Show that PA has a denumerable model that is not isomorphic to N

Question

I know that PA has inifnitely many models that can satisfy itself, but how we do prove that it has one particular denumerable model that is not isomorphic to its standard interpretation N?

Answer 1

Take any model of PA in a language $L$ (typically $(0, succ)$), call it $M$. Extend $L$ by a constant $c$ i.e., $L' = L \cup c$. Let $M'$ be an elementary extension of $M$ (a model which agrees with $M$) that also satisfies $\forall n\, c > n$. Show that $M'$ is finitely satisfiable ($c$ only needs to be greater than finitely many numerals $n$). By compactness $M'$ is satisfiable. By Löwenheim-Skolem there is an elementary countable sub-model of $M'$, call it $M*$. By transitivity of elementarity $M*$ is an elementary extension of $M$ so is a model of PA, it is countable and it isn't isomorphic to $\mathbb{N}$ because there is no element $c \in \mathbb{N}$ such that $\forall n\, c > n$.