# Commutative Algebra/Modules, submodules and homomorphisms

## Basics[edit]

**Definition 5.1 (modules)**:

Let be a ring. A **left -module** is an Abelian group together with a function

such that

- ,
- ,
- and
- .

Analogously, one can define right -modules with an operation ; the difference is only formal, but it will later help us define bimodules in a user-friendly way.

For the sake of brevity, we will often write *module* instead of *left -module*.

**Exercise 5.1.1**: Prove that every Abelian monoid together with an operation as specified in 1.) - 4.) of definition 5.1 is already a module.

## Submodules[edit]

**Definition 5.2 (submodules)**:

A subgroup which is closed under the module function (i.e. the left multiplication operation defined above) is called a **submodule**. In this case we write .

The following lemma gives a criterion for a subset of a module being a submodule.

**Lemma 5.3**:

A subset is a submodule iff

- .

**Proof**:

Let be a submodule. Then since since we have an Abelian group and further due to closedness under the module operation, also .

If is such that , then for any also .

**Definition and theorem 5.4 (factor modules)**:
If is a submodule of , the **factor module** by is defined as the group together with the module operation

- .

This operation is well-defined and satisfies 1. - 4. from definition 5.1.

**Proof**:

Well-definedness: If , then , hence and thus .

- analogous to 3. (replace by )

### Sum and intersection of submodules[edit]

We shall now ask the question: Given a module and certain submodules , which module is the smallest module containing all the ? And which module is the largest module that *is itself contained* within all ? The following definitions and theorems answer those questions.

**Definition and theorem 5.5 (sum of submodules)**:

Let be a module over a certain ring and let be submodules of . The set

is a submodule of , which is the smallest submodule of that contains all the . It is called the **sum of **.

**Proof**:

1. is a submodule:

- It is an Abelian subgroup since if , then

- .

- It is closed under the module operation, since

- .

2. Each is contained in :

This follows since for each and each .

3. is the smallest submodule containing all the : If is another such submodule, then must contain all the elements

due to closedness under addition and submodule operation.

**Definition and theorem 5.6 (intersection of submodules)**:

Let be a module over a ring , and let be submodules of . Then the set

is a submodule of , which is the largest submodule of containing all the . It is called the **intersection of the **.

**Proof**:

1. It's a submodule: Indeed, if , then for each and thus for each , hence .

2. It is contained in all by definition of the intersection.

3. Any set that contains all elements from each of the is contained within the intersection.

We have the following rule for computing with intersections and sums:

**Theorem 5.7 (modular law; Dedekind)**:

Let be a module and such that . Then

- .

**Proof**:

: Let . Since , and hence . Since also by assumption, .

: Let . Since , and since further , . Hence, .

More abstractly, the properties of the sum and intersection of submodules may be theoretically captured in the following way:

### Lattices[edit]

**Definition 5.8**:

A **lattice** is a set together with two operations (called the join or least upper bound) and (called the meet or greatest lower bound) such that the following laws hold:

- Commutative laws: ,
- Idempotency laws: ,
- Absorption laws: ,
- Associative laws: ,

There are some special types of lattices:

**Definition 5.9**:

A **modular lattice** is a lattice such that the identity

holds.

**Theorem 5.10 (ordered sets as lattices)**:

Let be a partial order on the set such that

- every set has a least upper bound (where a least upper bound of satisfies for all (i.e. it is an upper bound) and for every other upper bound of ) and
- every set has a greatest lower bound (defined analogously to least upper bound with inequality reversed).

Then , together with the joint operation sending to the least upper bound of that set and the meet operation analogously, is a lattice.

In fact, it suffices to require conditions 1. and 2. only for sets with two elements. But as we have shown, in the case that is the set of all submodules of a given module, we have the "original" conditions satisfied.

**Proof**:

First, we note that least upper bound and greatest lower bound are unique, since if for example are least upper bounds of , then and and hence . Thus, the joint and meet operation are well-defined.

The commutative laws follow from .

The idempotency laws from clearly being the least upper bound, as well as the greatest lower bound, of the set .

The first absorption law follows as follows: Let be the least upper bound of . Then in particular, . Hence, is a lower bound of , and any lower bound satisfies , which is why is the greatest lower bound of . The second absorption law is proven analogously.

