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

  1.  ,
  2.  ,
  3.   and
  4.  .

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  .

  1.  
  2.  
  3.  
  4. 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:

  1. Commutative laws:  ,  
  2. Idempotency laws:  ,  
  3. Absorption laws:  ,  
  4. 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

  1. 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
  2. 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

  1.   and
  2.  .

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:

  1.   is an isomorphism
  2.  
  3.   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  . 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  . 

Exercises edit