The group $\mathbb{Z}^n/A\mathbb{Z}^n$ is finite iff the determinant of A is nonzero

Question

Let $n\geq 1$ and $A$ be an $n\times n$ integral matrix with columns $c_1,\ldots, c_n\in\mathbb{Z}^n$. Write $A\mathbb{Z}^n$ as the set of all integral linear combinations of the columns of A, i.e., $$A\mathbb{Z}^n=\{x_1c_1+\cdots+x_nc_n: x_1, \ldots, x_n\in\mathbb{Z}\}. $$ The problem is to prove that the group $G=\mathbb{Z}^n/A\mathbb{Z}^n$ is a finite group if and only if $\det(A)\neq0$, and moreover that in the case $\det(A)\neq0$, the order of $G$ is precisely $|\det(A)|$.

The case when $\det(A)$ is a unit ($\pm1$) is somewhat trivial. But how about other nonzero values for $\det(A)$? I tried to define a homomorphism $f:\mathbb{Z}^n\to\mathbb{Z}^n, x\mapsto Ax$ for $x\in\mathbb{Z}^n$, and observed that $G$ is the quotient $\mathbb{Z}^n/\operatorname{im}(f)$. How can we proceed?

Thanks.

Answer 1

Assume first that $\det A = d \ne 0$. Using Cramer we see that $d \mathbb{Z}^n\subset A \cdot \mathbb{Z}^n$. So $\mathbb{Z}^n/A\mathbb{Z}^n$ is of order at most $d^n$.

Assume $\det A =0$. Then there exists $r\in \mathbb{Z}^n$, $r\ne 0$, such that $r^t\cdot A=0$. Consider the map $\mathbb{Z}^n\to \mathbb{Z}$, $c\mapsto r^t\cdot c$. The map has infinite image and the kernel of the map contains $A\mathbb{Z}^n$. We conclude that $\mathbb{Z}^n/A\mathbb{Z}^n$ is infinite.

Obs: In fact one can show that if $\det A\ne 0$ then the order of $\mathbb{Z}^n/A\mathbb{Z}^n$ is $|d|$. We can show this with geometry as follows:

Consider the lattice $L= A\mathbb{Z}^n$ inside the lattice $\mathbb{Z}^n$. Consider a basis $c_1$, $\ldots$, $c_n$ of $L$. The volume of the parallelotope $P$ consisting of $ \sum t_i c_i$, with $0\le t_i< 1$ equals on one hand $|\det A|$, on the other hand it equals the limit $$\lim_{N\to \infty} \frac{a_N}{N^n}$$ where $a_N$ is the cardinality of $P \cap \frac{1}{N}\mathbb{Z}^n$. However, $a_N$ also equals the cardinality of $N P \cap \mathbb{Z}^n$, and this equals the order of $\mathbb{Z}^n/N L$. Therefore we have $$|\det A |= \lim_{N\to \infty}\frac{|\mathbb{Z}^n/N L|}{N^n}=\lim_{N\to \infty}\frac{|\mathbb{Z}^n/N L|}{|\mathbb{Z}^n/N\mathbb{Z}^n|}=\lim |N\mathbb{Z}^n/NL|=\lim |\mathbb{Z}^n/L|= |\mathbb{Z^n}/L|$$

Answer 2

The volume of the parallelepiped spanned by the columns of $A$ has volume $|\det A|$ and contains exactly one representative of each coset. Thus the vloume is approximately the combined volume of the unit cubes cntered at these representatives. The approximation gets (reletively) better if you take a large number of copies of the epiped...

And of course, $\det A=0$ means that the column vectors do not span the full space, henc ewe find a lattice point iutside o f the span; this has infinite order also in the quotient.

Answer 3

This follows from the Smith normal form, which is an algorithmic procedure for writing $A=PDQ$, where $P,Q,D$ are integer matrices, $P,Q$ invertible, and $D$ diagonal. It follows that $|\det(A)|=|\det(D)|$ and $\mathbb{Z}^n/A\mathbb{Z}^n \cong \mathbb{Z}^n/D\mathbb{Z}^n$.

I'd love to see an elementary proof along the lines below. I wish I could make that argument work better.

Let $E$ be the canonical basis of $\mathbb Z^n$. Let $V$ be the image of $E$ under $A$. Then $V=AE$ and $\text{adj}(A) V = \text{adj}(A)AE = \det(A)IE=\det(A)E$ and so $\det(A)E \subseteq A\mathbb{Z}^n$. Therefore, if $\det(A)\ne0$, then $\mathbb{Z}^n/A\mathbb{Z}^n$ is finite and has size at most $|\det(A)|^n$.