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Let $L$ be a countable language. Let $T$ be a complete $L$ theory. We know that if $T$ is small, then there is a prime model of the theory. But $\text{Th}(\mathbb{N},+,\times,0,1)$ is not small but it has a prime model.

As far as I'm aware there is no necessary and sufficient condition for the existence of prime models. But I would like to see how to show certain theories do not have prime models. For example, how would you show that any consistent extension of ZFC doesn't have a prime model? (I strongly suspect that is the case).

Edit 1: To clarify: I'm asking if there is a "standard" argument that you can try when you suspect a theory doesn't have a prime model.

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  • $\begingroup$ What is a prime model? Is it something like a prime field? $\endgroup$ – bof Apr 1 '16 at 1:04
  • $\begingroup$ @bof It is a model of $T$ that embeds elementarily into any model of $T$. $\endgroup$ – user185596 Apr 1 '16 at 1:07
  • $\begingroup$ Doesn't ZFC have a model $M$ in which each element is definable? And isn't $M$ a prime model for the theory $\operatorname{Th}(M)$? $\endgroup$ – bof Apr 1 '16 at 1:26
  • $\begingroup$ @bof No, there are theories without prime models. So $M$ isn't always a prime model for $Th(M)$. Also the concepts only make sense for complete theories, and ZFC is not. That's why the question is phrased the way it is (.....every completion of ZFC.....) $\endgroup$ – user185596 Apr 1 '16 at 14:35
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    $\begingroup$ @bof I think your answer contributed a lot, I hope you'll undelete it! kav11, you misunderstood bof's comment: It doesn't say every $M$ is prime for $\text{Th}(M)$. Rather, if $M$ is a pointwise-definable model of ZFC, then $M$ is a prime model for $\text{Th}(M)$: $\text{Th}(M)$ is a complete extension of ZFC and $M$ is a model which realizes only isolated types. This is a nice observation, since it's the same reason $\mathbb{N}$ is an atomic model for true arithmetic: every element is definable (even given by a term, in the case of $\mathbb{N}$). $\endgroup$ – Alex Kruckman Apr 1 '16 at 14:40
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There is a simple necessary and sufficient condition for the existence of a prime model: "isolated types are dense". If $T$ is a complete theory in a countable language, then

  1. If $M\models T$, then $M$ is prime if and only if $M$ is countable and atomic (meaning that $M$ realizes only isolated types over the empty set).
  2. $T$ has a countable atomic model if and only if isolated types are dense (meaning that for every formula $\varphi(\overline{x})$, there is an isolated type $p(\overline{x})$ in $S^n(\varnothing)$ such that $\varphi\in p$).

These facts should be proven in every model theory textbook - they're Theorems 4.2.8 and 4.2.10 in Marker, for example. The situation is more subtle if $L$ is uncountable (largely because it's harder to omit types), and I suspect there's not a satisfying general criterion for the existence of prime models in that context.

One more comment: To prove that $T$ does not have a prime model, you need to come up with a formula $\varphi(\overline{x})$ which is not contained in any isolated type. This means $[\varphi(\overline{x})]$ is a perfect set in the Stone space, so there must be continuum-many types containing $\varphi(\overline{x})$. This is why small theories have prime models.

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  • $\begingroup$ Thank you. Do you have an idea for the extensions of ZFC? I'm not sure how to come up with such a formula. $\endgroup$ – user185596 Apr 1 '16 at 14:51
  • $\begingroup$ Given @bof s answer, things are probably very complicated. I took a look at the paper by Hamkins et al. It looks like you need to look at models of $ZFC+HOD\neq{V}$. $\endgroup$ – user185596 Apr 1 '16 at 23:18
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    $\begingroup$ Also, that should be "I" not "you" but grammar is strange at times :) $\endgroup$ – user185596 Apr 1 '16 at 23:25
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I address the part of your question which asks "how would you show that any consistent extension of ZFC doesn't have a prime model?" In fact (assuming of course that ZF is consistent), some consistent extensions of ZFC do have prime models.

Let $M$ be a model of ZFC in which each element is definable. (Such "pointwise definable" models of set theory were the subject of this question.) Then $T=\operatorname{Th}(M)$ is a consistent extension of ZFC, and $M$ is a prime model of $T.$

P.S. For the existence of pointwise definable models of ZFC, see the answers to this question, or consult the paper of J. D. Hamkins, D. Linetsky, and J. Reitz, Pointwise definable models of set theory, Journal of Symbolic Logic 78(1), pp. 139-156, 2013.

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  • $\begingroup$ Thank you for the answer! Apparently things are not as clean cut as I had thought. $\endgroup$ – user185596 Apr 1 '16 at 23:16

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