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I'm trying to understand the concept of dual lattices by solving a problem:

Prove that the dual lattice of $(\Bbb N_0,\text{divisibility})$ is isomorphic with the lattice $\text{sub}(\Bbb Z)$ where $(\Bbb Z,+)$ is the group of all integers.

I'm not really sure how the dual lattice is found for lattices like these (in order theory). I only know the dual lattice definition (in group theory):

The dual of a lattice $\wedge$ is the set $\hat{\wedge}$ of all vectors $\mathbf{x}\in\text{span}(\wedge)$ such that $\langle \mathbf{x,y} \rangle$ is an integer for $\mathbf{y}\in\wedge$

which doesn't seem to be applicable here.

Any idea how to find the dual lattice and then show the isomorphism?

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1 Answer 1

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We obtain the dual lattice if we simply reverse its partial order, or equivalently, if we exchange the lattice operations $\land$ and $\lor$.

For the statement, you basically have to prove that all subgroups of $\Bbb Z$ are of the form $n\Bbb Z$ for some $n\in\Bbb Z$.

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  • $\begingroup$ You mean I have to replace the partial order relation from "divides" to "is divisible by"? $\endgroup$
    – user563280
    Commented May 23, 2018 at 0:01
  • $\begingroup$ Yes. But you don't have to, just observe that the ordering is reversed by the mapping you define. $\endgroup$
    – Berci
    Commented May 23, 2018 at 0:03

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