[proofplan] We use the identity $\zeta_L(s) = \zeta_\mathbb{Q}(s) \cdot L(\chi_D, s)$, which factors the Dedekind zeta function of a quadratic field $L = \mathbb{Q}(\sqrt{d})$ into the Riemann zeta function times the quadratic $L$-function. Both $\zeta_L$ and $\zeta_\mathbb{Q}$ have simple poles at $s = 1$ with non-zero residues (the analytic class number formula for $\zeta_L$, and the classical result $\operatorname{Res}_{s=1} \zeta(s) = 1$ for $\zeta_\mathbb{Q}$). The factor $L(\chi_D, s)$ is holomorphic at $s = 1$ (from [Theorem 1620](/theorems/1620), since $\chi_D$ is non-trivial). Computing the limit $L(\chi_D, 1) = \lim_{s \to 1} \zeta_L(s)/\zeta_\mathbb{Q}(s)$ as the ratio of two non-zero residues gives a non-zero value. [/proofplan]

[step:Write the factorisation $\zeta_L(s) = \zeta_\mathbb{Q}(s) L(\chi_D, s)$]Let $L = \mathbb{Q}(\sqrt{d})$ with $d$ square-free, $d \neq 0, 1$. The Dedekind zeta function of $L$ factors as \begin{align*} \zeta_L(s) &= \zeta_\mathbb{Q}(s) \cdot L(\chi_D, s), \end{align*} where $\chi_D$ is the [Kronecker-symbol character](/theorems/1619) of modulus $D = \operatorname{disc}(L)$. This identity is a standard consequence of comparing Euler factors: for each rational prime $p$ (with $p \nmid D$ for simplicity), $p$ either splits, is inert, or ramifies in $\mathcal{O}_L$. In each case, the Euler factor at $p$ of $\zeta_L$ decomposes multiplicatively: \begin{align*} \text{Euler factor of } \zeta_L \text{ at } p &= \bigl(1 - p^{-s}\bigr)^{-1} \cdot \bigl(1 - \chi_D(p) p^{-s}\bigr)^{-1}, \end{align*} where the first factor matches $\zeta_\mathbb{Q}$ and the second matches $L(\chi_D, s)$. The ramified primes $p \mid D$ contribute compatibly by the extended Kronecker symbol setting $\chi_D(p) = 0$.[/step]

[guided]**Why this factorisation?** The Dedekind zeta function of a number field $L$ is the Euler product \begin{align*} \zeta_L(s) = \prod_{\mathfrak{p} \leq \mathcal{O}_L} \bigl(1 - N(\mathfrak{p})^{-s}\bigr)^{-1}, \end{align*} ranging over non-zero prime ideals $\mathfrak{p}$. For a quadratic field $L = \mathbb{Q}(\sqrt{d})$, each rational prime $p$ contributes either one or two prime ideals above it: - If $p$ **splits** ($p\mathcal{O}_L = \mathfrak{p}\mathfrak{q}$), then two prime ideals of norm $p$ contribute, giving Euler factor $(1 - p^{-s})^{-2}$. - If $p$ is **inert** ($p\mathcal{O}_L = \mathfrak{p}$ prime), one prime ideal of norm $p^2$ contributes, giving $(1 - p^{-2s})^{-1} = (1 - p^{-s})^{-1}(1 + p^{-s})^{-1}$. - If $p$ **ramifies** ($p\mathcal{O}_L = \mathfrak{p}^2$), one prime ideal of norm $p$ contributes, giving $(1 - p^{-s})^{-1}$. **Relating to $\chi_D$.** The Kronecker-symbol character $\chi_D$ is defined so that $\chi_D(p) = 1$ if $p$ splits, $\chi_D(p) = -1$ if $p$ is inert, and $\chi_D(p) = 0$ if $p$ ramifies. Then for $p \nmid D$: \begin{align*} \bigl(1 - p^{-s}\bigr)^{-1} \bigl(1 - \chi_D(p) p^{-s}\bigr)^{-1} &= \begin{cases} (1 - p^{-s})^{-2} & \text{if split } (\chi_D(p) = 1), \\ (1 - p^{-s})^{-1}(1 + p^{-s})^{-1} & \text{if inert } (\chi_D(p) = -1). \end{cases} \end{align*} The ramified case $\chi_D(p) = 0$ gives $(1 - p^{-s})^{-1} \cdot 1$, matching the single-prime contribution. **Euler product comparison.** \begin{align*} \zeta_L(s) = \prod_p \bigl(1 - p^{-s}\bigr)^{-1} \bigl(1 - \chi_D(p) p^{-s}\bigr)^{-1} = \zeta_\mathbb{Q}(s) \cdot L(\chi_D, s). \end{align*} Both products converge for $\operatorname{Re}(s) > 1$ and the identity continues by analytic continuation to the half-plane where both sides are holomorphic. **On the convention.** The character $\chi_D$ is defined in [Theorem 1619](/theorems/1619). Its values at ramified primes are set to zero and its Euler factor at those primes is $1$, which matches the actual Euler factor of $\zeta_L$ at ramified primes because the ramified prime contribution is $(1 - p^{-s})^{-1}$ on both sides.[/guided]

