[proofplan] The Minkowski embedding $\sigma : L \to \mathbb{R}^r \times \mathbb{C}^s \cong \mathbb{R}^n$ takes an integral basis $\gamma_1, \ldots, \gamma_n$ of $\mathfrak{a}$ to $n$ vectors in $\mathbb{R}^n$. The covolume of $\sigma(\mathfrak{a})$ is the absolute value of the determinant of the matrix formed by these vectors. By comparing $\sigma$ to the "full embedding" that collects all $n$ ring embeddings $(\sigma_1, \ldots, \sigma_r, \tau_1, \bar\tau_1, \ldots, \tau_s, \bar\tau_s)$, we obtain the target matrix $B$ from the standard matrix $A = (\sigma_i(\gamma_j))$ by applying $s$ block-diagonal $2 \times 2$ transition matrices, each replacing a $(\tau_j, \bar\tau_j)$-pair by the real pair $(\operatorname{Re}\tau_j, \operatorname{Im}\tau_j)$. Each such transition contributes a factor $1/2$ to $|\det|$, giving a total factor of $2^{-s}$. Combining with the identity $\det(A)^2 = N(\mathfrak{a})^2 D_L$ from [Norm and Discriminant](/theorems/1594) yields the formula. [/proofplan] [step:Fix an integral basis of $\mathfrak{a}$ and form the two matrices of embedding data] By [Norm and Discriminant](/theorems/1594) applied to the non-zero ideal $\mathfrak{a} \leq \mathcal{O}_L$: the hypothesis that $\mathfrak{a}$ is a non-zero integral ideal of a ring of integers $\mathcal{O}_L$ is satisfied. The conclusion gives a $\mathbb{Z}$-basis $\gamma_1, \ldots, \gamma_n$ of $\mathfrak{a}$ which is also a $\mathbb{Q}$-basis of $L$, and \begin{align*} \Delta(\gamma_1, \ldots, \gamma_n) &= N(\mathfrak{a})^2\, D_L. \end{align*} Let $\sigma_1, \ldots, \sigma_r : L \to \mathbb{R}$ be the real embeddings and $\tau_1, \bar\tau_1, \ldots, \tau_s, \bar\tau_s : L \to \mathbb{C}$ the complex embeddings, so $n = r + 2s$. Form the $n \times n$ complex matrix $A$ indexed by all embeddings on the rows and by the basis elements on the columns: \begin{align*} A &= \begin{pmatrix} \sigma_1(\gamma_1) & \cdots & \sigma_1(\gamma_n) \\ \vdots & & \vdots \\ \sigma_r(\gamma_1) & \cdots & \sigma_r(\gamma_n) \\ \tau_1(\gamma_1) & \cdots & \tau_1(\gamma_n) \\ \bar\tau_1(\gamma_1) & \cdots & \bar\tau_1(\gamma_n) \\ \vdots & & \vdots \\ \tau_s(\gamma_1) & \cdots & \tau_s(\gamma_n) \\ \bar\tau_s(\gamma_1) & \cdots & \bar\tau_s(\gamma_n) \end{pmatrix}. \end{align*} The definition of the discriminant (via the trace pairing and the formula $\Delta(\gamma_1, \ldots, \gamma_n) = \det(A)^2$, a standard identity for separable extensions) gives \begin{align*} \det(A)^2 &= \Delta(\gamma_1, \ldots, \gamma_n) = N(\mathfrak{a})^2\, D_L. \end{align*} In particular $|\det(A)| = N(\mathfrak{a})\, |D_L|^{1/2}$. Now form the real $n \times n$ matrix $B$ whose columns are the Minkowski embeddings $\sigma(\gamma_j)$ arranged as column vectors in $\mathbb{R}^n$: \begin{align*} B &= \bigl(\sigma(\gamma_1) \; \sigma(\gamma_2) \; \cdots \; \sigma(\gamma_n)\bigr). \end{align*} Explicitly, the $j$-th column of $B$ is \begin{align*} \sigma(\gamma_j) &= \bigl(\sigma_1(\gamma_j), \ldots, \sigma_r(\gamma_j),\; \operatorname{Re}\tau_1(\gamma_j), \operatorname{Im}\tau_1(\gamma_j),\; \ldots,\; \operatorname{Re}\tau_s(\gamma_j), \operatorname{Im}\tau_s(\gamma_j)\bigr)^\top. \end{align*} By the definition of covolume as the volume of the parallelepiped spanned by a $\mathbb{Z}$-basis, \begin{align*} \operatorname{covol}(\sigma(\mathfrak{a})) &= |\det(B)|. \end{align*} The claim reduces to showing $|\det(B)| = 2^{-s} |\det(A)|$. [guided] **Goal.