## 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

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:

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

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 $R$.