Past exam of the logic and set theory course of the University of Oxford 2008 A 3 c Solution 2026-10-31
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 ; an inverse function symbol is unnecessary because inverses are asserted to exist by a quantified formula.
For each positive integer , let be the first-order term recursively defined by and . 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.
Past exam of the logic and set theory course of the University of Oxford 2008 A 3 d Solution 2026-10-31
No finite axiomatization exists. Write for the axioms just constructed and suppose a finite set of first-order sentences had exactly the same first-order models. Its logical conjunction is a single first-order sentence, and . By the compactness theorem there is a finite subset with : otherwise would be finitely satisfiable and hence satisfiable, a contradiction.
Only finitely many of the integers in the torsion-free scheme occur in . Choose a prime number larger than all of them. The cyclic group satisfies all the group axioms and each of these finitely many conditions from the torsion-free abelian group axiom scheme: multiplication by an integer is invertible modulo , so implies . It consequently satisfies , then , and therefore .
But its nonzero elements have finite order , so it is not a torsion-free group, contradicting the assumed description of the first-order models of . 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).