings of $\mathbb{Q}$ that are not finitely generated. Exercise. Let $A$ be a subring of $\mathbb{Q}$ such that $\mathbb{Z}\subsetneqq A$. Then $A$ is not finitely generated as $\mathbb{Z}$-module.
I was trying in the following way, but unfortunately I cannot solve it. 
Suppose otherwise, that is, there are $\frac{a_{1}}{b_{1}}, \cdots, \frac{a_{n}}{b_{n}}\in \mathbb{Q}$ such that $A=\bigg\langle \frac{a_{1}}{b_{1}}, \cdots, \frac{a_{n}}{b_{n}} \bigg\rangle$ and consider $b=b_{1}...b_{n}\in\mathbb{Z}\setminus\{0\}\subseteq A$. Also, by hypothesis, there is $a=\frac{p}{q}\in A\setminus \mathbb{Z}$. Note that, the set $H=\{p\in A : p \hspace{0.1cm}\mbox{is prime and }\hspace{0.1cm} p \nmid  b\}$ is infinite. 
My question is : Is it possible to guarantee that there exists $p\in H$ such that $\frac{1}{p}\in A$? Note that if yes, we have to $\frac{1}{p}=\sum_{i=1}^{n} \frac{m_{1}a_{1}}{b_{1}}=\frac{m}{b}$, where $m,m_{i}\in \mathbb{Z}$, so $b=pm$, that is, $p|b$, contradiction.
Could someone tell me if I am on the right path for the resolution of the exercise? Otherwise, someone has an idea for the exercise.
Thanks 
 A: There must be some $\zeta \in \mathbb {A \backslash Z}$ and some prime number $p$ such that the power to which $p$ appears in the unique factorization of $\zeta $as a product of primes is $m<0.$ Suppose $A$ is generated as a $\mathbb Z-$module by $a_1,...,a_n \in A.$ Let the power to which $p$ appears in the factorization of $a_i$ into primes be $\mu_i, 1 \le i \le n.$ Let $\eta$ be a positive integer such that 
$\eta m<\min\{\mu_1,...,\mu_n\}.$ Then, since $A$ is a sub-ring of $\mathbb Q,$it follows that $\zeta^{\eta} \in A.$ But the power to which $p$ appears in the factorization of $\zeta^{\eta}$ into primes is $\eta m$ and the power to which $p$ appears in the factorization into primes of any member of $\mathbb Z[a_1,...,a_n]$ is $\ge \min\{\mu_1,...,\mu_n\}.$ Thus$ \zeta^{\eta} \in A \backslash \mathbb Z[a_1,...,a_n] .$
A: Suppose $\frac{a_1}{b_1},\ldots,\frac{a_n}{b_n} \in \mathbb{Q}$ generate A over $\mathbb{Z}$. Observe that any $\mathbb{Z}$-linear combination of these elements has denominator bounded by $\vert b_1 \cdots b_n \vert$. If $A$ contains a nonintegral element $x$, consider powers of $x$.
