MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$G$ is a group with a simple first order theory $T$ as defined by Shelah, hence equiped with a "nice" notion of independence. $G$ also has generic elements. I write $acl^n(A)$ for the set of elements in $M$ lying in a finite $A$-definable set of size at most $n$, and $acl(A)$ for the union of $acl^n(A)$ when $n$ ranges over all natural numbers. Let $a$ and $b$ be independent elements over $0$. One has

$$acl(a)\cap acl(b)=acl(\emptyset)$$

In particular, this holds if $a$ and $b$ are independent generic elements of $G$.

Question 1 : If $a$ and $b$ are generic, does the following equality hold? $$acl^n(a)\cap acl^n(b)=acl^n(\emptyset)$$

As I expect the answer to question 1 to be no, let me ask :

Question 2 : Is there a constant $k$ (depending only on $T$ and $n$) such that for all independent generics $a$ and $b$ over $0$, one would have $$acl^n(a)\cap acl^n(b)\subset acl^{k.n}(\emptyset)$$

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.