IB Groups, Rings and Modules (Full)

Part IB — Groups, Rings and Modules

Based on lectures by O. Randal-Williams

Notes taken by Dexter Chua

Lent 2016

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Groups

Basic concepts of group theory recalled from Part IA Groups. Normal subgroups,

quotient groups and isomorphism theorems. Permutation groups. Groups acting on

sets, permutation representations. Conjugacy classes, centralizers and normalizers.

The centre of a group. Elementary properties of finite

-groups. Examples of finite

linear groups and groups arising from geometry. Simplicity of A

Sylow subgroups and Sylow theorems. Applications, groups of small order. [8]

Rings

Definition and examples of rings (commutative, with 1). Ideals, homomorphisms,

quotient rings, isomorphism theorems. Prime and maximal ideals. Fields. The

characteristic of a field. Field of fractions of an integral domain.

Factorization in rings; units, primes and irreducibles. Unique factorization in principal

ideal domains, and in polynomial rings. Gauss’ Lemma and Eisenstein’s irreducibility

criterion.

Rings

[

] of algebraic integers as subsets of

and quotients of

[

]. Examples of

Euclidean domains and uniqueness and non-uniqueness of factorization. Factorization

in the ring of Gaussian integers; representation of integers as sums of two squares.

Ideals in polynomial rings. Hilbert basis theorem. [10]

Modules

Definitions, examples of vector spaces, abelian groups and vector spaces with an

endomorphism. Sub-modules, homomorphisms, quotient modules and direct sums.

Equivalence of matrices, canonical form. Structure of finitely generated modules over

Euclidean domains, applications to abelian groups and Jordan normal form. [6]

Contents

0 Introduction

1 Groups

1.1 Basic concepts

1.2 Normal subgroups, quotients, homomorphisms, isomorphisms

1.3 Actions of permutations

1.4 Conjugacy, centralizers and normalizers

1.5 Finite p-groups

1.6 Finite abelian groups

1.7 Sylow theorems

2 Rings

2.1 Definitions and examples

2.2 Homomorphisms, ideals, quotients and isomorphisms

2.3 Integral domains, field of factions, maximal and prime ideals

2.4 Factorization in integral domains

2.5 Factorization in polynomial rings

2.6 Gaussian integers

2.7 Algebraic integers

2.8 Noetherian rings

3 Modules

3.1 Definitions and examples

3.2 Direct sums and free modules

3.3 Matrices over Euclidean domains

3.4 Modules over F[X] and normal forms for matrices

3.5 Conjugacy of matrices*

0 Introduction

The course is naturally divided into three sections — Groups, Rings, and Modules.

In IA Groups, we learnt about some basic properties of groups, and studied

several interesting groups in depth. In the first part of this course, we will

further develop some general theory of groups. In particular, we will prove two

more isomorphism theorems of groups. While we will not find these theorems

particularly useful in this course, we will be able to formulate analogous theorems

for other algebraic structures such as rings and modules, as we will later find in

the course.

In the next part of the course, we will study rings. These are things that

behave somewhat like

, where we can add, subtract, multiply but not (necessar-

ily) divide. While

has many nice properties, these are not necessarily available

in arbitrary rings. Hence we will classify rings into different types, depending on

how many properties of

they inherit. We can then try to reconstruct certain IA

Numbers and Sets results in these rings, such as unique factorization of numbers

into primes and B´ezout’s theorem.

Finally, we move on to modules. The definition of a module is very similar

to that of a vector space, except that instead of allowing scalar multiplication

by elements of a field, we have scalar multiplication by elements of a ring. It

turns out modules are completely unlike vector spaces, and can have much more

complicated structures. Perhaps because of this richness, many things turn out

to be modules. Using module theory, we will be able to prove certain important

theorems such as the classification of finite abelian groups and the Jordan normal

form theorem.

1 Groups

1.1 Basic concepts

We will begin by quickly recapping some definitions and results from IA Groups.

Definition (Group). A group is a triple (

G, ·, e

), where

is a set,

G×G → G

is a function and e ∈ G is an element such that

(i) For all a, b, c ∈ G, we have (a · b) · c = a · (b · c). (associativity)

(ii) For all a ∈ G, we have a · e = e · a = a. (identity)

(iii)

For all

a ∈ G

, there exists

−1

∈ G

such that

a ·a

−1

·a

.(inverse)

Some people add a stupid axiom that says

g · h ∈ G

for all

g, h ∈ G

, but this

is already implied by saying

is a function to

. You can write that down as

well, and no one will say you are stupid. But they might secretly think so.

Lemma. The inverse of an element is unique.

Proof. Let a

−1

, b be inverses of a. Then

b = b · e = b ·a · a

−1

= e · a

−1

= a

−1

Definition (Subgroup). If (

G, ·, e

) is a group and

H ⊆ G

is a subset, it is a

subgroup if

(i) e ∈ H,

(ii) a, b ∈ H implies a · b ∈ H,

(iii) · : H × H → H makes (H, ·, e) a group.

We write H ≤ G if H is a subgroup of G.

Note that the last condition in some sense encompasses the first two, but we

need the first two conditions to hold before the last statement makes sense at all.

Lemma.

H ⊆ G

is a subgroup if

is non-empty and for any

, h

∈ H

, we

have h

−1

∈ H.

Definition (Abelian group). A group

is abelian if

a · b

b · a

for all

a, b ∈ G

Example. We have the following familiar examples of groups

(i) (Z, +, 0), (Q, +, 0), (R, +, 0), (C, +, 0).

(ii) We also have groups of symmetries:

(a)

The symmetric group

is the collection of all permutations of

{1, 2, ··· , n}.

(b) The dihedral group D

is the symmetries of a regular n-gon.

(c)

The group

(

) is the group of invertible

n × n

real matrices,

which also is the group of invertible

-linear maps from the vector

space R

to itself.

(iii) The alternating group A

≤ S

(iv) The cyclic group C

≤ D

(v)

The special linear group

(

)

≤ GL

(

), the subgroup of matrices of

determinant 1.

(vi) The Klein-four group C

× C

(vii)

The quaternions

{±

, ±i, ±j, ±k}

with

k, ji

−k

= −1, (−1)

= 1.

With groups and subgroups, we can talk about cosets.

Definition (Coset). If H ≤ G, g ∈ G, the left coset gH is the set

gH = {x ∈ G : x = g · h for some h ∈ H}.

For example, since

is a subgroup, we know

e ∈ H

. So for any

g ∈ G

, we

must have g ∈ gH.

The collection of

-cosets in

forms a partition of

, and furthermore,

all

-cosets

are in bijection with

itself, via

h 7→ gh

. An immediate

consequence is

Theorem (Lagrange’s theorem). Let G be a finite group, and H ≤ G. Then

|G| = |H||G : H|,

where |G : H| is the number of H-cosets in G.

We can do exactly the same thing with right cosets and get the same

conclusion.

We have implicitly used the following notation:

Definition (Order of group). The order of a group is the number of elements

in G, written |G|.

Instead of order of the group, we can ask what the order of an element is.

Definition (Order of element). The order of an element

g ∈ G

is the smallest

positive n such that g

= e. If there is no such n, we say g has infinite order.

We write ord(g) = n.

A basic lemma is as follows:

Lemma. If G is a finite group and g ∈ G has order n, then n | |G|.

Proof. Consider the following subset:

H = {e, g, g

, ··· , g

n−1

This is a subgroup of

, because it is non-empty and

−s

r−s

is on the list

(we might have to add

to the power of

to make it positive, but this is fine

since

). Moreover, there are no repeats in the list: if

, with wlog

i ≥ j

, then

i−j

. So

i − j < n

. By definition of

, we must have

i − j

= 0,

i.e. i = j.

Hence Lagrange’s theorem tells us n = |H| | |G|.

1.2

Normal subgroups, quotients, homomorphisms, iso-

morphisms

We all (hopefully) recall what the definition of a normal subgroup is. However,

instead of just stating the definition and proving things about it, we can try to

motivate the definition, and see how one could naturally come up with it.

Let

H ≤ G

be a subgroup. The objective is to try to make the collection of

cosets

G/H = {gH : g ∈ G}

into a group.