The first associative law follows since if is the least upper bound of and is the upper bound of , then (as is an upper bound for ) and if is the least upper bound of , then since is an upper bound and further, and . The same argument (with and swapped) proves that is also the least upper bound of the l.u.b. of and . Again, the second associative law is proven similarly.

From theorems 5.5-5.7 and 5.10 we note that the submodules of a module form a modular lattice, where the order is given by set inclusion.

### Exercises[edit]

**Exercise 5.2.1**: Let be a ring. Find a suitable module operation such that together with its own addition and this module operation is an -module. Make sure you define this operation in the simplest possible way. Prove further, that with respect to this module operation, the submodules of are exactly the ideals of .

## Homomorphisms[edit]

We shall now get to know the morphisms within the category of modules over a fixed ring .

**Definition 5.11 (homomorphisms)**:

Let be two modules over a ring . A **homomorphism** from to , also called an **-linear function** from to , is a function

such that

- and
- .

The kernel and image of homomorphisms of modules are defined analogously to group homomorphisms.

Since we are cool, we will often simply write morphisms instead of homomorphisms where it's clear from the context in order to indicate that we have a clue about category theory.

We have the following useful lemma:

**Lemma 5.12**:

is -linear iff

- .

**Proof**:

Assume first -linearity. Then we have

- .

Assume now the other condition. Then we have for

and

since due to ; since is an abelian group, we may add the inverse of on both sides.

**Lemma 5.13**:

If is -linear, then .

**Proof**:

This follows from the respective theorem for group homomorphisms, since each morphism of modules is also a morphism of Abelian groups.

**Definition 5.8 (isomorphisms)**:

An **isomorphism** is a homomorphism which is bijective.

**Lemma 5.14**:

Let be a morphism. The following are equivalent:

- is an isomorphism
- has an inverse which is an isomorphism

**Proof**:

**Lemma 5.15**:

The kernel and image of morphisms are submodules.

**Proof**:

1. The kernel:

2. The image:

The following four theorems are in complete analogy to group theory.

**Theorem 5.16 (factoring of morphisms)**:

Let be modules, let be a morphism and let such that . Then there exists a unique morphism such that , where is the canonical projection. In this situation, .

**Proof**:

We define . This is well-defined since . Furthermore, this definition is already enforced by . Further, .

**Corollary 5.17 (first isomorphism theorem)**:

Let be -modules and let be a morphism. Then .

**Proof**:

We set and obtain a homomorphism with kernel by theorem 5.11. From lemma 5.16 follows the claim.

**Corollary 5.18 (third isomorphism theorem)**:

Let be an -module, let and let . Then

- .

**Proof**:

Since and also by definition. We define the function

- .

This is well-defined since

- .

Furthermore,

and hence . Hence, by theorem 5.17 our claim is proven.

**Theorem 5.19 (second isomorphism theorem)**:

Let . Then

- .

**Proof**:

Consider the isomorphism

- .

Then , which is why the kernel of that homomorphism is given by . Hence, the theorem follows by the first isomorphism theorem.

And now for something completely different:

**Theorem 5.20**:

Let be a homomorphism of modules over and let . Then is a submodule of .

**Proof**:

Let . Then and hence . Let further . Then .

Similarly:

**Theorem 5.21**:

Let be a homomorphism of modules over and let . Then is a submodule of .

**Proof**: Let . Then and . Let further . Then .

### Exercises[edit]

**Exercise 5.3.1**: Let be rings regarded as modules over themselves as in exercise 5.2.1. Prove that the ring homomorphisms are exactly the module homomorphisms ; that is, every ring hom. is a module hom. and vice versa.

## The projection morphism[edit]

**Definition 5.22**:

Let be a module and . By the mapping we mean the canonical projection mapping to ; that is,

- .

The following two fundamental equations for and shall gain supreme importance in later chapters, , .

**Theorem 5.23**:

Let be a module and . Then for every set , . Furthermore, for every other submodule , .

**Proof**:

Let first . Then , since . Hence, . Let then . Then there exists such that , that is . Now means that . Hence, .

Let first , that is, for suitable , . Then , which is why by definition . Let then . Then , that is with , that is for a suitable , that is .

The following lemma from elementary set theory have relevance for the projection morphism and we will need it several times:

**Lemma 5.24**:

Let be a function, where are completely arbitrary sets. Then induces a function via , the image of , where . This function preserves inclusion. Further, the function , also preserves inclusion.

**Proof**:

If , let . Then for an . Similarly for .