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)

  1. 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.
  2. Homology with coefficients in a ring $R$: take the free chain complex over $R$, then take homology. It returns $R$-modules.
  3. 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.

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    $\begingroup$ I’m not able to parse what your question is. If it is whether the universal coefficient theorem is telling you how to calculate with coefficients in an R module from coefficients in R, then yes this is correct. As a result, as long as the underlying abelian group is the same, it doesn’t matter what ring you use to calculate if you only care about additive structure. $\endgroup$ Feb 26, 2020 at 1:26
  • $\begingroup$ In your example there is no meaningful distinction, since an abelian group admits at most one $\mathbb{F}_2$-module structure. $\endgroup$ Feb 26, 2020 at 4:44

1 Answer 1


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.

  • $\begingroup$ Thanks for your reply, but I feel like I didn't understand it. Let's denote by $H_n(X; M, R)$ the $n$-homology of $X$ with coefficient in a $R$-module $M$. Then $H(X; M, R)$ is an $R$-module. You say that if I want to compute $H_n(X; F_2, F_2)$, I can compute $H_n(X; F_2, Z)$ and then there is only one $F_2$-module structure. First of all: why? But then, even if I can put an unique $F_2$-module structure on $H_n(X; F_2, Z)$, how can I know that $H_n(X; F_2, Z)$ with this $F_2$-module structure is the same as $H_n(X; F_2, F_2)$? Probably my misunderstanding is about module theory. $\endgroup$
    – CNS709
    Feb 26, 2020 at 15:26
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    $\begingroup$ The definitions of $H_n(X; F_2, Z)$ and $H_n(X; F_2, F_2)$ are identical--you're taking homology of the exact same chain complex, namely $X\otimes_Z F_2$. It's just that for the first one you are thinking of it as a chain complex of $Z$-modules, and for the second you are thinking of it as a chain complex of $F_2$-modules. You're still just taking $\ker\partial_n/\operatorname{im}\partial_{n+1}$ either way so you'll get the same abelian groups. $\endgroup$ Feb 26, 2020 at 15:59
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    $\begingroup$ To define an $F_2$-module structure on an abelian group $A$, you have to define scalar multiplication by each element of $F_2$. The only elements are $0$ and $1$, and you have only one choice for how to define the multiplication: $0\cdot a$ must be $0$ and $1\cdot a$ must be $a$. $\endgroup$ Feb 26, 2020 at 16:02
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    $\begingroup$ Well, you can talk about homology of spaces, but in my answer $X$ is just a chain complex. I suppose strictly speaking $H_n(X;F_2,F_2)$ would be defined as the homology of $(X\otimes_Z F_2)\otimes_{F_2}F_2$ but the second tensor doesn't change anything since you are just tensoring with the ring itself. $\endgroup$ Feb 26, 2020 at 16:05
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    $\begingroup$ For any $Z$-module $A$, $A\otimes_Z R$ has a canonical $R$-module structure, using the multiplication on the second coordinate of the tensor product. $\endgroup$ Feb 26, 2020 at 16:16

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