Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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,


share|cite|improve this question
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
up vote 3 down vote accepted

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 $\mathbb Z$:

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.