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.