Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

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 $n \leq m$

share|improve this answer
    
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.