Having defined submodules, let’s proceed to quotient modules. Unlike the case of groups and rings, any submodule can give a quotient module without any additional condition imposed.

Definition. Let N be a submodule of M. By definition, it’s an additive subgroup which is normal since (M, +) is abelian. So take the group quotient M/N and define a module structure on it via the scalar multiplication:This is called the

module quotient.

The only worrying aspect is that scalar multiplication is well-defined: i.e. suppose *m*+*N* = *m’*+*N*; is it true that (*rm*)+*N* = (*rm’*)+*N* then? Well:

The remaining axioms of a module are then easily verified.

**Examples**

- Clearly,
*M*/{0} =*M*and*M*/*M*= {0}. - Let be the ring of
*n*×*n*matrices with integer entries. Then is a module and is a submodule. The resulting quotient is isomorphic to - Recall that if
*R*=**Z**, the class of*R*-modules is precisely the class of abelian groups. The submodules are precisely the subgroups and quotient modules are the quotient groups. - Let be an ideal of
*R*and*M*be an*R*-module. The module quotient*M*/*IM*is not only an*R*-module, but an (*R*/*I*)-module as well. Indeed, if then multiplying with gives This is well-defined since if and , then we have and which gives us and so*rm*+*I*=*r’m’*+*I*. Thus scalar multiplication by*R*/*I*is well-defined and it’s easy to show that*M*/*IM*is an*R*/*I*-module.

## Homomorphisms

The definition is straightforward.

Definition. Let M and N be R-modules. Amodulehomomorphismis a function f : M → N such that:and

for any and

Notice that since *f* is a homomorphism of the underlying additive groups, we automatically have *f*(0) = 0 and *f*(-*m*) = –*f*(*m*). As in the case of groups, an injective (resp. surjective / bijective) homomorphism is called a **monomorphism** (resp. **epimorphism** / **isomorphism**).

The following properties are hardly surprising:

Proposition. Let be a homomorphism of R-modules.

- If is a submodule, then f(M’) is a submodule of N.
- If is a submodule, then f
^{-1}(N’) is a submodule of M.In other words, the “push-forward” or “pull-back” of a submodule is also a submodule.

**Proof**.

The proof is easy: we know that *f*(*M’*) is an additive subgroup of *N*. Next, if and , then we can write *n* = *f*(*m*) for some *m* in *M’*. This gives *rn* = *r**·f*(*m*) = *f*(*rm*), which is in *f*(*M’*).

Likewise, *f*^{-1}(*N’*) is an additive subgroup of *M*. If and , then *f*(*m*) lies in *N’* so *f*(*rm*) = *r*·*f*(*m*) which lies in *N’*. Hence and we’re done. ♦

In particular, we have the following special cases.

Definition. For a module homomorphism , the pullback is called thekernelof f; this is a submodule of M, denoted ker(f).Likewise, is called the

imageof f; this is a submodule of N, denoted im(f).Finally the

cokernelof f, denoted coker(f), is the quotient .

[ Note: the cokernel is kind of unusual since we didn’t encounter it in the case of groups and rings. Its primary purpose is as a “dual” to the kernel. Roughly, the kernel satisfies some universal property; if we reverse the arrows around the resulting universal property describes the cokernel. ]

As before, we have the standard three isomorphism theorems.

First Isomorphism Theorem. Let be a homomorphism of R-modules. Then

**Proof**.

Construct a map which takes to *f*(*m*). Let’s show that *g* is injective and well-defined. Indeed:

Finally, *g* is surjective by the definition of im(*f*), so we’re done. ♦

Armed with the first isomorphism theorem, the remaining two are a piece of cake.

Second Isomorphism Theorem. Let N, N’ be submodules of M. Then

Third Isomorphism Theorem. Let be submodules. Then N/P is a submodule of M/P and

**Proof**.

For the first result, map *N* → (*N*+*N’*) → (*N*+*N’*)/*N’* by composing the inclusion with projection Apply the first isomorphism theorem. The kernel of this map is while the image is the whole of (*N*+*N’*)/*N’* since any element (*n*+*n’*)+*N’* (for and ) is basically just *n*+*N’*.

For the second result, map *M*/*P* → *M*/*N* by taking *m*+*P* to *m*+*N*. The map is well-defined since if *m*+*P* = *m’*+*P*, then *m*–*m’* lies in *P* and hence *N*, so *m*+*N* = *m’*+*N* also. The map is obviously surjective. Finally, the kernel is

Now apply the first isomorphism theorem. ♦

Let . As in the case of groups and rings, there’s a correspondence between submodules of *M* containing *N*, as well as submodules of *M*/*N*.

Theorem. Let N be a submodule of M. We have a bijection:The correspondence preserves inclusion.

**Sketch of Proof**.

Let *f* : *M* → *M*/*N* be the projection map.

- Any submodule
*P*of*M*gives a submodule*f*(*P*) of*M*/*N*. - On the other hand, any submodule
*Q*of*M*/*N*gives a submodule*f*-1(*Q*) of*M*containing*N*.

One then checks that always holds, and that if *P* contains *N*, then we have . ♦

**Exercise**.

Prove that in the above correspondence, the intersection (resp. sum) of submodules on one side corresponds to the intersection (resp. sum) of submodules on another.

[ *Hint for exercise: avoid “picking elements” from the submodules if possible; instead argue that the intersection or sum satisfies some unique property pertaining to inclusion of submodules.*]

## Direct Sum and Direct Product

It’s clear that given any two *R*-modules *M* and *N*, we can form the product *M* × *N* and let *R* act on it component-wise. On the other hand, with infinitely many modules, there’re two common ways to construct this “product”.

Definition. Consider any class of R-modules .

- The
direct productis the set-theoretic product . It’s given the structure of an R-module, by letting act on it component-wise.- The
direct sumis the subset of for which only finitely many m_{i}‘s are non-zero. This is clearly a submodule of the direct product.

For example, suppose *R *= **Z** and *M _{n}* =

**Z**for

*n*= 1, 2, 3, … .The direct product of the

*M*‘s is the set of all sequences of integers, while the direct sum is the set of sequences of integers in which only finitely many terms are non-zero.

_{n}

The difference may seem subtle, but these two modules turn out to be quite different in their behaviour. In fact, one can even say they’re *dual* to each other, thanks to their universal properties.

Universal Property of Direct Product. For each index j, let be the projection map. If N is a module and is any collection of module homomorphisms, then there’s a unique such that:for all j.

Universal Property of Direct Sum. For each index j, let be the natural inclusion map which takes to the tuple which comprises of all zeros except If N is a module and is any collection of module homomorphisms, then there’s a unique such that:for all j.

For direct product, the collection of maps gives rise to the map by taking:

for all .

One easily checks that this is a module homomorphism and that it is the unique map satisfies the desired requirement.

For direct sum, the maps give by taking

for any

This is well-defined since there are only finitely many non-zero *m _{i}*‘s so the RHS is a finite sum. ♦

**Exercise**

Consider the case where *R*=**Z** and . Explain why doesn’t satisfy the universal property for direct sum and why doesn’t satisfy that of direct product. [ Hint: in one case, the induced map fails to exist; in the other, the induced map is not unique. ]