I'm looking for something like a Grassmannian, but which parameterizes the submodules of a module rather than the subspaces of a vector space. Most specifically, I'm looking for something which parameterizes the submodules of specifically $\mathbb{Z}^n$. So another way to say it is that I'm looking for a space parameterizing for the subgroups of a free abelian group. (A moduli space?)


I've seen some references to the concept of a "Grassmannian of submodules" here and there (like the papers on the first page of <https://www.google.com/search?q=%22grassmannian+of+submodules%22>) but can't figure out if this handles modules like $\mathbb{Z}^n$ or not.


Does anyone know if such an object exists and if so, how to construct it? Where I can get more information on this?


EDIT: to give a bit more info, the only specific application I really care about is parameterizing the free subgroups of a free abelian group, or the "lattices" in the $\mathbb{Z}$-module $\mathbb{Z}^n$, etc. A solution which works only for that, but which doesn't handle more exotic modules would be fine for my purposes. (And if it doesn't work out for all free abelian groups in general, then having a solution for at least finitely generated free abelian groups would even be a great start.)


I framed the question in terms of the "submodules of a module" in general just because I saw some references to there being a "Grassmannian of submodules" before, so I thought such a construction might be widely known.



Generalized Grassmannians that parameterize the submodules of a module
Building on Donu Arapura's answer and Qiaochu Yuan's comment:


If the quotient is torsion then $k$ changes as a functioin of the characteristic, so no $Grass\_{k,n}$ will do. If it is non-torsion then classifying the $\mathbb Z$-module is equivalent to classifying the $\mathbb Q$-subsoace it generates, so a point on the Grassmanian will be a fine enough invariant. It's also clear that every rational subspace contains a rank-k $\mathbb Z$-submodule whose quotient is torsion-free, so for the torsion-free case that is sufficient.


The next-simplest case is probably when the quotient is finite, since every other case can be viewed as a combination of this and the other one (by splitting the quotient into rank and torsion.) As Quiaochu points out, this is choosing a surjection $\mathbb Z^n\to A$, which is equivalent to choosing a set of $n$ generators of $A$. But since only the kernel of the map matters, and composing with an automorphism of $A$ doesn't change the kernel, you need to quotient out by the automorphism group of $A$. (The automorphism group of $A$ is not the image of $GL\_n(\mathbb Z)$, for instance $GL\_n(\mathbb Z) \to GL\_n(\mathbb F\_p)$ is not a surjection.)


For instance, there is only one sublattice of $\mathbb Z^n$ whose quotient is $(\mathbb Z/p)^n$: that being $p\mathbb Z^n$.


We can break this finite quotient into a product of $p$-torsion parts and consider those separately. If the $p$-torsion part is $(\mathbb Z/p)^k$, then $Grass\_{k,n}(\mathbb F\_P)$ will classify the choices of generators. If the torsion involves prime powers then I am not sure what to do.


But a deeper problem is that I don't think one can make a scheme over $\mathbb Z$ whose $\mathbb Z$-points have a natural bijection to the $\mathbb F\_p$ points of an arbitrary scheme over $\mathbb F\_p$ like $Grass\_{k,n}(\mathbb F\_p)$, other then silly stuff like the disjoint union of one copy of $\textrm{Spec} \mathbb Z$ for each point on the $\mathbb F\_p$-scheme.


