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Let $ \zeta \in \Bbb C$ be a seventh root of $1$. Find the minimal polynomial of the element $ \alpha = \zeta ^{-1} + \zeta$ over $ \Bbb Q$ and show that if $K= \Bbb Q ( \alpha ) $, then $ K/ \Bbb Q$ is an Galois extension. Compute the Galois group of the extension.

I have no idea how to approach this problem, because i don't know what the $ \zeta $ is.

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  • $\begingroup$ What do you mean by "you don't know what the $\zeta$ is"? It tells you that $\zeta$ is a $7$th root of $1$. What else would you care to know? $\endgroup$ – Omnomnomnom Feb 15 '15 at 23:37
  • $\begingroup$ I mean I don't know how to compute minimal polynomials of those elements $\endgroup$ – Kuba Feb 15 '15 at 23:38
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    $\begingroup$ "$\zeta$ is a seventh root" means that $\zeta \neq 1$ satisfies $$ \zeta^7 = 1 $$ so, its minimal polynomial (over $\Bbb Q$) will be $$ (x^7 - 1)/(x - 1) = x^6 + \cdots + x + 1 $$ the same is true for $\zeta^{-1}$. $\endgroup$ – Omnomnomnom Feb 15 '15 at 23:40
  • $\begingroup$ But how do we know, that this is the minimal polynomial? And there are 7 roots, so which one should I pick? $e^{2/7 \pi i}$? $\endgroup$ – Kuba Feb 15 '15 at 23:58
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    $\begingroup$ Seventh implies somehow not first. That is, $\zeta$ is one of the numbers $e^{2k \pi i/7}$, for some $1\le k \le 6$. You have $6$ choices for $\zeta$. But you will get exactly $3$ possible $\zeta^{-1} + \zeta$. And all of them will have the same minimal polynomial. But I agree, it could feel like incomplete information. $\endgroup$ – Orest Bucicovschi Feb 16 '15 at 4:02
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Here is a transparent but a longish approach: There are totally seven 7th roots of unity out of which 1 is not primitive. The remaining 6 can made into 3 pairs of conjugates (same as reciprocals). Take the sum in each pairs obtaining 3 numbers: $a = \zeta +\zeta^6,\ b=\zeta^2 +\zeta^5,\ c=\zeta^3 +\zeta^4$. Note that $a,b,c$ are real numbers. You want to know the minimal polynomial for $a$ (which is the same for $b$ or $c$). It is the cubic polynomial $ (x-a)(x-b)(x-c)$. To get the coefficients and (see that they are rational numbers, actually integers) you have to calculate $a+b+c, ab+bc+ca $ and $abc$. This is a pleasant exercise. If you are comfortable working with cosine functions, note that $a= 2\cos\theta$ with $\theta=2\pi/7$ (this could equally well be $b$ or $c$, but it does not matter due to symmetry).

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  • $\begingroup$ hmm, I'm wondering.. If $a = \zeta + \zeta ^6 = \zeta + \zeta ^{-1} $, why isn't $a$ rational? I mean that I know it doesn't have to be, but I don't know why it is not ;) And if it's not rational, why can't $(x-a)(x-b)$ be the minimal polynomial? ;) $\endgroup$ – Kuba Feb 16 '15 at 0:14
  • $\begingroup$ Look at the definition of $\zeta$ in the first sentence of your question and multiply it with its sixth power! For other questions find out and see if those polynomials have rational or non-rational questions. $\endgroup$ – P Vanchinathan Feb 16 '15 at 0:18
  • $\begingroup$ OK, so the minimal polynomial is $M(x)=x^3+x^2-2x-1$ is that right? $\endgroup$ – Kuba Feb 16 '15 at 0:47
  • $\begingroup$ Have not done the computation. But your solution is not correct as it has negative roots. It is checked that the roots $a.b.c$ are all positive. $\endgroup$ – P Vanchinathan Feb 16 '15 at 1:05
  • $\begingroup$ I made a mistake: your polynomial is correct. Not all of $a,b,c$ are positive. $\endgroup$ – P Vanchinathan Feb 16 '15 at 2:28
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set $\theta =\frac{2\pi}7$ so $\zeta=e^{i\theta}$

the minimal polynomial $P(x)$ of $\zeta$ is $$ P(x) = \prod_{k=1}^6 (x-\zeta^k) = \sum_{k=0}^6 x^k $$ since $\zeta^{-k}=\zeta^{7-k}$ the product may be rewritten, grouping terms in pairs: $$ P(x) = \prod_{k=1}^3 (x^2 - 2x \cos k\theta +1) $$ which may be rearranged as: $$ \frac{P(x)}{x^3} =Q(x+\frac1{x})=\prod_{k=1}^3 (x+\frac1{x} - 2 \cos k\theta) $$ so the minimal polynomial for $\frac{\alpha}2$ has roots $\cos \theta, \cos 2\theta, \cos 3\theta$, which we may call $\beta_1, \beta_2, \beta_3$

from elementary trigonometry: $$ \beta_2 = 2\beta_1^2 -1 \\ \beta_3 = 4\beta_1^3-3\beta_1 $$ we also have $$ \sum_{k=1}^3 \beta_k = -\frac12 $$ since the sum of the $\beta$'s is half the sum of roots of the original polynomial $P(x)$. so, substituting, we have (omitting the suffix): $$ \beta + (2\beta^2 -1) + (4\beta^3 - 3 \beta) = -\frac12 $$ simplifying, and increasing the roots by a factor of $2$, we have the minimal polynomial of $\zeta + \frac1{\zeta} $ to be $$ x^3 + x^2 - 2x -1 = 0 $$ in the Galois group of $\mathbb{Q}(\zeta)$ the automorphism sending $\zeta$ to $\frac1{\zeta}$ fixes the three roots of this equation. since the automorphism has degree $2$ the Galois group of $\mathbb{Q}(\alpha)/\mathbb{Q}$ is cyclic of order $3$

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Here’s yet another way of looking at things:

You know that your seventh root of unity, $\zeta$, unequal to $1$, is a root of $f(X)=X^6+X^5+X^4 +X^3+X^2+X+1$. This is irreducible ’cause $f(X+1)$ satisfies Eisenstein.

Write the fundamental equation for $\zeta$ in the form $$ 0=\zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^3\,. $$ Now write $\xi=\zeta+\zeta^{-1}$, and calculate $\xi^3=\zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3}$, and subtract zero to get $\xi^3=-\zeta^2+2\zeta-1+2\zeta^{-1}-\zeta^{-2}$. Now add $\xi^2$ to get $\xi^3+\xi^2=2\zeta+1+2\zeta^{-1}=2\xi+1$, so that a polynomial satisfied by $\xi=\zeta+\zeta^{-1}$ is $g(X)=X^3+X^2-2X-1$, and you can check that $g$ is irreducible by calculating $g(X+2)$.

For the action of the Galois group, the group of $\Bbb Q(\zeta)$ over $\Bbb Q$ is generated by $\zeta\mapsto\zeta^3$, and so the group of $\Bbb Q(\xi)$ over $\Bbb Q$ is generated by $\xi\mapsto\zeta^3+\zeta^{-3}$, and I leave it to you to express this as a polynomial in $\xi$ by the same method I used above.

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