Rank and dimension dilemma for modules

Question

Let $R$ be a field of characteristic zero, and $M$ be a $\mathbb{Z}$-module. Knowing that $R\otimes M$ is a vector space, can we deduce that $M$ is a free $\mathbb{Z}$-module and $\text{rank}(M)=\dim_R(R\otimes M)$? And maybe finding a basis in $M$.

I came up with this problem trying to understand why

$$\text{rank}\ H_p(X,\mathbb{Z})=\dim_R H_p(X,R)$$.

It is known that $H_p(X,R)\simeq R\otimes H_p(X,\mathbb{Z})$ and $H_p(X,R)$ is a vector space over $R$.

By $H_p(X,R)$ I mean the $p$-th singular homology $R$-module (i.e. vector space when $R$ is a field) associated to $X$ and by $H_p(X,\mathbb{Z})$ the $p$-th homology $\mathbb{Z}$-module associated to the topological space $X$.

Answer 1

• For any abelian group $M$, $\mathbb{Q} \otimes M$ is a rational vector space (see Why is the tensor product of abelian group and $\mathbb Q$ a rational vector space?, for example).

• There are abelian groups which contain torsion.

So knowing that $\mathbb{Q} \otimes M$ is a rational vector space does not let you conclude that $M$ is torsion-free, because it need not be. It is, however, true that if $M$ is finitely generated, then its rank (meaning the rank of its free part) is equal to the dimension of $\mathbb{Q} \otimes M$. You can prove this using the classification of finitely generated abelian groups, combined with the fact that $A \otimes (B \oplus C) \cong (A \otimes B) \oplus (A \otimes C)$ and the isomorphisms $\mathbb{Q} \otimes \mathbb{Z} \cong \mathbb{Q}$ and $\mathbb{Q} \otimes \mathbb{Z}/n\mathbb{Z} \cong 0$ for $n > 0$.

Answer 2

take $M=\mathbb Z \oplus \mathbb Z/2$ and $R=\mathbb Q$ as a counterexample.

Maybe assume that $H_p(X, \mathbb Z)$ to be a finitely generated module and use the classification theorem for finitely generated abelian groups. Note that $\mathbb Q \subseteq R$. Can you show one side of the inequality?