[proofplan] The four statements of the theorem tie the geometric area of an ideal lattice in $\mathbb{R}^2$ to its algebraic invariants (norm, discriminant, field discriminant). The imaginary quadratic case $d < 0$ is convenient: the single pair of complex embeddings lets us identify $L = \mathbb{Q}(\sqrt{d})$ with a subfield of $\mathbb{C}$, and an ideal $\mathfrak{a} \leq \mathcal{O}_L$ sits inside $\mathbb{C} \cong \mathbb{R}^2$ as a real rank-$2$ lattice. Part (1) is a direct computation: $|\alpha|^2 = \alpha \bar\alpha = N(\alpha)$ because $|d| = -d$ when $d < 0$. Part (2) is a case-by-case determinant computation on the two possible integral bases. Part (3) relates the area of $\mathfrak{a}$ (a determinant of a real-basis matrix) to the discriminant $\Delta(\alpha_1, \alpha_2)$ (a determinant of a conjugate matrix) via the algebraic identity $\Delta = (\alpha_1 \bar\alpha_2 - \alpha_2 \bar\alpha_1)^2$, which expands to $-4(\operatorname{Im} \alpha_1 \operatorname{Re} \alpha_2 - \operatorname{Im} \alpha_2 \operatorname{Re} \alpha_1)^2$. Part (4) is immediate from (2), (3), and the identity $\Delta(\alpha_1, \alpha_2) = N(\mathfrak{a})^2 D_L$ from [Norm and Discriminant](/theorems/1594). [/proofplan]

[step:Identify $L = \mathbb{Q}(\sqrt{d})$ with a subfield of $\mathbb{C}$ and fix the embedding convention]Since $d < 0$, write $d = -|d|$ with $|d| > 0$, and choose once and for all the embedding \begin{align*} \iota : L &\to \mathbb{C} \\ a + b\sqrt{d} &\mapsto a + b\, i\sqrt{|d|}, \end{align*} where $a, b \in \mathbb{Q}$ and $i$ is a fixed square root of $-1$ in $\mathbb{C}$. This is a $\mathbb{Q}$-algebra monomorphism: $\iota(\sqrt{d})^2 = (i\sqrt{|d|})^2 = -|d| = d = \iota(d)$, so $\iota$ respects the defining relation. Identifying $\mathbb{C}$ with $\mathbb{R}^2$ via $z = a + bi \leftrightarrow (a, b)$, $\iota$ sends $a + b\sqrt{d}$ to the point $(a, b\sqrt{|d|}) \in \mathbb{R}^2$. Throughout this proof, "$\alpha \in L$" is silently identified with $\iota(\alpha) \in \mathbb{C}$, and "the area $A(\mathfrak{a})$" is the area of the fundamental parallelogram of $\iota(\mathfrak{a}) \subseteq \mathbb{C} \cong \mathbb{R}^2$. The two embeddings of $L$ into $\mathbb{C}$ are $\iota$ and $\bar\iota$, with $\bar\iota(a + b\sqrt{d}) = \overline{\iota(a + b\sqrt{d})} = a - bi\sqrt{|d|}$. For $\alpha \in L$, write $\bar\alpha := \bar\iota(\alpha) = \overline{\iota(\alpha)}$ — the complex conjugate inside $\mathbb{C}$ after the embedding.[/step]

