Covolume of an integral ideal lattice

Given: A number field $K$ of degree $n$, its ring of integers ${\mathcal{O}}_K$, discriminant $d_K\mathrm{\ne }0$, the unscaled Minkowski embedding $\sigma$, and a nonzero ideal $\mathfrak{a}\subseteq {\mathcal{O}}_K$.

The image $\sigma (\mathfrak{a})$ of a nonzero integral ideal is a full lattice, and $\mathfrak{a}$ has a $\mathbb{Z}$-basis ${\alpha }_1,\dots ,{\alpha }_n$; moreover $\sigma ({\mathcal{O}}_K)$ is a full lattice with integral basis ${\beta }_1,\dots ,{\beta }_n$ of ${\mathcal{O}}_K$ (Number-field integer rings and ideals are full lattices, Integral and power integral bases).

For an ordered $\mathbb{Q}$-basis ${\alpha }_1,\dots ,{\alpha }_n$ of $K$, the real $n\times n$ matrix $A$ with columns $\sigma ({\alpha }_j)$ satisfies $\mathrm{\mid }\det A\mathrm{\mid }=2^{-r_2}\sqrt{\mathrm{\mid }\mathrm{disc}({\alpha }_1,\dots ,{\alpha }_n)\mathrm{\mid }}$, the real determinant being obtained from the full complex embedding matrix by replacing each conjugate pair of rows by its real and imaginary parts (Unscaled Minkowski embedding).

$\mathrm{disc}({\alpha }_1,\dots ,{\alpha }_n)=\det ({\psi }_i({\alpha }_j){)}^2$ for the full list of embeddings ${\psi }_1,\dots ,{\psi }_n$, and this determinant is nonzero (Embedding determinant formula).

If ${\beta }_j={\sum }_i{a}_{ij}{\alpha }_i$ for two ordered bases, then $\mathrm{disc}({\beta }_1,\dots ,{\beta }_n)=\det (A{)}^2\mathrm{disc}({\alpha }_1,\dots ,{\alpha }_n)$ (Change of basis for discriminants).

For every integral basis ${\beta }_1,\dots ,{\beta }_n$ of ${\mathcal{O}}_K$ one has $\mathrm{disc}({\beta }_1,\dots ,{\beta }_n)=d_K\mathrm{\ne }0$ (Discriminant of a basis and order, Number-field discriminant is well-defined and nonzero).

$N\mathfrak{a}=\mathrm{\mid }{\mathcal{O}}_K\mathrm{/}\mathfrak{a}\mathrm{\mid }$ is finite (The absolute norm of an integral ideal, A nonzero number-field ideal has finite quotient).

For $C\in M_n(\mathbb{Z})$ with $\det C\mathrm{\ne }0$, the subgroup $C{\mathbb{Z}}^n\subseteq {\mathbb{Z}}^n$ has finite index $\mathrm{\mid }\det C\mathrm{\mid }$ (The index of a full-rank subgroup of ${\mathbb{Z}}^n$ is the absolute determinant of a generating matrix).

$\mathrm{covol}(\mathrm{\Lambda })=\mathrm{\mid }\det B\mathrm{\mid }$ for a $\mathbb{Z}$-basis $b_1,\dots ,b_n$ of a full lattice $\mathrm{\Lambda }$ with matrix $B=(b_1\cdots b_n)$ (Full Euclidean lattice and covolume).