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).


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