[guided]Before starting any computation we fix a single embedding $L \hookrightarrow \mathbb{C}$ so that geometric statements about $\mathfrak{a}$ as a subset of $\mathbb{R}^2$ are unambiguous. In the imaginary quadratic case, $L$ has exactly two embeddings into $\mathbb{C}$, and they are complex conjugates of each other — we call one of them $\iota$ and the other $\bar\iota$. **Fixing the embedding.** Since $d < 0$, the element $\sqrt{d}$ in $L$ (an abstract square root of $d$) corresponds under embedding to one of $\pm i \sqrt{|d|}$ in $\mathbb{C}$. Arbitrarily, we choose \begin{align*} \iota : a + b\sqrt{d} \mapsto a + b i\sqrt{|d|}. \end{align*} One verifies this is a ring homomorphism: the multiplicative structure is preserved because both $\sqrt{d} \in L$ and $i\sqrt{|d|} \in \mathbb{C}$ square to $d$, and $\iota$ is $\mathbb{Q}$-linear by construction. **Writing $L \subseteq \mathbb{C}$.** From here on we treat $L$ as a subfield of $\mathbb{C}$ via $\iota$. An element $\alpha = a + b\sqrt{d}$ with $a, b \in \mathbb{Q}$ becomes the complex number $a + b i \sqrt{|d|}$, with real part $a$ and imaginary part $b\sqrt{|d|}$. **The complex conjugate $\bar\alpha$.** The other embedding of $L$ into $\mathbb{C}$, namely $\bar\iota$, sends $\alpha$ to $a - b i\sqrt{|d|}$, which is exactly the complex conjugate of $\iota(\alpha)$ inside $\mathbb{C}$. We write $\bar\alpha = \bar\iota(\alpha) = \overline{\iota(\alpha)}$ for this. This conflation of "algebraic conjugate (Galois action)" with "complex conjugate (in $\mathbb{C}$)" is a *feature* of the imaginary quadratic case: the two coincide. We will use this repeatedly. **$\mathbb{R}^2$ identification.** We identify $\mathbb{C}$ with $\mathbb{R}^2$ via $a + bi \leftrightarrow (a, b)$. Lengths, areas, and the fundamental parallelogram construction all make sense under this identification. The term "area of the fundamental parallelogram of $\mathfrak{a}$" refers to $|\det[\alpha_1 \mid \alpha_2]|$ where $\alpha_i$ are written as column vectors in $\mathbb{R}^2$ — equivalently, the area of the parallelogram with vertices $0, \alpha_1, \alpha_2, \alpha_1 + \alpha_2$.[/guided]

[step:Prove part (1): $|\alpha|^2 = N(\alpha)$ for $\alpha = a + b\sqrt{d}$]Let $\alpha = a + b\sqrt{d}$ with $a, b \in \mathbb{Q}$. By [Norm and Trace via Embeddings](/theorems/1577) applied to the two embeddings $\iota, \bar\iota$: \begin{align*} N_{L/\mathbb{Q}}(\alpha) = \iota(\alpha) \cdot \bar\iota(\alpha) = \alpha \cdot \bar\alpha. \end{align*} In $\mathbb{C}$, the product $\alpha \bar\alpha$ is $|\alpha|^2$ by definition of the complex modulus: \begin{align*} \alpha \bar\alpha = (a + b i \sqrt{|d|})(a - b i \sqrt{|d|}) = a^2 - (b i \sqrt{|d|})^2 = a^2 + b^2 |d| = a^2 - b^2 d = |\alpha|^2, \end{align*} using $i^2 = -1$ and $|d| = -d$ (since $d < 0$). Hence $N(\alpha) = \alpha \bar\alpha = |\alpha|^2$.[/step]

