Intersecting the algebraic closure of independent elements - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:47:03Z http://mathoverflow.net/feeds/question/81327 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81327/intersecting-the-algebraic-closure-of-independent-elements Intersecting the algebraic closure of independent elements Drike 2011-11-19T09:08:06Z 2011-11-20T10:38:57Z <p>\$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 </p> <p>\$\$acl(a)\cap acl(b)=acl(\emptyset)\$\$</p> <p>In particular, this holds if \$a\$ and \$b\$ are independent generic elements of \$G\$.</p> <p><strong>Question 1 :</strong> If \$a\$ and \$b\$ are generic, does the following equality hold? \$\$acl^n(a)\cap acl^n(b)=acl^n(\emptyset)\$\$</p> <p>As I expect the answer to question 1 to be no, let me ask :</p> <p><strong>Question 2 :</strong> 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)\$\$</p>