Pedantic nitpick: it's not the dumbest possible multiplication; it's the dumbest possible bilinear multiplication. There are "freer" free products on vectors than tensor product; the freest possible one is their Cartesian product as sets, which just makes a tuple out of them ij = (i, j). If you regarded the resulting sets as a vector space, it would not be true that i0 + 0j = (i, 0) + (0, j) ≠ (i, j). The tensor product is the freest that is well-behaved as a vector space in the sense that it respects the underlying scalar operations from its arguments.