# Abstract Algebra/Modules

## Motivation[edit | edit source]

Let *G* be an abelian group under addition. We can define a sort of multiplication on *G* by elements of by writing for and . We can extend this to the case where *n* is negative by writing . We would, however, like to be able to
define a sort of multiplication of a group by an arbitrary ring.

## Definition[edit | edit source]

- Definition 1 (Module)
- Let
*R*be a ring and*M*an abelian group. We call*M*a**left**if there is a function , called a*R*-module*scalar multiplication*, satisfying- ,
- , and

- for all .
- We call
*R*the*ring of scalars*of*M*.

Note: We can also define a right *R*-module analogously by using a function . In particular the third property then reads:

Note that the two notions coincide if *R* is a commutative ring, and in this case we can simply say that *M* is an *R*-module.

**Definition 2:** Given any ring *R*, we can define it's *opposite ring*, , having the same elements and addition operation as *R*, but opposite multiplication. Their multiplication rules are related by . In contrast to group theory, there is no reason in general for a ring to be isomorphic to its opposite ring.

The obervant reader will have noticed that the scalar multiplication in a left *R*-module *M* is simply a ring homomorphism such that for all . We leave it as an exercise to verify that the scalar multiplication in a right *R*-module is a ring homomorphism . Thus a right *R*-module is simply a left *R ^{op}*-module. As a consequence of this, all the results we will formulate for left

*R*-modules are automatically true for right

*R*-modules as well. There are no assumptions that the module is unital, namely that 1m = m for all m in M.

## Examples of Modules[edit | edit source]

- Any ring
*R*is trivially an*R*-module over itself. More interestingly, any left ideal*I*of*R*is also a left*R*-module with the obvious scalar multiplication. In addition, if*I*is a two-sided ideal of*R*, then the quotient ring is an*R*-module with the induced scalar multiplication . - If
*R*is a ring, then the set of matrices with entries in*R*is an*R*-module under componentwise addition and scalar multiplication. More generally, for any set*X*, the set of function from*X*to*R*, with or without finite support, is an*R*-module in an obvious way. - The
*k*-modules over a field*k*are simply the*k*-vector spaces. - As was shown in the introduction of this chapter, any abelian group is a -module in a natural way. ("Natural" here has a rigorous mathematical meaning which will be explained later.
- Let
*S*be a subring of a ring*R*. Then*R*is an*S*-module in a natural way. We can extend this as follows. Let S,R be rings and a ring homomorphism. Then*R*is an*S*-module with scalar multiplication and for all . - Any matrix ring of a ring
*R*is a*R*-module under componentwise scalar multiplication. - If
*S*is a subring of a ring*R*, then any left*R*-module is also a left*S*-module with the restricted scalar multiplication. We will treat this more generally later.

## Submodules[edit | edit source]

**Definition 3: (Submodule)**

- Given a left -module a submodule of is a subset satisfying
- N is a subgroup of M, and
- for all and all we have .

The second condition above states that submodules are closed under left multiplication by elements of ; it is implicit that they inherit their multiplication from their containing module; must be the restriction of .

**Example 4:** Any module *M* is a submodule of itself, called the improper submodule, and the zero submodule consisting only of the additive identity of *M*, called the trivial submodule.

**Example 5:** A left ideal *I* is a submodule of *R* viewed as an *S*-module, where *S* is any (not necessarily proper) subring of *R*.

**Lemma 6:** Let *M* be a left *R*-module. Then the following are equivalent.

- i)
*N*is a submodule of*M* - ii) If and for all , then .
- iii) If and , then .

*Proof*: i) => iii): and are in by the second property, then by the first property of Definition 3.

iii) => ii): Follows by induction on .

ii) => i): Let , , then for arbitrary be have , proving is a subgroup. Now let , then for arbitrary , , proving property 2 in Definition 3. ∎

The lemma gives an alternative characterisation of submodules, and those sets closed under linear combinations of elements.

Analogously to the case of vector spaces, we have ways of creating new subspaces from old ones. The rest of this subsection will be concerned with this.

**Lemma 7:** Let *M* be a left *R*-module, and let *N* and *L* be submodules of *M*. Then is a submodule in *M*, and it is the largest submodule contained in both *N* in *L*.

*Proof*: Let and . Then and since *N* and *L* are submodules, so and is a submodule of *M*. Now, assume that *S* is a submodule of *M* contained in *N* and *L*. Then any must be in both *N* and *L* and therefore in such that , proving the lemma. ∎

Now, as the reader should expect at this point, given submodules *N* and *L* of *M*, the union is in general not a submodule. In fact, we have the following lemma:

**Lemma 8:** Let *M* be a left *R*-module and let *N* and *L* be submodules. Then is a submodule if and only if or .

*Proof*: The left implication is obvious. For the right implication, assume is a submodule of *M*. Then if and , then , which implies that or . Assume without loss of generality that . Then, since *N* is a submodule, we must have , proving . ∎

**Definition 9:** Let *M* be a left *R*-module, and let be submodules for . Then define their *sum*, .

Definition 9 has a straightforward extension to sums over arbitrary index sets. This definition is left for the reader to state. We will only need the finite case in this chapter.

