Let $K$ be number field of degree $d$. Suppose we are given module $ \mathcal{M}$ in form: \begin{equation}\label{key} \mathcal{M} = v_1 \cdot \mathfrak{a}_1 \oplus v_2 \cdot \mathfrak{a}_2 \oplus \ldots \oplus v_{n}\cdot \mathfrak{a}_{n}, \end{equation} where all $v_i \in K^m$ - vectors of length $m$. Therefore we are provided with the pseudobasis $(\boldsymbol{V},\mathfrak{A})$ where $\boldsymbol{V} \,$ is the $m \times n$ matrix with $v_i$ as columns and $\mathfrak{A} = \{ \mathfrak{a}_i \}_{i=\overline{1,n}}$ - set of corresponding fraction ideals of $K$. Wlog we can assume that pseudo-basis $(\boldsymbol{V},\mathfrak{A})$ is in HNF form (see Cohen H. - Advanced Topics in Computational Number Theory).

Since the arbitrary fraction ideal $\mathfrak{a} $ of $K$ is also a $\mathbb{Z}$-module, we can construct its $\mathbb{Z}$-basis $\bigoplus_{i=1}^{d} a_i \cdot \mathbb{Z}$ where all $a_i$ are integers (for example using pari gp). I think, we even can force all $v_i$ to be integers by finding equivalent pair $(v_i', \mathfrak{a}_i')$ such that $v_i' \mathfrak{a}_i' = v_i \mathfrak{a}_i$ so that we obtain a direct sum of $\mathbb{Z}$-modules which must be a $\mathbb{Z}$-module itself.

The question is: how to compute matrix that corresponds to the $\mathbb{Z}$-basis of $\mathcal{M}$?

Is it in the next form given by the block matrix: \begin{equation*} \begin{pmatrix} Z(v_{1,1}\cdot \mathfrak{a}_1) & Z(v_{1,2}\cdot \mathfrak{a}_2) & \ldots & Z(v_{1,n}\cdot \mathfrak{a}_n) \\ Z(v_{2,1}\cdot \mathfrak{a}_1) & Z(v_{2,2}\cdot \mathfrak{a}_2) & \ldots & Z(v_{2,n}\cdot \mathfrak{a}_n) \\ \vdots & \vdots & \ddots & \vdots \\ Z(v_{m,1}\cdot \mathfrak{a}_1) & Z(v_{m,2}\cdot \mathfrak{a}_2) & \ldots & Z(v_{m,n}\cdot \mathfrak{a}_n) \\ \end{pmatrix}, \end{equation*} where $Z(\mathfrak{a})$ - is matrix of $\mathbb{Z}$-basis of fraction ideal $\mathfrak{a}$?

  • $\begingroup$ $\mathbb{ZZ}$ in your title should be just $\mathbb Z$, right? Also, I am confused by your writing a fractional ideal as $\bigoplus_i a_i\cdot\mathbb Z$, where $a_i$ are integers … if you mean $a_i \in \mathbb Z$, then this can only give ideals of $\mathbb Z$; but, if you mean $a_i \in \mathcal O_K$, then it can only give ideals of $\mathcal O_K$, not fractional ideals, right? $\endgroup$
    – LSpice
    Aug 9, 2022 at 17:29
  • $\begingroup$ Yes! The $\mathbb{Z}$ is typo. I meant that $\mathfrak{a}$ is isomorphic to $\bigoplus_{i} a_i \cdot \mathbb{Z}$ because it is $\mathcal{O}_K$-module (for $\mathcal{O}_K$ - ring of integers of $K$) and therefore a $\mathbb{Z}$ module itself. I'll send an example in the next comment. $\endgroup$ Aug 9, 2022 at 18:39
  • $\begingroup$ If $K = \mathbb{Q}[\sqrt{-5}]$ and $\mathfrak{a}=(2,1+\sqrt{-5})$ then the $\mathfrak{a}$ as a module is isomorphic to $2\cdot\mathbb{Z} \oplus (1+\sqrt{-5})\cdot\mathbb{Z} \cong \mathbb{Z}^2$ and the elements in its integral basis are $2$ which is a vector $(2,0)^T$ and $1+\sqrt{-5}$ which is $(1,1)^T$. So basis of $\mathfrak{a}$ is $\begin{pmatrix} 2 & 1\\ 0 & 1 \end{pmatrix}$. $\endgroup$ Aug 9, 2022 at 18:39

1 Answer 1


Let $\mathcal{M} = \sum_{i=1}^{m} \mathfrak{b}_i \cdot b_i$ be a module. It is represented via pseudobasis therefore it is projective and has a Z-basis (subsect. 2.3 in https://perso.ens-lyon.fr/damien.stehle/downloads/OKLatRed.pdf). Let $\{ \beta_i^{(y)} \}_{y \in [1,d]}$ be integral basis for ideals $\mathfrak{b}_i$ and $d = [K:\mathbb{Q}]$. Then: \begin{multline*} \mathcal{M} = \begin{pmatrix} \sum_{i=1}^{m} \mathfrak{b}_i \cdot b_i^{(0)} \\ \vdots \\ \sum_{i=1}^{m} \mathfrak{b}_i \cdot b_i^{(n-1)} \end{pmatrix} := \left( \sum_{i=1}^{m} \mathfrak{b}_i \cdot b_i^{(x)} \right)_{x\in [1,n]} = \left( \sum_{i=1}^{n} \sum_{j=1}^{d} \mathbb{Z} \cdot \underbrace{\beta_j b_{i}^{(x)}}_{\in K} \right)_{x\in [1,n]} = \left( \sum_{i=1}^{n} \sum_{j=1}^{d} \mathbb{Z} \cdot ( \beta_j b_{i}^{(x)} )^{(y)} \right)_{\small{\begin{matrix} x \in [1,n],\\ y\in [1,d]\end{matrix} }}, \end{multline*}

which is a $\mathbb{Z}$-module with $ \sum_{j=1}^{d} ( \beta_j b_{i}^{(x)} )^{(y)} $ as the entries of its $i$-th vector. So to embedd $\mathcal{M}$ to $\mathbb{Z}^{nd}$ you need to compute all $( \beta_j b_{i}^{(x)} )^{(y)} $ for all $x$-th coordinates of vectors $b_i$ over $K$ and for all $y$-th coordinates of the $\beta_j b_{i}^{(x)} \in K$ which is a vector of degree $d$ over $\mathbb{Z}$ due to canonical embedding.


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