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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?...

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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 $xc$.

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    I didn't understand this part of the answer: `then any prime (in the sense of M) p>c is an (externally) infinite prime`. Could you elaborate a little bit on it? How do we know there are any p>c? It sounds to be the claim to be proved, not an assumption. – 2017-02-19
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    @A.Loc Because the statement "For all $x$, there is a prime $>x$" is in $Th(\mathbb{N})$, hence is true in $\mathcal{M}$. (Remember that $Th(\mathbb{N})$ is the set of all sentences true in $\mathbb{N}$ - well, the unboundedness of primes is certainly such a statement! So since $\mathcal{M}\models Th(\mathbb{N})$ . . .) – 2017-02-19
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    I'm curious, why the downvote? – 2017-02-22
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    @NoahSchweber: Maybe because your assertions are verbal, not mathematical. Could you formally present your proof? – 2017-02-28
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    @Roboticist See my edit. But *the proof in the first paragraph is entirely correct and rigorous already* - adding symbols doesn't make it more solid. – 2017-02-28
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    @NoahSchweber: I think it's a deserving answer. Let's hope downvoters do so! – 2017-02-28