**Lemma 10:** Let *M* be a left *R*-module and let *N* and *L* be submodules. Then is a submodule of *M*, and it is the smallest submodule containing both *N* and *L*.

*Proof*: It is straightforward to see that is a submodule. To see that it is the smallest submodule containing both *N* and *L*, let *S* be a submodule containing both *N* and *L*. Then for any and , we must have . But this is the same as saying that , proving the lemma. ∎

With Lemma 7 and Lemma 10 established, we can state the main result of this subsection.

**Definition 11:** Let *M* be a left *R*-module. Then let be the set of submodules ordered by set inclusion.

**Lemma 12:** Let *M* be a left *R*-module. Then forms a lattice, the join of being given by and their meet by .

*Proof*: Most of the work is already done. All that remains is to check assosiativity, the absorption axioms and the idempotency axioms. The associativity is trivially satisfied, and for all . As for absorption, We have to check and for all , but this is also trivially true. Lastly, we obviously have and for all , so we are done. ∎

**Corollary 13:** Let *M* be a left *R*-module. Then is a modular lattice.

**Note:** Recall that is modular if and only if whenever such that , we have .

*Proof*: Let such that . Since , we have for some , such that . Thus and . On the other hand, we have and , so . ∎

**Definition 14:** Let *M* be a left *R*-module. A submodule *N* is called *maximal* if whenever *L* is a submodule satisfying , then or .

**Theorem 15:** Every submodule of a finitely generated left *R*-module is contained in a maximal submodule.

*Proof*: Let *N* be a submodule, and let . Then *S* is a poset under set inclusion. Let be a chain in *S*, and note that is a submodule containing each , such that *U* is an upper bound for the chain. Then, since each chain in *S* has an upper bound, by Zorn's Lemma *S* has a maximal element, *P*, say. *P* is obviously an ideal containing *N*. By the definition of *S*, *P* is also a maximal submodule of *M*, proving the theorem. ∎

## Generating Modules[edit | edit source]

Given a subset of a left -module , we define the left submodule generated by to be the smallest submodule (w.r.t. set containment) of that contains . It is denoted by for a reason which will become clear in a moment.

The existence of such a submodule comes from the fact that an intersection of -modules is again an -module: Consider the set of all submodules of containing . Since contains , we see that is non-empty. The intersection of the modules in clearly contains and is a submodule of . Further, any submodule of containing also contains the intersection. Thus .

Assuming that is unitary, the elements of have a simple description;

- .

That is, every element of can be written as a finite left linear combination of elements of . This equality can be justified by double inclusion: First, any submodule containing must contain all left -linear combinations of elements of since modules are closed under addition and left multiplication by elements of . Thus, . Secondly, the set of all such linear combinations forms a submodule of containing (use and ) and hence it contains .

## Generating Submodules by Ideals[edit | edit source]

Consider any ring , left ideal , and left -module . One can think of as a subring of (non-unitary when ) and hence is an -module using the regular multiplication by elements of .

If we consider the set we obtain a submodule of . This follows from our discussion of generated submodules. However, since is not unitary, it is not necessary that .

Thus, we may consider the quotient module . Clearly this is an -module but it is also an module under the obvious action.

- Proposition
- Given an -module and ideal of , the module is an -module with multiplication .
- proof.
- To show that this is well defined, we observe that if then and hence
- since . Thus,
- which proves that the action of on is well defined. It follows now that is an -module simply because it is an -module.

## Quotient Modules[edit | edit source]

Recall that any subgroup of an abelian group allows one to construct an equivalence relation; for ,

- .

Cosets of , equivalence classes under the relation above, can then be endowed with a group structure, derived from the original group, and is given the name M/N. The sum of two cosets and is simply .

**Lemma 16** Let *M* be a left *R*-module and *N* be a submodule. Then *M/N*, defined above, is a left *R*-module.

*Proof:* *M/N* is obviously an abelian group, so we just have to check that it has a well-defined *R*-action. Let and . Then we define . The distributivity and associativity properties of the action are inherited from *M*, so we just need well-definedness. Let with . Then since *N* is a submodule, and we are done. ∎

## Module Homomorphisms[edit | edit source]

Like all algebraic structures, we can define maps between modules that preserve their algebraic operations.

- Definition (Module Homomorphism)
- An -module homomorphism is a function from to satisfying
- (it is a group homomorphism), and

When a map between two algebraic structures satisfies these two properties then it called an -linear map.

- Definition (Kernel, Image)
- Given a module homomorphism the kernel of is the set
- and the image of is the set
- .

The kernel of is the set of elements in the domain that are sent to zero by . In fact, the kernel of any module homomorphism is a submodule of . It is clearly a subgroup, from group theory, and it is also closed under multiplication by elements of : for .

Similarly, one can show that the image of is a submodule of .