[proofplan] Each ideal class $[\mathfrak{a}] \in \mathrm{Cl}_L$ has the inverse class $[\mathfrak{a}^{-1}]$ containing the fractional ideal $\mathfrak{a}^{-1}$. Any nonzero element $\alpha \in \mathfrak{a}^{-1}$ generates a principal fractional ideal $\langle \alpha \rangle$ with $\langle \alpha \rangle \subseteq \mathfrak{a}^{-1}$, and multiplying by $\mathfrak{a}$ yields an *integral* ideal $\mathfrak{b} = \langle \alpha \rangle \mathfrak{a} \subseteq \mathcal{O}_L$ with $[\mathfrak{b}] = [\mathfrak{a}]$. To make $N(\mathfrak{b})$ small, we want $|N(\alpha)|$ small relative to $N(\mathfrak{a}^{-1})$ — which is exactly what [Minkowski's Lemma (Quadratic Case)](/theorems/1602) provides when applied to the ideal lattice $\mathfrak{a}^{-1}$: it furnishes a non-zero $\alpha \in \mathfrak{a}^{-1}$ with $|\alpha|^2 \leq 4 A(\mathfrak{a}^{-1})/\pi$, and part (1) of [Areas of Ideal Lattices](/theorems/1603) turns $|\alpha|^2$ into $|N(\alpha)|$. The area formulas from Theorem 1603 rearrange to $|N(\alpha)| \leq c_L N(\mathfrak{a}^{-1})$, and the final bound $N(\mathfrak{b}) \leq c_L$ follows from $N(\mathfrak{b}) = |N(\alpha)| N(\mathfrak{a}) \leq c_L N(\mathfrak{a}^{-1}) N(\mathfrak{a}) = c_L$. [/proofplan] [step:Reduce to finding a small non-zero element of the fractional ideal $\mathfrak{a}^{-1}$] Fix a class $[\mathfrak{a}] \in \mathrm{Cl}_L$, represented by an integral ideal $\mathfrak{a} \leq \mathcal{O}_L$ (any class has such a representative by the definition of $\mathrm{Cl}_L$ as the quotient of fractional ideals by principal fractional ideals). The inverse fractional ideal $\mathfrak{a}^{-1}$ exists by [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586), and the product $\mathfrak{a} \mathfrak{a}^{-1} = \mathcal{O}_L$. Given any nonzero $\alpha \in \mathfrak{a}^{-1}$, define \begin{align*} \mathfrak{b} := \langle \alpha \rangle \mathfrak{a}. \end{align*} We claim $\mathfrak{b}$ is an integral ideal in the class $[\mathfrak{a}]$: - **Integrality.** $\langle \alpha \rangle \subseteq \mathfrak{a}^{-1}$ because $\alpha \in \mathfrak{a}^{-1}$ and $\mathfrak{a}^{-1}$ is an $\mathcal{O}_L$-module. Hence $\mathfrak{b} = \langle \alpha \rangle \mathfrak{a} \subseteq \mathfrak{a}^{-1} \mathfrak{a} = \mathcal{O}_L$, so $\mathfrak{b}$ is integral. - **Class.** In the class group $\mathrm{Cl}_L$, principal fractional ideals are trivial, so $[\langle \alpha \rangle] = [\mathcal{O}_L]$ and \begin{align*} [\mathfrak{b}] = [\langle \alpha \rangle][\mathfrak{a}] = [\mathcal{O}_L][\mathfrak{a}] = [\mathfrak{a}]. \end{align*} The norm of $\mathfrak{b}$ factors by multiplicativity of the ideal norm: \begin{align*} N(\mathfrak{b}) = N(\langle \alpha \rangle) N(\mathfrak{a}) = |N_{L/\mathbb{Q}}(\alpha)| \cdot N(\mathfrak{a}), \end{align*} using [Element Norm Equals Ideal Norm](/theorems/1596) for the last equality. Therefore the bound $N(\mathfrak{b}) \leq