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Given the following theorem

(Smith normal form) Let $R$ be a PID and $A$ an $m \times n$ matrix over $R$. Then there exist square invertible $R$-matrices $P$ and $Q$ such that $A'=PAQ$ is a diagonal matrix of the form $$A'= \left( \begin{matrix} d_1 & & & 0 \\ & \ddots & & \vdots \\ & & d_r & 0 \\ 0 & \dots & 0 & 0 \end{matrix} \right),$$ where each diagonal element divides the following, $d_1 | d_2 | \dots | d_r$. Moreover, the elements $d_1,\dots,d_r$ are uniquely determined apart from some units.

I would like to solve following exercise.

Let $\phi: \mathbb{Z}^k \to \mathbb{Z}^k$ be a morphism given by left multiplication with a matrix $A$.

(i) Prove that $\mathrm{im}(\phi)$ has finite index if and only if $A$ is non-singular.

(ii) Show that in this case the index is equal to $|\det(A)|$.

My approach: by changing the bases one can bring $A$ in Smith normal form. I believe that one can see that $\mathbb{Z}^k/im(\phi) \cong \mathbb{Z}/d_1\mathbb{Z} \oplus \dots \oplus \mathbb{Z}/d_r\mathbb{Z} \oplus \mathbb{Z}^{k-r}$ but I don't have a rigorous argument yet. The matrices $P$ and $Q$ don't affect the absolute value of the determinant of $A$ since they are unimodular. Therefore if A is non singular ($r=k$), $\det(A)=d_1 \cdots d_r$.

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You need to show that $A$ is non-singular if and only if $A'$ has no $0$'s on the diagonal. Since $R$ is a PID, there are no $0$ divisors. Thus $\det(A')\neq 0$ if it has all nonzero entries on the diagonal.

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  • $\begingroup$ Yes I know but that's not really my problem. I think I have a problem with the fact that one chooses two bases for $\mathbb{Z}^k$. Isn't there a possibility to choose one basis for $\mathbb{Z}^k$ in which the matrix changes to Smith normal form? $\endgroup$ – Michiel Van Couwenberghe Jul 29 '14 at 12:16
  • $\begingroup$ @mvcouwen This isn't even generally true for operators on vector spaces - not every linear operator is diagonalizable. The best you can do with a linear map $A\colon V\to V$ ($V$ finite dimensional), if you only allow yourself to choose one basis for $V$, is to pick a basis so that the matrix of $A$ is in Jordan normal form. $\endgroup$ – mdp Jul 29 '14 at 12:26
  • $\begingroup$ But how does one prove that the quotient has the form that I've given in my question? $\endgroup$ – Michiel Van Couwenberghe Jul 29 '14 at 12:27
  • $\begingroup$ @mvcouwen You don't. Since $P$ and $Q$ are isomorphisms, the images of $A$ and $A'$ are isomorphic. Thus they have the same index. $\endgroup$ – Joe Johnson 126 Jul 29 '14 at 12:41
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    $\begingroup$ @mvcouwen Well, $A'$ is still a homomorphism from $\mathbb{Z}^k$ to $\mathbb{Z}^k$. So, it is clearly true for $A'$. $\endgroup$ – Joe Johnson 126 Jul 29 '14 at 12:49

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