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Suppose for a contradiction that $m>n$. Compose the given injective module homomorphism with the standard injection $B^n\hookrightarrow B^m$ that appends $m-n$ zero coordinates. This gives an injective endomorphism $T$ of the finite free module $B^m$ whose matrix has a zero final row.

Its characteristic polynomial has zero constant term, so the Cayley-Hamilton theorem givesInjectivity lets us cancel $T$. Repeating this argument eventually gives the identity endomorphism equal to zero. That would imply $B^m=0$, contrary to $B\ne 0$. Hence $m\le n$. This proves the rank inequality for an injection of finite free modules.

An $A$-module $M$ is a flat module when the tensor functor $-{\otimes }_{A}M$ preserves injections, equivalently when it is exact.

Suppose first that $M$ is flat. For any nonzero $a\in A$, tensor the injection $A\stackrel{a}{\to }A$ with $M$. The resulting map $M\stackrel{a}{\to }M$ is injective, so $am=0$ implies $m=0$. Thus $M$ is a torsion-free module.

Conversely, suppose $M$ is torsion-free over the principal ideal domain $A$. Every finitely generated submodule of $M$ is a finitely generated torsion-free module over a PID, hence a finite free module and therefore flat. The module $M$ is the filtered colimit of these submodules. Tensor products commute with filtered colimits, and filtered colimits of modules preserve exact sequences, so $M$ is flat. This proves that a torsion-free module over a principal ideal domain is flat.