The universal coefficient theorem for cohomology reads:

$$0 \to Ext(H_{n-1}(C), R) \to H^n(C;R) \to Hom(H_n(C), R) \to 0,$$

where $C$ is a chain complex of free abelian groups and $R$ is a ring.

It is understood that the homology groups $H_i(C)$ are taken with respect to $\mathbb{Z}$ coefficients.

My question: what happens if you consider instead homology groups with $R$ coefficients? After trying some examples, it seems that you obtain the very same cohomology groups from the theorem, even though $Ext$ and $Hom$ may both change. However, I don't know if this is true in general.

  • Yes, there is a more general form of the theorem. Check out the Wikipedia article. Also this answer might be relevant. – Ayman Hourieh Apr 24 '14 at 20:34
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    The general form in Wikipedia applies only to principal ideal domains (or, more generally, rings of global dimension $1$). In the general general case one needs to take into account higher Exts and there is, in fact, a spectral sequence. – Mariano Suárez-Álvarez Apr 24 '14 at 20:43
up vote 5 down vote accepted

$\DeclareMathOperator{\Ext}{Ext}$The exact sequence mentioned in Wikipedia is only a corollary of a more general theorem; the corollary is only true for PIDs. The spectral sequence has the form: $$E_2^{p,q} = \Ext{}^q_S(H_p(X), R) \Rightarrow H^*(X; R)$$

I won't delve into the details if you don't know anything about spectral sequences, but when $S$ is a PID (so for example $S = \mathbb{Z}$) the higher $\Ext$ functors disappear (because every module has a projective resolution of length at most one). So the spectral sequence degenerates, and you're left with the exact sequence that you mention. But in full generality, you cannot get rid of the higher $\Ext$ terms, the differentials that appear in the SS, and the extension problems at the $E_\infty$ page.

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