c_L$ reduces to finding a non-zero $\alpha \in \mathfrak{a}^{-1}$ with \begin{align*} |N_{L/\mathbb{Q}}(\alpha)| \cdot N(\mathfrak{a}) \leq c_L, \qquad \text{i.e.,} \qquad |N_{L/\mathbb{Q}}(\alpha)| \leq c_L \cdot N(\mathfrak{a})^{-1} = c_L \cdot N(\mathfrak{a}^{-1}), \end{align*} where the last equality uses the multiplicativity of the norm on fractional ideals together with $N(\mathcal{O}_L) = 1$. [guided] The theorem asserts that every ideal class has a "small" representative (measured by ideal norm). The standard trick for producing such a representative is to find a small *element* of the **inverse** class — this exploits the fact that principal fractional ideals are trivial in the class group. **Setting up the inverse class.** Given a class $[\mathfrak{a}] \in \mathrm{Cl}_L$, we may pick an integral representative $\mathfrak{a} \leq \mathcal{O}_L$. The inverse fractional ideal $\mathfrak{a}^{-1}$ exists by [Theorem 1586](/theorems/1586) and satisfies $\mathfrak{a}\mathfrak{a}^{-1} = \mathcal{O}_L$. Note: $\mathfrak{a}^{-1}$ is a fractional ideal, not in general an integral ideal — it sits inside $L$ but may have denominators. **The "principal shift" construction.** If $\alpha \in \mathfrak{a}^{-1}$ is any non-zero element, consider the principal fractional ideal $\langle \alpha \rangle = \alpha \mathcal{O}_L$. Set \begin{align*} \mathfrak{b} := \langle \alpha \rangle \mathfrak{a}. \end{align*} Geometrically: we start from $\mathfrak{a}$ and twist by the principal ideal generated by $\alpha$. Because principal ideals are trivial in the class group, the twist does not change the class. **Verifying $\mathfrak{b}$ is integral.** We need $\mathfrak{b} \subseteq \mathcal{O}_L$. Since $\alpha \in \mathfrak{a}^{-1}$, the submodule $\alpha \mathcal{O}_L = \langle \alpha \rangle$ is contained in $\mathfrak{a}^{-1}$ (because $\mathfrak{a}^{-1}$ is an $\mathcal{O}_L$-module and $\alpha$ is one of its elements). Hence \begin{align*} \mathfrak{b} = \langle \alpha \rangle \mathfrak{a} \subseteq \mathfrak{a}^{-1} \mathfrak{a} = \mathcal{O}_L, \end{align*} so $\mathfrak{b}$ is integral. **Verifying $[\mathfrak{b}] = [\mathfrak{a}]$.** In $\mathrm{Cl}_L$, the principal class $[\mathcal{O}_L]$ is the identity, and every principal fractional ideal $\langle \alpha \rangle$ satisfies $[\langle \alpha \rangle] = [\mathcal{O}_L]$ (this is the definition of the class group as a quotient by the subgroup of principal fractional ideals). Hence \begin{align*} [\mathfrak{b}] = [\langle \alpha \rangle \mathfrak{a}] = [\langle \alpha \rangle][\mathfrak{a}] = [\mathcal{O}_L][\mathfrak{a}] = [\mathfrak{a}]. \end{align*} So $\mathfrak{b}$ is a valid representative of the class $[\mathfrak{a}]$. **Computing $N(\mathfrak{b})$.