[guided]We want to show that the squared complex modulus $|\alpha|^2$ equals the field-theoretic norm $N(\alpha)$. The norm is algebraic (defined via the minimal polynomial or determinant of the multiplication map); the modulus is geometric. They happen to coincide in the imaginary quadratic case, and this coincidence is the entire reason the geometric Minkowski machinery can be combined with the algebraic ideal theory. **Expressing the norm as a product of conjugates.** For $\alpha \in L$, [Theorem 1577](/theorems/1577) gives \begin{align*} N_{L/\mathbb{Q}}(\alpha) = \prod_{\sigma: L \hookrightarrow \mathbb{C}} \sigma(\alpha). \end{align*} Verifying hypotheses: $L = \mathbb{Q}(\sqrt{d})$ is a number field of degree $n = 2$, and Theorem 1577 requires exactly that. The $n = 2$ embeddings into $\mathbb{C}$ are $\iota, \bar\iota$, so \begin{align*} N(\alpha) = \iota(\alpha) \cdot \bar\iota(\alpha). \end{align*} **Using $\bar\iota(\alpha) = \overline{\iota(\alpha)}$.** After embedding via $\iota$ (and our convention to write $\alpha \in L$ for $\iota(\alpha) \in \mathbb{C}$), this reads $N(\alpha) = \alpha \cdot \bar\alpha$, where $\bar\alpha = \overline{\iota(\alpha)}$ is the complex conjugate inside $\mathbb{C}$ (established in Step 1). **Computing $\alpha \bar\alpha$ explicitly.** For $\alpha = a + b\sqrt{d}$, $\iota(\alpha) = a + b i \sqrt{|d|}$, so \begin{align*} \alpha \bar\alpha = (a + b i\sqrt{|d|})(a - b i\sqrt{|d|}) = a^2 - (b i \sqrt{|d|})^2 = a^2 - (- b^2 |d|) = a^2 + b^2 |d|. \end{align*} Using $|d| = -d$ (since $d < 0$): \begin{align*} \alpha \bar\alpha = a^2 - b^2 d. \end{align*} **Comparing with $|\alpha|^2$.** By the definition of complex modulus, $|\alpha|^2 = \alpha \bar\alpha$ — this is a basic identity in $\mathbb{C}$. Concretely, writing $\alpha = a + b i \sqrt{|d|}$ in terms of real and imaginary parts $(\operatorname{Re} \alpha, \operatorname{Im} \alpha) = (a, b\sqrt{|d|})$: \begin{align*} |\alpha|^2 = (\operatorname{Re} \alpha)^2 + (\operatorname{Im} \alpha)^2 = a^2 + b^2 |d| = a^2 - b^2 d. \end{align*} Both equal $N(\alpha)$. Hence \begin{align*} |\alpha|^2 = N_{L/\mathbb{Q}}(\alpha), \end{align*} proving part (1). This is the identity that lets us bound field-theoretic norms by geometric $|\cdot|^2$ norms — the fundamental bridge between Minkowski's lemma and algebraic number theory. **Positivity check.** Both sides are automatically $\geq 0$: $|\alpha|^2 \geq 0$ as a squared modulus, and $N(\alpha) = a^2 + b^2 |d| \geq 0$ (sum of squares since $|d| > 0$). Equality holds iff $a = b = 0$ iff $\alpha = 0$. This positivity is special to the imaginary quadratic case — in real quadratic fields, $N$ can be negative.[/guided]

[step:Prove part (2): $A(\mathcal{O}_L) = \tfrac{1}{2}\sqrt{|D_L|}$ by cases on $d \bmod 4$]By [Integers of Quadratic Fields](/theorems/1575), there are two cases. **Case $d \equiv 2, 3 \pmod 4$.