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Let $F$ be a commutative field, and let $U$, $V$, and $W$ be finite dimensional vector spaces over $F$. How can one prove $(U \otimes V) \otimes W \cong U \otimes (V \otimes W)$ without using the universal property?

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Why would you want to avoid the universal property? It gives a pretty explicit map. –  Dylan Moreland Jun 16 '12 at 8:32
@DylanMoreland Just for the sake of learning an alternative proof. –  Eugene Jun 16 '12 at 8:33
The snide answer is that they have the same dimension and they're vector spaces :) But you probably want more than that. –  Dylan Moreland Jun 16 '12 at 8:34
There's no reason to avoid the universal property. It tells you which isomorphism is the most important one! (It's the one that most naturally extends to an isomorphism of functors and that generalizes most naturally to all possible parenthesizations of $n$ factors.) What the universal property really tells you is that it is somewhat unnatural to regard the tensor property as a thing that takes two arguments; it is actually a collection of things that take $n$ arguments which satisfy a natural compatibility relation (see en.wikipedia.org/wiki/Multicategory ). –  Qiaochu Yuan Jun 16 '12 at 11:49
That is, there is a functor that ought to be called the ternary tensor product $U \otimes V \otimes W$ which has a universal property with respect to trilinear maps and which one can define without defining the usual tensor product, and similarly etc. –  Qiaochu Yuan Jun 16 '12 at 11:52
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1 Answer

up vote 3 down vote accepted

Note that you only need to show they have equal dimensions.

Then, you can use the fact that $U \otimes V$ has basis $(u_i \otimes v_j)$ when $U$ and $V$ are vector spaces where $(u_i)$ and $(v_i)$ are bases for $U$ and $V$.

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So if I were to change the question to modules would it still be possible to prove without the universal property? –  Eugene Jun 16 '12 at 8:40
This really makes me appreciate the universal property then! Thanks! –  Eugene Jun 16 '12 at 8:45
@Eugene You don't have "dimension" when you deal with modules over general commutative rings. –  user38268 Jun 16 '12 at 10:00
@Eugene This is an order from the supreme commander: Use universal properties. –  user38268 Jun 16 '12 at 10:11
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