The question I have is on how to find a Projective Resolution of $\mathbb{Q}$ over $\mathbb{Z}$.
I have found that $\mathbb{Q}$ is flat, but it is not projective. I mention this as I wonder if there is some sort of trick I can use knowing that it is flat.
I am ultimately wanting to figure out $\mathrm{Ext}_{\mathbb{Z}}^n(\mathbb{Q},B)$, respectively $\mathrm{Ext}_{\mathbb{Z}}^n(\mathbb{Q/Z},B)$ for some arbitrary $\mathbb{Z}$-module $B$.
Now I know how to calculate the rest of the steps i.e. $\mathrm{Hom}_{\mathbb{Z}}(_-,B)(P)$ where $P$ is the projective resolution of $\mathbb{Q}$ or $\mathbb{Q/\mathbb{Z}}$, but could use any tricks for getting the resolutions.
Thanks in advance,
Brian
The fact is that $\mathbb Z$ is a hereditary ring, that is, submodules of projectives are still projectives (or, equivalently, quotients of injectives are still injective). Thus for finding a projective resolution of $\mathbb Q$ you can proceed as follows:
(1) take a surjection $f:\mathbb Z^{(\mathbb Q)}\rightarrow \mathbb Q$;
(2) the kernel of $f$ is projective as it is a submodule of a direct sum of projectives;
(3) $0\to\ker(f)\to \mathbb Z^{(\mathbb Q)}\to \mathbb Q\to 0$ is a projective resolution.
For computing $\mathrm{Ext}$, you can notice by the above argument that higer ext's are always trivial on hereditary rings...
It is also easy to find an injective resolution for $\mathbb Z$ (and you can use that to compute the $\mathrm{Ext}$-groups):
$$0\to \mathbb Z\to \mathbb Q\to \mathbb Q/\mathbb Z\to 0$$