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 $\zeta_n^p$.

  • 2
    $\begingroup$ "caprice" is not a commonly recognized mathematical term. Might you mean "coprime"? $\endgroup$ Jun 3, 2016 at 9:32
  • $\begingroup$ Yes, I meant coprime, sorry for the confusion. The appropriate edit has been made. $\endgroup$
    – user319128
    Jun 3, 2016 at 9:38
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    $\begingroup$ Is $\zeta_n$ a primitive $n$th root of unity? If so, $\zeta_n^p$ is as well, and the $n$th cyclotomic polynomial is known to be the minimal polynomial of any primitive $n$th root of unity over $\mathbb{Q}$. $\endgroup$
    – Ben West
    Jun 3, 2016 at 9:42
  • $\begingroup$ Is it true if $\zeta_n$ is simply an $n$th root of unity? $\endgroup$
    – user319128
    Jun 3, 2016 at 9:45
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    $\begingroup$ what is important is that ${\zeta_n}^k$ is a primitive $n$th root of unity whenever $gcd(n,k) = 1$. i.e. $\zeta_n = e^{2i \pi a / n}$ where you can choose $a$ as you want with the only one constraint $gcd(n,a) =1$ $\endgroup$
    – reuns
    Jun 3, 2016 at 23:55

4 Answers 4


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 $P = Q$.
  • $\begingroup$ (if someone can check my answer, I'd appretiate, tks) $\endgroup$
    – reuns
    Jun 6, 2016 at 4:02
  • $\begingroup$ I believe in your representation of elements from $K(\zeta_n)$ in the second bullet point, the sum should only go up to $d-1$. Aside from that, your argument looks good. $\endgroup$
    – JasonM
    Jun 10, 2016 at 8:48
  • $\begingroup$ Ah, well I mostly pointed it out because I get the feeling it obscures the reason behind the representation, being that $K(\zeta_n)$ is a $d$-dimensional vector space over $K$. It's really a minor notational thing. I get the convenience though; $d$ is easier to type in LaTeX exponents than $d-1$. $\endgroup$
    – JasonM
    Jun 10, 2016 at 9:20
  • $\begingroup$ @JasonM : yes. the key point is that $\sigma(\zeta_n) = \zeta_n^a$ is an automorphism, and I'm not sure what are the minimal facts needed to prove it. note that if $K$ contains say $\zeta_{n/m}$, then $\sigma(\zeta_n) = \zeta_n^a$ should stay an automorphism, and the argument stays the same ? $\endgroup$
    – reuns
    Jun 10, 2016 at 9:24
  • $\begingroup$ Yes of course. I'm certainly impressed with what you've come up with. Is the goal to minimize assumptions about cyclotomic polynomials or assumptions in general? Because your argument uses a much deeper field-theoretic argument, which includes field-theoretic assumptions. $\endgroup$
    – JasonM
    Jun 10, 2016 at 9:31

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] $.


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 $\zeta_n^p$.

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.

  • $\begingroup$ How do you know that the nth cyclotomic polynomial is the minimal polynomial of all of the primitive roots of unity? What if they had distinct minimal polynomials? You are assuming the result you want to prove in your proof. Of course, if you proved first that the nth cyclotomic polynomial is always irreducible, there would be no problems. $\endgroup$
    – Ege Erdil
    Jun 10, 2016 at 10:26
  • $\begingroup$ @Starfall Like I said, it really depends on how much we are allowed to know. I was just using the assumptions the author used. If he is taking a class, he may have already learned facts like the irreducibility of $\Phi_n(x)$. I think the problem is that there are two ways to define $\Phi_n(x)$ that each have a nontrivial proof of equivalence with the other. $\endgroup$
    – JasonM
    Jun 10, 2016 at 16:40

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 $\text{lcm}(p, n)=pn$.


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