Is $Th(\mathbb{Z}[x])$ uncountably categorical?

Question

Consider $T=Th(\mathbb{Z}[x])$ in the language $L = \{0,1,+,\times,deg(), \circ\}$ where $0,1,+$ and $\times$ have their usual interpretations, $deg()$ is a unary function symbol which gives the degree of a polynomial and $\circ$ is a binary function symbols where (if $p(x)$ and $q(x)$ are polynomials) $$ p(x)\circ q(x) = p(q(x))$$ and if $p(x)$ is a "constant", then $$p(x) \circ q(x) = p(x)$$

Clearly, $T$ is not countably categorical since it has $\aleph_0$ many $1-$types definable without parameters. However, I cannot figure out whether the theory is uncountably categorical.

Thank you.

Answer 1

Since you contain the arithmetic of the integers you can pick some infinite family of primes and say that an element of degree $0$ is divisible by those primes and not any other primes. This gives uncountably many types over a countable set, so the theory is not $\aleph_0$ stable and thus not uncountably categorical.

Answer 2

This is a response to Rene Schipperus's comment asking whether or not we can define an ordering on the integers in this structure. It turns out that we can.

We say that $P(x)$, i.e. "x is positive", holds on an element of our structure if $(\exists y)(deg(y)=x)$.

Note that $x$ is an "integer" if and only if $deg(x) = 0$.

Now, we say that $x < y \iff (deg(x)=deg(y)=0)\wedge (\exists z)[(P(z)\wedge (z\neq 0) \wedge (x+z =y)]$