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.

I'm aware that over a commutative ring, two free finitely generated modules of finite rank are isomorphic if and only if they have the same rank.

I'm trying to understand a curious example of why this does not hold over a noncommutative ring.

Let's take $M=\oplus_{i\in\omega}\mathbb{Z}_i$. Then let $R=\text{End}(M)$, which is not commutative. Then $R\cong R^2$ as left $R$-modules.

It seems like one could possibly make a basis of $R$ using only endomorphisms which act on the even or odd coordinates of elements of $M$, which might suggest such an isomorphism. Is there a clear proof of why $R\cong R^2$?

share|improve this question
    
At first you seem only interested in finitely generated modules, but your noncommutative example is infinitely generated. But then you might as well look at $R=\mathbb Z^\omega$, which is a countable product of $\mathbb Z$'s and is isomorphic to $R^2$. –  Grumpy Parsnip Oct 15 '11 at 2:25
2  
Actually, you probably want to take $M=\oplus_{i\in\omega}\mathbb{Z}$, rather than the product. The advantage is that $M$ is a free abelian group, whereas your $M$ is not free, which makes describing the endomorphisms more difficult. –  Arturo Magidin Oct 15 '11 at 3:16
    
@ArturoMagidin OK, I've changed that if it makes it easier to explain. –  Danielle Intal Oct 15 '11 at 3:22

2 Answers 2

up vote 1 down vote accepted

One is tempted (as I was originally), to argue as follows: since $M\cong M\oplus M\cong M\times M$, we have $$R = \mathrm{Hom}(M,M) \cong \mathrm{Hom}(M,M\times M) \cong \mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M) = R\times R,$$ and stop there. The problem with the argument is that in some instances we are dealing with these objects as $\mathbb{Z}$-modules rather than as $R$-modules (specifically the isomorphisms are isomorphism from $\mathrm{Hom}(M,M)$ to $\mathrm{Hom}(M,M\times M)$ is, right now, just an isomorphism as $\mathbb{Z}$-modules), so some care needs to be exercised to make the argument actually work.

First, note that $M\cong M\times M$ as $\mathbb{Z}$-modules. Define a homomorphism $\varphi\colon M\to M\times M$ by maping $(m_i)$ to $\bigl((m_{2i-1}), (m_{2i})\bigr)$. That is, $(m_1,m_2,m_3,\ldots)$ maps to $\bigl( (m_1,m_3,m_5,\ldots),(m_2,m_4,m_6,\ldots)\bigr)$.

The elements of $R$ can be thought of as infinite "column-finite matrices"; that is, each endomorphism $M\to M$ corresponds to a family of functions $(\mathbf{f}_i)_{i\in\omega}$, with $\mathbf{f}_i\colon\mathbb{Z}\to M$; hence $\mathbf{f}_i = \sum_{j\in\omega}f_{ji}e_j$, where $e_j$ is the element of $M$ that has $1$ in the $j$th coordinate and $0$s elsewhere, $f_{ji}\in\mathbb{Z}$, and $f_{ji}=0$ for almost all $j$.

If we take an element of $\mathbf{f}=(f_{ij})$ of $R$, and compose it with the isomorphism $M\to M\times M$, we obtain a homomorphism $M\to M\times M$, given by $\bigl( (f_{i,2j-1}), (f_{i,2j})\bigr)$. So the map $\mathrm{Hom}(M,M)\to\mathrm{Hom}(M,M)\times \mathrm{Hom}(M,M)$ maps $(f_{ij})$ to $\bigl( (f_{i,2j-1}),(f_{i,2j})\bigr)$.

If we compose maps $\mathbf{f}=(f_{ij})$ and $\mathbf{g}=(g_{ij})$, we get the map $(h_{ij})$, where $$h_{ij} = \sum_{k=1}^{\infty}g_{ik}f_{kj}.$$ Since $f_{kj}=0$ for almost all $k$, the sum is finite and $h_{ij}$ makes sense.

This gives the action of $R$ on $\mathrm{Hom}(M,M)$. The action of $R$ on $\mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M)$ is then given by $$ \mathbf{g}\bigl((f_{ij}),(f'_{ij})\bigr) = \bigl( \mathbf{g}(f_{ij}), \mathbf{g}(f'_{ij})\bigr) = \bigl( (h_{ij}), (h'_{ij})\bigr)$$ where $$h_{ij} = \sum_{k=1}^{\infty}g_{ik}f_{kj},\quad\text{end}\quad h'_{ij}=\sum_{k=1}^{\infty}g_{ik}f'_{kj}.$$ To see that the map from $\mathrm{Hom}(M,M)$ to $\mathrm{Hom}(M,M\times M)$ respects the action of $R$ (that is, that we get an $R$-module homomorphism, not merely a $\mathbb{Z}$-module homomorphism), let $(f_{ij})\in\mathrm{Hom}(M,M)$ and $\mathbf{g}\in R$. The element $\mathbf{g}(f_{ij})$ is mapped to $$\left( \Bigl(\sum_{k=1}^{\infty}g_{ik}f_{k,2j-1}\Bigr),\Bigl( \sum_{k=1}^{\infty}g_{ik}f_{k,2j}\Bigr)\right)$$

On the other hand, if we let $\mathbf{g}$ act on $\bigl( (f_{i,2j-1}), (f_{i,2j})\bigr)$, we get $$\mathbf{g}\left( (f_{i,2j-1}), (f_{i,2j})\right) = \left(\Bigl( \sum_{k=1}^{\infty}g_{ik},f_{i,2j-1}\Bigr), \Bigl( \sum_{k=1}^{\infty}g_{ik}f_{i,2j}\Bigr)\right),$$ that is, the same as the image of $\mathbf{g}(f_{ij})$. So the homomorphism $\mathrm{Hom}(M,M)\to\mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M)$ is actually a homomorphism as $R$-modules. Since the original map as $\mathbb{Z}$-modules was a bijection, so is this one, hence $\mathrm{Hom}(M,M)$ and $\mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M)$ are isomorphic not only as $\mathbb{Z}$-modules, but also as $R$-modules. So we obtain: $$R = \mathrm{Hom}(M,M) \cong \mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M) =R\times R,$$ with the isomorphism being an isomorphism of $R$-modules, as desired.

share|improve this answer
    
One should be careful when following the module structures along those isomorphisms, though! –  Mariano Suárez-Alvarez Oct 15 '11 at 2:24
    
Thanks! Do you mind explaining what exactly you mean by interweaving the coordinates to get $M\cong M\times M$? –  Danielle Intal Oct 15 '11 at 2:29
    
@Danielle: There is a gap that needs explaining, but I'm going away for a bit. I'll delete and be back. –  Arturo Magidin Oct 15 '11 at 2:31
    
Amazing, thanks! –  Danielle Intal Oct 17 '11 at 9:06

You probably mean that "two FREE finitely generated $R$-modules $M_1$ and $M_2$ are isomorphic iff they have equal rank".

It's not so surprising this doesn't work in the infinite dimensional case. Just look at infinite-dimensional vector spaces.

You should be careful when you talk about modules over a non-commutative ring. You should indicate whether you're talking about left/right modules, etc.

Edit: Not sure if this is correct, but you can write down an explicit isomorphism once you choose a bijection from the set of integers to the set of even integers. (Or something like that.)

share|improve this answer
    
OK, I've edited it to try to tighten up the language. –  Danielle Intal Oct 15 '11 at 2:06

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.