Let M be the $n\times m$ matrix representing the injection $A^m \to A^n$. Define D_i(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 D_r(M) has no annihilator (i.e. there is no nonzero element a∈A such that aD_r(M)=0); I think r is usually called the McCoy rank of M.

Choose a nonzero a∈A such that aD_r+1(M)=0. By assumption, $a$ does not annihilate D_r(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. Let M₁, ..., M_r+1 be the cofactors of the upper-left (r+1)-by-(r+1) minor obtained by expanding along the bottom row. Note, in particular, we know M_r+1 is the determinant of the upper-left r-by-r minor, so aM_r+1≠0.

The claim is that the vector (aM₁,...,aM_r+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.

Let M be the $n\times m$ matrix representing the injection $A^m \to A^n$. Define D_i(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 D_r(M) has no annihilator (i.e. there is no nonzero element a∈A such that aD_r(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 aD_r+1(M)=0. By assumption, $a$ does not annihilate D_r(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, ..., M_r+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 M_r+1 is the determinant of the upper-left r-by-r minor, so aM_r+1≠0.

The claim is that the vector (aM₁,...,aM_r+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.

Define D_i(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 D_r(M) has no annihilator (i.e. there is no element a∈A such that aD_r(M)=0); I think r is usually called the McCoy rank of M.

Choose an a∈A such that aD_r+1(M)=0. By assumption, $a$ does not annihilate D_r(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. Let M₁, ..., M_r+1 be the cofactors of the upper-left (r+1)-by-(r+1) minor obtained by expanding along the bottom row. Note, in particular, we know M_r+1 is the determinant of the upper-left r-by-r minor, so aM_r+1≠0.

The claim is that the vector (aM₁,...,aM_r+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-right (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 $a$.

Let M be the $n\times m$ matrix representing the injection $A^m \to A^n$. Define D_i(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 D_r(M) has no annihilator (i.e. there is no nonzero element a∈A such that aD_r(M)=0); I think r is usually called the McCoy rank of M.

Choose a nonzero a∈A such that aD_r+1(M)=0. By assumption, $a$ does not annihilate D_r(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. Let M₁, ..., M_r+1 be the cofactors of the upper-left (r+1)-by-(r+1) minor obtained by expanding along the bottom row. Note, in particular, we know M_r+1 is the determinant of the upper-left r-by-r minor, so aM_r+1≠0.

The claim is that the vector (aM₁,...,aM_r+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.