Is the following true: if $K$ and $F$ are fields with the same characteristic and $X$ is a topological space, then for any $n$ there holds $$\dim_K H_n(X;K) = \dim_F H_n(X;F)\text{ and }\dim_K H^n(X;K) = \dim_FH^n(X;F),$$ where $H_n(-;-)$ and $H^n(-;-)$ are singular homology and cohomology with coefficients?
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1$\begingroup$ See the Universal Coefficient Theorem en.wikipedia.org/wiki/Universal_coefficient_theorem $\endgroup$– Andrea MoriJul 14, 2013 at 20:13
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Yes. Any field is an extension of $\mathbb{Q}$ or some $\mathbb{F}_p$, so it suffices to take $K = \mathbb{Q}$ or $\mathbb{F}_p$ and $F$ an extension of $K$. Then $F$ is flat over $K$, so using the universal coefficient theorem (or just tensoring $F$ with the singular chain complex with coefficients in $K$), you see that $H_n(X;F) \cong H_n(X;K) \otimes F$ and likewise for cohomology.
answered Jul 14, 2013 at 20:15
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$\begingroup$ Why does it suffice to take $K=\mathbb{Q},\mathbb{F}_p$? I know that these are the the smallest fields of chosen characteristic, but why do we have $$\dim_{\mathbb{F}_p}H_n(X;\mathbb{F}_p)= \dim_{\mathbb{F}_{p^k}}H_n(X;\mathbb{F}_{p^k})= \dim_{\mathbb{F}_p(t)}H_n(X;\mathbb{F}_p(t))?$$ $\endgroup$– LeoJul 14, 2013 at 20:23
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$\begingroup$ Aha, I think I get it: if $k\subseteq K$ are fields, then $K$ is free as a $k$-module, hence projective, hence flat, hence $\mathrm{Tor}_1^k(-,K)=0$, hence $H_n(X;K)\cong H_n(X;k)\otimes_kK\cong k^{(I)}\otimes_kK\cong K^{(I)}$, so the dimension $I$ is the same. However, if $K$ and $F$ have different characteristics, then the dimensions need not be the same, right? $\endgroup$– LeoJul 14, 2013 at 21:22
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1$\begingroup$ Right. For example, if you have $p$-torsion in your integral homology and you're picking a field for your coefficients, you won't see that $p$-torsion unless your field has characteristic $p$. $\endgroup$ Jul 16, 2013 at 0:03