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For a first-order theory $T$ and $A\models T_{\forall }$, an algebraically prime extension is a structure embedding $i:A\to P\models T$ such that every embedding $j:A\to N\models T$ factors as $j=h\circ i$ for some structure embedding $h:P\to N$. The embeddings need not be elementary or unique. A theory has algebraically prime models when every such $A$ has an extension of this kind.

An embedding of first-order structures is existential when every existential formula with parameters from the source has the same truth value in the source and target. Forward preservation follows from being a structure embedding; reflection is the additional requirement. Applying compactness theorem to the target's diagram of a structure and the source's elementary diagram places the target inside an elementary extension of the source.

An existential formula is an existential block followed by a quantifier-free formula, with remaining free variables serving as parameters. Truth is preserved by structure embeddings. Reflection of truth for every such formula defines an existential embedding.

Name every element $a\in A$ by a new constant $c_a$, distinct from the symbols already in $L^{\prime }$. Let $\mathrm{Diag}(A)$ be its full atomic and negated atomic diagram of a structure. A model of this diagram contains an isomorphic copy of $A$ as an $L$-substructure. We claimis satisfiable.

Otherwise compactness theorem gives a finite diagram conjunction $\delta (c_{a_1},\dots ,c_{a_m})$ inconsistent with $T^{\prime }$. Replace these new constants by variables. Since $T^{\prime }$ says nothing about the new names, it then entails the universal sentenceThe conjunction $\delta$ uses only symbols of $L$, so this is one of the universal $L$-consequences assumed true in $A$. But $A$ satisfies $\delta (a_1,\dots ,a_m)$, a contradiction. A model of the consistent enlarged theory gives an embedding of $A$ into the $L$-reduct of a model of $T^{\prime }$. Negated atomic facts and disequalities in the diagram ensure that this is a structure embedding, not merely a homomorphism.

Let $U$ be all universal sentences true in $C$. The theory $T\cup U$ is consistent. Indeed, if a finite conjunction $u$ of sentences from $U$ were inconsistent with $T$, then $T\models \neg u$. A finite conjunction of universal sentences is universal, so its negation is equivalent to an existential sentence. By the hypothesis $C$ would satisfy $\neg u$, whereas it satisfies $u$.

Choose $A\models T\cup U$. Now expand the language by names for every element of $C$ and consider its elementary diagram $\mathrm{ElDiag}(C)$. Its universal consequences in the original language are exactly $U$: any such consequence is true in the named copy of $C$, while any original-language sentence true in $C$ already belongs to its elementary diagram. Part (b), applied to $A$ and this expanded theory, gives a structure embedding of $A$ into a model $D$ of $\mathrm{ElDiag}(C)$. The named copy of $C$ is an elementary substructure of $D$. Identifying that copy with $C$, we obtainThus the required target is an elementary extension of $C$, although $C$ itself need not be a model of $T$.

A 1-embedding, or existential embedding, is a structure embedding $\alpha :A\to B$ satisfyingfor every quantifier-free $\theta$ and every parameter tuple $\mathbf{a}$ from $A$. The forward direction is automatic for an embedding; reflection of existential formulas is the extra property. For an inclusion this says that $A$ is simply closed in $B$. This is weaker than requiring an elementary embedding, which preserves all formulas.

Let $T_{\forall }$ denote the universal consequences of a theory: all universal first-order sentences entailed by $T$. A first-order structure satisfies $T_{\forall }$ exactly when it embeds into a model of $T$, by the compactness theorem applied to its diagram of a structure.

The theory has algebraically prime models if, for every $\mathcal{A}\models T_{\forall }$, there are $\mathcal{P}\models T$ and a structure embedding $i:\mathcal{A}\to \mathcal{P}$ such that every structure embedding $j:\mathcal{A}\to \mathcal{N}$, $\mathcal{N}\models T$, factors as $j=h\circ i$ for some structure embedding $h:\mathcal{P}\to \mathcal{N}$. Neither $i$ nor $h$ is required to be elementary.

For $\mathcal{M}\subseteq \mathcal{N}$, simple closure means that every existential quantifier-free formula over $M$ which has a witness in $N$ has one in $M$:Now take two models $\mathcal{M},\mathcal{N}\models T$ with common substructure $\mathcal{A}$. Since $A$ embeds into $M$, it satisfies $T_{\forall }$. Choose its algebraically prime extension $\mathcal{P}$, and embed $P$ into both $M$ and $N$ over $A$.

If $\exists x\theta (x,\bar{a})$ holds in $M$, the image of $P$ in $M$ is a model of $T$. The assumed simple closure of this image transfers a witness from $M$ into $P$. Its embedding into $N$ then transfers the quantifier-free formula and its witness into $N$. Thus the hypothesis of QET1 is satisfied. The second test follows: $T$ has quantifier elimination.

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 $a-b$, withThe 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 hullConcretely 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 $b\ne 0$; the map $q\mapsto qb$ embeds $P$ into $N$ over zero. Therefore DAG has algebraically prime models, including over the trivial base.

A first-order theory $T$ is model-complete if every structure embedding between models of $T$ is an elementary embedding. Equivalently, whenever $\mathcal{M}\subseteq \mathcal{N}$ and both satisfy $T$, one has $\mathcal{M}\preccurlyeq \mathcal{N}$:for every first-order formula $\phi$ and every finite tuple $\bar{a}\in M$. This is preservation of all formulas with parameters, rather than merely all first-order sentences.

Assume $T$ has quantifier elimination, and let $j:\mathcal{M}\to \mathcal{N}$ be a structure embedding between its models. For any first-order formula $\phi (\bar{x})$, choose a quantifier-free formula $\psi$ equivalent to it modulo $T$. A structure embedding preserves and reflects atomic formulas; induction through the Boolean connectives therefore preserves every quantifier-free formula. ThusThe embedding is elementary. Every theory with quantifier elimination is model-complete.

The universal consequences $T_{\forall }$ of a first-order theory $T$ are all universal first-order sentences entailed by $T$. Their model class is exactly the class of substructures of models of $T$, up to structure embedding. This characterization follows from the diagram embedding criterion for universal theories.