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Is it possible to show that there is a simple formula, preferably existential, that characterizes a nonstandard model of the ring of integers among the elements of $\prod_p \mathbb{F}_p(t)/\mathcal{U}$ where $\mathcal{U}$ is a nonprincipal ultrafilter over the prime numbers?

Thank you

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  • $\begingroup$ This is a duplicate of mathoverflow.net/questions/86905 . $\endgroup$
    – Emil Jeřábek
    Commented Feb 2, 2012 at 13:59
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    $\begingroup$ No this is not a duplicate because, this time, the ultraproduct is over $\mathbb{F}_p(t)$ and not over $\mathbb{F}_p$ $\endgroup$
    – user16974
    Commented Feb 2, 2012 at 14:06
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    $\begingroup$ Oh, I see. But the answer is still the same. By Joel’s argument, any subset of your field containing 1 and closed under + contains $\prod_p\mathbb F_p/\cal U$ (and possibly more stuff), hence it is not elementarily equivalent to the integers: e.g., it makes $-1$ (or any other integer for that matter) a sum of two squares. $\endgroup$
    – Emil Jeřábek
    Commented Feb 2, 2012 at 14:08
  • $\begingroup$ I meant: any definable subset etc. $\endgroup$
    – Emil Jeřábek
    Commented Feb 2, 2012 at 14:11
  • $\begingroup$ Emil, why not post that as an answer? $\endgroup$
    – Joel David Hamkins
    Commented Feb 2, 2012 at 14:12

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No, such a formula does not exist. If $R$ is a definable subset of $\prod_p\mathbb F_p(t)/\cal U$ which contains $1$ and is closed under addition, then by the argument Joel David Hamkins gave in his answer to your previous question, we must have $R\supseteq S:=\prod_p\mathbb F_p/\cal U$. In $S$, every element is a sum of two squares (because this holds in all finite fields). It follows that any integer (e.g., $-1$) is a sum of two squares in $R$, hence $R$ does not satisfy the universal theory of $\mathbb Z$.

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