What is the free abelian group on $M \times N$ where $M,N$ are modules.

Question

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 $M \times N$?

Answer 1

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 $F[M^2]$.

Answer 2

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 $[n]\mapsto X^n$).

Answer 3

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