Modules: Basics

How to think of modules

• Modules are to rings as vector spaces are to fields.
• Modules are to rings as sets with group actions are to groups.

Definition of (left) modules

Definition: Let $R$ be a ring (for now, not necessarily commutative and not necessarily having a unit). A left $R$-module is an abelian group $M$ together with a map $R\times M\to M$ (written $(r,m)\mapsto rm$) such that:

• $r(m_1+m_2)=r{m}_1+r{m}_2$
• $(r_1+r_2)m=r_{1}m+r_{2}m$
• $r_1(r_{2}m)=(r_1{r}_2)m$

If $R$ has a unit element $1$, we also require $1m=m$ for all $m\in M$.

Right modules

A right module is defined by a map $M\times R\to M$ and written $(m,r)\mapsto mr$ and satisfying the property $(m{r}_1)r_2=m(r_1{r}_2).$

If $R$ is not commutative, these really are different, since for a left module:

• $r_1{r}_2$ acts by “first $r_2$, then $r_1$

while for a right module

• $r_1{r}_2$ acts by “first $r_1$, then $r_2$.”

Left and Right modules

If $R$ is commutative, and $M$ is a left $R$-module, then we can define a right $R$ module $M^{\prime }$ with the same underlying abelian group $M$ and by defining $m^{\prime }r=(rm{)}^{\prime }$. This works because

$$(m^{\prime }{r}_1)r_2=(r_{1}m{)}^{\prime }{r}_2=(r_2(r_{1}m){)}^{\prime }=((r_2{r}_1)m{)}^{\prime }=((r_1{r}_2)m{)}^{\prime }=m^{\prime }(r_1{r}_2)$$

Remarks

Vector spaces

If $R$ is a field, then a left (or right) $R$-module is the same as a vector space.

Another definition

If $M$ is an abelian group, and $R$ is a ring, then a left $R$-module structure on $M$ is the same as a ring map

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

If ${\varphi }_r$ is the endomorphism associated to $r\in R$, then $rm={\varphi }_r(m)$. The associativity comes from defining the ring structure on $\mathrm{End}(M)$ as the usual composition of functions: ${\varphi }_{r_1{r}_2}={\varphi }_{r_1}\circ {\varphi }_{r_2}.$

Submodules

Definition: If $M$ is a left $R$-module, then a submodule $N$ of $M$ is a subgroup with the property that, if $n\in N$, then $rn\in N$ for all $r\in R$.

Observation: A ring $R$ is a left module over itself by ring multiplication. The (left) ideals of $R$ are exactly the left submodules of $R$.

Essential examples

Rings as modules over themselves

• Every ring $R$ is a left module over itself. The submodules of $R$ are the left ideals.
• $R$ is also a right module over itself, with the right ideals being the right submodules.

If $F$ is a field and $n>1$, let $R=M_n(F)$ be the $n\times n$ matrix ring over $F$. The matrices with arbitrary first column and zeros elsewhere form a left ideal $J$ and therefore a left submodule of $R$ as left $R$-module. But $J$ is not a right $R$-submodule.

A field $F$ is a one-dimensional vector space over itself, and a commutative ring $R$ is a module (left and right) over itself with the ideals of $R$ being the submodules.

Free modules

Let $R$ be a ring with unity and let $n\ge 1$ be a positive integer. Then

$$R^n=\{(r_1,\dots ,r_n):r_i\in R\mathrm{for}i=1,\dots ,n\}$$

is an $R$ module with componentwise addition and multiplication given by $r(r_1,\dots ,r_n)=(r{r}_1,\dots ,r{r}_n)$.

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

Free modules and vector spaces

• If $R$ is a field, the free $R$-module of rank $n$ is an $n$-dimensional vector space.
• The submodules of a finite dimensional vector space are all subspaces which are copies of $R^k$ for $k\le n$.
• For more general $R$ the picture is more complicated. Let $R=\mathbb{Z}$ and $M={\mathbb{Z}}^2$. Then:
  • $\{(n,0):n\in \mathbb{Z}\}$ is a submodule of $M$ which “looks like” a subspace.
  • $2M=\{(a,b):a,b\in 2\mathbb{Z}\}$ is a submodule of $M$ which does not.

Change of rings (restriction of scalars)

• An abelian group $M$ may be an $R$ module for different rings $R$. For example:
  • $\mathbb{Q}$ is a module over $\mathbb{Q}$, where it is a one dimensional vector space and its only $\mathbb{Q}$-submodules are $0$ and itself.
  • $\mathbb{Q}$ is a module over $\mathbb{Z}$, and it has many $\mathbb{Z}$-submodules, such as $\mathbb{Z}[1/2]$.

More generally, if $R\subset S$ is a subring, and $M$ is an $S$-module, then it is an $R$-module. This is called restriction of scalars.

$\mathbb{Z}$-modules are the same as abelian groups

Let $M$ be an abelian group. Then it is automatically a $\mathbb{Z}$-module where we define

$$nx=\stackrel{n}{\overbrace{x+x+\cdots +x}}.$$

Furthermore, given any $\mathbb{Z}$-module, it must be the case that

$$nx=(\stackrel{n}{\overbrace{1+1+\cdots +1}})x=\stackrel{n}{\overbrace{x+x+\cdots +x}}.$$

(Note: this is why we require $1x=x$ when $R$ is a ring with unity in the module axioms).

