Finding generators for a module

Question

Suppose, if $p$ is a prime natural number, that $R$ is the set of rational numbers such that $p$ is not a factor of the denominator after cancelling out common prime factors of numerator and denominator. Consider $\mathbb{Q}$ as an $R$-module. I know that $\mathbb{Q}$ cannot be finitely generated (Nakayama's lemma). But what would be a 'small' set of generators of $\mathbb{Q}$ over $R$ and does a minimal set of generators exist? I know that $\left \{1, \dfrac {1}{p}, \dfrac {1}{p^2}, \ldots \right \}$ is a set of generators, therefore, so are $\left \{\dfrac {1}{p}, \dfrac {1}{p^2},\dfrac {1}{p^3} \ldots \right \}$ (since $1 = p\dfrac {1}{p}$), $\left \{\dfrac {1}{p^2}, \dfrac {1}{p^3},\dfrac {1}{p^4} \ldots \right \}$ etc., which might suggest that there is no minimal set of generators or maybe not...

Answer 1

In the same way as you have already argued, you could say let $X$ be any set of generators and let $\frac{a}{bp^i}$ and $\frac{c}{dp^j}$ be two elements of $X$ where $b$ and $d$ are coprime to $p$.

We can suppose $i\le j$ and then $\frac{a}{bp^i}=(\frac{adp^{j-i}}{bc})\frac{c}{dp^j}$.

Therefore $X$ can by reduced by removal of $\frac{a}{bp^i}$ and so no minimal set can exist.