Given $R$ an integral domain with field of fractions $F$, I want to prove $F$ is a finitely generated $R$-module if and only if $R = F$.
For $(\Rightarrow)$: I tried to first define a generating set based on the definition i.e. finitely many $m_1,\dots,m_n$ such that $\sum_i Rm_i = M$. Then somehow show every element in $R$ is invertible. From where $D=F$ but how...
For $(\Leftarrow)$: If $R = F$, I am not sure, but I know by definition, the embedding of $R$ in $F$ maps each $n \in R$ to the fraction $\frac{an}{a}$ for nonzero $a$ in $R$. In further details, I am a bit lost though.
If $F=R$, then you have nothing to prove.
Suppose $F$ is finitely generated as $R$-module. Let $$ \left\{\frac{a_1}{c},\frac{a_2}{c},\dots,\frac{a_n}{c}\right\} $$ be a set of generators. It's not restrictive to take the same denominator, because we can always multiply numerator and denominator by the same factor.
In particular, there are $b_1,\dots,b_n\in R$ with $$ \frac{1}{c^2}=\frac{a_1}{c}b_1+\frac{a_2}{c}b_2+\dots+\frac{a_n}{c}b_n $$ so $$ \frac{1}{c}=a_1b_1+\dots+a_nb_n\in R $$ Therefore…