Further, submodules of $M$ (as $\mathbb{Z}$-module) are just the subgroups of $M$ (as abelian group).

Change of rings (quotients)

Suppose that $M$ is a left $R$ module and $I\subset R$ is a two-sided ideal with the property that, for all $y\in I$, and all $x\in M$, we have $yx=0$. In this case we say that $I$ annihilates $M$ or that $IM=0$.

With this hypothesis, we may view $M$ as an $R/I$ module by defining $(r+I)m=rm$ for any coset representative $r+I\in R/I$. This is well-defined since two different coset representatives $r,r^{\prime }$ satisfy $r^{\prime }=r+i$ for some $i\in I$ and therefore $r^{\prime }m=(r+i)m=rm$ since $im=0$.

If $M$ is an abelian group and $m\in Z$ is a positive integer such that $mM=0$, then $M$ can be viewed as a module over $\mathbb{Z}/m\mathbb{Z}$ by this process.

This operation is a special case of a general operation called base change or extension of scalars that we will study in more detail later.

Modules over $F[x]$

Basic construction

Let $F$ be a field, let $V$ be a vector space over $F$, and let $T:V\to V$ be an $F$-linear transformation. Define a homomorphism

$$F[x]\to \mathrm{End}(V)$$

by sending

$$x^k\mapsto T^k=\stackrel{n}{\overbrace{T\circ T\circ \cdots \circ T}}.$$

This construction makes $V$ into a module for $F[x]$ which depends on the choice of the linear transformation $T$.

Polynomials and linear transformations

Let $V=F^2$ and let $T$ be the linear transformation given by the matrix

$$T=\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$$

If $e_0$ and $e_1$ are the standard basis elements of $F^2$ then

$$\begin{array}{rlr}T{e}_0& =& e_1\\ T^2{e}_0=T{e}_1=e_0+e_1=e_0+T{e}_0& =& (1+T)e_0\end{array}$$

Polynomials and linear transformations continued

Therefore $(T^2-T-1)e_0=0$ and

$$(T^2-T-1)e_1=(T^2-T-1)T{e}_0=T(T^2-T-1)e_0=0$$

so the polynomial $x^2-x-1$ is in the kernel of the map from $F[x]\to \mathrm{End}(V)$.

By the base change construction above this means that $V$ can be viewed as a module over $F[x]/(x^2-x-1)$.

Characterization of $F[x]$ modules

We saw above that, given an $F$-vector space $V$ with a linear transformation $T$, we get an $F[x]$ module where $x$ acts on $V$ through $T$.

Conversely, suppose that $M$ is an module over $F[x]$. Then $M$ is an $F$ vector space (via the restriction of scalars from $F[x]$ to $F$). Furthermore, the element $x\in F[x]$ acts on $M$ as an $F$-linear transformation because that’s what the module axioms amount to.

Therefore there is an equivalence between

$$\{F[x]-\mathrm{modules}\}\{\mathrm{vector}\mathrm{spaces}V\mathrm{over}F\mathrm{with}\mathrm{a}\mathrm{given}\mathrm{linear}\mathrm{map}T:V\to V\}$$

Submodules of $F[x]$ modules

In the correspondence above, a submodule of an $F[x]$ module $M$ corresponds to a subspace $W\subset V$ that is preserved by $T$, meaning $TW\subset W$.

Thus, not all subspaces of $V$ correspond to submodules.

In the example given earlier, the only $T$-stable proper subspace of $V$ is the zero subspace.

If we consider instead the linear map on $F^2$ satisfying $U{e}_0=0$ and $U{e}_1=e_0$, then the one dimensional subspace spanned by $e_0$ is $U$-stable and $F^2$ viewed as an $F[x]$ module via $U$ has a submodule corresponding to that subspace.

Checking the submodule property

Proposition: A subset $N$ of a left $R$-module $M$ is a submodule if it is nonempty and, for all $x,y\in N$ and $r\in R$, we have $x+ry\in N$. Alternatively, if $N$ is a subgroup of the abelian group $M$ and $rN\subset N$ for all $r\in R$ then $N$ is a submodule.

Algebras

Definition: Let $R$ be a commutative ring with unity. An $R$-algebra is a (not necessarily commutative) ring $S$ with a ring homomorphism $f:R\to S$ carrying $1_R$ to $1_S$ such that $f(R)$ is in the center of $S$.

The polynomial ring $F[x]$ is an $F$-algebra, as is the matrix ring $M_n(F)$ where the homomorphism $f:F\to M_n(F)$ embeds $F$ as the diagonal matrices. More generally, any $F$-algebra $A$, where $F$ is a field, contains $F$ in its center and the identites of $A$ and $F$ are the same.

The ring $\mathbb{Z}/p\mathbb{Z}$ is a $\mathbb{Z}$-algebra. In fact any ring $S$ with $1$ is a $\mathbb{Z}$ algebra by the map sending $n\in \mathbb{Z}$ to $n{1}_S$.

