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$ζ_n = e^{2πi}/n$

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?

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  • $\begingroup$ For $\zeta_9$ you need to use $\zeta_{9}^3-\zeta_3=0$ and then you see a polynomial with coefficients in $\mathbb{Q}(\zeta_3)$ satisfied by $\zeta_9$. So now you can try the other outstanding cases ... $\endgroup$ – ancientmathematician Mar 30 '17 at 13:17
  • 2
    $\begingroup$ a general remark that may be helpful: the minimal polynomial of $\zeta_n$ over $\mathbb{Q}[\zeta_n]$ must divide the minimal polynomial over $\mathbb{Q}$, so for example in the case $x^4-x^2+1$ it can be instructive to factor it in $\mathbb{C}$. You can see immediately that by the change of variable $y=x^2$ and the ABC formula you can factor it into two second degree polynomials... This might help. $\endgroup$ – M. Van Mar 30 '17 at 13:25
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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]
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  • $\begingroup$ You seem to have an error for $\zeta_6$. I would think that the minimal polynomial for $\zeta_6$ over $\Bbb Q(\zeta_3)$ is $X+\zeta_3$. Or am I misunderstanding your notation? $\endgroup$ – Lubin Mar 30 '17 at 17:57
  • $\begingroup$ @Lubin $\zeta_6 \in \mathbb{Q}(\zeta_3)$ so its minimal polynomial is $x-\zeta_6$ (I don't have a notation : this is the output of a Sage program). It is quite hard to generate programmatically such a polynomial with the coefficients in $\mathbb{Q}[\zeta_3]$ as we'd like to $\endgroup$ – reuns Mar 30 '17 at 18:17
  • $\begingroup$ I see: I in fact I didn’t understand the notation. I was looking for a polynomial in $X$, with coefficients expressed in terms of $\zeta_3$. My apologies. $\endgroup$ – Lubin Mar 31 '17 at 13:44
  • $\begingroup$ @Lubin I edited. And do you think there is an easier way to find that $F_{n,m}(x) = \prod_{d=1,gcd(dn+1,l)=1}^{l/n} (x-\zeta_m^{dn+1})$ ? $\endgroup$ – reuns Mar 31 '17 at 18:27
  • $\begingroup$ Have never thought much about this question. If I get time… $\endgroup$ – Lubin Apr 1 '17 at 16:39

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