If *T* is a theory, a **model companion** of *T* is a theory *T'* with the following properties:

- Every model of
*T*embeds into a model of*T'.* - Every model of
*T'*embeds into a model of*T.* *T'*is model complete

Model companions are unique if they exist. A theory is **companionable** if it has a model companion.

- ACF is the model companion of the theory of fields, or even the theory of integral domains (in the ring language).
- RCF (in the language of ordered rings) is the model companion of ordered fields.
- RCF (in the pure language of rings) is the momdel companion of orderable fields.
- DLO is the model companion of ordered sets
- DLO is also the model companion of ordered sets in which every element has a successor and a predecessor (even though models of DLO do not have this property)
- The random graph is the model companion of graphs (i.e., sets with an antireflexive symmetric relation)
- ACFA is the model companion of difference fields (i.e., fields with an endomorphism)
- DCF is the model companion of differential fields (i.e., fields with a
**Z**-linear derivation) - ACVF is the model companion of valued fields.

Suppose *T*_{1} and *T*_{2} are two model companions of a theory *T*. Then *T*_{1} = *T*_{2} (or more precisely, *T*_{1} implies and is implied by *T*_{2}; they have the same models).

*Proof:* We will show that every model *M* of *T*_{1} is a model of *T*_{2}. Then by symmetry, *T*_{1} and *T*_{2} have the same models.

Note that every model of *T*_{1} embeds in a model of *T*, which in turn embeds in a model of *T*_{2}. Conversely, every model of *T*_{2} embeds in a model of *T*_{1}.

Now given *M*_{1} an arbitrary model of *T*_{1}, embed it in a model *M*_{2} of *T*_{2}. Then embed *M*_{2} in a model *M*_{3} of *T*_{1}. Continuing on in an alternating fashion yields an increasing chain of structures

with and . Then

is an increasing chain of models of *T*_{1}. By model completeness of *T*_{1}, this is an elementary chain

Similarly,

.

By the Tarski-Vaught Theorem

Thus *M*_{1} is elementarily equivalent to *M*_{2}, which is a model of *T*_{2}. In particular, *M*_{1} is a model of *T*_{2}. QED.

**Theorem:** Let *T* be an inductive theory. Then *T* is companionable if and only if the existentially closed model of *T* form an elementary class, in which case the model companion of *T* is exactly the class of existentially closed models of *T*.

To prove this, we need a few lemmas.

**Lemma:** Let *M* be a model of a model complete theory. Let *S* be a substructure of *M*. Then *S* is an elementary substructure of *M* if and only if *S* is existentially closed in *M*.

*Proof:* If , then for every formula and every tuple *a* from *S*, we have

.

In particular, this holds for existential formulas, which is what it means for *S* to be existentially closed in *M*.

Conversely, suppose *S* is existentially closed in *M*. To show that *S* is an elementary substructure of *M*, it suffices (by the Tarski-Vaught criterion) to show that every non-empty definable subset of *M*, definable over *S*, contains a point of *S*. Let be a non-empty definable subset of *M*, with *b* a tuple from *S*. By model completeness of *M*, we may assume that is an existential formula. Then so is the formula . Since is non-empty, we have

.

By existential closedness

.

So there is an *a* in *S* such that

Applying existential closedness another time,

Then , so the Tarski-Vaught criterion holds. QED.

**Lemma:** Suppose *M*_{1} ⊆ *M*_{2} ⊆ *M*_{3} and *M*_{1} is existentially closed in *M*_{3}. Then *M*_{1} is existentially closed in *M*_{2}.

*Proof:* Let be an existential formula. Then for any tuple *a* from *M*_{1}, we have

,

where the first two implications hold because is an existential formula, and the last holds because of existential closedness. From this we see

.

As this holds for arbitrary and *a*, *M*_{1} is existentially closed in *M*_{2}. QED.

Now we prove the theorem.

*Proof:* Suppose the existentially closed models of *T* form an elementary class. Then there is a theory whose models are exactly the existentially closed models of *T*. We claim that is a model companion of *T*. Every model of is a model of *T*, so trivially embeds into a model of *T*. Conversely, if *M* is a model of *T*, then *M* embeds into an existentially closed model of *T*, i.e., a model of , because there are enough existentially closed models in inductive theories. Finally, is model complete. Indeed, every model of is existentially closed as a model of *T*, so also existentially closed as a model of , since this is a weaker condition. Therefore is a model companion of *T*.

Conversely, suppose that *T* has a model companion . We must show that models of are exactly the same thing as existentially closed models of *T*.

