Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 120 / 2

For a first-order theory $T$ in a first-order language $L$, a complete $n$-type is a maximal $T$-consistent set $p(x_1,\dots ,x_n)$ of formulas whose free variables lie among $x_1,\dots ,x_n$. Equivalently, it chooses exactly one of $\phi$ and $\neg \phi$ for every such formula while remaining consistent with $T$.

An isolated type $p$ is isolated by a formula $\theta (\bar{x})$ when $T\cup \{\exists \bar{x}\theta (\bar{x})\}$ is consistent andfor every $\phi \in p$. The type is an omitted type in an $L$-structure $M$ when no tuple $\bar{a}\in M^n$ satisfies every formula in $p$.

The omitting types theorem states that if $T$ is a consistent theory in a countable language and $(p_i{)}_{i\in \mathbb{N}}$ is a countable family of nonisolated finite-arity types, then $T$ has a countable model omitting every $p_i$.

Let $\mathcal{U}$ be a nonprincipal ultrafilter on an infinite set $I$. It contains no finite set, so a finite $X\subseteq I$ does not belong to $\mathcal{U}$. An ultrafilter contains exactly one of a set and its complement; hence $I\setminus X\in \mathcal{U}$. Equivalently, every nonprincipal ultrafilter contains the cofinite filter.

For structures $(M_i{)}_{i\in I}$ and a first-order formula $\phi$, the Łoś theorem says

Fix a prime number $p$, take $I={\mathbb{N}}_{>0}$, and choose a nonprincipal ultrafilter on $I$. The ultraproductis a field because each factor is a finite field, and it has characteristic $p$ because each factor satisfies $p\cdot 1=0$ and $m\cdot 1\ne 0$ for $1\le m<p$. For every natural number $r$, all sufficiently large factors contain at least $r$ distinct elements. The first-order sentence asserting the existence of $r$ distinct elements therefore holds in $K$. Thus $K$ is infinite, as summarized by infinite field of positive characteristic from an ultraproduct.

The Ehrenfeucht-Mostowski theorem says that if a first-order theory $T$ has an infinite model, then for every total order $I$ there is a model $M\models T$ generated as the Skolem hull of distinct order indiscernibles $(a_i{)}_{i\in I}$, and every order automorphism of $I$ extends to an automorphism of $M$.

Given an infinite cardinal $\kappa$, let $I=\kappa \times \mathbb{Q}$ with the lexicographic order, viewed as $\kappa$ consecutive copies of the rational order. In each copy independently choose either the identity or a fixed nonidentity order automorphism of $\mathbb{Q}$. These choices give $2^{\kappa }$ distinct order automorphisms of $I$.

Apply the theorem to this order. Distinct order automorphisms act differently on the distinct generators $a_i$, so their extensions give an injection into the automorphism group of a first-order structure $\mathrm{Aut}(M)$. Therefore $|\mathrm{Aut}(M)|\ge 2^{\kappa }$, proving that $T$ has models with arbitrarily large automorphism groups.