Let $A$ be a commutative ring with $1$ not equal to $0$.  (The ring A is not necessarily a domain, and is not necessarily Noetherian.)  Assume we have an injective map of free $A$-modules $A^m \to A^n$.  Must we have $m \le n$?

I believe the answer is yes.  For instance, why is there no injective map from  $A^2 \to A^1$?  Say it's represented by a matrix $(a_1, a_2)$.  Then clearly $(a_2, -a_1)$ is in the kernel.  In the $A^{n+1} \to A^{n}$ case, we can look at the $n \times (n+1)$ matrix which represents it; call it $M$.  Let $M_i$ denote the determinant of the matrix obtained by deleting the $i$-th column.  Let $v$ be the vector $(M_1, -M_2, ..., (-1)^nM_{n+1})$.  Then $v$ is in the kernel of our map, because the vector $Mv^T$ has $i$-th component the determinant of the $(n+1) \times (n+1)$ matrix attained from $M$ by repeating the $i$-th row twice.

That almost finishes the proof, except it is possible that $v$ is the zero vector.  I would like to see either this argument finished, or, even better, a nicer proof.

Thank you!