Let $F$ be a real closed field. It is known$^{[1]}$ that $F$ has an integer part, that is, a subring $A$ such that $\forall x \in F, \exists ! a \in A, a \leq x < a+1$.
Are all integer parts over $F$ isomorphic as ordered rings?
$[1]$: M. H. Mourgues and J. P. Ressayre The Journal of Symbolic Logic Vol. 58, No. 2 (Jun., 1993), pp. 641-647
Real closed fields are fields in which $X^2 + 1$ is irreductible and $F[X] / (X^2 + 1)$ is algebraically closed. They are ordered by setting $x \geq 0$ iff $x$ is a square.
A few ideas:
-Every order preserving isomorphism between two integer parts in $F$ extends in a unique way as an automorphism of $F$.
-Integer parts of real closed fields are exactly models of open induction (PA with induction scheme restricted to quantifier-free formulas).
-If $G$ is a subgroup of $(F,+)$ such that $\forall x \in F, \exists ! g \in G, g \leq x < g+1$ then the group $(F,+) / G$ is isomorphic to the "torus" $([0;1[_F,\underline{+})$ where $x\underline{+}y = x+y \ \mod 1$. This torus need not be isomorphic to ${{\mathbb{S}}^1}_F = \{(x,y) \in F^2 \ | \ x^2 + y^2 = 1\}$ with $(x,y)\underline{.}(z,t) = (x.z-y.t,x.t+y.z)$.
-No integer part over $F$ is definable in $(F,+,.,0,1)$.
Gurgen Asatryan proved in 2008 in On Ordered Fields with Infinitely Many Integer Parts that some ordered fields such as ${\mathbb{R}((x))}^{\mathbb{Q}}$ have $2^{\aleph_0}$ integer parts with distinct existential theories so the answer to my question is no.