Difficulty with Matsumura Exercise 2.4 in Chapter 1 The exercise is as follows:

Let $C=(c_{ij})$ be an $n\times m$ matrix over $A$, and suppose that $C$ has a nonzero $r\times r$ minor, but that all $(r+1)\times(r+1)$ minors are 0. Show then that if $r<m$, the $m$ column vectors of $C$ are linearly dependent.

We can assume $m=r+1$. I have no idea where to start on this. If $A$ is a domain and $n=m$, then we the result follows from $\det(C)=0$, but if $A$ is not a domain I don't know whether this holds. Regardless, I have no idea how to approach the case of a nonsquare matrix.
I have tried looking for examples of when $A$ is not a domain and $\det(C)=0$ but the columns of $C$ are not linearly dependent, and I have found nothing whatsoever. I also cannot find anything regarding the linear dependence of columns of a nonsquare matrix, with the exception of $A$ a field (in which case this is all easy).
 A: We can—without loss of generality—assume that $m = r + 1$ as you have done. I will allow for the possibility that $n = r$, and that the hypothesis for the $(r + 1) \times (r + 1)$ minors is vacuously true; but in general, $n\geq r.$ From the hypotheses, $A$ is a nonzero commutative ring (otherwise there would not be a nonzero $r\times r$ minor), but we don’t need any additional restrictions on $A$ to proceed.
Let $\{i_1,\ldots,i_r\}$ be the row indices of an $r\times r$ submatrix of $C$ whose determinant is nonzero. If $w$ is a row of $C$, let $M_w$ be the $(r+1)\times(r+1)$ matrix whose first row is $w$ and whose successive rows are the rows of $C$ with indices $i_1, \ldots, i_r$ in that order.
The matrix $M_w$ is either a permutation of some $(r+1)\times(r+1)$ submatrix of $C$, or it has a repeated row. Either way, its determinant is $0$. Consider the adjugate matrix $M_w^*$ whose $(i,j)$-entry is $$(-1)^{i+j} (\det \text{of } M_w \text{ with row } j 
\text{ and column } i \text{ removed})$$ for $1\leq i,j\leq r+1$.
Up to a sign, the entries of the first column of $M_w^*$ will be determinants which only depend on the rows with indices $2,\ldots,r+1$ of $M_w$; so, it is independent of $w$. This column will have a nonzero entry. This is because $\{i_1,\ldots,i_r\}$ was chosen so that there is at least one $r\times r$ submatrix of $C$ with nonzero determinant which contains the rows (where one element in a fixed column is removed) with these indices, and this column contains all such $r\times r$ determinants up to a sign.
Call this column $v$. Since
$$
M_w M_w^* = (\det M_w) I = 0,
$$
and the $(1,1)$-entry of this product is $w\cdot v$, we must have that $w \cdot v = 0$.
As mentioned before, the choice of $v$ is independent of $w$ which means that the last equality is true for all rows of $C$. Hence, $v$ is a nonzero column for which $Cv = 0$, and the columns of $C$ are linearly dependent.
