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.

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 $M \times N$?

share|improve this question
1  
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
1  
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
1  
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
show 1 more comment

3 Answers

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 $F[M^2]$.

share|improve this answer
    
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
2  
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
add comment

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

share|improve this answer
add comment

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

share|improve this answer
    
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
add comment

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.