** The ideal norm $N: \{\text{nonzero fractional ideals}\} \to \mathbb{Q}_{>0}$ is a group homomorphism (multiplicative). Hence \begin{align*} N(\mathfrak{b}) = N(\langle \alpha \rangle) \cdot N(\mathfrak{a}). \end{align*} By [Element Norm Equals Ideal Norm](/theorems/1596), $N(\langle \alpha \rangle) = |N_{L/\mathbb{Q}}(\alpha)|$. (Verifying hypotheses of Theorem 1596: we need $\alpha \in \mathcal{O}_L$, but here $\alpha \in \mathfrak{a}^{-1}$ might not lie in $\mathcal{O}_L$. The theorem nevertheless extends to fractional principal ideals via multiplicativity of $N$ across non-zero fractional ideals; more carefully, $N(\langle \alpha \rangle)$ is by definition the index $[\mathcal{O}_L : \alpha \mathcal{O}_L]$ when $\alpha \in \mathcal{O}_L$, and extended multiplicatively to fractional principal ideals, it agrees with $|N_{L/\mathbb{Q}}(\alpha)|$ for any non-zero $\alpha \in L^\times$.) So \begin{align*} N(\mathfrak{b}) = |N_{L/\mathbb{Q}}(\alpha)| \cdot N(\mathfrak{a}). \end{align*} **The target inequality.** We want $N(\mathfrak{b}) \leq c_L = 2\sqrt{|D_L|}/\pi$, equivalently \begin{align*} |N_{L/\mathbb{Q}}(\alpha)| \leq \frac{c_L}{N(\mathfrak{a})} = c_L \cdot N(\mathfrak{a})^{-1}. \end{align*} Since $\mathfrak{a}\mathfrak{a}^{-1} = \mathcal{O}_L$ and $N$ is multiplicative with $N(\mathcal{O}_L) = 1$, we have $N(\mathfrak{a}) N(\mathfrak{a}^{-1}) = 1$, so $N(\mathfrak{a})^{-1} = N(\mathfrak{a}^{-1})$. Rewriting: \begin{align*} |N_{L/\mathbb{Q}}(\alpha)| \leq c_L \cdot N(\mathfrak{a}^{-1}). \end{align*} **Goal.** Exhibit a non-zero $\alpha \in \mathfrak{a}^{-1}$ satisfying this inequality. The next two steps carry out this construction using the quadratic Minkowski lemma. [/guided] [/step] [step:Apply Minkowski's Lemma to the lattice $\iota(\mathfrak{a}^{-1}) \subseteq \mathbb{R}^2$] By Step 1 of the proof of [Areas of Ideal Lattices](/theorems/1603), we identify $L$ with a subfield of $\mathbb{C} \cong \mathbb{R}^2$ via the chosen embedding $\iota$. Under this identification, every nonzero fractional ideal — and in particular $\mathfrak{a}^{-1}$ — is a lattice of full rank $2$ in $\mathbb{R}^2$. (The rank-$2$ structure on $\mathfrak{a}^{-1}$ follows from $\mathfrak{a}^{-1}$ being a free $\mathbb{Z}$-module of rank $n = 2$ by [Norm and Discriminant](/theorems/1594) part 1 applied to a suitable integer multiple of $\mathfrak{a}^{-1}$, and $\mathbb{R}$-linearly independent because $\mathfrak{a}^{-1} \otimes_{\mathbb{Z}} \mathbb{R} = L \otimes_{\mathbb{Q}} \mathbb{R} \cong \mathbb{C}$.) Let $A(\mathfrak{a}^{-1})$ denote the area of the fundamental parallelogram of $\iota(\mathfrak{a}^{-1})$ in $\mathbb{R}^2$. Part (4) of [Areas of Ideal Lattices](/theorems/1603), applied to the fractional ideal $\mathfrak{a}^{-1}$ (the proof of Theorem 1603 extends from integral ideals to non-zero fractional ideals by scaling: pick $m \in \mathbb{Z}_{>0}$ with $m \mathfrak{a}^{-1} \leq \mathcal{O}_L$, apply the integral result to $m \mathfrak{a}^{-1}$, and descend via $A(\mathfrak{a}^{-1}) = m^{-2} A(m\mathfrak{a}^{-1})$ and $N(\mathfrak{a}^{-1}) = m^{-n} N(m\mathfrak{a}^{-1})$ with $n = 2$), gives \begin{align*} A(\mathfrak{a}^{-1}) = N(\mathfrak{a}^{-1}) \cdot A(\mathcal{O}_L) = N(\mathfrak{a}^{-1}) \cdot \tfrac{1}{2}\sqrt{|D_L|}. \end{align*} Applying [Minkowski's Lemma (Quadratic Case)](/theorems/1602) to the lattice $\Lambda = \iota(\mathfrak{a}^{-1})$: there exists a non-zero $\alpha \in \mathfrak{a}^{-1}$ (identified with a non-zero point of $\Lambda$) with \begin{align*} |\alpha|^2 \leq \frac{4 A(\mathfrak{a}^{-1})}{\pi} = \frac{4 \cdot N(\mathfrak{a}^{-1}) \cdot \tfrac{1}{2}\sqrt{|D_L|}}{\pi} = \frac{2 N(\mathfrak{a}^{-1}) \sqrt{|D_L|}}{\pi} = c_L \cdot N(\mathfrak{a}^{-1}), \end{align*} where the last equality uses $c_L = 2\sqrt{|D_L|}/\pi$. [guided] We apply the quadratic Minkowski's lemma to the lattice $\iota(\mathfrak{a}^{-1}) \subseteq \mathbb{R}^2$. This requires (i) that $\iota(\mathfrak{a}^{-1})$ is actually a lattice of full rank, and (ii) knowledge of the area of its fundamental parallelogram, which we get from [Areas of Ideal Lattices](/theorems/1603). **Why $\iota(\mathfrak{a}^{-1})$ is a rank-$2$ lattice in $\mathbb{R}^2$.** By scaling, pick any positive integer $m$ with $m \mathfrak{a}^{-1} \subseteq \mathcal{O}_L$ (take $m = N(\mathfrak{a})$, for instance; then $N(\mathfrak{a}) \mathfrak{a}^{-1} \subseteq \mathfrak{a} \mathfrak{a}^{-1} = \mathcal{O}_L$ after multiplying by $\mathfrak{a}$; more directly, any denominator of $\mathfrak{a}^{-1}$ works). The rescaled ideal $m\mathfrak{a}^{-1}$ is integral, so by [Norm and Discriminant](/theorems/1594) part 1 it is a free $\mathbb{Z}$-module of rank $n = 2$. The scaling map $x \mapsto x/m$ is a $\mathbb{Q}$-linear isomorphism $m\mathfrak{a}^{-1} \to \mathfrak{a}^{-1}$, so $\mathfrak{a}^{-1}$ is also a free $\mathbb{Z}$-module of rank $2$. Its image under $\iota$ in $\mathbb{C} \cong \mathbb{R}^2$ spans $\mathbb{R}^2$ over $\mathbb{R}$ because $\iota(\mathfrak{a}^{-1}) \otimes_{\mathbb{Z}} \mathbb{R}$ fills up $\iota(L) \otimes_{\mathbb{Q}} \mathbb{R} = \mathbb{R} \cdot 1 + \mathbb{R} \cdot i\sqrt{|d|} = \mathbb{C}$. So $\iota(\mathfrak{a}^{-1})$ is a rank-$2$ lattice in $\mathbb{R}^2$, as required. **Computing $A(\mathfrak{a}^{-1})$.** Part (4) of [Theorem 1603](/theorems/1603) was stated and proved for *integral* ideals $\mathfrak{a} \leq \mathcal{O}_L$. The extension to non-zero fractional ideals is by scaling: - pick $m \in \mathbb{Z}_{>0}$ with $m\mathfrak{a}^{-1} \leq \mathcal{O}_L$; - apply part (4) to the integral ideal $m\mathfrak{a}^{-1}$: $A(m\mathfrak{a}^{-1}) = N(m\mathfrak{a}^{-1}) \cdot A(\mathcal{O}_L)$; - descend via the scaling laws $A(\lambda \Lambda) = \lambda^2 A(\Lambda)$ (a linear dilation of factor $\lambda$ scales areas by $\lambda^2$ in $\mathbb{R}^2$) and $N(\lambda \mathfrak{a}) = |N(\lambda)| N(\mathfrak{a}) = \lambda^n