2
$\begingroup$

Let $T$ be a first order theory $\mathcal{C}$ be its monster model. Assume that $T$ is stable, let $A \subseteq \mathcal{C}$. Is it true that if $\bar{c}$ and $\bar{d}$ are tuples such that $\bar{c} \smile_{A} \bar{d}$ (i.e.$ \bar{c}$ is independent from $\bar{d}$ over $A$) then $\bar{c}$ is independent from $\bar{d}$ over $acl(A)$ (i.e. over the algebraic closure of $A$)?

$\endgroup$
2
$\begingroup$

Yes, this is true, and even in simple theories (every stable theory is simple). This is already stated in this answer, but there is no proof there. So let's do that.

I am not sure what definition of independence you use, but we can work with just its properties (which hold in any simple theory):

  • Algebraic closure: for any $c$ and $A$ we have $c \overset{\vert}{\smile}_A \operatorname{acl}(A)$.
  • Monotonicity and transitivity: $c \overset{\vert}{\smile}_A BE$ if and only if $c \overset{\vert}{\smile}_A B$ and $c \overset{\vert}{\smile}_{AB} E$. From left to right is called monotonicity and the other direction is transitivity.

Claim. From the above properties we can deduce $$ c \overset{\vert}{\smile}_A D \Longleftrightarrow c \overset{\vert}{\smile}_A AD. $$

Proof. We always have $c \overset{\vert}{\smile}_A A$ (which follows from algebraic closure and then monotonicity). Then apply transitivity with $A$ in the roles of $A$ and $B$, and $D$ in the role of $E$. This gives the left to right direction, the other direction is just monotonicity.

Now, let's assume $c \overset{\vert}{\smile}_A d$, then using the claim we have $c \overset{\vert}{\smile}_A Ad$. By algebraic closure, we have $c \overset{\vert}{\smile}_{Ad} \operatorname{acl}(Ad)$. So by transitivity we find $c \overset{\vert}{\smile}_A \operatorname{acl}(Ad)$. Then applying monotonicity once gives $c \overset{\vert}{\smile}_{\operatorname{acl}(A)} \operatorname{acl}(Ad)$, and again finally yields $c \overset{\vert}{\smile}_{\operatorname{acl}(A)} d$.

Using similar techniques you can put the $\operatorname{acl}$ in different places (and in multiple places). In other words, independence in simple theories does not care about algebraic closure. Although you will also need the symmetry property for this (being able to 'swap' the left and right side of $\overset{\vert}{\smile}$), but that still holds in all simple theories (and is in fact equivalent to the theory being simple).

A proof of the properties I mentioned in terms of dividing can be found in, for example, A course in Model Theory by Katrin Tent and Martin Ziegler. The property I named algebraic closure is Remark 7.1.3 there and monotonicity and transitivity is corollary 7.2.17.

$\endgroup$
1
$\begingroup$

Here's another argument, working directly from the definitions of forking and dividing. So it holds in any theory (I won't assume stability or simplicity). Let $B = \text{acl}(A)$.

(1) A formula $\varphi(x,d)$ divides over $A$ if and only if it divides over $B$.

If $I = (d_i)_{i\in \omega}$ is a $B$-indiscernible sequence with $d_0 = d$ witnessing dividing over $B$, then $I$ is also $A$-indiscernible, so it witnesses dividing over $A$. Note we didn't use $B = \text{acl}(A)$ here, just $A\subseteq B$.

Conversely, if $I = (d_i)_{i\in \omega}$ is an $A$-indiscernible sequence with $d_0 = d$ witnessing dividing over $A$, then since an $A$-indiscernible sequence is automatically $\text{acl}(A)$-indiscernible, $I$ also witnesses dividing over $B$.

(2) A formula $\varphi(x,d)$ forks over $A$ if and only if it forks over $B$.

This follows immediately from (1).

(3) $\text{tp}(c/Ad)$ forks over $A$ if and only if $\text{tp}(c/Bd)$ forks over $B$.

Suppose $\text{tp}(c/Ad)$ forks over $A$. Then it contains a formula $\varphi(x,a,d)$ which forks over $A$. By (2), $\varphi(x,a,d)$ also forks over $B$, and $\varphi(x,a,d)\in \text{tp}(c/Bd)$, so this type forks over $B$.

Conversely, suppose $\text{tp}(c/Bd)$ forks over $B$. Then there is a formula $\psi(x,a,b,d)\in \text{tp}(c/Bd)$ (notation: $a\in A$, $b\in B\setminus A$) which forks over $B$, and hence forks over $A$ by (2). The problem is that $\psi(x,a,b,d)\notin \text{tp}(c/Ad)$, since it contains parameters $b\in B\setminus A$.

To fix this, let $b_1,\dots,b_n$ be the finitely many realizations of $\text{tp}(b/Ad)$ (finitely many because $b\in B = \text{acl}(A)$). These are the points in the orbit of $b$ under the action of $\text{Aut}(\mathbb{M}/Ad)$. Since forking is automorphism invariant, $\psi(x,a,b_i,d)$ forks over $A$ for all $i$. And since forking formulas are closed under disjunction, $$\varphi(x,a,b_1,\dots,b_n,d) = \bigvee_{i=1}^n \psi(x,a,b_i,d)$$ forks over $A$. But the set defined by $\varphi(x,a,b_1,\dots,b_n,d)$ is fixed by the action of $\text{Aut}(\mathbb{M}/Ad)$ (which just permutes the disjuncts), so it is equivalent to an $L(Ad)$-formula $\chi(x,a',d)$ (this is a compactness argument). It follows that $\chi$ forks over $A$ and is in $\text{tp}(c/Ad)$, so we're done.

$\endgroup$

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.