I have seen various definitions for the tensor product $V \otimes W$ of two finite dimensional vector spaces $V$ and $W$ over some common field $F$. However, they all strike me as unnecessarily complicated. For a seemingly simpler definition, let's say $V$ is dimension $m$ and $W$ is dimension $n$. Then we know that $V \cong F^m$ and $W \cong F^n$. I know ahead of time that we sensibly want the dimension of $V \otimes W$ to be the product of the dimensions of $V$ and $W$, and we know that there is only one vector space over $F$ of any given dimension up to isomorphism, so why don't we first define $F^m \otimes F^n := F^{mn}$? Then, if we have an isomorphism from $V$ to $F^m$, and from $W$ to $F^n$, we can just 'pass through' $F^m \otimes F^n$, and use the (inverse) isomorphisms to then define $V \otimes W$? For instance, take $V$ to be polynomials in the variable $x$ of degree at most 2, and $W$ to be polynomials in variable $y$ of degree at most 1, with the standard bases, and with isomorphisms $a + bx + cx^2 \mapsto (a, b, c)$, and $s + ty \mapsto (s, t)$. Then $(a, b, c) \otimes (s, t) = (as, bs, cs, at, bt, ct)$ which then defines the map $(a + bx + cx^2) \otimes (s + ty) \mapsto (as \cdot 1 \otimes 1, bs \cdot x \otimes 1, cs \cdot x^2 \otimes 1, at \cdot 1 \otimes y, bt \cdot x \otimes y, ct \cdot x^2 \otimes y)$, where $1 \otimes 1, x \otimes 1, \dots, x^2 \otimes y$ are just symbols.
I am aware there is apparently some subtlety regarding the existence of the tensor product, and the usual way of constructing it is to perform some complicated quotient of vector spaces. Why do we bother going through this, if we know at the end the vector space needs to be isomorphic to $F^{mn}$ if it is going to exist at all? Is there some extra structure in tensor products that is not found in $F^{mn}$? I.e. is $V \otimes W$ 'more' than just a vector space?
Best Answer
In terms of extra structure, there is a bilinear map $V\times W \to V\otimes W$. However, if $F^n$ has basis $e_1,\ldots,e_n$ and $F^m$ has basis $f_1,\ldots,f_m$ then you can just take the standard basis of $F^{mn}$ and label the basis elements $g_{ij}$ for $i=1,\ldots,n$ and $j=1,\dots,m$.
Then you can define the bilinear map to send $(e_i,f_j)\mapsto g_{ij}$.
So what you describe is in practice the same as the standard construction.
One objection someone might raise to yours is that it involves picking a basis for your vector spaces (that is you need isomorphisms $V\cong F^n, \,W\cong F^m$). You therefore need to show that these choices do not effect the result.
Again this is easy to fix: just show that your construction satisfies the universal property, and then any pair of sets of choices will lead to naturally isomorphic results.