Simplest proof/greatest generality in which $|R/(\det A) R| = |R^n/A R^n|$

Question

$\def\ZZ{\mathbb{Z}}$Let $A$ be an $n \times n$ integer matrix with nonzero determinant. Then the abelian group $\ZZ^n / A \ZZ^n$ has order $|\det A|$. I've known this for ages, but I just realized I don't know a really clean proof. For example, it can be deduced by putting $A$ into Smith Normal Form, but that's surely overkill.

Question 1 What is a quick elementary proof that $|\ZZ^n/A \ZZ^n| = |\det A|$?

We can rewrite this as $|\ZZ^n/A \ZZ^n| = |\ZZ/(\det A) \ZZ|$ and, in this formulation, it is a meaningful equation for any commutative ring $T$. Using Smith Normal Form, it is true for PID's and, using localization followed by Smith Normal Form, it is true for Dedekind domains. It is probably not reasonable to ask this question unless $R$ is a finite quotient domain, meaning an integral domain where $R/a R$ is finite for any nonzero $a \in R$.

Question 2 What is the most general class of commutative rings in which $|R^n/A R^n| = |R/(\det A) R|$?

For example, does $\ZZ[\sqrt{8}]$ have this property?

Answer 1

Here's a proof that likely has some generalization; I saw it on another similar post on this site which only asked Q1.

For $g$ with nonzero determinant, write $f(g) = |\mathbb Z^n/g\mathbb Z^n|$.

Note that $f$ is multiplicative : indeed if $g,h$ have nonzero determinant, $\mathbb Z^n /g\mathbb Z^n \cong (\mathbb Z^n/ gh\mathbb Z^n)/(g\mathbb Z^n/gh\mathbb Z^n)$ and $g\mathbb Z^n/gh\mathbb Z^n\cong \mathbb Z^n/h\mathbb Z^n$ so we have a short exact sequence $$0\to \mathbb Z^n/h\mathbb Z^n \to \mathbb Z^n/gh\mathbb Z^n \to \mathbb Z^n/g\mathbb Z^n\to 0$$

which implies $f(gh) = f(g)f(h)$.

Now extend this to invertible $\mathbb Q$- matrices by putting $f(g) := \frac{1}{k^n }f(kg)$ for $k$ such that $kg \in M_n(\mathbb Z)$.

Note that this is well defined : indeed if $kg, lg$ both have integer coefficients, then so does $lkg$ and by considering $lI_n$ and $kI_n$ we have that $f(lkg) = f(lI_n) f(kg) = f(kI_n) f(lg)$ and we can easily compute $f(mI_n)$ for any integer $m$.

Then $f : GL_n(\mathbb Q)\to \mathbb Q^\times$ is well-defined and multiplicative.

Now we can easily compute $f$ on elementary matrices, for which the result is clear, so since they generate $GL_n(\mathbb Q)$ we get the result for all matrices, in particular for integer-coefficient matrices.

The question is now : how far can this proof method get us ? I will stick with integral domains in what follows so I will not say it each time, but $R$ is always assumed to be integral.

Note that if at first I allowed infinite cardinals, I wouldn't really need to know beforehand that $\mathbb Z^n/g\mathbb Z^n$ is finite. I do use some finiteness assumptions in two places :

1- to define $f$ on $GL_n(\mathbb Q)$. I divide by $k^n$ which should reasonably be replaced with $|R/aR|^n$ for $a \in R$ such that $ag\in M_n(R)$ (taking $g\in GL_n(Frac(R))$), so for this to make sense I need to assume that $R/aR$ is finite.

Then the argument $f(lI_n)f(kg) = f(kI_n) f(lg)$ makes sense even in the presence of infinite cardinals : $f(lI_n), f(kI_n)$ aren't zero anyway, and they're finite by the assumption I just made, so that the equality imposes that $\frac{1}{|R/kR|^n} f(kg) = \frac{1}{|R/lR|^n} f(lg) $ holds even if $f(lg)$ or $f(kg)$ is infinite (if one of them is, both are, and they are thus equal, by nonzeroness) [of course I should add an "obvious convention" : an infinite cardinal divided by a finite number is defined to be the infinite cardinal itself]

2- To say that $f$ takes values in $\mathbb Q^\times$. But if you think about it, this one is not really necessary : indeed with size assumptions on $|Frac(R)|$ or $n$ (something like $\geq 2$ or $3$ perhaps, $4$ to be sure), $SL_n(Frac(R))$ is generated by matrices of the form $I_n + \alpha E_{ij}, i\neq j$ for which we can explicitly compute $f$, and see that it is indeed $1$.

Thus, by Gaussian elimination, we only need to compute $f$ on matrices of the form $\mathrm{diag} (1, ..., 1, d)$, for which we can, again, explicitly compute and see that we get the desired result (which is $|R/\det(A)R|$ as you suggested)

So unless I'm mistaken this proof works for any finite quotient domain. Now would there be a way to extend this to non integral rings with finite quotients ? I don't know , and am currently too tired to think about it.

EDIT : As noted by Lord Shark in the comments below, another possible generalization is to keep integral domains but remove the finiteness hypothesis. In which case, $f$ may take values in the Grothendieck group of finitely generated torsion $R$-modules, and the same proof goes mutatis mutandis to show that $f(g)$ is the class of $R/\det(g)R$. Of course for $\mathbb Z$ this Grothendieck group is $\mathbb Q^\times_+$ so we get the same thing as before. I still don't see how to make this interesting for nonintegral rings. One would first have to see which kind of matrices one wants to consider, and then perhaps one could do something with some localizations or quotient maps ?

Answer 2

How about this?

Let $\Lambda =A\Bbb Z^n$. Consider the natural map $\pi:\Bbb R^n/\Lambda\to\Bbb R^n /\Bbb Z^n$. This is a covering map with $N$ sheets where $N=|\Bbb Z^n:\Lambda|$. But $\pi$ is locally volume-preserving. The volume of the manifolds $\Bbb R^n/\Lambda$ and $\Bbb R^n/\Bbb Z^n$ are $|\det A|$ and $1$ respectively. So $N$ must equal $|\det A|$.