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Question is :

Let $K$ denote the field $\mathbb{Q}(\zeta)$ where $\zeta=e^{\frac{2\pi i}{5}}$

  • Find $[K:\mathbb{Q}]$
  • Show that the splitting field of $x^{10}-1$ over $\mathbb{Q}$ is $K$
  • Find the galois group $G(K/\mathbb{Q})$
  • Show that $K$ contains $\sqrt{5}$

What i have done so far is

Easy to see that $[K:\mathbb{Q}]=4$ as $1+\zeta+\zeta^2+\zeta^3+\zeta^4=0$ and $f(x)=a+x+x^2+x^3+x^4$ is irreducible in $\mathbb{Q}[x]$

See that $x^{10}-1=(x^5-1)(x^5+1)$ and also that $K$ is defined to be splitting field of $x^5-1$ so only thing we need to make sure is $x^5+1$ splits and that i am not able to proceed...

Galois group is Multiplicative group $(\mathbb{Z}/5\mathbb{Z})^*$

Now to show that $\sqrt{5}$ is in $K$ i have used a Gauss sum (see Is $\sqrt 7$ the sum of roots of unity? for an on-site version) and it turns out to be natural..

$S=\zeta+\zeta^4$ then $S^2=\zeta^2+2\zeta^5+\zeta^3=2+\zeta^2+\zeta^3=1-(\zeta+\zeta^4)=1-S$

So, $S^2=1-S\Rightarrow S^2+S=1\Rightarrow (2S+1)^2=4(S^2+S)+1=5$

So, $5$ is a square in $K$ so, $K$ contains $\sqrt{5}$

Please help me to see the other part and please let me know if there are any other better ways of doing this...

EDIT : Should i use same method as $5$ is square in $K$ to prove $-1$ is a $5$th power in $K$?

Community
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    The zeros of $x^5+1$ are very simply related to the zeros of $x^5-1$. – Gerry Myerson Aug 07 '14 at 07:28
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    Put all sharp objects away... and prove that $-1$ is a fifth power in $\Bbb{Z}$. After you're done facepalming you can reach for the pencil again. – Jyrki Lahtonen Aug 07 '14 at 09:41
  • @GerryMyerson : I was stupid... I was concentrating more on proving $\sqrt{5}\in K$ and neglected other simple things.... –  Aug 07 '14 at 10:12
  • @JyrkiLahtonen : I was stupid... I was concentrating more on proving $\sqrt{5}\in K$ and neglected other simple things.... –  Aug 07 '14 at 10:12
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    Don't worry. You just had a blind spot. Happens to all of us. Surely you remember this point in the future! Time to move on :-) – Jyrki Lahtonen Aug 07 '14 at 10:23
  • @JyrkiLahtonen : :) :) –  Aug 07 '14 at 12:51

1 Answers1

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Roots of $x^5+1$ are $-1,-\zeta,-\zeta^2,-\zeta^3,-\zeta^4$ and $\zeta\in \mathbb{Q}(\zeta)$

So, Splitiing field of $x^{10}-1$ is $\mathbb{Q}(\zeta)$