In chapter $6$ exercises section $6.5$ question $5$, in the solution provided, it states that $\lbrack \mathbb{Q}(\omega):F(\langle \sigma_6\rangle ) \rbrack =|\langle \sigma_6\rangle |$. Why is this so? I have been puzzled by this for an hour and I still can't figure out why. Can someone enlighten me?

EDIT: In the solution provided, the part I don't understand is that why $\lbrack \mathbb{Q}_7:K \rbrack=|\langle \sigma_6\rangle |=2$. To avoid any confusion, $K=F(\langle \sigma_6\rangle )$*strong text*

EDIT $2$: Usually, when we want to determine what is the degree of extension $\lbrack E:F \rbrack$, we determine what is the minimal polynomial of a root in $E$ over $F$ and take ite degree . But in this case, I don't know how to formulate a minimal polynomial of a root in $\mathbb{Q}_7$ over $K$.

  • $\begingroup$ I'm lost: exercise 5 in section 6.5 talks of $\;\cos\frac{2\pi}{7};$ ...? $\endgroup$ – DonAntonio Nov 8 '13 at 15:44
  • $\begingroup$ @DonAntonio:Yup $\endgroup$ – Idonknow Nov 8 '13 at 15:45
  • $\begingroup$ I'm still lost: why to use $\;F\; $ then? And where in that solution is written anything close to what you wrote in your question? $\endgroup$ – DonAntonio Nov 8 '13 at 15:47
  • $\begingroup$ what does the strong text mean? $\endgroup$ – Idonknow Nov 9 '13 at 4:11

If $G={\rm Gal}(E/F)$ and $H\le G$ has fixed field $K$ then $[E:K]=|H|$. See item (2)(d) in §6.2.1.

Note $\Bbb Q(\cos2\pi /n)$ is the fixed field of $\Bbb Q(\zeta_n)$ by $\langle\sigma\rangle$ where $\sigma\in G$ is conjugation ($\zeta\leftrightarrow\zeta^{-1}$).

  • $\begingroup$ In the question, is $\mathbb{Q}_7=\mathbb{Q}(\omega)$? or is it a quotient field $\mathbb{Q}(\omega) / \mathbb{Q}$? $\endgroup$ – Idonknow Nov 9 '13 at 5:00
  • 1
    $\begingroup$ @Idontknow The answer to your first question is yes, where $\omega$ is a primitive $7$th root of unity. The answer to your second question is no, because there is no such thing as a quotient field! When we write $E/F$, we don't mean quotient, we mean $E$ is an extension of $F$. $\endgroup$ – anon Nov 9 '13 at 5:02

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