Let *M* be an existentially closed model of *T*. Then we can embed *M* into a model *M*_{2} of . As is a model companion of *T*, we can embed *M*_{2} back into a model of *T*. This model of *T* can further be embedded into an existentially closed model of *T*, because *T* is inductive. So we have

with *M* and *M*_{3} existentially closed models of *T*, and . As *M* is existentially closed as a model of *T*, it is existentially closed in *M*_{3}. By the second Lemma, *M* is existentially closed in *M*_{2}. By the first Lemma, , so in particular , and *M* is a model of .

Conversely, let *M* be a model of . Then we can embed *M* in a model of *T*. This model of *T* can be further extended to be existentially closed. So we get *M ⊆ N* with *N* an existentially closed model of *T*. By the previous paragraph, *N* is a model of . By model completeness of , , so in particular, *M* is a model of *T*. To see existential closedness, let *N* be any model of *T* extending *M*. Then *N* can be embedded into a model *M*_{2} of . Then

.

By model completeness of , *M* is existentially closed in *M*_{2}. By the second lemma, *M* is existentially closed in *N*.

Thus, models of are the same thing as existentially closed models of *T*. QED.

Here are two examples of theories without model companions

- The theory of groups
- The theory of dense linear orders with an automorphism

*Proof:* As both theories are inductive, it suffices to show that the existentially closed models do not form an elementary class.

**Claim:** Let *G* be an existentially closed group, and *a* and *b* be two elements of *G*. If some automorphism of *G* sends *a* to *b*, then some inner automorphism sends *a* to *b*.

*Proof:* Let be the automorphism of *G* sending *a* to *b*. Consider the action of the integers on *G* by powers of . Let *H* be the semidirect product of and *G* from this action. Then *H* is a group extending *G*. In *H*, the equation *xax^{-1} = b* has a solution (namely, the generator of the copy of ). But since *G* is existentially closed in *H*, this equation also has a solution in *G*, which means that some inner automorphism sends *a* to *b*. QED.

Now suppose that the theory of groups had a model companion. Then there would be a theory *T* whose models are exactly the existentially closed groups. Let *G* be a -saturated and -strongly homogeneous model of *T*. Then two elements *a* and *b* of *G* have the same type over the empty set if and only if they are in the same orbit of Aut(*G*), by definition of strong homogeneity. In particular, *G*/Aut(*G*) has cardinality at most , because there are at most that many types over the empty set. (There are only 0-definable sets). On the other hand, by the claim, *G*/Aut(*G*) is the same as the quotient of *G* by inner automorphisms. This is an interpretable set, and *G* is saturated, so it must have size at least (a contradiction) or it must be finite.

Consequently *G* has finitely many conjugacy classes, and the conjugacy classes correspond to 1-types over the empty set. But *G* must also have the property that for every prime *p*, some element of *G* has order *p*. (It is easy to produce an extension of *G* which has this property, for example. Then use existential closedness.) If two elements have different orders, they cannot be conjugate. So *G* has infinitely many conjugacy classes, a contradiction.

Consequently the theory of groups is not companionable.

Next consider the theory of a dense linear order with an automorphism.

**Claim:** Let (M,<,) be an existentially closed DLO-with-automorphism. Let *a* < *b* be two elements of *M* and suppose . Then the following system of equations

can be solved if and only if for every *n*.

*Proof:* If *a < x < b* and *x* is fixed by , then

,

so one direction is clear. Conversely, suppose for every *n*. Let *U* be the set of elements in *M* which exceed every . Then *U* and *M*\*U* form a Dedekind cut on *M*. Also, *U* and *M*\*U* are each stabilized by . Let *N* be an extension of *M* obtained by adding a new element *c* in this Dedekind cut, and setting . Then *N* is a legimitate DLO-with-automorphism. The desired system of equations can now be solved in *N*, specifically by the new element *c*. By existential closedness, it can also be solved in *M*. QED.

Now suppose the theory of DLO-with-automorphism had a model companion *T*. Let *M* be a -saturated model of *T*. It is easy to create an extension of *M* in which the equation has a solution, so it must have a solution in *M*. In particular, we can find some *a* ∈ *M* such that .

Now consider the partial type *p(y)* over *a* which says the following:

and so on. Any finite subset of this is satisfiable, by taking for *n* sufficiently large (using the claim). By saturation, the type *p(y)* should have a realization *b* in *M*. But then *a* and *b* contradict the Claim above. QED.

::: ::: :::