Timeline for Defining $\mathbb{Z}^*$ in $\prod_p \mathbb{F}_p/\mathcal{U}$ (or pseudo-finite fields)
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 2, 2012 at 19:00 | vote | accept | CommunityBot | ||
| Feb 2, 2012 at 18:59 | vote | accept | CommunityBot | ||
| Feb 2, 2012 at 18:59 | |||||
| Feb 2, 2012 at 17:47 | vote | accept | CommunityBot | ||
| Feb 2, 2012 at 17:47 | |||||
| Jan 30, 2012 at 9:26 | comment | added | Thomas Scanlon | 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. | |
| Jan 29, 2012 at 22:59 | comment | added | Joel David Hamkins | 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. | |
| Jan 29, 2012 at 11:03 | vote | accept | CommunityBot | ||
| Feb 2, 2012 at 17:24 | |||||
| Jan 28, 2012 at 20:59 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |