# $\text{Ext}(\mathbb{Z}[\frac{1}{p}]/\mathbb{Z},\mathbb{Z})\cong\mathbb{Z}_p$

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.

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 $$\mathbb Z_p$$.

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.