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Topic: Cubics and roots
Replies: 24   Last Post: Oct 19, 2012 4:15 PM

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Posts: 137
Registered: 3/4/09
Re: Cubics and roots
Posted: Oct 18, 2012 9:30 AM
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On Tue, 16 Oct 2012 07:18:35 -0500, quasi <quasi@null.set> wrote:

>AP wrote:
>>P=X^3-3X+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, q dans Q


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_(i-1)] 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

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

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