Is a finitely generated module over the field of fractions is also finitely generated over the original integral domain?

Question

Let $R$ be an integral domain and $F$ its field of fractions. Let $M$ be a finitely generated $F$-module.

Question: Is $M$ also a finitely generated $R$-module?

I know that $M$ is an $R$-module since $R$ is a subring of $F$, cf. here. But how can I show that it is finitely generated as an $R$-module?

Could you please help me with this question? Thank you!

Answer 1

Suppose a nonzero finitely generated module $M$ over $F$ is also finitely generated over $R$. Since $M$ is a vector space, it contains an $F$-submodule $L$ isomorphic to $F$. On the other hand there exists a surjective homomorphism $M\to L$ of $F$-modules, which is also a homomorphism of $R$-modules. Hence $L$ is also finitely generated over $R$.

Then the problem is reduced to showing whether $F$ is a finitely generated $R$-module.

Suppose $F$ is generated by $x_1/y,x_2/y,\dots,x_n/y$ over $R$, with $x_1,\dots,x_n,y\in R$; note that it is not restrictive to assume the same denominator. Then, for every $z\in F$ there are $r_1,\dots,r_n\in R$ such that $$ z=\sum_{k=1}^n r_k\frac{x_k}{y} $$ If $R\ne F$, we conclude that $y$ is not invertible in $R$. However $$ \frac{1}{y^2}=\sum_{k=1}^n r_k\frac{x_k}{y} $$ implies $$ \frac{1}{y}=\sum_{k=1}^n r_kx_k\in R $$ a contradiction.