[proofplan] The ideal class $[\mathfrak{a}] \in \mathrm{Cl}_L$ is represented by an arbitrary fractional ideal $\mathfrak{a}$. The strategy is to apply the [Minkowski Bound](/theorems/1610) not to $\mathfrak{a}$ directly, but to its inverse $\mathfrak{a}^{-1}$. This produces a non-zero element $\alpha \in \mathfrak{a}^{-1}$ with small norm; the principal ideal $\langle \alpha \rangle$ then combines with $\mathfrak{a}$ to give an integral ideal $\mathfrak{b} := \langle \alpha \rangle \mathfrak{a} \subseteq \mathcal{O}_L$ in the same class as $\mathfrak{a}$. The multiplicativity of the ideal norm, together with the identity $N(\mathfrak{a}^{-1}) N(\mathfrak{a}) = 1$, cancels the $N(\mathfrak{a}^{-1})$ factor and leaves $N(\mathfrak{b}) \leq c_L$. [/proofplan] [step:Reduce to working with the inverse ideal $\mathfrak{a}^{-1}$] By definition of the ideal class group $\mathrm{Cl}_L$, an ideal class $[\mathfrak{a}]$ is represented by a non-zero fractional ideal $\mathfrak{a}$ of $\mathcal{O}_L$ (integral or not). Without loss of generality, we may assume $\mathfrak{a}$ itself is an integral ideal $\mathfrak{a} \leq \mathcal{O}_L$: indeed, if $\mathfrak{a}$ is fractional, we can multiply by a non-zero integer $d \in \mathbb{Z}$ with $d \mathfrak{a} \subseteq \mathcal{O}_L$, and $[d \mathfrak{a}] = [\mathfrak{a}]$ because $\langle d \rangle$ is a principal ideal. Since $\mathfrak{a}$ is a non-zero fractional ideal of the Dedekind domain $\mathcal{O}_L$, its inverse $\mathfrak{a}^{-1}$ exists as a fractional ideal, with \begin{align*} \mathfrak{a}^{-1} = \{ \beta \in L : \beta \mathfrak{a} \subseteq \mathcal{O}_L \}, \qquad \mathfrak{a} \cdot \mathfrak{a}^{-1} = \mathcal{O}_L. \end{align*} We will produce $\mathfrak{b}$ by applying the Minkowski bound to $\mathfrak{a}^{-1}$. [guided] **Why invert?** The Minkowski bound [Theorem 1610](/theorems/1610) produces a non-zero $\alpha \in \mathfrak{c}$ inside a given ideal $\mathfrak{c}$, with $|N(\alpha)| \leq c_L N(\mathfrak{c})$. If we applied it to $\mathfrak{a}$ directly, we would get $\alpha \in \mathfrak{a}$ with $|N(\alpha)| \leq c_L N(\mathfrak{a})$; the principal ideal $\langle \alpha \rangle$ is contained in $\mathfrak{a}$, so $\mathfrak{a}^{-1} \langle \alpha \rangle \subseteq \mathcal{O}_L$ has norm $N(\alpha)/N(\mathfrak{a}) \leq c_L$ — in a class $[\mathfrak{a}^{-1}]$ different from $[\mathfrak{a}]$. That is not what we want. Instead we apply the bound to $\mathfrak{a}^{-1}$: the $\alpha \in \mathfrak{a}^{-1}$ we extract satisfies $\alpha \mathfrak{a} \subseteq \mathcal{O}_L$ (by definition of $\mathfrak{a}^{-1}$), and the integral ideal $\mathfrak{b} = \langle \alpha \rangle \mathfrak{a}$ lies in the same class as $\mathfrak{a}$ (it equals $\mathfrak{a}$ multiplied by a principal ideal). **Reduction to integral $\mathfrak{a}$.** The definition of $[\mathfrak{a}]$ allows $\mathfrak{a}$ to be fractional. But multiplying by $d \in \mathbb{Z}$ with $d \mathfrak{a} \subseteq \mathcal{O}_L$ clears denominators without changing the class. We may freely assume $\mathfrak{a} \leq \mathcal{O}_L$ is integral. **Existence of $\mathfrak{a}^{-1}$.