Before we do that, we quickly come up with a criterion for when two cosets

and

are equal. Notice that if

, then

g ∈ g

. So

·h

for

some

. In other words, (

)

−1

· g

h ∈ H

. So if two elements represent the

same coset, their difference is in

. The argument is also reversible. Hence two

elements g, g

represent the same H-coset if and only if (g

)

−1

g ∈ H.

Suppose we try to make the set

G/H

{gH

g ∈ G}

into a group, by the

obvious formula

H) · (g

H) = g

However, this doesn’t necessarily make sense. If we take a different representative

for the same coset, we want to make sure it gives the same answer.

If g

H = g

H, then we know g

= g

· h for some h ∈ H. So

H) · (g

H) = g

H = g

hH = g

H = (g

H) · (g

H).

So all is good.

What if we change g

? If g

H = g

H, then g

= g

· h for some h ∈ H. So

H) · (g

H) = g

H = g

Now we are stuck. We would really want the equality

H = g

to hold. This requires

)

−1

∈ H.

This is equivalent to

−1

∈ H.

So for

G/H

to actually be a group under this operation, we must have, for any

h ∈ H and g ∈ G, the property g

−1

hg ∈ H to hold.

This is not necessarily true for an arbitrary

. Those nice ones that satisfy

this property are known as normal subgroups.

Definition (Normal subgroup). A subgroup

H ≤ G

is normal if for any

h ∈ H

and g ∈ G, we have g

−1

hg ∈ H. We write H C G.

This allows us to make the following definition:

Definition (Quotient group). If

H C G

is a normal subgroup, then the set

G/H

of left H-cosets forms a group with multiplication

H) · (g

H) = g

with identity eH = H. This is known as the quotient group.

This is indeed a group. Normality was defined such that this is well-defined.

Multiplication is associative since multiplication in

is associative. The inverse

of gH is g

−1

H, and eH is easily seen to be the identity.

So far, we’ve just been looking at groups themselves. We would also like to

know how groups interact with each other. In other words, we want to study

functions between groups. However, we don’t allow arbitrary functions, since

groups have some structure, and we would like the functions to respect the group

structures. These nice functions are known as homomorphisms.

Definition (Homomorphism). If (

G, ·, e

) and (

H, ∗, e

) are groups, a function

φ : G → H is a homomorphism if φ(e

) = e

, and for g, g

∈ G, we have

φ(g · g

) = φ(g) ∗ φ(g

If we think carefully,

(

) =

can be derived from the second condition,

but it doesn’t hurt to put it in as well.

Lemma. If φ : G → H is a homomorphism, then

φ(g

−1

) = φ(g)

−1

Proof. We compute φ(g · g

−1

) in two ways. On the one hand, we have

φ(g · g

−1

) = φ(e) = e.

On the other hand, we have

φ(g · g

−1

) = φ(g) ∗ φ(g

−1

By the uniqueness of inverse, we must have

φ(g

−1

) = φ(g)

−1

Given any homomorphism, we can build two groups out of it:

Definition (Kernel). The kernel of a homomorphism φ : G → H is

ker(φ) = {g ∈ G : φ(g) = e}.

Definition (Image). The image of a homomorphism φ : G → H is

im(φ) = {h ∈ H : h = φ(g) for some g ∈ G}.

Lemma. For a homomorphism

G → H

, the kernel

ker

(

) is a normal

subgroup, and the image im(φ) is a subgroup of H.

Proof. There is only one possible way we can prove this.

To see ker(φ) is a subgroup, let g, h ∈ ker φ. Then

φ(g · h

−1

) = φ(g) ∗ φ(h)

−1

= e ∗ e

−1

= e.

So gh

−1

∈ ker φ. Also, φ(e) = e. So ker(φ) is non-empty. So it is a subgroup.

To show it is normal, let

g ∈ ker

(

). Let

x ∈ G

. We want to show

−1

gx ∈ ker(φ). We have

φ(x

−1

gx) = φ(x

−1

) ∗ φ(g) ∗ φ(x) = φ(x

−1

) ∗ φ(x) = φ(x

−1

x) = φ(e) = e.

So x

−1

gx ∈ ker(φ). So ker(φ) is normal.

Also, if φ(g), φ(h) ∈ im(φ), then

φ(g) ∗ φ(h)

−1

= φ(gh

−1

) ∈ im(φ).

Also, e ∈ im(φ). So im(φ) is non-empty. So im(φ) is a subgroup.

Definition (Isomorphism). An isomorphism is a homomorphism that is also a

bijection.

Definition (Isomorphic group). Two groups

and

are isomorphic if there

is an isomorphism between them. We write G

∼

Usually, we identify two isomorphic groups as being “the same”, and do not

distinguish isomorphic groups.

It is an exercise to show the following:

Lemma. If φ is an isomorphism, then the inverse φ

−1

is also an isomorphism.

When studying groups, it is often helpful to break the group apart into smaller

groups, which are hopefully easier to study. We will have three isomorphism

theorems to do so. These isomorphism theorems tell us what happens when we

take quotients of different things. Then if a miracle happens, we can patch what

we know about the quotients together to get information about the big group.

Even if miracles do not happen, these are useful tools to have.

The first isomorphism relates the kernel to the image.

Theorem (First isomorphism theorem). Let

G → H

be a homomorphism.

Then ker(φ) C G and

ker(φ)

∼

im(φ).

Proof.

We have already proved that

ker

(

) is a normal subgroup. We now

have to construct a homomorphism

G/ ker

(

)

→ im

(

), and prove it is an

isomorphism.

Define our function as follows:

f :

ker(φ)

→ im(φ)

g ker(φ) 7→ φ(g).

We first tackle the obvious problem that this might not be well-defined, since we

are picking a representative for the coset. If

g ker

(

) =

ker

(

), then we know

−1

· g

∈ ker(φ). So φ(g

−1

· g

) = e. So we know

e = φ(g

−1

· g

) = φ(g)

−1

∗ φ(g

Multiplying the whole thing by

(

) gives

(

) =

(

). Hence this function is

well-defined.

Next we show it is a homomorphism. To see

is a homomorphism, we have

f(g ker(φ) · g

ker(φ)) = f(gg

ker(φ))

= φ(gg

)

= φ(g) ∗ φ(g

)

= f(g ker(φ)) ∗ f (g

ker(φ)).

So f is a homomorphism. Finally, we show it is a bijection.

To show it is surjective, let

h ∈ im

(

). Then

(

) for some

. So

h = f(g ker(φ)) is in the image of f.

To show injectivity, suppose

(

g ker

(

)) =

(

ker

(

)). So

(

) =

(

). So

(

−1

· g

) =

. Hence

−1

· g

∈ ker

(

), and hence

g ker

(

) =

ker

(

). So

done.

Before we move on to further isomorphism theorems, we see how we can use

these to identify two groups which are not obviously the same.

Example. Consider a homomorphism

C → C \ {

}

given by

z 7→ e

. We

also know that

z+w

= e

This means φ is a homomorphism if we think of it as φ : (C, +) → (C \ {0}, ×).

What is the image of this homomorphism? The existence of

log

shows that

φ is surjective. So im φ = C \ {0}. What about the kernel? It is given by

ker(φ) = {z ∈ C : e

= 1} = 2πiZ,

i.e. the set of all integer multiples of 2πi. The conclusion is that

(C/(2πiZ), +)

∼

(C \ {0}, ×).

The second isomorphism theorem is a slightly more complicated theorem.

Theorem (Second isomorphism theorem). Let

H ≤ G

and

K C G

. Then

HK = {h · k : h ∈ H, k ∈ K} is a subgroup of G, and H ∩ K C H. Moreover,

∼

H ∩ K

Proof. Let hk, h

∈ HK. Then

(hk)

−1

= h

−1

= (h

−1

)(hk

−1

The first term is in

, while the second term is

−1

∈ K

conjugated by

, which also has to be in

be normality. So this is something in

times

something in K, and hence in HK. HK also contains e, and is hence a group.

To show

H ∩K C H

, consider

x ∈ H ∩K

and

h ∈ H

. Consider

−1

. Since

x ∈ K

, the normality of

implies

−1

xh ∈ K

. Also, since

x, h ∈ H

, closure

implies h

−1

xh ∈ H. So h

−1

xh ∈ H ∩ K. So H ∩ K C H.

