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# polynomial equation with algebraic coefficients

If $\alpha_{1},\,\ldots,\,\alpha_{k}$ are algebraic numbers [resp. algebraic integers] and

$f_{1}(\alpha_{1},\,\ldots,\,\alpha_{k}),\;\ldots,\;f_{n}(\alpha_{1},\,\ldots,% \,\alpha_{k})$ |

polynomials in $\alpha_{1},\,\ldots,\,\alpha_{k}$ with rational [resp. integer] coefficients, then all complex roots of the equation

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

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

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.

## Mathematics Subject Classification

11R04*no label found*

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