** Then $\mathcal{O}_L = \mathbb{Z} \oplus \mathbb{Z}\sqrt{d}$. In the $\mathbb{R}^2$-identification of Step 1, $1 \leftrightarrow (1, 0)$ and $\sqrt{d} \leftrightarrow (0, \sqrt{|d|})$. The fundamental parallelogram of $\mathcal{O}_L$ is spanned by these two vectors, and its area is the absolute value of the determinant of the basis matrix: \begin{align*} A(\mathcal{O}_L) = \left|\det \begin{pmatrix} 1 & 0 \\ 0 & \sqrt{|d|} \end{pmatrix}\right| = \sqrt{|d|}. \end{align*} By the computation $D_L = 4d$ in this case (standard, established in the number-fields thread when computing field discriminants; see [Norm and Discriminant](/theorems/1594) applied with basis $\{1, \sqrt{d}\}$), $|D_L| = 4|d|$, so $\tfrac{1}{2}\sqrt{|D_L|} = \tfrac{1}{2} \cdot 2\sqrt{|d|} = \sqrt{|d|}$. Hence $A(\mathcal{O}_L) = \tfrac{1}{2}\sqrt{|D_L|}$. **Case $d \equiv 1 \pmod 4$.** Then $\mathcal{O}_L = \mathbb{Z} \oplus \mathbb{Z} \cdot \tfrac{1}{2}(1 + \sqrt{d})$. In $\mathbb{R}^2$, $1 \leftrightarrow (1, 0)$ and $\tfrac{1}{2}(1 + \sqrt{d}) \leftrightarrow (\tfrac{1}{2}, \tfrac{1}{2}\sqrt{|d|})$. So \begin{align*} A(\mathcal{O}_L) = \left|\det \begin{pmatrix} 1 & \tfrac{1}{2} \\ 0 & \tfrac{1}{2}\sqrt{|d|} \end{pmatrix}\right| = \tfrac{1}{2}\sqrt{|d|}. \end{align*} In this case $D_L = d$ (standard, from the basis $\{1, \tfrac{1}{2}(1+\sqrt{d})\}$ and minimal polynomial $x^2 - x + \tfrac{1-d}{4}$). So $|D_L| = |d|$ and $\tfrac{1}{2}\sqrt{|D_L|} = \tfrac{1}{2}\sqrt{|d|}$. Hence $A(\mathcal{O}_L) = \tfrac{1}{2}\sqrt{|D_L|}$ again. In both cases, $A(\mathcal{O}_L) = \tfrac{1}{2}\sqrt{|D_L|}$.[/step]

[guided]The computation of $A(\mathcal{O}_L)$ depends on which integral basis of $\mathcal{O}_L$ we work with, and by [Integers of Quadratic Fields](/theorems/1575), this depends on $d \bmod 4$. **Case 1: $d \equiv 2$ or $3 \pmod 4$.** The integral basis is $\{1, \sqrt{d}\}$. After the embedding $\iota$ of Step 1, these become \begin{align*} \iota(1) = 1 = (1, 0), \qquad \iota(\sqrt{d}) = i\sqrt{|d|} = (0, \sqrt{|d|}). \end{align*} The fundamental parallelogram is the parallelogram with vertices $0, (1, 0), (0, \sqrt{|d|}), (1, \sqrt{|d|})$ — in fact a rectangle. Its area is $1 \cdot \sqrt{|d|} = \sqrt{|d|}$. Equivalently, writing the basis vectors as columns of a $2 \times 2$ matrix and taking the absolute value of the determinant: \begin{align*} A(\mathcal{O}_L) = \left|\det\begin{pmatrix} 1 & 0 \\ 0 & \sqrt{|d|} \end{pmatrix}\right| = \sqrt{|d|}. \end{align*} **Field discriminant in Case 1.** Using the basis $\alpha_1 = 1, \alpha_2 = \sqrt{d}$ and [Norm and Trace via Embeddings](/theorems/1577) for the conjugate matrix: \begin{align*} \Delta(1, \sqrt{d}) = \det\begin{pmatrix} \iota(1) & \iota(\sqrt{d}) \\ \bar\iota(1) & \bar\iota(\sqrt{d}) \end{pmatrix}^2 = \det\begin{pmatrix} 1 & \sqrt{d} \\ 1 & -\sqrt{d} \end{pmatrix}^2 = (-2\sqrt{d})^2 = 4d. \end{align*} Since $\{1, \sqrt{d}\}$ is an integral basis, $D_L = \Delta(1, \sqrt{d}) = 4d$, so $|D_L| = 4|d|$ (we use absolute value because $d < 0$ makes $D_L$ negative). Hence \begin{align*} \tfrac{1}{2}\sqrt{|D_L|} = \tfrac{1}{2} \cdot 2\sqrt{|d|} = \sqrt{|d|} = A(\mathcal{O}_L). \checkmark \end{align*} **Case 2: $d \equiv 1 \pmod 4$.