Abstract Algebra/Modules

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[edit] 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.

[edit] 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

$m (r 1 + r 2, g 1) = m (r 1, g 1) + m (r 2, g 1)$ ,
$m (r 1, g 1 + g 2) = m (r 1, g 1) + m (r 1, g 2)$ , and
$m (r 1, m (r 2, g 1)) = m (r 1 r 2, g 1)$

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 (r 1, m (r 2, g)) = m (r 2 r 1, 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.

[edit] 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.

[edit] Submodules

Definition (Submodule)

Given a left

R

-module

M

a submodule of

M

is a subset $N \subseteq M$ satisfying

N is a subgroup of M, and
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 $m:R\times M \to M$ .

[edit] Module Homomorphisms

Like all agebraic 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

$φ(m + m') = φ(m) + φ(m')$ (it is a group homomorphism), and
$φ(r m) = r φ(m).$

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 $R$ : $φ(r m) = r φ(m) = r (0) = 0$ for $m \in \ker \phi$ .

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

[edit] Generating Modules

Given a subset $A$ of a left $R$ -module $M$ , we define the 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 $R A$ for a reason which will become clear in a moment.

The existance 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 $R A$ 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 $R A$ 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 $r 1 = 1 R$ ) and hence it contains $R A$ .

[edit] 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 $(g + G h) + H = [g + h] ˜$ .

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 $r (m + N) = r m + N$ 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.

[edit] 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 $I M = M$ .

Thus, we may consider the quotient module $M / I M$ . 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

R

module

M / I M

is an

R / I

-module with multiplication

(r + I)(m + I M) = r m + I M

.

proof.

To show that this is well defined, we observe that if

r + I = s + I

then $r-s \in I$ and hence

(r m + I M) - (s m + I M) = r m - s m + I M = (r - s) m + I M = 0 + I M

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

(r + I)(m + I M) = r m + I M = s m + I M = (s + I)(m + I M)

which proves that the action of

R / I

on

M / I M

is well defined. It follows now that

M / I M

is an

R / I

-module simply because it is an

R

-module.

Abstract Algebra/Modules

From Wikibooks, the open-content textbooks collection

Contents

[edit] Motivation

[edit] Definition

[edit] Equivalent Definition

[edit] Submodules

[edit] Module Homomorphisms

[edit] Generating Modules

[edit] Quotient Modules

[edit] Generating Submodules by Ideals

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