Let $R$ be a commutative local artin $k$-algebra,where $k$ is a field with characteristic $0$.I wonder whether universal coefficient theorem holds in this case.Namely,if $C$ is a chain of flat $R$-modules. Then does $H(C\otimes_R S)\cong H(C)\otimes_R S$ whenever $H(C)$ is flat $R$-module. $H(C)$ is homology of chain complex $C$.$S$ is commutative ring over $R$

I checked wikipedia and nLab,it seems that there is only the case when $R$ is field or Principal ideal domain.I dont know whether one can use so called Kunneth spectral sequence to make arguments.I just started to understand


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    $\begingroup$ When $C$ is bounded below, this appears to follow from the two hyperhomology spectral sequences, which both collapse in this case. All that is required is that every term of $C$ and of its homology is a flat $R$-module. For a description of the hyperhomology spectral sequences, see the wikipedia page on hyperhomology or the more detailed proposition 5.7.6 in Weibel's book "An introduction to homological algebra". $\endgroup$ Apr 3, 2014 at 3:23
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    $\begingroup$ More directly, the required Künneth spectral sequence for bounded below chain complexes of flat modules is given in theorem 5.6.4 of Weibel's book. $\endgroup$ Apr 3, 2014 at 4:00

1 Answer 1


It doesn't hold. Take $k$ any field (of characteristic zero if you wish), $R=k[\epsilon]/(\epsilon^2)$ its ring of dual numbers, $S=R/(\epsilon)=k$ its residue field, and $C$ the periodic complex $$\cdots\rightarrow R\stackrel{\epsilon}\longrightarrow R\rightarrow\cdots.$$ This is a complex of projective (hence flat) $R$-modules with trivial (hence flat) homology $H(C)=0$, so $H(C)\otimes_RS=0$, but $C\otimes_RS$ is $$\cdots\rightarrow S\stackrel{0}\longrightarrow S\rightarrow\cdots,$$ hence $H_n(C\otimes_RS)=S\neq 0$ in each degree $n\in\mathbb Z$.

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    $\begingroup$ If $C$ is bounded on the right, it is a particular case of the Künneth formula, see Bourbaki, Algebra X, § 4, no. 7, Corollary 4. $\endgroup$
    – abx
    Apr 3, 2014 at 5:51

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