TITLE: Resolutions of $\mathbb{Z}_{(p)}$ as $\mathbb{Z}$-module
QUESTION [5 upvotes]: Are there any interesting canonical (maybe unbounded) projective resolutions of $\mathbb{Z}_{(p)}$ over $\mathbb{Z}$, for instance by tensoring together all the $\mathbb{Z}[x] \stackrel{qx-1}\to \mathbb{Z}[x]$ for all primes $q$ different from $p$?

REPLY [6 votes]: Put $a_n=p^{n!}-1$.  It is easy to see that $a_n$ divides $a_{n+1}$, so we can define $b_n=a_{n+1}/a_n\in\mathbb{N}$.  Put $P=\bigoplus_n\mathbb{Z}$ and define $f\colon P\to P$ and $g\colon P\to\mathbb{Z}_{(p)}$ by $f(e_n)=b_ne_{n+1}-e_n$ and $g(e_n)=1/a_n$.  Then $f$ is injective and $g\circ f=0$ and the induced map $\text{cok}(f)\to\mathbb{Z}_{(p)}$ is an isomorphism, so we have a two-stage projective resolution of $\mathbb{Z}_{(p)}$.  The key point here is that ($m$ divides $a_n$ for some $n$) iff (some power of $p$ is one in $\mathbb{Z}/m$) iff ($p$ is a unit in $\mathbb{Z}/m$) iff $m$ is coprime with $p$.  All this works even when $p$ is composite, provided that we interpret $\mathbb{Z}_{(p)}$ as the ring of rationals with denominator coprime to $p$.
UPDATE: Here is another version with some advantages.  Put 
$$ c_n=\prod_{i=1}^n(p^i-1) = |GL_n(\mathbb{Z}/p)|/p^{n(n-1)/2}, $$
and let $R$ be the subgroup of $\mathbb{Q}[x]$ with basis $\{x^n/c_n\;:\;n\in\mathbb{N}\}$ over $\mathbb{Z}$.  Using the fact that $GL_n\times GL_m$ embeds in $GL_{n+m}$, we can see that $c_nc_m$ divides $c_{n+m}$,so $R$ is a subring of $\mathbb{Q}[x]$ containing $\mathbb{Z}[x]$.  There is a surjective ring map $h_1\colon\mathbb{Q}[x]\to\mathbb{Q}$ given by $h_1(x)=1$, and this restricts to give a surjective ring map $h\colon R\to\mathbb{Z}_{(p)}$.  One can check that $x-1$ is a regular element of $R$ that generates $\ker(h)$ as an ideal in $R$, so $R\xrightarrow{x-1}R\xrightarrow{h}\mathbb{Z}_{(p)}$ is a projective resolution of $\mathbb{Z}_{(p)}$ over $\mathbb{Z}$.  This can be regarded as a Koszul resolution and so has a multiplicative structure.