Definition:Divide (Model Theory)

Definition
Let $T$ be a complete $\mathcal{L}$-theory.

Let $\mathfrak{C}$ be a monster model for $T$.

Let $A$ be a subset of the universe of $\mathfrak{C}$.

Let $\phi (\bar{x}, \bar{b})$ be an $\mathcal{L}$-formula with free variables $\bar{x}$ and parameters $\bar{b}$ from the universe of $\mathfrak{C}$.

$\phi(\bar{x}, \bar{b})$ $k$-divides in $\mathfrak{C}$ over $A$ if:
 * There is a sequence $(\bar{b}_i)_{i\in \N}$ such that $\operatorname{tp}(\bar{b}_i / A) = \operatorname{tp}(\bar{b} / A)$ for each $i\in\N$, and
 * For any distinct $k$-many terms $\bar b_{i_1}, \dots, \bar b_{i_k}$ of the sequence, the set $\{\phi(\bar{x}, \bar{b}_{i_1}), \dots, \phi(\bar{x}, \bar{b}_{i_k})\}$ is not satisfiable in $\mathfrak{C}$.

Let $\pi(\bar{x}, \bar{b})$ be a set of formulas with parameters $\bar{b}$.

$\pi(\bar{x}, \bar{b})$ $k$-divides over $A$ if it implies a formula $\phi(x, \bar{c})$ which $k$-divides over $A$.

Formulas and sets of formulas are said to divide if they $k$-divide for some $k$.

Note
To help understand the definition, consider the following informal discussion.

We can think of $\phi(x, y)$ as a description of $x$ in terms of $y$.

If $b$ and $b'$ have the same type over $A$, then as far as $A$ can tell, $b$ and $b'$ are the same thing, in the sense that they both satisfy the same formulas with parameters from $A$.

So $\phi (x, b)$ $k$-dividing over $A$ means that we can find an infinite sequence of parameters $b_i$ that $A$ can't tell apart, but nevertheless, no $x$ can satisfy any $k$ of the descriptions $\phi(x, b_i)$ at the same time.

Example
Consider the language $\mathcal{L} = \{\sim\}$ with one binary relation symbol $\sim$.

Let $\mathcal{M} = (\R^2, P)$ be the $\mathcal{L}$-structure where $P$ is the equivalence relation defined by $(x_1,y_1) P (x_2, y_2)$ iff $x_1 = x_2$. That is, $P$ identifies points in the plane which project vertically to the same value on the $x$-axis.

Let $\mathfrak{C}$ be some monster model for $\mathcal{M}$

The formula $x \sim (0,0)$ $2$-divides over $\emptyset$, which can be seen by considering the sequence $((0,0),(1,0),(2,0),(3,0),\dots)$. The language is not very complicated; every $(n, m)$ has the same type over $\emptyset$, so in particular each term in this sequence has the same type as $(0,0)$. However, $x \sim (n,0)$ and $x \sim (m,0)$ cannot be simultaneously satisfied for $n\neq m$, since this would imply $(n,0) P (m,0)$, which is impossible.

The formula $x \sim (0,0)$ does not $k$-divide over $A=\{(0,y): y \in \R\}$ for any $k$ however, since only points of the form $(0, y)$ are of the same type as $(0,0)$ over $A$.