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Crossposted from math.stackexchange since I'm not getting any answer.

Let $W$ be the finite $\mathbb{Z}$-module obtained from $\mathbb{Z}_q^n$ with addition componentwise . where $\mathbb{Z}_q$ is the integers mod $q$. 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 ${(V^{\perp})}^{\perp} = V$ for all $q \geq 2$? If not, when is it the case?

3 deleted 86 characters in body

Crossposted from math.stackexchange since I'm not getting any answer.

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 ${(V^{\perp})}^{\perp} = V$as it would be the case in linear algebra (for example, it is true when $q$ is prime)?

2 edited title; added 2 characters in body

# Double orthogonal complement of a finite module

Crossposted from math.stackexchange since I'm not getting any answer.

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 $V^{\perp} = {w \{w \in W \, : \, \forall v \in V \quad v.w = 0 }$ \}$where "." is the dot product. Is it true that${(V^{\perp})}^{\perp} = V$as it would be the case in linear algebra (for example, it is true when$q\$ is prime)?

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