** The ring $\mathcal{O}_L$ is a Dedekind domain, so every non-zero fractional ideal has an inverse. For integral $\mathfrak{a} \neq 0$, \begin{align*} \mathfrak{a}^{-1} = \{\beta \in L : \beta \mathfrak{a} \subseteq \mathcal{O}_L\} \end{align*} is the largest fractional ideal with $\mathfrak{a}^{-1} \mathfrak{a} \subseteq \mathcal{O}_L$; in a Dedekind domain this inclusion is an equality: $\mathfrak{a}^{-1} \mathfrak{a} = \mathcal{O}_L$. [/guided] [/step] [step:Apply the Minkowski bound to $\mathfrak{a}^{-1}$ to extract a non-zero $\alpha$ of small norm] By [Minkowski Bound](/theorems/1610) applied to the non-zero ideal $\mathfrak{a}^{-1}$: the hypotheses are that $\mathfrak{a}^{-1} \leq \mathcal{O}_L$ is a non-zero ideal, but $\mathfrak{a}^{-1}$ may be fractional rather than integral, so we must be careful. The [Minkowski Bound](/theorems/1610) as proved uses only that $\sigma(\mathfrak{a}^{-1})$ is a full-rank lattice in $\mathbb{R}^n$ with covolume $2^{-s}|D_L|^{1/2} N(\mathfrak{a}^{-1})$, a statement that holds for every non-zero fractional ideal by [Covolume of Ideal Lattices](/theorems/1609). We apply the Minkowski bound to $\mathfrak{a}^{-1}$: there exists a non-zero $\alpha \in \mathfrak{a}^{-1}$ with \begin{align*} |N(\alpha)| \leq c_L \cdot N(\mathfrak{a}^{-1}). \end{align*} [guided] **Hypothesis check for Theorem 1610 applied to $\mathfrak{a}^{-1}$.** Theorem 1610 is stated for a non-zero ideal $\mathfrak{a} \leq \mathcal{O}_L$, i.e., an integral ideal. Strictly interpreted, $\mathfrak{a}^{-1}$ may fail this hypothesis when $\mathfrak{a}$ is a proper ideal (in which case $\mathfrak{a}^{-1} \supseteq \mathcal{O}_L$ strictly, so it is fractional but not integral). However, the *proof* of Theorem 1610 uses only that $\sigma(\mathfrak{a}^{-1}) \subset \mathbb{R}^n$ is a full-rank lattice with known covolume, namely $\operatorname{covol}(\sigma(\mathfrak{a}^{-1})) = 2^{-s}|D_L|^{1/2} N(\mathfrak{a}^{-1})$. By [Covolume of Ideal Lattices](/theorems/1609), this formula holds for every non-zero fractional ideal (the proof goes via $\mathbb{Z}$-bases of fractional ideals, which exist equally well). The same closed-convex-symmetric body $B_{r,s}(t)$ argument applies verbatim: we pick $t$ so that $\operatorname{vol}(B_{r,s}(t)) = 2^n \operatorname{covol}(\sigma(\mathfrak{a}^{-1}))$, invoke [Minkowski's Theorem](/theorems/1608) part (2b) to extract $\alpha \in \mathfrak{a}^{-1}$, and apply AM-GM. **Conclusion of Theorem 1610 applied to $\mathfrak{a}^{-1}$.** There exists $\alpha \in \mathfrak{a}^{-1} \setminus \{0\}$ with \begin{align*} |N(\alpha)| \leq c_L N(\mathfrak{a}^{-1}), \end{align*} where \begin{align*} c_L = \left(\frac{4}{\pi}\right)^s \frac{n!