[step:Note the pole structure of $\zeta_\mathbb{Q}$ and $\zeta_L$ at $s = 1$]The Riemann zeta function $\zeta_\mathbb{Q}(s)$ has a simple pole at $s = 1$ with residue $1$: \begin{align*} \operatorname{Res}_{s = 1} \zeta_\mathbb{Q}(s) = \lim_{s \to 1} (s - 1) \zeta_\mathbb{Q}(s) = 1. \end{align*} This is the classical analytic continuation of $\zeta_\mathbb{Q}$ (and can be verified from the basic estimate $\zeta_\mathbb{Q}(s) = 1/(s - 1) + \gamma + O(s - 1)$ near $s = 1$, where $\gamma$ is the Euler-Mascheroni constant). The Dedekind zeta function $\zeta_L(s)$ also has a simple pole at $s = 1$ with non-zero residue by the **analytic class number formula**: the residue is \begin{align*} \operatorname{Res}_{s = 1} \zeta_L(s) &= \frac{2^{r}(2\pi)^s h_L R_L}{w_L |D_L|^{1/2}}, \end{align*} where $(r, s)$ is the signature of $L$, $h_L$ is the class number, $R_L$ is the regulator, $w_L$ is the number of roots of unity in $L$, and $D_L$ is the discriminant. For our purposes the important fact is that this residue is strictly positive (all quantities in the formula are positive real numbers).[/step]

[guided]**Pole of $\zeta_\mathbb{Q}$.** The Riemann zeta function has the Laurent expansion near $s = 1$: \begin{align*} \zeta_\mathbb{Q}(s) = \frac{1}{s - 1} + \gamma + O(s - 1), \end{align*} where $\gamma \approx 0.5772$ is the Euler-Mascheroni constant. The residue is $\lim_{s \to 1}(s - 1)\zeta_\mathbb{Q}(s) = 1$. **Pole of $\zeta_L$.** For any number field $L$ of degree $n$, the Dedekind zeta function $\zeta_L(s)$ has a simple pole at $s = 1$, and the residue is given by the analytic class number formula: \begin{align*} \operatorname{Res}_{s=1} \zeta_L(s) = \frac{2^{r}(2\pi)^s h_L R_L}{w_L \sqrt{|D_L|}}, \end{align*} where $(r, s)$ is the signature of $L$ (number of real embeddings and pairs of complex embeddings), $h_L = |\mathrm{Cl}_L|$ is the class number (positive integer), $R_L > 0$ is the regulator, $w_L$ is the number of roots of unity in $L$ (a positive integer, at least $2$ since $\pm 1 \in \mathcal{O}_L^\times$), and $D_L$ is the discriminant. For $L = \mathbb{Q}(\sqrt{d})$ (quadratic, $n = 2$), the signature is $(2, 0)$ for $d > 0$ (real quadratic) or $(0, 1)$ for $d < 0$ (imaginary quadratic). In both cases all factors in the formula are positive, so the residue is strictly positive. **Key takeaway.** Both $\zeta_\mathbb{Q}$ and $\zeta_L$ have simple poles at $s = 1$ with positive non-zero residues: \begin{align*} \operatorname{Res}_{s=1} \zeta_\mathbb{Q}(s) = 1, \qquad \operatorname{Res}_{s=1} \zeta_L(s) > 0. \end{align*}[/guided]

