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If $T$ has algebraically prime models and every inclusion between models of $T$ is a simple closure, then $T$ has quantifier elimination. For a common base $A$, embed its algebraically prime extension into both models. Simple closure transfers a witness into that extension; the second embedding transfers it to the other model.

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.

Both assertions are false. For simple closure, take the inclusionBoth satisfy DAG prime, but the quantifier-free formula $x\ne 0$ has a witness in $N$ and none in $M$. The same example works in the reduced language $\{+,0\}$.

For quantifier elimination, the first-order sentence $\exists x(x\ne 0)$ distinguishes these two models. Every closed group term is zero, so every atomic closed equality is true in both models. Every quantifier-free sentence, being a Boolean combination of such equalities, has the same truth value in both. The distinguishing sentence therefore has no equivalent quantifier-free sentence modulo DAG prime. HenceIncluding the trivial group is exactly what makes the proposed simple closure condition fail.

Remove the nonzero-model axiom from DAG. The inclusion $\{0\}\subseteq \mathbb{Q}$ is then an inclusion of models but is not a simple closure, because $x\ne 0$ has a witness only in the larger model. The theory does not have quantifier elimination: all closed group terms are zero, so no quantifier-free sentence distinguishes the two models.