Determining if there are prime hyperintegers I have $F$ a fixed nonprincipal ultrafilter on $\omega$ and let $N^*$ be the ultrapower with domain $\mathbb{N}/F$ (the extended natural numbers). The language contains $0,1,+,\cdot,<$ and =. A normal model for this language is the standard interpretation of integers $N$. Now I need to determine if there are infinitely large prime number. 
In order to determine this I use Los's theorem that states that any sentence $B$ of $K$ is true in $\mathbb{N}/F$ if and only if $A = \{ n\in \mathbb{N} : B$ is true$\}\in F$. 
I made the following sentence $B$:
$(\forall x)(\forall y)( (xy = n) \Rightarrow ((x=1)\lor (y=1)))$
My conclusion is that we can't determine if $A$ is in $F$. 
For nonprincipal ultrafilters either $A$ or $\mathbb{N}\setminus A$ is in $F$, and there are no finite sets in $F$. 
But both $A$ and $\mathbb{N}\setminus A$ are infinite, so we can't determine if $A$ is in $F$, so we can't say if there are infinitely large prime numbers.
I have two questions:


*

*Is my sentence $B$ closed. The tutor gave similar sentences in class, but there is no quantifier for my $n$..

*Is my conclusion correct. 

 A: One way to see that there are nonstandard primes is by considering the formula $$\phi(x)= ``x\text{ is prime''}$$ (the formalization of that is easy)
and finding a nonstandard element satisfying it. Recall that Łoś's construction works for any formulas, not just sentences. 
So if we construct an element $[f]\in N^*$ (where $f:\mathbb N\rightarrow \mathbb N$ and $[f]$ is its equivalence class), we have $$N^*\models\phi([f])\quad\text{iff}\quad \{n\in\mathbb N\mid N\models \phi(f(n))\}\in F\,.$$
But now it's easy to see what we have to do: We just have to choose a function $f$ which takes on prime values often enough. To ensure that $[f]$ is not a standard natural number (and thus not a standard prime), it better not be equal to any particular standard number too often (recall that if $m$ is a standard natural number and $c_m$ is the function with constant value $m$, then $[c_m]$ represents $m$ in the ultrapower; so if $f(n)=m$ for $F$-many $n$, $[f]=[c_m]$).
Good that there are infinitely many primes. Let $(p_n)_{n\in\mathbb N}$ be an enumeration of the primes (it can really be any sequence consisting of primes without any prime appearing infinitely often, so you can construct several nonstandard primes) and consider the function $$f\colon\mathbb N\rightarrow\mathbb N, n\mapsto p_n\,.$$ Then $[f]$ is prime and not a standard natural number, as desired. 
A: Transfer the sentence $\forall n\; \exists p\colon (p \text{ is prime })\wedge (p>n)$ (being a prime is of course an elementary statement as pointed out in the question itself).
