Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cos/Sin as roots of rationals

  1. Apr 6, 2006 #1
    This is indirectly related to issues with cyclotomic polynomials and glaois groups.

    Is there some easy way to know if you are dealing with a cos or sin that is expressable in terms of roots of rationals? Like [itex]\pi/3[/itex] for example? If so, is there any straightforward way of figuring it out?

    I'm asking because I'm trying to do a problem involving the galois group for the 7th roots of unity. The fact that there is a order 2 and order 3 subgroup of the galois group makes me think that Q(i) must be a fixed field and then some cube root another one for the subgroups. This would further imply that [itex]cos(4\pi/7)[/itex] is expressable as a ratioal function plus a cube root (unless I'm way off in space, which is possible).

    Thanks
     
  2. jcsd
  3. Apr 7, 2006 #2
    A very interesting problem. I'm not very fluent with Galois theory these days, so please take my words with a grain of salt.

    For the 7th roots of unity we have the polynomial

    [tex](x-1)^7 \quad = \quad (x-1)(x^6+x^5+x^4+x^3+x^2+x+1) [/tex]

    So the interesting polynomial for these roots is:

    [tex]p \quad = \quad x^6+x^5+x^4+x^3+x^2+x+1[/tex]

    Now, it is obvious that this polynomial has as Galois group [tex]\mathbb{Z}_7[/tex]. If we call [tex]\omega[/tex] some seventh root of unity (not 1), the other roots of [tex]p[/tex] are the powers [tex]\omega^k,\, k=2,\ldots, 6[/tex]. Hence the field for this polynomial is [tex]\mathbb{Q}(\omega)[/tex], extension of dimension 7 over [tex]\mathbb{Q}[/tex], and its automorphisms are obviously the powers of the roots.

    The group [tex]\mathbb{Z}_7[/tex] has no nontrivial subgroups. I think you are erroneously taking [tex]\mathbb{Z}_6[/tex] as the Galois group of the polynomial p.

    Anyway, we come to the conclusion that any root of our polynomial p has the form [tex]a_0 + a_1\omega + a_2\omega^2 + a_3\omega^3 + a_4\omega^4 + a_5\omega^5 + a_6\omega^6[/tex], where the [tex]a_i[/tex] are rationals. Since [tex]\omega[/tex] is a radical (a seventh root of unity), this trivially means that all the roots of p are radicals.

    However, I think that in order to get better information on the structure of the roots, the coefficients [tex]a_i[/tex], etc., you could try reading modern elaborations of Gauss' works on cyclotomic groups. I hope this link helps : http://en.wikipedia.org/wiki/Gaussian_period

    ----

    If you take sixth roots of unity, you have [tex]\mathbb{Z}_6 \cong \mathbb{Z}_3\times\mathbb{Z}_2[/tex], so the Galois group should be generated by two radicals [tex]\mathbb{Q}(i,\xi)[/tex], with [tex]\xi^3 = 1[/tex].

    This means that the sixth roots of unity should be writeable in the form: [tex]a_0 + a_1 i + a_2 \xi + a_3 i\xi + a_4\xi^2 + a_5i\xi^2[/tex]. Again, more finese in the form of these roots should be achieved turning to Gauss' works.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?