= Solution
A <torsion-free divisible Abelian group> is naturally a <vector space over the rational numbers>: for $m\in\mathbb Z$ and $n>0$, define $(m/n)a$ as the unique $b$ with $nb=ma$. Divisibility supplies existence and torsion-freeness supplies uniqueness.
In the usual group <first-order language> $\{+, -,0\}$, a model $A$ of the universal part of <DAG> is a <torsion-free group> which is Abelian. If the language instead uses only $\{+,0\}$, a <substructure> can be merely a <torsion-free cancellative commutative monoid>. Handle this convention by first taking its <Grothendieck group> $G(A)$: its elements are formal differences $a-b$, with
$$
a-b=a'-b'\quad\Longleftrightarrow\quad a+b'=a'+b.
$$
The <cancellative commutative monoid> condition makes $A\to G(A)$ injective. If $n(a-b)=0$, then $na=nb$, and torsion-freeness gives $a=b$; hence $G(A)$ is a <torsion-free abelian group>. In the full group language simply take $G(A)=A$.
For $A\ne\{0\}$ form its <rational divisible hull>
$$
P=G(A)\otimes_{\mathbb Z}\mathbb Q.
$$
Concretely its elements are fractions $g/n$ with $n>0$, where $g/n=h/m$ if $mg=nh$. The canonical embedding of $A$ into $P$ is injective, and $P$ is nontrivial, divisible, Abelian and torsion-free, so it satisfies <DAG>.
Let $j:A\to N$ be any <structure embedding> into a model of <DAG>. Extend it first to formal differences if necessary. Its unique extension to the <rational divisible hull> sends $g/n$ to the unique element $b\in N$ with $nb=j(g)$. This is a <group homomorphism> fixing the given copy of $A$. It is injective: an element mapped to zero has $j(g)=0$, hence $g=0$. Thus every embedding into a <DAG> model factors through $P$.
The zero case must be treated separately: its <rational divisible hull> is zero and does not satisfy <DAG>. Instead choose $P=(\mathbb Q,+)$. Given any nontrivial <DAG> model $N$, choose $b\ne0$; the map $q\mapsto qb$ embeds $P$ into $N$ over zero. Therefore \b[<DAG> has <algebraically prime models>, including over the trivial base.]
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