Let $R$ be a unital commutative ring and $A\in M_{n\times m}(R)$. Under which appropriate invariant "rank" of modules discussed in "Ranks of Modules" one can say that the row rank of $A$ is equal to the column rank of $A$?
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5$\begingroup$ I think you can say this if you define the column rank as the minimum number of generators for the column space, and define the row rank accordingly. Indeed, in this case, both ranks equal the smallest nonnegative integer $r$ such that $A$ can be written as $A = BC$ with $B\in M_{n\times r}\left(R\right)$ and $C\in M_{r\times m}\left(R\right)$. Please check; I am distracted with a different proof right now. $\endgroup$– darij grinbergAug 1, 2019 at 8:01

2$\begingroup$ @darijgrinberg Is right, of course. As is discussed at that link, a more usual definition of rank, for $R$ a domain with fields of fractions $K$, is the rank of $M \otimes_R K$. This simply means considering $A$ as a matrix with entries in $K$, so the row rank is also column rank in that sense. $\endgroup$– David E SpeyerAug 1, 2019 at 10:10
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