# 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

freeA-module on I is a direct sum of copies of A, indexed by :More generally, an A-module M is

freeif 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 independentover A, if for any distinct ,If is linearly independent and generates the whole module, we call it a

basisof 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

projectiveif 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’).