AP
Posts:
137
Registered:
3/4/09


Re: Cubics and roots
Posted:
Oct 18, 2012 5:08 PM


On Thu, 18 Oct 2012 15:19:17 0500, quasi <quasi@null.set> wrote:
>AP wrote: >>quasi 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? >> >>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 > >Thank you very much for taking the time to translate that >exercise. > >And yes, if that exercise can be completed successfully, then >it _does_ prove that the roots of x^3  3x + 1 cannot be >obtained using just arithmetic operations and the taking of >real square roots and real cube roots. > >However I don't yet accept the validity of the logic of the >exercise. > >The author suggests a sequence of 5 steps. > >It's clear to me that if step (1) goes through successfully, >the rest is easy  steps (2),(3),(4),(5) follow smoothly. > >But I don't see how to accomplish step (1). > >Note that L = L(d) since, by hypothesis, K_1 = Q(d). d is in K_1, but why d would be in L ? ; L=Q[u_1,u_2,...u_n] >Also, by definition of L, we have a in L. > >The claim of step (1) is that b,c are also in L. > >I don't see how that can be justified. because Q(a,d)=Q(a,b,c) (see below) so , cf Q in L, Q(a,d) is in L(a,d) but a is in L and L(a,d)=L(a) so Q(a,b,c)=Q(a,d) is in L(a,d)=L(d)  proof of Q(a,d)=Q(a,b,c)
we have d=(ab)(bc)(ca) so Q(a,d) is in Q(a,b,c)
because P irreducible q=abc =/=0
(ab)(ca)=2a^2+q/a
bc=d/(2a^2+q/a) and bc is in K[a,d] and b+c=a is in K[a,d]
so, b and c also and K[a,b,c] is in Q[a,d]
> >quasi

