Modules can be likened to “vector spaces for rings”. To be specific, we shall see later that a vector space is precisely a module over a field (or in some cases, a division ring). This set of notes assumes the reader is reasonably well-acquainted with rings and groups, and has some prior experience with linear algebra, at least the computational aspects.

Throughout this article, we’ll let *R* denote a ring, possibly non-commutative.

Definition. A(left)-moduleover R is an abelian group (M, +) together with a binary operation R × M → M (denoted as r·m or just rm, for ) such that for any and we have:

- ;
- ;
- ;
- .
The map is often called

scalar multiplication, since one should think of elements of M as “vectors” and those of R as “scalars”, following the terminology of linear algebra.

**Exercise**.

Find a structure which satisfies conditions 1-3 but not 4.

Let’s begin with the elementary properties of modules. For any , we have:

- , where and ;
- (-
*r*)*m*= -(*rm*) =*r*(-*m*).

The proof is pretty straightforward:

- For the first equality, . Adding to both sides gives The other equality is left as an easy exercise.
- For the first equality, (-
*r*)*m*+ (*rm*) = ((-*r*) +*r*)*m*= 0_{R}·*m*. Hence adding -(*rm*) to both sides gives us the desired equality. The other is an exercise.

**Right Modules and Opposite Rings**

Instead of defining a left module, we could have applied the action of on the right. Thus, a right module *M* over *R* is an abelian group with a binary operation *M* × *R* → *M* such that for all and

*m*(*r*+*s*) =*mr*+*ms*;- (
*m*+*n*)*r*=*mr*+*nr*; *m*(*rs*) = (*mr*)*s*;*m**·1 = m*.

Note that if we turn the order of operation around to give *R* × *M* → *M*, then conditions 1, 2, 4 are exactly as above while condition 3 then becomes “(*rs*)*m* = *s*(*rm*)”. Hence, if *R* is a commutative ring, then a left module over *R* can easily be turned to a right module by flipping the action as above. Conclusion: if *R* is non-commutative, a right *R*-module cannot be easily turned into a left module.

However, we can define the **opposite ring**:

Definition. If R is a ring, then its opposite ring R^{op}is defined by taking the abelian group (R, +) with the product operation *, where r*s = sr, with the RHS product taken from R.

Then a right *R*-module can also be described as a left module over its opposite ring *R*^{op}. Obviously, if *R* is commutative, then its opposite ring is identical to itself. On the other hand, even if *R* is non-commutative, it’s possible for in which case one can turn a right *R*-module into a left one via this isomorphism. For example, if **H** denotes the division ring of all quaternions with real coefficients, then:

gives where * is the product operation for the opposite ring.

**Exercise**.

Prove that the ring *R* of *n* × *n* real matrices is isomorphic to its opposite.

Finally, one notes that there are ample examples of rings which are not isomorphic to their opposite.

**Examples of Modules**

- Every ring
*R*is a module over itself. - Every vector space over a field
*K*(e.g.**R**) is a*K*-module. - A
**Z**-module is precisely an abelian group (clearly, the underlying additive structure of a module gives us an abelian group; in the other direction, every abelian group*A*has a multiplication-by-*n*map which takes and gives us the structure of a**Z**-module). - Every ideal is an
*R*-module, where scalar multiplication is given by ring product in*R*. - The
*n*-dimensional**free***R*-module is , the Cartesian product of*n*copies of*R*. Product is given by - Consider the ring of Gaussian integers
**Z**[*i*]. Now**C**is a**Z**[*i*]-module, where scalar multiplication is by the usual ring multiplication map in**C**. - Let be the ring of matrices with integer entries. Then is an
*R*-module, via multiplying a matrix by vector. - In representation theory, a representation of a group
*G*over field*K*is precisely a*K*[*G*]-module.

**Exercise**

Let **H** be the ring of quaternions. Is **R** an **H**-module, if we define scalar multiplication as follows?

## Submodules

For now on, unless explicitly stated, all modules are assumed to be *left* modules.

Now submodules are simply subsets of a module *M* which can inherit the module structure from *M.*

Definition. A submodule of M is a subset satisfying the following:

- ;
- if then ;
- if , then .

Immediately, the second and third conditions give us: for any *m*–*n* = *m*+(-1)*n* also lies in *N*. Thus, we can replace the first condition with since we can then pick and infer that from the second and third conditions.

Given a collection of submodules of *M*, there’re at least two ways to create another submodule:

Proposition. If is a collection of submodules of M, then is a submodule of M;andis also a submodule of M. Thus is the set of all sums of finitely many terms from

**Proof**.

- For
*N*, clearly . Next, if , then for every*i*. Hence, for every*i*, which gives Finally, suppose and Then for each*i*so for each*i*. Hence - For
*N’*, again and clearly*N’*is closed under addition since if are sums of finitely many terms from , then*x*+*y*is obtained by concatenating the two sums. Finally where each ♦

**Generated Submodules**

Let be any *subset* and consider:

i.e. the collection of all submodules of *M* which contain *S*. This collection is non-empty since it at least contains *M*. Let’s take the intersection of all these modules:

Being an intersection of submodules of *M*, <*S*> is also a submodule.

Definition. The resulting <S> is called thesubmodule generated by S.

Observe that <*S*> satisfies the following two critical properties:

- : indeed, it’s an intersection of
*N*‘s, each of which contains*S*. - if
*N’*is a submodule of*M*such that , then : indeed since*N’*contains*S*, it’s found in the collection ∑. Since <*S*> is defined to be the intersection of*N’*and some other submodules, we have

It is for this reason that we often say:

The generated submodule <S> is the

smallest submodule containing S.

The case where *S* = {*m*} is a singleton set is of special interest: we claim that <*S*> is the product Indeed, since we clearly have for any scalar thus giving us the inclusion On the other hand, it’s easy to check that *Rm* is itself a submodule of *M* containing *m*. Hence, and the two sets are equal.

**Examples**

- Every module
*M*has two obvious submodules: {0} and*M*. Any other submodule is called a**proper submodule**. - Consider ring
*R*as a module over itself. Its submodules are called**left ideals**(compare this with the normal, two-sided ideals). Unwinding the definition, a left ideal of*R*is a subset such that- ;
- if , then ;
- if , then .

- Consider
**C**as a**Q**-module. Then the set*S*= {1+*i*, 3-2*i*, √2} generates the submodule comprising of all*a*+*bi*+*c*√2, for all - If
*I*is a left ideal of*R*and*M*is a module, then*IM*is a submodule of*M*. Here*IM*is defined as the set of all finite sums where and - Consider the ring of all
*n*×*n*integer matrices, with module Then is a proper submodule of*M*. - On the other hand, for the ring of
*n*×*n*real matrices and module one sees that there’s no proper submodule. In other words, any single non-zero generates the whole module*M*. Such modules are called**simple modules**and we’ll encounter them again later. - Let
*R*=**R**×**R**and consider*M*=*R*as a module over itself. Then*N*:= {0} ×**R**is a left*R*-module since*N*is an ideal of*M*, and in particular a left-ideal.