Modules

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:

$r(m_1+m_2)=rm_1+rm_2$
$(r_1+r_2)m = r_1 m + r_2 m$
$r_1 (r_2 m) = (r_1 r_2) m$

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 $M’$ with the same underlying abelian group $M$ and by defining $m’ r=(r m)’$. This works because

\[(m'r_1) r_2 = (r_1 m)'r_2 = (r_2(r_1 m))'=((r_2 r_1)m)' =((r_1 r_2)m)' = m'(r_1 r_2)\]

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

\[R\to \End (M).\]

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

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

\[R^{n}=\lbrace (r_1,\ldots, r_n) : r_{i}\in R\mathrm{\ for\ } i=1,\ldots,n\rbrace\]

is an $R$ module with componentwise addition and multiplication given by $r(r_1,\ldots, r_n)=(rr_1,\ldots, rr_n)$.

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 $R=\Z$ and $M=\Z^{2}$. Then:
- $\lbrace (n,0) : n\in \Z\rbrace$ is a submodule of $M$ which “looks like” a subspace.
- $2M=\lbrace (a,b) : a,b\in 2\Z\rbrace$ 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:
- $\Q$ is a module over $\Q$, where it is a one dimensional vector space and its only $\Q$-submodules are $0$ and itself.
- $\Q$ is a module over $\Z$, and it has many $\Z$-submodules, such as $\Z[1/2]$.

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.

$\Z$-modules are the same as abelian groups

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

\[nx=\overbrace{x+x+\cdots+x}^{n}.\]

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

\[nx = (\overbrace{1+1+\cdots+1}^{n})x = \overbrace{x+x+\cdots+x}^{n}.\]

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

Further, submodules of $M$ (as $\Z$-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 $r,r’$ satisfy $r’=r+i$ for some $i\in I$ and therefore $r’m=(r+i)m=rm$ 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 $\Zn{m}$ 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

\[F[x]\to \End(V)\]

by sending

\[x^{k} \mapsto T^{k}=\overbrace{T\circ T\circ\cdots\circ T}^{n}.\]

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

\[T = \begin{pmatrix} 0 & 1 \\ 1 & 1\end{pmatrix}\]

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

\[\begin{aligned} Te_0&=&e_1 \\ T^2e_0=Te_1=e_0+e_1=e_0+Te_0&=&(1+T)e_0\\ \end{aligned}\]

Polynomials and linear transformations continued

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

\[(T^2-T-1)e_1=(T^2-T-1)Te_0=T(T^2-T-1)e_0=0\]

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

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

\[\begin{xy}\xymatrix{ \lbrace F[x]-\mathrm{modules}\rbrace\ar@2{<->}[d]\\ \lbrace \mathrm{vector\ spaces\ }V\mathrm{\ over\ }F\mathrm{\ with\ a\ given\ linear\ map\ }T:V\to V\rbrace \\} \end{xy}\]

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 $Ue_0=0$ and $Ue_1=e_0$, 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 $\Zn{p}$ is a $\Z$-algebra. In fact any ring $S$ with $1$ is a $\Z$ algebra by the map sending $n\in\Z$ to $n 1_S$.

The ring $\Q[x]$ is a $\Z[x]$ 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 $\Z$-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 $\Hom_{R}(M,N)$ 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 $\ker(f)=\lbrace m\in M : f(m)=0\rbrace$ (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:

\[A+B=\lbrace a+b : a\in A, b\in B\rbrace\]

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 $\overline{f}:M/N\to K$ making this diagram commutative:

\[\begin{xy} \xymatrix { M\ar[rd]^{f}\ar[d]^{\pi} & \\ M/N\ar[r]_{\overline{f}} & K\\ } \end{xy}\]

Isomorphism theorems

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

If $f:M\to K$ is a homomorphism, then the map $\overline{f}$ 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 $R$.

$\Hom_{R}(M,N)$

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

If $R$ is commutative then $\Hom_{R}(M,N)$ 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.

$\Hom_{R}(M,M)$

The set $\Hom_{R}(M,M)$ 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 $\phi_r\in\Hom_{R}(M,M)$ given by $\phi_r(m)=rm$. This is a homomorphism because

\[\phi_r(sm)=rsm=srm=s\phi_r(m)\]

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

More on $\Hom_{R}(M,M)$

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

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