3.1. Definitions.

In group theory, we started with a set $G$ equipped with a bilinear operation $\cdot :G\times G\to G$ which mapped $G$ to itself. The operation was required to be associative, and there needed to be inverses and an identity element.

In ring theory, we went further to assume $R$ was not only an abelian group, we placed the group operation with $+:R\times R\to R$ and then defined a multiplication $\cdot :R\times R\to R$ which is was associative and left- and right- distributive.

Finally, we reach module theory, which considers again an abelian group $M$ with operation $+:M\times M\to M$ but lets a ring $R$ act on $M$, whose addition $+R\times R\to R$ agrees with the one which acts on $M$ but whose multiplication $\cdot :R\times M\to M$ acts on $R$ and $M$.

Note how abelian groups and rings are special cases of modules. This will be more clear once we introduce the axioms.

Let $R$ be a ring with identity, and $M$ an abelian group equipped with $+:M\times M\to M$. Then $M$ is an left $R$-module if we equip $R\times M$ with multiplication $\cdot :R\times M\to M$ and for all $m\in M$ and $a,b\in R$ \begin{enumerate} \item $a(m_1+m_2)=a{m}_1+a{m}_2$ \item $(a+b)m=am+bm$ \item $(ab)m=a(bm)$ \item $1_{R}m=m$ where $1_R$ is the identity of $R$. \end{enumerate}

Alternatively, an abelian group $M$ is a right $R$-module if we equip $M\times R\to M$ with multiplication $\cdot :M\times R\to M$ and for all $m\in M$ and $a,b\in R$ \begin{enumerate} \item $(m+n)a=ma+na$ \item $m(a+b)=ma+mb$ \item $m(ab)=(ma)b$ \item $m{1}_R=m$ where $1_R$ is the identity of $R$. \end{enumerate}

Notice that we can think of these products as a group action, or sort of a "ring action" acting on $M$. That is, an $R$-module $M$ is just an abelian group that a ring $R$ can act on. If you have an abelian group $N$ that $R$ simply cannot act on and satisfy the above axioms, then $N$ is not an $R$-module. \ For convenience, we will develop the theory of $R$-modules by solely working with left $R$-modules, since all proofs and statements will be equivalent up to a swap of variables for right $R$-modules. \ Examples.\

• 1. \textcolor{NavyBlue}{Note that if $R$ is commutative, then a left $R$-module coincides with a right $R$-module. To see this, let $M$ be a left $R$-module. Then construct the right $R$-module by defining the multiplication as

$$m\cdot r=rm.$$

Then we see that for all $m\in M$, $a,b\in R$, \begin{enumerate} * $(m_1+m_2)\cdot a=a(m_1+m_2)=a{m}_1+a{m}_2=m_1\cdot a+m_2\cdot a$ \checkmark * $m(a+b)=(a+b)m=am+bm=m\cdot a+m\cdot b$ \checkmark * $m\cdot (ab)=(ab)m=(ba)m=b(am)=b(m\cdot a)=(m\cdot a)\cdot b$ \checkmark * $m\cdot 1_R=1_{R}m=m$. \checkmark \end{enumerate} Note that in part $(c)$ is where we used the fact that $R$ is commutative. So whenever $R$ is commutative, the existence of a left $R$-module automatically implies that existence of a right $R$-module, and vice versa. } * 2. Let $R$ be a ring. Then if we substitute $M=R$ in the above definition, and let the multiplication $ \cdot$ be the multiplication on $R$ then $R$ is a left and a right $R$-module. This is because $R$ is an abelian group which is associative and left- and right-distributive. Hence, it satisfies all of the above axioms.

So keep in mind that a ring $R$ is just a left- and right-$R$ module that acts on $R$.

Here's another example which shows that abelian groups are simply $\mathbb{Z}$ modules.

Let $G$ be an abelian group. Then $G$ is a left and right $\mathbb{Z}$-module.

Let $\mathbb{Z}$ act on $G$ as follows. Define

$$ng=\{\begin{array}{ll}g+g+\cdots +g\text{ (}n\text{ times)}& \text{ if }n>0\\ 0& \text{ if }n=0\\ (-g)+(-g)\cdots (-g)\text{ (}n\text{ times) }& \text{ if }n<0\end{array}$$

and

$$gn=\{\begin{array}{ll}g+g+\cdots +g\text{ (}n\text{ times)}& \text{ if }n>0\\ 0& \text{ if }n=0\\ (-g)+(-g)\cdots (-g)\text{ (}n\text{ times) }& \text{ if }n<0.\end{array}$$

Then with this definition of multiplication, it is easy to show that the axioms (a)-(d) are satisfied.

