# A basis for the tensor product

Suppose are all free -modules of respective rank Then, from the distributivity of tensor products over direct sums, we can deduce that is the free -module of rank . Indeed, first recall that given a set of free generators (a basis) of the free -module of rank we can write just as in the case of vector spaces. In this case, we get, with a basis for (),

where the last sum is taken on all multi-indices such that is an element of the basis of for all and all (fixed ). This is the free -module on the set

which indeed has cardinality

I recommend writing down the particular case where to clarify this as I’m sure the indices are confusing. Another way of showing that this set is a basis for is to show that the free -module over that set satisfies the universal property of namely that every k-multilinear maps from to some -module factors uniquely through this set. Then since the universal property defines the tensor product of the ‘s uniquely up to isomorphism, we’re done.

As a corollary of this discussion, we get

Proposition:If is a free -module of rank (for a commutative ring), then, if is a set of free generators for the set of k-tensors is a basis for In particular, is free of rank

Note that since a vector space of dimension over a field can be identified with the free -module of rank this proposition (and the whole discussion) is applicable to vector spaces.

**Reference:** D. S. Dummit, R. M. Foote, Abstract Algebra, 3rd ed.