# Modules

Having dipped our toes into algebraic geometry, we are back in commutative algebra.

Next we would like to introduce “linear algebra” over a ring *A*. Most of the proofs should pose no difficulty to the reader so we will skip them. We recommend that the reader prove these properties at least once.

Definition.An

A-moduleis an abelian group together with a map , written as , such that the following hold for all and .

- .
- .
- .
- .

### Alternative Look

If we fix , we obtain a “multiplication-by-*a*” map defined by . The last property says is an (additive) homomorphism so we get a map , where is the set of all group homomorphisms . Now is not just any old set – it has a natural structure of a (non-commutative) ring if we introduce the following definitions:

In words, we say “addition is defined on the image and product is given by composition”.

Now that has a ring structure, the first three axioms of the definition says is a *ring homomorphism*.

### Easy Properties

Immediately, we can show the following:

E.g. *a*(-*m*) = -(*am*) because is an additive homomorphism.

### Examples

1. Since the homomorphism is unique, every abelian group is automatically a -module.

2. A module over a field *k* is the same as a vector space over *k*.

3. Any ring *A* is a module over itself. More generally if is a subring, then *B* is an *A*-module.

4. If *M* and *M’* are *A*-modules, then becomes an *A*-module via

.

5. From examples 3 and 4, we get with

.

6. Let be a product of two rings. Then any *B*-module *M* gives an *A*-module ; similarly, any *B’*-module *M’* gives an *A*-module .

7. More generally for , any *B*-module *N* and *B’*-module *N’* gives an *A*-module .

# Basic Constructions

Just as we have subspaces of vector spaces, we have:

Definition.If M is an A-module, a A-

submoduleis a subgroup N of such that for all we have .

And this gives:

Proposition.Let N be an A-submodule of M. The quotient of by the subgroup N has a natural structure of an A-module, given by:

.

Thus we obtain the

quotient module.

**Proof**. Easy exercise. ♦

Next, we have the linear maps.

Definition.Let M, N be A-modules. A

module homomorphismis a function such thatWe will also say f is an

A-linear map(or justlinearmap if the base ring is implicit).A

monomorphism(resp.epimorphism,isomorphism) of modules is an injective (resp. surjective, bijective) homorphism of modules.

Immediately we have the following.

First Isomorphism Theorem.Let be a linear map of A-modules.

- The
kernelof f, , is a submodule of M.- The
imageof f, , is a submodule of N.- The map f induces an isomorphism of modules

**Proof**. Easy exercise. ♦

**Exercise**

Let be a product of two rings. For a *B*-module *N* and *B’*-module *N’*, take the *A*-module . Must every submodule of *M* be of the form for *B*-submodule and *B’*-submodule ?

# Ideals as Modules

Recall that every non-trivial ring *A* comes with two modules.

- The
**zero module**is the trivial group (0, +), and for all . - The ring A is a module over itself, where is given by ring product .

Lemma.For any ring A, a subset of A is a submodule if and only if it is an ideal of A.

**Proof**. Easy exercise. ♦

## Some Notes

Although the lemma is a trivial observation, it highlights an important philosophy: *modules can be considered as a generalization of ideals*. For an ideal , it is geometrically more natural to associate it with as a module, although the reasons are not clear for now.

Note that an ambiguity may arise if we were sloppy with words: for , as a module, could mean the ring as a module over itself, or the quotient *A*-module of by the submodule . We will try to be extra careful in stating the base ring. Often, the base ring is used as a subscript in notation, e.g. means a certain construction from *M* by regarding it as an *A*-module.

*For readers with some experience with differentiable manifolds*, another geometrical application of modules is as follows. Recall that closed sets *V* correspond with their coordinate ring *k*[*V*]. Modules over *k*[*V*] then correspond to “vector bundles” over the set *V*. Of special interest are those bundles which are “locally free”, i.e. close to each point the vector bundle restricts to a trivial one.

# Operations on Submodules

Finally, the operations on ideals can be generalized.

Proposition.Let be a collection of submodules of M, over some fixed base ring.

- is a submodule of M.
- Let be the set of all finite sums , where . This is a submodule of M.
- If is an ideal of A, let be the set of all finite sums for and . This is a submodule of M.

**Proof**. Easy exercise. ♦

**Exercise**

An *A*-module is said to be **simple** if it is non-zero and its only submodules are 0 and itself. Prove that a simple *A*-module must be isomorphic to for some maximal ideal .

**Note**

When *M* = *A* and each is an ideal of *A*, these definitions are consistent with the earlier ones for ideals. Thus we can think of them as generalizations of the corresponding operations on ideals.

Note that product of submodules is not defined since we cannot multiply elements in a module. But as we shall see later, one can forcibly define a “product space” *P* for modules *M* and *N* so that and multiply to give an element in *P*. That leads us to tensor products.

The next result is important, so we will state it separately.

Proposition.If M is an A-module and is an ideal of A, then has a canonical structure of an -module, via

.

**Proof**. Easy exercise. ♦

The above construction is a “canonical” way of constructing a module over given a module over *A*. We will state more precisely what canonicity means in later articles, by universal properties.