be a commutative ring, a finite set and the free -module on . Prove there is an -module isomorphism . - Explicitly describe the isomorphism
. In other words, given an -module morphism , what is the corresponding element in ?
We have two options:
A very short proof
The shortest solution is to recall that if
and so (using another problem from the study guide with
If we use this solution to part (1), then for part (2) we need to chain together the actual maps behind the above isomorphisms. When we do so, we'll arrive at the same map described below.
A direct proof
We can also give a direct proof. Define a map
It is also compatible with the left
It is therefore an
As for the inverse map, define
Finally, observe that
and for every
Show that for every
-bimodule there is an -bimodule morphism defined on simple tensors by
Prove that if
is surjective, then so is .
Define a set map
by . This map is linear in : It is linear in
, since is additive: It is
-balanced, since is compatible with the left -actions: It is compatible with the left
-actions: It is compatible with the right
-actions, since is compatible with the right -actions: So, by the universal property of the tensor product the corresponding map
is a well-defined -bimodule morphism. -
is surjective, every element in is of the form for some . Since the simple tensors of the form generate as an -bimodule, it follows that the simple tensors of the form also generate as an -bimodule. But this implies that the elements generate as an -bimodule. Since those same elements generate the image of , this proves the image of is all of ; i.e., the morphism is surjective.
- Prove that every element of
can be written uniquely in the form where . - Let
be nonzero vectors. Prove that in if and only if for some .
First consider simple tensors
. We can write and then use the bilinearity properties of the tensor product to write This shows it is possible to express every simple tensor
in the desired form. For a general tensor we can use the above process to write each summand in the desired form, and then properties of the tensor product allow us to write the entire tensor in the desired form: This proves that every tensor can be written in the desired form.
It remains to prove uniqueness of each such expression. To that end, suppose
can be written both as and as for some vectors . We then have