Finite index submodules of finitely generated free modules.

Let $$R$$ be a ring of characteristic zero, not necessarily a commutative ring. Let $$F$$ be a finitely generated free left $$R$$-module, and let $$M$$ be a left $$R$$-submodule of $$F$$.

(1) Suppose $$M$$ is a finite index subgroup of $$F$$ (as abelian groups), is $$M$$ a finitely generated $$R$$-module?

The case that I am interested is when $$R$$ is the integral group ring $$\mathbb{Z}G$$ of a finitely generated group $$G$$. Note the answer is positive, for example, for a rational group ring $$\mathbb{Q}G$$ or for $$R$$ a field since in these cases such $$M$$ equals $$F$$.

As a remark, question (1) is equivalent to

(2) If $$A$$ is $$R$$-module with finite underlying set; is $$A$$ finitely presented as an $$R$$-module?

No, not in general. For instance, let $$R=\mathbb{Z}[x_1,x_2,\dots]$$ be a polynomial ring in infinitely many variables, let $$F=R$$, and let $$M$$ be the ideal $$(2,x_1,x_2,\dots)$$. Then $$M$$ has index $$2$$ but is not finitely generated.
It is true for the integral group ring $$\mathbb{Z}G$$ of a finitely generated group $$G$$. Indeed, suppose $$A$$ is a finite $$\mathbb{Z}G$$-module. Then the $$\mathbb{Z}G$$-module structure of $$A$$ can be completely described by writing down the addition table of $$A$$ and the action of each generator of $$G$$ on $$A$$. That's a finite set of relations from which the entire structure of $$A$$ can be deduced, so it gives a finite presentation of $$A$$ (with the entire underlying set of $$A$$ as the set of generators).
More generally, a similar argument shows that if $$S$$ is a commutative ring and $$R$$ is a finitely generated $$S$$-algebra, then every $$R$$-module that is finitely presented as an $$S$$-module is also finitely presented as an $$R$$-module (just start with a finite presentation over $$S$$ and add relations that tell you what each generator of $$R$$ does on each generator of the module).