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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.

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    $\begingroup$ Set $\langle v,u\rangle = e^{2\pi i(v \cdot u)/q}$. Then $v \cdot u = 0 \bmod q$ if and only if $\langle v,u\rangle = 1$. Each character of the finite abelian group $({\mathbf Z}/q{\mathbf Z})^n$ has the form $v \mapsto \langle v,u\rangle$ for a unique $u \in ({\mathbf Z}/q{\mathbf Z})^n$. (The term used here is that the group $({\mathbf Z}/q{\mathbf Z})^n$ is "self-dual": it's isomorphic to its own character group.) In any finite abelian group $G$ we have the property $(H^{\perp})^{\perp} = H$ for all subgroups $H$ of $G$. Try Terras, "Fourier Analysis on Finite Groups and Applications". $\endgroup$
    – KConrad
    Commented Nov 29, 2013 at 13:08

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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.

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    $\begingroup$ It's definitely written down somewhere. This is a special case of a fact about characters of finite abelian groups. See, for instance, chapter 2 of Larry Washington's Introduction to Cyclotomic Fields. $\endgroup$
    – KConrad
    Commented Nov 29, 2013 at 12:37
  • $\begingroup$ Thanks @KConrad, that's very helpful. If I understand correctly, I believe you are referring to Proposition 3.4 in Introduction to Cyclotomic Fields. $\endgroup$
    – David Wilding
    Commented Nov 29, 2013 at 13:47
  • $\begingroup$ I don't have Washington's book on me at the moment and can't check it in Google books, so I can't confirm (or deny) that's right just from the proposition number. Can you write down the content of that proposition? $\endgroup$
    – KConrad
    Commented Nov 29, 2013 at 13:54
  • $\begingroup$ Proposition 3.4 itself says that $(H^\perp)^\perp=H$, and the preceding setup is that $H$ is a subgroup of a finite abelian group $G$. $H^\perp$ is defined to be $\{\chi\in\widehat G:\chi(h)=1,\forall h\in H\}$, so I think this matches up with your comment on the question. $\endgroup$
    – David Wilding
    Commented Nov 29, 2013 at 14:02
  • $\begingroup$ OK. And earlier in that chapter Washington should have shown that $\widehat{A \times B} \cong \widehat{A} \times \widehat{B}$. Coupling the way this isomorphism works with the standard isomorphism of ${\mathbf Z}/q{\mathbf Z}$ with its character group by $a \bmod q \mapsto [b \bmod q \mapsto e^{2\pi iab/q}]$ shows $({\mathbf Z}/q{\mathbf Z})^n$ is isomorphic to its character group by the mapping I wrote down in my comment to Michael's question. $\endgroup$
    – KConrad
    Commented Nov 29, 2013 at 14:12
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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.

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