polynomial equation with algebraic coefficients

If ${\alpha }_1,\mathrm{\dots },{\alpha }_k$ are algebraic numbers [resp. algebraic integers] and

$$f_1({\alpha }_1,\mathrm{\dots },{\alpha }_k),\mathrm{\dots },f_n({\alpha }_1,\mathrm{\dots },{\alpha }_k)$$

polynomials in ${\alpha }_1,\mathrm{\dots },{\alpha }_k$ with rational [resp. integer] coefficients, then all complex roots of the equation

$x^n+f_1({\alpha }_1,\mathrm{\dots },{\alpha }_k)x^{n-1}+\mathrm{\dots }+f_n({\alpha }_1,\mathrm{\dots },{\alpha }_k)=\mathrm{\hspace{0.33em}0}$(1)

are algebraic numbers [resp. algebraic integers].

Proof.  Let the minimal polynomial $x^m+a_1{x}^{m-1}+\mathrm{\dots }+a_m$ of ${\alpha }_1$ over $\mathbb{Z}$ have the zeros (http://planetmath.org/ZeroOfAFunction)

$${\alpha }_1^{(1)}={\alpha }_1,{\alpha }_1^{(2)},\mathrm{\dots },{\alpha }_1^{(m)}$$

and denote by  $F(x;{\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_k)$  the left hand side of the equation (1).  Consider the equation

$G(x;{\alpha }_2,\mathrm{\dots },{\alpha }_k):={\displaystyle \prod _{i=1}^m}F(x;{\alpha }_1^{(i)},{\alpha }_2,\mathrm{\dots },{\alpha }_k)=\mathrm{\hspace{0.33em}0}.$(2)

Here, the coefficients of the polynomial $G$ are polynomials in the numbers

$${\alpha }_1^{(1)},{\alpha }_1^{(2)},\mathrm{\dots },{\alpha }_1^{(m)},{\alpha }_2,\mathrm{\dots },{\alpha }_k$$

with rational [resp. integer] coefficients.  Thus the coefficients of $G$ are symmetric polynomials $g_j$ in the numbers ${\alpha }_1^{(i)}$:

$$G=\sum _j{g}_j{x}^j$$

By the fundamental theorem of symmetric polynomials, the coefficients $g_j$ of $G$ are polynomials in${\alpha }_2,\mathrm{\dots },{\alpha }_k$ with rational [resp. integer] coefficients.  Consequently, $G$ has the form

$$G=x^h+a_1^{\prime }({\alpha }_2,\mathrm{\dots },{\alpha }_k)x^{h-1}+\mathrm{\dots }+a_h^{\prime }({\alpha }_2,\mathrm{\dots },{\alpha }_k)$$

where the coefficients $a_i^{\prime }({\alpha }_2,\mathrm{\dots },{\alpha }_k)$ are polynoms in the numbers ${\alpha }_j$ with rational [resp. integer] coefficients.  As one continues similarly, removing one by one also ${\alpha }_2,\mathrm{\dots },{\alpha }_k$ which go back to the rational [resp. integer] coefficients of the corresponding minimal polynomials, one shall finally arrive at an equation

$x^s+A_1{x}^{s-1}+\mathrm{\dots }+A_s=\mathrm{\hspace{0.33em}0},$(3)

among the roots of which there are the roots of (1); the coefficients $A_{\nu }$ do no more explicitely depend on the algebraic numbers ${\alpha }_1,\mathrm{\dots },{\alpha }_k$ but are rational numbers [resp. integers].
Accordingly, the roots of (1) are algebraic numbers [resp. algebraic integers], Q.E.D.