** The integral basis is $\{1, \tfrac{1}{2}(1 + \sqrt{d})\}$, and these embed as \begin{align*} \iota(1) = (1, 0), \qquad \iota\!\left(\tfrac{1}{2}(1 + \sqrt{d})\right) = \tfrac{1}{2}(1 + i\sqrt{|d|}) = \left(\tfrac{1}{2}, \tfrac{1}{2}\sqrt{|d|}\right). \end{align*} The fundamental parallelogram has vertices $0, (1, 0), (\tfrac{1}{2}, \tfrac{1}{2}\sqrt{|d|}), (\tfrac{3}{2}, \tfrac{1}{2}\sqrt{|d|})$, and its area is \begin{align*} A(\mathcal{O}_L) = \left|\det\begin{pmatrix} 1 & \tfrac{1}{2} \\ 0 & \tfrac{1}{2}\sqrt{|d|} \end{pmatrix}\right| = \tfrac{1}{2}\sqrt{|d|}. \end{align*} **Field discriminant in Case 2.** With $\alpha_1 = 1$, $\alpha_2 = \tfrac{1}{2}(1 + \sqrt{d})$: \begin{align*} \Delta\!\left(1, \tfrac{1}{2}(1 + \sqrt{d})\right) = \det\begin{pmatrix} 1 & \tfrac{1}{2}(1 + \sqrt{d}) \\ 1 & \tfrac{1}{2}(1 - \sqrt{d}) \end{pmatrix}^2 = (-\sqrt{d})^2 = d. \end{align*} Hence $D_L = d$, $|D_L| = |d|$, and \begin{align*} \tfrac{1}{2}\sqrt{|D_L|} = \tfrac{1}{2}\sqrt{|d|} = A(\mathcal{O}_L). \checkmark \end{align*} **Unified conclusion.** In both cases, \begin{align*} A(\mathcal{O}_L) = \tfrac{1}{2}\sqrt{|D_L|}. \end{align*} The factor $\tfrac{1}{2}$ is not a coincidence: it is the factor $2^{-s}$ in the general formula [Covolume of Ideal Lattices](/theorems/1609), specialized to $s = 1$ (one complex-conjugate pair in the imaginary quadratic case).[/guided]

[step:Prove part (3): $A(\mathfrak{a}) = \tfrac{1}{2}\sqrt{|\Delta(\alpha_1, \alpha_2)|}$ by comparing the real and conjugate determinants]Let $\alpha_1, \alpha_2$ be a $\mathbb{Z}$-basis of $\mathfrak{a}$. Write each $\alpha_k$ (after the embedding $\iota$) in Cartesian form as $\alpha_k = a_k + b_k i$ with $a_k = \operatorname{Re}(\alpha_k)$ and $b_k = \operatorname{Im}(\alpha_k) \in \mathbb{R}$. The fundamental parallelogram of $\mathfrak{a}$ in $\mathbb{R}^2$ is spanned by the columns $(a_1, b_1)^\top$ and $(a_2, b_2)^\top$, so \begin{align*} A(\mathfrak{a}) = \left|\det \begin{pmatrix} a_1 & a_2 \\ b_1 & b_2 \end{pmatrix}\right| = |a_1 b_2 - a_2 b_1|. \end{align*} On the other hand, by the definition of the discriminant and [Norm and Trace via Embeddings](/theorems/1577): \begin{align*} \Delta(\alpha_1, \alpha_2) = \det \begin{pmatrix} \iota(\alpha_1) & \iota(\alpha_2) \\ \bar\iota(\alpha_1) & \bar\iota(\alpha_2) \end{pmatrix}^2 = \det\begin{pmatrix} \alpha_1 & \alpha_2 \\ \bar{\alpha}_1 & \bar{\alpha}_2 \end{pmatrix}^2 = (\alpha_1 \bar\alpha_2 - \alpha_2 \bar\alpha_1)^2. \end{align*} We compute $\alpha_1 \bar\alpha_2 - \alpha_2 \bar\alpha_1$ using $\alpha_k = a_k + b_k i$, $\bar\alpha_k = a_k - b_k i$: \begin{align*} \alpha_1 \bar\alpha_2 - \alpha_2 \bar\alpha_1 &= (a_1 + b_1 i)(a_2 - b_2 i) - (a_2 + b_2 i)(a_1 - b_1 