# Axiomatizable $\exists \forall$ theory

I have been thinking the following problem proposed by my friends for a long time.

Let $\mathcal{L}$ be the first-order language of theory of rings and let $K$ be the class of algebraic number fields. Consider the set $T_{\exists \forall}(K)$ which collects all the $\mathcal{L}$-sentences $\exists x\forall y \varphi(x,y)$ true in all number fields.($\varphi(x,y)$ is quantifier-free). Is this set axiomatizable? That is, can we find a set of axiom $A\subseteq T_{\exists \forall}(K)$ such that (1) $A$ is recursive and (2) $A\models \exists x\forall y \varphi(x,y)$ for all $\exists x\forall y \varphi(x,y) \in T_{\exists \forall}(K)$?

I have googling for several times but I cannot found any useful methods since

most of the axiomatizable results are about elementary theory of some structures.

I try first to consider the simple case with $T_{\exists \forall}(K)$ contains $\exists \forall$ sentence with single variable $x$ and $y$ but still cannot get any answers.

Any suggestions or comments are welcomed! Thank you

• You should require also some constraints on $A$,otherwise $A = \{\bot\}$ works. So $A\subset T_{\exists \forall}(K)$ for instance, or $\forall F\in K, F\models A$ – Maxime Ramzi Apr 4 '18 at 8:26
• A cursory reading of arxiv.org/abs/1309.0441 suggests to me that this may be open. Some relevant results are that for any number field $F$, $\mathcal O_F$ is universally definable in $F$ by arxiv.org/abs/1202.6371 , and that conditional on the weak Tate–Shafarevich conjecture, the existential theory of $\mathcal O_F$ is undecidable (this is in the Koenigsmann paper), thus the $\exists\forall$ theory of $F$ is undecidable. It is not clear to me to what extent are these results uniform over $F$, so it does not really answer the question. – Emil Jeřábek Apr 4 '18 at 8:39
• Though even at the best of times, what I wrote could only show undecidability of the theory, not that it is not recursively axiomatizable, as the existential theory of $\mathcal O_F$ is uniformly r.e. So maybe it is not that relevant. – Emil Jeřábek Apr 4 '18 at 8:59
• @EmilJeřábek Thanks for your comments ! I found this problem is not clear even if we consider the simple case which there are only one variable x and y. I try to use the results in algebraic surface but still have no answer. – Max CYLin Apr 4 '18 at 11:25
• @MattF. One of the examples I found is $\exists x \forall y y^2\neq x$. This example is also mentioned in G.Cherlin's Model theoretic algbebra. Since each number field is of the form $\mathbb{Q}(\alpha)$, the above sentence is true in all number fields. – Max CYLin Apr 4 '18 at 13:24

I know this isn't an answer but you said you were looking for any thoughts. It might be bonkers because it's been awhile since my Galois theory but based on my fuzzy memory and wikipedia any number field is characterized by a degree $n$ and $q_0...q_{n-1}$ such that some primitive element $x$ satisfies $x^n = q_0 + q_1 x + q_2 x^2 + \ldots + q_{n-1} x^{n-1}$.
Given this you might want to try and prove (say by induction on the terms in some nice form in $\phi$) that it's uniformly computable whether or not there is a counterexample to a given $\Sigma_2$ statement in an algebraic number field of degree $n$ and then try and bound the minimum degree that the a counterexample must appear by if it exists.
Hmm, but this would establish the much stronger result that the $\Sigma_2$ theory was decidable if all those parts are true so there is probably something wrong about it. But I found it interesting enough that I'm hoping someone can tell me why this can't be right.