# 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.

Title | polynomial equation with algebraic coefficients |
---|---|

Canonical name | PolynomialEquationWithAlgebraicCoefficients |

Date of creation | 2013-03-22 19:07:37 |

Last modified on | 2013-03-22 19:07:37 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 10 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 11R04 |

Synonym | monic equation with algebraic coefficients |