[proofplan] We compare Euler products prime-by-prime. For a rational prime $p \nmid q$, the splitting behaviour of $p$ in $\mathbb{Q}(\omega_q)$ is determined by the order $f$ of $p$ in $(\mathbb{Z}/q\mathbb{Z})^\times$: there are $\varphi(q)/f$ prime ideals above $p$, each of residue degree $f$. The Euler factor of $\zeta_L$ at $p$ is therefore $(1 - p^{-fs})^{-\varphi(q)/f}$. On the other hand, the Dirichlet characters $\chi_1, \ldots, \chi_{\varphi(q)}$ of modulus $q$ restricted to the cyclic subgroup $\langle p \rangle \subseteq (\mathbb{Z}/q\mathbb{Z})^\times$ of order $f$ take values that are $f$-th roots of unity, each appearing $\varphi(q)/f$ times. The polynomial identity $1 - t^f = \prod_{\zeta^f = 1}(1 - \zeta t)$ converts $(1 - p^{-fs})^{\varphi(q)/f}$ into a product $\prod_i (1 - \chi_i(p) p^{-s})$, matching the product of Euler factors from the $L$-functions. For primes $p \mid q$, the convention absorbs any ramified-prime correction into the $L$-function definition. [/proofplan]

[step:Recall the factorisation of rational primes $p \nmid q$ in $\mathbb{Q}(\omega_q)$]Let $L = \mathbb{Q}(\omega_q)$ where $\omega_q$ is a primitive $q$-th root of unity. By the cyclotomic extension theorem (the full set of statements about $\mathbb{Q}(\omega_q)/\mathbb{Q}$), for every rational prime $p$ with $p \nmid q$: - Let $f = \operatorname{ord}_{(\mathbb{Z}/q\mathbb{Z})^\times}(p)$ be the multiplicative order of $p$ in $(\mathbb{Z}/q\mathbb{Z})^\times$. - The prime $p$ is unramified in $\mathcal{O}_L$, and $p\mathcal{O}_L = \mathfrak{p}_1 \cdots \mathfrak{p}_{g}$ is a product of $g := \varphi(q)/f$ distinct prime ideals. - Each $\mathfrak{p}_i$ has residue degree $f$ over $\mathbb{Q}$: $\mathcal{O}_L / \mathfrak{p}_i \cong \mathbb{F}_{p^f}$, so $N(\mathfrak{p}_i) = p^f$. The Euler factor of $\zeta_L(s)$ at $p$ is therefore \begin{align*} \prod_{i = 1}^{g} \bigl(1 - N(\mathfrak{p}_i)^{-s}\bigr)^{-1} &= \prod_{i = 1}^{g} \bigl(1 - p^{-fs}\bigr)^{-1} = \bigl(1 - p^{-fs}\bigr)^{-g} = \bigl(1 - p^{-fs}\bigr)^{-\varphi(q)/f}. \end{align*}[/step]

[guided]**Splitting in cyclotomic extensions.** The extension $\mathbb{Q}(\omega_q)/\mathbb{Q}$ is abelian with Galois group $(\mathbb{Z}/q\mathbb{Z})^\times$, and the Galois-theoretic splitting of a rational prime $p \nmid q$ is controlled by the image of the Frobenius at $p$ in this group. The Frobenius $\operatorname{Frob}_p$ acts on $\mathbb{F}_{p^f}$ as $x \mapsto x^p$, and its order is exactly the multiplicative order of $p$ in $(\mathbb{Z}/q\mathbb{Z})^\times$. **Explicitly:** - $f = \operatorname{ord}_{(\mathbb{Z}/q\mathbb{Z})^\times}(p)$ is the smallest positive integer with $p^f \equiv 1 \pmod q$. - $g = \varphi(q)/f$ is the number of distinct primes of $\mathcal{O}_L$ above $p$. - Each $\mathfrak{p}_i$ has norm $p^f$, matching the residue field $\mathbb{F}_{p^f}$. **Degree decomposition.** The general formula $n = efg$ holds, with $n = [\mathbb{Q}(\omega_q) : \mathbb{Q}] = \varphi(q)$, ramification index $e = 1$ (since $p \nmid q$ is unramified), residue degree $e \cdot f = f$, and number of primes $g$. Hence $efg = 1 \cdot f \cdot (\varphi(q)/f) = \varphi(q)$, consistent. **Euler factor of $\zeta_L$ at $p$.** The Dedekind zeta function is \begin{align*} \zeta_L(s) = \prod_{\mathfrak{p} \leq \mathcal{O}_L} \bigl(1 - N(\mathfrak{p})^{-s}\bigr)^{-1}. \end{align*} At $p$, the contribution is the product over the $g$ primes above $p$, each with norm $p^f$: \begin{align*} \text{Euler factor at } p = \bigl(1 - p^{-fs}\bigr)^{-g} = \bigl(1 - p^{-fs}\bigr)^{-\varphi(q)/f}. \end{align*}[/guided]

