# Given a non-singular rational matrix $Q \in \mathbb Q^{n\times n}$, what is $Q(\mathbb Z^{n}) \cap \mathbb Z^{n}$?

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.

• You can proceed from $Q(\mathbb{Q})$ to $Q(\mathbb{Z})$ by factorizing 1/LCM(denominators of entries of matrix $Q$)... Commented May 10, 2023 at 8:19

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 $$Q^{-1}\in M_n(\mathbb{Z})$$.