Now we can start to prove the second isomorphism theorem. To do so, we

apply the first isomorphism theorem to it. Define

φ : H → G/K

h 7→ hK

This is easily seen to be a homomorphism. We apply the first isomorphism

theorem to this homomorphism. The image is all

-cosets represented by

something in H, i.e.

im(φ) =

Then the kernel of φ is

ker(φ) = {h ∈ H : hK = eK} = {h ∈ H : h ∈ K} = H ∩ K.

So the first isomorphism theorem says

H ∩ K

∼

Notice we did more work than we really had to. We could have started by

writing down

and checked it is a homomorphism. Then since

H ∩ K

is its

kernel, it has to be a normal subgroup.

Before we move on to the third isomorphism theorem, we notice that if

K C G

, then there is a bijection between subgroups of

G/K

and subgroups of

containing K, given by

{subgroups of G/K} ←→ {subgroups of G which contain K}

X ≤

−→ {g ∈ G : gK ∈ X}

≤

←− K C L ≤ G.

This specializes to the bijection of normal subgroups:

{normal subgroups of G/K} ←→ {normal subgroups of G which contain K}

using the same bijection.

It is an elementary exercise to show that these are inverses of each other.

This correspondence will be useful in later times.

Theorem (Third isomorphism theorem). Let

K ≤ L ≤ G

be normal subgroups

of G. Then



∼

Proof. Define the homomorphism

φ : G/K → G/L

gK 7→ gL

As always, we have to check this is well-defined. If

, then

−1

∈

K ⊆ L. So gL = g

L. This is also a homomorphism since

φ(gK · g

K) = φ(gg

K) = gg

L = (gL) · (g

L) = φ(gK) · φ(g

K).

This clearly is surjective, since any coset

is the image

(

). So the image

is G/L. The kernel is then

ker(φ) = {gK : gL = L} = {gK : g ∈ L} =

So the conclusion follows by the first isomorphism theorem.

The general idea of these theorems is to take a group, find a normal subgroup,

and then quotient it out. Then hopefully the normal subgroup and the quotient

group will be simpler. However, this doesn’t always work.

Definition (Simple group). A (non-trivial) group

is simple if it has no normal

subgroups except {e} and G.

In general, simple groups are complicated. However, if we only look at abelian

groups, then life is simpler. Note that by commutativity, the normality condition

is always trivially satisfied. So any subgroup is normal. Hence an abelian group

can be simple only if it has no non-trivial subgroups at all.

Lemma. An abelian group is simple if and only if it is isomorphic to the cyclic

group C

for some prime number p.

Proof.

By Lagrange’s theorem, any subgroup of

has order dividing

Hence if

is prime, then it has no such divisors, and any subgroup must have

order 1 or

, i.e. it is either

{e}

itself. Hence in particular any normal

subgroup must be {e} or C

. So it is simple.

Now suppose

is abelian and simple. Let

e 6

g ∈ G

be a non-trivial element,

and consider

{··· , g

−2

, g

−1

, e, g, g

, ···}

. Since

is abelian, conjugation

does nothing, and every subgroup is normal. So

is a normal subgroup. As

is simple,

{e}

. Since it contains

g 6

, it is non-trivial. So we

must have H = G. So G is cyclic.

is infinite cyclic, then it is isomorphic to

. But

is not simple, since

2Z C Z. So G is a finite cyclic group, i.e. G

∼

for some finite m.

n | m

, then

m/n

generates a subgroup of

of order

. So this is a normal

subgroup. Therefore

must be

or 1. Hence

cannot be simple unless

has

no divisors except 1 and m, i.e. m is a prime.

One reason why simple groups are important is the following:

Theorem. Let G be any finite group. Then there are subgroups

G = H

B H

B ··· B H

= {e}.

such that H

i+1

is simple.

Note that here we only claim that

i+1

is normal in

. This does not say

that, say, H

is a normal subgroup of H

Proof. If G is simple, let H

= {e}. Then we are done.

is not simple, let

be a maximal proper normal subgroup of

. We

now claim that G/H

is simple.

G/H

is not simple, it contains a proper non-trivial normal subgroup

L C G/H

such that

L 6

{e}, G/H

. However, there is a correspondence

between normal subgroups of

G/H

and normal subgroups of

containing

must be

K/H

for some

K C G

such that

K ≥ H

. Moreover, since

non-trivial and not

G/H

, we know

is not

. So

is a larger normal

subgroup. Contradiction.

So we have found an

C G

such that

G/H

is simple. Iterating this process

gives the desired result. Note that this process eventually stops, as

i+1

< H

, and hence |H

i+1

| < |H

|, and all these numbers are finite.

1.3 Actions of permutations

When we first motivated groups, we wanted to use them to represent some

collection of “symmetries”. Roughly, a symmetry of a set

is a permutation

, i.e. a bijection

X → X

that leaves some nice properties unchanged. For

example, a symmetry of a square is a permutation of the vertices that leaves the

overall shape of the square unchanged.

Instead of just picking some nice permutations, we can consider the group of

all permutations. We call this the symmetric group.

Definition (Symmetric group). The symmetric group

is the group of all

permutations of {1, ··· , n}, i.e. the set of all bijections of this set with itself.

A convenient way of writing permutations is to use the disjoint cycle notation,

such as writing (1 2 3)(4 5)(6) for the permutation that maps

1 7→ 2 4 7→ 5

2 7→ 3 5 7→ 4

3 7→ 1 6 7→ 6.

Unfortunately, the convention for writing permutations is weird. Since permuta-

tions are bijections, and hence functions, they are multiplied the wrong way, i.e.

f ◦ g

means first apply

, then apply

. In particular, (1 2 3)(3 4) requires first

applying the second permutation, then the first, and is in fact (1 2 3 4).

We know that any permutation is a product of transpositions. Hence we

make the following definition.

Definition (Even and odd permutation). A permutation

σ ∈ S

is even if it

can be written as a product of evenly many transpositions; odd otherwise.

In IA Groups, we spent a lot of time proving this is well-defined, and we are

not doing that again (note that this definition by itself is well-defined — if a

permutation can be both written as an even number of transposition and an odd

number of transposition, the definition says it is even. However, this is not what

we really want, since we cannot immediately conclude that, say, (1 2) is odd).

This allows us to define the homomorphism:

sgn : S

→ ({±1}, ×)

σ 7→

(

+1 σ is even

−1 σ is odd

Definition (Alternating group). The alternating group

≤ S

is the subgroup

of even permutations, i.e. A

is the kernel of sgn.

This immediately tells us

C S

, and we can immediately work out its

index, since

∼

im(sgn) = {±1},

unless n = 1. So A

has index 2.

More generally, for a set X, we can define its symmetric group as follows:

Definition (Symmetric group of

). Let

be a set. We write

Sym

(

) for the

group of all permutations of X.

However, we don’t always want the whole symmetric group. Sometimes, we

just want some subgroups of symmetric groups, as in our initial motivation. So

we make the following definition.

Definition (Permutation group). A group

is called a permutation group if it

is a subgroup of

Sym

(

) for some

, i.e. it is given by some, but not necessarily

all, permutations of some set.

We say G is a permutation group of order n if in addition |X| = n.

This is not really a too interesting definition, since, as we will soon see, every

group is (isomorphic to) a permutation group. However, in some cases, thinking

of a group as a permutation group of some object gives us better intuition on

what the group is about.

Example.

and

are obviously permutation groups. Also, the dihedral

group

is a permutation group of order

, viewing it as a permutation of the

vertices of a regular n-gon.

We would next want to recover the idea of a group being a “permutation”.

G ≤ Sym

(

), then each

g ∈ G

should be able to give us a permutation of

in a way that is consistent with the group structure. We say the group

acts

on X. In general, we make the following definition:

Definition (Group action). An action of a group (

G, ·

) on a set

is a function

∗ : G × X → X

such that

(i) g

∗ (g

∗ x) = (g

· g

) ∗ x for all g

, g

∈ G and x ∈ X.

(ii) e ∗ x = x for all x ∈ X.

There is another way of defining group actions, which is arguably a better

way of thinking about group actions.

