Skip to main content Link Menu Expand (external link) Copy Copied

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×MM (written (r,m)rm) such that:

  • r(m1+m2)=rm1+rm2
  • (r1+r2)m=r1m+r2m
  • r1(r2m)=(r1r2)m

If R has a unit element 1, we also require 1m=m for all mM.

Right modules

A right module is defined by a map M×RM and written (m,r)mr and satisfying the property (mr1)r2=m(r1r2).

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

  • r1r2 acts by “first r2, then r1

while for a right module

  • r1r2 acts by “first r1, then r2.”

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 mr=(rm). This works because

(mr1)r2=(r1m)r2=(r2(r1m))=((r2r1)m)=((r1r2)m)=m(r1r2)

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

REnd(M).

If ϕr is the endomorphism associated to rR, then rm=ϕr(m). The associativity comes from defining the ring structure on End(M) as the usual composition of functions: ϕr1r2=ϕr1ϕr2.

Submodules

Definition: If M is a left R-module, then a submodule N of M is a subgroup with the property that, if nN, then rnN for all rR.

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=Mn(F) be the n×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 n1 be a positive integer. Then

Rn={(r1,,rn):riR for i=1,,n}

is an R module with componentwise addition and multiplication given by r(r1,,rn)=(rr1,,rrn).

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 Rk for kn.
  • For more general R the picture is more complicated. Let R=Z and M=Z2. Then:
    • {(n,0):nZ} is a submodule of M which “looks like” a subspace.
    • 2M={(a,b):a,b2Z} 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 RS 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=x+x++xn.

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

nx=(1+1++1n)x=x+x++xn.

(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 IR is a two-sided ideal with the property that, for all yI, and all xM, 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+IR/I. This is well-defined since two different coset representatives r,r satisfy r=r+i for some iI and therefore rm=(r+i)m=rm since im=0.

If M is an abelian group and mZ is a positive integer such that mM=0, then M can be viewed as a module over Z/mZ 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:VV be an F-linear transformation. Define a homomorphism

F[x]End(V)

by sending

xkTk=TTTn.

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=F2 and let T be the linear transformation given by the matrix

T=(0111)

If e0 and e1 are the standard basis elements of F2 then

Te0=e1T2e0=Te1=e0+e1=e0+Te0=(1+T)e0

Polynomials and linear transformations continued

Therefore (T2T1)e0=0 and

(T2T1)e1=(T2T1)Te0=T(T2T1)e0=0

so the polynomial x2x1 is in the kernel of the map from F[x]End(V).

By the base change construction above this means that V can be viewed as a module over F[x]/(x2x1).

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 xF[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]modules}{vector spaces V over F with a given linear map T:VV}

Submodules of F[x] modules

In the correspondence above, a submodule of an F[x] module M corresponds to a subspace WV that is preserved by T, meaning TWW.

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 F2 satisfying Ue0=0 and Ue1=e0, then the one dimensional subspace spanned by e0 is U-stable and F2 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,yN and rR, we have x+ryN. Alternatively, if N is a subgroup of the abelian group M and rNN for all rR 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:RS carrying 1R to 1S such that f(R) is in the center of S.

The polynomial ring F[x] is an F-algebra, as is the matrix ring Mn(F) where the homomorphism f:FMn(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 Z/pZ is a Z-algebra. In fact any ring S with 1 is a Z algebra by the map sending nZ to n1S.

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:AB is a ring homomorphism that is R-linear in the sense that f(ra)=rf(a) for all rR and aA.

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:MN 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 rR.

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 HomR(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:MN be a homomorphism.

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

Quotient modules

Let M be an R module and let NM 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 nN, and rx=ry+rn. Since N is a submodule, rnN 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 π:MM/N defined by π(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:aA,bB}

Mapping Properties

Let M, N, and K be R modules, and let f:MK be a homomorphism with Nker(f). Then there is a unique homomorphism f:M/NK making this diagram commutative:

MfπM/NfK

Isomorphism theorems

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

  • If f:MK is a homomorphism, then the map f gives an isomorphism between M/ker(f) and f(M)K.
  • (M+N)/N is isomorphic to M/(MN).
  • (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 KK/N.

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

HomR(M,N)

The set HomR(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 HomR(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.

HomR(M,M)

The set HomR(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 rR, we have an element ϕrHomR(M,M) given by ϕr(m)=rm. This is a homomorphism because

ϕr(sm)=rsm=srm=sϕr(m)

but this fails in general if R is not commutative. Thus, if R is commutative, HomR(M,M) is an R-algebra.

More on HomR(M,M)

If M=Rn, then HomR(M,M) is the ring of n×n matrices with entries from R.

View as slides