Finitely generated submodule of the fraction field of a PID

Question

$R$ is a PID with field of fractions $K$, and $M \subseteq K$ is a finitely generated $R$-submodule. I am trying to show $M$ is in fact generated by one element.

Answer 1

Consider a finite generating set and find a common denominator for its elements. Now, look at what you've got...

Alternatively, if $M$ is such a submodule, it has no torsion and we know from the structure theorem for f.g. modules over a PID that it must be free. Its rank can be computed by first tensoring with $K$ over $R$ and computing the dimension over $K$ of the resulting $K$-vector space. Because $M$ is contained in $K$, it is very easy to see that that dimension can only be $1$.

Answer 2

HINT $\: $ Fractions too have $\rm\: gcd\: \bigg(\!\!\dfrac{a}b,\:\dfrac{c}d\!\!\bigg)\ =\ \dfrac{gcd(a,b)}{lcm(b,d)}\ $ $\Rightarrow\:$ Bezoutness $\:\Rightarrow\:$ principality.