3
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ This is a duplicate of mathoverflow.net/questions/86905 . $\endgroup$ – Emil Jeřábek Feb 2 '12 at 13:59
  • 1
    $\begingroup$ 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 Feb 2 '12 at 14:06
  • 2
    $\begingroup$ 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 Feb 2 '12 at 14:08
  • $\begingroup$ I meant: any definable subset etc. $\endgroup$ – Emil Jeřábek Feb 2 '12 at 14:11
  • $\begingroup$ Emil, why not post that as an answer? $\endgroup$ – Joel David Hamkins Feb 2 '12 at 14:12
4
$\begingroup$

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 $\mathbb Z$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy