The subset of $\mathbb{C}$ with all roots of polynomials in $\mathbb{Q}[x]$ is a field? Let $\overline{\mathbb{Q}}$ be a subset of $\mathbb{C}$ where the elements are roots of the polynomials in $\mathbb{Q}[x]$ and let n be an integer.


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*Show that $\overline{\mathbb{Q}}$ is a field.

*Show that there are irreducibles polynomials of degree n with rational coefficients.


I have the feeling that $\overline{\mathbb{Q}}=\mathbb{C}$ but I really don't know how to do this exercise. 
 A: The set $\overline{\mathbb{Q}}$ cannot be equal to $\mathbb{C}$ as it is infinite denumerable, since it is a denumerable union of finite sets:
$$\overline{\mathbb{Q}}=\bigcup_{f\in\mathbb{Q}[X]\setminus\{0\}}\{x\in\mathbb{C}\textrm{ t.q. }f(x)=0\}.$$
For question $1.$, let $\alpha$ and $\beta$ in $\overline{\mathbb{Q}}$, then $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$ has finite dimensions over $\mathbb{Q}$ and so has $\mathbb{Q}(\alpha+\beta)$. Notice that $\mathbb{Q}(\alpha+\beta)\subseteq\mathbb{Q}(\alpha,\beta)$ which has finite degree using the telescopic basis theorem, hence: $$\alpha+\beta\in\overline{\mathbb{Q}}.$$ The same thing can be said about $\mathbb{Q}(\alpha\beta)$ but let us use a constructive approach. Let $\mu_x$, respectively, $\mu_y$ in $\mathbb{Q}[X]\setminus\{0\}$ such that $\mu_x(x)=0$, respectively $\mu_y(y)=0$, then notice that:
$$\textrm{res}_Y\left(\mu_x(X),X^{\deg(\mu_y)}\mu_y\left(\frac{Y}{X}\right)\right)$$
is a polynomial of $\mathbb{Q}[X]$ which admits $\alpha\beta$ as a root.
For question $2.$ use Eisenstein's criterion, for example on $X^n-2$.
A: A complex number $\alpha$ is algebraic (over $\mathbb{Q}$) iff there are some $a_n \in \mathbb{Q}$ such that $$\sum_{n=0}^d a_n \alpha^n=0 \qquad (a_0 \ne 0, a_d \ne 0)$$ ie. there is a linear dependence between the powers of $\alpha$. 
If $d$ is the minimum possible, then $\mathbb{Q}[\alpha]$ (the smallest ring containing $\mathbb{Q}$ and $\alpha$, whose elements are of the form $\sum_{n=0}^N c_n \alpha^n$) 
is a $d$ dimensional $\mathbb{Q}$-vector space.
Now take any (non-zero) element $\beta \in \mathbb{Q}[\alpha]$ then there is a linear dependence between the powers of $\beta$, because otherwise $\mathbb{Q}[\beta]$ would be an infinite dimensional $\mathbb{Q}$-vector space, impossible since $\mathbb{Q}[\beta] \subseteq \mathbb{Q}[\alpha]$.
Thus we have a linear dependence of the form
$$\sum_{n=0}^{d'} b_n \beta^n =0\quad \implies \quad \beta^{-1} = \sum_{n=1}^{d'} b_n \beta^{n-1}$$
and hence $\beta^{-1} \in \mathbb{Q}[\alpha]$.
Qed $\mathbb{Q}[\alpha]$ is a field.

Generalize this to $\mathbb{Q}[\alpha_1,\alpha_2,\ldots,\alpha_k]$ with $k$ arbitrary large to obtain that $\overline{\mathbb{Q}}$ (which is of the form $\lim_{k \to \infty}\mathbb{Q}[\alpha_1,\alpha_2,\ldots,\alpha_k]$) is a field.
