5
$\begingroup$

For a $\mathcal{L}$-structure $\mathcal{M}$, one can say two important finitary closure operators on $M$ are definable closure and algebraic closure.

The definable closure of a tuple $a$, $\operatorname{dcl}(a)$, is the set of elements $c\in M$ such that there exists a formula $\phi(x,a)$ that $c$ is the only solution in $M$, namely $c$ is the "unique solution" for $\phi$. Now instead of the only solution, if $\phi$ has "only finitely many solutions" in $M$ then we will have algebraic closure, $\operatorname{acl}(a)$.

About one of the important roles of algebraic closure in Model Theory, as an example, one can say this concept appears in strongly minimal theories effectively. Having the algebraic closure concept, in a strongly minimal theory under some conditions, we are allowed to define the notions of independence, basis, and dimension.

It is clear from the definition that algebraic closure always contains definable closure. Although one can guess algebraic closure generalizes the usual algebraic closure of fields in Field Theory I have no idea about the definable closure concept. I would like to know

  1. What idea is behind the definition of definable closure and what is the necessity of defining definable closure in Model Theory?

  2. In those theories that these closures do not coincide, using definable closure do we have some notions like independence, basis, and dimension, similar to linear algebra?

$\endgroup$
1
$\begingroup$
  1. An element is in the definable closure of a set $A$ iff it is the image under a definable function of a tuple of elements of $A$. So, in a sense, taking the definable closure of a set $A$ is taking the substructure generated by $A$, if we allowed all definable functions to have dedicated function symbols.

Another interesting fact is that $dcl(A)$ is exactly the set of points that are fixed by all automorphisms that fix $A$ pointwise, when the structure you're working in is sufficiently saturated. So, if you know Galois theory, you could say that the definable closure appears "naturally" when you study the fix points of automorphism groups.

  1. No, you do not in general have the exchange property, which is the generalization of "dividing by a coefficient" in vector spaces. The exchange property says "if $b \in cl(aA)\setminus cl(A)$, then $a \in cl(bA)\setminus cl(A)$", where $cl$ is the closure operator you're studying. This property is crucial in order to show that all bases of a set have the same cardinal.
$\endgroup$

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.