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Let $n, m$ be two natural numbers and $\Phi_n(q), \Phi_m(q)$ the $n$-th and $m$-th cyclotomic polynomials respectively. Define a function $c_{n,m} \colon \mathbb{N} \to \left\{0,1\right\}\cup \left\{ p \mid p \, \text{prime}\right\}$ by $$c_{m,n} = \begin{cases} 0, \, & \text{if $n = m$}\\ p, \, & \text{if $\frac{n}{m} = p^j$ where $p$ prime and $j \in \mathbb{Z}\setminus \{0\}$}\\ 1, \, & \text{if $\frac{n}{m}$ is not an integer power of a prime} \end{cases}$$

Show that $\Phi_m(q) \in \sqrt{\left( \Phi_n(q)\right) + (c_{m,n})[q]}$

This result should rely on properties of cyclotomic polynomials I am not aware of. Obviously if $c_{n,m} = 0$ the result is true since $n = m$ and $(c_{n,m}) = (0)$. If $c_{n,m} = 1$, then $(c_{m,n}) = (1) = \mathbb{Z}$, and then it is true that $\Phi_m(q) \in \sqrt{\left( \Phi_n(q)\right) +\mathbb{Z}[q]}$, correct?

Any answer or any reference to literature where this results or proven would be appreciated. Thank you in advance.

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  • $\begingroup$ Your notation is really unclear. Using the $\in$ symbol seems to indicate that you intend $\sqrt{\left( \Phi_n(q)\right) + (c_{m,n})[q]}$ to denote a set; what is the exact definition of this set? $\endgroup$ Jul 14, 2018 at 18:03
  • $\begingroup$ @GregMartin The notation $\sqrt{(\Phi_n(q))+ (c_{m,n})[q]}$ means the radical of $(\Phi_n(q))+ (c_{m,n})[q]$. I hope it is clear now. $\endgroup$
    – user313212
    Jul 14, 2018 at 18:39

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As you write, the statement in case $c_{m,n} = 1$ is immediate because clearly $\Phi_m(x) \in \mathbb{Z}[x] = \mathbb{Z}[x] + (\Phi_n(x)) = \sqrt{\mathbb{Z}[x] + (\Phi_n(x))}$. Since this is true for all $m,n$, the statement in this case is also not meaningful.

If $n = m p^i$ so that $c_{m,n} = p$, then we have $\Phi_n(x) = \Phi_m(x^{p^i})$ (elementary property of cyclotomic polynomials).

Therefore it suffices to prove that $$\Phi_m(x)^{p^i} \equiv \Phi_m(x^{p^i}) \mod p$$

But in fact this statement is more generally true of any polynomials over $\mathbb{Z}/p\mathbb{Z}$. It can be seen as an elaboration of the so-called "Freshman's Dream," that $(x+y)^{p} \equiv x^p + y^p \mod p$. I also found a reference here as lemma 4.3. The first property I mentioned is theorem 3.2 there.

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  • $\begingroup$ Thank you! The case where $c_{m,n} = 1$ is okay as I wrote on the question? $\endgroup$
    – user313212
    Jul 15, 2018 at 11:01
  • $\begingroup$ yep, i included a sentence about that. Curious, does this exercise have some application? $\endgroup$ Jul 15, 2018 at 14:48
  • $\begingroup$ Thank you, I've accepted your answer. And yes, though maybe not this part itself. I am trying to understand a paper about cyclotomic completions and this "obvious" result (as wrote by the author) wasn't so obvious to me at first... $\endgroup$
    – user313212
    Jul 15, 2018 at 18:09
  • $\begingroup$ Yes if I came across this in a paper I also would've appreciated a quick sentence for justification. $\endgroup$ Jul 16, 2018 at 18:26

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