Question

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

Best regards.

Answer 1

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 $U$.