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All I know is playing with this fact that $Q_{mp}(x) = \dfrac{x^{mp}-1}{\prod_{d|mp}Q_d(x)}$ and I really did many calculations but they went nowhere. Also I proved that $Q_{mp}(x) = Q_m(x^p)/Q_m(x)$ if prime $p$ doesn't divide $m$. Combination of latter result and induction didn't help me through and it seems I need more.

hnrzi
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2 Answers2

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Here is a proof based only on the definition of the cyclotomic polynomial:

I will denote by $\Phi_n(X)$ the $n$-th cyclotomic polynomial. If $\varepsilon = e^{\frac{2i\pi}{n}} = \cos(\frac{2\pi}{n})+i\sin(\frac{2\pi}{n})$, then $$\Phi_n(X) = \prod_{1\le i \le n \\ (i,n)=1} (X-\varepsilon^i)$$ Now consider $m, p$ such that $p$ is a prime number and $p \mid m$. Take $\xi = e^{\frac{2i\pi}{pm}} = \cos(\frac{2\pi}{mp})+i\sin(\frac{2\pi}{mp})$. Then: $$\Phi_{mp}(X) = \prod_{1\le i \le mp \\ (i,mp)=1} (X - \xi^i) = \prod_{1\le i \le mp \\ (i,m)=1} (X - \xi^i) = \prod_{1\le i \le m \\ (i,m)=1} \prod_{j=0}^{p-1}(X - \xi^{jm+i} )$$ Since $\xi^p = e^{\frac{2i\pi}{m}} = \cos(\frac{2\pi}{m})+i\sin(\frac{2\pi}{m})$ we have: $$\Phi_m(X) = \prod_{1\le i \le m \\ (i,m)=1} (X- \xi^{pi})$$ and $$\Phi_m(X^p) = \prod_{1\le i \le m \\ (i,m)=1} (X^p- \xi^{pi})$$ Fix some $1\le i \le m$ with $(i,m)=1$ and consider the polynomial $f(X) = X^p - \xi^{pi}$. It is very easy to observe that $f$ has $p$ distinct roots: $\xi^i, \xi^{m+i}, \ldots \xi^{(p-1)m+i}$. Since $f$ has degree $p$, these are all the roots of $f$. Therefore $$X^p - \xi^{pi} = f(X) = (X-\xi^i)(X-\xi^{m+i})\ldots(X - \xi^{(p-1)m+i}) = \prod_{j=0}^{p-1}(X - \xi^{jm+i} )$$ which finishes the proof.

cos_dm_math21
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  • I should have mentioned but I didn't specify base field. It's like you did it over field of complex numbers which I'm not sure could be carried to an arbitrary finite field. (Also title has been corrected) – hnrzi Aug 08 '21 at 13:31
  • How do you define the $7$-th cyclotomic polynomial over $\mathbb{F}_4$? – cos_dm_math21 Aug 08 '21 at 15:11
  • $\prod_{(7,s)=1} (x-\zeta^s)$ where $\zeta$ is a primitive root of $x^7-1$ in the extenion field of $F_4$. Also it can be proven that its coefficients should be in the prime subfield. Am I missing some point? It was supposed to be indefinable? – hnrzi Aug 08 '21 at 16:11
  • Ok. Do you find this post useful? – cos_dm_math21 Aug 08 '21 at 16:54
  • No, Should I? I feel dumb and frustrated now. We defined polynomial over some finite field, so I don't understand why image of zero-characteristic field would be beneficial for us. Cyclotomic polynomial of some degree inside finite field can be completely irrelevant to any zero-characteristic field. Can you explain more? – hnrzi Aug 08 '21 at 17:14
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    Consider the finite field $\mathbb{F}q$ with $q=r^t$ for some prime number $r$ and $t \ge 1$. Assume that $(r,m)=1$. I proved that $\Phi{mp}(X) = \Phi_{m}(X^p)$ over $\mathbb{Q}$. Reducing modulo $r$ we get that the same equality is true over $\mathbb{F}_q$. – cos_dm_math21 Aug 08 '21 at 17:29
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    Let us continue this discussion in chat. – cos_dm_math21 Aug 08 '21 at 18:55
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Big Hint:

Step $1$: Show that $Q_{mp}(x)$ and $Q_{m}(x^p)$ have the same degree

Proof: By assumption $mp = np^i$ where $\text{gcf}(n,p) = 1$, $i\geq 2$, and $m= np^{i-1}$.

The roots of $Q_{mp}$ are the primitive $(mp)^{th}$ roots of unity and there are $\phi(mp) = \phi(np^i)$ which is also the degree of $Q_{mp}$. But we also have,

$$\phi(np^i) = \phi(n)\phi(p^i) = \phi(n)(p^{i} - p^{i-1}) = p\phi(n)(p^{i-1} - p^{i-2}) = p\phi(m)$$

The right hand side is the degree of $Q_m(x^p)$. So $Q_{mp}$ and $Q_{m}(x^p)$ have the same degree.

Step $2$: Show that any root of $Q_{mp}$ is also a root of $Q_{m}(x^p)$. In other words, show that if $\zeta$ is a primitive $(mp)^{th}$ root of unity, then $\zeta^p$ is a primitive $m^{th}$ root of unity.

(You can do it in reverse and show that any root of $Q_{m}(x^p)$ is a root of $Q_{mp}$, but there are only like $5$ people in the world who would find that more natural and they're weird).

Dionel Jaime
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