§17 Modules

§17.1 Modules

Just as groups act on sets, rings act on Abelian groups. When a ring acts on an Abelian group, that Abelian group is called a module over the ring.

When a group acts on a set, it must act by bijections, and therefore it must preserve properties such as the cardinality of the set. But when a ring acts on an Abelian group, it respects the additive structure inside the Abelian group. This is condition (1) in the definition below.

In this setting, we will build all the definitions of module algebra in much the same way as we did for groups.

Definition 17.1 Let $R$ be a ring and let $M$ be an Abelian group. A left action of $R$ on $M$ is a function

\begin{aligned} R\times M&\to M,\\ (r,m)&\mapsto rm \end{aligned}

such that for all $r,s\in R$ and $m,m^{\prime}\in M$ :

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

(2) $(r+s)m=rm+sm$ .

(3) $s(rm)=(sr)m$ .

(4) $1m=m$ .

Once a left action of $R$ on $M$ is specified, we call $M$ a left $R$ -module.

Remark Here, “multiplication by $r$ ” may be interpreted as scaling by some element of the ring. Thus, a module is a set equipped with addition together with a notion of scaling by $r$ .

There is a more concise way to express this. Everything above is equivalent to giving a ring homomorphism

R\to\mathrm{End}(M).

A right $R$ -module is an Abelian group $M$ equipped with a function

M\times R\to M

satisfying analogues of conditions (1)–(4).

At first, this may seem like a large amount of data, since we are simultaneously dealing with both $R$ and $M$ .

In practice, however, we usually fix a ring $R$ and study only the relationships between different modules $M$ .

Exercise Let $M$ be a left $R$ -module. Then

0m=m\quad\text{and}\quad(-r)m=-(rm).

Proof: For clarity, write

0_R

for the zero element of $R$ .

By condition (2),

0_Rm = (0_R+0_R)m = 0_Rm+0_Rm.

Therefore, by the cancellation law,

0=0_Rm.

The $0$ on the left is the additive identity of $M$ .

Similarly,

rm+(-r)m = (r-r)m = 0_Rm = 0.

~\tag*{$\square$}

Example 17.1

(a) Let

R=\mathbb{R}, \qquad M=\mathbb{R}^n,

the $n$ -dimensional real vector space.

Define a function

\begin{aligned} \mathbb{R}\times M&\to M,\\ (t,v)&\mapsto t\vec{v} \end{aligned}

by scaling by $t$ .

\vec{v} = (v_1,\ldots,v_n),

then

t\vec{v} = (tv_1,\ldots,tv_n).

This satisfies all of the properties above.

(b) Every ring $R$ is naturally a left module over itself.

Let

\mathbb{R}[t]

be the ring of real-coefficient polynomials in the variable $t$ .

Choose an $m\times m$ matrix $T$ , viewed as a linear map

T:\mathbb{R}^m\to\mathbb{R}^m.

Then $\mathbb{R}^m$ is a left $\mathbb{R}[t]$ -module with action defined by

(a_0+a_1t+\cdots+a_kt^k)v := a_0v+a_1T(v)+\cdots+a_k(T\circ\cdots\circ T)(v),

where

T\circ\cdots\circ T

denotes $T$ composed with itself $k$ times.

§17.2 Submodules

Definition 17.2 Let $R$ be a ring and let $M$ be a left $R$ -module.

An Abelian subgroup

M^{\prime}\subset M

is called a submodule of $M$ if

x\in M^{\prime} \quad\Rightarrow\quad rx\in M^{\prime}

for all $r\in R$ .

Example 17.2

(a) If

R=\mathbb{R}

and

M=\mathbb{R}^n,

then a submodule is a subset closed under addition, additive inverses, and scalar multiplication.

This is exactly the same thing as a linear subspace of $\mathbb{R}^n$ .

(b) Let $R$ be a commutative ring. Then

I\subset R

is an ideal if and only if $I$ is a submodule of $R$ .

M=\mathbb{R}^m

be the $\mathbb{R}[t]$ -module defined by a linear transformation

T:\mathbb{R}^k\to\mathbb{R}^k.

Then a submodule is a linear subspace $V$ satisfying

T(V)\subset V.

In other words, it is a $T$ -invariant subspace.

§17.3 Module Homomorphisms

Definition 17.3 Let $M$ and $N$ be left $R$ -modules.

