Tensor product of $\mathbb{R}^m$ and $\mathbb{R}^n$ I read this definition of a tensor product on Wikipedia. I'm quite confused.. so the elements of $F(V\times W)$ are the pairs $(v,w)$, where $v\in V,w\in W$.
Now, if we take $V=\mathbb{R}^m,W=\mathbb{R}^n$, then the dimension of $V\times W$ is $mn$. Then by definition we have $V\otimes W=F(V\times W)/R$, where $R$ is generated by the specified elements.
So how can it be that the dimension of $V\otimes W$ is still $mn$?
What exactly are the elements of $F(V\times W)$ for $V=\mathbb{R}^m,W=\mathbb{R}^n$?
 A: The dimension of $V \times W$ is not $mn$, but $m + n$.
Then, when you take $F(V \times W)$, this space will generally be infinite dimensional, because each elements in $V \times W$ becomes a generator, so you get infinitely many linearly independent elements.
Then you mod out by those relations, and this drastically reduces the dimension to $mn$.
A 'generic' element of $F(V \times W)$ looks like this:
$$\sum_{i = 1}^n a_i(v_i, w_i),$$
with $v_i \in V$, $w_i \in W$ and $a_i \in \mathbb{R}$ (or whatever the field is).
The point is that in $F(V \times W)$, you have no relations. So you cannot in general combine their elements to get to the reduced form $(v, w)$.
A: Elements in $F(V\times W)$ are linear combinations of the type
$$\sum_{ij}a_{ij}(v_i,w_j)$$
where $a_{ij}$ are scalars and $(v_i,w_j)$ are in $V\times W$.
Elements in $V\otimes W$ are cosets $\sum_{ij}a_{ij}(v_i,w_j)+\langle R\rangle$.
In this way an element $(v,w)\in V\times W$ can be used to represent an element $$v\otimes w=(v,w)+\langle R\rangle,$$
in $V\otimes W$.
Note that 
$$a(v\otimes w)=(av)\otimes w=v\otimes(aw),$$ 
since $a(v,w)-(av,w)\in R$ as well $a(v,w)-(v,aw)\in R$.
