On countable isotypic structures

We obtain several results concerning the concept of isotypic structures. Namely we prove that any field of finite transcendence degree over a prime subfield is defined by types; then we construct isotypic but not isomorphic structures with countable underlying sets: totally ordered sets, fields, and groups. This answers an old question by B. Plotkin for groups.


Introduction
The concept of isotypicity for structures naturally arose within the framework of universal algebraic geometry and logical geometry [12], [7], [3].In the paper [15] it was formulated as a kind of Morita-type logical equivalence on algebras.Namely, Definition 1.1.Algebras  1 and  2 are logically similar if the categories of definable sets over  1 and  2 are isomorphic.This definition is related to the following one formulated by B.Plotkin in [10] for algebras: Definition 1.2.(see [10]).Let ℒ be a first order language and ,  be ℒ-structures.Then  and  are isotypic if for any finite tuple ā = ( 1 , . . .,   ) over  there exists a tuple b = ( 1 , . . .,   ) over  such that their types coincide, that is, tp  (ā) = tp  ( b), and vice versa, where for a -tuple ā we define tp  (ā) as a set of all first order formulas in  variables that hold true for ā.
The major problem behind all the considerations which remain widely open is whether two finitely generated isotypic groups are isomorphic, see [10].Our paper can be viewed as a step towards the clarification of this problem.It also answers the question about existence of countable isotypic but not isomorphic groups.Note that it was recently found that there are such groups in any cardinality bigger than contable, see [2,Remark 4.1].
The paper is structured as follows.In Section 2 we prove that any field of finite transcendence degree over a prime subfield is defined by types.In Section 3 we give an example of countable isotypic but not isomorphic totally ordered sets.In Section 4 we construct countable isotypic but not isomorphic fields and groups.

Acknowledgment
I am grateful to the anonymous reviewer of the first version of this paper, who pointed out references that allow to simplify the proofs, and suggested a way to extend one of the results from rings to fields.
I am also thankful to ISF grant 1994/20 for the support of the paper.

Isotypic fields of finite transcendence degree are isomorphic
In this Section we prove the following theorem.
Theorem 2.1.Any field of finite transcendence degree over a prime subfield is defined by types.
Proof.We will denote the transcendence degree of a field  by tr.deg .
Let  be a field of finite transcendence degree over a prime subfield, and  be a field that is isotypic to .Let us prove that  ≃  .
We may assume that   =   for all ; so that  and  share the common subfield  = ( 1 , . . .,   ), and both fields are algebraic over .Since tp  () = tp  (), it follows that any polynomial  ∈ [] has a root in  if and only if it has a root in  (existence of a root is an existential formula on  with one quantifier).Hence by [5][Lemma 20.6.3 (b)], we have  ≃  .
3. An example of countable isotypic but not isomorphic totally ordered sets.
In this Section we prove the following theorem.
Theorem 3.1.There exist two countable isotypic but not isomorphic totally ordered sets.
See Corollary 3.7 below for the concrete example.First we need some preparation.
So dist(, ) is an element of the set N 0 ∪ {∞}.
For totally ordered sets (, <  ) and (, <  ) we denote by ( × , <  ) the Cartesian product  ×  with the lexicographic order Let (Z, < Z ) be the set of integers with the usual order.Note that for a totally ordered set , if we consider the product ( × Z, < Z ), we have dist((, ), Proof.To prove that tp  ′ () = tp  ′ () is the same as prove that ( ′ , <  ′ , ) and ( ′ , <  ′ , ) are elementary equivalent in the language of ordered sets with  constants (constant can be understand as an unary predicate indicating that given element is this constant).So we can do it by showing that for any  ∈ N in the corresponding Ehrenfeucht-Fraisse game of length  the duplicator has a winning strategy.Fix the number  .By [16, Theorem 1.8] the tuples  and  are  -equivalent (see definition in [16, Section 1.1.])in the language of ordered sets.Let  + ∈  ′ and  + ∈  ′ be the elements chosen at -th turn of the game.The duplicator's winning strategy is to make choice in such a way that the tuples ( 1 , . . .,  + ) and ( 1 , . . .,  + ) be ( − )equivalent.He can do it by the definition of  -equivalence.In this case, in the end of the game the tuples ( 1 , . . .,  + ) and ( 1 , . . .,  + ) will be 0-equivalent, i.e the order on them will be the same.In particular that means that ( +1 , . . .,  + ) and ( +1 , . . .,  + ) are in the same order and for any 1 ⩽  ⩽  and  + 1 ⩽  ⩽  +  we have   =   ⇔   =   , which means that the duplicator won.
The proposition above allows us to construct many examples of isotypic totally ordered sets.Namely we have the following corollary.
Proof.It follows from the fact that (, <  ) is the quotient of ( × Z, < Z ) by equivalence relation  ∼  ′ ⇔ dist(,  ′ ) < ∞; and this quotient inherits the order.Now we can construct our example.
Corollary 3.7.The pair (Z × Z, < ZZ ) and (Q × Z, < QZ ) is an example of countable isotypic but not isomorphic totally ordered set.

