The difference between the free $A$-module $A^{(M\times N)}$, and $M\times N$

Question

Let $M$ and $N$ be two $A$-modules, where $A$ is a commutative ring. What is the difference between the free $A$-module $A^{(M\times N)}$, and $M\times N$?

In Atiyah-Macdonald, both of these seem to have the same description. Hence the confusion.

Answer 1

When you write $A^{(M \times N)}$ you're denoting a free $A$-module with basis indexed by $M \times N$. In general, you can expect this to be much bigger than $M \times N$.

For instance, if we consider the $\mathbb{Z}$-module $M = 2 \mathbb{Z}$, then $\mathbb{Z}^{(M)}$ is the free module indexed by the set $\{0, \pm 2, \pm 4, \dots\}$, i.e., $$ \mathbb{Z}^{(M)} = \mathbb{Z}_0 \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{-2} \oplus \dots $$

This is horrible notation that just means that I get a copy of $\mathbb{Z}$ for every element in my index set $M$. So the important point is that in this example, I not thinking of $M$ as a module at all. It's just a set over which I perform the sum. Another notation could be $$ \mathbb{Z}^{(M)} = \bigoplus_{\alpha \in M} \mathbb{Z}_\alpha $$ where $\mathbb{Z}_\alpha$ just means "the copy of $\mathbb{Z}$ corresponding to $\alpha$."

Here is @Ayman's example worked out: If $M = \mathbb{Z}/2 \mathbb{Z}$ is the abelian group with two elements, then $M \times M$ is the abelian group ($\mathbb{Z}$-module) $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, which is a finite group ($\mathbb{Z}$-module) with four elments, say $a,b,c,d$. Then $$\mathbb{Z}^{M \times M} = \bigoplus_{\alpha \in M\times M} \mathbb{Z}_\alpha = \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$$

... much bigger than $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$!

EDIT: I just looked at Atiyah-MacDonald (p.21 in my copy) and I think maybe the confusion is that they denoted a free $A$-module over a set $I$ by $A^{(I)}$? To be clear, they never mean for $I$ to be an ideal or module of $A$. It's just a set, like maybe $I = \{2,3,5\}$ so $A^{(I)}$ would be $A \oplus A \oplus A$.

Answer 2

What is the difference between $X$ and $A^X$ in general? Well, are $\{1,2,3\}$ and $\Bbb R^3$ any different? A more revealing rhetorical question: are $\{\vec{i},\vec{j},\vec{k}\}$ and $\Bbb R^3$ any different? Obviously, but the first spans the second: everything in the space is a unique $\Bbb R$-linear combination of things in the set. This is exactly how it works with free modules on a set: everything in $A^X$ is an $A$-linear combination of $X$'s things.