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There is a universal property for free modules, where for any map $f:B\rightarrow S$ there is a map $g:F(B)\rightarrow S$ such $g\cdot b=f$ where $b$ is the canonical map from the basis set into the free module $F(B)$.

This property is used in the proof (linked below) that the tensor product is the only module that satisfies the "universal property for tensor products."

http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf

The universal property for tensor products replaces $f$ above with a bilinear map. Now suddenly the free module isn't good enough; we the only module which satisfies the universal property is the tensor product. The free module can't also satisfy it, as the tensor product is unique up to isomorphism.

So in short, I don't understand why the free module $F(V\times W)$ doesn't satisfy the universal property for tensor products!

Thanks for any help!

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The canonical map $V \times W \to F(V \times W)$ is not bilinear. One way to construct the tensor product $V\otimes W$ is that it is the largest quotient of $F(V \times W)$ for which the composite of this canonical map with the map to the quotient is bilinear.

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  • $\begingroup$ Oh I see, thanks a lot. $\endgroup$ – James Machin May 14 '14 at 16:29
  • $\begingroup$ @JamesMachin In particular, you can define the quotient to be the intersection of all kernels of maps $F(V\times W)\to M$ induced from a bilinear map $V\times W\to M$. $\endgroup$ – mez May 14 '14 at 18:36

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