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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
up vote 0 down vote accepted

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:

That shows injectivity of my natural map. Surjectivity follows because the dimensions are right. (Again this is a field argument, or at least module over a PID or something like that, but definitely for fields.)

This works when the vector spaces are finite dimensional.

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Incidentally I have no idea why it thinks I'm a new person every time I post now. I never had this problem before. – Jeff Jun 10 '12 at 2:50
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
@Jeff: one of your accounts is unregistered, and unregistered accounts are managed entirely by cookies. If you've recently wiped your cookies, that would explain it. Your accounts have been merged. If you keep using the registered account things should be fine. – Qiaochu Yuan Jun 10 '12 at 11:02

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