N(\mathfrak{a}) = \lambda^2 N(\mathfrak{a})$ (for $\lambda = m \in \mathbb{Z}_{>0}$, $n = 2$, using $N(\langle \lambda \rangle) = |N_{L/\mathbb{Q}}(\lambda)| = \lambda^2$ for a rational integer $\lambda$); - to conclude, $A(\mathfrak{a}^{-1}) = m^{-2} A(m\mathfrak{a}^{-1}) = m^{-2} N(m\mathfrak{a}^{-1}) A(\mathcal{O}_L) = m^{-2} \cdot m^2 N(\mathfrak{a}^{-1}) \cdot A(\mathcal{O}_L) = N(\mathfrak{a}^{-1}) \cdot A(\mathcal{O}_L)$. Combined with part (2) of Theorem 1603 ($A(\mathcal{O}_L) = \tfrac{1}{2}\sqrt{|D_L|}$): \begin{align*} A(\mathfrak{a}^{-1}) = N(\mathfrak{a}^{-1}) \cdot \tfrac{1}{2}\sqrt{|D_L|}. \end{align*} **Applying Minkowski's Lemma.** [Theorem 1602](/theorems/1602) (the quadratic Minkowski lemma) states: for any lattice $\Lambda \subseteq \mathbb{R}^2$, there exists a non-zero $\alpha \in \Lambda$ with $|\alpha|^2 \leq 4 A(\Lambda)/\pi$. Verifying hypotheses: we need $\Lambda$ to be a lattice, which we verified above for $\Lambda = \iota(\mathfrak{a}^{-1})$. Theorem 1602 then yields a non-zero point of $\iota(\mathfrak{a}^{-1})$ — equivalently, a non-zero $\alpha \in \mathfrak{a}^{-1}$ (using the embedding to go between $L$ and its image in $\mathbb{C}$) — with \begin{align*} |\alpha|^2 \leq \frac{4 A(\mathfrak{a}^{-1})}{\pi}. \end{align*} **Simplifying the bound.** Substitute $A(\mathfrak{a}^{-1}) = N(\mathfrak{a}^{-1}) \cdot \tfrac{1}{2}\sqrt{|D_L|}$: \begin{align*} |\alpha|^2 \leq \frac{4 \cdot N(\mathfrak{a}^{-1}) \cdot \tfrac{1}{2}\sqrt{|D_L|}}{\pi} = \frac{2\sqrt{|D_L|}}{\pi} \cdot N(\mathfrak{a}^{-1}) = c_L \cdot N(\mathfrak{a}^{-1}), \end{align*} where the last equality uses the definition $c_L = 2\sqrt{|D_L|}/\pi$ from the theorem statement. **The $\alpha$ we have found.** A non-zero $\alpha \in \mathfrak{a}^{-1}$ with $|\alpha|^2 \leq c_L N(\mathfrak{a}^{-1})$. This is the ingredient we will feed into the principal-shift construction of Step 1. [/guided] [/step] [step:Translate $|\alpha|^2 \leq c_L N(\mathfrak{a}^{-1})$ into $|N(\alpha)| \leq c_L N(\mathfrak{a}^{-1})$ and conclude] By part (1) of [Areas of Ideal Lattices](/theorems/1603), $|\alpha|^2 = N_{L/\mathbb{Q}}(\alpha)$ for $\alpha \in L = \mathbb{Q}(\sqrt{d})$ with $d < 0$. Moreover, when $d < 0$ the norm $N_{L/\mathbb{Q}}(\alpha) = \alpha \bar\alpha = |\alpha|^2 \geq 0$ is always non-negative, so $|N_{L/\mathbb{Q}}(\alpha)| = N_{L/\mathbb{Q}}(\alpha) = |\alpha|^2$. Combining with the inequality of Step 2: \begin{align*} |N_{L/\mathbb{Q}}(\alpha)| = |\alpha|^2 \leq c_L \cdot N(\mathfrak{a}^{-1}). \end{align*} This is exactly the condition we derived in Step 1 for the ideal $\mathfrak{b} = \langle \alpha \rangle \mathfrak{a}$ to satisfy $N(\mathfrak{b}) \leq c_L$: \begin{align*} N(\mathfrak{b}) = |N_{L/\mathbb{Q}}(\alpha)| \cdot N(\mathfrak{a}) \leq c_L \cdot N(\mathfrak{a}^{-1}) \cdot N(\mathfrak{a}) = c_L \cdot N(\mathfrak{a}^{-1} \mathfrak{a}) = c_L \cdot