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 -> 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} -> A^{n} case, we can look at the n x (n+1) matrix which represents it; call it M. Let M_i denote the determinant of the matrix obtained by deleting the ith 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 M*v^T has ith component the determinant of the (n+1) x (n+1) matrix attained from M by repeating the ith 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!