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.
 A: 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 $\rm\:\mathbb Q\:.$
A: Let $p(x) = a_0+a_1x+\cdots +a_{n-1}x^{n-1} + x^n$ be a polynomial with coefficients in $\overline{\mathbb{Q}}$. For each $i$, $0\leq i\leq n-1$, let $a_i=b_{i1}, b_{i2},\ldots,b_{im_i}$ be the $m_i$ conjugates of $a_i$ (that is, the "other" roots of the monic irreducible polynomial with coefficients in $\mathbb{Q}$ that has $a_i$ as a root).
Now let $F = \mathbb{Q}[b_{11},\ldots,b_{n-1,m_{n-1}}]$. This field is Galois over $\mathbb{Q}$. Let $G=\mathrm{Gal}(F/\mathbb{Q})$. Now consider
$$q(x) = \prod_{\sigma\in G} \left( \sigma(a_0) + \sigma(a_1)x+\cdots + \sigma(a_{n-1})x^{n-1} + x^n\right).$$
This is a polynomial with coefficients in $F$, and any root of $p(x)$ is also a root of $q(x)$ (since one of the elements of $G$ is the identity, so one of the factors of $q(x)$ is $p(x)$).
The key observation is that if you apply any element $\tau\in G$ to $q(x)$, you get back $q(x)$ again:
$$\begin{align*}
\tau(q(x)) &= \tau\left(\prod_{\sigma\in G} \left( \sigma(a_0) + \sigma(a_1)x+\cdots + \sigma(a_{n-1})x^{n-1} + x^n\right)\right)\\
&= \prod_{\sigma\in G} \left( \tau\sigma(a_0) +\tau\sigma(a_1)x+\cdots + \tau\sigma(a_{n-1})x^{n-1} + x^n\right)\\
&= \prod_{\sigma'\in G} \left( \sigma'(a_0) + \sigma'(a_1)x+\cdots + \sigma'(a_{n-1})x^{n-1} + x^n\right)\\
&= q(x).
\end{align*}$$
That means that the coefficients of $q(x)$ must lie in the fixed field of $G$. But since $F$ is Galois over $\mathbb{Q}$, the fixed field is $\mathbb{Q}$. That is: $q(x)$ is actually a polynomial in $\mathbb{Q}[x]$. 
Thus, every root of $p(x)$ is the root of a polynomial with coefficients in $\mathbb{Q}$.
For an example of how this works, suppose you have $p(x) = x^3 - (2\sqrt{3}+\sqrt{5})x + 3$. The conjugate of $\sqrt{3}$ is $-\sqrt{3}$; the conjugate of $\sqrt{5}$ is $-\sqrt{5}$. The field $\mathbb{Q}[\sqrt{3},\sqrt{5}]$ already contains all the conjugates, and the Galois group over $\mathbb{Q}$ has four elements: the one that maps $\sqrt{3}$ to itself and $\sqrt{5}$ to $-\sqrt{5}$; the one the maps $\sqrt{3}$ to $-\sqrt{3}$ and $\sqrt{5}$ to itself; the one that maps $\sqrt{3}$ to $-\sqrt{3}$ and $\sqrt{5}$ to $-\sqrt{5}$; and the identity. So $q(x)$ would be the product of
$x^3 - (2\sqrt{3}+\sqrt{5})x + 3$, $x^3 - (-2\sqrt{3}+\sqrt{5})x+3$, $x^3 - (2\sqrt{3}-\sqrt{5})x + 3$, and $x^3 - (-2\sqrt{3}-\sqrt{5})x + 3$. If you multiply them out, you get
$$\begin{align*}
\Bigl( &x^3 - (2\sqrt{3}+\sqrt{5})x + 3\Bigr)\Bigl( x^3 + (2\sqrt{3}+\sqrt{5})x+3\Bigr)\\
&\times \Bigl(x^3 - (2\sqrt{3}-\sqrt{5})x + 3\Bigr)\Bigl( x^3 + (2\sqrt{3}-\sqrt{5})x + 3\Bigr)\\
&= \Bigl( (x^3+3)^2 - (2\sqrt{3}+\sqrt{5})^2x^2\Bigr)\Bigl((x^3+3)^2 - (2\sqrt{3}-\sqrt{5})^2x^2\Bigr)\\
&=\Bigl( (x^3+3)^2 - 17x^2 - 2\sqrt{15}x^2\Bigr)\Bigl( (x^3+3)^2 - 17x^2 + 2\sqrt{15}x^2\Bigr)\\
&= \Bigl( (x^3+3)^2 - 17x^2\Bigr)^2 - 60x^4,
\end{align*}$$
which has coefficients in $\mathbb{Q}$. 
A: 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 $\beta\in\mathbb A$.