N(\mathcal{O}_L) = c_L, \end{align*} using multiplicativity of the ideal norm and $\mathfrak{a}^{-1}\mathfrak{a} = \mathcal{O}_L$, $N(\mathcal{O}_L) = 1$. Hence $\mathfrak{b} \leq \mathcal{O}_L$ is an integral ideal with $[\mathfrak{b}] = [\mathfrak{a}]$ (by Step 1) and $N(\mathfrak{b}) \leq c_L$, as required. [guided] We have assembled all pieces; the last step is the elementary bookkeeping that translates $|\alpha|^2$ into $|N(\alpha)|$ (which requires part (1) of Theorem 1603) and then assembles the norm of $\mathfrak{b}$. **From $|\alpha|^2$ to $N(\alpha)$.** Part (1) of [Theorem 1603](/theorems/1603) states: for $\alpha \in L = \mathbb{Q}(\sqrt{d})$ with $d < 0$, \begin{align*} |\alpha|^2 = N_{L/\mathbb{Q}}(\alpha). \end{align*} This is the key "bridge identity" between geometric length-squared (complex modulus squared) and algebraic norm. **Sign of the norm.** In the imaginary quadratic case, $N_{L/\mathbb{Q}}(\alpha) = \alpha \bar\alpha = |\alpha|^2 \geq 0$ is non-negative, and it is strictly positive for $\alpha \neq 0$. So $|N(\alpha)| = N(\alpha) = |\alpha|^2$ — the absolute value does not actually do anything for $d < 0$ and non-zero $\alpha$. **Combining with Step 2's bound.** From Step 2, we have a non-zero $\alpha \in \mathfrak{a}^{-1}$ with $|\alpha|^2 \leq c_L \cdot N(\mathfrak{a}^{-1})$. Using $|N(\alpha)| = |\alpha|^2$: \begin{align*} |N_{L/\mathbb{Q}}(\alpha)| \leq c_L \cdot N(\mathfrak{a}^{-1}). \end{align*} **Assembling $N(\mathfrak{b})$.** Using Step 1's computation $N(\mathfrak{b}) = |N_{L/\mathbb{Q}}(\alpha)| \cdot N(\mathfrak{a})$: \begin{align*} N(\mathfrak{b}) = |N_{L/\mathbb{Q}}(\alpha)| \cdot N(\mathfrak{a}) \leq c_L \cdot N(\mathfrak{a}^{-1}) \cdot N(\mathfrak{a}). \end{align*} By multiplicativity of the ideal norm (on non-zero fractional ideals), $N(\mathfrak{a}^{-1}) N(\mathfrak{a}) = N(\mathfrak{a}^{-1} \mathfrak{a}) = N(\mathcal{O}_L) = 1$. Hence \begin{align*} N(\mathfrak{b}) \leq c_L \cdot 1 = c_L. \end{align*} **Ticking off all claims.** We have produced $\mathfrak{b} = \langle \alpha \rangle \mathfrak{a}$ such that: - $\mathfrak{b} \leq \mathcal{O}_L$ (integral), by Step 1; - $[\mathfrak{b}] = [\mathfrak{a}]$ (same class), by Step 1; - $N(\mathfrak{b}) \leq c_L = 2\sqrt{|D_L|}/\pi$, by the chain just computed. This is the statement of the theorem, and the argument is complete. **Remark on the constant $c_L$.** The constant $c_L = 2\sqrt{|D_L|}/\pi$ is the quadratic-field specialization of the general Minkowski bound $c_L = (4/\pi)^s (n!/n^n) |D_L|^{1/2}$ from [Minkowski Bound](/theorems/1610) (after substituting $n = 2, r = 0, s = 1$; then $(4/\pi)^1 \cdot 2!/2^2 = 4/(4\pi) \cdot 2 = 2/\pi$, giving $c_L = 2\sqrt{|D_L|}/\pi$). The factor $4/\pi$ arises from packing circles in squares — precisely the loss we incurred in Step 2 when converting a disc-area hypothesis into a point of small modulus. [/guided] [/step]