# Generated Submodule

Since the intersection of an arbitrary family of submodules of *M* is a submodule, we have the concept of a *submodule generated by a subset*.

Definition.Given any subset , let denote the set of all submodules of M containing S. Note that is non-empty since . Now, the

submodule of M generated by Sis:

This is the “smallest submodule of *M* containing *S*“; to be precise, (a) it is a submodule of *M* containing *S*, and (b) any submodule *N* of *M* containing *S* must also contain (*S*). With this, we see that given a family of submodules , the sum is simply the submodule of *M* generated by their union .

Now (*S*) has a very simple description.

Lemma.For any subset S of M, the submodule consists of the set of all finite linear combinations

Thus we may sometimes call (*S*) the **span** of *S* (over the base ring *A*).

Definition.A module M is said to be

finitely generated(or justfinite) if for some finite subset S of M.A

finitely generated idealis an ideal which is finitely generated as an A-module.

It is not true that every submodule of a finitely generated module must be finitely generated. In fact, there exist rings with ideals which are not finitely generated, so this gives where *A* = (1) is finitely generated but is not.

E.g. let ; more explicitly

Let be the set of all sums with only positive . This ideal is not finitely generated, since any finite subset is contained in for some *n* > 0 so lies outside (*S*).

# Linear Dependence

Note that if is a finite subset, there is a unique *A*-linear

In fact, such a map exists for any finite *sequence* , i.e. some of the elements may repeat.

This map is surjective if and only if *S* generates *M*. Thus *M* is finitely generated if and only if there is a surjective *A*-linear map for some *n* > 0. When is *f* injective?

Proposition.Let and f be as above; f is injective if and only if:

- for any , if , then all .
When that happens, we say that are

linearly independent.

**Proof**. Easy exercise. ♦

We make two important definitions here, inherited from linear algebra.

Definition.

- If S is linearly independent and generates M, we say that S is a
basis.- We say M is
finite freeif it has a finite subset which is a basis.- The
rankof a finite free M is the size of S for any basis .

Note: the zero module is a free module of rank zero since generates M.

Subtle question: is the rank well-defined? In other words, if finite subsets are both bases, must we have ? This question will be answered a few articles later.

# Two More Isomorphism Theorems

Just like the case of group theory, we have the following.

Theorem.Let M be any A-module.

- (
Second Isomorphism Theorem) For any submodules we have

- (
Third Isomorphism Theorem) If we have submodules , then treating as a submodule of we have

**Note**

We have written as to avoid cluttering up the notation.

**Proof**

For the first claim, map by .

- The kernel of this map is .
- The map is surjective since any for is just .

Hence by the first isomorphism theorem, we get our result.

The second claim is similar: map via , which is a well-defined homomorphism. Clearly the map is surjective; the kernel is just *N*/*P*; now apply the first isomorphism theorem. ♦

# Correspondence of Submodules

Finally, for any submodule *P* of *M*, there is a *bijective* correspondence between

- submodules of
*M*containing*P*, and - submodules of
*M*/*P*.

Specifically, if *N* satisfies in the first case, it corresponds to in the second.

[Diagram: correspondence between submodules – arrows indicate inclusion.]

The correspondence respects:

- containment: if and only if ;
- intersection: ;
- sum: .

**Tip**: you can prove all these directly, or you can deduce the second and third properties from the first by noting: the intersection of submodules is the largest submodule contained in every . A similar property exists for the sum.

Finally the case for product is tricky: even if *N* contains *P*, may not contain *P*. However, does and we have:

**Exercise**. Prove this result.

# Hom Module

For *A*-modules *M* and *N*, let be the set of all *A*-linear maps .

Proposition.is an A-module, under the following operations. For A-linear maps and , we have

- ,

**Proof**. Easy exercise. ♦

This generalizes the case of vector spaces, where the set of all linear maps of vector spaces, forms a vector space.

The Hom construction thus creates a new *A*-module out of two existing ones. We have:

Proposition.Let and be A-linear maps of A-modules. These induce A-linear maps

**Proof**. Easy exercise. ♦

## Example

Suppose for some , considered as an *A*-module (not a ring!). Then corresponds to the submodule of *N*. Indeed any *A*-linear map is determined by the image . This can be any element as long as .

[We usually say *n* is **annihilated** by . This will be covered in greater detail later.]

E.g. if and , then picks out the 2-torsion subgroup of *N*.

Now if is an *A*-linear map, the above map can be interpreted as follows:

where the composed map is simply *g* restricted to .

*The upshot is: the process of taking the a-torsion elements of a module can be expressed by Hom. *We write this process as , or simply . All these can be formalized in the language of *category theory*.