Is a prime model for a theory homogeneous?

Question

Consider a theory which admits a prime model $M$ of cardinality $\kappa$. Is it true that it is $\kappa$-omogeneous?

As an example, $\mathbb Q$ works for the theory of fields of characteristic $0$ (it should be $\omega$-homogeneous, right?).

Is the statement true in the general case?

Thank you in advance.

Answer 1

If the language is countable, then prime models are homogeneous. In this case, a model is prime if and only if it is atomic (every tuple realizes an isolated type over the empty set), a theory has a prime model if and only if isolated types are dense (every consistent formula is implied by a consistent formula which isolates a complete type), and prime models (if they exist) are countable, unique up to isomorphism, and $\omega$-homogeneous. You can find a proof of these facts in any basic model theory text book.

But all of these nice properties can fail in uncountable languages. For a prime model which is not homogeneous in its own cardinality, consider the language with a single unary predicate $P$ and an uncountable family of constants $(c_i)_{i\in \kappa}$ with $\kappa >\aleph_0$.

Let $T$ be the theory asserting that:

1. $P(c_i)$ for all $i\in \kappa$.
2. $c_i\neq c_j$ for all $i\neq j$.
3. There are infinitely many elements satisfying $\lnot P$.

Then $T$ has a prime model $M$ of cardinality $\kappa$. In $M$, an element satisfies $P$ if and only if it is named by a constant, and there are only countably many elements $(a_i)_{i\in \omega}$ satisfying $\lnot P$.

Now $M$ is not $\kappa$-homogeneous. The infinite tuples $a = (a_i)_{i\geq 1}$ and $a' = (a_i)_{i\geq 0}$ have the same type, since any finite subtuple of $a$ can be mapped to the corresponding finite subtuple of $a'$ by an automorphism of $M$. But the type $\text{tp}(a_0/a)$ is not realized over $a'$, since every element satisfying $\lnot P$ is already in the tuple $a'$.