# Tensor products of maps

Let $V, W, U, X$ be $R$-modules where R is a ring. At what level of generality, if any is it true that the maps (I always mean linear) from $V \otimes W$ to $U \otimes X$ can be identified with $L(V, U)\otimes L(W, X)$ where $L(., .)$ is the space of maps, via the mapping that takes a tensor product of maps to the map that acts on elementary tensors "componentwise?" I can see that this natural map that might establish the identification is a homomorphism from $L(V, U)\otimes L(W, X)$ to $V \otimes W$ to $U \otimes X$. When it makes sense to speak of dimensions, I can also see that the dimension is suggestive that perhaps it is an isomorphism. But is it one at any level of generality of $R$? This is to justify the usual notation of $f \otimes g$ to refer to the map between tensor products, when the same symbol already refers to the element in the tensor product of $L(. , .)$ spaces. I suppose even without the isomorphism, and only a homomorphism, the notation is already well-defined, but I'd like to know anyway.

Edit: To clarify for the reader, the universal property has been used twice. Once to establish that $f \otimes g$ defines a map, and a second time to show that the map taking (f, g) to $f \otimes g$ the map defines yet another map, which is the homomorphism in question.

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Suppose R is commutative. Your map is well-defined. If one of the following conditions hold, your map is an isomorphism. (1) $V$ and $W$ are finitely generated projective modules over $R$. (2) $V$ and $U$ are finitely generated projective modules over $R$. (3) $W$ and $X$ are finitely generated projective modules over $R$. For the proof, see Bourbaki, Algebra II. – Makoto Kato Jun 9 '12 at 6:04
Thank you, but is there an "obvious" proof when R is a field, or even the field of complex numbers? I happen to be interested in this question mostly for the purpose of understanding Von Neumann Algebras. – Jeff Jun 9 '12 at 19:23
Hint: Suppose $V = R^n$. Then $L(V, U)$ is isomorphic to $U^n$. – Makoto Kato Jun 9 '12 at 19:44

Ah there is a "bare hands" proof as I desired in the field case. You just say: suppose $f \otimes g$ considered as a map is 0. Then it's 0 applied to the fundamental tensors. Suppose f is not 0. I will even prove that g is 0. Let $v$ be a member of $B$ which is a basis for $V$ such that $f(v)$ doesn't vanish. Now for any $w$ in $W$ we find that $f(v)\otimes g(w)=0$. Suppose toward a contradiction that $g(w)\neq0$. Then extend $g(w)$ to a basis of X. Extend $f(v)$ to a basis of U. (I am critically using here that $R$ is a field. Not even a free module would suffice because I need to know I can extend to a basis, not just that I can find one.) Then define the bilinear map that takes $(f(v), g(w))$ to 1 and pairs all other basis vector combinations together to 0. (It maps into $R$ the field.) Then by page 10 characterization (2) here we are done: http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf
A free module would suffice if you're working over an integral domain, even if $f(v)$ is not part of a basis. See Theorem 4.21 at the link you posted in your answer. – KCd Jun 10 '12 at 5:14