Question on determinants of matrices changing between integer matrices [duplicate]

The following problem came up from a though I had while reading:

Let's say we have $M=\mathbb{Z}^n$ and we have another free $\mathbb{Z}$-module, $N$, inside of $M$ also with rank $n$.

We know we can make a matrix $A$ that changes $N$ to its representation in $M$ (ie has as columns the basis $N$ expressed in coordinates coming from the basis of $M$). If the index of $N$ in $M$ is $m$, I am pretty sure just from tinkering that the determinant of $A$ should also be $m$, however, I have not been able to show this.

It is unclear to me how to proceed. I though maybe I could do something related to the points in $\mathbb{Z}^n$ that are within the unit box of $N$, but I cannot really make sense of this.

Thank you for any help or direction.

Solution 1:

For a $\mathbb{Z}$-linear map $f : \mathbb{Z}^n \to \mathbb{Z}^n$ we have either $\mathrm{det}(f)=0$ and the cokernel $\mathrm{coker}(f) := \mathbb{Z}^n / \mathrm{im}(f)$ is countably infinite, or $\mathrm{det}(f) \neq 0$ and the cokernel has order $|\mathrm{det}(f)|$.

Proof. The Smith Normal Form shows that $f$ is represented (in appropriate bases) by a diagonal matrix $\mathrm{diag}(e_1,\dotsc,e_n)$, so that $\mathrm{coker}(f) \cong \mathbb{Z}/e_1 \oplus \dotsc \oplus \mathbb{Z}/e_n$, which has order $e_1 \cdot \dotsc \cdot e_n = \mathrm{det}(f)$ if all $e_i \neq 0$, and otherwise is countably infinite. $\square$