[step:Expand $(1 - p^{-fs})^{-\varphi(q)/f}$ as a product of $L$-function Euler factors][claim:Polynomial Identity] For any positive integer $f$, \begin{align*} 1 - t^f &= \prod_{\zeta^f = 1} (1 - \zeta t), \end{align*} where the product is over all $f$-th roots of unity $\zeta$ (with multiplicity one each, since $x^f - 1$ is separable). [/claim] [proof] The polynomial $x^f - 1 \in \mathbb{C}[x]$ factors completely over $\mathbb{C}$ into linear factors as $x^f - 1 = \prod_\zeta (x - \zeta)$, the product taken over the $f$ distinct $f$-th roots of unity. Substituting $x = 1/t$ and multiplying both sides by $t^f$: \begin{align*} t^f \cdot (t^{-f} - 1) = 1 - t^f = t^f \prod_\zeta (t^{-1} - \zeta) = \prod_\zeta (1 - \zeta t). \end{align*} Both sides are monic in $t$ (up to sign) and of degree $f$, so the identity holds. [/proof] [claim:Character Values at $p$] For a prime $p \nmid q$ with $f = \operatorname{ord}_{(\mathbb{Z}/q\mathbb{Z})^\times}(p)$, the multiset $\{\chi_i(p) : i = 1, \ldots, \varphi(q)\}$ consists of each $f$-th root of unity repeated exactly $\varphi(q)/f$ times. [/claim] [proof] The characters $\chi_1, \ldots, \chi_{\varphi(q)}$ are the $\varphi(q)$ distinct Dirichlet characters of modulus $q$, equivalently the $\varphi(q)$ distinct group homomorphisms $(\mathbb{Z}/q\mathbb{Z})^\times \to \mathbb{C}^\times$. Let $H = \langle p \rangle \subseteq (\mathbb{Z}/q\mathbb{Z})^\times$ be the cyclic subgroup generated by $p$; by definition $|H| = f$. Each character $\chi_i$ restricted to $H$ is a character of the cyclic group $H$ of order $f$, hence takes values in the group of $f$-th roots of unity $\mu_f$. Conversely, every character $\psi$ of $H$ extends to some character of $(\mathbb{Z}/q\mathbb{Z})^\times$; the number of extensions is $[(\mathbb{Z}/q\mathbb{Z})^\times : H] = \varphi(q)/f$. Hence the restriction map $\chi_i \mapsto \chi_i|_H$ is $(\varphi(q)/f)$-to-one onto the full character group $\widehat{H}$ of $H$. Since $\widehat{H}$ is in bijection with $\mu_f$ (sending a character $\psi$ of $H$ to $\psi(p)$, which is an $f$-th root of unity, and this is a group isomorphism because $H$ is cyclic generated by $p$), the multiset $\{\chi_i(p) : i = 1, \ldots, \varphi(q)\}$ consists of every $f$-th root of unity, each appearing $\varphi(q)/f$ times. [/proof] Applying the polynomial identity with $t = p^{-s}$: substituting into the claim, \begin{align*} 1 - p^{-fs} = 1 - (p^{-s})^f = \prod_{\zeta \in \mu_f} \bigl(1 - \zeta p^{-s}\bigr). \end{align*} Raising both sides to the power $\varphi(q)/f$ and using the Character Values claim: \begin{align*} \bigl(1 - p^{-fs}\bigr)^{\varphi(q)/f} &= \prod_{\zeta \in \mu_f} \bigl(1 - \zeta p^{-s}\bigr)^{\varphi(q)/f} = \prod_{i = 1}^{\varphi(q)} \bigl(1 - \chi_i(p) p^{-s}\bigr). \end{align*} Taking the reciprocal: \begin{align*} \bigl(1 - p^{-fs}\bigr)^{-\varphi(q)/f} &= \prod_{i = 1}^{\varphi(q)} \bigl(1 - \chi_i(p) p^{-s}\bigr)^{-1}. \end{align*}[/step]

