Integer points spanned by real, rational and integer combination of integer vectors

Question

We are given a rank $r$ matrix $B\in\Bbb Z^{k\times n}$ where $0\leq r\leq k\leq n$ holds.

We have $\mathcal L_\Bbb Z\subseteq \mathcal L_\Bbb Q\subseteq\mathcal L_\Bbb R$ where $$\mathcal L_\Bbb Z=\{uB\in\Bbb Z^n:u\in\Bbb Z^k\},\quad\mathcal L_\Bbb Q=\{uB\in\Bbb Z^n:u\in\Bbb Q^k\},\quad\mathcal L_\Bbb R=\{uB\in\Bbb Z^n:u\in\Bbb R^k\}$$ holds.

When can we have

1. $\mathcal L_\Bbb Z\subsetneq \mathcal L_\Bbb Q\subsetneq\mathcal L_\Bbb R$?

2. $\mathcal L_\Bbb Z\subsetneq \mathcal L_\Bbb Q=\mathcal L_\Bbb R$?

3. $\mathcal L_\Bbb Z= \mathcal L_\Bbb Q\subsetneq\mathcal L_\Bbb R$?

4. $\mathcal L_\Bbb Z= \mathcal L_\Bbb Q=\mathcal L_\Bbb R$?

Answer 1

We always have $\mathcal L_\Bbb Q=\mathcal L_\Bbb R:$ For any particular $v \in\Bbb Z^n$ consider how we determine $\{u \in \Bbb R^n \mid uB=v\}.$ The method, if the set is nonempty, will yield one or more solutions with $u \in \Bbb Q^n.$

The Smith Normal Form of $B$ is a certain integer diagonal matrix. I'd expect that we have $\mathcal L_\Bbb Z\subsetneq \mathcal L_\Bbb Q$ exactly when this matrix has a diagonal entry greater than $1.$

Answer 2

This is only a partial answer.

$\mathcal{L}_\mathbb{Q}=\mathcal{L}_\mathbb{R}$ for all $B$. Decompose $\mathbb{R}$ as a $\mathbb{Q}$-vector space into $\mathbb{Q}\oplus V$. Any $u\in\mathbb{R}^k$ can be written $u=q+v$ where $q\in\mathbb{Q}^k$ and $v\in V^k$. If $uB\in\mathbb{Q}^n$, then $vB=uB-qB\in\mathbb{Q}^n$, yet each of the $n$ entries of $vB$ is a $\mathbb{Z}$-linear combination of elements of $V$, so $vB\in\mathbb{Q}^n\cap V^n=0$. Hence $uB=qB$.