minimal polynomial $\zeta_n$ and $\zeta_n^p$ is the same for any prime $p$ not dividing $n$

Question

I want to prove that for any prime $p$ not dividing $n$, $\zeta_n$ and $\zeta_n^p$ have the same minimal polynomial over $\mathbb{Q}$.

My proposed proof,

Suppose $\zeta_n$ is a primitive $n$th root of unity, then the minimal polynomial of $\zeta_n$ is the $n$th cyclotomic polynomial. Namely, $$\Phi_n(x) = \prod_{\zeta_n \in \mu_n^{\times}} (x - \zeta_n) \implies \Phi_n(\zeta_n) = \prod_{\zeta_n \in \mu_n^{\times}} (\zeta_n - \zeta_n) =0.$$ We claim that $\Phi_n(x)$ is also the minimal polynomial of $\zeta_n^p$ for some prime $p$ not dividing $n$. Consider that $$\Phi_n(\zeta_n^p) = \prod_{\zeta_n \in \mu_n^{\times}} (\zeta_n^p - \zeta_n)= (\zeta_n^p - \zeta_n) \cdots (\zeta_n^p - \zeta_n^p) =0.$$ Thus, $\Phi_n(x)$ has both $\zeta_n$ and $\zeta_n^p$ as a root. Given that $\Phi_n(x)$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion, we have that $\Phi_n(x)$ is the minimal polynomial of both $\zeta_n$ and $\zeta_n^p$.

Answer 1

Let $K$ being a field, and $\zeta_n$ a primitive root of unity in $K$, that is $\zeta_n$ is a root of the polynomial $X^n-1 \in K[X]$ but not of $X^m-1$ when $m < n$.

• Let $\gcd(a,n) = 1$ so that $\zeta_n^a$ is also a primitive root of unity.

It is clear that the field extension $K(\zeta_n) = K(\zeta_n^a)$ since $\zeta_n \in K(\zeta_n^a)$ and $\zeta_n^a \in K(\zeta_n)$.

• Let $P \in K[X], \ \ P(X) = \sum_{k=0}^d c_k X^k$ being the minimal polynomial of $\zeta_n$.

We need the theorem that $K(\zeta_n)\simeq K[X]/(P(X))$, so that every element of $K(\zeta_n)$ is of the form $\sum_{k=0}^{d-1} \alpha_k \zeta_n^k$ for some coefficients $\alpha_k \in K$, in order to prove that $\sigma : K(\zeta_n) \to K(\zeta_n)$, $ \sigma\left(\sum_{k=0}^{d-1} \alpha_k \zeta_n^k\right) = \sum_{k=0}^{d-1} \alpha_k \zeta_n^{ak}$ is an automorphism, defined by $\sigma(\zeta_n) = \zeta_n^a$.

• Hence $$\textstyle 0 = \sigma(0) = \sigma(P(\zeta_n)) = \sigma\left(\sum_{k=0}^d c_k \zeta_n^k\right) =\sum_{k=0}^d c_k \zeta_n^{ak} = P(\zeta_n^a)$$
• From $K[X]/(P(X)) \simeq K(\zeta_n) \simeq K(\zeta_n^a) \simeq K[X]/(Q(X))$ with $Q$ the minimal polynomial of $\zeta_n^a$ we know that $\text{deg}(P) = \text{deg}(Q)$, hence together with $\zeta_n^a$ is a root of $P$ we conclude that $P = Q$.

Answer 2

The following proof is due to Dedekind.

Assume the contrary, i.e that $ \zeta_n $ and $ \zeta_n^p $ have distinct minimal polynomials over $ \mathbb{Q} $, say $ f $ and $ g $. Then, both of these minimal polynomials divide $ x^n - 1 $, so we have an equality

$$ x^n - 1 = fgh $$

Gauss' lemma implies that we have $ f, g, h \in \mathbb{Z}[x] $. Then, we may reduce both sides of this equation modulo $ p $. We find

$$ x^n - 1 = \bar{f} \bar{g} \bar{h} $$

where $ \bar{f} \equiv f \pmod{p} $. The formal derivative of the LHS does not vanish by the assumption that $ p $ does not divide $ n $, therefore the RHS has distinct roots. This implies that $ \bar{f} $ and $ \bar{g} $ are coprime.

