# Weibel 1.4.1: Show that acyclic bounded below chain complexes of free R-modules are always split exact.

I can't quite figure out this problem

Weibel 1.4.1: Show that acyclic bounded below chain complexes of free R-modules are always split exact.

I can see that at the end of the chain, we have $$\cdots \to C_1 \xrightarrow{d_1} C_0 \to 0$$ with $C_0$ free, so the short exact sequence $$0 \to {\ker d_1}\to C_1 \xrightarrow{d_1} C_0 \to 0$$ splits.

But without assuming, say that R is a PID (so that $\ker d_1$ is free itself), I don't see how I can keep splitting every such short exact sequence for larger n.

Also, for the second part of the question, Weibel asks

• Show that an acyclic chain complex of finitely generated free abelian groups is always split exact, even when it is not bounded below.

I think the proof relies on that submodules of f.g. free abelian groups are free. But in that case, we can just assume we are working with R-modules for a PID R? In addition, is the f.g. restriction just to avoid using well-ordering/AC?

For the first question: $\text{ker}(d_1)$ might not be free, but as a summand of $C_1$, it is projective. Now use $\text{ker}(d_1)=\text{im}(d_2)$ and continue the splitting process.
The simplest example of an acyclic complex of free modules that is not contractible is probably $$...\xrightarrow{\cdot x} k[x]/(x^2)\xrightarrow{\cdot x} k[x]/(x^2)\xrightarrow{\cdot x}...$$ over the ring $k[x]/(x^2)$.