[proofplan] The strategy combines two ingredients: (a) the existence of a small-norm representative in each class (a consequence of the Minkowski bound applied via [Every Class Has a Small Representative](/theorems/1611)), and (b) the fact that for any fixed $M > 0$ there are only finitely many integral ideals $\mathfrak{b} \leq \mathcal{O}_L$ with $N(\mathfrak{b}) \leq M$. Part (a) says every class is hit by at least one ideal of norm $\leq c_L$, and (b) says the number of such ideals is finite. Together they force $\mathrm{Cl}_L$ to be finite. For the generating set, we observe that any representative of norm $\leq c_L$ factors uniquely into primes each of norm $\leq c_L$, and any such prime divides the rational prime it sits above, which must itself be $\leq c_L$. [/proofplan] [step:Recall that every class has a representative of norm at most $c_L$] By [Every Class Has a Small Representative](/theorems/1611), the hypothesis that $L$ is a number field and $[\mathfrak{a}] \in \mathrm{Cl}_L$ is any ideal class: for every class $[\mathfrak{a}] \in \mathrm{Cl}_L$ there exists an integral ideal $\mathfrak{b} \leq \mathcal{O}_L$ with \begin{align*} [\mathfrak{b}] &= [\mathfrak{a}] \quad \text{and} \quad N(\mathfrak{b}) \leq c_L, \end{align*} where $c_L = (4/\pi)^s (n!/n^n) |D_L|^{1/2}$ is the [Minkowski constant](/theorems/1610) of $L$. In particular, the assignment \begin{align*} \Phi: \mathrm{Cl}_L &\to \mathcal{P}(\{\mathfrak{b} \leq \mathcal{O}_L : N(\mathfrak{b}) \leq c_L\}) \\ [\mathfrak{a}] &\mapsto \{\mathfrak{b} \leq \mathcal{O}_L : [\mathfrak{b}] = [\mathfrak{a}],\ N(\mathfrak{b}) \leq c_L\} \end{align*} has non-empty image on every class. [guided] **Recap of the class group.** The ideal class group $\mathrm{Cl}_L = I_L / P_L$ is the quotient of the group of non-zero fractional ideals $I_L$ by the subgroup $P_L$ of non-zero principal fractional ideals. A class $[\mathfrak{a}]$ consists of all fractional ideals $\mathfrak{a} \cdot \langle \lambda \rangle$ for $\lambda \in L^\times$; two fractional ideals are in the same class if they differ by a principal fractional ideal. **The Minkowski-bound representative.** [Theorem 1611](/theorems/1611) states that for every class $[\mathfrak{a}] \in \mathrm{Cl}_L$, the class contains an integral ideal $\mathfrak{b} \leq \mathcal{O}_L$ with $N(\mathfrak{b}) \leq c_L$. The proof proceeds by applying the [Minkowski Bound](/theorems/1610) to $\mathfrak{a}^{-1}$, producing $\alpha \in \mathfrak{a}^{-1} \setminus \{0\}$ with $|N(\alpha)| \leq c_L N(\mathfrak{a}^{-1})$, and setting $\mathfrak{b} := \langle \alpha \rangle \mathfrak{a}$, which lies in the same class as $\mathfrak{a}$ and has $N(\mathfrak{b}) = |N(\alpha)| N(\mathfrak{a}) \leq c_L N(\mathfrak{a}^{-1}) N(\mathfrak{a}) = c_L$. **Hypothesis check for Theorem 1611.** The hypothesis is that $L$ is a number field and $[\mathfrak{a}] \in \mathrm{Cl}_L$ is any class. Both are satisfied. The conclusion produces the required representative $\mathfrak{b}$. **Rephrasing as a map from classes to bounded-norm ideals.** Every class $[\mathfrak{a}]$ has at least one representative $\mathfrak{b}$ with $N(\mathfrak{b}) \leq c_L$. Hence there is a surjection \begin{align*} \Psi: \{\mathfrak{b} \leq \mathcal{O}_L : N(\mathfrak{b}) \leq c_L\} &\twoheadrightarrow \mathrm{Cl}_L \\ \mathfrak{b} &\mapsto [\mathfrak{b}]. \end{align*} The surjectivity is precisely the content of Theorem 1611. The domain of $\Psi$ is a set of integral ideals with bounded norm; if we can show this set is finite, then so is $\mathrm{Cl}_L$ (finite surjection). [/guided] [/step] [step:Show that the set of integral ideals of bounded norm is finite] [claim:Bounded-Norm Ideals] For any real $M > 0$, the set \begin{align*} \{\mathfrak{b} \leq \mathcal{O}_L : N(\mathfrak{b}) \leq M\} \end{align*} is finite. [/claim] [proof] By [Norm Lies in the Ideal](/theorems/1592) applied to each non-zero