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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?

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2  
Where did you hear that such models are isomorphic? Are you sure you didn't actually hear that they are elementarily equivalent? –  Chris Eagle May 29 '12 at 15:28
    
Precisely what do you mean by a model of the "theory of reals"? –  Bill Dubuque May 29 '12 at 15:30
    
I feel that I have answered this before on this site, but I cannot find the post. –  Asaf Karagila May 29 '12 at 15:35

4 Answers 4

up vote 5 down vote accepted

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.

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+1 for the comment about the 2nd-order theory of the reals, which is probably the source of the OP's confusion –  Alex Kruckman May 29 '12 at 17:31

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 $\kappa$.

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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.

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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.

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