An $R$ -module homomorphism, or $R$ -module map, is a function

f:M\to N

such that $f$ is a group homomorphism and

f(rx)=rf(x)

for all $r\in R$ and $x\in M$ .

Definition 17.4 An $R$ -module isomorphism is a bijective homomorphism.

Definition 17.5 The kernel and image of an $R$ -module homomorphism

f:M\to N

are respectively the kernel and image of $f$ viewed as a group homomorphism.

Thus

\ker(f) = \{x\mid f(x)=0\},

and

\mathrm{im}(f) = \{y\in N\mid y=f(x),~x\in M\}.

Example 17.3

(a) If

M\cong\mathbb{R}^n

and

N\cong\mathbb{R}^m

with their usual $\mathbb{R}$ -module structures, then a linear map

M\to N

is an $\mathbb{R}$ -module homomorphism.

(b) (There is no obvious example corresponding to (b) above.)

M=\mathbb{R}^m

be a left module defined by a linear transformation $T$ .

Let

N=\mathbb{R}^n

be a left module defined by a linear transformation

S:\mathbb{R}^n\to\mathbb{R}^n.

Then an $R$ -module homomorphism is a linear map

f:\mathbb{R}^m\to\mathbb{R}^n

with the property

f(T(v)) = S(f(v)).

Definition 17.6 Let $M$ and $N$ be left $R$ -modules. The set of all $R$ -module homomorphisms is denoted

\mathrm{Hom}_R(M,N).

Proposition 17.1

\mathrm{Hom}_R(M,N)

is a left $R$ -module.

Proof: Definition of scalar multiplication:

For $r\in R$ and $f\in\mathrm{Hom}_R(M,N)$ , define the scalar multiple $r\cdot f$ by

(r\cdot f)(m) = rf(m), \qquad \forall m\in M.

Verify that $r\cdot f$ is an $R$ -module homomorphism:

Additivity:

\begin{aligned} (r\cdot f)(m_1+m_2) &= rf(m_1+m_2)\\ &= r[f(m_1)+f(m_2)]\\ &= rf(m_1)+rf(m_2)\\ &= (r\cdot f)(m_1) + (r\cdot f)(m_2). \end{aligned}

Thus $r\cdot f$ is additive.

Compatibility with scalar multiplication:

For all $s\in R$ and $m\in M$ ,

\begin{align*} (r\cdot f)(sm) &= rf(sm)\\ &= r[sf(m)]\\ &= s[rf(m)] \tag{since $R$ is commutative}\\ &= s(r\cdot f)(m). \end{align*}

Thus $r\cdot f$ is $R$ -linear.

Verify the module axioms:

Distributivity with respect to addition in the ring:

\begin{aligned} ((r+s)\cdot f)(m) &= (r+s)f(m)\\ &= rf(m)+sf(m)\\ &= (r\cdot f+s\cdot f)(m). \end{aligned}

Distributivity with respect to addition in the module:

\begin{aligned} (r\cdot(f+g))(m) &= r(f+g)(m)\\ &= r[f(m)+g(m)]\\ &= rf(m)+rg(m)\\ &= (r\cdot f)(m) + (r\cdot g)(m). \end{aligned}

Associativity of scalar multiplication:

\begin{aligned} ((rs)\cdot f)(m) &= (rs)f(m)\\ &= r(sf(m))\\ &= r\cdot(s\cdot f)(m). \end{aligned}

Identity:

(1_R\cdot f)(m) = 1_Rf(m) = f(m).

Conclusion:

Since all the module axioms are satisfied,

\mathrm{Hom}_R(M,N)

is a left $R$ -module under the scalar multiplication defined above.

~\tag*{$\square$}

§17.4 Direct Sums and Free Modules

Definition 17.7 Let $M$ and $N$ be left $R$ -modules.

The direct sum

M\oplus N

is defined to be the group

M\times N

equipped with the $R$ -module structure

r(m,n) := (rm,rn).

Proposition 17.2

M\oplus N

is an $R$ -module.

Proof: We already know that

M\oplus N

is an Abelian group.

On the other hand,

1(m,n) := (1m,1n) = (m,n)

because both $M$ and $N$ are modules.

