# 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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Probable typo: $\subseteq_c$ should be $\subseteq_e$, where $e$ is for initial or end. (The typo is in the OP's question, not in Kaye's book). – jeq Dec 3 '15 at 15:41