The theory of modules over a PID is very interesting since it is simple enough to understand while giving a feel for deeper ring theory questions.
PID’s show up everywhere and many common examples of modules are over PID’s so this is a good place to start anyways.
Throughout these notes consider a ring to be a PID.
The main goal (and final theorem) is to classify all modules over .
A notion of dimension
Let be a free R-module, and say that for some finite set .
The cardinality of is well-defined (that is if then .)
Thus, we write for the cardinality of .
This builds in a canonical way off the idea of dimension for a vector space.
We know that dimension is invariant for any choice of basis in a vector space, and that it “plays nice” with subspaces, quotienting, etc.
Proof
Let be a prime element.
Then is a module over the field (prime ideals are maximal in PID).
A module over a field is just a vector space, which has a well defined-notion of dimension.
Now we can note that is a basis for .
Finitely generated submodules
Let be a (finitely generated) free -module.
Then any submodule is free and
Proof
Let be finitely generated by , so that .
For each define the submodule
We can use these submodules to prove the statement inductively.
For the base case, since so for some .
So is free with dimension 1 or 0.
Next, assume that is free with dimension at most .
Let
(these are the elements of that allow “extending” into the next submodule).
Since is a -module, then is an ideal.
is principal, so for some .
If then and we’re done.
If then let be such that the coefficient on is .
If then the coefficient on is divisible by (by the above).
This means that , but then
so the induction works.
Torsion and free modules
A finitely generated torsion free -module is free.
Proof
Let be finitely generated with generateors and torsion free.
Let be a maximal subset of linearly independent elements such that
For every generator there exists (not all 0) such that
otherwise it would contradict the maximality condition on .
Also, otherwise would not be linearly independent.
This gives that .
We get an for each , so let since is a PID.
So
Therefore, the map
is an injective module homomorphism with the image contained in a free module, so the image is free.
Pulling back generators
Let be a surjective module homomorphism.
If is free, then there exists a free submodule such that is an isomorphism and
Proof
Let be a basis of .
Let be elements such that and consider the submodule .
Say
since is free, this means that for all and is linearly independent and is free.
Now let .
can be written using the basis as
This implies that
so .
By construction so .
Quotienting out torsion
Let be a finitely generated -module.
Then is free.
Moreover, there exists a free submodule such that
Lastly, is well-defined.
Proof
For , we can denote the equivalence class (image in ) by .
Let be non-zero, and .
Then , so .
In other words, for some non-zero .
is a PID, so .
This means that , i.e. .
This gives that is torsion free.
Since it is a quotient of finitely generated modules, is finitely generated.
Using the propositions above, we can see that splits into
using the projection map to .
Classifying torsion submodules
Let be a finitely generated torsion -module.
Then
Moreover, for each prime (up to a multiplying by a unit), there is an isomorphism
for unique integers
Proof
First statement
Let be an exponent of .
Then say for relatively prime.
Claim:
where
Proof of claim: By Bezout’s theorem,
for some .
Let , then
Since , then .
By the same idea, .
Now let , we have
Therefore,
Thus, by induction on the number of distinct prime factors in ,
Classification of finitely generated modules
Let be a finitely generated -module.
then
for a unique , which is called the rank or Betti number of , and unique primes and natural numbers (these are unique up to permutation and multiplication by units of ).
Moreover, the isomorphism type of is unique determined by its rand and elementary divisors.
Proof
Let be finitely generated -modules and a morphism.
restricts to a morphism
Thus, for the diagram
the morphism descends to a morphism
If is an isomorphism, then and are also isomorphisms.
This means and have the same Betti numbers.
and have the same elementary divisors, this is proven using induction by quotienting off submodules of maximal order.