Showing that every polynomial over the Algebraic Numbers has a $0$ in the Algebraic Numbers.

Question

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.

Answer 1

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 $a\in\mathbb A$.

Answer 2

Let

$$p(x)=\sum_{k=0}^n a_kx^k\in\Bbb A[x]\;,\;\;\text{and define}\;\;K:=\Bbb Q(a_0,a_1,...,a_n)$$

Observe that $\;[K:\Bbb Q]<\infty\;$ .

Let $\;\alpha\;$ be a root of $\;p(x)\in K[x]\;$ in some extension of $\;K\;$, say $\;L:=K(\alpha)\;$ , but then also $\;[L:\Bbb Q]<\infty\;$ , and thus $\;\alpha\;$ is algebraic over $\;\Bbb Q\;\implies \alpha\in\Bbb A$ and we've finished.