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

Question

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
$\frac{z^{11} - 1}{z - 1} = z^{10} + z^{9} + z^{8} + z^{7} + z^{6} + z^{5} + z^{4} + z^{3} + z^{2} + z + 1$ 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 $ζ^{1}, ζ^{2}, \dots, ζ^{10}$ , 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].
$\begin{aligned} A_{0} & = x_{1} + x_{2} + x_{3} + x_{4} + x_{5} \\ A_{1} & = x_{1} + ζ x_{2} + ζ^{2} x_{3} + ζ^{3} x_{4} + ζ^{4} x_{5} \\ A_{2} & = x_{1} + ζ^{2} x_{2} + ζ^{4} x_{3} + ζ x_{4} + ζ^{3} x_{5} \\ A_{3} & = x_{1} + ζ^{3} x_{2} + ζ x_{3} + ζ^{4} x_{4} + ζ^{2} x_{5} \\ A_{4} & = x_{1} + ζ^{4} x_{2} + ζ^{3} x_{3} + ζ^{2} x_{4} + ζ x_{5} \end{aligned}$
Once one has $A_{0}, \dots, A_{4}$ one easily gets $x_{1}, \dots, x_{5}$ . It's easy to find $A_{0}$ . The point is that $τ$ takes $A_{j}$ to $ζ^{- j} A_{j}$ and so takes $A_{j}^{5}$ to $A_{j}^{5}$ . Thus $A_{j}^{5}$ 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 $A_{1}^{5}$ 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 $A_{1}$ as a fifth root of a certain explicit complex number. Then one can express the other $A_{j}$ in terms of $A_{1}$ . 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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Landyn Whitney

Answered 2022-04-26 Author has 19 answers

The solution of a solvable quintic,

x^{5} + a x^{4} + b x^{3} + c x^{2} + d x + e = 0

can be given by,

x = \frac{1}{5} (- a + z_{1}^{\frac{1}{5}} + z_{2}^{\frac{1}{5}} + z_{3}^{\frac{1}{5}} + z_{4}^{\frac{1}{5}})

and the

z_{i}

are the roots of the quartic. However, there is also a form where one takes a fifth root only once.

p = 11

Let,

x^{5} + x^{4} - 4 x^{3} - 3 x^{2} + 3 x + 1 = 0

The five roots

x_{k}

for

k = 0, 1, 2, 3, 4

are,

x_{k} = \frac{- 1}{5} (\frac{1}{β_{k}^{- 1}} + \frac{1}{β_{k}^{0}} + \frac{11}{β_{k}^{1}} + \frac{a}{β_{k}^{2}} + \frac{b}{β_{k}^{3}})

where,

a = t \frac{11}{4} (- 1 + 5 \sqrt{5} + \sqrt{- 10 (5 + \sqrt{5})})

b = t \frac{11}{4} (- 31 + 5 \sqrt{5} - \sqrt{- 10 (85 + 31 \sqrt{5})})

β_{k} = ζ_{5}^{k}, {(\frac{a b}{11})}^{\frac{1}{5}}

ζ_{5} = e^{2 π, \frac{i}{5}}

p = 31

As discussed in this post,

x^{5} + x^{4} - 12 x^{3} - 21 x^{2} + x + 5 = 0

x_{k} = \frac{1}{5} (\frac{1}{β_{k}^{- 1}} - \frac{1}{β_{k}^{0}} + \frac{31}{β_{k}^{1}} + \frac{a}{β_{k}^{2}} + \frac{b}{β_{k}^{3}})

where,

a = t \frac{31}{4} (11 + 5 \sqrt{5} + \sqrt{- 10 (25 - 11 \sqrt{5})})

b = t \frac{31}{4} (- 1 - 5 \sqrt{5} + \sqrt{- 10 (1525 - \sqrt{5})})

β_{k} = ζ_{5}^{k}, {(\frac{a b}{31})}^{\frac{1}{5}}

and so on for other '

p = 10 n + 1

.

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Jayla Matthews

Answered 2022-04-27 Author has 18 answers

Step 1

ω_{1} = \sqrt{5} {(\frac{66}{3125} ζ_{5} + \frac{451}{3125} ζ_{5}^{2} + \frac{176}{3125} ζ_{5}^{3} + \frac{286}{3125} ζ_{5}^{4})}

ω_{2} = \sqrt{2} {- \frac{11}{20} + (\frac{1}{4} ζ_{5}^{4}) ω_{1} + (- \frac{5}{44} ζ_{5} + \frac{15}{44} ζ_{5}^{2} + \frac{5}{44} ζ_{5}^{3} + \frac{5}{44} ζ_{5}^{4}) ω_{1}^{2} + (\frac{25}{121} ζ_{5} - \frac{75}{242} ζ_{5}^{2} - \frac{75}{484} ζ_{5}^{3} + \frac{75}{242} ζ_{5}^{4}) ω_{1}^{3} + (- \frac{375}{2662} ζ_{5} + \frac{625}{1331} ζ_{5}^{2} + \frac{625}{2662} ζ_{5}^{3} + \frac{4375}{5324} ζ_{5}^{4}) ω_{1}^{4}}

Step 2
the roots of the polynomial are given by

- \frac{1}{10} + \frac{1}{2} ω_{1} + (- \frac{5}{22} ζ_{5}^{2} - \frac{5}{11} ζ_{5}^{3} + \frac{5}{11} ζ_{5}^{4}) ω_{1}^{2} + (\frac{75}{242} ζ_{5} + \frac{150}{121} ζ_{5}^{2} + \frac{75}{121} ζ_{5}^{3} + \frac{125}{121} ζ_{5}^{4}) ω_{1}^{3} + (\frac{1625}{1331} ζ_{5} + \frac{1000}{1331} ζ_{5}^{2} + \frac{5125}{2662} ζ_{5}^{3} + \frac{375}{1331} ζ_{5}^{4}) ω_{1}^{4} - ω_{2}

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