## models of PA which are isomorphic but not elementarily equivalent?

On page 164 of his book Models of Peano Arithmetic, Kaye states Friedman's Theorem:

Let $M{\vDash}PA$ be nonstandard and countable, let $a\in M$ and let $n\in {\mathbb N}$. Then there is a proper initial segment $I\subseteq_c M$ ($\subseteq_c$ means "cofinal in") containing $a$ such that $I\cong M$ and $I<_{\Sigma_n} M$.

However, on the next page he writes "Nor can we expect in general to get initial segments $I$ with $M\cong I < M$ and $M\neq I$, i.e., elementary for all formulas. For example if $M=K_T$ (where $T\neq Th({\mathbb N})$ is a complete extension of PA) then $M$ has no proper elementary substructures, and so certainly has no proper elementary initial segments!"

I am confused. If two models in the same language are isomorphic, are they not elementarily equivalent? Of course the converse need not be true.

An isomorphism of models is a bijective homomorphism of the language's algebraic portion (constants and functions) which preserves and reflects all relations of the language (Hodges, Model Theory, p5). Therefore by induction on the structure of any ${\mathcal L}_{\omega,\omega}$ formula, the isomorphism will both preserve and reflect it. So an isomorphism preserves all formulas (Hodges, Theorem 2.4.3(c)).

If the proper initial segment is isomorphic as a model to the entire model, how could any first-order sentence possibly be true in one and not in the other?

Thanks,

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There is a major difference between elementary equivalence and elementary embedding. Moreover, in this case, the actual embedding is somewhat ambiguous. First, let me recap some often confused terminology.

Two models are elementary equivalent if they satisfy the same first-order sentences. Any two isomorphic models are always elementary equivalent. An elementary embedding is a map j:A→B such that, for all first-order formulas φ(v1,...,vk) and all a1,...,ak ∈ A, A ⊧ φ(a1,...,ak) iff B ⊧ φ(j(a1),...,j(ak)). An isomorphism is always an elementary embedding.

The notation A ≺ B means that A is an elementary submodel of B, i.e. the inclusion map A ⊆ B is an elementary embedding from A into B. In your context, the isomorphism (or its inverse) is not the proposed elementary embedding, it is the inclusion map which is in question: it is elementary for Σn formulas, but not elementary for all first-order formulas.

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Ah, thanks. I stupidly took "≺" to mean "elementarily embedded in" rather than "elementary submodel of". Must memorize more symbols... – Adam May 30 2010 at 21:31
It can be very confusing! I've seen j:A ≺ B to mean that j is an elementary embedding... – François G. Dorais May 30 2010 at 21:36