If $k$ is a field with 2 elements and $\mathbb{Q}$ is the additive group of rational numbers, is there a finite resolution of $k$ as $R=k[\mathbb{Q}]$ module?
By "finite" I mean a chain complex $$X= R^{n_1}\rightarrow R^{n_{2}}\rightarrow \dots \rightarrow R^{n_{j}}\rightarrow k $$
where $R^{n}$ is a free generated R-module of dimension $n$.
Edit: It will be fine if one can find a finite resolution of finitely generated projective $R$-module.
Best Answer
No. Indeed, $k$ is not even finitely presented, so there does not exist any exact sequence $$R^m\to R^n\to k\to 0$$ where $m$ and $n$ are finite. (Note that the existence of such a sequence with finitely generated free modules is equivalent to the existence of one with finitely generated projective modules, since a finitely generated projective module is a summand of a finitely generated free module.)
More generally, let us show that if $k$ is any nonzero commutative ring and $G$ is any group that is not finitely generated, then $k$ is not finitely presented as a $k[G]$-module. It suffices to show that the kernel $I$ of the augmentation $k[G]\to k$ is not finitely generated. Since $I$ is generated by the elements $g-1$ for $g\in G$, if it were finitely generated then it would be generated by finitely many such elements $g_1-1,\dots,g_n-1$. Let $H$ be the subgroup of $G$ generated by $g_1,\dots,g_n$. Let $M$ be the free $k$-module on the coset space $G/H$. The left action of $G$ on $G/H$ gives a $k[G]$-module structure on $M$, with the quotient map $G\to G/H$ of $k$-sets inducing a surjective $k[G]$-module homomorphism $f:k[G]\to M$. Moreover, $h-1$ is in the kernel of $f$ for all $h\in H$, so the kernel of $f$ contains $I$ so $f$ induces a surjective homomorphism $k\cong k[G]/I\to M$. Since $k$ is free of rank $1$, this means $M$ is free of rank at most $1$. But $M$ is free on $G/H$, so we must have $H=G$. Since $H$ was generated by $g_1,\dots,g_n$, this means $G$ is finitely generated.