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?
 A: 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).
