row rank = column rank over a PID?

Question

Suppose I have some matrix $M$ with entries from a PID. Define the column rank and row rank as you would expect: the rank of the module generated by the columns (resp. rows).

Is it always true that the column rank is equal to the row rank?

Answer 1

Yes, due to the existence/uniqueness of Smith normal form:

For any PID $R$ and any $(n\times m)$-dimensional matrix $A$ with entries in $R$, it holds that $A$ is equivalent to a diagonal matrix $B$ with diagonal entries $\delta_1,\delta_2,\ldots,\delta_k,0,\ldots,0$, where the $\delta_i$ are nonzero elements of $R$ and $\delta_i|\delta_{i+1}$ for all $i<k$.

Equivalence of $A$ and $B$ here means there are invertible square matrices $P\in M_n(R)$ and $Q\in M_m(R)$ with $B=PAQ$. In particular the rank of the images of $A$ and $B$ agree, and similarly for the null spaces.