Dual lattices up to a q scaling factor

Question

In this paper : https://eprint.iacr.org/2011/501.pdf There is an equality page 10, in the second paragraph considered by the authors as "easy to check". If someone could explain to me why the set at the left side is included in the set at the right side, I will be the happiest man on earth. All the definitions are given in the same page.

Answer 1

We prove $\frac{1}{q}\Lambda(A^t)=\Lambda^\perp(A)^*$.

(1) $\frac{1}{q}\Lambda(A^t)\subset\Lambda^\perp(A)^*$: If $z=A^ts$ (mod $q$) then for any $y\in\Lambda^\perp(A)$, $\langle \frac{1}{q}z,y\rangle\in\frac{1}{q}\langle A^ts,y\rangle+\mathbb{Z}=\frac{1}{q}\langle s,Ay\rangle+\mathbb{Z}\in\mathbb{Z},$ since $Ay=0$ (mod $q$). So, $\frac{1}{q}z\in\Lambda^\perp(A)^*$.

(2) $ \Bigl(\frac{1}{q}\Lambda(A^t)\Bigr)^*\subset\Lambda^\perp(A)$: Suppose $\langle y,\frac{1}{q}\Lambda(A^t)\rangle\in\mathbb{Z}$. Then $\langle y,A^ts\rangle=\langle Ay,s\rangle=0$ (mod $q$), for all $s$, and so $Ay=0$ (mod $q$).

Finally, $\Lambda^\perp(A)^*\subset\frac{1}{q}\Lambda(A^t)$ follows from (2) and two simple general facts about dual lattices: 1) $\Lambda\subset\Gamma$ iff $\Gamma^*\subset\Lambda^*$ and 2) $\Lambda^{**}=\Lambda$.