Linear independence, rank and the structure of free modules

Linear dependence in modules

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 M be a left R-module. A set of elements {m1,,mk}M is R-linearly dependent if there exist r1,,rkR (not all zero) such that

r1m1++rkmk=0M.

(If there is no cause for confusion, we will simply say "linearly dependent" or "linearly independent" without explicit reference to the ring R.)

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 R be an integral domain and M be a free R-module on k elements; i.e., MF({x1,,xk}). Let the corresponding set of generators in M be denoted {m1,,mk}.

Then the set {m1,,mk} is linearly independent, and any set of more than k elements in M is linearly dependent.

Let's prove this result. The assumption that R is an integral domain will allow us to embed M into an F-vector space, at which point the result will quickly follow.

As usual, let π:F({x1,,xk})M be the R-module morphism that maps ximi. This is the isomorphism that gives M the structure of a free R-module. The kernel of this morphism (which is trivial) is the set of all formal sums i=1krixi for which i=1krimi=0; i.e., its the set of all relations on the set {m1,,mk}. The fact that this kernel is trivial exactly corresponds to the fact that the set {m1,,mk} is linearly independent.

Now let SM be any subset of more than n elements and let F=Frac(R) be the field of fractions of R. We have an R-module isomorphism Mi=1kRRk and also an injective R-module morphism RF (since R is an integral domain), so we also have an injective R-module morphism Mi=1kFFk. Then (the image of ) S is a set of more than k elements in the k-dimensional F-vector space Fk and hence must be F-linearly dependent. For any nontrivial F-linear dependence relation among the elements in S, clearing denominators yields a nontrivial R-linear dependence relation among the elements in S. Thus, the set S is R-linearly dependent.


A better definition?

One could argue that the proof above suggests a "better" definition of linear dependence. Let SU(M) be any subset of M and let π:F(S)M be the R-module morphism corresponding to that inclusion SU(M). The image of this morphism is exactly the submodule of M generated by S, while the kernel is exactly the set of "relations" on the set S. So, the set S is linearly independent exactly when ker(π) 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 M is a torsion module (i.e., for every mM there exists some nonzero rR with rm=0), then every nonempty set in M is linearly dependent.

Even if M is torsion free, it still might not be a free R-module. For example, you can show that when a ring R is considered as an R-module, an ideal IR is a free R-module exactly when it is principal. So for example, in the ring Z[x] the ideal I=2,x is not principal and hence not a free Z[x]-module. The ring Z[x] is an integral domain, though, so I 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 R be an integral domain and M be an R-module. The rank of M is the maximum[1] number of R-linearly independent elements in M.

When M is a free R-module, our result above proves this notion of rank agrees with our previous notion; i.e., if MF({x1,,xk}) then M has rank k. When R=F is a field, this notion of rank matches the dimension of an F-module M 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 R be a principal ideal domain and M be a free R-module of finite rank k. For every submodule N of M:

  1. N is free of rank lk; and

  2. there exists a basis {m1,,mk} for M and nonzero elements a1,,alR such that {a1m1,,alml} is a basis for N and[2]

    a1a2al.

Let's walk through the proof of this one. If N is the trivial submodule then its rank is 0 and its basis is the empty set, so there's nothing to prove. Now suppose N 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 M=m1mk that also induces a direct sum decomposition N=a1m1alml with the prescribed properties. One of those properties (the divisibility condition on the ai) tells us that a1 should be the "smallest" element among the ai; i.e., correspond to the largest ideal a1R among the ideals aiR. So that's where we begin: by looking for a projection from MR for which the image of N is as large as possible.

We begin by fixing a temporary basis {m1,,mk} for M. This is equivalent to fixing an R-module isomorphism MF({x1,,xk})RRRk. This also allows us to define the R-module projection morphisms πi:MR. Using these projections, each element mM can be written uniquely as

m=i=1kπi(m)mi.

We will use this later in the proof.

Finding the element a1

Note that for every R-module morphism ϕ:MR the image ϕ(N) of N is a submodule of R, i.e., an ideal of R. Since R is a PID this ideal is principal, say ϕ(N)=aϕ for some aϕR. Now consider the collection S of all such principal ideals in R that are also nontrivial:

S={aϕϕHomR(M,R),aϕ0}.

We first note that this collection is nonempty: since N is not the trivial submodule, for at least one of the projection morphisms πi:MR the image πi(N) must be nontrivial, otherwise we would have for all nN

n=i=1kπi(n)mi=i=1k0Rmi=0.

Since R is Noetherian the collection S has at least one maximal element. In other words there is an R-module morphism ν:MR so that the principal ideal ν(N)=aν is not properly contained in any other element of S.[3] Let a1=aν and n1N be any element with ν(n1)=a1. Note that a10 by the definition of S.

Constructing the element m1

Our next goal is to construct an element m1M so that ν(m1)=1R. Intuitively, we already have ν(n1)=a1 and so it would be nice to simply take m1=a11n1. We would then have ν(m1)=ν(a11n1)=a11ν(n1)=a11a1=1R. However, there is no guarantee that the element a1R is actually invertible. We only know that it is nonzero and that R is a PID, but not necessarily a field. So we need to be a little bit tricky.

We first show a1 divides ϕ(n1) for every R-module morphism ϕ:MR. To see this, fix some R-module morphism ϕ:MR and let I=a1,ϕ(n1) be the ideal generated by a1 and ϕ(n1). Since R is a PID this ideal is principal, so I=d for some dR. We can then write d=r1a1+r2ϕ(n1) for some r1,r2R. But now consider the R-module morphism ψ:MR defined by ψ=r1ν+r2ϕ. By construction we have ψ(n1)=r1ν(n1)+r2ϕ(n1)=r1a1+r2ϕ(n1)=d, so that dψ(N) and hence dψ(N). But we also have a1dψ(N) so by the maximality of a1 we must have equality: a1=d=ψ(N). This prove a1=d and hence ϕ(n1)a1; i.e., a1 divides ϕ(n1).

We now apply the above property to the projection morphism πi:MR, and so we see that a1 divides πi(n) for each i=1,,k. Write πi(n1)=a1bi for some biR and define

m1=i=1kbimi.

By construction we have

a1m1=i=1ka1bimi=i=1kπi(n1)mi=n1.

We therefore have that a1=ν(n1)=ν(a1m1)=a1ν(m1) and hence (since a1 is nonzero and R is an integral domain)

ν(m1)=1R.

Verifying m1 can be part of a basis for M

We will now verify that m1 can be taken as one element in a basis for M and that a1m1 can be taken as one element in a basis for N. First, let mM be an arbitrary element and write

m=ν(m)m1+(mν(m)m1).

Note that

ν(mν(m)m1)=ν(m)ν(m)ν(m1)=ν(m)ν(m)1R=0R

and so mν(m)m1 is in the kernel of ν:MR. This shows that we at least have M=m1+ker(ν). To see that this is a direct sum decomposition, suppose rm1ker(ν) for some rR. Then

0R=ν(rm1)=rν(m1)=r1R=r.

Thus, we do indeed have m1ker(ν)=(0) and hence we have a direct sum decomposition M=m1ker(ν). This implies that m1 can indeed be taken as one element in a basis for M.

Verifying a1m1 can be part of a basis for N

Observe that for every nN the element ν(n) is divisible by a1 (since a1 generates the ideal ν(N)). So given any nN we can write ν(n)=ba1 for some bR. Then we can write

n=ν(n)m1+(nν(n)m1)=ba1m1+(nba1m1).

By the same computation as above, the second term in the above sum is an element of N that is in the kernel of ν: