$\text{Ext}(\mathbb{Z}[\frac{1}{p}]/\mathbb{Z},\mathbb{Z})\cong\mathbb{Z}_p$ I'm beginning to read Riehl's "Category Theory in Context" and she mentions MacLane's calculation of $\text{Ext}(\mathbb{Z}[\frac{1}{p}]/\mathbb{Z},\mathbb{Z})\cong\mathbb{Z}_p$. I've been trying to find a proof of this but have been unable to. Any resources would be appreciated.
 A: This can be done by hand. Consider the following free resolution of $\mathbb Z[\frac 1p]/\mathbb Z$:
$$0\to\mathbb Z^{\oplus\mathbb N}\xrightarrow{d_1}\mathbb Z^{\oplus\mathbb N}\xrightarrow{d_0} \mathbb Z[\frac1p]/\mathbb Z\to 0,$$
where $d_0(e_i):=p^{-i}$ for $i\ge0$. The kernel of $d_0$ consists of $e_i-pe_{i+1}$ for $i\ge 0$ and $e_0$. Thus, let $d_1(e_i):=e_{i-1}-pe_{i}$ for $i\ge1$ and $d_1(e_0):=e_0$.
Now, take $\hom_{\mathbb Z}(-,\mathbb Z)$:
$$\mathbb Z^{\times\mathbb N}\xrightarrow{d_1^\vee}\mathbb Z^{\times\mathbb N}.$$
Here, $d_1^\vee$ sends $e_i^\vee$ to $e_i^\vee\circ d_1$, which can be calculated as:
$$\begin{align*}e_0^\vee&\mapsto e_0^\vee+e_1^\vee,\\
e_0^\vee&\mapsto -pe_i^\vee+e_{i+1}^\vee (i\ge1).\end{align*}$$
By definition, $Ext^1(\mathbb Z[\frac1p]/\mathbb Z,\mathbb Z)$ is the cokernel of $d_1^\vee$. That is,
$$Ext^1(\mathbb Z[\frac1p]/\mathbb Z,\mathbb Z)\cong\prod_{i\ge0}\mathbb Ze_i^\vee/(e_0^\vee+e_1^\vee,-pe_i^\vee+e_{i+1}^\vee)\cong\mathbb Z_p,$$
via the map $e_0^\vee\mapsto-1,e_i^\vee\mapsto p^{i-1} (i\ge1)$.

[Slightly suspicious???] Alternatively,taking direct limits is exact, as mentioned here. Thus, in particular, $Ext$ "commutes" with direct limits:
$$Ext^1(\lim_\rightarrow M_i,N)\cong\lim_\leftarrow Ext^1(M_i,N).$$
Thus,
$$\begin{align*}Ext^1(\mathbb Z[\frac1p]/\mathbb Z,\mathbb Z)&=Ext^1(\lim_\rightarrow\frac1{p^n}\mathbb Z/\mathbb Z,\mathbb Z)\\
&=\lim_\leftarrow Ext^1(\frac1{p^n}\mathbb Z/\mathbb Z,\mathbb Z)\\
&\cong\lim_\leftarrow\mathbb Z/p^n\mathbb Z,\end{align*}$$
and one may check that the induced maps $\mathbb Z/p^n\leftarrow\mathbb Z/p^{n+1}$ are multiplication by $p$. Thus, this is isomorphic to $\mathbb Z_p$.
A: This is an example of application of the derived functor of inverse limit. See [Weibel,Application 3.5.10].
As $p^{-i}\mathbb{Z}/\mathbb{Z}$ is an incresing sequence of abelian groups in $\mathbb{Z[p^{-1}]}/\mathbb{Z}$, there is a short exact sequence
$$0\rightarrow {\varprojlim}^1 \text{Ext}^{q-1}(p^{-i}\mathbb{Z}/\mathbb{Z},\mathbb{Z})\rightarrow \text{Ext}^{q}(\mathbb{Z}[p^{-1}]/\mathbb{Z})\rightarrow \varprojlim \text{Ext}^q(p^{-i}\mathbb{Z}/\mathbb{Z})\rightarrow 0.$$
For $q=1,$ we get $ {\varprojlim}^1 \text{Hom}(p^{-i}\mathbb{Z}/\mathbb{Z},\mathbb{Z})$ and this is clearly zero, hence we get the desired result.