Lemma. An action of

is equivalent to a homomorphism

G →

Sym(X).

Note that the statement by itself is useless, since it doesn’t tell us how to

translate between the homomorphism and a group action. The important part

is the proof.

Proof.

Let

∗

G × X → X

be an action. Define

G → Sym

(

) by sending

to the function

(

) = (

g ∗ ·

X → X

). This is indeed a permutation —

−1

∗ ·

is an inverse since

φ(g

−1

)(φ(g)(x)) = g

−1

∗ (g ∗ x) = (g

−1

· g) ∗ x = e ∗x = x,

and a similar argument shows

(

)

◦φ

(

−1

) =

. So

is at least a well-defined

function.

To show it is a homomorphism, just note that

φ(g

)(φ(g

)(x)) = g

∗ (g

∗ x) = (g

· g

) ∗ x = φ(g

· g

)(x).

Since this is true for all

x ∈ X

, we know

(

)

◦ φ

(

) =

(

· g

). Also,

(

)(

) =

e ∗x

. So

(

) is indeed the identity. Hence

is a homomorphism.

We now do the same thing backwards. Given a homomorphism

G →

Sym

(

), define a function by

g ∗x

(

)(

). We now check it is indeed a group

action. Using the definition of a homomorphism, we know

(i) g

∗

(

∗ x

) =

(

)(

(

)(

)) = (

(

)

◦ φ

(

))(

) =

(

· g

)(

) =

· g

) ∗ x.

(ii) e ∗ x = φ(e)(x) = id

(x) = x.

So this homomorphism gives a group action. These two operations are clearly in-

verses to each other. So group actions of

are the same as homomorphisms

G → Sym(X).

Definition (Permutation representation). A permutation representation of a

group G is a homomorphism G → Sym(X).

We have thus shown that a permutation representation is the same as a

group action.

The good thing about thinking of group actions as homomorphisms is that

we can use all we know about homomorphisms on them.

Notation. For an action of

given by

G → Sym

(

), we write

= im(φ) and G

= ker(φ).

The first isomorphism theorem immediately gives

Proposition. G

C G and G/G

∼

In particular, if G

= {e} is trivial, then G

∼

≤ Sym(X).

Example. Let

be the group of symmetries of a cube. Let

be the set of

diagonals of the cube.

Then

acts on

, and so we get

G → Sym

(

). What is its kernel? To pre-

serve the diagonals, it either does nothing to the diagonal, or flips the two vertices.

So G

= ker(φ) = {id, symmetry that sends each vertex to its opposite}

∼

How about the image? We have

(

)

≤ Sym

(

)

∼

. It is an

exercise to show that

(

) =

Sym

(

), i.e. that

is surjective. We are not

proving this because this is an exercise in geometry, not group theory. Then the

first isomorphism theorem tells us

∼

G/G

|G| = |G

||G

| = 4! · 2 = 48.

This is an example of how we can use group actions to count elements in a

group.

Example (Cayley’s theorem). For any group

, we have an action of

itself via

g ∗ g

= gg

It is trivial to check this is indeed an action. This gives a group homomorphism

G → Sym

(

). What is its kernel? If

g ∈ ker

(

), then it acts trivially on

every element. In particular, it acts trivially on the identity. So

g ∗ e

, which

means g = e. So ker(φ) = {e}. By the first isomorphism theorem, we get

∼

G/{e}

∼

im φ ≤ Sym(G).

So we know every group is (isomorphic to) a subgroup of a symmetric group.

Example. Let

be a subgroup of

, and

G/H

be the set of left cosets of

H. We let G act on X via

g ∗ g

H = gg

It is easy to check this is well-defined and is indeed a group action. So we get

φ : G → Sym(X).

Now consider

ker

(

). If

g ∈ G

, then for every

∈ G

, we have

g ∗ g

H = g

H. This means g

−1

∈ H. In other words, we have

g ∈ g

−1

This has to happen for all g

∈ G. So

⊆

∈G

−1

This argument is completely reversible — if

g ∈

∈G

−1

, then for each

∈ G, we know

−1

∈ H,

and hence

H = g

g ∗ g

H = g

So g ∈ G

. Hence we indeed have equality:

ker(φ) = G

∈G

−1

Since this is a kernel, this is a normal subgroup of

, and is contained in

Starting with an arbitrary subgroup

, this allows us to generate a normal

subgroup, and this is indeed the biggest normal subgroup of

that is contained

in H, if we stare at it long enough.

We can use this to prove the following theorem.

Theorem. Let

be a finite group, and

H ≤ G

a subgroup of index

. Then

there is a normal subgroup

K C G

with

K ≤ H

such that

G/K

is isomorphic to

a subgroup of S

. Hence |G/K| | n! and |G/K| ≥ n.

Proof.

We apply the previous example, giving

G → Sym

(

G/H

), and let

be the kernel of this homomorphism. We have already shown that

K ≤ H

. Then

the first isomorphism theorem gives

G/K

∼

im φ ≤ Sym(G/H)

∼

Then by Lagrange’s theorem, we know

|G/K| | |S

!, and we also have

|G/K| ≥ |G/H| = n.

Corollary. Let

be a non-abelian simple group. Let

H ≤ G

be a proper

subgroup of index

. Then

is isomorphic to a subgroup of

. Moreover, we

must have n ≥ 5, i.e. G cannot have a subgroup of index less than 5.

Proof.

The action of

G/H

gives a homomorphism

G → Sym

(

Then

ker

(

)

C G

. Since

is simple,

ker

(

) is either

{e}

. We first show

that it cannot be

. If

ker

(

) =

, then every element of

acts trivially on

G/H

. But if

g ∈ G \ H

, which exists since the index of

is not 1, then

g ∗H

gH 6

. So

does not act trivially. So the kernel cannot be the whole

of G. Hence ker(φ) = {e}.

Thus by the first isomorphism theorem, we get

∼

im(φ) ≤ Sym(X)

∼

We now need to show that G is in fact a subgroup of A

We know

C S

. So

(

)

∩A

C im

(

)

∼

. As

is simple,

(

)

∩A

is either

{e}

(

). We want to show that the second thing happens,

i.e. the intersection is not the trivial group. We use the second isomorphism

theorem. If im(φ) ∩ A

= {e}, then

im(φ)

∼

im(φ)

im(φ) ∩ A

∼

im(φ)A

≤

∼

(

) is a subgroup of

, i.e. either

{e}

itself. Neither of these are

non-abelian. So this cannot be the case. So we must have im(φ) ∩ A

= im(φ),

i.e. im(φ) ≤ A

The last part follows from the fact that

, S

have no non-abelian

simple subgroups, which you can check by going to a quiet room and listing out

all their subgroups.

Let’s recall some old definitions from IA Groups.

Definition (Orbit). If G acts on a set X, the orbit of x ∈ X is

G · x = {g ∗ x ∈ X : g ∈ G}.

Definition (Stabilizer). If G acts on a set X, the stabilizer of x ∈ X is

= {g ∈ G : g ∗ x = x}.

The main theorem about these concepts is the orbit-stabilizer theorem.

Theorem (Orbit-stabilizer theorem). Let

act on

. Then for any

x ∈ X

there is a bijection between G · x and G/G

, given by g · x ↔ g · G

In particular, if G is finite, it follows that

|G| = |G

||G · x|.

It takes some work to show this is well-defined and a bijection, but you’ve

done it in IA Groups. In IA Groups, you probably learnt the second statement

instead, but this result is more generally true for infinite groups.

1.4 Conjugacy, centralizers and normalizers

We have seen that every group acts on itself by multiplying on the left. A group

G can also act on itself in a different way, by conjugation:

g ∗ g

= gg

−1

Let

G → Sym

(

) be the associated permutation representation. We know,

by definition, that

(

) is a bijection from

as sets. However, here

not an arbitrary set, but is a group. A natural question to ask is whether

(

)

is a homomorphism or not. Indeed, we have

φ(g)(g

· g

) = gg

−1

= (gg

−1

)(gg

−1

) = φ(g)(g

)φ(g)(g

(

) is a homomorphism from

. Since

(

) is bijective (as in any

group action), it is in fact an isomorphism.

Thus, for any group

, there are many isomorphisms from

to itself, one

for every g ∈ G, and can be obtained from a group action of G on itself.

