The elementary tensors of two bases form the product basis of the tensor product

Let $R$ be a commutative ring. If $M$ is free with basis $(e_i{)}_{i\in I}$ and $N$ is free with basis $(f_j{)}_{j\in J}$, then $M{\otimes }_{R}N$ is free with basis

$(e_i\otimes f_j{)}_{(i,j)\in I\times J}\mathrm{.}$

Equivalently, the canonical map $R^{(I\times J)}\to M{\otimes }_{R}N$ sending the standard basis vector at $(i,j)$ to $e_i\otimes f_j$ is an isomorphism. This includes an empty basis in either factor.

The triangle illustrates the free extension of the product-indexed family. Write $u_{(i,j)}$ for the standard vector of $R^{(I\times J)}$, and let $\iota (i,j)=u_{(i,j)}$ and $b(i,j)=e_i\otimes f_j$. The diagonal is a set map; $\mathrm{\Phi }$ is the module isomorphism of the statement.

The triangle commutes because $\mathrm{\Phi }(u_{(i,j)})=e_i\otimes f_j$, as traced in proof step 3.1. More generally, for every finitely supported coefficient family $(c_{ij}{)}_{(i,j)\in I\times J}$,

$\mathrm{\Phi }\left({\sum }_{(i,j)\in I\times J}{c}_{ij}{u}_{(i,j)}\right)={\sum }_{(i,j)\in I\times J}{c}_{ij}(e_i\otimes f_j)\mathrm{.}$

These sums are finite even when $I$ and $J$ are infinite. If either index set is empty, the top-left set is empty and both module vertices are zero, as in step 3.2. The given bases suffice; no further choice of basis is involved. The proof above is independent of the illustration.