Dear CJD, if you are still interested in your problem, already solved three weeks ago by Anton, here is another point of view.

Let M:A^m--->A^n be injective. Let B=Z[(m_ij)] be the subring of A generated by all the entries of the matrix M; this B is a noetherian ring and we have (by restriction) an injective linear map M:B^m--->B^n. In other words we may assume that A is noetherian. Now we localize at a prime P of B of height zero and we get an injective map (localization is exact and thus preserves injections) L:C^m--->C^n (the ring C is the ring B localized at P).

Ah,but now C is noetherian of dimension zero, hence artinian and we can talk about lengths. Since lengths are additive in exact sequences (Atiyah-MacDonald, Proposition 6.9) we get m.length(C) + length(coker L)= n.length(C), hence m<=n.

Friendly greetings, Georges.

Dear CJD, if you are still interested in your problem, already solved three weeks ago by Anton, here is another point of view.

Let $M:A^m \to A^n$ be injective. Let $B=\mathbb Z [\ldots,m_{ij},\ldots]$ be the subring of $A$ generated by all the entries of the matrix $M$ ; this $B$ is a noetherian ring and we have (by restriction) an injective linear map $M:B^m \to B^n$. In other words we may assume that $A$ is noetherian. Now we localize at a prime $\mathfrak p$ of $B$ of height zero and we get an injective map (localization is exact and thus preserves injections) $L:C^m \to C^n$ (the ring $C$ is the ring $B$ localized at $\mathfrak p$).

Ah, but now $C$ is noetherian of dimension zero, hence artinian and we can talk about lengths. Since lengths are additive in exact sequences (Atiyah-MacDonald, Proposition 6.9) we get $m.length(C) + length(coker L)= n.length(C)$, hence $m\leq n$.

Friendly greetings, Georges.