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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 $M=\oplus_{i\in\omega}\mathbb{Z}$. 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 $R\cong R^2$?

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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
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
up vote 3 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 $$h_{ij} = \sum_{k=1}^{\infty}g_{ik}f_{kj},\quad\text{and}\quad h'_{ij}=\sum_{k=1}^{\infty}g_{ik}f'_{kj}.$$ 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.

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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.)

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

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