We can, of course, take the collection of all isomorphisms of

, and form a

new group out of it.

Definition (Automorphism group). The automorphism group of G is

Aut(G) = {f : G → G : f is a group isomorphism}.

This is a group under composition, with the identity map as the identity.

This is a subgroup of

Sym

(

), and the homomorphism

G → Sym

(

) by

conjugation lands in Aut(G).

This is pretty fun — we can use this to cook up some more groups, by taking

a group and looking at its automorphism group.

We can also take a group, take its automorphism group, and then take its

automorphism group again, and do it again, and see if this process stabilizes, or

becomes periodic, or something. This is left as an exercise for the reader.

Definition (Conjugacy class). The conjugacy class of g ∈ G is

ccl

(g) = {hgh

−1

: h ∈ G},

i.e. the orbit of g ∈ G under the conjugation action.

Definition (Centralizer). The centralizer of g ∈ G is

(g) = {h ∈ G : hgh

−1

= g},

i.e. the stabilizer of

under the conjugation action. This is alternatively the set

of all h ∈ G that commute with g.

Definition (Center). The center of a group G is

Z(G) = {h ∈ G : hgh

−1

= g for all g ∈ G} =

g∈G

(g) = ker(φ).

These are the elements of the group that commute with everything else.

By the orbit-stabilizer theorem, for each

x ∈ G

, we obtain a bijection

ccl(x) ↔ G/C

(x).

Proposition. Let G be a finite group. Then

|ccl(x)| = |G : C

(x)| = |G|/|C

(x)|.

In particular, the size of each conjugacy class divides the order of the group.

Another useful notion is the normalizer.

Definition (Normalizer). Let H ≤ G. The normalizer of H in G is

(H) = {g ∈ G : g

−1

Hg = H}.

Note that we certainly have

H ≤ N

(

). Even better,

HCN

(

), essentially

by definition. This is in fact the biggest subgroup of G in which H is normal.

We are now going to look at conjugacy classes of

. Now we recall from IA

Groups that permutations in

are conjugate if and only if they have the same

cycle type when written as a product of disjoint cycles. We can think of the

cycle types as partitions of

. For example, the partition 2

1 of 5 corresponds

to the conjugacy class of (1 2)(3 4)(5). So the conjugacy classes of

are exactly

the partitions of n.

We will use this fact in the proof of the following theorem:

Theorem. The alternating groups

are simple for

n ≥

5 (also for

= 1

3).

The cases in brackets follow from a direct check since

∼

{e}

and

∼

, all of which are simple. We can also check manually that

has

non-trivial normal subgroups, and hence not simple.

Recall we also proved that

is simple in IA Groups by brute force — we

listed all its conjugacy classes, and see they cannot be put together to make a

normal subgroup. This obviously cannot be easily generalized to higher values

of n. Hence we need to prove this with a different approach.

Proof. We start with the following claim:

Claim. A

is generated by 3-cycles.

As any element of

is a product of evenly-many transpositions, it suffices

to show that every product of two transpositions is also a product of 3-cycles.

There are three possible cases: let a, b, c, d be distinct. Then

(i) (a b)(a b) = e.

(ii) (a b)(b c) = (a b c).

(iii) (a b)(c d) = (a c b)(a c d).

So we have shown that every possible product of two transpositions is a product

of three-cycles.

Claim. Let H C A

. If H contains a 3-cycle, then we H = A

We show that if

contains a 3-cycle, then every 3-cycle is in

. Then we

are done since

is generated by 3-cycles. For concreteness, suppose we know

(a b c) ∈ H, and we want to show (1 2 3) ∈ H.

Since they have the same cycle type, so we have

σ ∈ S

such that (

a b c

) =

(1 2 3)

−1

. If

is even, i.e.

σ ∈ A

, then we have that (1 2 3)

∈ σ

−1

Hσ

by the normality of H and we are trivially done.

is odd, replace it by

¯σ

σ ·

(4 5). Here is where we use the fact that

n ≥ 5 (we will use it again later). Then we have

¯σ(1 2 3)¯σ

−1

= σ(4 5)(1 2 3)(4 5)σ

−1

= σ(1 2 3)σ

−1

= (a b c),

using the fact that (1 2 3) and (4 5) commute. Now

¯σ

is even. So (1 2 3)

∈ H

above.

What we’ve got so far is that if

H C A

contains any 3-cycle, then it is

Finally, we have to show that every normal subgroup must contain at least one

3-cycle.

Claim. Let H C A

be non-trivial. Then H contains a 3-cycle.

We separate this into many cases

(i)

Suppose

contains an element which can be written in disjoint cycle

notation

σ = (1 2 3 ···r)τ,

for

r ≥

4. We now let

= (1 2 3)

∈ A

. Then by normality of

, we know

−1

σδ ∈ H

. Then

−1

σδ ∈ H

. Also, we notice that

does not contain

3. So it commutes with

, and also trivially with (1 2 3

··· r

). We

can expand this mess to obtain

−1

σδ = (r ··· 2 1)(1 3 2)(1 2 3 ··· r)(1 2 3) = (2 3 r),

which is a 3-cycle. So done.

The same argument goes through if

= (

··· a

)

for any

, ··· , a

(ii)

Suppose

contains an element consisting of at least two 3-cycles in disjoint

cycle notation, say

σ = (1 2 3)(4 5 6)τ

We now let δ = (1 2 4), and again calculate

−1

σδ = (1 3 2)(4 6 5)(1 4 2)(1 2 3)(4 5 6)(1 2 4) = (1 2 4 3 6).

This is a 5-cycle, which is necessarily in

. By the previous case, we get a

3-cycle in H too, and hence H = A

(iii)

Suppose

contains

= (1 2 3)

, with

a product of 2-cycles (if

contains

anything longer, then it would fit in one of the previous two cases). Then

= (1 2 3)

= (1 3 2) is a three-cycle.

(iv)

Suppose

contains

= (1 2)(3 4)

, where

is a product of 2-cycles. We

first let δ = (1 2 3) and calculate

u = σ

−1

σδ = (1 2)(3 4)(1 3 2)(1 2)(3 4)(1 2 3) = (1 4)(2 3),

which is again in

. We landed in the same case, but instead of two

transpositions times a mess, we just have two transpositions, which is nicer.

Now let

v = (1 5 2)u(1 2 5) = (1 3)(4 5) ∈ H.

Note that we used

n ≥

5 again. We have yet again landed in the same case.

Notice however, that these are not the same transpositions. We multiply

uv = (1 4)(2 3)(1 3)(4 5) = (1 2 3 4 5) ∈ H.

This is then covered by the first case, and we are done.

So done. Phew.

1.5 Finite p-groups

Note that when studying the orders of groups and subgroups, we always talk

about divisibility, since that is what Lagrange’s theorem tells us about. We

never talk about things like the sum of the orders of two subgroups. When it

comes to divisibility, the simplest case would be when the order is a prime, and

we have done that already. The next best thing we can hope for is that the order

is a power of a prime.

Definition (

-group). A finite group

is a

-group if

|G|

for some prime

number p and n ≥ 1.

Theorem. If

is a finite

-group, then

(

) =

{x ∈ G

gx for all g ∈ G}

is non-trivial.

This immediately tells us that for n ≥ 2, a p group is never simple.

Proof.

Let

act on itself by conjugation. The orbits of this action (i.e. the

conjugacy classes) have order dividing

|G|

. So it is either a singleton, or

its size is divisible by p.

Since the conjugacy classes partition

, we know the total size of the conjugacy

classes is |G|. In particular,

|G| = number of conjugacy class of size 1

order of all other conjugacy classes.

We know the second term is divisible by

. Also

|G|

is divisible by

. Hence

the number of conjugacy classes of size 1 is divisible by

. We know

{e}

is a

conjugacy class of size 1. So there must be at least

conjugacy classes of size 1.

Since the smallest prime number is 2, there is a conjugacy class {x} 6= {e}.

But if

{x}

is a conjugacy class on its own, then by definition

−1

for

all g ∈ G, i.e. xg = gx for all g ∈ G. So x ∈ Z(G). So Z(G) is non-trivial.

