Is there a finite or at worst recursive axiomatization of the theory of the rationals and reals combined?

Question

Consider the 2-element set $\{(\mathbb{Q};+,-,*,0,1,<),(\mathbb{R};+,-,*,0,1,<)\}$. What is an axiomatization of the theory of that set of structures? Meaning, what is an axiomatization of the set of sentences true of both the rational and the real ordered field? Is there a finite or at worst recursive axiomatization?

Answer 1

Given structures $\mathfrak{A},\mathfrak{B}$, the intersection of theories $Th(\mathfrak{A})\cap Th(\mathfrak{B})$ is always at least as complicated as each structure's theory individually: WLOG we have $\mathfrak{A}\not\equiv\mathfrak{B}$, but in that case (for example) fixing $\varphi$ such that $\mathfrak{A}\models\varphi$ and $\mathfrak{B}\models\neg\varphi$ we have $$\varphi\rightarrow\psi\in Th(\mathfrak{A})\cap Th(\mathfrak{B})\iff\psi\in Th(\mathfrak{A}).$$ In fact, the many-one degree of $Th(\mathfrak{A})\cap Th(\mathfrak{B})$ is the least upper bound of the many-one degrees of $Th(\mathfrak{A})$ and $Th(\mathfrak{B})$. (The same result holds with "$\cup$" in place of "$\cap$," by a similar construction.)

Now in your particular example, since $\mathbb{N}$ is definable in the field of rationals we have that $Th(\mathbb{Q};+,\times)$ is not computable; consequently $Th(\mathbb{Q};+,\times)\cap Th(\mathbb{R};+,\times)$ is not computable. (On the other hand $Th(\mathbb{R};+,\times)$ is computable by Tarski.)