# degree of algebraic number

Theorem. The degree (http://planetmath.org/DegreeOfAnAlgebraicNumber) of any algebraic number^{} $\alpha $ in the number field^{} $\mathbb{Q}(\vartheta )$ divides the degree of $\vartheta $. The zeroes of the characteristic polynomial^{} $g(x)$ of $\alpha $ consist of the algebraic conjugates of $\alpha $, each of which having equal multiplicity^{} as zero of $g(x)$.

*Proof.* Let the minimal polynomial^{} of $\alpha $ be

$$a(x):={x}^{k}+{a}_{1}{x}^{k-1}+\mathrm{\dots}+{a}_{k}$$ |

and all zeroes of this be ${\alpha}_{1}=\alpha ,{\alpha}_{2},\mathrm{\dots},{\alpha}_{k}$. Denote the canonical polynomial of
$\alpha $ with respect to the primitive element^{} (http://planetmath.org/SimpleFieldExtension) $\vartheta $ by $r(x)$; then

$$a(r(\vartheta ))=a(\alpha )=\mathrm{\hspace{0.33em}0}.$$ |

If $a(r(x)):=\phi (x)$, then the equation

$$\phi (x)=\mathrm{\hspace{0.33em}0}$$ |

has rational coefficients and is satisfied by $\vartheta $. Since the minimal polynomial $f(x)$ of $\vartheta $ is irreducible^{} (http://planetmath.org/IrreduciblePolynomial), it must divide $\phi (x)$ and all algebraic conjugates
${\vartheta}_{1}=\vartheta ,{\vartheta}_{2},\mathrm{\dots},{\vartheta}_{n}$ of $\vartheta $ make $\phi (x)$ zero. Hence we have

$$a({\alpha}^{(i)})=a(r({\vartheta}_{i}))=\mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}}\text{for}\mathit{\hspace{1em}}i=\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots},n$$ |

where the numbers ${\alpha}^{(i)}$ are the
$\mathbb{Q}(\vartheta )$-conjugates^{} (http://planetmath.org/CharacteristicPolynomialOfAlgebraicNumber) of $\alpha $. Thus these
$\mathbb{Q}(\vartheta )$-conjugates are roots of the irreducible equation $a(x)=0$, whence $a(x)$ must divide the characteristic polynomial $g(x)$. Let the power (http://planetmath.org/GeneralAssociativity)
${[a(x)]}^{m}$ exactly divide $g(x)$, when

$$g(x)={[a(x)]}^{m}b(x),a(x)\nmid b(x).$$ |

Antithesis: $\text{deg}(b(x))\geqq \mathrm{\hspace{0.17em}1}$ and $b(\beta )=\mathrm{\hspace{0.17em}0}$.

This implies that $g(\beta )=0$, i.e. $\beta $ is one of the numbers ${\alpha}^{(i)}$. Therefore, $\beta $ were a zero of $a(x)$ and thus $a(x)\mid b(x)$, which is impossible. Consequently,the antithesis is wrong, i.e. $b(x)$ is a constant, which must be 1 because $g(x)$ and $a(x)$ are monic polynomials. So, $g(x)={[a(x)]}^{m}$. Since

$$a(x)=(x-{\alpha}_{1})(x-{\alpha}_{2})\mathrm{\cdots}(x-{\alpha}_{k}),$$ |

it follows that

$$g(x)={(x-{\alpha}_{1})}^{m}{(x-{\alpha}_{2})}^{m}\mathrm{\cdots}{(x-{\alpha}_{k})}^{m}.$$ |

Hence $km=n$ and $k$ divides $n$, as asserted. Moreover, each ${\alpha}_{j}$ is a zero of order $m$ of $g(x)$, i.e. appears among the roots ${\alpha}^{(1)},{\alpha}^{(2)},\mathrm{\dots},{\alpha}^{(n)}$ of the equation $g(x)=0$ $m$ times.

Title | degree of algebraic number^{} |
---|---|

Canonical name | DegreeOfAlgebraicNumber |

Date of creation | 2013-03-22 19:08:51 |

Last modified on | 2013-03-22 19:08:51 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 12 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 11R04 |

Classification | msc 11C08 |

Classification | msc 12F05 |

Classification | msc 12E05 |

Related topic | KConjugates |

Related topic | CharacteristicPolynomialOfAlgebraicNumber |