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

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$\Lambda_q(A)$ 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.

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You should say where you found this, and what $q$ is.

Meanwhile, this resembles Lemma 2(ii) in WATSON. The main differences are Watson has your $m=n$ and he allows your $q$ to be any positive integer, which may matter as far as a possible factor of 2. If you give more detail of why you think this is a theorem and why you care, I might put in more time.

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I'm sorry,thanks for your reference. –  tiansong Nov 12 '11 at 2:33
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