Just how to address a cyclic quintic in radicals?

Galois concept informs 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 addressed in radicals due to the fact that its team is understandable. In fact executing the estimation is past me, though - below what I have actually obtained until now:

Allow the origins be $\zeta^1,\zeta^2,\ldots,\zeta^{10}$, adhering to Gauss we can divide the trouble right into addressing quintics and also quadratics by considering subgroups of the origins. Given that 2 is a generator of the team $[2,4,8,5,10,9,7,3,6,1]$ we can dividing right into the 5 subgroups of conjugate sets $[2,9]$, $[4,7]$, $[8,3]$, $[5,6]$, $[10,1]$.

Currently placed $q_1 = \zeta^2+\zeta^9$, $q_2 = \zeta^4+\zeta^7$, $q_3 = \zeta^8+\zeta^3$, $q_4 = \zeta^5+\zeta^6$, $q_5 = \zeta^{10}+\zeta^1$. So if we can address 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 certainly simply be entrusted to address a couple of square formulas.

Currently pari/gp informs me this quintic has the cyclic team 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 resolved instances of addressing square and also cubic formulas based upon the galois team yet when it involves this quintic I'm entirely puzzled so any kind of suggestions would certainly be enormously handy! Many thanks.

Modify: Many Thanks to Robin Chapman the trouble is lowered dramatically. Allow $\omega$ be a primitive 5th origin of unity (which is very easy to share in radicals), it just continues to be to share $(q_1 + \omega q_2 + \omega^2 q_3 + \omega^3 q_4 + \omega^4 q_5)^5$ in regards to rationals and also powers of $\omega$ (and afterwards every little thing can be replaced back and also addressed conveniently). We understand this is feasible due to the fact that the term is dealt with by the quintics galois team, just how to in fact execute this escapes me yet I will certainly search for a means.

2019-05-04 16:26:02
Source Share
Answers: 4

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

2019-05-09 03:43:17

I advise you go and also consider this reference. www.passhema.org/proceedings/2007/2007DavisGupta.pdf

2019-05-08 10:08:15

Just for enjoyable : Gap+ RadiRoot informs me that if $\zeta_n$ is a primitive $n$th origin of unity, and also 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 origins of the polynomial are offered 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 created this (in a gap4 session, with RadiRoot effectively mounted) as adheres to :

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

That developed a $\LaTeX$ documents,/ tmp/tmp. sfoZ6C/Nst. tex, having the solutions

2019-05-08 07:14:29

For complete information of this and also extra, the most effective area to look is the. adhering to paper :

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

The main point (which includes any kind of formula with a cyclic Galois team ) is to take into consideration Lagrange resolvents. Allow the formula have origins $x_1,\ldots,x_5$. with a component $\tau$ of the Galois team permuting them as $x_i\mapsto x_{i+1}$. Allow $\zeta=\exp(2\pi i/5)$ be the typical 5th origin of unity. After that 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 conveniently obtains $x_1,\ldots,x_5$. It's very easy to locate $A_0$ :- ) The factor is that $\tau$ takes $A_j$ to $\zeta^{-j}A_j$. therefore takes $A_j^5$ to $A_j^5$. Hence $A_j^5$ can be listed in terms. of rationals (if that's your beginning area ) and also powers of $\zeta$. Sadly, below is where the algebra comes to be hard. The coefficients. of powers of $\zeta$ in $A_1^5$ are made complex. They can be shared. in regards to an origin of a "resolvent polynomial" which will certainly have a sensible. origin as the formula is cyclic. As soon as one has actually done this, you have $A_1$. as a 5th origin of a particular specific facility number. After that one can. share the various other $A_j$ in regards to $A_1$. The information are not really positive,. yet Dummit skilfully browses via the intricacies, and also generates. solutions which are not as made complex as they could be. Sadly, I do not have. the moment neither the power to give even more information.

2019-05-07 23:28:40