Showing that every polynomial over the Algebraic Numbers has a $0$ in the Algebraic Numbers. Let $\mathbb{A}$ denote the field of Algebraic Numbers: the field of all complex numbers that are algebraic over $\mathbb{Q}$. Assuming that every polynomial over $\mathbb{C}$ has a $0$ in $\mathbb{C}$ how would you go about proving that every polynomial over $\mathbb{A}$ has a $0$ in $\mathbb{A}$? I've tried contradiction but all that has lead me to is concluding that the $0$ of some $p(x)\in \mathbb{A}[x]$, say $\alpha$, is transcendental over $\mathbb{Q}$, which doesn't help much. I'd appreciate a hint as where to start the proof.

Let $P\in\mathbb A[X]$. Then there is $a\in\mathbb C$ such that $P(a)=0$. This shows that $a$ is algebraic over $\mathbb A$, so over $\mathbb Q$ (why?). Thus we get $a\in\mathbb A$.
