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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.

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What is the point that makes you uncertain? Of course, this holds.. –  Berci Dec 8 '12 at 12:54
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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
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up vote 2 down vote accepted

(I realise that it was answered in the comments, but I'm posting the answer so as to keep the question from staying in the unanswered pool.)

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).

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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
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@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
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