finding a nonstandard arithmetic model containing an infinite prime number p I'm just thinking about the problem of finding a nonstandard arithmetic model like $\mathcal{M}$ (let's say an $\mathcal{L}_{NT}$ structure with $Th(\mathcal{M}) = Th(\mathbb{N})$ such that $\mathcal{M}$ contains an infinite prime number $p$).
My semantic interpretation of the argument by the structure is as following (I'm not sure they're sufficient!):


*

*$\forall v \in \mathbb{N}: v^{\mathcal{M}} <^{\mathcal{M}} p$ where $v^{\mathcal{M}}$ stands for applying successor function $n$ times to constant $0$, like: $S^{\mathcal{M}}(S^{\mathcal{M}}(...S^{\mathcal{M}}(0^{\mathcal{M}})..))$, meaning: "$p$ is infinite".

*"p is prime", by $p = .^{\mathcal{M}}(x,y)$, implying $x = 1^{\mathcal{M}}$ or $y = 1^{\mathcal{M}}$ $~~~~\forall x,y \in \mathcal{M}$. (Operator $<.>$ represents production.)
Any idea?...
 A: The sentence "For every $n$, there is a prime number $p>n$" can be expressed in the language of arithmetic, and is true in $\mathbb{N}$; so is also true in any nonstandard model $\mathcal{M}$. Let $c\in\mathcal{M}$ be any infinite element (which exists since $\mathcal{M}$ is nonstandard); then any prime (in the sense of $\mathcal{M}$) $p>c$ is an (externally) infinite prime, and such a $p$ exists for the reason above.
More generally, we can show by the same logic:

If $P$ is any definable property - primeness, squareness, perfectness, etc. - which holds of infinitely many natural numbers, then any nonstandard model $\mathcal{M}$ of $Th(\mathbb{N}$) contains infinite "natural numbers" $c$ with property $P$ (that is, such that $\mathcal{M}\models P(c)$; note that for this to make sense, $P$ has to be definable).


A commenter has asked for a more formal proof. Frankly, the argument in the first paragraph is perfectly rigorous, but here's another go:
The sentence $\varphi=$"$\forall x\exists y(y$ is prime and $x<y)$" is a sentence in the language of arithmetic (well, "is prime" isn't in the language of arithmetic, but we know how to express it in that language), and is true in the model $\mathbb{N}$, so is in $Th(\mathbb{N})$; since $\mathcal{M}\models Th(\mathbb{N})$, $\varphi$ is true in $\mathcal{M}$. Take an infinite element $c\in\mathcal{M}$; since $\mathcal{M}\models\varphi$, there is a prime in $\mathcal{M}$ which is $>c$.
