Is the nonprincipal ultraproduct of finite fields $\prod_p \mathbb{F}_p/\sim$ a nonstandard model of the rationals $\mathbb{Q}$?
EDIT: Can we realize $\mathbb{Q}^*$ as an ultraproduct?
Is the nonprincipal ultraproduct of finite fields $\prod_p \mathbb{F}_p/\sim$ a nonstandard model of the rationals $\mathbb{Q}$? EDIT: Can we realize $\mathbb{Q}^*$ as an ultraproduct? 


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. 


A (non principal) ultraproduct of finite fields of unbounded cardinalities is a pseudofinite field :
Indeed, these are firstorder properties, hence are inherited by ultraproducts : the first two ones hold for finite fields, while, it follows from LangWeil estimates that the last one is satisfied if the cardinality of the field is large enough, depending on the degrees of the equations. Observe that the field $\mathbf Q$ of rational numbers does not satisfy the last two properties. In fact, it is a theorem of Ax (1968, The elementary theory of finite fields, Ann. Math. 88 (1968) 239271) that conversely, pseudofinite fields are elementary equivalent to ultraproduct of finite fields. 


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. 

