When is an acyclic chain complex contractible? 


I know an acyclic chain complex of free modules over a PID (or field) are always contractible, but what about over a more complicated ring, like a graded algebra over Z/p (for instance the mod p Steenrod Algebra)? 


EDIT: I want to assume the chain complex is bounded below, but not necessarily that the ring is commutative. I suspect that it isn't always true for a non-commutative ring, but I don't really have a counter example. 


Thanks everybody! 



When is an acylic chain complex contractible
There is a useful characterization in Brown: Cohomology of Groups, Prop. 0.3: 



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> A chain complex $C$ over any ring is contractible iff it is acyclic and each short exact sequence $0 \to \ker(d\_n) \to C\_n \to \operatorname{im}(d\_n) \to 0$ splits. 
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This immediately explains the OP's PID example: If $C$ is a complex of free modules over a PID, then $\operatorname{im}(d\_n) \le C\_{n-1}$ is also free and thus the sequence splits. 


This observation can be axiomized as follows: A ring (with unit) is called *hereditary*, if each submodule of a projective module is again projective. As a corollary: 



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> Each acyclic chain complex of projective modules over a hereditary ring is contractible. 
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An example of an non-commutative hereditary ring is given by the upper-triangular matrices over a field. 


BTW: Tom's remark also follows easily from the criterion: Let all $C\_n$ be free and $C\_n=0$ for $n < 0$. Since $C\_0$ is free, the short exact sequence $0 \to \ker(d\_1) \to C\_1 \to C\_0 \to 0$ splits. Hence $\ker(d\_1)=\operatorname{im}(d\_2)$ is a direct summand of a free module and therefore projective. By induction then all of the short exact sequences split. 


