Ultraproduct $\prod_p \mathbb{F}_p/\sim$ and $\mathbb{Q}^*$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:09:53Z http://mathoverflow.net/feeds/question/86889 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86889/ultraproduct-prod-p-mathbbf-p-sim-and-mathbbq Ultraproduct $\prod_p \mathbb{F}_p/\sim$ and $\mathbb{Q}^*$ Math-player 2012-01-28T09:18:33Z 2012-01-28T13:23:37Z <p>Is the non-principal ultraproduct of finite fields $\prod_p \mathbb{F}_p/\sim$ a nonstandard model of the rationals $\mathbb{Q}$?</p> <p>EDIT: Can we realize $\mathbb{Q}^*$ as an ultraproduct?</p> http://mathoverflow.net/questions/86889/ultraproduct-prod-p-mathbbf-p-sim-and-mathbbq/86893#86893 Answer by a-fortiori for Ultraproduct $\prod_p \mathbb{F}_p/\sim$ and $\mathbb{Q}^*$ a-fortiori 2012-01-28T10:47:20Z 2012-01-28T10:47:20Z <p>It is easy to see that at least one of $-1,2,-2$ is a square in that field: the set of primes where neither $-1$ nor $2$ is a quadratic residue is contained in the set of primes where $-2$ is a quadratic residue.</p> http://mathoverflow.net/questions/86889/ultraproduct-prod-p-mathbbf-p-sim-and-mathbbq/86898#86898 Answer by Joel David Hamkins for Ultraproduct $\prod_p \mathbb{F}_p/\sim$ and $\mathbb{Q}^*$ Joel David Hamkins 2012-01-28T13:13:22Z 2012-01-28T13:13:22Z <p>Regarding your edit, of course any ultrapower $\Pi\mathbb{Q}/U$ of $\mathbb{Q}$ itself is a nonstandard model of the theory of $\mathbb{Q}$ (in whatever language you choose). So this version of $\mathbb{Q}^\ast$ is an ultraroduct. Indeed, ultraproducts are one of the principal methods of constructing nonstandard models. </p> http://mathoverflow.net/questions/86889/ultraproduct-prod-p-mathbbf-p-sim-and-mathbbq/86900#86900 Answer by ACL for Ultraproduct $\prod_p \mathbb{F}_p/\sim$ and $\mathbb{Q}^*$ ACL 2012-01-28T13:23:37Z 2012-01-28T13:23:37Z <p>A (non principal) ultraproduct of finite fields of unbounded cardinalities is a pseudo-finite field : </p> <ul> <li><p>It is perfect (if its characteristic is a prime number $p$, then any element is a $p$th power) ;</p></li> <li><p>For any positive integer $n$, it has exactly one extension of degree $n$ (equivalently, its Galois group is the profinite completion of the integers $\mathbf Z$);</p></li> <li><p>It is pseudo-algebraically closed (any geometrically irreducible algebraic variety has a rational point).</p></li> </ul> <p>Indeed, these are first-order properties, hence are inherited by ultraproducts : the first two ones hold for finite fields, while, it follows from Lang-Weil estimates that the last one is satisfied if the cardinality of the field is large enough, depending on the degrees of the equations.</p> <p>Observe that the field $\mathbf Q$ of rational numbers does not satisfy the last two properties.</p> <p>In fact, it is a theorem of Ax (1968, <em>The elementary theory of finite fields</em>, Ann. Math. 88 (1968) 239-271) that conversely, pseudo-finite fields are elementary equivalent to ultraproduct of finite fields.</p>