if $a\in acl(b)$, then there's $\varphi(x;b)\in tp_\mathcal{M}(a;b)$ algebraic and $\varphi(x;c)$ algebraic for all $c\in\mathcal{M}$.

Question

For given theory $T$ and $\mathcal{M}\models T$, if $a\in acl(b)$, then there's $\varphi(x;b)\in tp_\mathcal{M}(a;b)$ algebraic and $\varphi(x;c)$ algebraic for all $c\in\mathcal{M}$ in $\mathcal{M}$.

Corresponding definitions : ($A\subseteq\mathcal{M}\models T$ is given)

$\cdot$$acl(A)$:=$\{a\in\mathcal{M}\mid tp_{\mathcal{M}} (a/A) \text{ is algebraic}\}$

$\cdot$type $p$ in $\mathcal{M}$ is algebraic if it contains algebraic formulas

$\cdot$$\varphi(x)$ is algebraic in $\mathcal{M}$ if it has only finitely many realizations in $\mathcal{M}$.

I think $\varphi(x;b)$ isolating $tp(a/b)$ will be the right one but don't know how to deal with this. I can't relate $\varphi(x;b)$ to $\varphi(x;c)$

Answer 1

Isolating $\text{tp}(a/b)$ is not sufficient in general. For example, consider an equivalence relation $E$ with two classes: The first class $C_1$ has only two elements, $a$ and $b$. The second class $C_2$ has infinitely many elements. Let $\varphi(x,y)$ be the formula $(xEy)\land (x\neq y)$. Then $\varphi(x,b)$ is algebraic and isolates $\text{tp}(a/b)$. But if $c\in C_2$, the formula $\varphi(x,c)$ is not algebraic.

Going back to the general case, note that if $a\in \text{acl}(b)$ and $\varphi(x,b)$ isolates $\text{tp}(a/b)$, then there is some natural number $k$ such that $\varphi(x,b)$ has at most $k$ solutions. The problem is that $\varphi(x,y)$ might not be "strong enough" so that this upper bound is true of $\varphi(x,c)$ for all $c$.

To fix this, we can just beef up $\varphi(x,y)$ to assert this upper bound! Let $\varphi'(x,y)$ be the formula $$\varphi(x,y)\land (\exists^{\leq k} x\, \varphi(x,y)).$$

Now $\varphi'(x,b)\in \text{tp}(a/b)$, and for any $c$, either

1. $\exists^{\leq k} \varphi(x,c)$ is true, in which case $\varphi(x,c)$ is algebraic, and hence $\varphi'(x,c)$ is algebraic, or

2. $\exists^{\leq k} \varphi(x,c)$ is false, in which case $\varphi'(x,c)$ has no solutions, and hence is algebraic.

In the case of the equivalence relation example above, $\varphi'(x,a)$ defines $\{b\}$, $\varphi'(x,b)$ defines $\{a\}$, and $\varphi'(x,c)$ defines $\varnothing$ for all $c\in C_2$.