Minimum polynomial of a root involving the 7th root of unity Let ω be a primitive 7th root of unity in $\Bbb C$ and set $α := ω + ω
^6$
. Determine, with
justification, the minimum polynomial of α over Q.
Would one use logs in such a question , or how should one begin? it's the roots of unity that are throwing me off, I know how to these minimal polynomial questions for radicals etc..
 A: One has the following $\omega^7=1$ and $\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1=0$
Now we compute
$$\alpha=\omega+\omega^6$$
$$\alpha^2=\omega^2+\omega^5+2$$
$$\alpha^3=\omega^3+\omega^4+3\alpha$$
Adding the three identities and rearranging one gets
$$\alpha^3+\alpha^2-2\alpha-1=0$$
And $X^3+X^2-2X-1$ is irreducible over $\Bbb{Q}$ because it has no integer roots and therefore no rational roots (it is monic). So we have the minimal polynomial of $\alpha$ over $\Bbb{Q}$
A: 
Let ω be a primitive $7^{th}$ root of unity in $\Bbb C$.

Hint: $\;\omega^7=1, \omega \ne 1\,$, so $\,\omega^6 = \dfrac{1}{\omega}\,$ and $\,\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1=0\,$. Dividing by $\,\omega^3\,$:
$$
\left(\omega^3 + \dfrac{1}{\omega^3}\right) + \left(\omega^2 + \dfrac{1}{\omega^2}\right) + \left(\omega + \dfrac{1}{\omega}\right) + 1 = 0 \tag{1}
$$

and set $α := ω + ω
^6$
  . Determine, with
  justification, the minimum polynomial of α over Q.

$\alpha=\omega+\omega^6 = \omega + \dfrac{1}{\omega}\,$, then:
$$
\alpha^2 = \omega^2+\dfrac{1}{\omega^2}+2 \\
\alpha^3 =  \omega^3+\dfrac{1}{\omega^3}+3\left(\omega + \dfrac{1}{\omega}\right) 
$$
Express $\,(1)\,$ in terms of $\,\alpha\,$ using the above, and you get the cubic equation that $\,\alpha\,$ satisfies. Then show that's the minimal polynomial, indeed.
A: It's important to see the easy reason you expect a cubic, since that also gives you a direct way to compute the polynomial.  The Galois group for $\omega$ is $\mathbb{Z}/6\Bbb{Z}$, generated by sending $\omega$ to $\omega^{3}$.  Applying this once, $\omega^{3} + \omega^{-3}$ is a conjugate of $\omega + \omega^{-1}$.  Applying it again yields $\omega^{2} + \omega^{-2}$ since 3*3 = 2 mod 7.  Applying a third time gets you back where you started since 3*3*3 = 27 = -1 mod 7.  Thus, the minimal polynomial must be $(x - (\omega + \omega^{-1})) + (x - (\omega^{2} + \omega^{-2})) + (x - (\omega^{3} + \omega^{-3}))$, and the rest is simple arithmetic (and remembering the all 7 7th roots of 1 sum to 0, so that the six of them other than 1 sum to -1).
