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Is it true that if M is existentially closed in N then N can be embedded in an ultraproduct of M ?

Best regards.

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Should one of those $N$'s be an $M$? – Donu Arapura Oct 8 '12 at 11:19
Yes I've edited it. Thank you. – student Oct 8 '12 at 11:24
up vote 2 down vote accepted

Yes; isn't this in standard textbooks like Chang-Keisler? Anyway, here's a construction. Adjoin, to the vocabulary (= signature = language) of $M$ and $N$, new constant symbols $\dot n$ for all the elements of $N$, and let $D$ be the set of all atomic sentences and negations of atomic sentences that are true in $N$ (with the obvious interpretation of the new constants). The index set for the desired ultraproduct will be the set $I$ of all finite subsets $p$ of $D$; the ultrafilter $U$ will be any ultrafilter on $I$ that contains all the sets of the form $A_p=\{q\in I:p\subseteq q\}$. For each $p\in I$, let $M_p$ be the structure $M$, with the obvious interpretation of $\dot m$ for $m\in M$, and with constants $\dot n$ for $n\in N-M$ interpreted so as to make all the sentences in $p$ true. (Note that, if $n\in N-M$ and $\dot n$ doesn't occur in $p$, then the interpretation of $\dot n$ is entirely arbitrary.) Such an interpretation for the $\dot n$'s exists because $M$ is existentially closed in $N$. Now let $Z$ be the ultraproduct of the $M_p$'s with respect to $U$. If we ignore the interpretation of the new constants, this is an ultrapower of $M$. It remains to embed $N$ into it. Send each element $n\in N$ to the element $[f_n]_U$ of $Z$ where $f_n(p)$ is defined as the interpretation of $\dot n$ in $M_p$. That this is an embedding follows from Los's theorem and the fact that every atomic or negated atomic sentence in $D$ is true in $M_p$ for $U$-almost all $p$, by our choice of $U$.

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An alternative answer, shorter but overkill: A compactness argument, applied to the union of the diagram of $N$ and the elementary diagram of $M$, embeds $N$ into an elementary extension $M'$ of $M$. Then invoke Shelah's theorem to see that $M$ and $M'$, being elementarily equivalent, have isomorphic ultrapowers. – Andreas Blass Oct 8 '12 at 12:25
Thank you very much. – student Oct 8 '12 at 13:16
Shelah’s theorem is an overkill, it suffices to use the basic fact that if $M$ and $M'$ are elementarily equivalent, then $M'$ elementarily embeds into an ultrapower of $M$. However, this is proved in essentially the same way as the direct construction in your answer. – Emil Jeřábek Oct 8 '12 at 14:17

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