Sufficient conditions for homology to be free module Let $H_*$ be homology with coefficients in a commutative ring $R$.
In general, $H_*$ will be a $R$-module but not necessarily free.
Are there any conditions that can ensure that the homology will be a free $R$-module? 
I only know of one: if $R$ is a field, then the homology is a vector space, so it is free.
However this condition seems too restrictive. Any other such conditions?
Thanks a lot.
 A: Recall that by the universal coefficient theorem we have a split exact sequence
$$0 \to H_i(X, \mathbb{Z}) \otimes R \to H_i(X, R) \to \text{Tor}_1(H_{i-1}(X, \mathbb{Z}), R) \to 0.$$
The Tor term, if nonzero, is always torsion, so let's think about when it vanishes. The Tor term vanishes if either $H_{i-1}(X, \mathbb{Z})$ or $R$ are torsion-free (equivalently, flat over $\mathbb{Z}$), and more generally if there is no prime $p$ for which both $H_{i-1}(X, \mathbb{Z})$ and $R$ have $p$-torsion. 
Next, we want $H_i(X, \mathbb{Z}) \otimes R$ to be a free $R$-module. It's a bit tricky to say exactly when this is possible because $H_i(X, \mathbb{Z})$ could be an arbitrary abelian group, so let's restrict to the case that $H_i(X, \mathbb{Z})$ is finitely generated for simplicity. Then $H_i(X, \mathbb{Z})$ is a free module plus a torsion module by the structure theorem, so the tensor product with $R$ is a free $R$-module if (but not only if) there is no prime $p$ at which both $H_i(X, \mathbb{Z})$ and $R$ have $p$-torsion.
So as long as you're only interested in spaces with finitely generated integral homology, a sufficient condition is that $R$ is torsion-free. For example, $R$ could be a $\mathbb{Q}$-algebra. But also $R$ could be an $\mathbb{F}_p$-algebra for any $p$. 
A: A necessary and sufficient condition on $R$ for $H_*(X;R)$ to be free for all spaces $X$ is that either $R$ contains a field or $R$ is a zero ring.  Indeed, if $R$ contains a field $K$, then $H_*(X;R)\cong R\otimes_K H_*(X;K)$ since the singular chain complex with coefficients in $R$ is just obtained by applying $R\otimes_K -$ to the singular chain complex with coefficients in $K$, and this is an exact functor since $R$ is a free $K$-module.  Since $H_*(X,K)$ is free over $K$, it follows that $R\otimes_K H_*(X;K)$ is free over $R$.  The case where $R$ is the zero ring is trivial.
Conversely, suppose $R$ does not contain a field and is not the zero ring.  If $R$ has characteristic $0$, this means there is some nonzero $n\in\mathbb{Z}$ which does not have an inverse in $R$.  Letting $X$ be a space with $H_i(X;\mathbb{Z})\cong\mathbb{Q}$ and $H_{i-1}(X;\mathbb{Z})=0$ for some $i$ (this is possible for any $i>0$), the universal coefficients theorem gives $H_i(X;R)\cong \mathbb{Q}\otimes_\mathbb{Z} R$.  Since $R$ has characteristic $0$, $\mathbb{Q}\otimes_\mathbb{Z} R$ is nonzero, and it cannot be free as an $R$-module since multiplication by $n$ is an isomorphism on it.
If $R$ has positive characteristic $n$, then since $R$ does not contain a field $n$ must not be prime.  Since $R$ is not the zero ring, $n>1$.  Let $p$ be a prime factor of $n$ and let $X$ be a space with $H_i(X;\mathbb{Z})\cong\mathbb{Z}/(p)$ and $H_{i-1}(X;\mathbb{Z})=0$ for some $i$ (again this is possible for any $i>0$.  The universal coefficients theorem then gives $H_i(X;R)\cong \mathbb{Z}/(p)\otimes_\mathbb{Z} R\cong R/pR$.  Since $p\mid n$, $R/pR$ is nonzero (if $1\in R$ were divisible by $p$, then $p$ would be a unit in $R$), and $R/pR$ cannot be free over $R$ since $p$ annihilates it but does not annihilate $R$.