integral ideal $\mathfrak{b}$: the hypotheses (that $\mathfrak{b}$ is a non-zero ideal of $\mathcal{O}_L$) are satisfied, and the conclusion gives $N(\mathfrak{b}) \in \mathfrak{b} \cap \mathbb{Z}$. In particular, $\mathfrak{b}$ contains the principal ideal $\langle N(\mathfrak{b}) \rangle$. By [Divisibility Equals Containment](/theorems/1587) applied with $\mathfrak{a} = \mathfrak{b}$, $\mathfrak{b}' = \langle N(\mathfrak{b}) \rangle$: since $\langle N(\mathfrak{b}) \rangle \subseteq \mathfrak{b}$ we have $\mathfrak{b} \mid \langle N(\mathfrak{b}) \rangle$ as ideals of $\mathcal{O}_L$. Now, for a fixed integer $m$ with $1 \leq m \leq M$, the number of divisors of $\langle m \rangle$ in $\mathcal{O}_L$ is finite: by [Unique Factorization of Ideals](/theorems/1589), $\langle m \rangle$ factors uniquely into a product of finitely many prime ideals, and the set of divisors of a finite product of primes (each with bounded multiplicity) is finite. Therefore \begin{align*} \{\mathfrak{b} \leq \mathcal{O}_L : N(\mathfrak{b}) \leq M\} &\subseteq \bigcup_{1 \leq m \leq M} \{\mathfrak{b} \leq \mathcal{O}_L : \mathfrak{b} \mid \langle m \rangle\}. \end{align*} The right-hand side is a finite union of finite sets, hence finite. [/proof] Applying the claim with $M = c_L$ gives that the set of integral ideals of norm $\leq c_L$ is finite. Combined with Step 1, the surjection \begin{align*} \Psi: \{\mathfrak{b} \leq \mathcal{O}_L : N(\mathfrak{b}) \leq c_L\} &\twoheadrightarrow \mathrm{Cl}_L, \qquad \mathfrak{b} \mapsto [\mathfrak{b}], \end{align*} has finite domain, hence finite image. Therefore $\mathrm{Cl}_L$ is finite. [guided] **Finiteness of ideals with bounded norm.** The key step is to reduce the number of ideals of norm $\leq M$ to the number of divisors of the principal ideals $\langle m \rangle$ for $m = 1, 2, \ldots, \lfloor M \rfloor$. **Step 1: $\mathfrak{b}$ divides $\langle N(\mathfrak{b}) \rangle$.** By [Norm Lies in the Ideal](/theorems/1592), for any non-zero integral ideal $\mathfrak{b}$, the integer $N(\mathfrak{b}) = |\mathcal{O}_L / \mathfrak{b}|$ lies in $\mathfrak{b}$: the image $1 + \mathfrak{b} \in \mathcal{O}_L/\mathfrak{b}$ has order dividing the group order $N(\mathfrak{b})$, so $N(\mathfrak{b}) \cdot 1 \equiv 0 \pmod{\mathfrak{b}}$, i.e., $N(\mathfrak{b}) \in \mathfrak{b}$. Hence $\langle N(\mathfrak{b}) \rangle \subseteq \mathfrak{b}$. By [Divisibility Equals Containment](/theorems/1587), this is equivalent to $\mathfrak{b} \mid \langle N(\mathfrak{b}) \rangle$ in the ideal monoid of $\mathcal{O}_L$. **Step 2: finitely many divisors.** For a fixed $m \in \mathbb{N}$, the principal ideal $\langle m \rangle$ factors uniquely by [Unique Factorization of Ideals](/theorems/1589) into a finite product of prime ideals: $\langle m \rangle = \mathfrak{p}_1^{a_1} \cdots \mathfrak{p}_k^{a_k}$. The divisors of $\langle m \rangle$ correspond to the $\prod_i (a_i + 1)$ sub-products $\mathfrak{p}_1^{b_1} \cdots \mathfrak{p}_k^{b_k}$ with $0 \leq b_i \leq a_i$. This is finite. **Step 3: union over $m = 1, \ldots, \lfloor M \rfloor$.** Every ideal with norm $\leq M$ has integer norm in $\{1, 2, \ldots, \lfloor M \rfloor\}$, and each such ideal divides $\langle N(\mathfrak{b}) \rangle$ with $N(\mathfrak{b}) \leq M$. Hence \begin{align*} \{\mathfrak{b} \leq \mathcal{O}_L : N(\mathfrak{b}) \leq M\} \subseteq \bigcup_{m = 1}^{\lfloor M \rfloor} \{\mathfrak{b} : \mathfrak{b} \mid \langle m \rangle\}. \end{align*} The right side is a finite union of finite sets, hence finite. **Conclusion of finiteness of $\mathrm{Cl}_L$.