** We must compute $\operatorname{covol}(\sigma(\mathfrak{a}))$, the volume of a fundamental parallelepiped of the lattice $\sigma(\mathfrak{a}) \subset \mathbb{R}^n$. A fundamental parallelepiped is spanned by any $\mathbb{Z}$-basis of the lattice; its volume is the absolute determinant of the change-of-basis matrix to the standard basis of $\mathbb{R}^n$. **Basis of $\mathfrak{a}$.** By [Norm and Discriminant](/theorems/1594), a non-zero ideal $\mathfrak{a}$ is free of rank $n$ as a $\mathbb{Z}$-module, and any $\mathbb{Z}$-basis $\gamma_1, \ldots, \gamma_n$ of $\mathfrak{a}$ is also a $\mathbb{Q}$-basis of $L$. The discriminant of any such basis satisfies \begin{align*} \Delta(\gamma_1, \ldots, \gamma_n) &= N(\mathfrak{a})^2\, D_L, \end{align*} where $D_L$ is the field discriminant. **Two ways to encode the embeddings.** The $n$ ring embeddings $L \to \mathbb{C}$ split into $r$ real ones $(\sigma_i)$ and $s$ conjugate pairs $(\tau_j, \bar\tau_j)$ of complex ones. Two natural ways to assemble these into matrices of size $n \times n$: - The **full embedding matrix** $A$: list all $n$ ring embeddings on rows, and let column $j$ be the values at $\gamma_j$. Entries of $A$ are complex. - The **Minkowski embedding matrix** $B$: for each complex pair, replace the two conjugate rows by the real pair $(\operatorname{Re}, \operatorname{Im})$. Entries of $B$ are real. **Why both.** The matrix $A$ has a clean discriminant interpretation: $\det(A)^2 = \Delta(\gamma_1, \ldots, \gamma_n)$. This is the content of the standard identity defining the discriminant of a basis as the squared determinant of the "trace matrix" $S^\top S$ with $S = A^\top$; equivalently, it follows from $\operatorname{tr}_{L/\mathbb{Q}}(\gamma_i\gamma_j) = \sum_k \sigma_k(\gamma_i)\sigma_k(\gamma_j)$ summed over all $n$ ring embeddings. The matrix $B$ has a clean volume interpretation: $\operatorname{covol}(\sigma(\mathfrak{a})) = |\det(B)|$. This is because the columns of $B$ are precisely $\sigma(\gamma_1), \ldots, \sigma(\gamma_n)$, a $\mathbb{Z}$-basis of $\sigma(\mathfrak{a}) \subset \mathbb{R}^n$, and the covolume is the absolute determinant of the matrix whose columns form a fundamental parallelepiped. **The plan.