Is the direct product of two vector spaces (over the same field) just the tensor product of two vector spaces over $\mathbb{Z}$?
Am I right in thinking that essentially we would use the tensor product to add together pairs of the Cartesian product of our vector spaces over some ring with 'the same simplifications we would ask for in direct product viewed as a vector space', and if this ring happens to be $\mathbb{Z}$ then we have exactly the direct product?
Best Answer
No, certainly not. If $V$ is $m$-dimensional (say $e_1, \ldots, e_m$ is a basis) and $W$ is $n$-dimensional (with basis $f_1, \ldots, f_n$), then $V \otimes W$ is $mn$-dimensional, with basis $e_i \otimes f_j$. On the other hand, $V \oplus W$ (what I and most of the world understand by "direct product") is $(m+n)$-dimensional, effectively with basis $e_1, \ldots, e_m, f_1, \ldots, f_n$.
Tensors $V \otimes W$ are universal recipients of bilinear maps: there is a bilinear map $i: V \times W \to V \otimes W$ such that for any bilinear map $g:V \times W \to U$, there is a unique linear map $\phi: V \otimes W \to U$ making the evident triangle commute.