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Homedegree of algebraic number

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# degree of algebraic number

Theorem. The degree 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

$a(x)\;:=\;x^{k}+a_{1}x^{{k-1}}+\ldots+a_{k}$ |

and all zeroes of this be $\alpha_{1}=\alpha,\,\alpha_{2},\,\ldots,\,\alpha_{k}$. Denote the canonical polynomial of $\alpha$ with respect to the primitive element $\vartheta$ by $r(x)$; then

$a(r(\vartheta))\;=\;a(\alpha)\;=\;0.$ |

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

$\varphi(x)\;=\;0$ |

has rational coefficients and is satisfied by $\vartheta$. Since the minimal polynomial $f(x)$ of $\vartheta$ is irreducible, it must divide $\varphi(x)$ and all algebraic conjugates $\vartheta_{1}=\vartheta,\,\vartheta_{2},\,\ldots,\,\vartheta_{n}$ of $\vartheta$ make $\varphi(x)$ zero. Hence we have

$a(\alpha^{{(i)}})\;=\;a(r(\vartheta_{i}))\;=\;0\quad\mbox{for}\quad i\,=\,1,\,% 2,\,\ldots,\,n$ |

where the numbers $\alpha^{{(i)}}$ are the $\mathbb{Q}(\vartheta)$-conjugates 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 $[a(x)]^{m}$ exactly divide $g(x)$, when

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

Antithesis: $\mbox{deg}(b(x))\,\geqq\,1$ and $b(\beta)\,=\,0$.

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

$a(x)\;=\;(x\!-\!\alpha_{1})(x\!-\!\alpha_{2})\cdots(x\!-\!\alpha_{k}),$ |

it follows that

$g(x)\;=\;(x\!-\!\alpha_{1})^{m}(x\!-\!\alpha_{2})^{m}\cdots(x\!-\!\alpha_{k})^% {m}.$ |

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 $\alpha^{{(1)}},\,\alpha^{{(2)}},\,\ldots,\,\alpha^{{(n)}}$ of the equation $g(x)=0$ $m$ times.

## Mathematics Subject Classification

11R04*no label found*11C08

*no label found*12F05

*no label found*12E05

*no label found*

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