# injective $R$-module homomorphism vs. injective ring homomorphism

The following question has been lingering in my mind for months.

Let $R$ be a non-zero commutative ring with $1$. Consider $\phi : R^n \rightarrow R^m$,

1) as an injective $R$-module homomorphism.

2) as an injective ring homomorphism. (by definition $\phi(1)=1$.)

In which of the above cases, we can deduce that $n \leq m$? and why?

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Do you have a particular $\phi$ in mind? If so, which one? –  Pete L. Clark Apr 2 '11 at 18:38
No, I don't. I'm just trying to find the complete solution for the first case. –  Ehsan M. Kermani Apr 3 '11 at 18:07

Let $R = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \ldots \times \mathbb{Z}/2\mathbb{Z}$ (infinitely many times). Then as a ring $R^m = R^n = R$ $\forall m, n \in \mathbb{N}$. So the answer is negative in the case 2. But in the first case the answer is YES. But proof for general commutative ring is complicated. Here I am giving an easy proof assuming $R$ is commutative noetherian ring. After localizing at a minimal prime ideal we may assume that $R$ is a zero dimensional local ring ie artinian ring and $\phi : R^n \rightarrow R^m$ is an injective $R$ module homomorphism. Now length of $R$ as a $R$ module is finite and is equal to $l$ (say). Then comparing the length of both sides we have $ln \leq lm$. This means that $n \leq m$

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Interesting. I always assumed you needed $R$ to be an integral domain or something for case 1. –  Matt Mar 31 '11 at 16:53
No. I donot need $R$ to be an integral domain. Localization preserves exactness. So first I have simplified the problem. Then comparison of length has done job. –  A.G Mar 31 '11 at 17:08
Thanks. I appreciate it. It is a nice and important special case, but do you know the outline of the proof for the general case? –  Ehsan M. Kermani Apr 2 '11 at 8:19
Give me some time. I have to search my old notes for the solution. –  A.G Apr 2 '11 at 18:25

Let $S$ be your favorite non-zero commutative ring with $1$ and let $R$ be the product of countably many copies of $S$. Then $R^n$ is the product of countably many copies of $S$ for any $n\in\mathbb{N}$, since the union of finitely many countable sets of countable. Therefore $R^n$ and $R^m$ are isomorphic as rings for all $m,n\in\mathbb{N}$, so without further assumptions we cannot deduce $n\leq m$ in case $(2)$.

EDIT: I had originally written that this argument also applies in case $(1)$, but it does not; see the comments below.

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I have managed to confuse myself a bit: I believe my answer above is correct and one needs some additional assumption like $R$ being an integral domain to get the conclusion that the OP wants. However the wikipedia article on invariant basis number says that the counterexample I gave does not exist among commutative rings $R$. Would anyone care to correct me or Wikipedia? –  Noah Stein Mar 31 '11 at 12:09
Concerning the IBN question (which is about $R$-modules): you need to look at $(S^\infty)^n$ as a module over $S^\infty$ -- not just over $S$. –  Rasmus Mar 31 '11 at 12:34
@Rasmus: Ah, thanks. Shall I go ahead and delete the answer, then? –  Noah Stein Mar 31 '11 at 13:01
Well, as Anjan Gupta notes, your reasoning is correct for question (ii). For question (i) it's nice to have your link to the relevant wikipedia article. –  Rasmus Mar 31 '11 at 16:04