TITLE: Extension of group with Ext$^{1} (A, B) = 0.$
QUESTION [6 upvotes]: Are there any infinite torsion free abelian groups $A$ and $B,$ with $A$ is not projective and $B$ is not divisible but  $$\text{Ext} ^{1}(A, B) = 0.$$ Thanks

REPLY [4 votes]: Let $\mathbb{Z}_p$ denote the $p$-adic integers. Then 

$$Ext(\mathbb{Q},\mathbb{Z}_p)=0$$

satisfies the requirements in the question. Before giving the proof, let's put the example into a broader context: An abelian group $G$ is called cotorsion, if $Ext(F,G)=0$ for all torsion-free abelian groups $F$ (actually this is equivalent to $Ext(\mathbb{Q},G)=0)$. It turns out that the torsion-free cotorsion groups are just the algebraically compact groups [1, Corollary 54.5]. These groups can be completely classified [1, Proposition 40.1, Theorem 40.2]. 
Proof: [1, §38 Ex. 2] In the following I'll give a self-contained proof. 
By the short exact seq. $0 \to \mathbb{Z}_p \to \mathbb{Q}_p \xrightarrow[]{\kappa} \mathbb{Q}_p/\mathbb{Z}_p \to 0$, it's enough to show  the surjectivity of $\kappa_\ast:Hom(\mathbb{Q},\mathbb{Q}_p) \to Hom(\mathbb{Q},\mathbb{Q}_p/\mathbb{Z}_p)$. 
For a hom. $f: \mathbb{Q} \to \mathbb{Q}_p/\mathbb{Z}_p$ let $a_n\in\mathbb{Q}_p$ be a representative of $f(1/p^ n), \,\,n\ge 0$. By additivity of $f$, $a_n - p^{m-n}a_m \in \mathbb{Z}_p$ for $m \ge n$. 
Let $|\cdot|_p$ denote the $p$-adic value. Using $\mathbb{Z}_p=\{ a \in \mathbb{Q}_p\mid\,|a|_p \le 1\,\}$ and $|p^ n|_p  = p^{-n}$, we find 
$$|p^ na_n - p^ ma_m|_p \le \frac{1}{p^n}\tag{1}$$
Hence $(p^na_n)$ is a Cauchy sequence in $\mathbb{Q}_p$ and since $\mathbb{Q}_p$ is complete w.r.t. $|\cdot|_p$, there is $a = \lim_n\; p^na_n\in \mathbb{Q}_p$. Taking the limit $m\to \infty$ in $(1)$ for fixed $n$ yields $|p^na_n -a|_p \le 1/p^n$ for all $n \ge 0$. Hence 
$$a_n - \frac{a}{p^n} \in \mathbb{Z}_p\quad(n \ge 0),\tag{2}$$
showing that $f$ agrees with $\kappa_\ast(g)$ for $g: \mathbb{Q} \to \mathbb{Q}_p,\,\, q \mapsto qa$ on $1/p^n,\,n\ge 0$. But as $f$ is uniquely determined by these values, $\kappa_\ast(g)=f$ follows. qed
[1] L. Fuchs, Infinite Abelian Groups I.