I heard somewhere that models of theory of real closed field are isomorphic.
However, there is also a statement in Internet which seems to say the opposite.
Are the models of theory of reals isomorphic?
The real numbers have a second-order theory, namely an ordered field which is both Archimedean and complete. This is a categorical theory and as such all its models are isomorphic.
However we can consider the first-order theory of real-closed fields. This theory do not specify that the fields are complete, because we cannot express this in a first-order one-sorted theory.
The theory of real-formal fields is a first-order theory which has no finite models, and therefore it has a model of any cardinality: countable, continuum, larger or smaller.
The theory RCF of real-closed fields is complete. But there is no infinite cardinal $\kappa$ such that RCF is $\kappa$-categorical. This means that there are non-isomorphic models of RCF of cardinality $\kappa$ for any infinite $\kappa$.
The result is reasonably clear for $\kappa=\omega$, since the real algebraic numbers are a model of RCF, as is the real-closure of the field obtained by adding one real transcendental to the real algebraic numbers.
It is also not hard to see that RCF has non-isomorphic models of cardinality $c$, since the reals are a model, and it is not hard to construct a non-Archimedean model of cardinality $c$. It follows from general theory that RCF is therefore not $\kappa$-categorical for any uncountable $\kappa$.
The theory of algebraically closed fields of characteristic $0$ is better-behaved. It is not $\omega$-categorical, but it is $\kappa$-categorical for every uncountable $\kappa$.
No. As a countable first order theory with an infinite model, the theory of real-closed fields has models of every infinite cardinality by the Löwenheim–Skolem theorem. Clearly models of different cardinalities cannot be isomorphic.
It has been pointed out that there are real closed fields in any cardinality, and of course models of different cardinalities cannot be isomorphic.
But there are also many non-isomorphic countable real closed fields. A first example: Let $M$ be the set of all real algebraic numbers.
A second model: Now let $t\in \mathbb R \setminus M$ be transcendental, and let $M_t$ be the set of all real numbers which are algebraic over $M (t)$. The model $M_t$ is real closed, but certainly not isomorphic to $M $. (If $f:M_t\to M$ is an isomorphism, what should $f(t)$ be?)
A third model: Now let $s\notin M_t$ be some other real, and let $M_s$ be the set of all reals algebraic over $M(s)$. The model $M_s$ is not isomorphic to $M$. The model $M_s$ is also not isomorphic to $M_t$, for the same reason as above. (Note that any field isomorphism between real closed fields is also an order isomorphism.)
Also note that the three models I have constructed so far are all Archimedean. Of course there are also non-Archimedean real closed fields.