# polynomial equation with algebraic coefficients

 $f_{1}(\alpha_{1},\,\ldots,\,\alpha_{k}),\;\ldots,\;f_{n}(\alpha_{1},\,\ldots,% \,\alpha_{k})$
 $\displaystyle x^{n}+f_{1}(\alpha_{1},\,\ldots,\,\alpha_{k})x^{n-1}+\ldots+f_{n% }(\alpha_{1},\,\ldots,\,\alpha_{k})\;=\;0$ (1)

are algebraic numbers [resp. algebraic integers].

Proof.  Let the minimal polynomial $x^{m}+a_{1}x^{m-1}+\ldots+a_{m}$ of $\alpha_{1}$ over $\mathbb{Z}$ have the zeros (http://planetmath.org/ZeroOfAFunction)

 $\alpha_{1}^{(1)}=\alpha_{1},\;\alpha_{1}^{(2)},\;\ldots,\;\alpha_{1}^{(m)}$

and denote by  $F(x;\,\alpha_{1},\,\alpha_{2},\,\ldots,\,\alpha_{k})$  the left hand side of the equation (1).  Consider the equation

 $\displaystyle G(x;\,\alpha_{2},\,\ldots,\,\alpha_{k})\;:=\;\prod_{i=1}^{m}F(x;% \,\alpha_{1}^{(i)},\,\alpha_{2},\,\ldots,\,\alpha_{k})\;=\;0.$ (2)

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

 $\alpha_{1}^{(1)},\;\alpha_{1}^{(2)},\;\ldots,\;\alpha_{1}^{(m)},\;\alpha_{2},% \;\ldots,\;\alpha_{k}$

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 $\alpha_{2},\,\ldots,\,\alpha_{k}$ with rational [resp. integer] coefficients.  Consequently, $G$ has the form

 $G\;=\;x^{h}+a_{1}^{\prime}(\alpha_{2},\,\ldots,\,\alpha_{k})x^{h-1}+\ldots+a_{% h}^{\prime}(\alpha_{2},\,\ldots,\,\alpha_{k})$

where the coefficients $a_{i}^{\prime}(\alpha_{2},\,\ldots,\,\alpha_{k})$ are polynoms in the numbers $\alpha_{j}$ with rational [resp. integer] coefficients.  As one continues similarly, removing one by one also $\alpha_{2},\,\ldots,\,\alpha_{k}$ which go back to the rational [resp. integer] coefficients of the corresponding minimal polynomials, one shall finally arrive at an equation

 $\displaystyle x^{s}+A_{1}x^{s-1}+\ldots+A_{s}\;=\;0,$ (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 $\alpha_{1},\,\ldots,\,\alpha_{k}$ but are rational numbers [resp. integers].
Accordingly, the roots of (1) are algebraic numbers [resp. algebraic integers], Q.E.D.

Title polynomial equation with algebraic coefficients PolynomialEquationWithAlgebraicCoefficients 2013-03-22 19:07:37 2013-03-22 19:07:37 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 11R04 monic equation with algebraic coefficients