Take some stable theory $T$ with elimination of imaginaries, all sets appearing are small subsets of the monster model of $T$, elementary maps are restrictions of automorphism of the monster model of $T$.
Is the following statement correct?

  1. For all algebraic closed sets $A,B$, with additionaly that these sets are independent over $acl(\emptyset)$. And all elementary $f:acl(AB)\rightarrow acl(AB)$ fixing $A$ and $B$. And all $C,D$ two (closed) sets independent over $acl(\emptyset)$ with $A\subset C$ and $B\subset D$. We can extend $f$ to some automorphism fixing $C$ and $D$.

  2. Or equivalently, there is no $c\in acl(A\cup B)-dcl(A\cup B)$, such that $Ac$ independent of $B$ over $acl( \emptyset)$ and $A$ not independent of $Bc$ over $acl(\emptyset)$ (or vice versa).

  • 1
    $\begingroup$ Tim, I would encourage you to edit your question so that people can fully understand what you are asking. First, let's call maps which are not necessarily between models partial elementary. Now, first thing: you ask about extension to an isomorphism. Of what model? Arbitrary? Or do you simply mean a partial el map of the alg closure of some sets. Next, when you say two sets are independent, you should say over what base. If not, people might think over the empty set. I don't think you mean that. Notation: generally lower case letter are used for tuples or singletons. This is a matter of taste. $\endgroup$ Oct 14 '11 at 16:38

Martin Ziegler provided this solution.

We got that $C$ is independent of $B$ over $A$. Then $C$ is independent of $acl(A\cup B)$ over $A$. So by stationarity (since acl(A)=A) we can extend $f$ to elementary $g:C\cup acl(A\cup B)\rightarrow C\cup acl(A\cup B)$ with $g\vert C=id$.

Since $C\cup acl(A\cup B)\subset acl(CB)$ we have then that $C\cup acl(A\cup B)$ is independent of $D$ over $B$. So by stationarity again ($acl(B)=B$) we can extend $g$ to $h:D\cup C\cup acl(A\cup B)$ with $h\vert D=id$.


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