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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.

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Well, the determinant can also be seen (up to sign) as the volume of the parallelepiped generated by the basis for $N$. And the index is some count of the points $M$ on or inside that parallelepiped, with some effort made to avoid double-counting the points that on the borders. So you can express this relationship purely geometrically. Not sure if that helps. –  Thomas Andrews Feb 28 '13 at 18:21
    
Possible duplicate of math.stackexchange.com/questions/245733/… –  user26857 Feb 28 '13 at 18:53

1 Answer 1

up vote 1 down vote accepted

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. $\square$

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