The third problem on this problem sheet from Oxford asks for the construction of a chain complex of free $k[x,y]$-modules ($k$ any field) which have a homology group isomorphic to $k$. What is an example of such a chain complex?
I expect that we should be able to do this with a short chain complex $$0 \rightarrow S \xrightarrow{i} M \xrightarrow{\pi} M/N \rightarrow 0.$$ Then the conditions reasonably reduce to finding $S \le N \le M$ such that
- $M$, $S$, and $M/N$ are free, and
- $N/S \cong k$.
I then attempted to set $M = k[x,y]$ (as a module over itself). We then know that $S$ must be a principal ideal with non-zero-divisor generator, like $(x)$ or $(y)$. Noting that $$k[x,y]/(x)/(y) \cong k,$$ I tried $$0 \rightarrow (y) \xrightarrow{i} k[x,y] \xrightarrow{\pi} k[x,y]/(x) \rightarrow 0,$$ which has the correct homology groups. However, $k[x,y]/(x) \cong k[y]$ is not free over $k[x,y]$.
I suspect that the scope of my trial is too small, and that it is necessary to leverage either an infinitely generated free module over $k[x,y]$ as $M$ or an infinitely generated ideal of $k[x,y]$ as $N$. Maybe it is necessary to leverage an infinite complex in order to observe the behavior.
Since this is purely for my own interest, I'll happily accept any hints, feedback, or outright solutions.