# independence of characteristic polynomial on primitive element

The simple field extension $\mathbb{Q}(\vartheta )/\mathbb{Q}$ where $\vartheta $ is an algebraic number^{} of degree (http://planetmath.org/DegreeOfAnAlgebraicNumber) $n$ may be determined also by using another primitive element^{} $\eta $. Then we have

$$\eta \in \mathbb{Q}(\vartheta ),$$ |

whence, by the entry degree of algebraic number^{}, the degree of $\eta $ divides the degree of $\vartheta $. But also

$$\vartheta \in \mathbb{Q}(\eta ),$$ |

whence the degree of $\vartheta $ divides the degree of $\eta $. Therefore any possible primitive element of the field extension has the same degree $n$. This number is the degree of the number field^{} (http://planetmath.org/NumberField), i.e. the degree of the field extension, as comes clear from the entry canonical form of element of number field.

Although the characteristic polynomial^{}

$$g(x):=\prod _{i=1}^{n}[x-r({\vartheta}_{i})]=\prod _{i=1}^{n}(x-{\alpha}^{(i)})$$ |

of an element $\alpha $ of the algebraic number field $\mathbb{Q}(\vartheta )$ is based on the primitive element $\vartheta $, the equation

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

in the entry http://planetmath.org/node/12050degree of algebraic number shows that the polynomial^{} is fully determined by the algebraic conjugates of $\alpha $ itself and the number $m$ which equals the degree $n$ divided by the degree $k$ of
$\alpha $.

The above stated makes meaningful to define the norm and the trace functions in an algebraic number field as follows.

Definition. If $\alpha $ is an element of the number field $\mathbb{Q}(\vartheta )$, then the *norm*
$\text{N}(\alpha )$ and the *trace* $\text{S}(\alpha )$ of $\alpha $ are the product and the sum, respectively, of all http://planetmath.org/node/12046$\mathbb{Q}\mathtt{}\mathrm{(}\mathrm{\vartheta}\mathrm{)}$-conjugates^{} ${\alpha}^{(i)}$ of $\alpha $.

Since the coefficients of the characteristic equation of $\alpha $ are rational, one has

$$\text{N}:\mathbb{Q}(\vartheta )\to \mathbb{Q}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\text{S}:\mathbb{Q}(\vartheta )\to \mathbb{Q}.$$ |

In fact, one can infer from (1) that

$\text{N}(\alpha )={a}_{k}^{m},\text{S}(\alpha )=-m{a}_{1},$ | (2) |

where ${x}^{k}+{a}_{1}{x}^{k-1}+\mathrm{\dots}+{a}_{k}$ is the minimal polynomial^{} of $\alpha $.

Title | independence of characteristic polynomial on primitive element |

Canonical name | IndependenceOfCharacteristicPolynomialOnPrimitiveElement |

Date of creation | 2014-02-04 8:07:18 |

Last modified on | 2014-02-04 8:07:18 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 10 |

Author | pahio (2872) |

Entry type | Topic |

Classification | msc 11R04 |

Classification | msc 12F05 |

Classification | msc 11C08 |

Classification | msc 12E05 |

Synonym | norm and trace functions in number field |

Related topic | Norm |

Related topic | NormAndTraceOfAlgebraicNumber |

Related topic | PropertiesOfMathbbQvarthetaConjugates |

Related topic | DiscriminantInAlgebraicNumberField |

Defines | norm in number field |

Defines | trace in number field |

Defines | norm |

Defines | trace |