# Find $\mathbb{Z}$-basis of module over Dedekind domain provided its pseudobasis

Let $$K$$ be number field of degree $$d$$. Suppose we are given module $$\mathcal{M}$$ in form: $$$$\label{key} \mathcal{M} = v_1 \cdot \mathfrak{a}_1 \oplus v_2 \cdot \mathfrak{a}_2 \oplus \ldots \oplus v_{n}\cdot \mathfrak{a}_{n},$$$$ where all $$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 $$\mathfrak{a}$$?

• $\mathbb{ZZ}$ 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? Aug 9, 2022 at 17:29
• 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. Aug 9, 2022 at 18:39
• 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 $\begin{pmatrix} 2 & 1\\ 0 & 1 \end{pmatrix}$. Aug 9, 2022 at 18:39

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.