## On the factors of a cyclotomic polynomial over ZZ/(p)

Let $\ell$ be a prime and let $\Phi_\ell(x) = \frac{x^\ell-1}{x-1} \in \mathbb{Z}[x]$ be the $\ell$th cyclotomic polynomial. (Remember that $\Phi_\ell$ is irreducible over $\mathbb{Z}$.) Let $p$ be a prime and let $\zeta$ denote any fixed primitive $\ell$th root of unity.

1. Show that if $p=\ell$ then $\Phi_\ell(x) = (x-1)^{\ell-1} \in \mathbb{F}_p[x]$.
2. Suppose $p \neq \ell$ and let $f$ denote the order of $p$ in $\mathbb{F}_\ell$. (That is, $f$ is minimal such that $p^f \equiv 1$ mod $\ell$.) Show that $f$ is minimal such that $\zeta \in \mathbb{F}_{p^f}$. Conclude that the minimal polynomial of $\zeta$ over $\mathbb{F}_p$ has degree $f$.
3. Show that $\mathbb{F}_p(\zeta) = \mathbb{F}_p(\zeta^a)$ for any integer $a$ not divisible by $\ell$. Conclude that, in $\mathbb{F}_p[x]$, $\Phi_\ell(x)$ is the product of $\frac{\ell-1}{f}$ distinct irreducible polynomials of degree $f$.
4. As an example, find the degrees of the irreducible factors of $\Phi_7(x)$ over $\mathbb{F}_p$.

1. Mod $p=\ell$, we have $x^\ell-1 = (x-1)^\ell$. So $\Phi_\ell(x) = (x-1)^{\ell-1}$ as desired.
2. Note that $\zeta \in \mathbb{F}_{p^f}$ precisely when $\zeta$ is a root of $x^{p^f-1}-1$. Since $p^f \equiv 1$ mod $\ell$, we have that $\ell$ divides $p^f-1$. By this previous exercise, $x^\ell - 1$ divides $x^{p^f-1}-1$. Since $\zeta$ is a $\ell$th root of unity, $\zeta \in \mathbb{F}_{p^f}$. Conversely, suppose $\zeta \in \mathbb{F}_{p^t}$. Now $\zeta^\ell \equiv 1$, so $\ell$ divides $p^t-1$ by Lagrange’s theorem in the group $\mathbb{F}_{p^t}^\times$. Thus $p^t \equiv 1$ mod $\ell$, and so $f$ divides $t$. So $f$ is minimal such that $\zeta \in \mathbb{F}_{p^f}$.

Next we claim that $\mathbb{F}_p(\zeta) = \mathbb{F}_{p^f}$. The $(\subseteq)$ inclusion is clear since $\zeta \in \mathbb{F}_{p^f}$. $(\supseteq)$ $\mathbb{F}_p(\zeta) = \mathbb{F}_{p^t}$ for some $t$, since finite fields are essentially unique. By the preceding argument, $f|t$, so $\mathbb{F}_{p^f} \subseteq \mathbb{F}_{p^t} = \mathbb{F}_p(\zeta)$ using this previous exercise.

3. It is clear that $\mathbb{F}_p(\zeta^a) \subseteq \mathbb{F}_p(\zeta)$. Conversely, since $\ell$ does not divide a\$, by Bezout’s identity there exist $x,y \in \mathbb{Z}$ such that $ax + \ell y = 1$. Now $(\zeta^a)^x = \zeta^{ax}$ $= \zeta^{1-\ell y}$ $= \zeta$. So $\mathbb{F}_p(\zeta) = \mathbb{F}_p(\zeta^a)$ provided $\ell$ does not divides $a$.

Now consider $\Phi_\ell(x)$ as a polynomial over $\mathbb{F}_p$. Let $\zeta_i$ be the distinct primitive $\ell$th roots of unity. By part (b), $\mathbb{F}_p(\zeta_i)$ has degree $f$ over $\mathbb{F}_p$ for each $i$, so the minimal polynomial of $\zeta_i$ over $\mathbb{F}_p$ has degree $f$. So the irreducible factors of $\Phi_\ell$ have degree $f$, and there are $\frac{\ell-1}{f}$ such factors. Moreover, since $\Phi_\ell$ is separable, its factors are distinct.

4. If $p \equiv 1$ mod 7, then $\Phi_7(x) = (x-1)^6$ by part (a). Evidently 2 and 4 have order 3 mod 7, so if $p \in \{2,4\}$ mod 7 then $\Phi_7(x)$ factors into two irreducible cubics. Similarly, 3 and 5 have order 6 mod 7, so that if $p \in \{3,5\}$ mod 7 then $\Phi_7(x)$ is irreducible. Since 6 has order 2 mod 7, if $p \equiv 6$ mod 7, then $\Phi_7(x)$ factors into three irreducible quadratics.