This originates from Pinter's Abstract Algebra, Chapter 29, Exercise G5.
Any complex number which is algebraic over $\Bbb{Q}$ is called an algebraic number. The set of all the algebraic numbers is a field, which we shall designate by $\Bbb{A}$.
Let $a(x)=a_0+a_1x+\cdots+a_nx^n$ be in $\Bbb{A}[x]$, and let $c$ be any root of $a(x)$. Note $\Bbb{Q}(a_0, a_1, \cdots, a_n)$ is a finite extension of $\Bbb{Q}$.
Let $\Bbb{Q}(a_0, \cdots, a_n) = \Bbb{Q}_1$. Since $a(x) \in \Bbb{Q}_1[x], c$ is algebraic over $\Bbb{Q}_1$.
Prove: $c\in\Bbb{A}$
Given $c$ is algebraic over $\Bbb{Q}_1$, $\Bbb{Q}_1(c)$ is a finite extension of $\Bbb{Q}_1$, which in turn is a finite extension of $\Bbb{Q}$. Hence $\Bbb{Q}_1(c)$ is a finite extension of $\Bbb{Q}$.
$[\Bbb{Q}_1(c):\Bbb{Q}] = [\Bbb{Q}_1(c):\Bbb{Q}_1][\Bbb{Q}_1:\Bbb{Q}]$. This implies the existence of a minimal polynomial over $\Bbb{Q}$ such that $c$ is a root. Hence $c$ is algebraic over $\Bbb{Q}$ and therefore $c\in\Bbb{A}$.
Is this correct?