The theorem allows us to prove interesting things about

-groups by induction

— we can quotient

(

), and get a smaller

-group. One way to do this is

via the below lemma.

Lemma. For any group G, if G/Z(G) is cyclic, then G is abelian.

In other words, if

G/Z

(

) is cyclic, then it is in fact trivial, since the center

of an abelian group is the abelian group itself.

Proof.

Let

g ∈ Z

(

) be a generator of the cyclic group

G/Z

(

). Hence every

coset of

(

) is of the form

(

). So every element

x ∈ G

must be of the

form

for

z ∈ Z

(

) and

r ∈ Z

. To show

is abelian, let

¯x

¯r

¯z

be another

element, with

¯z ∈ Z

(

)

, ¯r ∈ Z

. Note that

and

¯z

are in the center, and hence

commute with every element. So we have

x¯x = g

¯r

¯z = g

¯r

z¯z = g

¯r

¯zz = g

¯r

¯zg

z = ¯xx.

So they commute. So G is abelian.

This is a general lemma for groups, but is particularly useful when applied

to p groups.

Corollary. If p is prime and |G| = p

, then G is abelian.

Proof.

Since

(

)

≤ G

, its order must be 1,

. Since it is not trivial, it

can only be

. If it has order

, then it is the whole group and the group

is abelian. Otherwise,

G/Z

(

) has order

. But then it must be cyclic,

and thus G must be abelian. This is a contradiction. So G is abelian.

Theorem. Let

be a group of order

, where

is a prime number. Then it

has a subgroup of order p

for any 0 ≤ b ≤ a.

This means there is a subgroup of every conceivable order. This is not true

for general groups. For example,

has no subgroup of order 30 or else that

would be a normal subgroup.

Proof.

We induct on

. If

= 1, then

{e}, G

give subgroups of order

and

So done.

Now suppose

a >

1, and we want to construct a subgroup of order

. If

b = 0, then this is trivial, namely {e} ≤ G has order 1.

Otherwise, we know

(

) is non-trivial. So let

x 6

e ∈ Z

(

). Since

ord

(

)

| |G|

, its order is a power of

. If it in fact has order

, then

c−1

has

order

. So we can suppose, by renaming, that

has order

. We have thus

generated a subgroup

hxi

of order exactly

. Moreover, since

is in the center,

hxi

commutes with everything in

. So

hxi

is in fact a normal subgroup of

This is the point of choosing it in the center. Therefore G/hxi has order p

a−1

Since this is a strictly smaller group, we can by induction suppose

G/hxi

has

a subgroup of any order. In particular, it has a subgroup

of order

b−1

. By

the subgroup correspondence, there is some

K ≤ G

such that

K/hxi

and

H C K. But then K has order p

. So done.

1.6 Finite abelian groups

We now move on to a small section, which is small because we will come back to

it later, and actually prove what we claim.

It turns out finite abelian groups are very easy to classify. We can just write

down a list of all finite abelian groups. We write down the classification theorem,

and then prove it in the last part of the course, where we hit this with a huge

sledgehammer.

Theorem (Classification of finite abelian groups). Let

be a finite abelian

group. Then there exist some d

, ··· , d

such that

∼

× C

× ··· × C

Moreover, we can pick

such that

i+1

| d

for each

, and this expression is

unique.

It turns out the best way to prove this is not to think of it as a group, but

as a Z-module, which is something we will come to later.

Example. The abelian groups of order 8 are C

, C

× C

, C

× C

Sometimes this is not the most useful form of decomposition. To get a nicer

decomposition, we use the following lemma:

Lemma. If n and m are coprime, then C

∼

× C

This is a grown-up version of the Chinese remainder theorem. This is what

the Chinese remainder theorem really says.

Proof.

It suffices to find an element of order

×C

. Then since

×C

has order nm, it must be cyclic, and hence isomorphic to C

Let

g ∈ C

have order

;

h ∈ C

have order

, and consider (

g, h

)

∈ C

×C

Suppose the order of (

g, h

) is

. Then (

g, h

)

= (

e, e

). Hence (

, h

) = (

e, e

So the order of

and

divide

, i.e.

m | k

and

n | k

. As

and

are coprime,

this means that mn | k.

ord

((

g, h

)) and (

g, h

)

∈ C

× C

is a group of order

, we must

have k | nm. So k = nm.

Corollary. For any finite abelian group G, we have

∼

× C

× ··· × C

where each d

is some prime power.

Proof.

From the classification theorem, iteratively apply the previous lemma to

break each component up into products of prime powers.

As promised, this is short.

1.7 Sylow theorems

We finally get to the big theorem of this part of the course.

Theorem (Sylow theorems). Let

be a finite group of order

· m

, with

prime and p - m. Then

(i) The set of Sylow p-subgroups of G, given by

Syl

(G) = {P ≤ G : |P | = p

is non-empty. In other words, G has a subgroup of order p

(ii) All elements of Syl

(G) are conjugate in G.

(iii)

The number of Sylow

-subgroups

|Syl

(

)

satisfies

≡

1 (

mod p

)

and n

| |G| (in fact n

| m, since p is not a factor of n

These are sometimes known as Sylow’s first/second/third theorem respec-

tively.

We will not prove this just yet. We first look at how we can apply this

theorem. We can use it without knowing how to prove it.

Lemma. If n

= 1, then the Sylow p-subgroup is normal in G.

Proof.

Let

be the unique Sylow

-subgroup, and let

g ∈ G

, and consider

−1

P g

. Since this is isomorphic to

, we must have

−1

P g|

, i.e. it is also

a Sylow

-subgroup. Since there is only one, we must have

−1

P g

. So

normal.

Corollary. Let

be a non-abelian simple group. Then

|G| |

for every prime

p such that p | |G|.

Proof.

The group

acts on

Syl

(

) by conjugation. So it gives a permutation

representation

G → Sym

(

Syl

(

))

∼

. We know

ker φ C G

. But

simple. So ker(φ) = {e} or G. We want to show it is not the whole of G.

If we had

ker

(

), then

−1

P g

for all

g ∈ G

. Hence

is a normal

subgroup. As

is simple, either

{e}

, or

. We know

cannot be

trivial since

p | |G|

. But if

, then

is a

-group, has a non-trivial center,

and hence G is not non-abelian simple. So we must have ker(φ) = {e}.

Then by the first isomorphism theorem, we know

∼

im φ ≤ S

. We have

proved the theorem without the divide-by-two part. To prove the whole result,

we need to show that in fact

(

)

≤ A

. Consider the following composition

of homomorphisms:

G S

{±1}.

φ sgn

If this is surjective, then

ker

(

sgn ◦φ

)

C G

has index 2 (since the index is the size

of the image), and is not the whole of

. This means

is not simple (the case

where |G| = C

is ruled out since it is abelian).

So the kernel must be the whole

, and

sgn ◦φ

is the trivial map. In other

words, sgn(φ(g)) = +1. So φ(g) ∈ A

. So in fact we have

∼

im(φ) ≤ A

So we get |G| |

Example. Suppose

|G|

= 1000. Then

|G|

is not simple. To show this, we need

to factorize 1000. We have

|G|

= 2

. We pick our favorite prime to be

= 5.

We know

∼

1 (

mod

5), and

= 8. The only number that satisfies this

is n

= 1. So the Sylow 5-subgroup is normal, and hence G is not normal.

Example. Let

|G|

= 132 = 2

11. We want to show this is not simple. So

for a contradiction suppose it is.

We start by looking at

= 11. We know

≡

1 (

mod

11). Also

12.

As G is simple, we must have n

= 12.

Now look at

= 3. We have

= 1 (

mod

3) and

44. The possible

values of n

are 4 and 22.

= 4, then the corollary says

|G| |

= 12, which is of course nonsense.

So n

= 22.

At this point, we count how many elements of each order there are. This is

particularly useful if

p | |G|

but

- |G|

, i.e. the Sylow

-subgroups have order

and hence are cyclic.

As all Sylow 11-subgroups are disjoint, apart from

{e}

, we know there are

(11

−

1) = 120 elements of order 11. We do the same thing with the Sylow

3-subgroups. We need 22

−

1) = 44 elements of order 3. But this is more

elements than the group has. This can’t happen. So G must be simple.

