Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 27 / 9 / Solution

An ultrapower $A^I/\mathcal{U}$ comes with its diagonal elementary embedding $a\mapsto [i\mapsto a]$, by the Łoś theorem. A model-theoretic ultralimit is the direct limit of successive ultrapowers along these diagonal embeddings, with elementary unions at limit stages. It is not a topological ultrafilter limit. We prove the needed embedding lemma with parameters and then build the isomorphism.

If $A\equiv B$, name every element $b\in B$ by a constant $c_b$ and let $D(B)$ be its elementary diagram of a structure: all sentences true in this fully named structure. For each finite $F\subseteq D(B)$, the conjunction mentions finitely many constants. Its existential closure in the original language is true in $B$, hence in $A$. Choose an interpretation of those constants in $A$ making $F$ true, and give unmentioned constants arbitrary values. Index these interpretations by $I=[D(B){]}^{<\omega }$. Extend the filter of finite-diagram cones to an ultrafilter $\mathcal{U}$.

Map each $b$ to the class of its coordinate interpretations. Every diagram formula holds on the cone containing it, so the Łoś theorem says this map preserves every formula, including negated equalities between distinct named elements. It is therefore an elementary embeddingThis is the Frayne embedding lemma. Its relative version is crucial: if $A\prec C$, interpret all constants naming elements of $A$ constantly by those same elements. Every finite fragment of the diagram of $C$ with these fixed parameters is realized in $A$, since its existential closure with parameters from $A$ is true in $C$ and elementarity transfers it to $A$. The same construction embeds $C$ into an ultrapower of $A$ while extending exactly the diagonal embedding of $A$.

Now start with $A\equiv B$. Embed $B$ into an ultrapower $A_1$ of $A$. Apply the relative lemma to $B\prec A_1$ to embed $A_1$ into an ultrapower $B_1$ of $B$, extending the diagonal map on $B$. Apply it to $A_1\prec B_1$ to embed $B_1$ into an ultrapower $A_2$ of $A_1$, extending the diagonal map on $A_1$. Continue alternately. Replace structures by isomorphic copies to turn these elementary embeddings into inclusions; the commuting identities supplied by the relative lemma giveIn this common chain the map $A_n\to A_{n+1}$ is its ultrapower diagonal, and so is $B_n\to B_{n+1}$. The initial $A\to A_1$ is also diagonal. The union of the $A$-stages and the union of the $B$-stages are cofinal unions of the same chain, hence are the same structure after these coherent identifications. The elementary chain theorem follows by induction on formulas, with every finite tuple and every existential witness lying in some sufficiently late stage. It shows this union is elementary over all its stages. Thus the two constructed ultralimits are isomorphic.

Conversely every successive ultrapower and elementary union preserves the original complete theory. If ultralimits of $A$ and $B$ are isomorphic, they satisfy the same sentences, and their elementary embeddings of the original structures give $A\equiv B$. ThereforeThis proves Frayne theorem by an explicit alternating construction; it does not assume the stronger theorem about isomorphic single ultrapowers or any cardinal-arithmetic hypothesis.