1

Let $f(x)=x^{5}-1 \in \mathbb{Q[x]}$. We know that $G=\operatorname{Gal}(\mathbb{Q(w)/\mathbb{Q)}}\simeq U(\mathbb{Z}_5)$ where $U(\mathbb{Z}_5)=\{\bar{1},\bar{2},\bar{3},\bar{4}\}$ is cyclic group. So |$\operatorname{Gal}(\mathbb{Q(w)/\mathbb{Q)}}$|$=4$ also and $U(\mathbb{Z}_5)=\langle\bar{2}\rangle$. So we get that $G=\langle\sigma_2\rangle$ where $\sigma_2: w \mapsto w^2$ and also $\operatorname{ord}(\sigma_2)=4$ because $G$ is cyclic.

The question is to prove that $E^{\langle\sigma^2\rangle}=\mathbb{Q}(w+w^{-1})$ where $w=e^{\frac{2πi}{5}}$

Also find $\operatorname{Irr}_\mathbb{Q}(\cos(\frac{2π}{5}))(x)$.

Thank you for your time!

Bernard
  • 175,478
argiriskar
  • 413
  • A hint: What does $\sigma_2^2$ do to $w$? – Jyrki Lahtonen May 29 '18 at 20:18
  • A hint for the last question is buried in here. See the Example in my answer :-) – Jyrki Lahtonen May 29 '18 at 20:20
  • $(\sigma_2)^2(w)=\sigma_2(\sigma_2(w))=(w^2)^2=w^4$ is this correct ? – argiriskar May 29 '18 at 20:23
  • Yes. But also remember that $w^5=1$ so $w^4=$____? – Jyrki Lahtonen May 29 '18 at 20:24
  • $w^4=w^{-1}$. I think i get your point. But the taught us a way in which in the end you get to a system of the coefficients. You solve the system and you get for example a $u\in E$ which tells you which field you get. I'm not sure if this is understandable. – argiriskar May 29 '18 at 20:28
  • Correct again. If I got it right the system you were given was to write elements of $\Bbb{Q}(w)$ w.r.t. the basis $1,w,w^2,w^3$ and check which linear combinations are invariant under $\sigma_2^2$. You can do it that way! But you will be needing the minimal polynomial relation $1+w+w^2+w^3+w^4=0.$ – Jyrki Lahtonen May 29 '18 at 20:39
  • I gues you need the minimal polynomial to express $w^4=-1-w-w^2-w^3$ and then solve the system of coefficients of the basis. Thank you very much proffesor. Also what is the way that you mention above. I guess it's an easier way. – argiriskar May 29 '18 at 20:59

0 Answers0