TITLE: Proof that Ext$^n_\mathbb{Z}(M, \mathbb{Q})=0$ and Baer's Criterion
QUESTION [8 upvotes]: 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)?

REPLY [6 votes]: 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 $\mathrm{Ext}_{\mathbb Z}^1(M,\mathbb Q)=0$.