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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 5:08 PM
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On Thu, 18 Oct 2012 15:19:17 -0500, quasi <quasi@null.set> wrote:

>AP wrote:
>>quasi 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
>>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

>Thank you very much for taking the time to translate that
>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
>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 ? ;
>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=(a-b)(b-c)(c-a)
so Q(a,d) is in Q(a,b,c)

because P irreducible q=-abc =/=0


b-c=d/(-2a^2+q/a) and b-c 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]


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