det(B) is a zero divisor if and only if the set of row vectors of B is dependent., this is more or less the Mc Coy theorem ( see e.g. Balcerzak, Jozefiak, Commutative rings).

No need for finitness. The question has been asked and answered here:

http://math.stackexchange.com/questions/71740/necessary-and-sufficient-condition-for-trivial-kernel-of-a-matrix-over-a-commuta

The determinant results go through just fine as long as the coefficients come from a commutative ring. Thus

Adj[A] *A= Det[A] *I

A*Adj[A] = Det[A]*I

The Adjoint of A is polynomials in the entries of A and can be computed in any commutative ring.

So if X A = 0, then X Det[A}=0

and Det[A] is a zero divisor or X = 0.

otherwise look at the beatiful book of Brown: "Matrices over Commutative Rings", chapt. 4 (where the notion of rank is discussed). In chapt. 5 you find the aforementioned Theorem of McCoy. Another good book on the topic is McDonald, "Linear Algebra over Commutative Rings", 1st chapter.

Please go through my paper entitled

THE INJECTIVE HULL OF A MODULE WITH FGD

published in INDIAN JOURNAL OF PURE AND APPLIED MATHEMATICS 1989

The full paper is available in research gate

You may get an idea about

because the definition of linearly independent elements in Module theory was available

We may consider every ring as a module over itself

Thank you

Thanks all very much. That was quite helpful! I've enjoyed reading the William C. Brown text, that was just the level of math that I needed. The references to McCoy's theorem were right on (I hadn't heard of it) as was the statement "No need for finiteness."