# Projective Resolution of $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$

The question I have is on how to calculate Projective Resolution of $\mathbb{Q}$ over $\mathbb{Z}$. I have found that $\mathbb{Q}$ is flat, but that 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 the $Ext_{\mathbb{Z}}^n(\mathbb{Q},B) \text{ respectively } Ext_{\mathbb{Z}}^n(\mathbb{Q/Z},B)$ for some arbitrary B module over $\mathbb{Z}$

Now I know how to calculate the rest of the steps i.e. $Hom_{\mathbb{Z}}(_-,B)(P)$ 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

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Check for a typo in your Ext. – Jack Schmidt Jul 10 '13 at 22:36
I don't understand? If you want to calculate $\mbox{Ext}_\mathbb{Z}^n(\mathbb{Z}; B)$, why don't you construct a projective resolution for $\mathbb{Z}$? Since $\mathbb{Z}$ is a free $\mathbb{Z}-$module (every unitary ring $R$ can be consider a free module over itself, with base $\{ 1 \}$), hence $\mathbb{Z}-$projective. So you can choose the 'trivial' projective resolution: $0 \to 0 \to ... \to 0 \to \mathbb{Z} \xrightarrow{1_{\mathbb{Z}}} \mathbb{Z} \to 0$, shouldn't it be easier, no? :) – user49685 Jul 12 '13 at 16:05
@user49685: as Jack Schmidt noticed there is probably a typo and what Relativeo whants to compute is $Ext^n(B,\mathbb Z)$. Interpreting litteraly the question, your comment completely answers the question... – Simone Jul 12 '13 at 16:25
Uhm, I think I get what you mean. Thanks. The question is not very clear itself, even if it reads as $\text{Ext}(\mathbb{Z}; B)$, or $\text{Ext}(B; \mathbb{Z})$, projective resolutions for $\mathbb{Q}$, and $\mathbb{Q} / \mathbb{Z}$ are not very relevant at all. – user49685 Jul 12 '13 at 17:22
yep, you are right... – Simone Jul 13 '13 at 7:06

## 1 Answer

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 resoluzion 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 $Ext$, you can notice by the above argument that higer ext's are always trivial on hereditary rings...

Notice also that it is quite easy to find an injective resolution for $\mathbb Z$:

$$0\to \mathbb Z\to \mathbb Q\to \mathbb Q/\mathbb Z\to 0$$

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I see that $0\to \mathbb Z\to \mathbb Q\to \mathbb Q/\mathbb Z\to 0$ is an injective resolution but am unsure as to why I can take the injective resolution instead of the projective resolution of $\mathbb Q/\mathbb Z$. – Relative0 Jul 11 '13 at 10:04
Well, in principle $Hom_{\mathbb Z}(-,-)$ is a functor from $Ab\times Ab$ to $Ab$, in particular it sends an ordered pair $(A,B)$ of Abelian groups to an Abelian group $Hom(A,B)$. Now, in order to construct its derived functors you just fix one of the two entries (say $A$) obtaining a functor $Hom(A,-):Ab\to Ab$. This gives you a left exact functor, to compute its derived functors you have to take an injective resolution of $B$ and proceed as you probably know. Of course there is another possibility, that is, fixing $B$ you have a right exact contravariant functor $Hom(-,B):Ab\to Ab$. [...] – Simone Jul 12 '13 at 9:57
To construct the derived functors of this second functor you shuold construct a projective resolution of $A$ and then proceed as you know. Now, it is a standard result (but not completely trivial... you should check it) that the two constructions give you the same out-put when you use them to construct $Ext^{n}(A,B)$. As Jack Schmidt was telling you in his comment, there is a mistake in your question, in the sense that, being $\mathbb Z$ a projective object, the $Ext$-groups $Ext^n(\mathbb Z,B)$ that you want to construct are always trivial [...] – Simone Jul 12 '13 at 10:01
(a part $n=0$ when $Ext^0(\mathbb Z,B)=Hom(\mathbb Z,B)\cong B$). The interesting thing is to compute $Hom(B,\mathbb Z)$, for which you can use the injective resolution I have written in my answer. – Simone Jul 12 '13 at 10:02