3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.