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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!

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Doesn't this fact about a free and acyclic chain complex being contractible hold over a general commutative ring? I mean, it's mentioned in Spanier (Theorem 5 in section 3 of chapter 4) and I can't see anywhere in there that he's assumed he's working over a PID or a field. Here's a link: books.google.com/… –  David White Jul 25 '12 at 2:14
    
Also worth noting that of course without the hypothesis of freeness this fails badly. For example, the complex $\dots 0 \rightarrow \mathbb{Z} \stackrel{x2}{\rightarrow} \mathbb{Z} \rightarrow \mathbb{Z}/2 \rightarrow 0\dots$ is acyclic because it's exact, but is not contractible because if it were then there would be a right-inverse to the "mod 2" map, i.e. a nonzero $\mathbb{Z}/2 \rightarrow \mathbb{Z}$. –  David White Jul 25 '12 at 2:16
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David, it's true if the complex is bounded below, but not in general. A simple example is when the ring is $Z/4Z$, every module is free of rank one, and every boundary map is multiplication by $2$. –  Tom Goodwillie Jul 25 '12 at 2:24
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@David: Oh, cool. In my homological algebra class we just proved it for PIDs I think, but that's helpful. On the other hand, I'm afraid I care about a non-commutative ring. But thanks though. @Tom: Good point; I'll make an edit about the bounded condition and the fact that I care about noncommutative cases. –  Joseph Victor Jul 25 '12 at 2:39
    
@David: Spanier takes as base ring the integers: See Chap. 4, sect. 1: "Thus a chain complex $C$ consists of a sequence of abelian groups". In the proof of the quoted theorem 5, he uses that a submodule of a free modules is again free ("Because $C_{q-1}$ is free, so is $Z_{q-1}(C)$". This holds for PID but not for general comm. rings. –  Ralph Jul 25 '12 at 5:58

1 Answer 1

up vote 11 down vote accepted

There is a useful characterization in Brown: Cohomology of Groups, Prop. 0.3:

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.

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:

Each acyclic chain complex of projective modules over a hereditary ring is contractible.

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.

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I couldn't have hoped for a simpler answer. Thank you. –  Joseph Victor Jul 25 '12 at 8:03

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