}{n^n} |D_L|^{1/2}. \end{align*} This constant $c_L$ depends only on the number field $L$ (its degree, its signature, and its discriminant), not on the ideal. [/guided] [/step] [step:Define $\mathfrak{b} := \langle \alpha \rangle \mathfrak{a}$ and verify $\mathfrak{b}$ is integral] Let $\mathfrak{b} = \langle \alpha \rangle \mathfrak{a}$, the product of the principal fractional ideal $\langle \alpha \rangle$ with $\mathfrak{a}$. We verify $\mathfrak{b} \subseteq \mathcal{O}_L$. Since $\alpha \in \mathfrak{a}^{-1}$ and $\mathfrak{a}^{-1} = \{\beta \in L : \beta \mathfrak{a} \subseteq \mathcal{O}_L\}$, by definition $\alpha \mathfrak{a} \subseteq \mathcal{O}_L$. Equivalently, the fractional ideal $\langle \alpha \rangle \mathfrak{a}$, which coincides with $\alpha \mathfrak{a}$ as a set (since $\mathfrak{a}$ is an $\mathcal{O}_L$-module and $\langle \alpha \rangle \mathfrak{a} = \{\alpha \beta : \beta \in \mathfrak{a}\}$ as a set), is contained in $\mathcal{O}_L$. Hence $\mathfrak{b}$ is an integral ideal of $\mathcal{O}_L$. Furthermore, $\mathfrak{b} \neq 0$ because $\alpha \neq 0$ and $\mathfrak{a} \neq 0$. [guided] **Constructing $\mathfrak{b}$.** We multiply the extracted element $\alpha \in \mathfrak{a}^{-1}$ against $\mathfrak{a}$ to form $\mathfrak{b} = \langle \alpha \rangle \mathfrak{a}$. This is an operation in the ideal monoid: principal $\langle \alpha \rangle$ times $\mathfrak{a}$ gives a fractional ideal. We claim it is integral. **Proof that $\mathfrak{b} \subseteq \mathcal{O}_L$.** By definition of the inverse ideal, \begin{align*} \mathfrak{a}^{-1} = \{\beta \in L : \beta \mathfrak{a} \subseteq \mathcal{O}_L\}. \end{align*} Since $\alpha \in \mathfrak{a}^{-1}$, we have $\alpha \mathfrak{a} \subseteq \mathcal{O}_L$. But as a set, $\langle \alpha \rangle \mathfrak{a} = \alpha \mathcal{O}_L \cdot \mathfrak{a} = \alpha \mathfrak{a}$ (since $\mathcal{O}_L \cdot \mathfrak{a} = \mathfrak{a}$), so $\mathfrak{b} = \alpha \mathfrak{a} \subseteq \mathcal{O}_L$. This makes $\mathfrak{b}$ an integral ideal. **Non-zero.** Both $\alpha$ and $\mathfrak{a}$ are non-zero, and $\mathcal{O}_L$ is an integral domain, so $\mathfrak{b} = \alpha \mathfrak{a}$ is non-zero. **Key insight.** The integrality of $\mathfrak{b}$ is the entire reason we applied [Minkowski Bound](/theorems/1610) to $\mathfrak{a}^{-1}$ rather than $\mathfrak{a}$: the defining property $\alpha \mathfrak{a} \subseteq \mathcal{O}_L$ is exactly membership in $\mathfrak{a}^{-1}$, so we needed $\alpha \in \mathfrak{a}^{-1}$ (not $\alpha \in \mathfrak{a}$) for this product to be integral. [/guided] [/step] [step:Verify $[\mathfrak{b}] = [\mathfrak{a}]$ in the class group] In the class group $\mathrm{Cl}_L = I_L / P_L$, the principal fractional ideals form the identity class $[P_L] = [\mathcal{O}_L]$. Since $\mathfrak{b} = \langle \alpha \rangle \mathfrak{a}$ with $\langle \alpha \rangle$ principal: \begin{align*} [\mathfrak{b}] = [\langle \alpha \rangle \mathfrak{a}] = [\langle \alpha \rangle] \cdot [\mathfrak{a}] = [\mathcal{O}_L] \cdot [\mathfrak{a}] = [\mathfrak{a}]. \end{align*} [/step] [step:Bound $N(\mathfrak{b})$ using multiplicativity of the norm] By multiplicativity of the ideal norm (a standard property of the ideal