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!



Atiyah-MacDonald, exercise 2.11
Let M be the $n\times m$ matrix representing the injection $A^m \to A^n$. Define Di(M) to be the ideal generated by the determinants of all i-by-i minors of M. Let r be the largest possible integer such that Dr(M) has no annihilator (i.e. there is no nonzero element a∈A such that aDr(M)=0); I think r is usually called the *McCoy rank* of M.


Assume that $m>n$. We shall get a contradiction.


Choose a nonzero a∈A such that aDr+1(M)=0. By assumption, $a$ does not annihilate Dr(M), so there is some r-by-r minor that is not killed by $a$; we may assume it is the upper-left r-by-r minor. Thus, $r \leq n$, so that $r+1 \leq n+1\leq m$. Let M1, ..., Mr+1 be the cofactors of the upper-left (r+1)-by-(r+1) minor obtained by expanding along the bottom row. (This is well-defined even if the $r+1$-th row of $M$ does not exist, because these cofactors use only the first $r$ rows and the first $r+1$ columns of $A$, and $A$ has both since $r \leq n$ and $r+1 \leq m$.) Note, in particular, we know Mr+1 is the determinant of the upper-left r-by-r minor, so aMr+1≠0.


The claim is that the vector (aM1,...,aMr+1,0,0,...) (which we've already shown is non-zero) is in the kernel of M. To see that, note that when you dot this vector with the k-th row of M, you get $a$ times the determinant of the matrix obtained by replacing the bottom row of the upper-left (r+1)-by-(r+1) minor with the first r+1 entries in the k-th row. If k≤r, this determinant is zero because a row is repeated, and if k>r, this determinant is the determinant of some (r+1)-by-(r+1) minor, so it is annihilated by $a$. But since A is injective, the kernel of M is 0, and we have a contradiction.


