Can someone describe explicitly an abelian group $A$ such that the extension $$0 \to \mathbb{Z} \to A \to \mathbb{Q} \to 0$$ doesn't split ? 


Background: The Stein-Serre theorem (Hilton, Stammbach: A course in homol. algebra, Theorem 6.1) states that if $A$ is abelian of countable rank (=maximal number of linear independent elements), then $Ext(A,\mathbb{Z})=0$ implies $A$ free. When applied to $A= \mathbb{Q}$, I conclude $Ext(\mathbb{Q},\mathbb{Z})\neq 0$. Moreover, by interpreting $A \in Ext(\mathbb{Q},\mathbb{Z})$ as extension $0 \to \mathbb{Z} \to A \to \mathbb{Q} \to 0$ of abelian groups, there must be an $A$ such that the extension doesn't split. The problem is that the proof of the theorem isn't constructive and doesn't show how to construct such an $A$. 



Non-split extension of the rationals by the integers 
Nice question. For a prime $p$ let $\mathbb{Z}\_{(p)} = \lbrace \frac{a}{b}\in \mathbb{Q}\mid p \nmid b\rbrace$ and $\mathbb{Z}[p^{-1}] = \lbrace \frac{a}{p^n}\in \mathbb{Q}\mid n \ge 0 \rbrace$. Then


$$A := \lbrace (x,y) \in \mathbb{Q} \times \mathbb{Z}[p^{-1}] \mid x-y \in \mathbb{Z}\_{(p)} \rbrace$$ 
has the desired non-split extension. Informal $A$ consists of all pairs $(x,y) \in \mathbb{Q} \times \mathbb{Q}$ where $y$ is the $p$-part in the partial fraction decomposition of $x$ (up to an integer summand). 


**Proof:** Let $\rho: A \to \mathbb{Q}$ be projection onto the first factor and $i: \mathbb{Z} \hookrightarrow A$ inclusion into the second factor. By partial fraction decomposition of the rationals, $\rho$ is surjective and $$\ker(\rho)=0 \times \big(\mathbb{Z}[p^{-1}] \cap \mathbb{Z}\_{(p)}\big) = 0 \times \mathbb{Z} = \operatorname{im}(i).$$


Next, let $j: \mathbb{Q} \to A$ be a splitting hom. of $\rho$. Composing $j$ with the projection onto the second factor yields a hom. $f: \mathbb{Q} \to \mathbb{Z}[p^{-1}] \le \mathbb{Q}$. Since each endomorphism of $\mathbb{Q}$ is multiplication with some $q \in \mathbb{Q}$ we have $f(x)=qx \in \mathbb{Z}[p^{-1}]$ for all $x \in \mathbb{Q}$ which is only possible for $q=0$. Hence $j(x)=(x,0) \in A$ for all $x \in \mathbb{Q}$. Setting $x=1/p$ we obtain $1/p \in \mathbb{Z}\_{(p)}$ which is the desired contradiction. **QED**


