Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 23 / 2

Here is the existential common-substructure test for quantifier elimination. Use a first-order language with a constant symbol, as in the applications below, or the usual conventions for empty generated substructures and truth constants. Let $T$ be a first-order theory. Suppose that whenever $\mathcal{M},\mathcal{N}\models T$ have a common substructure $\mathcal{A}$, then, for every quantifier-free formula $\theta (x,\bar{y})$ and every finite tuple $\bar{a}\in A$,The condition is symmetric because it applies also with $\mathcal{M}$ and $\mathcal{N}$ exchanged. The conclusion is quantifier elimination for $T$: for every first-order formula $\phi (\bar{y})$ there is a quantifier-free formula $\psi (\bar{y})$ such thatThe common substructure need not itself satisfy $T$, and witnesses are sought in the whole models, not necessarily in $A$. This test is the QET1 version of the common-substructure test for quantifier elimination.

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.

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.