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?
Proving $(U \otimes V) \otimes W \cong U \otimes (V \otimes W)$ without the universal property
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$\begingroup$
linear-algebra
abstract-algebra
commutative-algebra
tensor-products
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0That 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. – 2012-06-16
1 Answers
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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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3@Eugene $T$his is an order $f$rom the supreme commander: Use universal properties. – 2012-06-16