what is the minimal polynomial of $\alpha=\rho+\rho^4+\rho^{16}$ , $\rho^{21}=1$? here is a question that I recently tried to solve:
Is it possible to construct with Compass-and-straightedge the number $$\alpha=\rho+\rho^4+\rho^{16}$$ , while $\rho^{21}=1$ over $\mathbb{Q}$ ?
also to find the minimal polynomial of $\alpha$.
I know that it is possible to construct $\alpha$ because: $$|gal(Q[\alpha]/Q)|=4=2^2$$ moreover:
because $gal(Q[\rho]/Q)=z2*z6$ i deduced that $gal(Q[\alpha]/Q)=z2*z2$
I am still having trouble finding the minimal polynomial of $\alpha$.
can someone please help?
 A: The conjugates of $y$ are $y_2 = x^2 + x^8 + x^{11}, y_3 = x^5 + x^{17} + x^{20}$ and $y_4 = x^{10} + x^{13} + x^{19}$.
The standard way to get the minimal polynomial would be to compute $\prod (X - y_i)$. So just compute the elementary symmetric polynomials in the $y_i$ in terms of $x$, then simplify using the minimal polynomial of $x$ to get to an integer.
Becuase I don't particularly like expanding products with $3^4$ terms, let's see if we can find out who $y$ is with a completely different method.
First, the Galois group of $K = \Bbb Q(x)$ is $(\Bbb Z/21\Bbb Z)^\times = \{1,2,4,5,8,10,11,13,16,17,19,20\}$, and $\Bbb Q(y) = K^{\{1,4,16\}}$. Galois theory says that the intermediate fields between $\Bbb Q$ and $\Bbb Q(y)$ correspond to subgroups containing $\{1,4,16\}$.
It helps to know those subfields really are, and those are all quadratic extensions of $\Bbb Q$ :
$\Bbb Q(\sqrt {-3}) = K^{\{1,4,10,13,16,19\}}$,
$\Bbb Q(\sqrt {-7}) = K^{\{1,2,4,8,11,16\}} $,
$\Bbb Q(\sqrt{21}) = K^{\{1,4,5,16,17,20\}}$
Now let's go back to $y$ and its conjugates. They are algebraic integers because $x$ is one and algebraic integers form a ring. Moreover, $y+y_2+y_3+y_4 = \sum_{i=0}^{20} x^i - \sum_{i=0}^6 x^{3i} - \sum_{i=0}^2 x^{7i} + 1 = 1$.
Then by looking at the Galois action on $y$ and its conjugates, we get that $y+y_2 = \frac 12 (1 + n_2\sqrt {-7})$, $y + y_3 = \frac 12 (1 + n_3\sqrt {21})$, and $y+y_4 = \frac 12 (1 + n_4 \sqrt{-3})$, where $n_2,n_3,n_4$ are odd integers.
The triangle inequality then shows that the absolute values of each of those quantity is less than $6$, which leaves few possibilities for those coefficients. In particular we can quickly find that $y + y_3 = \frac 12 (1 + \sqrt {21})$, $|n_2| \le 3$ and $|n_4| \le 5$.
Eyeballing a unit circle, I would guess that $y + y_2 = \frac 12(1+ \sqrt {-7})$ and $y + y_4 = \frac 12(1- \sqrt{-3})$.
By summing everyone and substracting $y+y_2+y_3+y_4$ one gets $y = \frac 14 (1 + \sqrt{21}+\sqrt{-7}-\sqrt{-3})$
Hopefully you can get the construction and the minimal polynomial from there.
