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 *1*_{M}:M-->M. But
as *FGg=g* another solution is *FG*. By the uniqueness in
the universal property we deduce that *FG=1*_{M}. By
a symmetric argument we see that *GF=1*_{M}. 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
*V***x***W* (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 *e*_{i} : *1<=i<=n* and
*f*_{j} : *1<=j<=m* bases
for *V* and *W*. Now let *L* be some *mn*-dimensional
vector space with basis *g*_{i,j} for *1<=i<=n*
and *1<=j<=m*. Define a map *f:VxW-->L* by

*
f(a*_{1}e_{1}+...+ a_{n}e_{n}, b_{1}f_{1}+...+ b_{m}e_{m})
= Sum(1<=i<=n,(1<=j<=m) a_{i}b_{j}g_{i,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 *V***x***W* = (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.