On injective module homomorphism $M^m \to M^n$ for a faithful, finitely generated, non-zero module $M$ over a commutative ring Let $M$ be a non-zero finitely generated faithful module over a commutative ring with unity $R$. If $m,n$ are positive integers such that there exists an injective module homomorphism from $M^m$ to $M^n$, then does that imply $m \le n$ ? 
 A: [The following argument is adapted from this argument by Georges Elencwajg for the case $M=R$.]
Let us first reduce to the case that $R$ is Noetherian. Picking a finite set $G$ of generators for $M$, we can represent our map $f:M^m\to M^n$ with a matrix of elements of $R$.  Let $S\subseteq R$ be the subring generated by the entries of this matrix.  Let $N$ be the $S$-submodule of $M$ generated by $G$.  Then $f$ restricts to an $S$-module homomorphism $N^m\to N^n$.  Since $M$ is nonzero, so is $N$, and since $f$ is injective, so is $g$.  Since the ring $S$ is finitely generated, it is Noetherian.  So, we may replace $R$ with $S$, $M$ with $N$, and $f$ with $g$ and thus assume $R$ is Noetherian.
So from now on we assume $R$ is Noetherian.  I now claim that if $P\subset R$ is any prime ideal, then $M_P$ is nonzero.  Indeed, this follows from the fact that $M$ is finitely generated and faithful.  Letting $x_1,\dots,x_k$ generate $M$, we have $\operatorname{Ann}(x_1)\dots\operatorname{Ann}(x_n)\subseteq \operatorname{Ann}(M)=0\subseteq P$, and so $\operatorname{Ann}(x_i)\subseteq P$ for some $i$ since $P$ is prime.  That means that the image of $x_i$ in $M_P$ is nonzero, so $M_P$ is nonzero.
In particular, now let $P$ be a minimal prime of $R$ (since $M$ is nonzero and thus $R$ is nonzero, we know $R$ has a minimal prime ideal).  We then have an injective homomorphism $M_P^m\to M_P^n$ of $R_P$-modules.  But the ring $R_P$ is a zero-dimensional Noetherian ring and thus is Artinian, so $M_P$ has finite length over $R_P$.  If $M_P$ has length $\ell$, then $M_P^m$ has length $m\ell$ and $M_P^n$ has length $n\ell$, and so our injective homomorphism implies $m\ell\leq n\ell$.  Since $M_P$ is nonzero, $\ell>0$, so we conclude that $m\leq n$.
As a final remark, the assumption that $M$ is faithful is unnecessary, since we can always replace $R$ with the quotient $R/\mathrm{Ann}(M)$ over which $M$ is faithful.