We now get to prove our big theorem. This involves some non-trivial amount

of trickery.

Proof of Sylow’s theorem. Let G be a finite group with |G| = p

m, and p - m.

(i)

We need to show that

Syl

(

)

∅

, i.e. we need to find some subgroup of

order p

. As always, we find something clever for G to act on. We let

Ω = {X subset of G : |X| = p

We let G act on Ω by

g ∗ {g

, g

, ··· , g

} = {gg

, gg

, ··· , gg

Let Σ ≤ Ω be an orbit.

We first note that if

, ··· , g

} ∈

Σ, then by the definition of an orbit,

for every g ∈ G,

−1

∗ {g

, ··· , g

} = {g, gg

−1

, ··· , gg

−1

} ∈ Σ.

The important thing is that this set contains

. So for each

, Σ contains

a set X which contains g. Since each set X has size p

, we must have

|Σ| ≥

|G|

= m.

Suppose

. Then the orbit-stabilizer theorem says the stabilizer

any

, ··· , g

} ∈

Σ has index

, hence

|H|

, and thus

H ∈ Syl

(

So we need to show that not every orbit Σ can have size

> m

. Again, by

the orbit-stabilizer, the size of any orbit divides the order of the group,

|G|

. So if

| > m

, then

p | |

. Suppose we can show that

p - |

Ω

Then not every orbit Σ can have size

> m

, since Ω is the disjoint union of

all the orbits, and thus we are done.

So we have to show p - |Ω|. This is just some basic counting. We have

|Ω| =



|G|







−1

j=0

m − j

− j

Now note that the largest power of

dividing

m −j

is the largest power

dividing

. Similarly, the largest power of

dividing

−j

is also the

largest power of

dividing

. So we have the same power of

on top and

bottom for each item in the product, and they cancel. So the result is not

divisible by p.

This proof is not straightforward. We first needed the clever idea of letting

act on Ω. But then if we are given this set, the obvious thing to do

would be to find something in Ω that is also a group. This is not what we

do. Instead, we find an orbit whose stabilizer is a Sylow p-subgroup.

(ii)

We instead prove something stronger: if

Q ≤ G

is a

-subgroup (i.e.

|Q|

, for

not necessarily

), and

P ≤ G

is a Sylow

-subgroup, then

there is a

g ∈ G

such that

−1

Qg ≤ P

. Applying this to the case where

is another Sylow

-subgroup says there is a

such that

−1

Qg ≤ P

, but

since g

−1

Qg has the same size as P , they must be equal.

We let Q act on the set of cosets of G/P via

q ∗ gP = qgP.

We know the orbits of this action have size dividing

|Q|

, so is either 1 or

divisible by

. But they can’t all be divisible by

, since

|G/P |

is coprime

. So at least one of them have size 1, say

{gP }

. In other words, for

every

q ∈ Q

, we have

qgP

. This means

−1

qg ∈ P

. This holds for

every element q ∈ Q. So we have found a g such that g

−1

Qg ≤ P .

(iii)

Finally, we need to show that

∼

1 (

mod p

) and

| |G|

, where

|Syl

(G)|.

The second part is easier — by Sylow’s second theorem, the action of

Syl

(

) by conjugation has one orbit. By the orbit-stabilizer theorem,

the size of the orbit, which is

|Syl

(

)

, divides

|G|

. This proves the

second part.

For the first part, let

P ∈ Syl

(

). Consider the action by conjugation

Syl

(

). Again by the orbit-stabilizer theorem, the orbits each

have size 1 or size divisible by

. But we know there is one orbit of size 1,

namely

{P }

itself. To show

|Syl

(

)

∼

1 (

mod p

), it is enough to

show there are no other orbits of size 1.

Suppose {Q} is an orbit of size 1. This means for every p ∈ P , we get

−1

Qp = Q.

In other words,

P ≤ N

(

). Now

(

) is itself a group, and we

can look at its Sylow

-subgroups. We know

Q ≤ N

(

)

≤ G

. So

| |N

(

)

| | p

. So

is the biggest power of

that divides

(

)

So Q is a Sylow p-subgroup of N

(Q).

Now we know

P ≤ N

(

) is also a Sylow

-subgroup of

(

). By Sylow’s

second theorem, they must be conjugate in

(

). But conjugating

anything in

by something in

(

) does nothing, by definition of

(

). So we must have

. So the only orbit of size 1 is

{P }

itself.

So done.

This is all the theories of groups we’ve got. In the remaining time, we will

look at some interesting examples of groups.

Example. Let

(

Z/p

), i.e. the set of invertible

n × n

matrices with

entries in

Z/p

, the integers modulo

. Here

is obviously a prime. When we do

rings later, we will study this properly.

First of all, we would like to know the size of this group. A matrix

A ∈

(

Z/p

) is the same as

linearly independent vectors in the vector space

(

Z/p

)

. We can just work out how many there are. This is not too difficult,

when you know how.

We can pick the first vector, which can be anything except zero. So there

are

−

1 ways of choosing the first vector. Next, we need to pick the second

vector. This can be anything that is not in the span of the first vector, and this

rules out

possibilities. So there are

− p

ways of choosing the second vector.

Continuing this chain of thought, we have

|GL

(Z/p)| = (p

− 1)(p

− p)(p

− p

) ···(p

− p

n−1

What is a Sylow

-subgroup of

(

Z/p

)? We first work out what the order of

this is. We can factorize that as

|GL

(Z/p)| = (1 · p · p

· ··· · p

n−1

)((p

− 1)(p

n−1

− 1) ···(p − 1)).

So the largest power of

that divides

|GL

(

Z/p

)

(

)

. Let’s find a subgroup

of size p

(

)

. We consider matrices of the form

U =

















1 ∗ ∗ ··· ∗

0 1 ∗ ··· ∗

0 0 1 ··· ∗

0 0 0 ··· 1







∈ GL

(Z/p)











Then we know

|U|

(

)

as each

∗

can be chosen to be anything in

Z/p

, and

there are





∗s.

Is the Sylow

-subgroup unique? No. We can take the lower triangular

matrices and get another Sylow p-subgroup.

Example. Let’s be less ambitious and consider GL

(Z/p). So

|G| = p(p

− 1)(p − 1) = p(p − 1)

(p + 1).

Let

be another prime number such that

` | p −

1. Suppose the largest power of

` that divides |G| is `

. Can we (explicitly) find a Sylow `-subgroup?

First, we want to find an element of order

. How is

p −

1 related to

(apart

from the obvious way)? We know that

(Z/p)

= {x ∈ Z/p : (∃y) xy ≡ 1 (mod p)}

∼

p−1

So as

` | p −

1, there is a subgroup

≤ C

p−1

∼

(

Z/p

)

. Then we immediately

know where to find a subgroup of order `

: we have

× C

≤ (Z/p)

∗

× (Z/p)

≤ GL

(Z/p),

where the final inclusion is the diagonal matrices, identifying

(a, b) ↔



a 0

0 b



So this is the Sylow `-subgroup.

2 Rings

2.1 Definitions and examples

We now move on to something completely different — rings. In a ring, we are

allowed to add, subtract, multiply but not divide. Our canonical example of a

ring would be Z, the integers, as studied in IA Numbers and Sets.

In this course, we are only going to consider rings in which multiplication

is commutative, since these rings behave like “number systems”, where we can

study number theory. However, some of these rings do not behave like

. Thus

one major goal of this part is to understand the different properties of

, whether

they are present in arbitrary rings, and how different properties relate to one

another.

Definition (Ring). A ring is a quintuple (

, ·,

) where 0

∈ R

and +, · : R × R → R are binary operations such that

(i) (R, +, 0

) is an abelian group.

(ii) The operation · : R × R → R satisfies associativity, i.e.

a · (b · c) = (a · b) · c,

and identity:

· r = r · 1

= r.

(iii) Multiplication distributes over addition, i.e.

· (r

+ r

) = (r

· r

) + (r

· r

)

+ r

) · r

= (r

· r

) + (r

· r

Notation. If

is a ring and

r ∈ R

, we write

−r

for the inverse to

in (

This satisfies r + (−r) = 0

. We write r − s to mean r + (−s) etc.

