algebraic closures (model -theory)

Question

I try to prove this lemma : Let be N a saturated model. Let be $ \phi(x) $ a consistent formula with parameters in $A$ and $b\in N$ . If for all $a$ such that $ \phi(a) $ I have $b\in acl(A,a), $then $b\in acl(A) $ .Some help to prove this lemma?

Answer 1

I assume the question lives in a large saturated model. By your assumptions the type $\{\phi(x)\wedge\psi(b,x)\wedge\exists^{=n+1}y\ \psi(y,x)\ :\ \psi(y,x)\in L(A),\ n\in\omega\}$ is inconsistent. Apply compactness. The formula that witnesses $b\in Acl(A)$ is obtained by (lengthy) straightforward manipulations.

There is also a much shorter argument (but it requires some background): every model containing $A$ contains a solution of $\phi(x)$ hence it contains $b$.