TITLE: On the meaning of being algebraically closed
QUESTION [28 upvotes]: 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.

REPLY [4 votes]: HINT $\ $ It boils down to the transitivity of algebraic extensions, which boils down to  
$\rm\qquad\quad\ K\ =\   \mathbb Q(\alpha)\: =\ \mathbb Q\langle1,\alpha,\:\ldots\:,\alpha^{m-1}\rangle,\quad\ \ F\ =\ K(\beta)\: =\ K\langle1,\beta,\:\ldots\:,\beta^{\:n-1}\rangle $  
$\rm\quad \Rightarrow\ \ F\ =\ \mathbb Q(\alpha)(\beta)\: =\ \mathbb{Q}  \big\langle\{1,\alpha,\:\ldots\:,\alpha^{m-1} \}\cdot \{1,\beta,\:\ldots\:,\beta^{\:n-1}\}\big\rangle\: =\ \mathbb Q\big\langle\alpha^i\:\beta^j\big\rangle_{i\:<\:m,\ j\:<\:n}$  
which follows simply be employing the minimal polynomials of $\rm\:\alpha,\:\beta\:$ as rewrite rules $\rm\:\alpha^m \to\ f(\alpha),\ \ \beta^{\:n}\to\ g(\beta)\:$ to reduce all powers of $\rm\:\alpha,\:\beta\:$ to powers $\rm\:< m,\:n\:,\:$ respectively.  
Therefore, since $\rm\:[F:\mathbb Q]\:$ is finite, we deduce that $\rm\:\beta\:$ is algebraic over $\rm\:\mathbb Q\:.$ More explicitly, since, by above, $\rm\:dim\ F/\mathbb Q\ =\ m\:n\:,\: $ the $\rm\ m\:n+1\ $ elements $\rm\:1,\:\beta,\:\ldots\:,\beta^{\:m\:n}$ are linearly dependent over $\mathbb Q\:,\:$ yielding a polynomial $\rm\:0\ne h(x)\in \mathbb Q[x]\:$ such that $\rm\:h(\beta) = 0\:.\:$ Therefore $\rm\:\beta\:$ is algebraic over $\rm\:\mathbb Q\:.$