degree of algebraic number
Theorem. The degree (http://planetmath.org/DegreeOfAnAlgebraicNumber) of any algebraic number in the number field divides the degree of . The zeroes of the characteristic polynomial of consist of the algebraic conjugates of , each of which having equal multiplicity as zero of .
Proof. Let the minimal polynomial of be
If , then the equation
has rational coefficients and is satisfied by . Since the minimal polynomial of is irreducible (http://planetmath.org/IrreduciblePolynomial), it must divide and all algebraic conjugates of make zero. Hence we have
where the numbers are the -conjugates (http://planetmath.org/CharacteristicPolynomialOfAlgebraicNumber) of . Thus these -conjugates are roots of the irreducible equation , whence must divide the characteristic polynomial . Let the power (http://planetmath.org/GeneralAssociativity) exactly divide , when
Antithesis: and .
This implies that , i.e. is one of the numbers . Therefore, were a zero of and thus , which is impossible. Consequently,the antithesis is wrong, i.e. is a constant, which must be 1 because and are monic polynomials. So, . Since
it follows that
Hence and divides , as asserted. Moreover, each is a zero of order of , i.e. appears among the roots of the equation times.
|Title||degree of algebraic number|
|Date of creation||2013-03-22 19:08:51|
|Last modified on||2013-03-22 19:08:51|
|Last modified by||pahio (2872)|