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

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!

• Is $\mathbb{Q}$ finitely generated as a $\mathbb{Z}$ module? – Dirk May 20 at 13:46
• And if you don't want to think for yourself: math.stackexchange.com/questions/19941/… – Dirk May 20 at 13:47

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.
• You could skip the part about $L$ being a submodule of $M$ and just use the fact that every non-zero vector space over a field $F$ admits a linear surjection to $F$. – Andreas Blass May 20 at 16:45