# 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

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

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 (http://planetmath.org/SimpleFieldExtension) $\vartheta$ 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 (http://planetmath.org/IrreduciblePolynomial), 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 (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

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

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.

Title degree of algebraic number DegreeOfAlgebraicNumber 2013-03-22 19:08:51 2013-03-22 19:08:51 pahio (2872) pahio (2872) 12 pahio (2872) Theorem msc 11R04 msc 11C08 msc 12F05 msc 12E05 KConjugates CharacteristicPolynomialOfAlgebraicNumber