Ext group of $\mathbb{Z}[1/p]$ I'm trying to compute $\operatorname{Ext}_{\mathbb{Z}}^1(\mathbb{Z}[1/p],\mathbb{Z})\cong \mathbb{Z}_p/\mathbb{Z}$.
Now I have the projective resolution
$$0\rightarrow \bigoplus_{i>1}\mathbb{Z}\xrightarrow{\alpha} \mathbb{Z}\oplus \bigoplus_{i>1}\mathbb{Z}\xrightarrow{\beta} \mathbb{Z}[1/p]\rightarrow  0 .$$ The map $\alpha$ is given by $(a_i)_{i>0}\mapsto (-\Sigma a_i, a_ip^i)$ and $\beta$ is given by  $(b_i)_{i\geq 0}\mapsto \Sigma_{i\geq 0} b_i/p^i$.
Now apply the $\operatorname{Hom}(-,\mathbb{Z})$, I want to calculate the kernel of the dualised map $\prod_{i>1}\mathbb{Z}\xleftarrow{\alpha^*} \mathbb{Z}\prod (\prod_{i>1} \mathbb{Z})$ which is given by $(f_0,0,\dots)\mapsto f_0'$  and $(0,\dots,f_i,\dots )\mapsto (0,\dots,p^if_i,\dots )$, where $f_0': \prod_{i>1}\mathbb{Z}\to \mathbb{Z}$, $f_0'((a_i))=f_0(\Sigma a_i)$. Is there any way to see what is this kernel and how the quotient of $\prod_{i>1} \mathbb{Z}$ by this kernel is $\mathbb{Z}_p/\mathbb{Z}$?
 A: Let's do this by injective resolutions instead. I'll write $A$ for $\Bbb Z[1/p]$.
Then
$$0\to\Bbb Z\to\Bbb Q\to\Bbb Q/\Bbb Z\to0$$
gives an injective resolution of $\Bbb Z$. Therefore $\text{Ext}^1(A,\Bbb Z)$ is the
cokernel of
$$\text{Hom}(A,\Bbb Q)\to\text{Hom}(A,\Bbb Q/\Bbb Z).$$
It's easy to see that $\text{Hom}(A,\Bbb Q)\cong\Bbb Q$ via $f\mapsto f(1)$.
What is an element of $\text{Hom}(A,\Bbb Q/\Bbb Z)$? It is described completely
by $f(1/p^k)=a_k+\Bbb Z$ where $a_k\in\Bbb Q$ and $pa_{k+1}-a_k\in\Bbb Z$.
The image of $\text{Hom}(A,\Bbb Q)$ consists of those $f$ where $f(1/p^k)=a/p^k$
for some $a\in\Bbb Q$. We can subtract one of these from general $f$ and
assume that $a_0=0+\Bbb Z$ and then still $pa_{k+1}-a_k\in\Bbb Z$. Then $a_k=b_k/p^k$
where $b_k\in\Bbb Z$ and $b_k$ is defined modulo $p^k$; also $b_{k+1}\equiv b_k
\pmod{p^k}$. Thus the $(b_k)$ represents an element $b$ of the $p$-adic integers $\Bbb Z_p$.
We still have some freedom in choosing $a$; we need $a+\Bbb Z=f(0)+\Bbb Z$, so we
can still change $a$ by an integer, which changes $b$ by an integer. So the cokernel
is isomorphic to $\Bbb Z_p/\Bbb Z$.
I'm sure all of this can be done by direct and inverse limits....
A: $\newcommand\ZZ{\mathbb{Z}}$
$\DeclareMathOperator\Hom{Hom}$
$\DeclareMathOperator\Ext{Ext}$
Another approach using a free resolution of $\ZZ[1/p]$.
First recall that we have a ring isomorphism:
$$\mathbb Z[1/p]\cong\frac{\mathbb Z[x]}{(px-1)\mathbb Z[x]}$$
where $x$ is an indeterminate.
Then we get the following exact sequence:
$$\{0\}\to\mathbb Z[x]\to\mathbb Z[x]\to\mathbb Z[1/p]\to\{0\}$$
where the map $\mathbb Z[x]\to\mathbb Z[x]$ is the multiplication by $px-1$.
This is, in fact, a free resolution of $\ZZ[1/p]$.
The we get the exact sequence:
$$\Hom(\ZZ[x],\ZZ)\to\Hom(\ZZ[x],\ZZ)\to\Ext(\ZZ[1/p],\ZZ)\to\{0\}$$
We claim that also the following sequence is exact:
$$\Hom(\ZZ[x],\ZZ)\to\Hom(\ZZ[x],\ZZ)\xrightarrow\zeta\hat\ZZ_p/\ZZ\to\{0\}$$
Here $\zeta$ is esplicity given by
\begin{align}
\zeta&:\Hom(\ZZ[x],\ZZ)\to\hat\ZZ_p/\ZZ&
\psi&\mapsto\sum_{n=0}^\infty\psi(x^n)p^n+\ZZ
\end{align}
If $\psi\in\Hom(\ZZ[x],\ZZ)$ belongs to the image of $\Hom(\ZZ[x],\ZZ)\to\Hom(\ZZ[x],\ZZ)$, then there exists $\varphi\in\Hom(\ZZ[x],\ZZ)$ such that $\psi(f)=\varphi((xp-1)f)$
for every polynomial $f\in\ZZ[x]$.
In particular, for every $k\in\mathbb N$, we have:
$$\psi(x^k)=p\varphi(x^{k+1})-\varphi(x^k)\tag{1}$$
Consequently, we obtain:
$$\sum_{k=0}^{n-1}\psi(x^k)p^k=p^n\varphi(x^n)-\varphi(1)\xrightarrow{n\to\infty}-\varphi(1)$$
in $\hat\ZZ_p$, hence $\psi\in\operatorname{Ker}\zeta$.
Conversely, assume $\psi\in\operatorname{Ker}\zeta$.
We claim that $\psi$ is the image of $\varphi\in\Hom(\ZZ[x],\ZZ)$ satisfying:
$$\varphi(x^k)=\sum_{n=k}\psi(x^n)p^{n-k}$$
For every $k\in\mathbb N$.
By assumption, $\varphi(1)\in\ZZ$.
Then $\varphi(x^k)\in\ZZ$ for every $k\in\mathbb N$ follows by induction on $k$, and since $(1)$ is clearly satisfied, this proves the assertion.