• 3. If $R$ is a ring and $I$ is a left (right) ideal of $R$, then $I$ is a left (right) $R$-module.
• 4. Let $V$ be a vector space defined over a field $F$. Then $V$ is an $F$-module. (Now it is clear why there are a million axioms in the definition of a vector space!)

With $R$-modules introduced and understood, we can jump right into homomorphisms.

Let $R$ be a ring and $M$ and $N$ be $R$-modules. We define $f:M\to N$ to be an $R$-module homomorphism if

• 1. $f(m_1+m_2)=f(m_1)+f(m_2)$ for any $m_1,m_2\in M$
• 2. $f(am)=af(m)$ for all $a\in R$ and $m\in M$.

If $f$ is a bijective $R$-module homomorphism, then we say that $f$ is an isomorphism and that $M\cong N$. Thus we see that $R$-module homomorphisms must not only be linear over the elements of $M$, but they must also pull out scalar multiplication by elements of $R$.

Recall earlier that we said a vector space $V$ over a field $F$ is an $F$-module. Now if $W$ is another vector space and $T:V\to W$ is a linear transformation, then we see that $T$ is also an $F$-module homomorphism!

In the language of linear algebra, a linear transformation is usually defined as a function $T:V\to W$ such that for any ${\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\mathbf{,}\mathbf{v}\in \mathbf{V}$ and $\alpha \in F$ we have that

• 1. $T({\mathbf{v}}_{\mathbf{1}}\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{T}\mathbf{(}{\mathbf{v}}_{\mathbf{1}}\mathbf{)}\mathbf{+}\mathbf{T}\mathbf{(}{\mathbf{v}}_{\mathbf{2}}\mathbf{)}$
• 2. $T(\alpha\bf{v}) = $ $\alpha$$T(\mathbf{v}\mathbf{)}$.

As we will see, linear algebra is basically a special case of module theory.

Let $R$ be a ring and $M$ and $N$ a pair of $R$-modules. Then ${\mathrm{hom}}_R(M,N)$ is the set of all $R$-module homomorphisms from $M$ to $N$.

\textcolor{MidnightBlue}{It turns out we can turn ${\mathrm{hom}}_R(M,N)$ into an abelian group, and under special circumstances it can actually be an $R$-module itself. It will be the case that ${\mathrm{hom}}_R$ will actually be an important functor, but that is for later.} \ \indent To turn this into an abelian group, we define addition of the elements to be

$$(f+g)(m)=f(m)+g(m)$$

for all $f,g\in {\mathrm{hom}}_R(M,N)$. We let the identity be the zero map, and realize associativity and closedness are a given to conclude that this is in fact an abelian group. \

Suppose we want to make $R$, our ring, act on ${\mathrm{hom}}_R(M,N)$ in order for it to be an $R$-module. Then we define scalar multiplication to be $(af)(m)=a(f(m))$; a pretty reasonable definition for scalar multiplication.

\textcolor{Red}{This issue with this is that ${\mathrm{hom}}_R(M,N)$ will not be closed under scalar multiplication of elements of $R$ unless $R$ is a commutative ring. }

We'll demonstrate this as follows. Let $b\in R$ and $f\in {\mathrm{hom}}_R(M,N)$. Then the second property of an $R$-module homomorphism tells us that $f(bm)=bf(m)$ for all $m\in M$. Now suppose we try to use our definition of scalar multiplication, and consider $af$ where $a\in R$. Then if we try to see if $af$ will pass the second criterion for being an $R$-module homomorphism, we see that

$$(af)(bm)=a(f(bm))=a(bf(m))=abf(m).$$

That is, we see that $af$ isn't an $R$-module homomorphism because $(af)(bm)\ne b(af)(m)$ (which is required for an $R$-module homomorphism); rather, $(af)(bm)=abf(m).$ Now if $R$ is a commutative ring, then

$$abf(m)=baf(m)$$

so we can then say that $(af)(bm)=b(af)(m)$, in which case $af$ passes the test for being an $R$-module homomorphism.

This proves the following propsition, which will be useful for reference for later.

Let $M$ and $N$ be $R$-modules. Then ${\mathrm{hom}}_R(M,N)$ is an abelian group. Furthermore, it is an $R$-module if and only if $R$ is a commutative ring. Next, we make the following definitions for completeness.

Let $R$ be a ring and $M$ and $N$ be $R$-modules. If $f:M\to N$ is an $R$-module homomorphism, then

• 1. The set $\ker (f)=\{m\in M\mid f(m)=0\}$ is the kernal of $f$
• 2. The set $\text{Im}(f)=\{f(m)\mid m\in M\}$ is the image of $f$.