norm on a Dedekind domain): for fractional ideals $\mathfrak{c}, \mathfrak{d}$ of $\mathcal{O}_L$, $N(\mathfrak{c} \mathfrak{d}) = N(\mathfrak{c}) N(\mathfrak{d})$. For a principal ideal, $N(\langle \alpha \rangle) = |N(\alpha)|$ (absolute value of the field norm of the generator). Applying these to $\mathfrak{b} = \langle \alpha \rangle \mathfrak{a}$: \begin{align*} N(\mathfrak{b}) = N(\langle \alpha \rangle) \cdot N(\mathfrak{a}) = |N(\alpha)| \cdot N(\mathfrak{a}). \end{align*} From Step 2, $|N(\alpha)| \leq c_L N(\mathfrak{a}^{-1})$, so \begin{align*} N(\mathfrak{b}) \leq c_L N(\mathfrak{a}^{-1}) N(\mathfrak{a}). \end{align*} Using multiplicativity once more, $N(\mathfrak{a}^{-1}) N(\mathfrak{a}) = N(\mathfrak{a}^{-1} \mathfrak{a}) = N(\mathcal{O}_L) = 1$. Therefore \begin{align*} N(\mathfrak{b}) \leq c_L \cdot 1 = c_L. \end{align*} Combining the outputs of Steps 3–5: $\mathfrak{b} \leq \mathcal{O}_L$ is an integral ideal in the same class as $\mathfrak{a}$ with $N(\mathfrak{b}) \leq c_L$. This proves the claim. [guided] **Multiplicativity of the ideal norm.** For any non-zero fractional ideals $\mathfrak{c}, \mathfrak{d}$ of $\mathcal{O}_L$, \begin{align*} N(\mathfrak{c} \mathfrak{d}) = N(\mathfrak{c}) N(\mathfrak{d}). \end{align*} This is a standard property — on a Dedekind domain, the ideal norm is a completely multiplicative map from the group of fractional ideals to $\mathbb{Q}^\times_{>0}$. For integral ideals it gives the number of cosets $|\mathcal{O}_L / \mathfrak{c}\mathfrak{d}| = |\mathcal{O}_L/\mathfrak{c}| \cdot |\mathfrak{c}/\mathfrak{c}\mathfrak{d}|$ via the Chinese Remainder factorization into primes. For fractional ideals it extends multiplicatively. **Norm of a principal ideal.** For $\alpha \in L^\times$, $N(\langle \alpha \rangle) = |N_{L/\mathbb{Q}}(\alpha)|$; this is the standard identification of the ideal norm of a principal ideal with the absolute value of the field norm of its generator. **Computing $N(\mathfrak{b})$.** \begin{align*} N(\mathfrak{b}) &= N(\langle \alpha \rangle \mathfrak{a}) = N(\langle \alpha \rangle) N(\mathfrak{a}) = |N(\alpha)| N(\mathfrak{a}). \end{align*} Substituting the Minkowski bound $|N(\alpha)| \leq c_L N(\mathfrak{a}^{-1})$: \begin{align*} N(\mathfrak{b}) \leq c_L N(\mathfrak{a}^{-1}) N(\mathfrak{a}). \end{align*} **Cancellation.** The product $N(\mathfrak{a}^{-1}) N(\mathfrak{a})$ equals $N(\mathfrak{a}^{-1} \mathfrak{a}) = N(\mathcal{O}_L)$ by multiplicativity. But $\mathcal{O}_L / \mathcal{O}_L = 0$ is the zero module, so by the norm-as-index convention we take $N(\mathcal{O}_L) = 1$ (the norm of the trivial ideal, which has index $1$ inside itself). Hence \begin{align*} N(\mathfrak{a}^{-1}) N(\mathfrak{a}) = 1, \qquad N(\mathfrak{b}) \leq c_L. \end{align*} **Conclusion.** We have produced $\mathfrak{b} \leq \mathcal{O}_L$ with $[\mathfrak{b}] = [\mathfrak{a}]$ (Step 4) and $N(\mathfrak{b}) \leq c_L$ (this step), completing the proof that every class has a representative of norm at most $c_L$. [/guided] [/step]