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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 $\prod_p \mathbb{F}_p/\mathcal{U}$?

Thank you

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up vote 8 down vote accepted

There is no first-order definable subring of the ultraproduct $\Pi_p\mathbb{F}_p/U$ satisfying the theory of $\mathbb{Z}$. Indeed, every definable subset of $\Pi_p\mathbb{F}_p/U$ containing $1$ and closed under addition is the whole of $\Pi_p\mathbb{F}_p/U$, regardless of the complexity of the definition. To see this, suppose that $R\subset \Pi_p\mathbb{F}_p/U$ is closed under addition, contains $1$ and $x\in R\iff \Pi_p\mathbb{F}_p/U\models \varphi(x)$. Every such $x$ has the form $x=[f]_U$, for a function $f$ for which $f(p)\in \mathbb{F}_p$. For example, the $1$ of $\Pi_p\mathbb{F}_p/U$ is represented by the constant function $f(p)=1$. Since $\Pi_p\mathbb{F}_p/U\models\varphi(1)$, it follows by the Los theorem on ultraproducts that $U$-almost every $\mathbb{F}_p\models \varphi(1)$. Since $R$ is closed under addition, it follows again by Los that $U$-almost every $\mathbb{F}_p$ satisfies that the extension of $\varphi$ is closed under addition. Putting these two facts together, it follows for the $\mathbb{F}_p$ that satisfy both $\varphi(1)$ and the assertion that $\varphi$ is closed under addition, that they each also satisfy $\varphi(2)$ and $\varphi(3)$ and so on. Thus, in fact, they each satisfy $\forall n\varphi(n)$. Thus, by the Los theorem again, we get that $\Pi_p\mathbb{F}_p/U\models\forall x\varphi(x)$, and so $R$ is the whole of $\Pi_p\mathbb{F}_p/U$.

The argument also works just the same when the definition has parameters.

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Thank you for your nice answer. – user16974 Jan 29 '12 at 11:05
Another way to say it is: the ultraproduct satisfies the axiom scheme of induction, since each $\mathbb{F}_p$ does, namely, any definable subset containing $1$ and closed under $x\mapsto x+1$ is the whole set. – Joel David Hamkins Jan 29 '12 at 22:59
As an aside, even if we take an ultraproduct over all of the prime powers, there are no proper infinite definable subrings in $\prod_q {\mathbb F}_q / U$. This follows, for instance, from the work of Chatzidakis, van den Dries and Macintyre (Definable sets over finite fields. J. reine angew. Math. 427 (1992), 107-135) showing that the definable sets in pseudofinite fields satisfy a rational variant of the Weil bounds for the number of points on algebraic varieties. Of course, the field $\prod_p {\mathbb F}_p/U$ is a proper subfield of $\prod_p {\mathbb F}_{p^2}/U$, but it is not definable. – Thomas Scanlon Jan 30 '12 at 9:26

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