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?


1 Answer 1


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.

For the second question: You're completely right. The point is that all syzygies of the complex are projective again, and the restriction to f.g. modules is probably only to avoid use of AC. Note also that one can do with much weaker assumptions: it suffices to have a ring of finite global dimension.

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 $k[x]/(x^2)$.

  • $\begingroup$ Thank you! I'm also into Weibel and your answer helped me greatly. Coincidentally I ate a bunch of weebles in my Raisin Bran the other day. Weibels may wobble but they don't fall down. :) $\endgroup$ Commented Feb 9, 2023 at 5:06

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