In Hatcher's algebraic topology book, section 3.F, there is this exercise which seems really neat and I wanted to try. The exercise is to show that $Ext^1_\mathbb{Z}(A, \mathbb{Q})=0$ using the homology with $\mathbb{Q}$ coefficients of a Moore space $M(A,n)$.
I have tried gathering information to latter lead to a proof of this result, but am not succeeding. The main idea I have so far is noticing
$$H^{n+1}(M(A,n);\mathbb{Q})=Ext^1_\mathbb{Z}(A, \mathbb{Q})$$
From this if I could calculate by some other method that this second cohomology group is $0$, we'd be done.
Any help would be appreciated, thank you very much.
As you note, we want to show that $H^{n+1}(M(A,n);\mathbb{Q}) = 0$. By the universal coefficient theorem with field coefficients (Hatcher, bottom of p. 198), it suffices to show that $H_{n+1}(M(A,n);\mathbb{Q}) = 0$. The cellular chain complex to compute the homology of this space comes from the free resolution of the abelian group $A$: $$ 0 \to \bigoplus \mathbb{Z} \to \bigoplus \mathbb{Z} \to A \to 0 $$ We obtain the homology of $M(A,n)$ by removing $A$ and computing the homology of the resulting chain complex $$ 0 \to \bigoplus \mathbb{Z} \to \bigoplus \mathbb{Z} \to 0 $$ with nonzero terms in dimensions $n+1$ and $n$. The map is an inclusion, so we only get homology (equal to $A$) in dimension $n$.
To do this with coefficients in $\mathbb{Q}$, tensor this chain complex with $\mathbb{Q}$. Since $\mathbb{Q}$ is flat, tensoring with it preserves infectivity, so the chain complex $$ 0 \to \bigoplus \mathbb{Q} \to \bigoplus \mathbb{Q} \to 0 $$ only has homology (equal to $A \otimes \mathbb{Q}$) in dimension $n$. That is, $H_{n+1}(M(A,n);\mathbb{Q})=0$, which is what we need.