I am trying to find an elementary proof that if $\phi$ is a linear map from $\mathbb{Z}^n\rightarrow \mathbb{Z}^m$ represented by an $m \times n$ matrix $A$, then the map is surjective iff the gcd of the determinants of all the $m\times m$ minors of $A$ is $1$.
I know that for there to be surjectivity between $\mathbb{Z}^n$ and $\mathbb{Z}^m$ $n$ must be greater than or equal to $ m$ and for there to even be $m \times m$ minors $n$ again must be greater than or equal to $ m$, so I just assume this throughout.
I sort of have one direction $\Leftarrow$
i) Greatest Common Divisor =1 implies surjectivity: First observe that the if $ | \mathbb{Z}^m/ Im(M)| < \infty$ then $|\det M| = | \mathbb{Z}^m/ Im(M) |$ otherwise $\det(M) = 0$ where $M$ is an $m\times m$ matrix. We can consider the $n$ columns of $A$ as column vectors $v_1, v_2, \ldots, v_n$. These $n$ column vectors live in $\mathbb{Z}^m$. Let $S'' = \{ v_i\}$ and then let $S'$ be subsets of $S''$ of cardinality $m$ and lastly let $S$ be the elements of $S'$ such that when the $m$ $v_i$ vectors are considered as $m\times m$ matrices, the determinant is not zero, thus $S$ consists of all $m\times m$ minors of $A$ with non-zero determinant (we ignore zeroes since they do not affect gcd). For each $s\in S$ define a map $i_s: \mathbb{Z}^m \rightarrow \mathbb{Z}^n$ that maps the standard basis of $\mathbb{Z}^m$ to the basis elements $e_k \mathbb{Z}^n$ such that $v_k \in s$. That is, $\phi \circ i_s$ gives the matrix created by the column vectors of $s$. Let $\Lambda$ be the lattice Im$\phi \supset \sum_{s\in S}$ Im $\phi\circ i_s =\sum_{s \in S} \Lambda_s$. Thus $\forall s \in S$, $\Lambda_s \subset \Lambda \subset \mathbb{Z}^m$. Thinking in terms of group theory, we have that $\Lambda$ is a subgroup of $\mathbb{Z}^m$ and all the $\Lambda_s$ are subgroups of $\Lambda$. Thus by Lagrange's Theorem, we have $|\mathbb{Z}/\Lambda| \Big\vert |\mathbb{Z}^m/\Lambda_s|$ Since $|\mathbb{Z}^m/\Lambda_s|$ are the determininants of the $m\times m$ minors and the definition of the common divisor of several integers is the greatest positive integer dividing all of them. Thus by hypothesis $|\mathbb{Z}/\Lambda| \leq 1$ and so $|\mathbb{Z}/\Lambda| =1$ and we have that Im$A=\Lambda = \mathbb{Z}^m$ so the map is surjective.
I was hoping to get a more elementary proof that doesn't rely on the observation that the if $ | \mathbb{Z}^m/ Im(M)| < \infty$ then $|\det M| = | \mathbb{Z}^m/ Im(M) |$ otherwise $\det(M) = 0$ where $M$ is an $m\times m$ matrix or normal forms.
Thanks!