elementary substructure in a satureted model

Question

Suppose you have a saturated model N of a complete theory T without finite models. How is it possibile to construct a proper saturated elementary substructure of N of the same cardinality of N ?

Answer 1

HINT. In a saturated structure $N$ there is always a non algebraic element $c$. (This requires a compactness argument.)

A saturated model $M\preceq N$ such that $c\notin M$ is constructed collecting one by one all the necessary elements. Let $A_\alpha$ be the set of elements collected after $\alpha$ steps and suppose that $c$ is not algebraic over $A_\alpha$. Let $p(x)\in S(A)$ for some $A\subseteq A_\alpha$ of cardinality $<|N|$. Prove that $p(x)$ is realized by some $b\in N$ such that $c$ is not algebraic over $A_{\alpha+1}=A_\alpha,b$. (This also requires a compactness argument.)

After at most $|N|^{<|N|}$ steps in this construction you obtain what you want.

Answer 2

The result follows from the fact that saturated models are universal: if $\overline{M}$ is saturated of cardinality $\kappa$, then every model of size $\leq \kappa$ embeds elementarily in $\overline{M}$.

Say you have a saturated model $\overline{M}$. By compactness and downward Löwenheim-Skolem, take $N \succ \overline{M}$ a proper elementary extension with $|N| = |\overline{M}|$. By universality, take $f : N \to \overline{M}$ an elementary embedding, and consider $f(\overline{M}) \prec \overline{M}$. We now have $f(\overline{M})$ an isomorphic copy of $\overline{M}$ as a proper elementary substructure.