Calculate $\text{Ext}_{\mathbb{Z}}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z}) $? I'm calculating this cohomology group. 
First I was trying to construct a projective resolution for $\mathbb{Q}/\mathbb{Z}$ but since $\mathbb{Q}/\mathbb{Z}$ has infinite many generators over as $\mathbb{Z}$ module so the free objects (I think) will be the infinite direct product of $\mathbb{Z}$. 
Thus I have 
$$ \cdots \prod \mathbb{Z} \to \prod \mathbb{Z} \to \mathbb{Q}/\mathbb{Z}.$$
But now I'm having a little trouble determine what the maps should be. 
In particular, I'm not sure what the kernel of the surjective map is (in order to determine the image of the second map). 
Any help will be appreciated! 
 A: I think it is easier to choose an injective resolution to compute this group, as $$0\longrightarrow\mathbb Z\longrightarrow\mathbb Q\longrightarrow\mathbb Q/\mathbb Z\longrightarrow 0$$ is an injective resolution of $\mathbb Z$. Then consider the derived functor $R\operatorname{Hom}_{\mathbb Z}(\mathbb Q/\mathbb Z,-)$ and we have the long exact sequence
\begin{align}
\cdots\to\operatorname{Hom}_{\mathbb Z}(\mathbb Q/\mathbb Z,\mathbb Q)\to\operatorname{Hom}_{\mathbb Z}(\mathbb Q/\mathbb Z,\mathbb Q/\mathbb Z)
\to\operatorname{Ext}_{\mathbb Z}^1(\mathbb Q/\mathbb Z,\mathbb Z)\to\operatorname{Ext}_{\mathbb Z}^1(\mathbb Q/\mathbb Z,\mathbb Q)
\to\cdots
\end{align}
Note that $\mathbb Q/\mathbb Z$ is a torsion $\mathbb Z$-module while $\mathbb Q$ is torsion-free and thus $\operatorname{Hom}_{\mathbb Z}(\mathbb Q/\mathbb Z,\mathbb Q)=0$. Since $\mathbb Q$ is injective, $\operatorname{Ext}_{\mathbb Z}^1(\mathbb Q/\mathbb Z,\mathbb Q)=0$. Hence we arrive at $$\operatorname{Ext}_{\mathbb Z}^1(\mathbb Q/\mathbb Z,\mathbb Z)\simeq \operatorname{Hom}_{\mathbb Z}(\mathbb Q/\mathbb Z,\mathbb Q/\mathbb Z)\simeq\widehat{\mathbb Z},$$
for the last equal sign I found many post explaining it in MSE, for example, here.
