# 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 $$m \le n$$ ?

• Faithfulness is irrelevant since you can always just mod out the annihilator. Perhaps you mean to instead assume $M$ is nonzero? Jan 2, 2019 at 6:11

## 1 Answer

[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.

• The result follows easily from Orzech Theorem. (Of course, the Noetherian reduction is the method of proof for the theorem, too.) Jan 2, 2019 at 23:06