i) \\ &= \bigl[a_1 a_2 + b_1 b_2 + i(a_2 b_1 - a_1 b_2)\bigr] - \bigl[a_1 a_2 + b_1 b_2 + i(a_1 b_2 - a_2 b_1)\bigr] \\ &= 2 i (a_2 b_1 - a_1 b_2). \end{align*} Squaring and using $i^2 = -1$: \begin{align*} \Delta(\alpha_1, \alpha_2) = (2i(a_2 b_1 - a_1 b_2))^2 = -4(a_1 b_2 - a_2 b_1)^2. \end{align*} Taking absolute values: $|\Delta(\alpha_1, \alpha_2)| = 4(a_1 b_2 - a_2 b_1)^2 = 4 A(\mathfrak{a})^2$. Therefore \begin{align*} A(\mathfrak{a})^2 = \tfrac{1}{4}|\Delta(\alpha_1, \alpha_2)|, \qquad A(\mathfrak{a}) = \tfrac{1}{2}\sqrt{|\Delta(\alpha_1, \alpha_2)|}. \end{align*}[/step]

[guided]Part (3) relates two different determinants built from the same basis $\{\alpha_1, \alpha_2\}$: - **Geometric determinant:** the matrix has real rows $(a_k, b_k)$, and its determinant absolute value is the area $A(\mathfrak{a})$. - **Algebraic determinant (discriminant):** the matrix has rows given by the two embeddings $\iota, \bar\iota$, and the square of its determinant is $\Delta(\alpha_1, \alpha_2)$. The goal is to derive the exact relation $A(\mathfrak{a}) = \tfrac{1}{2}\sqrt{|\Delta|}$ by computing both sides in terms of the real-and-imaginary-part coordinates $(a_k, b_k)$. **Real-and-imaginary coordinates.** After embedding via $\iota$, each basis vector becomes a complex number $\alpha_k = a_k + b_k i \in \mathbb{C}$, where $a_k, b_k \in \mathbb{R}$. Explicitly, if $\alpha_k = p_k + q_k \sqrt{d}$ with $p_k, q_k \in \mathbb{Q}$, then \begin{align*} a_k = p_k, \qquad b_k = q_k \sqrt{|d|}. \end{align*} **Geometric area.** In the $\mathbb{R}^2$ identification of Step 1, $\alpha_k = (a_k, b_k)$. The fundamental parallelogram of the lattice $\mathfrak{a} = \mathbb{Z}\alpha_1 + \mathbb{Z}\alpha_2$ has area equal to the absolute value of the basis-matrix determinant: \begin{align*} A(\mathfrak{a}) = \left|\det\begin{pmatrix} a_1 & a_2 \\ b_1 & b_2\end{pmatrix}\right| = |a_1 b_2 - a_2 b_1|. \end{align*} **Algebraic discriminant.** By the definition of the discriminant of a basis and [Theorem 1577](/theorems/1577) (which expresses embeddings in $\mathbb{C}$): \begin{align*} \Delta(\alpha_1, \alpha_2) = \det\begin{pmatrix}\sigma_1(\alpha_1) & \sigma_1(\alpha_2) \\ \sigma_2(\alpha_1) & \sigma_2(\alpha_2)\end{pmatrix}^2 \end{align*} where $\sigma_1 = \iota, \sigma_2 = \bar\iota$. After embedding: \begin{align*} \Delta(\alpha_1, \alpha_2) = \det\begin{pmatrix} \alpha_1 & \alpha_2 \\ \bar\alpha_1 & \bar\alpha_2\end{pmatrix}^2 = (\alpha_1 \bar\alpha_2 - \alpha_2 \bar\alpha_1)^2. \end{align*} **Computing $\alpha_1 \bar\alpha_2 - \alpha_2 \bar\alpha_1$.