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In a paper I am reading involving simplicial homology, I have been told to think about certain Abelian groups(the boundary group and cycle group) as Z-modules so we can allow alternate ground rings of coefficients(in order to make some claims about structure if we choose a PID for the ground ring). I'm not sure what this means to allow alternate ground rings of coefficients for these modules. Could someone explain this to me?


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You should give a precise quote and reference. –  Martin Brandenburg Mar 25 '12 at 21:35
Paper is "Computing Persistent Homology" by Zomorodian and Carlsson, section 2.4 –  Andrew Mar 25 '12 at 21:46
Quote is:"The kth homology group is Hk = Zk / Bk . Its elements are classes of homologous cycles. To describe its structure, we view the Abelian groups we have defined so far as modules over the integers. This view allows alternate ground rings of coefficients, including fields" –  Andrew Mar 25 '12 at 21:56

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In homology you often have to consider the free $\mathbb{Z}$-module generated by stuff. These are formal linear combinations with coefficients in $\mathbb{Z}$.

However, there is nothing special about $\mathbb{Z}$ here...we could have taken our coefficients from any ring, for example $\mathbb{R}$ or $\mathbb{F}_2$ etc. Algebraically this is not a big leap but geometrically, changing the coefficient ring tells you different things about the space.

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Got it, this is what I was thinking but I was unsure. The paper later talks about using fields to treat the module as a vector space, and this interpretation makes sense. Thanks! –  Andrew Mar 25 '12 at 22:18
Well a module is a structure that generalises the notion of vector space. Instead of using a field for your scalars you can now use an arbitrary ring. –  fretty Mar 25 '12 at 22:58

This simply mean that since your homologygroups $H^i(X)$ are $\mathbb{Z}$-modules, you can tensor it by any abelian group and thus have "simpler" cohomology groups.

For example, it appears that sometimes if you consider homology groups with coefficients in $\mathbb{F}_p$ (i.e. $H^i(X)\otimes \mathbb{F}_p$) things become simpler, but you only get information on the $p$-torsion.

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