Atiyah-MacDonald, exercise 2.11
Here is another solution using only the Cayley-Hamilton Theorem for finitely generated modules (Proposition 2.4. in Atiyah-Macdonald) which, even though looks quite innocent, is a very powerful statement.
Assume by contradiction that there is an injective map $\phi: A^m \to A^n$ with $m>n$. The first idea is that we regard $A^n$ as a submodule of $A^m$, say the submodule generated by the first $n$ coordinates. Then, by the Cayley-Hamilton Theorem, $\phi$ satisfies some polynomial equation \begin{equation} \phi^k + a_{k-1} \phi^{k-1} + \cdots + a_1 \phi + a_0 = 0. \end{equation} Using the injectivity of $\phi$ it is easy to see that if this polynomial has the minimal possible degree, then $a_0 \ne 0$. But then, applying this polynomial of $\phi$ to $(0,\ldots,0, 1)$, the last coordinate will be $a_0$ which is a contradiction as it should be zero.
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.
Let M be the $n\times m$ matrix representing the injection $A^m \to A^n$. Define Di(M) to be the ideal generated by the determinants of all i-by-i minors of M. Let r be the largest possible integer such that Dr(M) has no annihilator (i.e. there is no nonzero element a∈A such that aDr(M)=0); I think r is usually called the McCoy rank of M.
Assume that $m>n$. We shall get a contradiction.
Choose a nonzero a∈A such that aDr+1(M)=0. By assumption, $a$ does not annihilate Dr(M), so there is some r-by-r minor that is not killed by $a$; we may assume it is the upper-left r-by-r minor. Thus, $r \leq n$, so that $r+1 \leq n+1\leq m$. Let M1, ..., Mr+1 be the cofactors of the upper-left (r+1)-by-(r+1) minor obtained by expanding along the bottom row. (This is well-defined even if the $r+1$-th row of $M$ does not exist, because these cofactors use only the first $r$ rows and the first $r+1$ columns of $A$, and $A$ has both since $r \leq n$ and $r+1 \leq m$.) Note, in particular, we know Mr+1 is the determinant of the upper-left r-by-r minor, so aMr+1≠0.
The claim is that the vector (aM1,...,aMr+1,0,0,...) (which we've already shown is non-zero) is in the kernel of M. To see that, note that when you dot this vector with the k-th row of M, you get $a$ times the determinant of the matrix obtained by replacing the bottom row of the upper-left (r+1)-by-(r+1) minor with the first r+1 entries in the k-th row. If k≤r, this determinant is zero because a row is repeated, and if k>r, this determinant is the determinant of some (r+1)-by-(r+1) minor, so it is annihilated by $a$. But since A is injective, the kernel of M is 0, and we have a contradiction.