Quotient group $\mathbb Z^n/\ \text{im}(A)$

Question

Let $A$ be an $n \times n$ matrix with integer coefficients and nonzero determinant. Can we say something about $ \mathbb{Z}^n /\ \text{im}( \phi )$ (here $\phi : v \mapsto Av$ )?

This problem has arised as I was solving some problem in homology theory.

Answer 1

$[\mathbb{Z}^n:\mathrm{im}\,\phi]=|\det A|$. The image under $\phi$ of the lattice generated by the standard basis in $\mathbb{Z}^n$ is another lattice with unit cell having volume $|\det A|$. Since every triple of basis elements in $\mathbb{Z}$ is linearly independent and $A$ is nonsingular, their images are also, and no lattice reductions can occur.

Answer 2

This can be seen from applying Smith's normal form to the matrix $A$ to get an equivalent matrix $$A'={\rm diag}\;(d_1,\ldots,d_n)$$

Then each $d_i\neq 0$ since the determinant is nonzero. Now one can produce distinct linear combinations $\lambda_1 \overline{e_1}+\cdots+\lambda_n\overline{e_n}$ in $d_1d_2\cdots d_n=\det A$ ways, and one can show these are all, which gives Eric's result.