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.


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 $n = m.$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.