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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 $m \le n$ ?

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  • $\begingroup$ Faithfulness is irrelevant since you can always just mod out the annihilator. Perhaps you mean to instead assume $M$ is nonzero? $\endgroup$ – Eric Wofsey Jan 2 '19 at 6:11
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[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.

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  • $\begingroup$ The result follows easily from Orzech Theorem. (Of course, the Noetherian reduction is the method of proof for the theorem, too.) $\endgroup$ – user26857 Jan 2 '19 at 23:06

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