Question: Given a non-singular rational matrix $Q \in \mathbb Q^{n\times n}$, is there any necessary condition such that $Q(\mathbb Z^{n}) \supseteq \mathbb Z^{n}$.
My thoughts so far: When Q is an integer matrix, it's clear that $Q(\mathbb Z^{n}) = Z^{n}$ if and only if $Q \in GL(n, \mathbb{Z})$. I am not sure if we can say something similar for rational matrices. For example, take $Q=\begin{bmatrix} 0 & 2 \\ 0.5 & 0 \end{bmatrix} \in GL(n, \mathbb{Q})$. This matrix has finite order, the characteristic polynomial has integer coefficients. However, $Q(\mathbb Z^{n}) \nsupseteq \mathbb Z^{n}$.
Any reference/ideas would be really appreciated.
Let $\{e_1,...,e_n\}$ and $\{f_1,...,f_n\}$ two bases of $\mathbb{Q}^n$ and put $L=\bigoplus_{i=1}^n\mathbb{Z}e_i$ and $M=\bigoplus_{i=1}^n\mathbb{Z}f_i$. Then we have $L\supset M$ precisely when the $f_i$ are linear combination with coefficients in $\mathbb{Z}$ of the $e_i$.
Let's apply this to $M=\mathbb{Z}^n$ with the standard basis and $L=Q(M)$. Then as $e_i$ we can take the columns of $Q$ and to compute back the $f_i$ from the $e_i$ you need the inverse matrix $Q^{-1}$ then the condition is that $Q^{-1}\in M_n(\mathbb{Z})$.