Question / exercise goes as follows: $M'$ is said to be finitely generated if there exists a finite $|X|\subset |M'|$ such that $M'$ is the least substructure of $M'$ whose universe $|M'|$ contains $|X|$. We say $M$ is $\omega$-saturated iff every finite $|Y|\subset |M|$, every set of formulas $\Gamma(x)$ from $L(Y)$ (language restricted to that of $Y$) consistent with $\text{Th}(M(Y))$ (i.e. the theory of $M$ restricted to the domain $Y$) is realized in $M(Y)$ (more generally will be realized in the full model $M$).

Show that $M$ is $\omega$-saturated iff given structures $M'$ and $N$ with $M'$ finitely generated and $N$ countable, and given $f:M'\rightarrow M$ and $g:M'\rightarrow N$ elementary embeddings, there exists an elementary embedding $h:N\rightarrow M$ such that $h\circ g = f$.

Now here's my proof of left to right:

Suppose $M$ is $\omega$-saturated, and let $f$, $g$ be defined as above. Now we are given $|N|\le \omega$. Two cases eventuate:

Case 1. $|N|< \omega$. Then given the theory is complete, and two finite models, we have that they are isomorphic, and so that we can set $h = f\circ g^{-1}$.

Case 2. Suppose for the more general case that $|N| = \omega$. Given $|M'| = \omega$ together with $g: M' \rightarrow N$ s.t. $g$ is elementary embedding and $|N| = \omega$, $M' \prec N$ (i.e. $N$ is an elementary extension of $M'$). By our hypothesis that $f: M' \rightarrow M$ is an elementary embedding, we know for any formula $\phi$, $M'\models \phi(m'_1,\dots, m'_n)$, where $m'\in |M'|$ iff $M\models\phi(f(m'_1),\dots,f(m'_n))$, where again $m'\in |M'|$. But $M \prec N$, which means $N \models \phi(f(m'_1),\dots,f(m'_n))$; hence there is an $h: N \rightarrow M$ s.t. $h$ is e.m.


  • $\begingroup$ Woah. Hold on. Need to do some edits -- for some reason many of the expressions aren't going thru. Hmm. $\endgroup$ – لويس العرب Feb 1 '14 at 19:10
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    $\begingroup$ Fixed your post. $\endgroup$ – Shahar Feb 1 '14 at 19:15
  • $\begingroup$ Thanks Shahar -- for whatever reason what I typed wasn't showing up on thru mathjax. $\endgroup$ – لويس العرب Feb 1 '14 at 19:16
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    $\begingroup$ I just added lots of LaTeX improvements (to this and your other question). I recommend taking a look at the style (press the edit button), and trying to use it in the future. One could attribute the number of views with no corrections to people opening the question but choosing not to read it due to the messy typesetting! $\endgroup$ – Alex Kruckman Feb 15 '14 at 21:12
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    $\begingroup$ For the first case, $h=f$ doesn't make sense, it has a wrong domain. Rather, you should take for $h$ the isomorphism. For the second case, the cardinality of $M'$ can't be finite, as being infinite is an elementary property! In any case, it doesn't even seem true as stated (at least not the right to left part): an algebraically closed field is never finitely generated, so the right hand side condition is vacuously true, but there are certainly algebraically closed fields which are not saturated (namely those with finite transcendence degree). $\endgroup$ – tomasz Feb 15 '14 at 21:15

Your proof isn't complete. Here are some comments:

  • In Case 1, you can't literally set $h = f$ ($f$ has domain $M'$, and you want $h$ to have domain $N$!). Rather, you should set $h = f\circ g^{-1}$, which you can do since you know $g$ is an isomorphism.

  • In Case 2, you say $|M|<\omega$, but you only know $M$ is finitely generated. In fact, there can be no elementary embedding from a finite structure into an infinite one, so we must have $|M| = \omega$.

  • You need to do more work in Case 2. How do you define $h$?

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    $\begingroup$ Maybe I should have said "you need to do a lot more work in Case 2". For example, you're going to have to use the fact that $M$ is $\omega$-saturated somewhere.... $\endgroup$ – Alex Kruckman Feb 15 '14 at 21:21
  • $\begingroup$ Thanks -- first bullet was me making a blunder, second bullet answers two questions of mine, and third bullet I'll have to think about. $\endgroup$ – لويس العرب Feb 16 '14 at 5:38

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