Moreover,

\begin{align*} r((m,n)+(m^{\prime},n^{\prime})) &= r(m+m^{\prime},n+n^{\prime})\\ &= (r(m+m^{\prime}), r(n+n^{\prime}))\\ &= (rm+rm^{\prime}, rn+rn^{\prime}) \tag{3}\\ &= (rm,rn) + (rm^{\prime},rn^{\prime})\\ &= r(m,n) + r(m^{\prime},n^{\prime}). \end{align*}

In (3), we used the fact that both $M$ and $N$ are left $R$ -modules.

~\tag*{$\square$}

Example 17.4 If

R=\mathbb{R},

and both $M$ and $N$ are also $\mathbb{R}$ viewed as modules over themselves, then

\mathbb{R}\oplus\mathbb{R} \cong \mathbb{R}\times\mathbb{R} \cong \mathbb{R}^2

as groups, with the usual scalar multiplication

r(x_1,x_2) = (rx_1,rx_2).

Remark There is an obvious isomorphism

(M\oplus N)\oplus O \cong M\oplus(N\oplus O), \quad (m,n,o)\mapsto(m,n,o).

Definition 17.8 Let $R$ be a ring. The direct-sum module

R^n := R\oplus\cdots\oplus R

is called the free $R$ -module of rank $n$ .

§17.5 Universal Property of Free Modules

Question: Why is it called a free $R$ -module?

Proposition 17.3 Let $M$ be an $R$ -module.

Any $n$ -tuple of elements

x_1,\ldots,x_n\in M

uniquely determines an $R$ -module homomorphism

X:R^n\to M

defined by

(0,\ldots,0,1,0,\ldots,0) \mapsto x_i,

where the $i$ th coordinate is $1$ .

Remark This is the same property as that of a free group on $n$ generators: any $n$ -tuple of elements of a group $G$ uniquely determines a map

F_n\to G.

Proof: Given

(x_1,\ldots,x_n),

define

X:R^n\to M

X(a_1,\ldots,a_n) := a_1x_1+\cdots+a_nx_n \in M.

This is a group homomorphism because

\begin{aligned} X((a_1,\ldots,a_n)+(b_1,\ldots,b_n)) &= (a_1+b_1)x_1 +\cdots + (a_n+b_n)x_n\\ &= (a_1x_1+\cdots+a_nx_x) + (b_1x_1+\cdots+b_nx_n)\\ &= X(a_1,\ldots,a_n) + X(b_1,\ldots,b_n). \end{aligned}

The middle equality uses the fact that $M$ is an $R$ -module.

It is also an $R$ -module homomorphism because

\begin{aligned} X(r(a_1,\ldots,a_n)) &= X((ra_1,\ldots,ra_n))\\ &= (ra_1)x_1+\cdots+(ra_n)x_n\\ &= r(a_1x_1+\cdots+a_nx_n)\\ &= rX((a_1,\ldots,a_n)). \end{aligned}

Again, the penultimate equality uses the fact that $M$ is an $R$ -module.

~\tag*{$\square$}

§17.6 The Free Module on 0 Generators

Exercise Let $M$ be a left $R$ -module. Prove that

r0_M=0_M, \quad\text{and}\quad r(-x)=-rx.

Proof: An action of $R$ on $M$ is equivalent to a ring homomorphism

R\to\mathrm{End}(M).

In particular, every element $r\in R$ determines an Abelian-group homomorphism.

Therefore, scaling by $r$ preserves the additive identity and additive inverses of $M$ .

If you prefer a more computational proof, observe that

r0_M+r0_M = r(0_M+0_M) = r0_M.

Therefore, by the cancellation law in the Abelian group, we may subtract $r0_M$ from both sides to obtain

r0_M = 0_M.

Hence

r(-x)+rx = r(-x+x) = r0_M = 0_M.

This shows that $r(-x)$ is the additive inverse of $rx$ .

~\tag*{$\square$}

Remark We understand

R^{\oplus n}

for $n\geqslant1$ .

But what about

n=0?

We should look for an $R$ -module

R^{\oplus0}

such that there is a bijection

\mathrm{Hom}_R(R^{\oplus0},M) \cong \mathrm{Map}_{\text{Sets}}(\varnothing,M).

But there is a unique function from the empty set to any set.

Therefore, we must find an $R$ -module

R^{\oplus0}

such that there is exactly one module homomorphism from it to every module $M$ .

The unique module with this property is the zero module, i.e. the trivial Abelian group satisfying

r0=0.