Image of the transpose of a matrix over a ring

Question

Let $R$ be a commutative ring with 1. Let $A$ be an $n\times n$ matrix defined over $R$. Let's say the determinant of $A$ is not zero (I am not sure whether this would be relevant to my question).

The columns and the rows of $A$ defines two submodules of $R^n$. I would like to know whether there is anything one can say about these two modules. Are they isomorphic? If $R$ was a field, then we would have two vector spaces of the same dimension. For a general $R$, we would have two modules of the same rank. Can we say something about their torsion parts?

Edit: I forgot to say that I am actually more interested in the quotients of $R^n$ by the column space and the row space. Are there any relations between the two quotients? Thanks.

Answer 1

For a simple example where the determinant is zero, take $A = \begin{bmatrix} x & x \\ y & y \end{bmatrix}$ (where $x, y \in R$ but we haven't decided how they should behave yet). Then the column space is generated by $(x, y)$, but the row space is generated by $(x, x), (y, y)$. These are not isomorphic modules in general; we can take $R = K[x, y]$. Let $I$ be the ideal of polynomials of positive degree; then the row space $M$ has the property that $M/IM$ is $2$-dimensional (spanned by $(x, x)$ and $(y, y)$). this means it can't be generated by one element, so can't be isomorphic to the column space.

The quotients $R^2/N$ and $R^2/M$ by the column and row space are also not isomorphic. $R^2/M$ has torsion (namely $(1, 1)$ is both $x$-torsion and $y$-torsion) but $R^2/N$ is torsion-free; if an element $(f, g) \in R^2/N$ is nonzero, meaning it is not of the form $(f, g) = (xh, yh)$ for some $h \in R$, then the same is true of any multiple of it.

This example can be modified into a more complicated example where the determinant is nonzero. Now we take

$$A = \begin{bmatrix} x & y \\ z & w \end{bmatrix}$$

where $x, y, z, w \in R$ but again we haven't decided how they should behave yet. The column space $N$ is generated by $(x, z), (y, w)$ and the row space $M$ is generated by $(x, y), (z, w)$. Now we can take, for example, $R = K[x, y, z, w, u]/(xu = zu = 0)$; then by construction $uN$ is generated by one element, namely $(uy, uw)$, but $uM$ is generated by two elements, namely $(0, uy), (0, uw)$; moreover an inspection of degrees shows that $uM$ can't be generated by one element, so $uN, uM$ are not isomorphic, hence neither are $N, M$.

Similarly the quotients $R^2/N, R^2/M$ are not isomorphic; the kernel of multiplication by $u$ on $R^2/N$ is generated by $(x, 0), (z, 0), (0, x), (0, z)$ but there is a relation $(x, 0) + (0, z) = 0$ in this quotient, so it is generated by $3$ elements. But the kernel of multiplication by $u$ on $R^2/M$ is also generated by $(x, 0), (z, 0), (0, x), (0, z)$, and these are linearly independent. A degree argument should again be able to show that this kernel can't be generated by $3$ elements.

(The degree arguments are a little shakier than I'd like them to be but they should all boil down to quotienting by the action of the ideal $I$ of elements of positive degree as in the previous case.)