Abstract Algebra/Modules

Motivation

Let G be an abelian group under addition. We can define a sort of multiplication on G by elements of $\mathbb{Z}$ by writing $ng=\underbrace{g+g+\cdots+g}_n$ for $n\in\mathbb{Z}^+$ and $g\in G$. We can extend this to the case where n is negative by writing $(-n)g=\underbrace{-g-g-\cdots-g}_n$. We would, however, like to be able to define a sort of multiplication of a group by an arbitrary ring.

Definition

Definition (Module)

Let R be a ring and G an abelian group. We call G a left R-module if there is a function $m:R\times G\to G$ satisfying

1. m(r₁ + r₂,g₁) = m(r₁,g₁) + m(r₂,g₁),
2. m(r₁,g₁ + g₂) = m(r₁,g₁) + m(r₁,g₂), and
3. m(r₁,m(r₂,g₁)) = m(r₁r₂,g₁)

for all $r_1,r_2\in R,\ g_1,g_2\in G$.

For convenience, we usually write rg for m(r,g). We can also define a right R-module analogously, if the last property reads:

m(r₁,m(r₂,g)) = m(r₂r₁,g)

In this case we usually write gr for m(r,g).

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

Definition (Unitary Module)

If R is a ring with unity 1 and 1g=g for all $g\in G$, then G is called a unitary R-module.

Equivalent Definition

Recall that the endomorphisms of an abelian group form a unitary ring. That is, given homomorphisms $f,g:G \to G$ one can show that $-f,f+g, f \circ g$ and the identity map are all homomorphisms from G to itself. They also satisfy the necessary associativity and distributivity properties needed to make a ring.

A ring homomorphism $\phi:R \to \hom(G,G)$ induces a map $m:R \times G \to G$ where m(r,g) = ϕ(r)(g) for any $r \in R, g \in G$. Indeed, the induced map m gives G a left R-module structure; the first and third conditions follow as ϕ is a ring homomorphism while the second condition follows as ϕ(r) is a group homomorphism.

One can also take a map $m:R \times G \to G$ and construct a ring homomorphism from R into the endomorphisms of G. Thus, we have an equivalent definition for a left R-module.

If we require, as commutative algebraists often do, that our ring is unitary and that ring homomorphisms preserve identity, then ϕ must send the identity of our ring to the identity homomorphism and hence G is a unitary module.

Submodules

Definition (Submodule)

Given a left R-module M a submodule of M is a subset $N \subseteq M$ satisfying

1. N is a subgroup of M, and
2. for all $r \in R$ and all $n \in N$ we have $rn \in N$.

The second condition above states that submodules are closed under left multiplication by elements of R; it is implicit that they inherit their multiplication from their containing module; $m':R \times N \to N$ must be the restriction of $m:R\times M \to M$.

Module Homomorphisms

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

Definition (Module Homomorphism)

An R-module homomorphism $\phi:M \to N$ is a function from M to N satisfying

1. ϕ(m + m') = ϕ(m) + ϕ(m') (it is a group homomorphism), and
2. ϕ(rm) = rϕ(m).

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

Definition (Kernel, Image)

Given a module homomorphism $\phi:M \to N$ the kernel of ϕ is the set

$\ker \phi = \{ m \in M \mid \phi(m) = 0 \}$

and the image of ϕ is the set

$\phi(M) = \{ n \in N \mid \exists m \in M, \phi(m) = n \}$.

The kernel of ϕ is the set of elements in the domain that are sent to zero by ϕ. In fact, the kernel of any module homomorphism is a submodule of M. It is clearly a subgroup, from group theory, and it is also closed under multiplication by elements of R: ϕ(rm) = rϕ(m) = r(0) = 0 for $m \in \ker \phi$.

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

Generating Modules

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

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

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

$RA = \left \{ \sum_{i=1}^n r_i a_i \mid n \in \mathbb N, r_i \in R, a_i \in A \right \}$.

That is, every element of RA can be written as a finite left linear combination of elements of A. This equality can be justified by double inclusion: First, any submodule containing A must contain all left R-linear combinations of elements of A since modules are closed under addition and left multiplication by elements of R. Thus, $RA \supseteq \{ \sum_{i=1}^n r_i a_i \mid n \in \mathbb N, r_i \in R, a_i \in A \}$. Secondly, the set of all such linear combinations forms a submodule of M containing A (use n = 1 and r₁ = 1_R) and hence it contains RA.

Quotient Modules

Recall that any subgroup H of an abelian group G allows one to construct an equivalence relation; for$g,h \in G$

$g \sim h \iff g-h \in H.$

Cosets of G --- equivalence classes under the relation above --- can then be endowed with a group structure, derived from the original group, and is given the name G/H. The sum of two cosets g + H = [g]_∼ and h + H = [h]_∼ is simply (g + _Gh) + H = [g + h]_∼.

Since modules are abelian groups, we can take any subgroup N of an R-module M and form the group quotient M / N. If N is a submodule of M, and not simply a subgroup, then M / N is an R-module.

To prove this, we define r(m + N) = rm + N and want to show that this is well defined. But clearly if we have two elements m,m' in the same coset, then $m-m' \in N$ and hence $r(m-m') \in N$ as N is an R-module. Using distributivity, we see that

$r(m+ N) = rm + N = rm' + N = r(m'+N) \,$

and hence this action of R on M / N endows M / N with a well defined module structure.

Quotient modules come with a so-called natural map; $\iota: M \to M/N$ given by $m \mapsto m +N$. This map is a module homomorphism and is often denoted by $\bar{ }$ rather than ι. In this case $\bar{m}$ is understood to be a coset of M.

There are many properties of a module that translate directly to its quotients. For instance, quotients of a finitely generated module are finitely generated. Also, the submodules of a quotient M/N are in correspondence (via the natural map) with submodules of M containing N. This will be covered in the section on module isomorphism theorems.

Generating Submodules by Ideals

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

If we consider the set $IM =\{ \sum_{i=1}^n r_i m_i \mid n \in \mathbb N, r_i \in I, m_i \in M \}$ we obtain a submodule of M. This follows from our discussion of generated submodules. However, since I is not unitary, it is not necessary that IM = M.

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

Proposition

Given an R-module M and ideal I of R, the Rmodule M / IM is an R / I-module with multiplication (r + I)(m + IM) = rm + IM.

proof.

To show that this is well defined, we observe that if r + I = s + I then $r-s \in I$ and hence

(rm + IM) − (sm + IM) = rm − sm + IM = (r − s)m + IM = 0 + IM

since $(r-s)m \in IM$. Thus,

(r + I)(m + IM) = rm + IM = sm + IM = (s + I)(m + IM)

which proves that the action of R / I on M / IM is well defined. It follows now that M / IM is an R / I-module simply because it is an R-module.