** Applying Theorem 1611 (every class has a representative of norm $\leq c_L$) together with the finiteness of such ideals (just proved with $M = c_L$): the quotient map from $\{N \leq c_L\}$-ideals to classes is surjective by 1611 with finite domain, so its image $\mathrm{Cl}_L$ is finite. [/guided] [/step] [step:Show $\mathrm{Cl}_L$ is generated by primes of norm at most $c_L$] Let $\mathcal{S} = \{\mathfrak{p} \text{ prime ideal of } \mathcal{O}_L : N(\mathfrak{p}) \leq c_L\}$. We show that $\mathrm{Cl}_L$ is generated by the classes $\{[\mathfrak{p}] : \mathfrak{p} \in \mathcal{S}\}$. Fix a class $[\mathfrak{a}] \in \mathrm{Cl}_L$. By Step 1, pick an integral representative $\mathfrak{b} \in [\mathfrak{a}]$ with $N(\mathfrak{b}) \leq c_L$. By [Unique Factorization of Ideals](/theorems/1589) applied to $\mathfrak{b}$ (the hypothesis being that $\mathfrak{b}$ is a non-zero ideal of $\mathcal{O}_L$, which holds), write \begin{align*} \mathfrak{b} &= \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_r^{e_r} \end{align*} with $\mathfrak{p}_1, \ldots, \mathfrak{p}_r$ distinct prime ideals and $e_i \geq 1$. By multiplicativity of the ideal norm (a standard property of the ideal norm on a Dedekind domain, used in the proof of [Every Class Has a Small Representative](/theorems/1611)): \begin{align*} N(\mathfrak{b}) &= N(\mathfrak{p}_1)^{e_1} \cdots N(\mathfrak{p}_r)^{e_r}. \end{align*} Each $N(\mathfrak{p}_i) \geq 2$ (because $\mathcal{O}_L/\mathfrak{p}_i$ is a non-trivial field, hence has at least two elements), so \begin{align*} N(\mathfrak{p}_i) \leq N(\mathfrak{p}_i)^{e_i} \leq N(\mathfrak{b}) \leq c_L, \end{align*} which means $\mathfrak{p}_i \in \mathcal{S}$ for all $i$. Taking classes: \begin{align*} [\mathfrak{a}] &= [\mathfrak{b}] = [\mathfrak{p}_1]^{e_1} \cdots [\mathfrak{p}_r]^{e_r}, \end{align*} where the exponents are in $\mathbb{Z}_{\geq 1}$ and each $[\mathfrak{p}_i]$ lies in the subgroup generated by $\{[\mathfrak{p}] : \mathfrak{p} \in \mathcal{S}\}$. Since $[\mathfrak{a}]$ was an arbitrary class, the classes of primes in $\mathcal{S}$ generate $\mathrm{Cl}_L$. [guided] **From small-norm representatives to small-norm primes.** Step 1 gives a representative $\mathfrak{b}$ with $N(\mathfrak{b}) \leq c_L$; Step 2 gives finitely many such representatives; now we must show each representative factors through primes of norm $\leq c_L$. **Unique factorization and multiplicativity.** By [Theorem 1589](/theorems/1589), $\mathfrak{b} = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_r^{e_r}$ for distinct primes $\mathfrak{p}_i$ and positive integers $e_i$. The ideal norm is multiplicative on products in a Dedekind domain: \begin{align*} N(\mathfrak{b}) &= N(\mathfrak{p}_1^{e_1}) \cdots N(\mathfrak{p}_r^{e_r}) = N(\mathfrak{p}_1)^{e_1} \cdots N(\mathfrak{p}_r)^{e_r}. \end{align*} **Bounding each $N(\mathfrak{p}_i)$.** Each factor on the right is at least $2$ (a prime ideal has a non-trivial residue field with at least $2$ elements), and their product equals $N(\mathfrak{b}) \leq c_L$. In particular, each factor individually satisfies $N(\mathfrak{p}_i) \leq N(\mathfrak{p}_i)^{e_i} \leq N(\mathfrak{b}) \leq c_L$. **Class group multiplication.** In the class group, multiplication is induced by ideal multiplication. Taking classes of the factorization: \begin{align*} [\mathfrak{a}] &= [\mathfrak{b}] = [\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_r^{e_r}] = [\mathfrak{p}_1]^{e_1} \cdots [\mathfrak{p}_r]^{e_r}. \end{align*} **Generation.** Every class is expressed as a product of powers of small-norm-prime classes. Hence $\mathrm{Cl}_L$ is generated as a group by $\{[\mathfrak{p}] : \mathfrak{p} \text{ prime},\ N(\mathfrak{p}) \leq c_L\}$. **Connection to rational primes (remark).** Any prime ideal $\mathfrak{p}$ of $\mathcal{O}_L$ lies above a unique rational prime $p \in \mathbb{Z}$, namely $p = \mathfrak{p} \cap \mathbb{Z}$, and $N(\mathfrak{p}) = p^f$ for some $f \geq 1$ (the residue field of $\mathfrak{p}$ is an extension of degree $f$ of $\mathbb{F}_p$). In particular $N(\mathfrak{p}) \geq p$, so the condition $N(\mathfrak{p}) \leq c_L$ forces $p \leq c_L$. This refines the generating set to primes lying above rational primes $p \leq c_L$. [/guided] [/step]