How to solve a cyclic quintic in radicals? Galois theory tells us that \frac{z^{11}-1}{z-

Dashawn Clark 2022-04-25 Answered

How to solve a cyclic quintic in radicals?
Galois theory tells us that
z111z1=z10+z9+z8+z7+z6+z5+z4+z3+z2+z+1 can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be ζ1,ζ2,,ζ10, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].
A0=x1+x2+x3+x4+x5A1=x1+ζx2+ζ2x3+ζ3x4+ζ4x5A2=x1+ζ2x2+ζ4x3+ζx4+ζ3x5A3=x1+ζ3x2+ζx3+ζ4x4+ζ2x5A4=x1+ζ4x2+ζ3x3+ζ2x4+ζx5
Once one has A0,,A4 one easily gets x1,,x5. It's easy to find A0. The point is that τ takes Aj to ζjAj and so takes Aj5 to Aj5. Thus Aj5 can be written down in terms of rationals (if that's your starting field) and powers of ζ. Alas, here is where the algebra becomes difficult. The coefficients of powers of ζ in A15 are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have A1 as a fifth root of a certain explicit complex number. Then one can express the other Aj in terms of A1. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.

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Answers (2)

Landyn Whitney
Answered 2022-04-26 Author has 19 answers
The solution of a solvable quintic,
x5+ax4+bx3+cx2+dx+e=0
can be given by, x=15(a+z115+z215+z315+z415)
and the zi are the roots of the quartic. However, there is also a form where one takes a fifth root only once.
p=11
Let, x5+x44x33x2+3x+1=0
The five roots xk for k=0,1,2,3,4 are,
xk=15(1βk1+1βk0+11βk1+aβk2+bβk3)
where, a=t114(1+55+10(5+5))
b=t114(31+5510(85+315))
βk=ζ5k,(ab11)15
ζ5=e2π,i5
p=31
As discussed in this post,
x5+x412x321x2+x+5=0
xk=15(1βk11βk0+31βk1+aβk2+bβk3)
where, a=t314(11+55+10(25115))
b=t314(155+10(15255))
βk=ζ5k,(ab31)15
and so on for other ' p=10n+1.
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Jayla Matthews
Answered 2022-04-27 Author has 18 answers
Step 1
ω1=5{(663125ζ5+4513125ζ52+1763125ζ53+2863125ζ54)}
ω2=2{1120+(14ζ54)ω1+(544ζ5+1544ζ52+544ζ53+544ζ54)ω12+(25121ζ575242ζ5275484ζ53+75242ζ54)ω13+(3752662ζ5+6251331ζ52+6252662ζ53+43755324ζ54)ω14}
Step 2
the roots of the polynomial are given by
110+12ω1+(522ζ52511ζ53+511ζ54)ω12+(75242ζ5+150121ζ52+75121ζ53+125121ζ54)ω13+(16251331ζ5+10001331ζ52+51252662ζ53+3751331ζ54)ω14ω2
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