Show that if \(H_n(X; \mathbb{Z})\) is free for each \(n\), then \(H^*(X; \mathbb{Z}_p)\) and \(H^*(X; \mathbb{Z}) \otimes \mathbb{Z}_p\) are isomorphic as rings, so in particular the ring structure with \(\mathbb{Z}\) coefficients determines the ring structure with \(\mathbb{Z}_p\) coefficients.
Solution
1. By the universal coefficient theorem, \(H^n(X; \mathbb{Z}_p) \cong \operatorname{Hom}(H_n(X), \mathbb{Z}_p) \oplus \operatorname{Ext}(H_{n-1}(X), \mathbb{Z}_p)\).
universal coefficient theorem for cohomology.
2. Since \(H_{n-1}(X)\) is free, \(\operatorname{Ext}(H_{n-1}(X), \mathbb{Z}_p) = 0\).
\(\operatorname{Ext}\) of a free abelian group vanishes.
3. Hence \(H^n(X; \mathbb{Z}_p) \cong \operatorname{Hom}(H_n(X), \mathbb{Z}_p)\).
1 and 2.
4. Since \(H_n(X)\) is free, \(\operatorname{Hom}(H_n(X), \mathbb{Z}_p) \cong \operatorname{Hom}(H_n(X), \mathbb{Z}) \otimes \mathbb{Z}_p \cong H^n(X; \mathbb{Z}) \otimes \mathbb{Z}_p\).
\(\operatorname{Hom}(F, \mathbb{Z}_p) \cong \operatorname{Hom}(F, \mathbb{Z}) \otimes \mathbb{Z}_p\) for free \(F\), and \(H^n(X; \mathbb{Z}) \cong \operatorname{Hom}(H_n(X), \mathbb{Z})\) (since \(\operatorname{Ext}(H_{n-1}, \mathbb{Z}) = 0\) by freeness).
5. Hence \(H^n(X; \mathbb{Z}_p) \cong H^n(X; \mathbb{Z}) \otimes \mathbb{Z}_p\) for each \(n\).
3 and 4.
6. This isomorphism is compatible with the cup product (it is induced by the natural map \(\mathbb{Z} \to \mathbb{Z}_p\) on coefficients, which is a ring homomorphism), so it is an isomorphism of rings.
the cup product is natural with respect to coefficient maps, and the reduction map \(\mathbb{Z} \to \mathbb{Z}_p\) induces a ring homomorphism \(H^*(X; \mathbb{Z}) \to H^*(X; \mathbb{Z}_p)\).
7. Hence \(H^*(X; \mathbb{Z}_p) \cong H^*(X; \mathbb{Z}) \otimes \mathbb{Z}_p\) as rings.
5 and 6.
8. Q.E.D.
7.