# Image of the transpose of a matrix over a ring

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.

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.)