Given a first order theory $T$ a formula $\phi(x,b)$ and a set of parameters $A$, $\phi(x,b)$ divides over $A$ if there is an $A$-indiscernible sequence $(b_i:i\in \omega)$ with $b_0=b$ such that $\{\phi(x,b_i):i\in\omega\}$ is inconsistent (Def. 7.1.2, Lemma 7.1.4, Tent-Ziegler, A course in model theory, Lecture notes in Logic, CUP).

Let $L$ be a field of whatever characteristic $p$ and $\phi(x,b)$ a first order formula in the language for the theory of algebraically closed fields of characteristic $p$. What are the conditions on $\phi(x,b)$ in order that $\phi(x,b)$ divides over $L$? Can one give a concrete example of such a formula $\phi(x,b)$, possibly also in case $x$ is a tuple of variables and not just a variable?

  • $\begingroup$ I would be happy with an example for characteristic $p=0$ and $L$ being the rationals, I just don't get clearly what is the meaning of dividing for algebraically closed fields. $\endgroup$ – curious on mathematics May 14 '18 at 15:00

Intuitively, $\varphi(x,b)$ divides over the field $L$ if and only if it imposes some positive polynomial condition on the variables $x$, but such that this polynomial condition is not already definable over $L$.

Here are some examples of dividing formulas (where $a,b\notin L$, but $1\in L$):

  • $x = b$
  • $ax + by = 0$
  • $(y = bx^2 \lor ax+by = 0)$

In the first case, if you instantiate the formula along a nonconstant indiscernible sequence, you get a bunch of different points, and $x$ can't be equal to $2$ of them at once ($2$-inconsistency). In the second case, assuming you pick the indiscernible sequence to be independent enough, you get a bunch of random lines, no three of which have a common point of intersection ($3$-inconsistency). The third case is a bit more complicated, but you can see it by a similar argument.

But the following formulas do not divide:

  • $x\neq b$
  • $x + y = 1$
  • $x + y = 1\lor y = bx^2$

In the first case, $\{x \neq b_i\mid i\in \omega\}$ is satisfied by any element not among the $b_i$. In the second case, any indiscernible sequence in $\text{tp}(1/L)$ is constant, so you just get a single instance of the formula, satisifed by $x = 1$ and $y = 0$. In the second case, the $b$ will vary in an indiscernible sequence, but because of the disjunction, $x = 1$ and $y = 0$ still always satisfies the partial type.

You can play around with similar examples. I only wrote down formulas without quantifiers, since ACF$_p$ has quantifier-elimination.

To really characterize dividing in algebraically closed fields, it helps to know some general facts:

  • In a simple theory, a formula divides over $A$ if and only if it forks over $A$ (Tent & Ziegler Prop. 7.2.15).
  • In any theory, a formula $\varphi(x,b)$ forks over $A$ if and only if every type containing $\varphi(x,b)$ forks over $A$ (by definition and Tent & Ziegler Prop. 7.1.11).
  • In a totally transcendental theory, a type $p(x)$ over $B\supseteq A$ does not fork over $A$ if and only if $\text{MR}(p) = \text{MR}(p|_A)$ (Tent & Ziegler Cor. 8.5.11).
  • In a strongly minimal theory, the Morley Rank of a type over $A$ is equal to the algebraic dimension over $A$ of a tuple realizing the type (Tent & Ziegler Thm 6.4.2).
  • In ACF$_p$ (which is strongly minimal, hence totally transcendental, stable, and simple!), the algebraic dimension of a tuple $a$ over a subfield $L$ is the transcendence degree of $L(a)/L$.

Putting this all together, $\varphi(x,b)$ divides over $L$ if and only if for any tuple $a$ satisfying $\varphi(x,b)$, the transcendence degree of $L(b)(a)/L(b)$ is strictly less than the transcendence degree of $L(a)/L$.

You can probably prove this characterization directly with some algebraic work, but it's easier to use the general theory.

  • $\begingroup$ thanks! perfect answer, very illuminating! $\endgroup$ – curious on mathematics May 15 '18 at 10:01
  • $\begingroup$ just I guess that Cor. 8.5.11 says that p does not fork over A if p has the same Morley rank it has its restriction to A. Many thanks again in any case! $\endgroup$ – curious on mathematics May 15 '18 at 10:05
  • $\begingroup$ @curiousonmathematics ah right, hanks for catching the typo $\endgroup$ – Alex Kruckman May 15 '18 at 12:56

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