Cyclotomic Polynomials

Let $\mathbb F$ be a field.

Definition 1. Define the n-th cyclotomic extension of $\mathbb F$ by $\mathcal Z(f_n) \subseteq \bar{\mathbb F}$ , where $f_n(x) =x^n - 1$ . We call $\alpha \in \mathcal Z(f_n)$ a primitive n-th root of unity if $\alpha^r \neq 1$ for any $1 \leq r < n$ .

Problem 1. Show that $\mathcal Z(f_n)$ contains a primitive $n$ -th root of unity if and only if $\mathrm{char}(\mathbb F) \nmid n$ . In this case, there are $\varphi(n)$ primitive n-th roots of unity, where $\varphi$ denotes the Euler totient function.

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Solution. If $\mathrm{char}(\mathbb F) = 0$ , then $\mathbb F$ contains $\mathbb Q$ as a subfield, so that $\bar{\mathbb F} \supseteq \bar{\mathbb Q}$ contains the usual primitive $n$ -th root of unity $\zeta_n := e^{2\pi i/n}$ . Otherwise, suppose $\mathrm{char}(\mathbb F) = p$ for some prime $p$ .

If $p \mid n$ , then $n = kp$ for some integer $k$ . Then for any $\alpha \in \mathcal Z(f_n)$ ,

$(\alpha^k-1)^p = \alpha^n - 1 = 0,$

so that $\alpha^k = 1$ , and $\alpha$ is not a primitive $n$ -th root of unity.

Suppose $p \nmid n$ . By Euler’s totient theorem,

$[p^{\varphi(n)}]_n = [p]_n^{\varphi(n)} = [1]_n.$

Therefore, $n \mid (p^{\varphi(n)} - 1) = |\mathbb F_{p^{\varphi(n)}}^*|$ , so that there exists an integer $k$ such that

$p^{\varphi(n)} - 1 = k \cdot n.$

Recall that $\mathbb F_{p^{\varphi(n)}}^*$ is cyclic, so that there exists $\alpha \in \mathbb F_{p^{\varphi(n)}}^*$ such that $\mathbb F_{p^{\varphi(n)}}^* = \langle \alpha \rangle$ . In particular,

$(\alpha^k)^n = \alpha^{k \cdot n} = \alpha^{p^{\varphi(n)}-1} = 1.$

In particular, $\alpha^k$ is a primitive $n$ -th root of unity.

Problem 2. Let $\zeta_n$ be a primitive $n$ -th root of unity. Show that $\mathbb F(\zeta_n) \supseteq \mathbb F$ is Galois, and $\mathrm{Gal}(\mathbb F(\zeta_n)/\mathbb F)$ is isomorphic to some subgroup of $\mathbb Z_n^*$ .

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Solution. Let $\zeta_n$ be a primitive $n$ -th root of unity and write $\mathcal Z(f_n) = \langle \zeta_n \rangle$ . Then $\mathbb F(\zeta_n)$ is the splitting field of the separable polynomial $f_n$ , and thus $\mathbb F(\zeta_n) \supseteq \mathbb F$ is Galois. For any $\sigma \in \mathrm{Gal}(\mathbb F(\zeta_n)/\mathbb F)$ , $\sigma$ is uniquely determined by $\sigma(\zeta_n) = \zeta_n^{i_\sigma}$ for $i_\sigma \in \mathbb Z_n^*$ . Define the desired monomorphism by $\sigma \mapsto i_\sigma$ .

Definition 2. Let $\{\alpha_1,\dots,\alpha_{\varphi(n)}\}$ denote the primitive n-th roots of unity. Define the n-th cyclotomic polynomial $\Phi_n$ by

$\displaystyle \Phi_n(x) = \prod_{i=1}^{\varphi(n)} (x-\alpha_i).$

We remark that $x^n - 1 = \prod_{d \mid n} \Phi_d(x)$ .

