The fundamental structure theorem for modules over a PID is a direct sum decomposition of each module into a free part and a torsion part. To understand both parts, it helps to look more closely at the ideas of linear (in)dependence and rank.
Let's start with linear dependence, which is exactly as you might guess.
Definition of linear dependence
Let be a left -module. A set of elements is -linearly dependent if there exist (not all zero) such that
(If there is no cause for confusion, we will simply say "linearly dependent" or "linearly independent" without explicit reference to the ring .)
This matches the usual definition of linear dependence in vector spaces. Moreover, if we think of vector spaces as the model example of free modules, then we also have the usual result about dependence:
Rank bounds the number of linearly independent elements
Let be an integral domain and be a free -module on elements; i.e., . Let the corresponding set of generators in be denoted .
Then the set is linearly independent, and any set of more than elements in is linearly dependent.
Let's prove this result. The assumption that is an integral domain will allow us to embed into an -vector space, at which point the result will quickly follow.
As usual, let be the -module morphism that maps . This is the isomorphism that gives the structure of a free -module. The kernel of this morphism (which is trivial) is the set of all formal sums for which ; i.e., its the set of all relations on the set . The fact that this kernel is trivial exactly corresponds to the fact that the set is linearly independent.
Now let be any subset of more than elements and let be the field of fractions of . We have an -module isomorphism and also an injective -module morphism (since is an integral domain), so we also have an injective -module morphism . Then (the image of ) is a set of more than elements in the -dimensional -vector space and hence must be -linearly dependent. For any nontrivial -linear dependence relation among the elements in , clearing denominators yields a nontrivial -linear dependence relation among the elements in . Thus, the set is -linearly dependent.
A better definition?
One could argue that the proof above suggests a "better" definition of linear dependence. Let be any subset of and let be the -module morphism corresponding to that inclusion . The image of this morphism is exactly the submodule of generated by , while the kernel is exactly the set of "relations" on the set . So, the set is linearly independent exactly when is trivial.
Rank of a module
When working with vector spaces, we are used to measuring "size" by the number of elements in a basis. This is the definition of the dimension of a vector space. The two key properties that allow us to use this language are: 1) every vector space has a basis; and 2) two bases for the same vector space always have the same cardinality.
If we were to work strictly with free modules, then we might consider measuring the size of the module by the size of a "basis" for that free module. When working with general modules, however, it is no longer the case that every module will have a basis. In other words, we will not always be able to find a linearly independent set of generators for a given module. For example, if is a torsion module (i.e., for every there exists some nonzero with ), then every nonempty set in is linearly dependent.
Even if is torsion free, it still might not be a free -module. For example, you can show that when a ring is considered as an -module, an ideal is a free -module exactly when it is principal. So for example, in the ring the ideal is not principal and hence not a free -module. The ring is an integral domain, though, so is torsion free.
Is there still some way to measure the "size" of a module? Yes, there's at least one way:
Definition of rank of a module
Let be an integral domain and be an -module. The rank of is the maximum[1] number of -linearly independent elements in .
When is a free -module, our result above proves this notion of rank agrees with our previous notion; i.e., if then has rank . When is a field, this notion of rank matches the dimension of an -module as a vector space.
The structure of free modules over a PID
Before we can determine the general structure of modules over a PID, we must first understand the structure of free modules over a PID. More specifically, we would like to know understand the submodules of a free module and how bases for submodules relate to bases for the module.
The structure of free modules over a PID
Let be a principal ideal domain and be a free -module of finite rank . For every submodule of :
is free of rank ; and
there exists a basis for and nonzero elements such that is a basis for and[2]
Let's walk through the proof of this one. If is the trivial submodule then its rank is 0 and its basis is the empty set, so there's nothing to prove. Now suppose is nontrivial. We'll break this long proof into manageable subsections.
The general idea
The general idea of the proof is to create a direct sum decomposition that also induces a direct sum decomposition with the prescribed properties. One of those properties (the divisibility condition on the ) tells us that should be the "smallest" element among the ; i.e., correspond to the largest ideal among the ideals . So that's where we begin: by looking for a projection from for which the image of is as large as possible.
We begin by fixing a temporary basis for . This is equivalent to fixing an -module isomorphism . This also allows us to define the -module projection morphisms . Using these projections, each element can be written uniquely as
We will use this later in the proof.
Finding the element
Note that for every -module morphism the image of is a submodule of , i.e., an ideal of . Since is a PID this ideal is principal, say for some . Now consider the collection of all such principal ideals in that are also nontrivial:
We first note that this collection is nonempty: since is not the trivial submodule, for at least one of the projection morphisms the image must be nontrivial, otherwise we would have for all
Since is Noetherian the collection has at least one maximal element. In other words there is an -module morphism so that the principal ideal is not properly contained in any other element of .[3] Let and be any element with . Note that by the definition of .
Constructing the element
Our next goal is to construct an element so that . Intuitively, we already have and so it would be nice to simply take . We would then have . However, there is no guarantee that the element is actually invertible. We only know that it is nonzero and that is a PID, but not necessarily a field. So we need to be a little bit tricky.
We first show divides for every -module morphism . To see this, fix some -module morphism and let be the ideal generated by and . Since is a PID this ideal is principal, so for some . We can then write for some . But now consider the -module morphism defined by . By construction we have , so that and hence . But we also have so by the maximality of we must have equality: . This prove and hence ; i.e., divides .
We now apply the above property to the projection morphism , and so we see that divides for each . Write for some and define
By construction we have
We therefore have that and hence (since is nonzero and is an integral domain)
Verifying can be part of a basis for
We will now verify that can be taken as one element in a basis for and that can be taken as one element in a basis for . First, let be an arbitrary element and write
Note that
and so is in the kernel of . This shows that we at least have . To see that this is a direct sum decomposition, suppose for some . Then
Thus, we do indeed have and hence we have a direct sum decomposition . This implies that can indeed be taken as one element in a basis for .
Verifying can be part of a basis for
Observe that for every the element is divisible by (since generates the ideal ). So given any we can write for some . Then we can write
By the same computation as above, the second term in the above sum is an element of that is in the kernel of :