AP
Posts:
137
Registered:
3/4/09


Re: Cubics and roots
Posted:
Oct 18, 2012 9:30 AM


On Tue, 16 Oct 2012 07:18:35 0500, quasi <quasi@null.set> wrote:
>AP wrote: >> >>P=X^33X+1 is irreducible on Q >> >>his roots are 2cos(pi/9) ; 2cos(7pi/9) ; 2cos(5pi/9) >>not obtainable via the four basic operations and >>squre, cube,.. roots of positifs numbers. > >The cubic formula expresses those roots using imaginary >numbers, but how can you _prove_ that there is no way >to express those roots using only real square roots, real >cube roots, and the usual arithmetic operations? > >quasi
cf book de Jean Paul Escofier P(X)=X^3+pX+q p, q dans Q
D=4p^327q^2>0
and d real such as d^2=D P irréducible on Q
a,b,c, roots of P : all are real
one suppose
there exist u_1, u_2, ..u_n real >0 p_1 p_2 ....p_n prime numbers
such as u_1^p_1 is in Q ; so u_1 is a radical of a real >0 on Q u_2^p_2 is in Q[u_1 ] ; so u_2 is a radical of a real >0 on Q[u_1] u_i^p_i is in Q[ u_1, u_2,.., u_(i1)] for i=2,..,n
and a is in L=Q[u_1,u_2,...u_n]
(so , a is express using only radicals of reals>0 and usual operations)
then prove : 1) Q[a,b,c] in L[d]
and , be the suit K_i :
K_0=Q K_1=Q[d] and K_(i+1)=Q[d,u_1,...,u_i] for i=1,..,n
and r the least integer i >=0 such that a in K_(i+1]
then prove 2) r>0
3) Q[a,b,c] in K_(r+1) and P irreducible on K_r
4) p_r=3, K_(r+1) is a normal extension of K_r
5) find a contradiction

