Cos/Sin as roots of rationals

1. Apr 6, 2006

gonzo

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 $\pi/3$ 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 $cos(4\pi/7)$ is expressable as a ratioal function plus a cube root (unless I'm way off in space, which is possible).

Thanks

2. Apr 7, 2006

leach

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

$$(x-1)^7 \quad = \quad (x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$$

So the interesting polynomial for these roots is:

$$p \quad = \quad x^6+x^5+x^4+x^3+x^2+x+1$$

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

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

Anyway, we come to the conclusion that any root of our polynomial p has the form $$a_0 + a_1\omega + a_2\omega^2 + a_3\omega^3 + a_4\omega^4 + a_5\omega^5 + a_6\omega^6$$, where the $$a_i$$ are rationals. Since $$\omega$$ 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 $$a_i$$, 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

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

This means that the sixth roots of unity should be writeable in the form: $$a_0 + a_1 i + a_2 \xi + a_3 i\xi + a_4\xi^2 + a_5i\xi^2$$. Again, more finese in the form of these roots should be achieved turning to Gauss' works.