Confusion about homology with coefficients in a ring $R$ and with coefficients in a $R$-module $M$ I used Universal coefficients theorem a lot, but now it seems to me that I have never understood it. My problem is the probably misunderstanding of homology with arbitrary coefficients. For what I have understood there are three types (ordinary and two successive generalization of it)


*

*Homology with coefficients in $\mathbb{Z}$: take the free chain complex over $\mathbb{Z}$, then take homology. It returns $\mathbb{Z}$-modules, i.e. abelian groups.

*Homology with coefficients in a ring $R$: take the free chain complex over $R$, then take homology. It returns $R$-modules.

*Homology with coefficients in a $R$-module $M$: take the free chain complex over $R$, tensor with $M$, then take homology. It returns again $R$-modules.


Universal coefficients theorem says that if $R$ is a PID then
$H_n(X; M) \cong H_n(X; R) \otimes_R M \oplus_R \text{Tor}_1^R(H_{n-1}(X; R), M)$
Now, if my understanding is correct, since it is an isomorphism of $R$-modules this means that we can use it to calculate 3 from 2, i.e. if we change module over the same ring, but not if we change ring. Is this correct?
For example, I always read "take homology over a field, for example $\mathbb{F}_2$". Does it mean $H(X; \mathbb{F}_2)$, i.e. a $\mathbb{F}_2$-vector space, or $H(X; \mathbb{Z}_2)$, i.e. an abelian group? I have always used the universal coefficient theorem in this case, but now I think that it is correct only if intended in the second sense. 
 A: If I understand correctly, here is a clearer rephrasing of your question.  Suppose $R$ is a PID, $X$ is a chain complex of free $R$-modules, and $S$ is an $R$-algebra.  Using the universal coefficients theorem, you can compute the homology $$H_n(X; S) \cong H_n(X; R) \otimes_R S \oplus \text{Tor}_1^R(H_{n-1}(X; R), S)$$ as an $R$-module.  However, $H_n(X;S)$ is not just an $R$-module but an $S$-module, and you are pointing out that the universal coefficients theorem does not tell you what $H_n(X;S)$ is as an $S$-module.
You are correct that the universal coefficients theorem as usually stated does not tell you the $S$-module structure.  However, you actually still can figure out the $S$-module structure from it.  First of all, let me remark that in many cases, the $S$-module structure is actually automatically uniquely determined by the $R$-module structure.  For instance, if $S$ is a quotient or localization of $R$, then  any $R$-module has at most one $S$-module structure.  This in particular covers the usual cases where $R=\mathbb{Z}$ and $S$ is $\mathbb{Z}/(n)$ or $\mathbb{Q}$.
Even when the $S$-module structure is not automatically determined, though, you can still figure it out by the naturality of the universal coefficients theorem.  Specifically, the short exact sequence $$0\to H_n(X; R) \otimes_R S \to H_n(X; S) \to \text{Tor}_1^R(H_{n-1}(X; R), S)\to 0$$ is natural in the coefficient module $S$.  For any $s\in S$, there is an $R$-module homomorphism $S\to S$ given by (right) multiplication by $S$, and naturality of the above sequence with respect to these homomorphisms says exactly that the maps in the sequence above are homomorphisms of (right) $S$-modules, not just of $R$-modules.  Moreover, if you examine the proof of the universal coefficients theorem, you can actually choose the splitting of this exact sequence to also be natural in $S$, for any fixed chain complex $X$ (the splitting comes from a choice of splitting of the inclusion of the $n$-cycles into $X_n$, and once you fix that splitting everything else is natural).  So, the isomorphism $$H_n(X; S) \cong H_n(X; R) \otimes_R S \oplus \text{Tor}_1^R(H_{n-1}(X; R), S)$$ is actually an isomorphism of $S$-modules, not just of $R$-modules.