On the other hand, the polynomial $ g(x^p) $ has $ \zeta_n $ as a root, therefore $ f(x) | g(x^p) $ in $ \mathbb{Q}[x] $. We then have

$$ g(x^p) = f(x) h_2 (x) $$

where, once again, we may assume $ h_2 \in \mathbb{Z}[x] $ by Gauss' lemma. Reducing this equation modulo p and using the identity $ (a+b)^p = a^p + b^p $ yields

$$ \bar{g}(x^p) = (\bar{g}(x))^p = \bar{f} \bar{h_2} $$

But then, any irreducible factor of $ \bar{f} $ has to divide $ \bar{g} $, contradicting coprimality.

The following lemma makes our use of Gauss' lemma precise:

Lemma. Let $ f = gh $ with $ f \in \mathbb{Z}[x] $ and $ g, h \in \mathbb{Q}[x] $, with all polynomials monic. Then, $ g, h \in \mathbb{Z}[x] $.

Proof. We may multiply $ g $ and $ h $ by appropriate integers $ \alpha $ and $ \beta $ so that $ \alpha g $ and $ \beta h $ are in $ \mathbb{Z}[x] $. We have

$$ \alpha \beta f = (\alpha g)(\beta h) $$

The content of the LHS is clearly $ \alpha \beta $, since $ f $ is monic and in $ \mathbb{Z}[x] $. Therefore, $ \alpha \beta = c(\alpha g) c(\beta h) $ by Gauss' lemma. On the other hand, since $ g $ and $ h $ are monic, $ c(\alpha g) $ divides $ \alpha $ and $ c(\beta h) $ divides $ \beta $. These facts imply that $ c(\alpha g) = \alpha $ and $ c(\beta h) = \beta $, therefore $ g $ and $ h $ are in $ \mathbb{Z}[x] $.

Answer 3

We know the minimal polynomial of $\zeta_n$ is the $n$th cyclotomic polynomial $\Phi_n (x)$, which is defined as $$\Phi_n (x)=\prod_{k \in \mathbb{Z}_n^{\times}}(x-\zeta_n^k)$$ where $\mathbb{Z}_n^{\times}$ is the distinct residue classes coprime to $n$. If $p$ does not divide $n$, then $p\equiv k \mod n$ for some $k \in \mathbb{Z}_n^{\times}$, say $p=nq+k$. Thus, $$\Phi_n (\zeta_n^p)=\Phi_n (\zeta_n^{nq+k})=\Phi_n (\zeta_n^{nq} \zeta_n^k)=\Phi_n (\zeta_n^k)=0$$

Since $\Phi_n(x)$ is irreducible and monic, it is identically the minimal polynomial of $\zeta_n^p$.

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Basically, there's no need to mention that $\Phi_n(x)$ is irreducible by Eisenstein's Criterion because we already assumed it was irreducible when we said it was the minimal polynomial of $\zeta_n$. Also, I'm pretty sure Eisenstein's Criterion only proves $\Phi_n(x)$ is irreducible when $n$ is a prime.

Furthermore, the $p$th power of $\zeta_n$ doesn't necessarily appear in the factorization of $\Phi_n(x)$ (as it did in your proof). Instead, the $p$th power gets reduced down to a coprime residue class, that does appear in its factorization as it is defined (it's a minor qualm, but you did say "airtight").

Since we already know $\Phi_n$ is monic and irreducible and since minimal polynomials are unique to their roots (from basic field theory), the proof boils down to showing $\zeta_n^p$ is a root of $\Phi_n$.

I think your proof is fine, but it also depends on how many facts about $\Phi_n(x)$ we are allowed to know. I just used the same assumptions you did, but with less assumptions it would probably be longer.

Answer 4

You only need to show $\zeta_n^p \in \mu_n^{\times}$. This can be seen as follows. Notice $\text{lcm}(p, n)=pn$. We have $(\zeta_n^p)^k=1$ if and only if $pk$ is a multiple of $n$. The smallest such $k$ is the order of $\zeta_n^p$. But the smallest $k$ such that $pk$ is a multiple of $n$ is exactly $pn$ because $\text{lcm}(p, n)=pn$.