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Suppose $ \zeta$ is a primitive $ 11$-th root of unity and $ \alpha=\zeta+\zeta^3+\zeta^4+\zeta^5+\zeta^9 $

Find $ [\mathbb{Q(\alpha)}: \mathbb{Q}] $

Could someone please give me a hint how too do that?

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    $\begingroup$ How about computing $\alpha^2$ for a start $\endgroup$ – Hagen von Eitzen Apr 13 '16 at 16:14
  • $\begingroup$ Thanks, why $\alpha^2$ and not, say, $\alpha^6$? $\endgroup$ – amiz9 Apr 14 '16 at 14:09
  • $\begingroup$ Knowing (or guessing) that $\alpha$ is quadratic over $\Bbb Q$, you get the minimal polynomial for $\alpha$ by looking at $1$, $\alpha$, and $\alpha^2$ and finding a $\Bbb Q$-linear relation of linear dependence. $\endgroup$ – Lubin Apr 14 '16 at 14:35
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You may notice that $$ \alpha = \sum_{n=1}^{5}\zeta^{n^2} $$ from which it follows that (just square the previous sum): $$ \alpha^2 = -\frac{5+i\sqrt{11}}{2} $$ and $\alpha=\pm \frac{1}{2}(1-i\sqrt{11})$. Gauss sums hence prove that $[\mathbb{Q}(\alpha):\mathbb{Q}]=\color{red}{2}$.

You may also use the general identity: $$ \sum_{k=1}^{p-1}\left(\frac{k}{p}\right)\exp\left(\frac{2\pi i k}{p}\right) = \sqrt{\pm p}$$ where the sign in the RHS just depends on $p\pmod{4}$.

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  • $\begingroup$ Thank you, this looks good. However, I have no idea how you would be able to 'notice' that $\alpha$ can also be expressed in summation form as you have given. Please could you let me know how you knew that? $\endgroup$ – amiz9 Apr 14 '16 at 14:10
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Here’s another method: Consider the Galois group of $\Bbb Q(\zeta_{11})$ over $\Bbb Q$. It’s cyclic, isomorphic to $(\Bbb Z/11\Bbb Z)^\times$, which is generated by $2$, that is, $\sigma:\zeta\mapsto\zeta^2$ generates the group. There’s only one subgroup of order five, generated by $\sigma^2:\zeta\mapsto\zeta^4$. Since the powers of $4$ in $(\Bbb Z/11\Bbb Z)^\times$ are $1$, $4$, $5$, $9$, and $3$, your number is the trace from the big field down the fixed field of that group of order five.

That should be enough for you to finish the argument.

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  • $\begingroup$ Thanks, I am doing a course on Galois theory so this looks good. How did you know that there is only one subgroup of order $5$? $\endgroup$ – amiz9 Apr 14 '16 at 14:12
  • $\begingroup$ There’s only one subgroup of order five because the original group is cyclic. These have precisely one group of each order dividing the order of the main group. $\endgroup$ – Lubin Apr 14 '16 at 14:33

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