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

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 ${\zeta }^{1},{\zeta }^{2},\dots ,{\zeta }^{10}$, 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].
$\begin{array}{rl}{A}_{0}& ={x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}\\ {A}_{1}& ={x}_{1}+\zeta {x}_{2}+{\zeta }^{2}{x}_{3}+{\zeta }^{3}{x}_{4}+{\zeta }^{4}{x}_{5}\\ {A}_{2}& ={x}_{1}+{\zeta }^{2}{x}_{2}+{\zeta }^{4}{x}_{3}+\zeta {x}_{4}+{\zeta }^{3}{x}_{5}\\ {A}_{3}& ={x}_{1}+{\zeta }^{3}{x}_{2}+\zeta {x}_{3}+{\zeta }^{4}{x}_{4}+{\zeta }^{2}{x}_{5}\\ {A}_{4}& ={x}_{1}+{\zeta }^{4}{x}_{2}+{\zeta }^{3}{x}_{3}+{\zeta }^{2}{x}_{4}+\zeta {x}_{5}\end{array}$
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 $\tau$ takes ${A}_{j}$ to ${\zeta }^{-j}{A}_{j}$ 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 $\zeta$. Alas, here is where the algebra becomes difficult. The coefficients of powers of $\zeta$ 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
The solution of a solvable quintic,
${x}^{5}+a{x}^{4}+b{x}^{3}+c{x}^{2}+dx+e=0$
can be given by, $x=\frac{1}{5}\left(-a+{z}_{1}^{\frac{1}{5}}+{z}_{2}^{\frac{1}{5}}+{z}_{3}^{\frac{1}{5}}+{z}_{4}^{\frac{1}{5}}\right)$
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, ${x}^{5}+{x}^{4}-4{x}^{3}-3{x}^{2}+3x+1=0$
The five roots ${x}_{k}$ for $k=0,1,2,3,4$ are,
${x}_{k}=\frac{-1}{5}\left(\frac{1}{{\beta }_{k}^{-1}}+\frac{1}{{\beta }_{k}^{0}}+\frac{11}{{\beta }_{k}^{1}}+\frac{a}{{\beta }_{k}^{2}}+\frac{b}{{\beta }_{k}^{3}}\right)$
where, $a=t\frac{11}{4}\left(-1+5\sqrt{5}+\sqrt{-10\left(5+\sqrt{5}\right)}\right)$
$b=t\frac{11}{4}\left(-31+5\sqrt{5}-\sqrt{-10\left(85+31\sqrt{5}\right)}\right)$
${\beta }_{k}={\zeta }_{5}^{k},{\left(\frac{ab}{11}\right)}^{\frac{1}{5}}$
${\zeta }_{5}={e}^{2\pi ,\frac{i}{5}}$
$p=31$
As discussed in this post,
${x}^{5}+{x}^{4}-12{x}^{3}-21{x}^{2}+x+5=0$
${x}_{k}=\frac{1}{5}\left(\frac{1}{{\beta }_{k}^{-1}}-\frac{1}{{\beta }_{k}^{0}}+\frac{31}{{\beta }_{k}^{1}}+\frac{a}{{\beta }_{k}^{2}}+\frac{b}{{\beta }_{k}^{3}}\right)$
where, $a=t\frac{31}{4}\left(11+5\sqrt{5}+\sqrt{-10\left(25-11\sqrt{5}\right)}\right)$
$b=t\frac{31}{4}\left(-1-5\sqrt{5}+\sqrt{-10\left(1525-\sqrt{5}\right)}\right)$
${\beta }_{k}={\zeta }_{5}^{k},{\left(\frac{ab}{31}\right)}^{\frac{1}{5}}$
and so on for other ' $p=10n+1$.
###### Not exactly what you’re looking for?
Jayla Matthews
Step 1
${\omega }_{1}=\sqrt{5}\left\{\left(\frac{66}{3125}{\zeta }_{5}+\frac{451}{3125}{\zeta }_{5}^{2}+\frac{176}{3125}{\zeta }_{5}^{3}+\frac{286}{3125}{\zeta }_{5}^{4}\right)\right\}$
${\omega }_{2}=\sqrt{2}\left\{-\frac{11}{20}+\left(\frac{1}{4}{\zeta }_{5}^{4}\right){\omega }_{1}+\left(-\frac{5}{44}{\zeta }_{5}+\frac{15}{44}{\zeta }_{5}^{2}+\frac{5}{44}{\zeta }_{5}^{3}+\frac{5}{44}{\zeta }_{5}^{4}\right){\omega }_{1}^{2}+\left(\frac{25}{121}{\zeta }_{5}-\frac{75}{242}{\zeta }_{5}^{2}-\frac{75}{484}{\zeta }_{5}^{3}+\frac{75}{242}{\zeta }_{5}^{4}\right){\omega }_{1}^{3}+\left(-\frac{375}{2662}{\zeta }_{5}+\frac{625}{1331}{\zeta }_{5}^{2}+\frac{625}{2662}{\zeta }_{5}^{3}+\frac{4375}{5324}{\zeta }_{5}^{4}\right){\omega }_{1}^{4}\right\}$
Step 2
the roots of the polynomial are given by
$-\frac{1}{10}+\frac{1}{2}{\omega }_{1}+\left(-\frac{5}{22}{\zeta }_{5}^{2}-\frac{5}{11}{\zeta }_{5}^{3}+\frac{5}{11}{\zeta }_{5}^{4}\right){\omega }_{1}^{2}+\left(\frac{75}{242}{\zeta }_{5}+\frac{150}{121}{\zeta }_{5}^{2}+\frac{75}{121}{\zeta }_{5}^{3}+\frac{125}{121}{\zeta }_{5}^{4}\right){\omega }_{1}^{3}+\left(\frac{1625}{1331}{\zeta }_{5}+\frac{1000}{1331}{\zeta }_{5}^{2}+\frac{5125}{2662}{\zeta }_{5}^{3}+\frac{375}{1331}{\zeta }_{5}^{4}\right){\omega }_{1}^{4}-{\omega }_{2}$