# $\mathbb Q[\cos(\frac{2\pi}{75})]$ is a Galois extension over $\mathbb Q$.

a) Prove that $$\mathbb Q[\cos(\frac{2\pi}{75})]$$ is a Galois extension over $$\mathbb Q$$.

b) Describe Galois group for this extension

First, it can be seen that $$\cos(\frac{2\pi}{75}) = \frac{e^{\frac{2\pi i}{75}}+e^{-\frac{2\pi i}{75}}}{2}$$. Thus $$\mathbb Q[\cos(\frac{2\pi}{75})] \subset \mathbb Q[e^{\frac{2\pi i}{75}}]$$.

$$\mathbb Q[e^{\frac{2\pi i}{75}}]$$ is clearly a Galois extension with $$Gal = (\mathbb Z_{75})^* = \mathbb Z_{5} \times \mathbb Z_4 \times \mathbb Z_2$$.

$$Gal$$ is abelian and therefore every subgroup is normal. So we can conclude that $$\mathbb Q[\cos(\frac{2\pi}{75})]$$ is also a Galois extension.

So now I need to understand which subgroup of $$\mathbb Z_{5} \times \mathbb Z_4 \times \mathbb Z_2$$ is a Galois group for $$\mathbb Q[\cos(\frac{2\pi}{75})]$$.

Thanks!

Edit: Calculated correct Galois group for $$\mathbb Q[e^{\frac{2\pi i}{75}}]$$ according to reuns and Lord Shark the Unknown.

• $\mathbb{Z}_{75}^\times \simeq \mathbb{Z}_{5^2}^\times \times \mathbb{Z}_{3}^\times \simeq \mathbb{Z}_{5} \times \mathbb{Z}_{4} \times \mathbb{Z}_{2}$. The (totally) real subfield is $K= \mathbb{Q}(\zeta_{75})^{\langle \sigma \rangle}$ where $\sigma(\zeta_{75}^{n})=\zeta_{75}^{-n}$ and $K = \mathbb{Q}(\zeta_{75}+\zeta_{75}^{-1})$ by comparing the degree. So $\text{Gal}(K/\mathbb{Q}) = \mathbb{Z}_{75}^\times / \{1,-1\}$. Oct 16, 2017 at 22:46
• And $\text{Gal}(K/\mathbb{Q})$ embeds into $\text{Gal}(\mathbb{Q}(\zeta_{75})/\mathbb{Q})$ as $Id_{\mathbb{Q}(\zeta_{75})/K} \times \text{Gal}(K/\mathbb{Q}) \simeq \mathbb{Z}_5 \times H$ where $H$ is the 3rd index $2$ subgroup of $\mathbb{Z}_{4} \times \mathbb{Z}_{2}$ Oct 16, 2017 at 23:12
• For a more concrete description of the field, I think it has basis $\{ \cos(\frac{2\pi k}{75}) : 1 \le k \le 37, \gcd(k,75)=1 \}$ and the automorphisms are maps $\sigma_{\ell}$ sending $\cos(\frac{2\pi k}{75})$ to $\cos(\frac{2\pi k \ell}{75})$. And the minimal polynomial of $\cos(\frac{2\pi k}{75})$ would be $\frac{(P_{38}(x) - P_{37}(x)) (P_3(x) - P_2(x))}{(P_{13}(x) - P_{12}(x)) (P_8(x) - P_7(x))}$ where $P_n(x)$ is the Chebyshev polynomial such that $P_n(\cos \theta) = \cos(n\theta)$. Oct 16, 2017 at 23:55

The minimal polynomial of $\cos(\dfrac{2\pi}{75})$ is of degree $20$ and a form to explicit it in a shorter way than the polynomial of degree $20$ is $$W(Z)=16Z^4-8Z^3-16Z^2+8Z+1\text{ where }\space Z(x)=16x^5-20x^3+5x$$ (Numerical verification:$\space\space \space Z(\cos(\dfrac{2\pi}{75}))\approx0.913545457643\\\space \space W(0.913545457643)\approx4.0003556023\cdot10^{-12}\approx 0)$
It follows that $G$ is a finite group of order $20=2^2\cdot5$ so we have the series $$G\triangleright H \triangleright \{e\}$$ where $H$ is the only $5$-Sylow subgroup of $G$ (which is normal) and the corresponding quotient groups are $G/H$ of order $4$ and $H/\{e\}$ of order $5$ then both cyclic.