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 $p$?



Resolutions of $\mathbb{Z}_{(p)}$ as $\mathbb{Z}$-module
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. 