Examples of countable isotypic but not isomorphic fields and groups.
In this Section we prove the following theorems.Theorem 4.1.There exist two countable isotypic but not isomorphic rings.
See Corollary 4.6 below for the concrete example.Theorem 4.2.There exist two countable isotypic but not isomorphic groups.

See Corollary 4.7 below for the concrete example.
Here is how our construction goes.Definition 4.3.For a totally ordered set (, <  ) we define the ordered abelian group Γ  as ⨁︀ ∈ (Q, +) with the lexicographic order (the positive elements are those with the highest non-zero component being positive).Then we construct the valued field   as follows: take the field Q(  :  ∈ ), where   are algebraically independent variables; add all the rational powers of the variables   ; define the non-archimedean valuation  with values in Γ  to be zero on Q ∖ {0} and such that for all  ∈  and  ∈ Q the valuation (   ) has  in the component that correspond to  and zero in all the other components; define   to be the henselization of this field with respect to the valuation , with the natural extension of that valuation (for the notion of henselization see, for example, [4, Section 5.2]).Proposition 4.4.Let (, <  ), (, <  ) be infinite totally ordered sets.Then the fields   and   are isotypic.
The field   has a unique valuation preserving embedding in the completion of the field Q(   :  ∈ ,  ∈ Q).Thus each of the elements   can be presented as a power series in    and clearly those series involve only    with  ∈ .Now choose elements  1 , . . .,   ∈  such that  1 < . . .<   .Then define   to be the same power series as   with each    replaced by    .Note that   are algebraic over Q(  :  ∈ ) and hence belong to   , because the residue field has characteristic zero and hence by [4, Theorem 4.1.10]the henselization is algebraically maximal.1:5 Now let us prove that tp   () = tp   ().Consider the triples (  , Γ  , Q) and (  , Γ  , Q) as structures in the three-sorted Pas language: it contains ring operations for the first and third sorts, group operation and order relation for the second sort, the valuation map  from the first sort to the second one, and the angular component map ac from the first sort to the third one.We define the angular component map for   and   as the map that takes the lowest coefficient in the corresponding power series.Now we prove an even stronger statement: any first order formula  ( 1 , . . .,   ) in the Pas language in  variables from the first sort has the same value on  and .It follows from [9] that in the theory of henselian valued fields with angular components where the residue field has characteristic zero any formula is equivalent to a boolean combination of formulas of the following types: • (), where  is a quantifier free formula in the language of rings; • (( 1 ()), ..., (  ())), where  is a formula in the language of ordered groups and   are terms in the ring language.
Moreover, by [6, Corollary 3.1.17] the theory of ordered divisible abelian groups has quantifier elimination; hence in the second item  can be taken quantifier free.Now the first two types of formulas have the same value on  and  because  and  lie in the isomorphic substructures, with isomorphism that takes  to  and the formulas are quantifier free.The third type gives the same value because the corresponded angular components are equal.Note that the sets of Archimedean classes of Γ  and Γ  are precisely  and .Therefore, we have (, <  ) ≃ (, <  ).
Corollary 4.6.The pair  Z and  Q is an example of countable isotypic but not isomorphic fields.

Proposition 4 . 5 .
Let (, <  ) and (, <  ) be totally ordered sets such that   ≃   (as fields).Then (, <  ) ≃ (, <  ).Proof.First let us prove that the isomorphism must preserve the valuation ring.Assume it does not.Then the field   has two different valuation rings O 1 and O 2 with maximal ideals M 1 and M 2 , such that both fraction fields are Q and   is henselian with respect to both valuation.It follows then by[4, Theorem 4.4.2]  that O 1 and O 2 are comparable, i.e. one ring is contained in the other and the maximal ideals are included in the opposite direction: say, O 1 ⩽ O 2 and M 2 ⩽ M 1 .Therefore, the quotient ring O 1 /M 2 has both fraction field and residue field isomorphic to Q, which is only possible ifO 1 /M 2 ≃ Q, i.e.O 1 = O 2 .Now for a valued field  with the valuation ring O one can recover the valuation group as  * /O * and the order can be recovered as[ 1 ] < [ 2 ] ⇔  2 / 1 ∈ O.Therefore, Γ  and Γ  must be isomorphic as ordered groups.Given an ordered abelian group Γ and elements ,ℎ ∈ Γ, we write  ∼ ℎ if and only if either |ℎ| ⩽ || and for some natural number  we have || ⩽ |ℎ|; or || ⩽ |ℎ| and for some natural number  we have |ℎ| ⩽ ||, where || = max{, −}.An Archimedean class of Γ is an equivalence class of ∼.The set of Archimedean classes inherits the order from Γ.