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The free abelian group (equivalently the free $\mathbb{Z}$-module) $F(M \times N)$ is defined as the set of all linear combinations of elements of $M \times N$,

$$F(M \times N) = \{n_1x_1 + \cdots +n_kx_k : n_i \in \mathbb{Z}, x_i \in M \times N, k \in \mathbb{Z}_{\ge0}\}$$

Wouldn't these linear combinations already be in $M \times N$? How is the free abelian group different from $M \times N$?

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What is the free abelian group on the set $\{\clubsuit,\spadesuit,\diamondsuit,\heartsuit\}$? – Mariano Suárez-Alvarez Feb 7 '13 at 5:58
Once you answer that: what is the free abelian group on the set $\mathbb Z_4$? – Mariano Suárez-Alvarez Feb 7 '13 at 6:02
Okay, so when considering the free abelian group $F(M \times N)$ the group structure of $M \times N$ is irrelevant? – user61323 Feb 7 '13 at 6:08
If it were relevant, would it make any sense to construct the free abelian group on a set which is not a group? – Mariano Suárez-Alvarez Feb 7 '13 at 6:10
Do you mean $R=\mathbb{Z}$? – Julian Kuelshammer Feb 8 '13 at 7:10

No. For instance, suppose you took the $\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}=M$. Then, in $M^2$ you have that $2(1,0)=(0,0)$ but in $F(M^2)$ one has that $2(1,0)\ne (0,0)$. The relations that hold amongst elements of $M^2$ DO NOT hold for elements of $F[M^2]$.

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Okay, so the free abelian group on Z_2 is the same as the free abelian group on any two element set? – user61323 Feb 7 '13 at 6:10
Exactly. You're just looking at sums of pairs of elements with no formal relations beteween them. – Alex Youcis Feb 7 '13 at 6:27

First of all, for arbitrary modules $M, N$ you don't need to have that $M\times N$ is a free module, so $M\times N$ cannot be the free module generated by something. But also if $M, N$ were free, these two modules would not be equal.

I take the following definition: $F_\mathbb{Z}(M\times N)$ is the free module with basis the set $M\times N$. (If you have another definition, comment and I should be able to prove equivalence.) This means that we forget the module structure of $M\times N$ and just make it into a free basis of a module, so for example we have: $$[0]+[0]=2\cdot [0]\neq [0]$$ where the elements in brackets denote the basis elements.

So for example we have $F_\mathbb{Z}(0\times 0)=F_\mathbb{Z}(0)\cong \mathbb{Z}$. And some non-product examples: $F_\mathbb{Z}(\mathbb{Z}/2)\cong \mathbb{Z}^2$ and $F_\mathbb{Z}(\mathbb{Z})\cong \mathbb{Z}[X, X^{-1}]$ (by the map $[n]\mapsto X^n$).

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A right $\mathbb Z$-module is also a left $\mathbb Z$-module, if that helps.

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I guess my question is what exactly is $F_\mathbb{Z} (M \times N)$? – user61323 Feb 7 '13 at 4:32
The Free abelian group generated by $M \times N$ – Severus Snape Feb 7 '13 at 4:33
How is it different from $M \times N$ itself? – user61323 Feb 7 '13 at 4:37

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