If $\text{Ext}^1(\mathbb{Q}/ \mathbb{Z}, D ) = 0$ then $D$ is divisible

Question

This is Exercise 7.15(ii) from Rotman's book, Introduction to homological algebra that I'm doing.

If $D$ is an abelian group and $\text{Ext}^1(\mathbb{Q}/ \mathbb{Z}, D ) = 0$, prove that $D$ is divisible.

Does it hold if we replace $\mathbb{Z}$ by a domain $R$ and $\mathbb{Q}/ \mathbb{Z}$ by Frac$(R) / R$?

My toughts: If $\text{Ext}^1(\mathbb{Q}/ \mathbb{Z}, D ) = 0$ then every exact sequence $$0 \to D \to E \to \mathbb{Q}/ \mathbb{Z} \to 0$$ splits, i.e. $E \cong D \oplus \mathbb{Q}/ \mathbb{Z}$.

How to proceed?

Answer 1

For every group $D$, $\def\Hom{\operatorname{Hom}}\Hom(\mathbb{Q},D)$ is a divisible group, because it is a vector space over $\mathbb{Q}$. Since $D\cong\Hom(\mathbb{Z},D)$, the piece of the long exact sequence $$\def\Q{\mathbb{Q}}\def\Z{\mathbb{Z}} \Hom(\Q,D)\to\Hom(\Z,D)\to\operatorname{Ext}^1(\Q/\Z,D) $$ shows that $D$ is an epimorphic image of a divisible group.

If $R$ is a domain and $Q$ is its field of fractions, the same reasoning holds, because again $\Hom_R(Q,D)$ can be given the structure of $Q$-vector space.

Answer 2

Following the suggestion of Jeremy Rickard:

I consider the exact sequence $$0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q} / \mathbb{Z} \to 0 $$

Then we obtain the long exact sequence $$0 \to \hom(\mathbb{Q} /\mathbb{Z},D) \to \hom(\mathbb{Q}, D) \to \hom( \mathbb{Z}, D) \to \text{Ext}_1(\mathbb{Q}/ \mathbb{Z}, D ) \to \ldots $$ i.e. by the hypotheses the following sequence is exact $$0 \to \hom(\mathbb{Q} /\mathbb{Z},D) \to \hom(\mathbb{Q}, D) \to \hom( \mathbb{Z}, D) \to 0$$ Thus let $d \in D $ , $n \in \mathbb{N}$, consider $$\phi \in \hom( \mathbb{Z}, D)$$ $$\phi(1) = d$$ Then due to the exactness of the previous sequence we can extend $\phi $ to $\mathbb{Q}$, and so $$d = \phi(1) = \phi(n \cdot \frac{1}{n})= n \cdot \phi(\frac{1}{n}) = ng$$ with $g =\phi(\frac{1}{n})$.

This implies that $D$ is divisible.