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 $ng=\underset{n}{\underbrace{g+g+\cdots +g}}$ for $n\in {\mathbb{Z}}^{+}$ and $g\in G$. We can extend this to the case where n is negative by writing $(-n)g=\underset{n}{\underbrace{-g-g-\cdots -g}}$. 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. $m(r_1+r_2,g_1)=m(r_1,g_1)+m(r_2,g_1)$,
2. $m(r_1,g_1+g_2)=m(r_1,g_1)+m(r_1,g_2)$, and
3. $m(r_1,m(r_2,g_1))=m(r_1{r}_2,g_1)$

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.

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 $\varphi :R\to \mathrm{hom}(G,G)$ induces a map $m:R\times G\to G$ where $m(r,g)=\varphi (r)(g)$ 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 $\varphi$ is a ring homomorphism while the second condition follows as $\varphi (r)$ 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 $\varphi$ 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; $m^{\prime }:R\times N\to N$ must be the restriction of $m:R\times M\to M$.

Module Homomorphisms

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

Definition (Module Homomorphism)

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

1. $\varphi (m+m^{\prime })=\varphi (m)+\varphi (m^{\prime })$ (it is a group homomorphism), and
2. $\varphi (rm)=r\varphi (m).$

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 $\varphi :M\to N$ the kernel of $\varphi$ is the set

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

and the image of $\varphi$ is the set

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

The kernel of $\varphi$ is the set of elements in the domain that are sent to zero by $\varphi$. 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 $R$: $\varphi (rm)=r\varphi (m)=r(0)=0$ for $m\in \ker \varphi$.

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

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 $RA$ 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 $RA$ have a simple description;

$RA=\{\sum _{i=1}^n{r}_i{a}_i\mid n\in \mathbb{N},r_i\in R,a_i\in A\}$.

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 $r_1=1_R$) 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$g,h\in G$

$g\sim h⟺g-h\in H.$

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 $g+H=[g{]}_{\sim }$ and $h+H=[h{]}_{\sim }$ is simply $(g{+}_{G}h)+H=[g+h{]}_{\sim }$.

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 $m,m^{\prime }$ in the same coset, then $m-m^{\prime }\in N$ and hence $r(m-m^{\prime })\in N$ as $N$ is an $R$-module. Using distributivity, we see that

$r(m+N)=rm+N=r{m}^{\prime }+N=r(m^{\prime }+N)$

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 $\stackrel{}{¯}$ rather than $\iota$. In this case $\overline{m}$ 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 $I\ne R$) 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 $R$module $M/IM$ 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.