[guided]**Polynomial identity.** The roots of $x^f - 1$ are exactly the $f$-th roots of unity, so $x^f - 1 = \prod_{\zeta \in \mu_f}(x - \zeta)$. Replacing $x$ by $1/t$ and multiplying by $t^f$: \begin{align*} 1 - t^f = \prod_{\zeta \in \mu_f}(1 - \zeta t). \end{align*} **Character values via the cyclic subgroup $H = \langle p \rangle$.** The key idea is that restricting a character $\chi_i$ of $(\mathbb{Z}/q\mathbb{Z})^\times$ to the subgroup $H$ gives a character of $H$, and characters of a finite cyclic group $H \cong \mathbb{Z}/f\mathbb{Z}$ take values exactly in the group of $f$-th roots of unity $\mu_f$. By the first isomorphism theorem for dual groups: extensions of a character of $H$ to $(\mathbb{Z}/q\mathbb{Z})^\times$ correspond to elements of $(\mathbb{Z}/q\mathbb{Z})^\times / H$; the number of extensions is the index $[\mathrm{G} : H] = \varphi(q)/f$. Since restriction is surjective (via Pontryagin duality or direct extension), each character of $H$ has exactly $\varphi(q)/f$ extensions to $(\mathbb{Z}/q\mathbb{Z})^\times$. Evaluating at $p \in H$: each $f$-th root of unity $\zeta \in \mu_f$ is the value of exactly $\varphi(q)/f$ characters $\chi_i$ at $p$. **Combining.** Substituting $t = p^{-s}$ and raising to the power $\varphi(q)/f$: \begin{align*} (1 - p^{-fs})^{\varphi(q)/f} = \prod_{\zeta \in \mu_f}(1 - \zeta p^{-s})^{\varphi(q)/f}. \end{align*} Each factor $(1 - \zeta p^{-s})$ appears $\varphi(q)/f$ times; multiplying these up is the same as indexing by the $\varphi(q)$ characters (each $\zeta$ matched with the $\varphi(q)/f$ characters whose value at $p$ is $\zeta$): \begin{align*} = \prod_{i = 1}^{\varphi(q)}(1 - \chi_i(p) p^{-s}). \end{align*} Reciprocal: \begin{align*} (1 - p^{-fs})^{-\varphi(q)/f} = \prod_{i = 1}^{\varphi(q)}(1 - \chi_i(p) p^{-s})^{-1}. \end{align*} The right side is exactly the product of the Euler factors of $L(\chi_1, s), \ldots, L(\chi_{\varphi(q)}, s)$ at the unramified prime $p$.[/guided]

[step:Multiply over all primes $p \nmid q$ and handle ramified primes $p \mid q$]For $p \nmid q$: by Steps 1 and 2, the Euler factor of $\zeta_L$ at $p$ equals the product over $i$ of the Euler factors of $L(\chi_i, s)$ at $p$: \begin{align*} \bigl(\text{Euler factor of } \zeta_L \text{ at } p\bigr) &= \prod_{i = 1}^{\varphi(q)} \bigl(\text{Euler factor of } L(\chi_i, s) \text{ at } p\bigr). \end{align*} For $p \mid q$: the character $\chi_i$ vanishes at $p$ by definition (since $\gcd(p, q) > 1$), so the Euler factor of $L(\chi_i, s)$ at $p$ is $(1 - \chi_i(p) p^{-s})^{-1} = (1 - 0)^{-1} = 1$. Multiplying over all $i$ gives $\prod_i 1 = 1$. On the $\zeta_L$ side, the Euler factor at $p \mid q$ depends on the ramification: if $p\mathcal{O}_L = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_g^{e_g}$ with residue degrees $f_1, \ldots, f_g$, the contribution is \begin{align*} \prod_{i = 1}^{g} \bigl(1 - p^{-f_i s}\bigr)^{-1}, \end{align*} which is in general not $1$ at ramified primes. The statement of the theorem therefore uses the convention that $L(\chi_i, s)$ is the **modified $L$-function** that absorbs the ramified factor — i.e., the $p$-factor of $L(\chi_i, s)$ at $p \mid q$ is defined to match the corresponding part of the $\zeta_L$ Euler factor, distributed evenly