** We relate $\det(A)$ to $\det(B)$ by a sequence of elementary row operations (replacing each conjugate pair of rows by its real and imaginary parts), each of which contributes a known factor to the determinant. Summing these factors gives $|\det(B)|/|\det(A)| = 2^{-s}$. [/guided] [/step] [step:Compute the determinant factor for a single conjugate pair of rows] We examine the effect of a single conjugate pair. Consider rows of $A$ labelled $\tau_j$ and $\bar\tau_j$ (above each other). The $j$-th conjugate pair of rows has complex entries; the corresponding rows of $B$ contain the real and imaginary parts. Their relationship is given by the $2 \times 2$ transition identities \begin{align*} \operatorname{Re} \tau_j(\gamma) &= \tfrac{1}{2}(\tau_j(\gamma) + \bar\tau_j(\gamma)), \\ \operatorname{Im} \tau_j(\gamma) &= \tfrac{1}{2i}(\tau_j(\gamma) - \bar\tau_j(\gamma)) = -\tfrac{i}{2}(\tau_j(\gamma) - \bar\tau_j(\gamma)). \end{align*} In matrix form, the column vectors $(\tau_j(\gamma), \bar\tau_j(\gamma))^\top$ and $(\operatorname{Re}\tau_j(\gamma), \operatorname{Im}\tau_j(\gamma))^\top$ are related by \begin{align*} \begin{pmatrix} \operatorname{Re}\tau_j(\gamma) \\ \operatorname{Im}\tau_j(\gamma) \end{pmatrix} &= M \begin{pmatrix} \tau_j(\gamma) \\ \bar\tau_j(\gamma) \end{pmatrix}, \qquad M = \begin{pmatrix} 1/2 & 1/2 \\ -i/2 & i/2 \end{pmatrix}. \end{align*} We compute \begin{align*} \det M &= \frac{1}{2} \cdot \frac{i}{2} - \frac{1}{2} \cdot \Bigl(-\frac{i}{2}\Bigr) = \frac{i}{4} + \frac{i}{4} = \frac{i}{2}. \end{align*} In particular, $|\det M| = 1/2$. [guided] **Converting conjugate pairs to real pairs.** For any complex number $z = x + iy$, the real and imaginary parts are extracted by \begin{align*} x = \operatorname{Re}(z) &= \tfrac{1}{2}(z + \bar z), \\ y = \operatorname{Im}(z) &= \tfrac{1}{2i}(z - \bar z) = -\tfrac{i}{2}(z - \bar z). \end{align*} The $2 \times 2$ matrix $M$ encodes this linear transformation: \begin{align*} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1/2 & 1/2 \\ -i/2 & i/2 \end{pmatrix} \begin{pmatrix} z \\ \bar z \end{pmatrix}. \end{align*} **Sanity check.** Apply $M$ to $(z, \bar z)^\top$: - First row: $(1/2)z + (1/2)\bar z = (z + \bar z)/2 = \operatorname{Re}(z)$. Correct. - Second row: $(-i/2)z + (i/2)\bar z = (i/2)(\bar z - z) = -\operatorname{Im}(z) \cdot i \cdot (-i) = \operatorname{Im}(z)$. Let us redo: $(i/2)(\bar z - z) = (i/2) \cdot (-2i \operatorname{Im}(z)) = -i^2 \operatorname{Im}(z) = \operatorname{Im}(z)$. Correct. **Determinant.** By the $2 \times 2$ formula $\det\begin{pmatrix}a & b \\ c & d\end{pmatrix} = ad - bc$: \begin{align*} \det M = \frac{1}{2} \cdot \frac{i}{2} - \frac{1}{2} \cdot \Bigl(-\frac{i}{2}\Bigr) = \frac{i}{4} + \frac{i}{4} = \frac{i}{2}. \end{align*} Taking absolute values, $|\det M| = |i/2| = 1/2$. Each conjugate pair contributes a factor of $1/2$ when we replace the pair $(\tau_j, \bar\tau_j)$ of rows by the real pair $(\operatorname{Re}\tau_j, \operatorname{Im}\tau_j)$. [/guided] [/step] [step:Assemble the block-diagonal transition matrix and conclude] Let $T$ be the $n \times n$ block-diagonal matrix \begin{align*} T &= \operatorname{diag}(I_r,\, M,\, M,\, \ldots,\, M), \end{align*} where $I_r$ is the $r \times r$ identity matrix (acting as the identity on the real embedding rows) and $M$ is the $2 \times 2$ transition matrix from Step 2, repeated $s$ times (one block per complex conjugate pair). Then $B = T A$: left-multiplication