Under what circumstances is $H^p(B,H^q(F,R))\cong H^p(B,R)\otimes H^q(F,R)$?

Question

The above question comes from the Serre spectral sequence in cohomology. One would like to write something like $H^p(B,H^q(F,R))\cong Hom(H_p(B;\mathbb{Z});Hom(H_q(F;\mathbb{Z});R))\cong Hom(H_p(B;\mathbb{Z})\otimes H_q(F;\mathbb{Z});R)\cong Hom(H_p(B;\mathbb{Z});R)\otimes Hom(H_q(F;\mathbb{Z});R)\cong H^p(B,R)\otimes H^q(F,R)$ but we are invoking the UCT a few times and would get some Ext-terms and we are interchanging $Hom$ with $\otimes$. This works well if $R$ is a field and everything is finitely generated. But what if for example $R=\mathbb{Z}$ and everything is finitely generated? Is it still possible? If necessary one can assume, that cohomology only appears in degrees far appart.

Answer 1

Let us assume that all the homology groups of $B$ are finitely generated and that $B$ is simply connected. There is a universal coefficient theorem for cohomology which involves $\mathrm{Tor}$. It follows that $$ H^p(B,H^q(F,R))\cong H^p(B,\mathbb{Z})\otimes H^q(F,R)\oplus \mathrm{Tor}\,(H^{p+1}(B,\mathbb{Z}),H^q(F,R)) $$

So if $H^{p+1}(B,\mathbb{Z})$ or $H^q(F,R)$ does not contain torsion you get a nice formula. I'm not sure how this interacts with the cup products in the SSS.