My aim is to find an infinite prime number in ultraproduct of N**N/U, where U - nonprincipal ultrafilter on natural numbers N (it contains all cofinite sets). We have also multiplication and relation "<" on naturals and therefore inheritated to ultraproduct.
I took a=[2,3,5,7,11,13...].
I have to prove 2 things: 1st is clear - for every natural m, a is greater than embedding of m into that ultraproduct. It is trivial. I did this. 2nd thing - for any b,c in ultraporduct, a=bc implies b=1 or c=1 in the sense of ultraproduct. Then for every coordinate i, bici=ith-prime number. Hence one of them is 1 and one of them is ith-prime number. But why cant we take b=[2,1,5,1,11,1...] and c=[1,3,1,7,1,13....]? Then a=bc. Is it related to our relation c in ultraproduct but nor a
Let $b = [2,1,5,1,11,1,\dots]$ and $c = [1,3,1,7,1,13\dots]$. Then $\{n\in \mathbb{N}\mid b = 1\} = O$, the set of odd numbers. And $\{n\in \mathbb{N}\mid c = 1\} = E$, the set of even numbers. Since $O$ and $E$ are complements, one of them is in the ultrafilter $\mathcal{U}$. If $O\in \mathcal{U}$, then $b = 1$ in the ultraproduct. If $E\in \mathcal{U}$, then $c = 1$ in the ultraproduct.
The best way to answer this problem is to use Łoś's theorem. Consider the first-order formula $\varphi(x)$ expressing "$x$ is prime": $$x\neq 1\land \forall y\forall z(yz = x \rightarrow (y = 1\lor z = 1)).$$
Since $\varphi(x)$ is true in $\mathbb{N}$ for each component of $[2,3,5,7,11,13,\dots]$, Łoś's theorem tells us that $\varphi(x)$ is true in the ultraproduct for the equivalence class $[2,3,5,7,11,13,\dots]$.