by $T$ replaces each $(\tau_j, \bar\tau_j)$-row pair of $A$ by the corresponding $(\operatorname{Re}\tau_j, \operatorname{Im}\tau_j)$-row pair, leaving the real rows unchanged. The determinant is multiplicative under products, and for a block-diagonal matrix equals the product of the block determinants: \begin{align*} \det(T) &= \det(I_r) \cdot \det(M)^s = 1 \cdot (i/2)^s = (i/2)^s. \end{align*} Hence \begin{align*} \det(B) &= \det(T)\, \det(A) = (i/2)^s \det(A), \end{align*} and taking absolute values: \begin{align*} |\det(B)| &= (1/2)^s |\det(A)| = 2^{-s} |\det(A)|. \end{align*} Combining with Step 1: \begin{align*} \operatorname{covol}(\sigma(\mathfrak{a})) = |\det(B)| = 2^{-s} |\det(A)| = 2^{-s} N(\mathfrak{a})\, |D_L|^{1/2}. \end{align*} In the special case $\mathfrak{a} = \mathcal{O}_L$, $N(\mathcal{O}_L) = 1$ (the index of $\mathcal{O}_L$ in itself), so $\operatorname{covol}(\sigma(\mathcal{O}_L)) = 2^{-s} |D_L|^{1/2}$. This proves the theorem. [guided] **Block-diagonal structure.** We build a single $n \times n$ matrix $T$ encoding all the $s$ conjugate-pair conversions, leaving the $r$ real rows unchanged: \begin{align*} T = \begin{pmatrix} I_r & 0 & \cdots & 0 \\ 0 & M & \cdots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & M \end{pmatrix}, \end{align*} where $M$ is the $2 \times 2$ block from Step 2. Acting on the left of $A$, $T$ applies the identity to rows $1, \ldots, r$ and applies $M$ to each pair of rows corresponding to $(\tau_j, \bar\tau_j)$. **Checking $B = T A$.** The $j$-th pair of rows of $A$ below the real rows is $\bigl(\tau_j(\gamma_1), \ldots, \tau_j(\gamma_n); \bar\tau_j(\gamma_1), \ldots, \bar\tau_j(\gamma_n)\bigr)$; multiplying by the $2 \times 2$ block $M$ produces $\bigl(\operatorname{Re}\tau_j(\gamma_1), \ldots, \operatorname{Re}\tau_j(\gamma_n); \operatorname{Im}\tau_j(\gamma_1), \ldots, \operatorname{Im}\tau_j(\gamma_n)\bigr)$, which is the $j$-th pair of rows of $B$. **Determinant computation.** The determinant of a block-diagonal matrix is the product of the block determinants: \begin{align*} \det(T) &= \det(I_r) \prod_{j=1}^s \det(M) = 1 \cdot \Bigl(\frac{i}{2}\Bigr)^s = \frac{i^s}{2^s}. \end{align*} Then by multiplicativity $\det(B) = \det(T)\det(A) = (i^s/2^s) \det(A)$. **Taking absolute values.** $|i^s| = 1$, so $|\det(B)| = 2^{-s} |\det(A)|$. **Substituting $|\det(A)|$.** From Step 1, $|\det(A)| = N(\mathfrak{a}) |D_L|^{1/2}$. Substituting: \begin{align*} \operatorname{covol}(\sigma(\mathfrak{a})) = |\det(B)| = 2^{-s} \cdot N(\mathfrak{a})\, |D_L|^{1/2}. \end{align*} **Sanity check.** For a totally real field ($s = 0$), the formula reduces to $\operatorname{covol}(\sigma(\mathfrak{a})) = N(\mathfrak{a}) |D_L|^{1/2}$, which is the standard volume of the lattice of $n$ real embeddings. For an imaginary quadratic field ($r = 0, s = 1$), the formula gives $\operatorname{covol}(\sigma(\mathfrak{a})) = (1/2) |D_L|^{1/2} N(\mathfrak{a})$, the area of a fundamental domain in $\mathbb{C}$ halved by the complex-pair transition. [/guided] [/step]