Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 144 / 3

Suppose $a\in \mathrm{acl}(A)$ in the sense of model-theoretic algebraic closure. Some formula $\phi (x,\bar{c})$ over $A$ has exactly $n<\infty$ realizations in the monster model and includes $a$. Every model $M$ containing $A$ is an elementary substructure of the monster and therefore contains $n$ distinct realizations. Since there are only $n$ in the monster, it contains all of them, including $a$. Hence

Suppose $a\notin \mathrm{acl}(A)$. Choose a small model $M_0$ containing $A$. The complete type $\mathrm{tp}(a/A)$ is nonalgebraic, so choose a realization $a^{\prime }\notin M_0$. Strong homogeneity of the monster model gives an automorphism $\sigma$ fixing $A$ with $\sigma (a)=a^{\prime }$. Then $M={\sigma }^{-1}(M_0)$ contains $A$, but $a\in M$ would imply $a^{\prime }=\sigma (a)\in M_0$. Taking the contrapositive proves

Put $B=N\setminus \mathrm{acl}(A)$. We seek a model containing $A$ and omitting every element of $B$. Every finite set of these omission requirements is satisfiable: if a finite $C\subseteq B$ met every model containing $A$, the supplied result would imply $C\cap \mathrm{acl}(A)\ne \varnothing$, a contradiction.

Apply the compactness theorem to the elementary-diagram formulation of these requirements, using the Tarski-Vaught test to axiomatize the selected elementary submodel. It gives a model $M$ containing $A$ and omitting all of $B$. Part (a) gives $\mathrm{acl}(A)\subseteq M\cap N$, while omission of $B=N\setminus \mathrm{acl}(A)$ gives the reverse inclusion. Therefore