For any matrix $A \in Z^{n\times m}$, Let $$\wedge\_q(A)=\{ y\in Z^m\mathpunct{:}\exists s\in Z^n,\text{ s.t. }y=A^ts \pmod q \},$$ $$\wedge\_q^\bot(A)=\{x\in Z^m: Ax=0 \pmod q\}.$$ There is a result stating that $$q(\wedge\_q(A))^\ast=\wedge\_q^\bot(A),$$ where$$(\wedge\_q(A))^\ast=\{y\in R^m\mathpunct{:} (y,z)\in Z \text{ for any } z\in \wedge\_q(A)\}.$$
If given $x\in(\wedge\_q(A))^\ast$, how to prove $qx\in Z^m$?



About lattice $\pmod q$
$\Lambda\_q(A)$ is the set of vectors of the form $A^ts+qu$ where $s\in\mathbf{Z}^n$ and $u\in\mathbf{Z}^m$. If $x\in(\Lambda\_q(A))^\*$ then $(x,A^ts+qu)\in\mathbf{Z}$ for all choices of $s\in\mathbf{Z}^n$ and $u\in\mathbf{Z}^m$. Suppose $qx\notin\mathbf{Z}^m$. Then for some $1\le j\le m$, we have $qx\_j\notin\mathbf{Z}$. But this is impossible since $(x,A^ts)$ and $(x,A^ts+qe\_j)$ must both be integer.