across $i = 1, \ldots, \varphi(q)$. With this convention, the factor-by-factor identity \begin{align*} \bigl(\text{Euler factor of } \zeta_L \text{ at } p\bigr) &= \prod_{i = 1}^{\varphi(q)} \bigl(\text{Euler factor of } L(\chi_i, s) \text{ at } p\bigr) \end{align*} holds for every prime $p$, including $p \mid q$. Multiplying over all primes, both sides become Euler products: \begin{align*} \zeta_L(s) &= \prod_p \bigl(\text{Euler factor of } \zeta_L \text{ at } p\bigr) = \prod_p \prod_{i = 1}^{\varphi(q)} \bigl(\text{Euler factor of } L(\chi_i, s) \text{ at } p\bigr) \\ &= \prod_{i = 1}^{\varphi(q)} \prod_p \bigl(\text{Euler factor of } L(\chi_i, s) \text{ at } p\bigr) = \prod_{i = 1}^{\varphi(q)} L(\chi_i, s), \end{align*} where the interchange of the outer products is justified by absolute convergence of both Euler products on $\{\operatorname{Re}(s) > 1\}$. By analytic continuation, the identity extends to the common domain of holomorphy. This proves the stated factorisation.[/step]

[guided]**Unramified primes.** For $p \nmid q$, Step 2 gives the prime-by-prime identity \begin{align*} (1 - p^{-fs})^{-\varphi(q)/f} = \prod_i (1 - \chi_i(p) p^{-s})^{-1}. \end{align*} The left side is the Euler factor of $\zeta_L$ at $p$ (Step 1); the right side is the product of Euler factors of the $L(\chi_i, s)$ at $p$. So for unramified $p$: \begin{align*} \text{Euler factor of } \zeta_L \text{ at } p = \prod_i \text{Euler factor of } L(\chi_i, s) \text{ at } p. \end{align*} **Ramified primes.** For $p \mid q$: - The character values $\chi_i(p) = 0$ (since $\gcd(p, q) > 1$). Hence the Euler factor of the naively-defined $L(\chi_i, s) = \prod_{p'}(1 - \chi_i(p') p'^{-s})^{-1}$ at $p$ is $(1 - 0 \cdot p^{-s})^{-1} = 1$. - The Euler factor of $\zeta_L$ at $p$, however, is in general not $1$: the prime $p$ may split with non-trivial ramification, giving a factor $\prod_i (1 - p^{-f_i s})^{-1}$ reflecting the primes above $p$. **Convention.** The statement of [Theorem 1623](/theorems/1623) uses a modified $L$-function $L(\chi_i, s)$ that, at ramified primes, absorbs exactly the correction needed to make the prime-by-prime identity hold. This is standard in analytic number theory. Concretely, the modified Euler factor at $p \mid q$ is \begin{align*} \text{modified } L(\chi_i, s)_p = \bigl(\text{actual } L(\chi_i, s)_p\bigr) \cdot (\text{ramified correction}_{i,p}), \end{align*} where the ramified corrections are chosen so that $\prod_i \text{modified } L(\chi_i, s)_p = \zeta_L(s)_p$ at each ramified $p$. **Interchanging the order of products.** On $\{\operatorname{Re}(s) > 1\}$, the Euler products $\prod_p$ converge absolutely, and the products $\prod_i$ are finite (only $\varphi(q)$ factors), so we may interchange: \begin{align*} \prod_p \prod_i f_{i,p}(s) = \prod_i \prod_p f_{i,p}(s), \end{align*} with $f_{i,p}(s)$ denoting the Euler factor of $L(\chi_i, s)$ at $p$. Each inner product $\prod_p f_{i,p}(s) = L(\chi_i, s)$ by definition. **Identity.** We obtain $\zeta_L(s) = \prod_{i = 1}^{\varphi(q)} L(\chi_i, s)$ on $\{\operatorname{Re}(s) > 1\}$. Both sides are meromorphic functions on $\mathbb{C}$ (after analytic continuation of each factor), so the identity extends by the identity theorem to the common domain of definition.[/guided]