Are all countable, nonstandard models of arithmetic given by ultrapowers? Countable models of PA fall into two categories: the standard one $(\omega, S)$ and the nonstandard ones (all the rest).  The only way I've seen to construct a nonstandard model is through taking an ultraproduct or, equivalently, using the compactness theorem.  My question is wether or not these are all the models there are?  There are continuum many ultrafilters and continuum many nonstandard, countable models, but I don't know if there's a surjective correspondence.
 A: It could be mentioned perhaps that Skolem's non-standard model of arithmetic (1933-1934) is countable and is a kind of a "definable" version of the ultraproduct construction.  Namely, Skolem only uses definable sequences in his construction.  The advantage of his model is that it is constructed without using the axiom of choice.
A: An ultrapower will never yield a countable nonstandard model of PA --- either you will recover the standard model or the result will be uncountable.
As far as the construction of a model is concerned, due to Tennenbaum's theorem (see http://en.wikipedia.org/wiki/Tennenbaum's_theorem) you will never see a recursive nonstandard model of PA.  Hence, in some sense, you will never construct a countable model of PA other than the standard model.    
On the other hand, if you consider the Henkin construction to be constructive enough for you, then by running his construction relative to the theory in ${\mathcal L}(+,\times,0,1,c,<)$ consisting of PA together with all the assertions $c > n$ for each $n \in {\mathbb N}$, then you would obtain a nonstandard model of PA.
