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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 $\zeta^1,\zeta^2,\ldots,\zeta^{10}$, 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 $[2,9]$,$[4,7]$,$[8,3]$,$[5,6]$,$[10,1]$.

Now put $q_1 = \zeta^2+\zeta^9$,$q_2 = \zeta^4+\zeta^7$,$q_3 = \zeta^8+\zeta^3$,$q_4 = \zeta^5+\zeta^6$,$q_5 = \zeta^{10}+\zeta^1$. So if we can solve the quintic $(q - q_1)(q - q_2)(q - q_3)(q - q_4)(q - q_5) = q^5 + q^4 - 4q^3 - 3q^2 + 3q + 1 = 0$ we would just be left to solve a few quadratic equations.

Now pari/gp tells me this quintic has the cyclic group C(5):

? polgalois(x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1)
%1 = [5, 1, 1, "C(5) = 5"]

I've worked through examples of solving quadratic and cubic equations based on the galois group but when it comes to this quintic I'm completely stumped so any advice would be tremendously helpful! Thanks.

Edit: Thanks to Robin Chapman the problem is reduced significantly. Let $\omega$ be a primitive 5th root of unity (which is easy to express in radicals), it only remains to express $(q_1 + \omega q_2 + \omega^2 q_3 + \omega^3 q_4 + \omega^4 q_5)^5$ in terms of rationals and powers of $\omega$ (and then everything can be substituted back and solved easily). We know this is possible because the term is fixed by the quintics galois group, how to actually perform this evades me but I will try to find a way.

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GAP and its Radiroot extension package will find the roots for you explicitely (and even typeset them in latex!) – Mariano Suárez-Alvarez Aug 2 '10 at 22:52
up vote 13 down vote accepted

For full details of this and more, the best place to look is the following paper:

D. S. Dummit, Solving solvable quintics. Math. Comp. 57 (1991), 387-401.

The main idea (which extends to any equation with a cyclic Galois group) is to consider Lagrange resolvents. Let the equation have roots $x_1,\ldots,x_5$ with an element $\tau$ of the Galois group permuting them as $x_i\mapsto x_{i+1}$. Let $\zeta=\exp(2\pi i/5)$ be the standard fifth root of unity. Then the Lagrange resolvents are

$\begin{align*} A_0&=x_1+x_2+x_3+x_4+x_5\\ A_1&=x_1+\zeta x_2+\zeta^2 x_3+\zeta^3 x_4+\zeta^4 x_5\\ A_2&=x_1+\zeta^2 x_2+\zeta^4 x_3+\zeta x_4+\zeta^3 x_5\\ A_3&=x_1+\zeta^3 x_2+\zeta x_3+\zeta^4 x_4+\zeta^2 x_5\\ A_4&=x_1+\zeta^4 x_2+\zeta^3 x_3+\zeta^2 x_4+\zeta x_5 \end{align*}$

Once one has $A_0,\ldots,A_4$ one easily gets $x_1,\ldots,x_5$. 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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Thanks very much! – anon Aug 2 '10 at 17:11

Just for fun: Gap+RadiRoot tells me that if $\zeta_n$ is a primitive $n$th root of unity, and we set

$\omega_1 = \sqrt[5]{\left(\frac{66}{3125}\zeta_{5} + \frac{451}{3125}\zeta_{5}^{2} + \frac{176}{3125}\zeta_{5}^{3} + \frac{286}{3125}\zeta_{5}^{4}\right)}$,

$\omega_2 = \sqrt[2]{ - \frac{11}{20} + \left(\frac{1}{4}\zeta_{5}^{4}\right)\omega_1 + \left( - \frac{5}{44}\zeta_{5} + \frac{15}{44}\zeta_{5}^{2} + \frac{5}{44}\zeta_{5}^{3} + \frac{5}{44}\zeta_{5}^{4}\right)\omega_1^2 + \left(\frac{25}{121}\zeta_{5} - \frac{75}{242}\zeta_{5}^{2} - \frac{75}{484}\zeta_{5}^{3} + \frac{75}{242}\zeta_{5}^{4}\right)\omega_1^3 + \left( - \frac{375}{2662}\zeta_{5} + \frac{625}{1331}\zeta_{5}^{2} + \frac{625}{2662}\zeta_{5}^{3} + \frac{4375}{5324}\zeta_{5}^{4}\right)\omega_1^4}$,

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$

NB: I generated this (in a gap4 session, with RadiRoot properly installed) as follows:

gap> LoadPackage("radiroot");
Loading  RadiRoot 2.4 (Roots of a Polynomial as Radicals)
by Andreas Distler (
gap> g := UnivariatePolynomial( Rationals, [1,1,1,1,1,1,1,1,1,1,1]);
gap>  RootsOfPolynomialAsRadicals(g, "latex");

That created a $\LaTeX$ file, /tmp/tmp.sfoZ6C/Nst.tex, containing the formulas

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Solving $x^{11} = 1$ in radicals was first accomplished by Vandermonde, whose method is described on p.24-26 of Edwards' book on Galois theory

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I recommend you go and look at this reference.

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excellent! The entire process is clear to me, thank you! – anon Aug 2 '10 at 17:11

The solution of a solvable quintic,

$$x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0$$

can be given by,

$$x = \frac{1}{5}\left(-a+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\right)$$

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



$$x^5 +x^4 −4x^3 −3x^2 +3x+1 = 0$$

The five roots $x_k$ for $k=0,1,2,3,4$ are,

$$x_k = \frac{-1}{5}\left(\frac{1}{\beta_k^{-1}}+\frac{1}{\beta_k^0}+\frac{11}{\beta_k^1}+\frac{a}{\beta_k^2}+\frac{b}{\beta_k^3} \right)$$




$$\beta_k = \zeta_5^k\,\left(\frac{ab}{11}\right)^{1/5}$$

$$\zeta_5 = e^{2\pi\,i/5}$$


As discussed in this post,

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

$$x_k = \frac{1}{5}\left(\frac{1}{\beta_k^{-1}}-\frac{1}{\beta_k^0}+\frac{31}{\beta_k^1}+\frac{a}{\beta_k^2}+\frac{b}{\beta_k^3} \right)$$




$$\beta_k = \zeta_5^k\,\left(\frac{ab}{31}\right)^{1/5}$$

and so on for other prime $p=10n+1$.

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