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Let $n$ be a positive integer, $p$ a prime and $e \geq 1$. Denote by $\Phi_n(x)$ the $n$-th cyclotomic polynomial. Then $$\Phi_{p^en}(x) \equiv \, \Phi_n(x)^d \mathrm{mod}\,{(p)}$$ where $$d = \frac{\deg(\Phi_{p^en}(x))}{\deg(\Phi_n(x))}$$

I am looking for a proof or a reference where a proof of this fact appears since I haven't been able to prove this by myself. Thank you.

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First, assume that $p$ and $n$ are relatively prime. Let $\xi$ be a primitive $n$th root of unity, let $\eta$ be a primitive $p^e$th root of unity, and let $\zeta$ be a primitve $p^e n$th root of unity. Then $$ \Phi_{p^e n}(x) = \prod_{i\in\left(\Bbb Z/n\Bbb Z\right)^\times,\,j\in\left(\Bbb Z/p^e\Bbb Z\right)^\times}(x - \xi^i\eta^j), $$ because as $i,j$ vary as in the product, $\xi^i\eta^j$ varies over all primitive $p^en$th roots of unity.

Let $\mathfrak{p}$ be a prime ideal lying over $(p)$ in $\Bbb Z[\zeta] = \mathcal{O}_{\Bbb Q(\zeta)}.$ Now, $x^{p^e} - 1\equiv (x - 1)^{p^e}\pmod{p},$ so that $\eta^i\equiv 1\pmod{\mathfrak{p}}$ for any $i.$ It then follows that \begin{align*} \Phi_{p^e n}(x) &= \prod_{i\in\left(\Bbb Z/n\Bbb Z\right)^\times,\,j\in\left(\Bbb Z/p^e\Bbb Z\right)^\times}(x - \xi^i\eta^j)\\ &\equiv \prod_{i,j}(x - \xi^i)\pmod{\mathfrak{p}}\\ &\equiv \left(\prod_i(x - \xi^i)\right)^{\varphi(p^e)}\pmod{\mathfrak{p}}\\ &\equiv \Phi_n(x)^{\varphi(p^e)}\pmod{\mathfrak{p}}. \end{align*}

From this, we may deduce that this is in fact a congruence of polynomials modulo $p.$ The claim about degrees follows from the observation that $\deg\Phi_m = \varphi(m)$ and elementary properties of the Euler $\varphi$ function.

In general, write $n = p^r m,$ $(p,m) = 1$. The above argument shows that $$ \Phi_{p^e n}(x)\equiv \Phi_m(x)^{\varphi(p^{e + r})}\pmod{p}. $$ It remains to be shown that $$ \Phi_m(x)^{\varphi(p^{e + r})} = \Phi_m(x)^{p^{e + r - 1}(p - 1)}\equiv\Phi_{p^r m}(x)^{p^{e}}\pmod{p}, $$ as $$ \frac{\deg\Phi_{p^e n}}{\deg\Phi_{n}} = \frac{\varphi(p^e n)}{\varphi(n)} = \frac{\varphi(p^{e + r})\varphi(m)}{\varphi(p^r)\varphi(m)} = \frac{p^{e + r - 1}(p - 1)}{p^{r - 1}(p - 1)} = p^e. $$ Moreover, it is enough to show that $$ \Phi_m(x)^{\varphi(p^r)} = \Phi_m(x)^{p^{r - 1}(p - 1)}\equiv\Phi_{p^r m}(x)\pmod{p}, $$ as the desired congruence then follows by raising both sides to the $p^e$th power. However, this congruence is exactly the first congruence we deduced in the relatively prime setting, with $e = r$ and $n = m.$

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  • $\begingroup$ This is excellent. Thank you very much! $\endgroup$ – user313212 Sep 3 '18 at 17:24

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