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First, I'm talking about lattices like $\mathbb{Z}^n$, not the order theory one.

The following definition is used frequently within Cryptography (definition from here):

Let $\beta_i\in\mathbb{R}^d$, and let $B = [\beta_1,\dots,\beta_k]\in\mathbb{R}^{d\times k}$. The lattice generated by $B$ is defined as the set: $$\mathcal{L}(B) = \{Bx : x\in\mathbb{Z}^k\} = \text{span}_{\mathbb{Z}}(\beta_1,\dots,\beta_k)$$

My understandng is that this definition is essentially "A lattice is a $\mathbb{Z}$-module", which seems like a fine definition.

My issue is that it's well-known that $\mathbb{Z}$-modules and abelian groups are equivalent (see here), which suggest that the above definition is "A lattice is an abelian group". This suddenly seems wrong (and looking at the definition here, it appears that it is wrong), as it includes things like $(\mathbb{C}^\times,\times)$ and $(\mathbb{R},+)$, which aren't discrete, and "feel" like they're missing a fundamental property to be a lattice.

It could very well be that $B = [\beta_1,\dots,\beta_k]$ being finite is the missing element here, which would suggest the definition of:

A lattice is a finitely-generated abelian group

This could be a reasonable definition, but I'd like to check with others before I internalize it.

So, what is the correct way to define a lattice as an abelian group?

Mark Schultz-Wu
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    Every integer lattice is a $\mathbb{Z}$-module, however not every $\mathbb{Z}$-module is an integer lattice so your proposed definition does not work. (To see this, note that all non-zero elements in an integer lattice have infinite order; however, $\mathbb{Z}$-modules can have elements of finite order.) It may be possible to fix this by placing additional conditions on the group. An obvious condition is that the group cannot have any elements of finite order. – Qudit Feb 23 '18 at 07:32
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    I recall seeing it defined as a finitely generated free $\mathbb{Z}$-module (which just means it is isomorphic to $\mathbb{Z}^n$ for some $n$). – Tobias Kildetoft Feb 23 '18 at 08:41

1 Answers1

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As Tobias Kildetoft says. A lattice is the same thing as a finitely generated free $\mathbb{Z}$-module, i.e. a finitely generated free abelian group.

This is equivalent to the definition you gave:

  • If I have a free $\mathbb{Z}$-module it is isomorphic to $\mathbb{Z}^d$ for some $d\in{\mathbb N}$. It is clear that $\mathbb{Z}^d\subset \mathbb{R}$ is $\mathcal{L}(B)$, where $B$ is the standard basis of $\mathbb{R}$.

  • Conversely, every $\mathcal{L}(B)$ is clearly finitely generated by definition. To see that it is free, observe that it is torsion-free (as a sub-$\mathbb{Z}$-module of $\mathbb{R}^d$ which is torsion-free). It is a general fact that torsion-free finitely generated $\mathbb Z$-modules are free.

Tashi Walde
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