On the meaning of being algebraically closed The definition of algebraic number is that $\alpha$ is an algebraic number if there is a nonzero polynomial $p(x)$ in $\mathbb{Q}[x]$ such that $p(\alpha)=0$.
By algebraic closure, every nonconstant polynomial with algebraic coefficients has algebraic roots; then, there will be also a nonconstant polynomial with rational coefficients that has those roots.
I feel uncomfortable with the idea that the root of a polynomial with algebraic coefficients is again algebraic; why are we sure that for every polynomial in $\mathbb{\bar{Q}}[x]$ we could find a polynomial in $\mathbb{Q}[x]$ that has the same roots?
I apologize if I'm asking something really trivial or my question comes from a big misunderstanding of basic concepts.

Define $\mathbb A$ as the set of all complex roots of rational polynomials; then we want to prove that $\mathbb A$ is algebraically closed.
Let $f=X^n+a_{n-1}X^{n-1}+\cdots+a_0$ be a polynomial with coefficients in $\mathbb A$. I assume that we already know that $\mathbb A$ is a field, so taking the leading coefficient to be $1$ does not lose generality. Let $S$ be $\mathbb Q[a_0,\ldots,a_{n-1}]$, the smallest ring extension of $\mathbb Q$ that contains all of the coefficients of $f$. $S$ is a finite-dimensional vector space over $\mathbb Q$, because it is spanned by all products of powers of the $a_i$'s up to the degree of the rational polynomial each $a_i$ is a root in.
Now let $\beta\in\mathbb C$ be a root of $f$. Then $S[\beta]$, the smallest ring extension of $S$ that contains $\beta$, is a finite-dimensional vector space over $S$ (actually a finitely generated module, except that it turns out that $S$ is in fact a field), because it is spanned by powers of $\beta$ from $1$ up to $\beta^{n-1}$. Thus, in particular $S[\beta]$ is also a finite-dimensional vector space over $\mathbb Q$.
Now take sufficiently many powers of $\beta$, enough that there are more of them than the dimension of $S[\beta]$ over $\mathbb Q$. They must then be linearly dependent. But a nontrivial rational linear relation between powers of $\beta$ is exactly a rational polynomial that has $\beta$ as a root. So $\beta\in\mathbb A$.
