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(1) Ext$^n_\mathbb{Z}(M, \mathbb{Q})=0$ for every module $M$

follows easily from the fact that

(2) $\mathbb{Q}$ is injective.

However, the only proof I have seen of the injectivity of $\mathbb{Q}$ relies on Baer's Criterion. While the proof of Baer's Criterion is not difficult, it seems stronger than the injectivity of $\mathbb{Q}$ (for example, the proof uses Zorn's Lemma). Is there a different (and preferably simpler) proof of (1) or (2)?

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If there is a good reason why Baer's Criterion is the most direct way to approach this problem and there is no reason to hope for a simpler proof, please let me know too. – Vitaly Lorman Feb 8 '11 at 0:31
up vote 6 down vote accepted

Let $M$ be an abelian group and consider an extension $$0\to\mathbb Q\to E\to M\to0$$ of abelian groups. Since $\mathbb Q$ is flat, there is an induced exact sequence $$0\to\mathbb Q\otimes_{\mathbb Z}\mathbb Q\to \mathbb Q\otimes_{\mathbb Z} E\to Q\otimes_{\mathbb Z} M\to 0$$ The latter, being a short exact sequence of rational vector spaces, splits and there is a map $\mathbb Q\otimes_{\mathbb Z}E\to\mathbb Q\otimes_{\mathbb Z}\mathbb Q$ such that the composition $$\mathbb Q\otimes_{\mathbb Z}\mathbb Q\to\mathbb Q\otimes_{\mathbb Z}E\to\mathbb Q\otimes_{\mathbb Z}\mathbb Q$$ is the identity. Now the composition $$E\to \mathbb Q\otimes_{\mathbb Z}E\to\mathbb Q\otimes_{\mathbb Z}\mathbb Q\to\mathbb Q$$ with the first and the last maps being the obvious morphisms, splits the original extension.

It follows that $\mathrm{Ext}_{\mathbb Z}^1(M,\mathbb Q)=0$.

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That's funny. I posted the exact same thing at almost the exact same time. – Vitaly Lorman Feb 9 '11 at 0:54
@Vitaly: while the two arguments are of course related, they also pretty different, I think. In particular I'm using the definition of Ext using extensions, while you use resolutions. – Mariano Suárez-Alvarez Feb 9 '11 at 3:00
Yes, I didn't notice, but you're right. Tensoring with $\mathbb{Q}$ to work with vector spaces is what they have in common. – Vitaly Lorman Feb 9 '11 at 17:19

The main use of Baer's criterion in this case is that injective modules over $\mathbb{Z}$ are the same as divisible groups. Since $Ext$ is closely related to injectivity, there isn't much reason to avoid a clean description of injective modules (via Baer's criterion).

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Sure, in this case Baer's criterion gives a quick and clear answer, but I wonder if there isn't a way to get from divisibility in the P.I.D case to injectivity without using the Zorn's lemma argument necessary to prove Baer's criterion. Maybe I am making more out of using that argument than is there, but it would be nice if some elementary properties of $\mathbb{Q}$ as the field of fractions of a P.I.D. could lead directly to the construction of the lift. – Vitaly Lorman Feb 8 '11 at 1:29
As for (1), perhaps the characterization of $\mathbb{Q}$ as the localization of $\mathbb{Z}$ or as a limit of localizations could be helpful. Direct limits can't be pulled out of the second coordinate of Ext in general, but I was wondering if there is some way around this. Also, (1) could perhaps be proved by giving some projective resolution of $\mathbb{Q}$, although this is probably hard. – Vitaly Lorman Feb 8 '11 at 1:33
oh i see, so you want to avoid Zorn's lemma? – Soarer Feb 8 '11 at 1:55
Yeah, a more constructive proof would be nice, if one exists. To me, there is no obvious simplification of the proof of Baer's criterion one can make even if one knows a priori that the module is $\mathbb{Q}$ and not just any module that has the property in the statement of Baer's criterion. – Vitaly Lorman Feb 8 '11 at 2:14

Well, I'm not sure that this is what you really want.

The injective hull of an integral domain is its quotient field. See Lam "Lectures on Modules and Rings", Example 3.35

Hence, $\mathbb{Q}$ is injective because it is injective hull of an integral domain $\mathbb{Z}$.

Anyway, isn't Baer's crieterion is equivalent to the injectivity?

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Thanks, I will look this up. Baer's criterion is equivalent to injectivity, but in the case of some modules, it ought to be possible to construct a lift explicitly. I am wondering if this is the case for $\mathbb{Q}$. – Vitaly Lorman Feb 8 '11 at 2:20

I got an answer from a professor in person today, so I will post it. Please let me know if something is wrong with this argument, or if you don't think it is a more direct proof of (2) than using Baer's criterion. Since $\mathbb{Z}$ is a P.I.D., we may take a two stage projective (and free) resolution of any module $M$: $$0 \rightarrow P_1 \rightarrow P_0 \rightarrow M \rightarrow 0$$ Suppose we have a homomorphism $\phi: P_1 \rightarrow \mathbb{Q}$. We will tensor everything with $\mathbb{Q}$. We require the fact that $\mathbb{Q}$ is flat, but it seems to me that the proof of this fact is more basic than the proof of Baer's criterion. After tensoring, we have the following exact sequence $$0 \rightarrow P_1 \otimes_\mathbb{Z} \mathbb{Q} \rightarrow P_0 \otimes_\mathbb{Z} \mathbb{Q} \rightarrow ...$$ With a homomorphism $P_1 \otimes_\mathbb{Z} \mathbb{Q} \rightarrow \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q} \cong \mathbb{Q}$. Now treating $P_1 \otimes_\mathbb{Z} \mathbb{Q}$ as a $\mathbb{Q}$-subspace of $P_0 \otimes_\mathbb{Z} \mathbb{Q}$, we have a homomorphism on the subspace, which may then be extended on the whole space (just take a basis for the subspace, and extend it to the whole space, sending all of the nonsubspace basis vectors to 0). We have thus found the map $P_0 \rightarrow P_0 \otimes_\mathbb{Z} \mathbb{Q}\rightarrow \mathbb{Q}$ that we were looking for.

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