The ring $\mathbb{Q}[x]$ is a $\mathbb{Z}[x]$ algebra.

We typically omit the explicit map $f$ and just think of $R$ as “contained in” $A$; this can be misleading since $f$ doesn’t need to be injective, but it works in practice.

Algebra morphisms

Definition: A map of $R$-algebras $f:A\to B$ is a ring homomorphism that is $R$-linear in the sense that $f(ra)=rf(a)$ for all $r\in R$ and $a\in A$.

Any homomorphism of rings with unity is a $\mathbb{Z}$-algebra morphism.

Modules Homomorphisms, Quotient Modules, and Mapping Properties

Module homomorphisms

Definition: Let $R$ be a ring and let $M$ and $N$ be (left) $R$-modules. A function $f:M\to N$ is an $R$-module homomorphism if:

• it is a homomorphism between the abelian group structures on $M$ and $N$
• it is $R$-linear, meaning $f(rm)=rf(m)$ for all $r\in R$.

Note that, if $R$ is a field, then $M$ and $N$ are vector spaces and an $R$-module homomorphism is just a linear map.

A module isomorphism is a bijective homomorphism.

We let ${\mathrm{Hom}}_R(M,N)$ denote the set of $R$-module homomorphisms from $M$ to $N$.

Kernels and images

Let $R$ be a ring and let $M$ and $N$ be $R$-modules. Let $f:M\to N$ be a homomorphism.

• Let $\ker (f)=\{m\in M:f(m)=0\}$ (the kernel of $f$). This is a submodule of $M$.
• Let $f(M)\subset N$ be the image of $f$. Then $f(M)$ is a submodule of $N$.

Quotient modules

Let $M$ be an $R$ module and let $N\subset M$ be a submodule.

Definition: Let $M/N$ be the quotient abelian group. Then $M/N$ is an $R$-module where $R$ acts on cosets by

$$r(x+N)=rx+N.$$

This is called the quotient module of $M$ by $N$.

The $R$-module structure is well defined because if $x+N=y+N$, then $x=y+n$ for some $n\in N$, and $rx=ry+rn$. Since $N$ is a submodule, $rn\in N$ so $rx+N=ry+N$.

Notice that $N$ can be any submodule, there is no “normality” condition like for groups.

There is always a “projection” homomorphism $\pi :M\to M/N$ defined by $\pi (m)=m+N$ which has kernel $N$.

Sums of modules

If $A$ and $B$ are submodules of a module $M$, then $A+B$ is the smallest submodule of $M$ containing both $A$ and $B$. Alternatively it is:

$$A+B=\{a+b:a\in A,b\in B\}$$

Mapping Properties

Let $M$, $N$, and $K$ be $R$ modules, and let $f:M\to K$ be a homomorphism with $N\subset \ker (f)$. Then there is a unique homomorphism $\bar{f}:M/N\to K$ making this diagram commutative:

$$Mf\pi M/N\bar{f}K$$

Isomorphism theorems

The isomorphism theorems for abelian groups give isomorphism theorems for modules.

• If $f:M\to K$ is a homomorphism, then the map $\bar{f}$ gives an isomorphism between $M/\ker (f)$ and $f(M)\subset K$.
• $(M+N)/N$ is isomorphic to $M/(M\cap N)$.
• $(M/A)/(N/A)$ is isomorphic to $M/N$.
• There is a bijection between the lattice of submodules of $M/N$ and submodules of $M$ containing $N$ given by $K\leftrightarrow K/N$.

The proofs of all of these facts are found by checking that the group isomorphisms respect the action of the ring $R$.

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

The set ${\mathrm{Hom}}_R(M,N)$ is an abelian group: $(f+g)(m)=f(m)+g(m)$ and the zero map is the identity.

If $R$ is commutative then ${\mathrm{Hom}}_R(M,N)$ is an $R$-module if we set $(rf)$ to be the function $(rf)(m)=r(f(m))=f(rm)$. We need $rf$ to be a module homomorphism, which means we need:

$$(rf)(sm)=s(rf)(m).$$

This works out ok if $R$ is commutative since

$$(rf)(sm)=f(rsm)=f(srm)=s(f(rm))=s((rf)(m))$$

but it fails if $R$ is not commutative.

${\mathrm{Hom}}_R(M,M)$

The set ${\mathrm{Hom}}_R(M,M)$ is a ring with multiplication given by composition. The identity map gives an identity for this ring.

If $R$ is commutative then, given $r\in R$, we have an element ${\varphi }_r\in {\mathrm{Hom}}_R(M,M)$ given by ${\varphi }_r(m)=rm$. This is a homomorphism because

$${\varphi }_r(sm)=rsm=srm=s{\varphi }_r(m)$$

but this fails in general if $R$ is not commutative. Thus, if $R$ is commutative, ${\mathrm{Hom}}_R(M,M)$ is an $R$-algebra.

More on ${\mathrm{Hom}}_R(M,M)$

If $M=R^n$, then ${\mathrm{Hom}}_R(M,M)$ is the ring of $n\times n$ matrices with entries from $R$.

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