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I'm having difficulty finding this result in the standard texts.

Theorem. Let $T$ be a theory in a language $\mathcal{L}$. TFAE:

1) $T$ has quantifier elimination,

2) Whenever $M, N$ are $\aleph_1$-saturated models of $T$, $A \subset M$, $B \subset N$ are countable nonempty substructures and $f : A \rightarrow B$ an $\mathcal{L}$-isomorphism, then for any $a \in M$ there exists an extension $f' \supset f$ with $a$ in the domain of $f'$ and $f'$ still an $\mathcal{L}$-isomorphism. In addition, for any $b \in N$ there exists an extension $f'' \supset f$ with $b$ in the range of $f''$ and $f''$ still an $\mathcal{L}$-isomorphism.

I'm obtaining this result from these notes by Chatzidakis (see 2.27): http://www.math.ens.fr/~zchatzid/papiers/CUP-MT.pdf. Any suggestions are appreciated.

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    $\begingroup$ You really mean to replace $\aleph_1$-saturated by $|{\mathcal L}|^+$-saturated. Also, it suffices to consider just the forth direction. That is, the last sentence beginning "In addition" may be dropped. $\endgroup$ – Thomas Scanlon Jul 7 at 18:15
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I'm not sure if this formulation with $\omega_1$-saturated models is in any of the "main standard" texts. But there is a complete treatment in these course notes by Pillay (see Proposition 2.29).

If you're looking for something in a standard published text, Corollary 3.1.6 of Model Theory: An Introduction (Marker) is along the same lines. It says:

Let $T$ be an $L$-theory. Suppose that for all quantifier-free formulas $\phi(\bar{v},w)$, if $M,N\models T$, $A$ is a common substructure of $M$ and $N$, $\bar{a}\in A$, and there is $b\in M$ such that $M\models\phi(\bar{a},b)$, then there is $c\in N$ such that $N\models \phi(\bar{a},c)$. Then $T$ has quantifier elimination.

It is not too difficult to deduce the theorem you quoted from this.

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    $\begingroup$ Also, see Proposition 4.3.28 in my book {\em Model Theory: an Introduction}. $\endgroup$ – Dave Marker 2 days ago
  • $\begingroup$ I stand corrected...there it is. And the proof does reduce to 3.1.6. Thanks Dave! $\endgroup$ – Gabe Conant 2 days ago

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