Problem 3. Show that $\Phi_n \in \mathbb Z[x]$ is irreducible in $\mathbb Q[x]$ . Furthermore, $\mathrm{Gal}(\mathbb Q(\zeta_n)/\mathbb Q) \cong \mathbb Q$ .

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Solution. Firstly, $\Phi_n \in \mathbb Q[x]$ . Fix a root $u$ of $\phi_n$ . Then for any prime $p \nmid n$ , $f_{\mathbb Q}^u = f_{\mathbb Q}^{u^p}$ . For any $1 \leq k\leq n-1$ , write

$\displaystyle k = \prod_{i=1}^r p_i^{\alpha_i},$

In particular,

$\begin{aligned} f_{\mathbb Q}^u(u) = 0 \quad &\Rightarrow \quad f_{\mathbb Q}^u(u^{p_i}) = f_{\mathbb Q}^{u^p}(u^{p_i}) = 0 \\ &\Rightarrow \quad f_{\mathbb Q}^{u}(u^k) = f_{\mathbb Q}^{u}(u^{p_i}) = 0.\end{aligned}$

Therefore, there are at least $\varphi(n)$ roots of $f$ , so that $\deg(f) = \varphi(n)$ . Therefore, $\Phi_n = f$ is irreducible, and $\mathbb Q(\zeta_n) = \mathcal Z(\Phi_n)$ . Using Problem 2, the monomorphism $\eta : \mathrm{Gal}(\mathbb Q(\zeta_n)/\mathbb Q) \to \mathbb Z_n^*$ defined by $\eta(\sigma) = i_\sigma$ has kernel $\ker(\eta) = \{\mathrm{id}_{\mathbb Q}\}$ , so that $\mathrm{Gal}(\mathbb Q(\zeta_n)/\mathbb Q) \cong \mathbb Q$ .

Problem 4. Let $\Omega$ be an algebraically closed field with $\mathrm{char}(\Omega) = p \neq 0$ . Suppose $n$ is a positive integer with $p \nmid n$ . Show that $\mathcal Z(\Phi_n)$ contains exactly the primitive $n$ -th roots of unity in $\Omega$ .

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Solution. Let $\mathcal Z_n \subseteq \mathcal Z(f_n)$ denote the subset of the primitive $n$ -th roots of unity in $\Omega$ . We aim to show that $\mathcal Z(\Phi_n) = \mathcal Z_n$ . Fix $u \in \mathcal Z(\Phi_d)$ for any $d \mid n$ . Then $\Phi_d(x) \mid (x^d - 1)$ implies that $u^d - 1 = 0$ , so that $u \in \mathcal Z_d$ . Setting $d = n$ , $\mathcal Z(\Phi_n) \subseteq \mathcal Z_n$ . Furthermore,

$\varphi(n) \leq |\mathcal Z(\Phi_n)| \leq \deg(\Phi_n) = \varphi(n) = |\mathcal Z_n|.$

Therefore, $\mathcal Z(\Phi_n) = \mathcal Z_n$ .

Problem 5. Prove Dirichlet’s theorem: for every positive integer $n$ , there exists infinitely many prime numbers $p$ such that $p \equiv_n 1$ .

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Solution. Fix $k \in \mathbb N$ . Since $\Phi_n(x) \to \infty$ as $x \to \infty$ , there exists $m_k \in \mathbb N$ such that the integer $\Phi_n(m_k k n) > 1$ . Hence, there exists a prime

$p_k \mid \Phi_n(m_k k m) \mid (m_k k n)^n - 1,$

so that $[m_k k n]_{p_k}^n = [1]_{p_k}$ . In particular, $\Phi_n(m_k k n) = 0$ so that $\mathbb Z_{p_k}^*$ is cyclic with order $p_k - 1$ . Therefore, $n \mid p_k - 1$ , so that $[p_k]_n = [1]_n$ . Define the infinite set $\{p_k : k \in \mathbb N \}$ as desired.