[step:Conclude $L(\chi_D, 1) \neq 0$ by computing a ratio of residues]By Step 1, near $s = 1$: \begin{align*} L(\chi_D, s) &= \frac{\zeta_L(s)}{\zeta_\mathbb{Q}(s)}. \end{align*} By [Holomorphicity of Non-Trivial $L$-Functions](/theorems/1620): the hypothesis is that $\chi_D$ is a non-trivial Dirichlet character, which holds (the character $\chi_D$ is non-trivial precisely because $\chi_D(p) = -1$ for any inert prime $p$, which exists for every proper quadratic extension by Chebotarev-type density arguments, or simply because a trivial $\chi_D$ would imply $L = \mathbb{Q}$). The conclusion gives that $L(\chi_D, s)$ is holomorphic on $\{\operatorname{Re}(s) > 0\}$, in particular at $s = 1$. Taking the limit as $s \to 1$ in the identity $\zeta_L(s) = \zeta_\mathbb{Q}(s) L(\chi_D, s)$, multiplied by $(s - 1)$: \begin{align*} \lim_{s \to 1} (s - 1) \zeta_L(s) &= \lim_{s \to 1} (s - 1) \zeta_\mathbb{Q}(s) \cdot \lim_{s \to 1} L(\chi_D, s). \end{align*} The left side is $\operatorname{Res}_{s = 1} \zeta_L(s) > 0$ by Step 2. The first factor on the right is $\operatorname{Res}_{s = 1} \zeta_\mathbb{Q}(s) = 1$ by Step 2. The second factor is $L(\chi_D, 1)$, which is a well-defined complex number since $L(\chi_D, s)$ is holomorphic at $s = 1$. Therefore \begin{align*} L(\chi_D, 1) &= \lim_{s \to 1} L(\chi_D, s) = \frac{\operatorname{Res}_{s = 1} \zeta_L(s)}{\operatorname{Res}_{s = 1} \zeta_\mathbb{Q}(s)} = \operatorname{Res}_{s = 1} \zeta_L(s) > 0. \end{align*} In particular, $L(\chi_D, 1) \neq 0$, as claimed.[/step]

[guided]**Strategy.** The factorisation $\zeta_L = \zeta_\mathbb{Q} \cdot L(\chi_D, \cdot)$ is a product of three meromorphic functions. At $s = 1$, the left side has a simple pole with positive residue, the factor $\zeta_\mathbb{Q}$ has a simple pole with residue $1$, and the factor $L(\chi_D, \cdot)$ is holomorphic by [Theorem 1620](/theorems/1620). A pole can arise from the product only if one of the factors has a pole; cancellation of a pole with a zero is possible, so to guarantee $L(\chi_D, 1) \neq 0$ we must verify that $L(\chi_D, \cdot)$ does not cancel. **Residue matching.** Multiplying $\zeta_L = \zeta_\mathbb{Q} L(\chi_D, \cdot)$ by $(s - 1)$: \begin{align*} (s - 1)\zeta_L(s) = (s - 1)\zeta_\mathbb{Q}(s) \cdot L(\chi_D, s). \end{align*} All three quantities on this line are holomorphic at $s = 1$ (the first two by removing the simple poles, the third by Theorem 1620). Taking $s \to 1$: \begin{align*} \operatorname{Res}_{s=1}\zeta_L = \operatorname{Res}_{s=1}\zeta_\mathbb{Q} \cdot L(\chi_D, 1) = 1 \cdot L(\chi_D, 1) = L(\chi_D, 1). \end{align*} **Non-vanishing.** Since $\operatorname{Res}_{s=1}\zeta_L > 0$ (Step 2, via the analytic class number formula), we conclude \begin{align*} L(\chi_D, 1) = \operatorname{Res}_{s=1}\zeta_L > 0, \end{align*} in particular non-zero. **Non-triviality of $\chi_D$.** We used that $\chi_D$ is non-trivial as a Dirichlet character. This is true for any proper quadratic extension $L = \mathbb{Q}(\sqrt{d})$ with $d \neq 1$: if $\chi_D$ were trivial, the Euler factors of $L(\chi_D, s)$ would all equal $(1 - p^{-s})^{-1}$, and $\zeta_L = \zeta_\mathbb{Q}^2$, which does not happen (the two zeta functions have different pole/zero structures, or more directly, $L \neq \mathbb{Q}$ has primes that are inert, for which $\chi_D = -1 \neq 1$). **Formula for $L(\chi_D, 1)$.** Combining the formula of Step 2 and the identity $L(\chi_D, 1) = \operatorname{Res}_{s=1}\zeta_L$: \begin{align*} L(\chi_D, 1) = \frac{2^r (2\pi)^s h_L R_L}{w_L |D_L|^{1/2}}. \end{align*} This is the **Dirichlet class number formula** for quadratic fields. Its non-vanishing is central in many arithmetic applications.[/guided]