Let $K/\mathbb Q$ be quadratic with discriminant $d_K$, and let
$$
\chi_{d_K}(n)=\left(\frac{d_K}{n}\right)
$$
be its primitive quadratic <Kronecker symbol>[Dirichlet character]. For an unramified prime $p$, the Euler factor of $\zeta_K$ is
$$
\begin{cases}
(1-p^{-s})^{-2},&p\text{ splits},\\
(1-p^{-2s})^{-1},&p\text{ is inert}.
\end{cases}
$$
Since $\chi_{d_K}(p)$ is respectively $1$ or $-1$, this equals the product of the Euler factors of $\zeta(s)$ and $L(s,\chi_{d_K})$. At a ramified prime $\chi_{d_K}(p)=0$, and the same equality holds. Thus
$$
\zeta_K(s)=\zeta(s)L(s,\chi_{d_K}).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 123 / 2 / 2 / 4 / Solution

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The Kronecker symbol extends the <Jacobi symbol> to all integer denominators. For a fundamental discriminant $D$, the function $\chi_D(n)=\left(\frac Dn\right)$ is the primitive quadratic <Dirichlet character> associated with $\mathbb Q(\sqrt D)$.

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 Incoming links: Kronecker symbol