Basis of tensor product of vector spaces Suppose $V,W$ are vector spaces with bases $\{e_1,e_2,\dots ,e_m\}$ and $\{f_1,f_2,\dots ,f_n\}$. I know that $V\otimes W=F(V\times W)/H$, where $F(V\times W)$ is the free vector space on $V\times W$ and $H$ is the subspace generated by all elements satisfying certain relations such that bi-linearity holds in the quotient space. I'm writing $v\otimes w$ for the congruency class of the pair $(v,w)\in V\times W$. I can easily show that the pairs $e_i\otimes f_j$ span $V\otimes W$, but I'm having trouble showing linear independence i.e. that if $\sum_{i,j}\lambda_{ij}e_i\otimes f_j=0$ then for all $i,j$, $\lambda_{ij}=0$. Could anybody point me in the right direction? Thanks for any replies!
 A: $\def\Hom{\mathrm{Hom}}$
So, since you already know that the product of the bases spans $V \otimes W$, you just need to show that the dimension of $V \otimes W$, which is the dimension of its dual, is the product of the dimension of $V$ and the dimension of $W$. 
We first reduce the problem to showing that
$$(V \otimes W)^{*} = \Hom_{k}(V \otimes W, k) \cong \Hom_{k}(V, \Hom_{k}(W, k)) = \Hom_{k}(V, W^{*}).$$
Note that if the above relationship holds, then the dimension of $V \otimes W$ is the dimension of the space of linear maps from $V$ to $W^{*}$, which is just the product of the dimensions of $V$ and $W^{*}$, and is hence the product of the dimensions of $V$ and $W$.
So, to show the above equalities, note that the leftmost and rightmost equalities are by definition of the dual. So we need to show that
$$\Hom_{k}(V \otimes W, k) \cong \Hom_{k}(V, \Hom_{k}(W, k)).$$
This is actually true if $k$ is any ring and $V$ and $W$ are any $k$-modules. This is the adjunction between $\Hom$ and tensor product. But let's go through the details for the vector space case.
Suppose $\phi$ is a linear map from $V \otimes W$ to $k$. We want to define $f(\phi)$ which takes a vector in $v$ and spits out a linear map from $W$ to $k$. To do so, we define
$$(f(\phi) (v)) (w) = \phi(v \otimes w).$$
You should check that this map is linear and well-defined. It is not too hard. We next construct an inverse map. Given $\psi$ which to each $v$ assigns a linear map from $W$ to $k$, we define $g(\psi)$ as a linear map from $V \otimes W$ to $k$ by first defining it on the simple tensors as
$$g(\psi)(v \otimes w) = \psi(v)(w)$$
noting that this is bilinear in the $v$ and $w$ and then noting that bilinear maps extend to all of $V \otimes W$.
Again, $g$ can be checked to be linear and you can also easily see that $f$ and $g$ are inverses.
