TITLE: Why does this nonstandard model of Robinson Arithmetic fail to be a model of PA?
QUESTION [4 upvotes]: As talked about in this question, there is a nonstandard model of Robinson arithmetic of the form:
$z_0 + z_1\omega + z_2\omega^2 + z_3\omega^3 + ... + z_n\omega^n$
in which $\omega$ is a formal variable which we treat as greater than any natural, the $z_n$ are integers, the leading coefficient is positive, and we include the element $0$. It is easy to see that this fails to be a model of PA because there is no element $x$ such that $x + x = \omega$ or $x + x + 1 = \omega$. If memory serves, another objection is that any nonstandard model will have some number that all of the naturals divide into, which clearly doesn't exist here. So in some sense we need to add more elements.
We can derive a related construction which extends the above by allowing for positive rational exponents and real coefficients for all but the constant term. We do this rigorously by starting with either the Levi-Civita field, or the field of Newton-Puiseux series of a rational variable in indeterminate $\epsilon$, letting $\omega = 1/\epsilon$, and looking at the strictly positive elements of the form
$z_0 + r_1\omega^{q_1} + r_2\omega^{q_2} + r_3\omega^{q_3} + ... + r_n\omega^{q_n}$
Where $z_0 \in \Bbb Z, r_n \in \Bbb R, q_n \in \Bbb Q$, and $0 < q_1 < q_2 < ... < q_n$. This gives us a ring of some sort of generalized "integers" within these fields. It is also fairly easy to see that the Robinson arithmetic axioms hold here for the non-negative elements, which satisfy the basic properties of how successor, addition, and multiplication should work. But now we have the problem that both $\omega$ and $\omega\sqrt2$ are integers, so $\sqrt2$ is "rational," which is provably not true in PA. Thus we now have too many elements.
Both of these problems would seem to be trivially fixed by requiring the coefficients to be rational instead of real (and keeping rational exponents). This also has order type $\Bbb N + \Bbb Z \Bbb Q$, which is the order type of any countable nonstandard model of PA. So that all checks out. But now the issue is that, unlike with real coefficients, all of the elements have a finite, computable representation, and thus addition and multiplication must be computable. Tennenbaum's theorem tells us is impossible for any nonstandard model of arithmetic, so somehow this must fail induction.
So the question:
With rational coefficients and rational exponents, how does this fail induction?
There must be some theorem which is true of the naturals that doesn't hold in this. What is it?

REPLY [4 votes]: Third try's the charm, hopefully:
Say that $x$ is a power of two iff whenever $y\vert x$ we have $y=1$ or $y=2$ or $2\vert y$. $\mathsf{PA}$ proves that there are arbitrarily large powers of two. However, no nonstandard element of your structure is a power of $2$: a nonstandard element with constant term $0$ will be divisible by $3$ (and a lot of other things too), while a nonstandard element with nonzero constant term will have some greatest standard power of two dividing it.
(In particular, the set of elements bounded by a power of two is exactly the standard cut.)

EDIT: Here's another point of difference, again gotten by focusing on the constant term:
Every nonstandard prime number (in your structure) has to have constant term $1$. But this means that there are no nonstandard primes which are $3$ mod $4$, and this (+ a standard exercise) means that the standard elements are characterized as those elements which are bounded by a prime which is $3$ mod $4$.

YET ANOTHER EDIT: Finally, it's worth noting that the standard definition of exponentiation in terms of addition and multiplication (via the $\beta$-function) has to break in your structure: $\omega^\omega$ is "too big." This is much less specific than the issues above, but it's also easier to see.
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