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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? |
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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. 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)? |
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Double orthogonal complement of a finite moduleCrossposted 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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