# Cyclotomic polynomials and Galois groups

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?

• Use Galois theory: the degree of an element $\alpha$ over $\mathbf Q$ is the index of the subgroup $H$ fixing it: $[\mathbf Q(\alpha):\mathbf Q] = [G:H]$. For each element compute that subgroup. – KCd Jan 31 '15 at 13:02

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.

• Hello, thanks for the answer. Can you explain me where this idea comes from? I understand what you are doing, but I dont understand where this comes from:) – Epsilondelta Jan 31 '15 at 13:17
• You are welcome. It is a general fact that for a (separable [over the rationals everything is separable though]) element the minimal polynomial is given in the way say. (It is likely you can find this somewhere in the documentation you have available posibly using a slightly different phrasing.) It is also a general fact that in a Galois extension the images under the Galois group of an element yield all the conjugates. (This is more or less the definition of being a Galois extension.) – quid Jan 31 '15 at 13:27
• For the first: I am doing galois theory for 2 weeks now and I thought I understood it. But its the first time that I am seeing this. What you mean with "The conjugates.."? Not the complex conjugate, right? Maybe the orbit of $\zeta+\zeta^{-1}$ under all elements of the Galois group? If you know a source where I can read about this fact, can you tell me the name please?:) For the second: We say that a field extension $L/K$ is a Galois extension iff it is normal and separable, where is the connection between this and your statement? Hopefully you can answer this question, I would appreciate it:) – Epsilondelta Jan 31 '15 at 13:45
• The conjugates of an element (over some base field) are by definition the roots of the minimal polynomial (since in the spearable case the roots are certainly simple roots one gets what I said). For this, and also the second question see en.wikipedia.org/wiki/Conjugate_element_%28field_theory%29 – quid Jan 31 '15 at 13:52
• Ah, okay. I got confused because in my language we wont call this "conjugate of an element". Thanks a lot. That does make sense :) – Epsilondelta Jan 31 '15 at 14:01

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.

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 $(\mathbb{Z}/n)^{\times}$ $\tiny{\text{(Gauss sums)}}$