Free Modules
All modules are over a fixed ring A.
We already mentioned finite free modules earlier. Here we will consider general free modules.
Definition.
Let
be any set. The free A-module on I is a direct sum of copies of A, indexed by
:
More generally, an A-module M is free if
for some I. Note that M is finite free if we can pick I to be finite.
Exercise A
Prove that if A is a non-trivial ring and , then
. [Hint: pick a maximal ideal of A; reduce this to a statement in linear algebra.]
For each , let
which takes zeros at all entries except i, where it takes .
Why do we use the direct sum and not the product? Well, the direct sum allows us to have the following nice properties. Following the case of linear algebra, we define:
Definition.
Let M be an A-module. A collection of elements
of M is said to be linearly independent over A, if for any distinct
,
If
is linearly independent and generates the whole module, we call it a basis of M.
The following result comes at no surprise.
Proposition 1.
The
form a basis of
, called the standard basis.
M is free if and only if it has a basis indexed by I.
Proof
Easy exercise. ♦
The free module also satisfies the following universal property.
Universal Property of Free Module.
Let M be an A-module, and
be a set of elements in M, indexed by I; we write
. Then there is a unique A-linear map
Note
In other words, we have a bijection of sets:
Exercise B
- Prove the universal property.
- Prove that
is surjective if and only if the
generate M.
This universal property is a special case of adjoint functors in category theory, which we will see later.
Projective Modules
One can imagine projective modules as a generalization of free modules. Geometrically, if we think of modules over a coordinate ring as vector bundles over V, then projective modules correspond to locally trivial bundles. To get acquainted with this concept, first we give the abstract definition before going through its implications.
Recall that for any A-module M, the functor is left-exact.
Definition.
The module M is said to be projective if
is exact.
Thus, M is projective if and only if: for any surjective , the map
is surjective. Equivalently, any
lifts to some
such that
:
Note
We only demand existence, not uniqueness, of h, i.e. this is not a universal property.
Lemma 1.
A free module is projective.
Proof
Without loss of generality let and
be an A-linear map. Let
, where
are elements of the standard basis. Since
is surjective, for each
there exists
such that
. By the universal property of the free module, there is a unique
which maps
for each
. Then
. Hence
. ♦
Splitting Lemma
Recall that for a submodule we do not have
in general. However, suppose we can find another
such that:
then we get an isomorphism by mapping
. The proof for this is a straight-forward exercise.
Splitting Lemma.
Suppose we have the following short exact sequence of A-modules.
If there is an A-linear map
such that
, then
by mapping
.
Note
For any A-modules N and P, we always have the canonical sequence
The map then satisfies
. The splitting lemma effectively states the converse: existence of such an i means the exact sequence is isomorphic to this form.
Proof
Note that since is injective so is
. Now we prove the following:
For the first claim, let and
. Let
. Note that
so we have
for some
. This gives
.
For the seecond claim, suppose for some
and
. Then
. ♦
Exercise C
Prove the other splitting lemma: if there is an A-linear map such that
, then
.
Projective vs Free
The splitting lemma gives us the following classification of projective modules.
Theorem 1.
An A-module M is projective if and only if there exists an A-module N such that
is free.
Furthermore, if M is finitely generated and projective, we can pick N such that
is finite free.
Proof
Consider the equivalence in the first statement.
(⇒) Let M be projective. Pick any generating subset (e.g. the whole module) which gives a surjective
mapping
. If M is finitely generated, we pick S to be finite. Since M is projective, the identity map
lifts to a map
such that
. By the splitting lemma we have
for some N. Note that this also proves the second statement.
(⇐) Suppose is free, and hence projective. It suffices to show: if
is projective so is M. But we have a natural isomorphism
of functors in P. Since is projective, the LHS is an exact functor, hence so is the RHS, and it follows that
is exact. ♦
Examples and Consequences
At first glance, it is not entirely clear there are projective modules that are not free. Here are two examples.
Example 1
Let . Then
and
are B-modules such that
. Hence M and N are projective even though they are clearly not free.
Example 2
A more interesting example occurs for , a non-PID. Recall that the prime ideals
are non-principal. However, they are projective because of the following.
Exercise D
Prove that we have an isomorphism of A-modules:
Example 3
Let and
be a maximal ideal of A. Then
is a projective module because of the following.
Exercise E
Prove that we have an isomorphism of A-modules
Now we discuss some consequences of the above results.
Corollary 1.
If
is a collection of projective A-modules, then
is projective.
Proof
For each i let be an A-module such that
is free. Then
is also free. Hence
is projective. ♦
Corollary 2.
If M is a projective A-module, then
is a projective
-module.
Proof
Indeed, if then
because localization commutes with taking direct sums. Hence is also projective over
. ♦
Is projective a local property then? In other words, suppose M is an A-module such that is projective over
for all maximal
, must M be projective? We will answer this in a later article (spoiler: the answer is a conditional ‘yes’).