Let $A$ be a commutative ring with $1$ not equal to $0$. (The ring A is not necessarily a domain, and is not necessarily noetherianNoetherian.) 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 $A^2 \to A^1$? Say it's represented by a matrix (a_1 a_2)$(a_1, a_2)$. Then clearly (a_2, -a_1)$(a_2, -a_1)$ is in the kernel. In the A^{n+1} -> A^{n}$A^{n+1} \to A^{n}$ case, we can look at the n x (n+1)$n \times (n+1)$ matrix which represents it; call it M$M$. Let M_i$M_i$ denote the determinant of the matrix obtained by deleting the ith$i$-th column. Let v$v$ be the vector (M_1 -M_2 ... (-1)^nM_{n+1})$(M_1, -M_2, ..., (-1)^nM_{n+1})$. Then v$v$ is in the kernel of our map, because the vector M*v^T$Mv^T$ has ith$i$-th component the determinant of the (n+1) x (n+1)$(n+1) \times (n+1)$ matrix attained from M$M$ by repeating the ith$i$-th row twice.
That almost finishes the proof, except it is possible that v$v$ is the zero vector. I would like to see either this argument finished, or, even better, a nicer proof.
Thank you!
