Theorem. The degree 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
and all zeroes of this be . Denote the canonical polynomial of with respect to the primitive element by ; then
If , then the equation
has rational coefficients and is satisfied by . Since the minimal polynomial of is irreducible, it must divide and all algebraic conjugates of make zero. Hence we have
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.