Cyclotomic polynomials and Galois groups

Question

According to this question I want to extend the question from there.

Lets consider again the galois extension $\mathbb Q(\zeta)/\mathbb Q$ where $\zeta$ is a primitive root of the $7^{th}$ cyclotomic polynomial.

I want to determine the minimal polynomial of $\zeta+\zeta^{-1}$ and $\zeta+\zeta^{2}+\zeta^{-3}$.

I know that one of the minimal polynomial has degree 2 and the other one degree 3, because $|Gal(L/K)|=6$.

Well, started with squaring the first one, which yields $\zeta^2+2+\zeta^{-2}$, but how to continue?

Answer 1

A good way to find the minimal polynomial of an element when knowing the Galois group is to compute all the conjuagtes of the element and compute $ \prod_j (X -a_j)$ where $a_j$ are the conjugates.

The conjugates in the first case are $\zeta + \zeta^{-1}$, $\zeta^2 + \zeta^{-2}$, and $\zeta^{3} + \zeta^{-3}$. Note the others just repeat, for example, $\zeta^4 + \zeta^{-4} = \zeta^3 + \zeta^{-3}$, so we do not consider them. So the minimal polynomial is $\prod_{j=1}^3 (X - (\zeta^j + \zeta^{-j}))$. You can further expand and simplify if you want.

You can do about the same for the other case.

Answer 2

Here’s my method for calculating the minimal polynomial of $\zeta+\zeta^{-1}$.

Write down the three powers of $\rho=\zeta+\zeta^{-1}$, as well as the fundamental equation \begin{align} 0&=\zeta^3+&\zeta^2+&\zeta+&1+&\zeta^{-1}+&\zeta^{-2}+&\zeta^{-3}\\ \rho&=&&\zeta+&&\zeta^{-1}\\ \rho^2&=&\zeta^2&&+2&&+\zeta^{-2}\\ \rho^3&=\zeta^3&&+3\zeta&&+3\zeta^{-1}&&+\zeta^{-3}\\ &=&-\zeta^2&+2\zeta&-1&+2\zeta^{-1}&-\zeta^{-2}\\ \rho^3+\rho^2&=&&2\zeta&+1&+2\zeta^{-1}\\ &=2\rho+1\,, \end{align} where you’ve gotten the fifth line by subtracting zero from the fourth line. (And I apologize for the bad alignment.) But you see that $\rho^3+\rho^2-2\rho-1=0$, there it is.

Answer 3

The $7$th cyclotomic polynomial is

$$\Phi_7(X)= X^6 + X^5 + X^4 + X^3 + X^2 + X + 1$$

and so, after dividing by $X^3$ and rearranging terms we get: $$\frac{\Phi_7(X)}{X^3} = X^3 + \frac{1}{X^3} + X^2 + \frac{1}{X^2} + X + \frac{1}{X} + 1 $$ The expression on the RHS can be expressed in terms of $X+\frac{1}{X}$. To do that, calculate:

$$(X+ \frac{1}{X})^3 = X^3 + \frac{1}{X^3} + 3( X + \frac{1}{X})\\ (X+\frac{1}{X})^2 = X^2 + \frac{1}{X^2} + 2$$

and therefore $$\frac{\Phi_7(X)}{X^3} = (X+\frac{1}{X})^3 +(X+\frac{1}{X})^2 - 2(X + \frac{1}{X}) -1$$

or $$\frac{\Phi_7(X)}{X^3} = \Psi(\,X+\frac{1}{X})$$ where $$\Psi(Y) = Y^3 + Y^2 - 2 Y -1$$

Therefore, if $\lambda$ is a root of $\Phi_7$ then $\lambda+ \frac{1}{\lambda}$ is a root of $\Psi$. We also notice that $\Psi$ is irreducible over $\mathbb{Q}$. Indeed, the only potential rational roots are $\pm 1$ and these are not roots. Therefore, the minimal polynomial of $\lambda+ \frac{1}{\lambda}$ is $\Psi$.

Obs: The complex roots of $\Phi_7$ are $e^{\frac{2 k \pi i}{7}}$, for $1 \le k \le 6$. They group in $3$ pairs of inverse numbers $e^{\frac{2 \pi i}{7}}$ and $e^{\frac{2 \cdot 6 \pi i}{7}}$, $e^{\frac{2\cdot 2 \pi i}{7}}$ and $e^{\frac{2\cdot 5 \cdot 6 \pi i}{7}}$, $e^{\frac{2 \cdot 3 \pi i}{7}}$ and $e^{\frac{2\cdot 4 \cdot 6 \pi i}{7}}$. We conclude that the roots of the polynomial $\Psi$ are $2\cos (\frac{2 \pi}{7}), 2\cos (\frac{4 \pi}{7}), 2\cos (\frac{6 \pi}{7})$

Notice the equality

$$\{2, 4, 1\} = \{x \in (\mathbb{Z}/7)^{\times}\ | x \ \text{is a square}\}$$

Therefore, for evey $a \in (\mathbb{Z}/7)^{\times}$ we have \begin{eqnarray} a \cdot \{2, 4, 1\} &=&\{2, 4, 1\} \text{ if } a \ \text{is a square} \\ a \cdot \{2, 4, 1\} &=& \{3,5,6\} \text{ if } a \ \text{is not a square} \end{eqnarray}

We conclude that the automorphism $\rho_a$ of $\mathbb{Q}(\zeta)$, $\zeta\mapsto \zeta^a$ invariates each of $\zeta + \zeta^2 + \zeta^ 4$, $\zeta ^3 + \zeta^5 + \zeta^6$ or switches them. Hence these elements are conjugate. Let's find their sum and product.

$$(\zeta + \zeta^2 + \zeta^ 4) +( \zeta ^3 + \zeta^5 + \zeta^6)=-1 \\ (\zeta + \zeta^2 + \zeta^ 4) ( \zeta^3 + \zeta^5 + \zeta^6) = \zeta^4+\zeta^6+1+\zeta^5+1+\zeta+ 1+\zeta^2+\zeta^3=\\ 2 + (1 + \zeta + \zeta^2+\zeta^3+\zeta^4+\zeta^5+\zeta^6)=2 $$

Therefore $\zeta + \zeta^2 + \zeta^ 4$ and $\zeta^3 + \zeta^5 + \zeta^6$ are the roots of $$Z^2 + Z + 2$$

Obs: For general $n$ and $\zeta$ a primitive root of unity of order $n$, and t $H$ is a subgroup of $(\mathbb{Z}/n)^{\times}$, consider the sum $$\sum_{g \in H} \zeta^{g}$$ Then its conjugates will be $$\sum_{g \in aH} \zeta^{g}$$ where $aH$ are the cosets of $H$ in $ (\mathbb{Z}/n)^{\times}$ $\tiny{\text{(Gauss sums)}}$