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Let $\alpha = \zeta_7 + \zeta_7^{-1}$. I want to find the splitting field of $\alpha$ over $\mathbb{Q}$. I know that $\operatorname{minpoly}_Q (\alpha)$ is $x^3+x^2-2x-1$. So $\mathbb{Q}(\alpha)$ is degree $3$ extension of $\mathbb{Q}$. I can factor $x^3+x^2-2x-1 = \left(x - (\zeta_7 + \frac{1}{\zeta_7})\right) \left(x^2 + (1+\zeta_7 + \frac{1}{\zeta_7})x + \frac{1}{\zeta_7 + \frac{1}{\zeta_7}}\right)$ but then im unable to further factorize the RHS term. What can I do from here?

eatfood
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    This cubic is one of those "famous" polynomials with the property that if $r$ is a zero, then $r^2-2$ is another. It follows that $\Bbb{Q}(\alpha)$ is the splitting field. The trick comes from the fact that if $r=2\cos x$ then $r^2-2=4\cos^2x-2=2\cos2x$. See also Jack's answer. – Jyrki Lahtonen Oct 12 '19 at 19:04
  • Approach0 gives many older questions where this polynomial makes an appearance. Usually its zeros are considered also. I don't have the time to pick a good duplicate target here. – Jyrki Lahtonen Oct 12 '19 at 19:08
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    if $f(r)=0$, then $f(r^2-2)=r^6-5r^4+6r^2-1= (r^3-r^2-2r+1)(r^3+r^2-2r-1)=0$. – eatfood Oct 13 '19 at 04:28
  • @JyrkiLahtonen thanks for the insight! I can understand this part now. – eatfood Oct 13 '19 at 04:29
  • A bit more 1, 2. There really is a lot of repetition within the available material. Not clear that I picked the objectively best ones, so not voting to close. – Jyrki Lahtonen Oct 13 '19 at 05:12
  • Thanks for the links, I will take a look at them! – eatfood Oct 14 '19 at 09:58
  • @JyrkiLahtonen about your first comment: I am quite curious if it also follows that $\mathbb{Q}(r)$ is the splitting field of $x^3 + x^2 -2x -1$ over $\mathbb{Q}$. Is it possible to show this without Galois Theory? I am only a beginner in abstract algebra and hence I don't know any Galois theory. – hello Jun 10 '20 at 13:58
  • @hello See the calculation eatfood made a few comments up. If $r$ is a zero, so is $r^2-2$. Repeating the dose, so is $(r^2-2)^2-2$. These are obviously in the field $\Bbb{Q}(r)$. If you can show that they are all distinct, then you are done. Hint: if $r=r^2-2$, then $r=$____, and these are not roots. If you are familiar with Vieta relations, you get an easier formula for the third root. – Jyrki Lahtonen Jun 10 '20 at 14:03
  • @JyrkiLahtonen hmm okay, but if the roots are all distinct, why is the splitting field of $x^3 + x^2 -2x -1$ over $\mathbb{Q}$ then equal to $\mathbb{Q}(r)$? – hello Jun 10 '20 at 15:22
  • @hello A cubic has three zeros, so they are all in there. – Jyrki Lahtonen Jun 10 '20 at 15:44

2 Answers2

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The conjugated roots of $\alpha=2\cos\frac{2\pi}{7}$ are $2\cos\frac{4\pi}{7}$ and $2\cos\frac{6\pi}{7}$, so $\mathbb{Q}(\alpha)$ is the splitting field of $\alpha$ over $\mathbb{Q}$, by the cosine duplication and triplication formulas.

Jack D'Aurizio
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  • Hi @Jack, thanks for the answer. I don't understand how you know what the other roots of $\alpha$ are. (Right now I am just taking an introductory course of Galois theory.) But if you could include the stuff mentioned in Jyrki's comments, I'd be happy to mark this as the accepted answer. – eatfood Oct 13 '19 at 04:32
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For an odd prime $p$, let $\alpha_p =\zeta_p + \zeta_p^{-1}$. It is known that the cyclotomic field $K=\mathbf Q(\zeta_p)$ is a normal extension of $\mathbf Q$, with cyclic Galois group $G\cong (\mathbf Z/p)^*$. The subgroup of $G$ generated by the automorphism $\zeta_p \to \zeta_p^{-1}$ ("complex conjugation") is the unique subgroup of order $2$, and its fixed field $K$ is the unique subfield $L$ s.t. $K/L$ is quadratic. By definition, $\alpha_p \in L$. Moreover, $\zeta_p$ is obviously a root of the polynomial $X^2 -\alpha_p X+1$, so $K/\mathbf Q(\alpha_p)$ is quadratic, and $L=\mathbf Q(\alpha_p)$ and is the splitting field of $\alpha_p$. This is a very classical result from Galois theory, so I wonder whether there isn't some additional question.

nguyen quang do
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