I have been trying to prove the following:
Let $A,B$ be abelian groups. If $A$ is divisible and $B$ is torsion-free, then the group $Ext_{\mathbb{Z}}^1(A,B)$ is torsion-free.
So, I study homological algebra and in my class we have come as far as to define the $Ext$ functor. I have tried proving this result using long exact sequences, since I Know that $Ext_{\mathbb{Z}}^i(A,B)=0$ for all $i \geq 2$ and all abelian groups $A,B$. After that, I am completely stumped. I should also mention that in my class we did not define the $Tor$ functor and we defined the $Ext$ using projective resolutions.
I tried looking in the following books:
Basic Homological Algebra by M. Scott Osborne
An Introduction to Homological Algebra by Joseph J. Rotman
to no avail. Of course, it is possible that the answer is in one of these books and I may have missed it.
Best Answer
Since $A$ is divisible, the map $n:A\to A$ is surjective for any nonzero integer $n$. There is thus a short exact sequence $$0\to K\to A\stackrel{n}\to A\to 0$$ which gives a long exact sequence $$0\to \operatorname{Hom}(A,B)\stackrel{n}\to\operatorname{Hom}(A,B)\to \operatorname{Hom}(K,B)\to \operatorname{Ext}^1(A,B) \stackrel{n}\to \operatorname{Ext}^1(A,B)\to \operatorname{Ext}^1(K,B)\to 0.$$
But since $K$ is the kernel of multiplication by $n$, it is $n$-torsion, so $\operatorname{Hom}(K,B)=0$ since $B$ is torsion-free. Exactness of the sequence above now says that $n:\operatorname{Ext}^1(A,B)\to \operatorname{Ext}^1(A,B)$ is injective. Since $n$ is an arbitrary nonzero integer, this says that $\operatorname{Ext}^1(A,B)$ is torsion-free.