# If $H_n(X;\mathbb{Z})$ are all f. g. free abelian, then $H^*(X;\mathbb{Z}) \otimes \mathbb{Z}_p \cong H^*(X; \mathbb{Z}_p)$?

An exercise in Hatcher's book asks to prove that whenever $X$ is a space with the homology groups $H_n(X; \mathbb{Z})$ finitely generated free abelian for each $n \geq 0$, then $H^*(X; \mathbb{Z}) \otimes \mathbb{Z}_p$ and $H^*(X;\mathbb{Z}_p)$ are isomorphic as graded rings.

I've only started to learn about the cup product and I have no idea how to proceed. I see that the two objects in question are isomorphic as groups (from the universal coefficients theorem), but that's about it. So the question is: how to prove the statement in the title of the question?

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Hint: $H^n(X)\to H^n(X;\mathbb{Z}_p)$ is a ring homomorphism. Putting that in a well-chosen commutative diagram, you ultimately need to show that a map $\prod\mathbb{Z}\to\prod\mathbb{Z}_p$ induces an isomorphism $(\prod\mathbb{Z})\otimes\mathbb{Z}_p\to\prod\mathbb{Z}_p$.
The universal coefficient theorem says not only that the two are isomorphic as groups but that a natural map between them, namely the reduction $\bmod p$ map $H^{\bullet}(X, \mathbb{Z}) \otimes \mathbb{Z}_p \to H^{\bullet}(X, \mathbb{Z}_p)$, is an isomorphism. This map always respects cup products (with no hypotheses).
Since $H_n(X;Z)$ finitely generated free abelian for each $n≥0$ So the Tor part will vanish in the short exact sequence for homology.Another observation: Since homology free this implies $H_n(X) = H^n(X)$.Hence the result follows.