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 \ellth cyclotomic polynomial. (Remember that \Phi_\ell is irreducible over \mathbb{Z}.) Let p be a prime and let \zeta denote any fixed primitive \ellth 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 \ellth 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 \ellth 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.
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