Looking for a reference: double orthogonal complement in $(\mathbb{Z}/q\mathbb{Z})^n$ I'm using the following result in a computer science paper:
Let $V$ be a submodule of $(\mathbb{Z}/q\mathbb{Z})^n$ (n-tuples with addition and multiplication mod $q$). Let
$$V^\perp = \{u \in (\mathbb{Z}/q\mathbb{Z})^n : \forall v \in V \quad v \cdot u = 0\}$$
where $v \cdot u = v_1 u_1 + \ldots + v_n u_n \text{ mod } q$.
Propositon: $(V^\perp)^\perp = V$.
I can't find any reference with this exact result even though I'm quite sure it is basic, mostly because I'm not from the mathematics community. I'd be glad to have a reference for $q$ prime, even though I'd rather have the more general result. Do any of you know a good reference for this result? Thank you very much.
 A: The results remains true for lattices in locally compact abelian groups, for example finite abelian groups. 
Let $A$ be an l.c.a. group and $\hat{A}$ its dual group, let $B \subset A$ be a subgroup.
Then you can define $B^\bot= \{ \psi \in \widehat{A}: \psi|_{B} = 1 \}$ (the so called dual lattice). You can do this again an obtain $( B^\bot)^\bot \subset \hat{\hat{A}}$. Now, $\hat{\hat{A}}$ is isomorphic to $A$ and gives an identification $(B^\bot)^\bot = B$.
Let's get to your specific example. The submodule structure gives rise to an isomorphism $A\cong \hat{A}= Z/qZ^n$. Note that every additive character $u \mapsto \psi_u$ given as $\psi_u(x) = \exp(2 \pi i u x/q)$ embeds $Z/qZ^n$ in to the space of one-dimensional additive characters (Pontriyagin dual).
Now, $V^\bot = \{ u \in V : \psi_u|_V = 1 \}$ is a dual lattice.
A: As explained in this answer, the property $V=(V^\perp)^\perp$ holds for submodules $V$ of $(\mathbb Z/q\mathbb Z)^n$ because you can use Smith normal form to find a suitable generating matrix for $V^\perp$. If you just want a reference for the result, I can only suggest the following paper:
D. Wilding, M. Johnson, M. Kambites. Exact rings and semirings.
Journal of Algebra, 388 (2013), 324–337; arXiv:1212.5358.
When writing this paper we faced the same problem as you. That $(V^\perp)^\perp=V$ follows quite easily from known results, but we could not find it written down explicitly in this formulation.
