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