Proof that Ext$^n_\mathbb{Z}(M, \mathbb{Q})=0$ and Baer's Criterion That   
(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)?
 A: 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 $\mathrm{Ext}_{\mathbb Z}^1(M,\mathbb Q)=0$.
A: 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).
A: 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?
A: 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.  
