A is prime over B, then A is also atomic over B

Question

The definition of $A$ is atomic over $B$ is:

Let $A$ be a model and $B \subset A$. An element $a \in A$ is atomic over $B$ if the type $\text{tp}(a/B)$ is isolated. If each element $a \in A$ is atomic over $B$, we say that $A$ is atomic over $B$.

The question is about the following exercise:

Let $A$ be a model of some theory $T$ and $B$ be a subset of $A$. We will say that $A$ is prime over $B$, if any partial elementary map $B \to M$ extends to an elementary map $A \to M$.

Show that if $T$ is totally transcendental and $A$ is prime over $B$, then $A$ is also atomic over $B$.

I don't understand why you need $T$ totally transcendental. Can't we just say: for $a \in A$ suppose that $p = \text{tp}(a/B)$ is not isolated. Then by the omitting types theorem there is a model $M$ of $\text{Th}_B(A)$ that omits $p$. Since $A \equiv M$ in the language $L_B$ there is an elementary map $f : B \to M$. By the assumption $A$ is prime over $B$, we can extend this map to an elementary map $f' :A \to M \supset f$. Therefore we see that $p$ must be omitted in $A$ as well, but by definition it is realized by $a \in A$. Contradiction: $p$ must be isolated.

This is basically the same proof as for $A$ is prime $\Rightarrow$ $A$ is atomic.

Answer 1

The discussion in the comments settles the question of why your argument doesn't work. Note that it works in the countable case (and the statement is true without the totally transcendental hypothesis in this case), but the theorem is also true for uncountable $B$ when $T$ is totally transcendental.

To prove it in the uncountable case, we use the notion of constructibility. Here are the three crucial facts, which come together to easily solve the exercise:

1. If $T$ is totally transcendental, then constructible extensions exist: for every subset $B$ of a model of $T$, there is a model $M\models T$ such that $M$ is constructible over $B$. (You probably proved this in your course. If not, this is the hard part of the exercise.)

2. If $M$ is constructible over $B$, then $M$ is atomic over $B$. (In the comments, you said that you know this fact).

3. If $M$ is atomic over $B$, and $B\subseteq N\preceq M$, then $N$ is atomic over $B$. (This is easy).