polynomial equation with algebraic coefficients
are algebraic numbers [resp. algebraic integers].
Proof. Let the minimal polynomial of over have the zeros (http://planetmath.org/ZeroOfAFunction)
and denote by the left hand side of the equation (1). Consider the equation
Here, the coefficients of the polynomial are polynomials in the numbers
with rational [resp. integer] coefficients. Thus the coefficients of are symmetric polynomials in the numbers :
By the fundamental theorem of symmetric polynomials, the coefficients of are polynomials in with rational [resp. integer] coefficients. Consequently, has the form
where the coefficients are polynoms in the numbers with rational [resp. integer] coefficients. As one continues similarly, removing one by one also which go back to the rational [resp. integer] coefficients of the corresponding minimal polynomials, one shall finally arrive at an equation
among the roots of which there are the roots of (1); the coefficients do no more explicitely depend on the algebraic numbers but are rational numbers [resp. integers].
Accordingly, the roots of (1) are algebraic numbers [resp. algebraic integers], Q.E.D.
|Title||polynomial equation with algebraic coefficients|
|Date of creation||2013-03-22 19:07:37|
|Last modified on||2013-03-22 19:07:37|
|Last modified by||pahio (2872)|
|Synonym||monic equation with algebraic coefficients|