How to find the irreducible polynomial in Q[ζ3]?

Question

$ζ_n = e^{2πi}/n$

The question is to determine the irreducible polynomial over $\mathbb{Q}$ then over $\mathbb{Q}[ζ_3]$

of the following:

$ζ_4$

$ζ_6$

$ζ_8$

$ζ_9$

$ζ_{10}$

$ζ_{12}$

My attempt:

$ζ_4$ is a root of $x^2+1$ so that's the irreducible polynomial over $\mathbb{Q}$ but when we adjoin $ζ_3 = \frac{1 + i \sqrt{3}}{2}$ we don't obtain $i$ so it's still irreducible.

$ζ_6$ is a root of $x^2-x-1$ in $\mathbb{Q}$ but when we adjoin $ζ_3$ we do get the root. So the question is, how do I find the new irreducible polynomial?

$ζ_8$ = $\frac{\sqrt{2} -i \sqrt{2}}{2}$ and the irreducible polynomial is $x^4+1$ and even when we adjoin $ζ_3$ it's still irreducible

$ζ_9$ the irreducible polynomial is $x^6 + x^3 + 1$ (i'm completely lost on the rest)

$ζ_{10}$ again, not sure where to proceed, but the polynomial is $x^4-x^3+x^2-x+1$

$ζ_{12}$ the polynomial is $x^4 - x^2 + 1$ and $ζ_{12}$ is $\frac{-i + \sqrt{3}}{2}$ this looks like it would reduce in $ζ_3$ but how can I find the new polynomial?

Answer 1

This is what I get with the program below for the minimal polynomial over $\mathbb{Q}(\zeta_3)$ of :

$$\begin{array}{lll}\zeta_4 \qquad & x^2 + 1\\ \zeta_6& x - \zeta_6 &=x - 1-\zeta_3\\ \zeta_8& x^4 + 1&\\ \zeta_9& x^3 - \zeta_9^3&=x^3 - \zeta_3\\ \zeta_{10}& x^4 - x^3 + x^2 - x + 1&\\ \zeta_{12}& x^2 - \zeta_{12}^2&=x^2 - 1-\zeta_3\end{array}$$

---

Let $l = lcm(n,m)$ so that $\mathbb{Q}(\zeta_n,\zeta_m) = \mathbb{Q}(\zeta_l)$

We are looking at $\mathbb{Q}(\zeta_l) /\mathbb{Q}(\zeta_n)$.

The elements of $Gal(\mathbb{Q}(\zeta_l)/\mathbb{Q})$ are $\sigma_k(\zeta_l) = \zeta_l^k$ with $gcd(k,l) = 1$.

So that $\sigma_k(\zeta_n) = \sigma_k(\zeta_l^{l/n}) =\zeta_n^{k}$, which means $\sigma_k \in Gal(\mathbb{Q}(\zeta_l) /\mathbb{Q}(\zeta_n))$ iff $k \equiv 1 \bmod n$.

With $H = \{ \sigma_k, k\equiv 1 \bmod n\}$, $\ \ \mathbb{Q}(\zeta_n)$ is the field fixed by $H$.

Since the extensions are Galois then $H = Gal(\mathbb{Q}(\zeta_l) /\mathbb{Q}(\zeta_n))$.

Hence the minimal polynomial of $\zeta_m$ over $\mathbb{Q}(\zeta_n)$ is $$F_{m,n}(x) = \prod_{\sigma \in H} (x-\sigma(\zeta_m)) = \prod_{k \in \mathbb{Z}_l^\times,\ k \equiv 1 \bmod n} (x-\zeta_m^k)=\prod_{d=1,gcd(dn+1,l)=1}^{l/n} (x-\zeta_m^{dn+1})$$

    % sage math
    tab = [4,6,8,9,10,12]
    n = 3
    for m in tab :
        l = lcm(n,m)
        K.<w> = CyclotomicField(l)
        R.<x> = K[]
        P = 1
        a = w^(l/m)
        for d in [1..l/n] :
            k = n*d+1
            if gcd(l,k) == 1 :
                P = P * (x-a^k)
        [m,P]