Existentially closed substructure and ultraproducts Is it true that if M is existentially closed in N then N can be embedded in an ultraproduct of M ?
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 A: 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 $U$.