** Expanding with $\alpha_k = a_k + b_k i$: \begin{align*} \alpha_1 \bar\alpha_2 &= (a_1 + b_1 i)(a_2 - b_2 i) = a_1 a_2 - a_1 b_2 i + b_1 a_2 i - b_1 b_2 i^2 \\ &= (a_1 a_2 + b_1 b_2) + (b_1 a_2 - a_1 b_2) i, \end{align*} \begin{align*} \alpha_2 \bar\alpha_1 &= (a_2 + b_2 i)(a_1 - b_1 i) = a_1 a_2 - a_2 b_1 i + a_1 b_2 i - b_1 b_2 i^2 \\ &= (a_1 a_2 + b_1 b_2) + (a_1 b_2 - a_2 b_1) i. \end{align*} Subtracting: \begin{align*} \alpha_1 \bar\alpha_2 - \alpha_2 \bar\alpha_1 &= (a_1 a_2 + b_1 b_2) + (b_1 a_2 - a_1 b_2) i - (a_1 a_2 + b_1 b_2) - (a_1 b_2 - a_2 b_1) i \\ &= (b_1 a_2 - a_1 b_2) i - (a_1 b_2 - a_2 b_1) i \\ &= 2(a_2 b_1 - a_1 b_2) i. \end{align*} (We used $b_1 a_2 = a_2 b_1$.) **Squaring.** Using $i^2 = -1$: \begin{align*} \Delta(\alpha_1, \alpha_2) = [2(a_2 b_1 - a_1 b_2) i]^2 = 4(a_2 b_1 - a_1 b_2)^2 \cdot (-1) = -4(a_1 b_2 - a_2 b_1)^2, \end{align*} where the last equality squares $(a_2 b_1 - a_1 b_2)^2 = (-(a_1 b_2 - a_2 b_1))^2 = (a_1 b_2 - a_2 b_1)^2$. **Taking absolute value.** Since $(a_1 b_2 - a_2 b_1)^2 \geq 0$: \begin{align*} |\Delta(\alpha_1, \alpha_2)| = 4(a_1 b_2 - a_2 b_1)^2 = 4|a_1 b_2 - a_2 b_1|^2 = 4 A(\mathfrak{a})^2. \end{align*} (The middle equality uses that $(a_1 b_2 - a_2 b_1)^2 = |a_1 b_2 - a_2 b_1|^2$ for real numbers, and the last uses the expression for $A(\mathfrak{a})$ derived above.) **Extracting the square root.** Both $A(\mathfrak{a})$ and $|\Delta|$ are non-negative, so we may take square roots: \begin{align*} A(\mathfrak{a}) = \tfrac{1}{2}\sqrt{|\Delta(\alpha_1, \alpha_2)|}, \end{align*} proving part (3). The sign of $\Delta$ in the imaginary quadratic case is negative (as we saw, $\Delta = -4(a_1 b_2 - a_2 b_1)^2$), which is consistent with the general fact that discriminants of fields with pairs of complex embeddings have sign $(-1)^s$ times a non-negative quantity, here with $s = 1$.[/guided]

[step:Prove part (4): $A(\mathfrak{a}) = N(\mathfrak{a}) A(\mathcal{O}_L)$ from parts (2), (3) and the identity $\Delta = N(\mathfrak{a})^2 D_L$]By [Norm and Discriminant](/theorems/1594) part 2, for any $\mathbb{Z}$-basis $\alpha_1, \alpha_2$ of a nonzero ideal $\mathfrak{a} \leq \mathcal{O}_L$, \begin{align*} \Delta(\alpha_1, \alpha_2) = N(\mathfrak{a})^2 D_L. \end{align*} Taking absolute values and square roots (both $N(\mathfrak{a}) \geq 1$ and $|D_L| \geq 0$, so both sides are non-negative): \begin{align*} \sqrt{|\Delta(\alpha_1, \alpha_2)|} = N(\mathfrak{a}) \sqrt{|D_L|}. \end{align*} Multiplying both sides by $\tfrac{1}{2}$ and invoking parts (2) and (3): \begin{align*} A(\mathfrak{a}) = \tfrac{1}{2}\sqrt{|\Delta(\alpha_1, \alpha_2)|} = N(\mathfrak{a}) \cdot \tfrac{1}{2}\sqrt{|D_L|} = N(\mathfrak{a}) \cdot A(\mathcal{O}_L). \end{align*} This is part (4), completing the proof.[/step]

