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Crossposted to mathoverflow

Let $W$ be the finite $\mathbb{Z}$-module obtained from $\mathbb{Z}_q^n$ with addition componentwise. Let $V$ be a submodule of $W$. Let $V^{\perp} = \{w \in W \, : \, \forall v \in V \quad v.w = 0 \}$ where "." is the dot product. Is it true that ${(V^{\perp})}^{\perp} = V$?

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I would use the notation $V^{\perp}$. – Qiaochu Yuan Sep 12 '11 at 22:26
I don't think you waited long enough before crossposting. I would give it at least a day. – Qiaochu Yuan Sep 13 '11 at 2:04

This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Chris Godsil below.

The answer is yes. The easiest way for me is to appeal to character theory. If $\zeta$ is a complex $q$-th root of 1 then the map from $\mathbb{Z}_q^n$ to $\mathbb{C}$ given by $$ x \mapsto \zeta^{a^Tx},\qquad (a\in\mathbb{Z}_q^n) $$ is a character of the abelian group $W=\mathbb{Z}_q^n$. The set of characters obtained as $a$ varies over the elements of $W$ is the character group $W^*$ of $W$. If $V$ is a subgroup of $W$, then $V^\perp=V^*$ is isomorphic to the subgroup $(W/V)^*$ of $W^*$.

A convenient source for the relevant character theory is from our own KConrad: (under characters of finite abelian groups).

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