Cartesian product distributes over second factor in tensor product? I was thinking: a linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$ is nothing other than $m$ linear maps from $\mathbb{R}^n$ to $\mathbb{R}$. A linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$ is also an element of $\mathbb{R}^{* n} \otimes \mathbb{R}^m$. So does that mean that in general, for finite-dimensional vector spaces $A$ and $B$, $(A\otimes B)^n = \left(\mathcal{L}(A^*,B)\right)^n=\mathcal{L}(A,B^n)=A\otimes B^n$?
 A: As has been mentioned in comments, we always have a natural isomorphism $(A\otimes B)^n\cong A\otimes B^n$ where $A,B$ are arbitrary modules over a ring: the isomorphism $A\otimes (B\oplus C)\to (A\oplus B)\otimes(A\oplus C)$ just looks like $ a\otimes (b,c) \mapsto (a\otimes b,a\otimes c), (a\otimes b,a'\otimes c)\mapsto a\otimes (b,0)+a'\otimes (0,c)$. 
There are two reasons this may be a valuable approach to take rather than passing through the isomorphism $A\otimes B\cong \mathcal{L}(A^*,B)$. If $A$ is infinite-dimensional and $B$ is positive-dimensional the latter isomorphism always fails: using $|\cdot |$ for dimension, $|A\otimes B|=|B||A|<|\mathcal{L}(A^*,B)|=|B||A^*|$. If $A$ and $B$ are not free modules, it may fail even if they're "finite-dimensional," i.e. finitely generated. For instance finite abelian groups all have trivial duals, but have nontrivial tensor products except when their orders are relatively prime. So in general we have a map $A\otimes B\to \mathcal{L}(A^*,B)$ using the canonical map $A\to A^{**}$, but it need be neither injective nor surjective.