[guided]Part (4) links the geometric area of an ideal lattice to the area of the ring-of-integers lattice via the *norm* of the ideal. It is the quadratic-field instance of the general formula \begin{align*} \operatorname{covol}(\sigma(\mathfrak{a})) = N(\mathfrak{a}) \cdot \operatorname{covol}(\sigma(\mathcal{O}_L)) \end{align*} from [Covolume of Ideal Lattices](/theorems/1609). Here we derive it from the three ingredients we have just established. **Ingredients.** - Part (2): $A(\mathcal{O}_L) = \tfrac{1}{2}\sqrt{|D_L|}$. - Part (3): $A(\mathfrak{a}) = \tfrac{1}{2}\sqrt{|\Delta(\alpha_1, \alpha_2)|}$ for any $\mathbb{Z}$-basis $\{\alpha_1, \alpha_2\}$ of $\mathfrak{a}$. - [Norm and Discriminant](/theorems/1594) part 2: $\Delta(\alpha_1, \alpha_2) = N(\mathfrak{a})^2 D_L$. **Verifying hypotheses of Theorem 1594.** Theorem 1594 applies to a nonzero ideal $\mathfrak{a} \unlhd \mathcal{O}_L$ in a number field of degree $n = [L : \mathbb{Q}]$. Our hypothesis gives $\mathfrak{a} \leq \mathcal{O}_L$ nonzero (the theorem statement) in the quadratic field $L$ with $n = 2$. Moreover Theorem 1594 part 1 guarantees an integral $\mathbb{Z}$-basis $\alpha_1, \ldots, \alpha_n$ of $\mathfrak{a}$, which is the one we used in part (3). All hypotheses are satisfied. **Chain of equalities.** Starting from part (3): \begin{align*} A(\mathfrak{a}) = \tfrac{1}{2}\sqrt{|\Delta(\alpha_1, \alpha_2)|}. \end{align*} Substitute the Norm-and-Discriminant identity $\Delta = N(\mathfrak{a})^2 D_L$, taking absolute value and square root. Since $N(\mathfrak{a}) \in \mathbb{Z}_{\geq 1}$ (ideal norms are positive integers for nonzero ideals), $N(\mathfrak{a})^2 \geq 0$ is non-negative; multiplying by $|D_L| \geq 0$ and taking absolute value: \begin{align*} |\Delta(\alpha_1, \alpha_2)| = |N(\mathfrak{a})^2 D_L| = N(\mathfrak{a})^2 |D_L|. \end{align*} (We can drop the absolute value on $N(\mathfrak{a})^2$ because it is already non-negative.) Taking the square root: \begin{align*} \sqrt{|\Delta(\alpha_1, \alpha_2)|} = \sqrt{N(\mathfrak{a})^2 |D_L|} = N(\mathfrak{a}) \sqrt{|D_L|}, \end{align*} using $\sqrt{N(\mathfrak{a})^2} = |N(\mathfrak{a})| = N(\mathfrak{a})$ (positive). **Assembling the final equation.** Multiply by $\tfrac{1}{2}$ and use part (2): \begin{align*} A(\mathfrak{a}) = \tfrac{1}{2} \sqrt{|\Delta(\alpha_1, \alpha_2)|} = N(\mathfrak{a}) \cdot \tfrac{1}{2}\sqrt{|D_L|} = N(\mathfrak{a}) \cdot A(\mathcal{O}_L). \end{align*} This is part (4). **Geometric meaning.** The relation $A(\mathfrak{a}) = N(\mathfrak{a}) A(\mathcal{O}_L)$ says that when you shrink an ideal from the full $\mathcal{O}_L$ to a sub-ideal of norm $N(\mathfrak{a})$, the fundamental parallelogram's area *grows* by a factor of $N(\mathfrak{a})$. This is because $\mathcal{O}_L / \mathfrak{a}$ has size $N(\mathfrak{a})$ (the definition of ideal norm), so $\mathfrak{a}$ is an index-$N(\mathfrak{a})$ sublattice of $\mathcal{O}_L$ — and the covolume of a sublattice scales by the index. The algebraic identity $\Delta = N^2 D_L$ is the quantitative restatement of this geometric fact, squared (determinants and areas are squares of discriminants and linear volumes).[/guided]