Tensor products are rather useful gadgets that are ubiquitous in applications, for example in quantum mechanics.

Definition Let V,W be vector spaces over a field k. A bilinear function f:VxW-->L to a vector space L is called a tensor product of V and W if it satisfies the following universal property:

Whenever g:VxW-->M is a bilinear function to a vector space then there exists a unique linear transformation F:L-->M such that Ff=g.

At this point we don't know that there can be any such tensor products. But we will show that if there is a tensor product then it is unique. So suppose that f:VxW-->L and g:VxW-->M both satisfy the universal property. By the universal property for f (with the other bilinear map being g) there exists a linear transformation F:L-->M such that Ff=g. Likewise using the universal property for g (with f the other bilinear map) there exists a linear transformation G:M-->L such that Gg=f. From these two equations it follows that FGg=g. But now think about the universal property for g when the other bilinear map is just g. Clearly a solution to the universal property is the the identity function 1M:M-->M. But as FGg=g another solution is FG. By the uniqueness in the universal property we deduce that FG=1M. By a symmetric argument we see that GF=1M. So we see that F is an isomorphism of vector spaces. By the uniqueness of the universal property it is the unique isomorphism with Ff=g. Putting it all together.

Theorem If a tensor product exists then it is unique up to unique isomorphism.

Notation Thus we can speak of the tensor product of V and W. It is usual to abuse notation (as I did in the theorem) and refer to the vector space as the tensor product with the bilinear map being implicit. The notation is VxW (actually usually the x is denoted by a circle with a x in it).

That just leaves one tiny detail. Proving that there is a tensor product. I'll do the finite dimensional case to keep the notation simpler but the same idea works in general. Fix ei : 1<=i<=n and fj : 1<=j<=m bases for V and W. Now let L be some mn-dimensional vector space with basis gi,j for 1<=i<=n and 1<=j<=m. Define a map f:VxW-->L by

f(a1e1+...+ anen, b1f1+...+ bmem) = Sum(1<=i<=n,(1<=j<=m) aibjgi,j.
It is routine to check that f satisfies the universal property and so is the tensor product of V and W. Note that the above shows:

Proposition If V,W are finite dimensional vector spaces then dim VxW = (dim V).(dim W)

More generally, if M and N are modules for a commutative ring R then a tensor product of M and N is a bilinear map f:MxN-->L to an R-module that is universal. That is, whenever g:MxN-->K is a bilinear map to a module there is a unique R-module homomorphism F:L-->K such that Ff=g. Exactly as above one shows that such a tensor product is unique up to unique isomorphism. The construction is a little more involved though.