Re: Relationship between polynomial of Galois group "[2^4]E(4)" and a polynomial of degree 16.

On Nov 21, 7:34 pm, Gerry <Gerry...@xxxxxxxxx> wrote:
On Nov 21, 6:30 pm, Gerry <Gerry...@xxxxxxxxx> wrote:





Hi all,

the irreducible polynomial P1 (see below) of degree 8 has Galois
group :"[2^4]E(4)".
What is the Galois group of the irreucible polynomial P2 of degree 16?
Is there a relationship between the two Galois groups?

 P1= a8*x^8 + a7*x^7 + a6*x^6 + a5*x^5 + a4*x^4 + a3*x^3 + a2*x^2 +
a1*x^1 + a0

 coefficients of P1 with b,c in Q :

 a0= 1
 a1= 0
 a2= 0
 a3= 0
 a4=     c^4
 a5= 4* c^3 * x
 a6= 6* c^2 * x^2
 a7= 4* c   *  x^3
 a8=             x^4

 P2= a16*x^16 + a15*x^15 + a14*x^14 + a13*x^13 + a12*x^12 + a11*x^11 +
a10*x^10 + a9*x^9
    + a8*x^8 + a7*x^7 + a6*x^6 + a5*x^5 + a4*x^4 + a3*x^3 + a2*x^2 +
a1*x^1 + a0

 coefficients of P2 with b,c in Q :

 a0=   1
 a1=   0
 a2=   0
 a3=   0
 a4= -      c^4
 a5= - 4 * c^3   * b
 a6= - 6 * c^2   * b^2
 a7= - 4 * c      * b^3
 a8=          c^8 - b^4
 a9=     8 * c^7 * b
 a10= 28 * c^6 * b^2
 a11= 56 * c^5 * b^3
 a12= 70 * c^4 * b^4
 a13= 56 * c^3 * b^5
 a14= 28 * c^2 * b^6
 a15=   8 * c   *  b^7
 a16=                b^8

Regards

Gerry

Correction

coefficients of P1 with b,c in Q :

The P1 coefficients  should be:

 a0= 1
 a1= 0
 a2= 0
 a3= 0
 a4=    c^4
 a5= 4* c^3 * b
 a6= 6* c^2 * b^2
 a7= 4* c   * b^3
 a8=          b^4- Hide quoted text -

- Show quoted text -

Maybe an example helps :

Consider the polynomial of degree 24 which i generated using f1 and
f2 for b=1 and c=1:

H(x)=x^24 + 12*x^23 + 66*x^22 + 220*x^21 + 495*x^20 + 792*x^19 +
924*x^18 + 792*x^17 + 495*x^16 + 220*x^15 + 66*x^14 + 12*x^13 + x^12 +
1

H*x) factors into f1, an irreducible polynomial of degree 8 of galois
group 2^4E(4)
and f2 an irreducible polynomial of degree 16 of a galois group i
don't know at the moment :

Now f1 and f2 are:

f1 = x^8 + 4*x^7 + 6*x^6 + 4*x^5 + x^4 + 1

this polynomial has the roots numerically :

rf1[1]=-0.06131851429853062278308 + 0.805945548460810102917548*I
rf1[2]=-0.06131851429853062278308 - 0.805945548460810102917548*I
rf1[3]=0.536119490512812745849395 + 0.341228394823735525030110*I
rf1[4]=0.536119490512812745849395 - 0.341228394823735525030110*I
rf1[5]=-1.53611949051281274584939 + 0.341228394823735525030110*I
rf1[6]=-1.53611949051281274584939 - 0.341228394823735525030110*I
rf1[7]=-0.93868148570146937721691 + 0.805945548460810102917548*I
rf1[8]=-0.93868148570146937721691 - 0.805945548460810102917548*I

f2 = x^16 + 8*x^15 + 28*x^14 + 56*x^13 + 70*x^12 + 56*x^11 + 28*x^10 +
8*x^9 - 4*x^7 - 6*x^6 - 4*x^5 - x^4 + 1

this polynomial has the roots numerically :

rf2[1]=0.608849023263346073043136 + 0.116706169944045006043875*I
rf2[2]=0.608849023263346073043136 - 0.116706169944045006043875*I
rf2[3]=0.191790083745827679902514 + 0.698135062198984557684683*I
rf2[4]=0.191790083745827679902514 - 0.698135062198984557684683*I
rf2[5]=0.394585136800061869958176 + 0.539873616581755778414127*I
rf2[6]=0.394585136800061869958176 - 0.539873616581755778414127*I
rf2[7]=-1.39458513680006186995817 + 0.539873616581755778414127*I
rf2[8]=-1.39458513680006186995817 - 0.539873616581755778414127*I
rf2[9]=-1.60884902326334607304313 + 0.116706169944045006043875*I
rf2[10]=-1.6088490232633460730431 - 0.116706169944045006043875*I
rf2[11]=-1.1917900837458276799025 - 0.698135062198984557684683*I
rf2[12]=-1.1917900837458276799025 + 0.698135062198984557684683*I
rf2[13]=-0.6505780632485201430201 + 0.859418163306282245877204*I
rf2[14]=-0.6505780632485201430201 - 0.859418163306282245877204*I
rf2[15]=-0.3494219367514798569798 + 0.859418163306282245877204*I
rf2[16]=-0.3494219367514798569798 - 0.859418163306282245877204*I

The exact definition for the roots of f1 and f2 is:

Let: a=1 ,b=1 ,c=1 ,n=12 and e=exp((2*i-1+n%2)*Pi*I/n

Then the roots for f1 are :

x1(i)=(-c-(c^2-4*a*b*a*e)^(1/2))/(2*a*b) for i=[2,5,8,11]
x2(i)=(-c+(c^2-4*a*b*a*e)^(1/2))/(2*a*b) for i=[2,5,8,11]


and the roots for f2 are :

x1(i)=(-c-(c^2-4*a*b*a*e)^(1/2))/(2*a*b) for i=[1,3,4,6,7,9,10,12]
x2(i)=(-c+(c^2-4*a*b*a*e)^(1/2))/(2*a*b) for i=[1,3,4,6,7,9,10,12]

So as you see there is a relationship between the two polynomials f1
and f2.
Both multiplied give a complete set of indexes i from 1 to 12 for the
roots.

I tried to find the galois group of f2 using KASH3 but unfortunately i
hat to interrupt the routine because it was taking too long.

Regards

Gerry
.

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