$F^m \otimes F^n \cong F^{mn}$, so what is the bilinear map?

Question

Let $F$ be a field and $m, n \in \mathbb{N}$. By definition of the tensor product, there is a bilinear map $B : F^m \times F^n \to F^m \otimes F^n$ such that if $V$ is an $F$-vector space and $C : F^m \times F^n \to V$ is bilinear, then there exists a unique linear map $\tilde{C} : F^m \otimes F^n \to V$ such that $C = \tilde{C} \circ B$.

Since $\dim F^m \otimes F^n = \dim F^{mn}$, it follows that $F^m \otimes F^n \cong F^{mn}$. Is there a bilinear map $B' : F^m \times F^n \to F^{mn}$ that behaves just like $B$?

Answer 1

The universal bilinear map $F^m \times F^n\to F^m\otimes F^n$ is just $$(\sum_j a_j e_j,\sum_i b_i e_i)\to (\sum_j a_j e_j)\otimes (\sum_i b_i e_i)=\sum_j\sum_i a_j b_i e_j\otimes e_i$$ Sending $e_j\otimes e_i\to e_{ji}$ you get your universal bilinear map $F^m \times F^n\to F^{mn}$.

It is the same as sending $(u,v)$ to the matrix $u v^\top$.