Past exam of the logic and set theory course of the University of Oxford / 2008 / A / 3

An $L$-structure consists of a nonempty domain $A$, an element $c^{\mathcal{A}}\in A$ for each constant symbol, a function $f^{\mathcal{A}}:A^n\to A$ for each $n$-ary function symbol, and a relation $R^{\mathcal{A}}\subseteq A^n$ for each $n$-ary predicate symbol. When equality is part of the logic it is interpreted as actual identity on $A$.

An assignment in a first-order structure is a function $s$ from logical variables to $A$. It induces a term valuation of first-order terms recursively:For an atomic formula, substitute these first-order term values into the interpreted relation, or compare them for equality. Extend satisfaction for a set structure through the truth tables of the logical connectives, and through the quantifier clausesThus the first-order structure and assignment in a first-order structure together determine truth values for all logical formulas. A first-order sentence has no free variables, so its truth does not depend on the assignment in a first-order structure.

The assertion that $\mathcal{A}$ is a first-order model of $\Gamma$ means that every first-order sentence belonging to $\Gamma$ is true in the first-order structure:Equivalently each first-order sentence is true under every assignment in a first-order structure, since first-order sentences have no free variables. The first-order sentences must all hold in the same interpretation of the language; one cannot select a different first-order structure for each member of $\Gamma$.

Use the following group axioms in the given language:Commutativity supplies the other identity and inverse equations. These first-order sentences therefore specify precisely an abelian group with the named identity $0$; an inverse function symbol is unnecessary because inverses are asserted to exist by a quantified formula.

For each positive integer $n$, let $t_n(x)$ be the first-order term recursively defined by $t_1(x)=x$ and $t_{n+1}(x)=t_n(x)+x$. Add the torsion-free abelian group axiom schemeEvery first-order model of these first-order sentences is an abelian group with no nonidentity element of finite order. Conversely every torsion-free abelian group satisfies each of the displayed first-order sentences. This is a countably infinite first-order axiomatization of exactly the required class.

No finite axiomatization exists. Write $\Gamma$ for the axioms just constructed and suppose a finite set $\Delta$ of first-order sentences had exactly the same first-order models. Its logical conjunction $\delta$ is a single first-order sentence, and $\Gamma \models \delta$. By the compactness theorem there is a finite subset ${\Gamma }_0\subseteq \Gamma$ with ${\Gamma }_0\models \delta$: otherwise $\Gamma \cup \{\neg \delta \}$ would be finitely satisfiable and hence satisfiable, a contradiction.

Only finitely many of the integers $n$ in the torsion-free scheme occur in ${\Gamma }_0$. Choose a prime number $p$ larger than all of them. The cyclic group $\mathbb{Z}/p\mathbb{Z}$ satisfies all the group axioms and each of these finitely many conditions from the torsion-free abelian group axiom scheme: multiplication by an integer $1\le n<p$ is invertible modulo $p$, so $nx=0$ implies $x=0$. It consequently satisfies ${\Gamma }_0$, then $\delta$, and therefore $\Delta$.

But its nonzero elements have finite order $p$, so it is not a torsion-free group, contradicting the assumed description of the first-order models of $\Delta$. This establishes the nonfinite axiomatizability of torsion-free abelian groups, ruling out every finite set of defining first-order sentences, not just finite subsets of the particular scheme in part (c).