1
$\begingroup$

My question is about the posibility of to construct a parameter space of models in a first order theory, finitely presented, with out existencial quantifiers (parameter space in the sense of nonconmutative algebraic geometry over $\mathbb{F}_2$). I explain: if we take as elements of our category $\mathcal{C}$ as the types containing a fixed theory $\mathcal{T}$. We define the functor $Val( )$ to the $\mathbb{F}_2$-vector spaces given for the valuations in every type. Every monadic term induct on every type $T$ a endomorphism in $Val(T)$. The space of endomorphism form a non-conmutative ring acting over the types. Specifically my question is if this ring parametrize all models of $\mathcal{T}$ (modulo elementery equivalence). I want know about some reference in this direction.

$\endgroup$
2
  • $\begingroup$ Can you clarify a few points for me? 1. What exactly do you mean by a "finitely presented theory without existential quantifiers"? Do you allow universal quantifiers and negations? Maybe you mean a finite universal theory (axiomatized by finitely many sentences of the form $\forall \overline{x}\,\varphi(\overline{x})$ with $\varphi$ quantifier-free)? 2. When you say "types containing $T$", do you mean full first-order types or types in a restricted logic without existentials? Types in how many variables? 3. What are the morphisms of $C$? Can you explain what the functor $\text{Val}$ does? $\endgroup$ Mar 24, 2015 at 17:22
  • $\begingroup$ Let $L$ the language of our theory, Let $M$ the monoid inducted by the monadic terms. With finetelly presented I mean that this monoid is finetelly generated and there exist $P_1(x),...,P_n(x)$ such that every propositin is inducted by $M$ and boolean combintion. I suspect that $L$ is a $\mathbb{F}_2[M][P_1,P_2,...,P_n]/<P_i^2=P_i>$ algebra. My question is that the spectrum of this ring say someting of the set of first order structures in L. $\endgroup$
    – camilo
    Mar 25, 2015 at 23:55

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.