Can a pair of complex numbers be "completely" algebraically independent? Two complex numbers $\alpha,\beta$ are called algebraically independent if there is no polynomial $p(x,y)\in\mathbb{Q}[x,y]$ such that $p(\alpha,\beta)=0$.
For a complex number $x$, let $A(x)=\{z\in\mathbb{C}:x \text{ is algebraic over } \mathbb{Q}(z)\}$
Can there exist a pair of complex numbers $x,y$ such that $A(x)\cap A(y)$ is empty? Such a pair would also be algebraically independent.
A necessary condition for it to be empty is that both of them must be transcendental over the rationals. For all $x\in\mathbb{C}$, $x$ is contained in $A(x)$, so $A(x)$ is never empty. If $x$ is algebraic over the rationals, then $A(x)$ would be the whole of $\mathbb{C}$. It follows that if either $x$ or $y$ is algebraic over the rationals, then the intersection would not be empty.
 A: It's easy to show that, for each specific polynomial $p$ with coefficients in $\mathbb{Q}(t)$ with $t$ indeterminate, the set of pairs $(x,z)\in\mathbb{C}^2$ such that $x$ is algebraic over $\mathbb{Q}(z)$ via $p$ is nowhere-dense in $\mathbb{C}^2$. Consequently, the set of pairs $(x,z)\in\mathbb{C}^2$ such that $x$ is not algebraic over $\mathbb{Q}(z)$ at all and $z$ is not algebraic over $\mathbb{Q}(x)$ at all is comeager, so by the Baire category theorem nonempty.
A similar argument shows that the set of pairs $(x,z)\in\mathbb{C}^2$ such that neither $x$ nor $z$ is algebraic over the other has full measure (= its complement is null). So in multiple senses, "most" pairs of complex numbers have the property you ask about.
Similarly, for a fixed transcendental $x\in\mathbb{C}$, the set of $z\in\mathbb{C}$ such that $x$ and $z$ are each transcendental over the other is comeager and of full measure.
A: If $x$ is algebraic over $\mathbb{Q}(z)$ but not over $\mathbb{Q}$ then $z$ is algebraic over $\mathbb{Q}(x)$. Hence your notion of "complete" algebraic independence is the same as the classical https://en.wikipedia.org/wiki/Algebraic_independence
and there are plenty of such pairs $(x,y)$.
