# Is one structure elementary equivalent to its elementary extension?

Let $\mathfrak A,\mathfrak A^*$ be $\mathcal L$-structures and $\mathfrak A \preceq \mathfrak A^*$. That implies forall n-ary formula $\varphi(\bar{v})$ in $\mathcal L$ and $\bar{a} \in \mathfrak A^n$ $$\models_{\mathfrak A}\varphi[\bar{a}] \iff \models_{\mathfrak A^*}\varphi[\bar{a}]$$

Therefore forall $\mathcal L$-sentence $\phi$

$$\models_{\mathfrak A}\phi \iff \models_{\mathfrak A^*}\phi$$

,which implies $\mathfrak A \equiv \mathfrak A^*$

But I haven't found this result in textbook, so I'm not sure.

-
What is the point that makes you uncertain? Of course, this holds.. – Berci Dec 8 '12 at 12:54
In particular, this holds either by setting $n=0$ or (if you don't allow that) setting $n=1$ and $\varphi[a] \equiv \phi\land (a=a)$ (where $a$ is not free in $\phi$ because the latter is a sentence). – Henning Makholm Dec 8 '12 at 13:57
I see, thank you. – Popopo Dec 8 '12 at 14:10

This is, of course, true, an $\mathcal L$-sentence without parameters is an $\mathcal L$-sentence with parameters, that happens not to use any parameters, so elementary extension is a stronger condition. :)
To put it differently, $M\preceq N$ is equivalent to saying that $M$ is a substructure of $N$ and that $(M,m)_{m\in M}\equiv (N,m)_{m\in M}$, which is certainly stronger than mere $M\equiv N$ (to see that the converse does not hold, consider, for instance, $M=(2{\bf Z},+)$, $N=({\bf Z},+)$ -- $M$ is a substructure of $N$ and is e.e. (even isomorphic!) to it, but is still not an elementary substructure).
The last conclusion is impressive, there is an elementary embedding from $M$ onto $N$ but not the inclusion... – Popopo Dec 9 '12 at 15:48
@Popopo: This is related to another thing: the integers are the prime model, but not the minimal model of its own theory. Whenever such a situation arises, we can construct an infinite ascending, elementary chain of $\aleph_1$ isomorphic prime models, and the union of the chain is an atomic model of cardinality $\aleph_1$, and iirc, the existence of such a model is actually equivalent to the existence of a prime model but not a minimal model. (If a theory has a prime model and a minimal one, they are the same and unique.) – tomasz Dec 9 '12 at 18:44