Some people don’t insist on the existence of the multiplicative identity, but

we will for the purposes of this course.

Since we can add and multiply two elements, by induction, we can add and

multiply any finite number of elements. However, the notions of infinite sum and

product are undefined. It doesn’t make sense to ask if an infinite sum converges.

Definition (Commutative ring). We say a ring

is commutative if

a · b

b · a

for all a, b ∈ R.

From now onwards, all rings in this course are going to be commutative.

Just as we have groups and subgroups, we also have subrings.

Definition (Subring). Let (

, ·,

) be a ring, and

S ⊆ R

be a subset.

We say

is a subring of

if 0

∈ S

, and the operations +

, ·

make

into a

ring in its own right. In this case we write S ≤ R.

Example. The familiar number systems are all rings: we have

Z ≤ Q ≤ R ≤ C

under the usual 0, 1, +, ·.

Example. The set

[

] =

a, b ∈ Z} ≤ C

is the Gaussian integers, which

is a ring.

We also have the ring Q[

√

2] = {a + b

√

2 ∈ R : a, b ∈ Q} ≤ R.

We will use the square brackets notation quite frequently. It should be clear

what it should mean, and we will define it properly later.

In general, elements in a ring do not have inverses. This is not a bad thing.

This is what makes rings interesting. For example, the division algorithm would

be rather contentless if everything in

had an inverse. Fortunately,

only has

two invertible elements — 1 and −1. We call these units.

Definition (Unit). An element

u ∈ R

is a unit if there is another element

v ∈ R

such that u · v = 1

It is important that this depends on

, not just on

. For example, 2

∈ Z

not a unit, but 2 ∈ Q is a unit (since

is an inverse).

A special case is when (almost) everything is a unit.

Definition (Field). A field is a non-zero ring where every

u 6

= 0

∈ R

is a unit.

We will later show that 0

cannot be a unit except in a very degenerate case.

Example. Z is not a field, but Q, R, C are all fields.

Similarly, Z[i] is not a field, while Q[

√

2] is.

Example. Let

be a ring. Then 0

+ 0

= 0

, since this is true in the group

(R, +, 0

). Then for any r ∈ R, we get

r · (0

+ 0

) = r · 0

We now use the fact that multiplication distributes over addition. So

r · 0

+ r · 0

= r · 0

Adding (−r · 0

) to both sides give

r · 0

= 0

This is true for any element

r ∈ R

. From this, it follows that if

R 6

{

}

, then

6= 0

— if they were equal, then take r 6= 0

. So

r = r · 1

= r · 0

= 0

which is a contradiction.

Note, however, that

{

}

forms a ring (with the only possible operations

and identities), the zero ring, albeit a boring one. However, this is often a

counterexample to many things.

Definition (Product of rings). Let

R, S

be rings. Then the product

R × S

is a

ring via

(r, s) + (r

, s

) = (r + r

, s + s

), (r, s) · (r

, s

) = (r · r

, s · s

The zero is (0

, 0

) and the one is (1

, 1

We can (but won’t) check that these indeed are rings.

Definition (Polynomial). Let

be a ring. Then a polynomial with coefficients

in R is an expression

f = a

+ a

X + a

+ ··· + a

with a

∈ R. The symbols X

are formal symbols.

We identify f and f + 0

· X

n+1

as the same things.

Definition (Degree of polynomial). The degree of a polynomial

is the largest

m such that a

6= 0.

Definition (Monic polynomial). Let

have degree

. If

= 1, then

called monic.

Definition (Polynomial ring). We write

[

] for the set of all polynomials

with coefficients in

. The operations are performed in the obvious way, i.e. if

···

and

···

are polynomials,

then

f + g =

max{n,k}

r=0

+ b

and

f · g =

n+k

i=0





j=0

i−j





We identify

with the constant polynomials, i.e. polynomials

with

= 0 for

i >

0. In particular, 0

∈ R

and 1

∈ R

are the zero and one of

[

This is in fact a ring.

Note that a polynomial is just a sequence of numbers, interpreted as the

coefficients of some formal symbols. While it does indeed induce a function in

the obvious way, we shall not identify the polynomial with the function given by

it, since different polynomials can give rise to the same function.

For example, in

[

is not the zero polynomial, since its

coefficients are not zero. However,

(0) = 0 and

(1) = 0. As a function, this is

identically zero. So f 6= 0 as a polynomial but f = 0 as a function.

Definition (Power series). We write

[[

]] for the ring of power series on

i.e.

f = a

+ a

X + a

+ ··· ,

where each

∈ R

. This has addition and multiplication the same as for

polynomials, but without upper limits.

A power series is very much not a function. We don’t talk about whether

the sum converges or not, because it is not a sum.

Example. Is 1

−X ∈ R

[

] a unit? For every

···

(with

= 0),

we get

(1 − X)g = stuff + ··· − a

n+1

which is not 1. So

cannot be the inverse of (1

− X

). So (1

− X

) is not a unit.

However, 1 − X ∈ R[[X]] is a unit, since

(1 − X)(1 + X + X

+ X

+ ···) = 1.

Definition (Laurent polynomials). The Laurent polynomials on

is the set

R[X, X

−1

], i.e. each element is of the form

f =

i∈Z

where

∈ R

and only finitely many

are non-zero. The operations are the

obvious ones.

We can also think of Laurent series, but we have to be careful. We allow

infinitely many positive coefficients, but only finitely many negative ones. Or

else, in the formula for multiplication, we will have an infinite sum, which is

undefined.

Example. Let

be a set, and

be a ring. Then the set of all functions on

i.e. functions f : X → R, is a ring with ring operations given by

(f + g)(x) = f(x) + g(x), (f · g)(x) = f(x) · g(x).

Here zero is the constant function 0 and one is the constant function 1.

Usually, we don’t want to consider all functions

X → R

. Instead, we look at

some subrings of this. For example, we can consider the ring of all continuous

functions

R → R

. This contains, for example, the polynomial functions, which

is just R[X] (since in R, polynomials are functions).

2.2 Homomorphisms, ideals, quotients and isomorphisms

Just like groups, we will come up with analogues of homomorphisms, normal

subgroups (which are now known as ideals), and quotients.

Definition (Homomorphism of rings). Let

R, S

be rings. A function

R → S

is a ring homomorphism if it preserves everything we can think of, i.e.

(i) φ(r

+ r

) = φ(r

) + φ(r

(ii) φ(0

) = 0

(iii) φ(r

· r

) = φ(r

) · φ(r

(iv) φ(1

) = 1

Definition (Isomorphism of rings). If a homomorphism

R → S

is a bijection,

we call it an isomorphism.

Definition (Kernel). The kernel of a homomorphism φ : R → S is

ker(φ) = {r ∈ R : φ(r) = 0

Definition (Image). The image of φ : R → S is

im(φ) = {s ∈ S : s = φ(r) for some r ∈ R}.

Lemma. A homomorphism φ : R → S is injective if and only if ker φ = {0

Proof.

A ring homomorphism is in particular a group homomorphism

(

)

→

(

) of abelian groups. So this follows from the case of

groups.

In the group scenario, we had groups, subgroups and normal subgroups,

which are special subgroups. Here, we have a special kind of subsets of a ring

that act like normal subgroups, known as ideals.

Definition (Ideal). A subset I ⊆ R is an ideal, written I C R, if

(i)

It is an additive subgroup of (

), i.e. it is closed under addition and

additive inverses. (additive closure)

(ii) If a ∈ I and b ∈ R, then a · b ∈ I. (strong closure)

We say I is a proper ideal if I 6= R.

Note that the multiplicative closure is stronger than what we require for

subrings — for subrings, it has to be closed under multiplication by its own

elements; for ideals, it has to be closed under multiplication by everything in

the world. This is similar to how normal subgroups not only have to be closed

under internal multiplication, but also conjugation by external elements.

Lemma. If φ : R → S is a homomorphism, then ker(φ) C R.

Proof.

Since

: (

)

→

(

) is a group homomorphism, the kernel is

a subgroup of (R, +, 0

For the second part, let

a ∈ ker

(

b ∈ R

. We need to show that their

product is in the kernel. We have

φ(a · b) = φ(a) · φ(b) = 0 · φ(b) = 0.

So a · b ∈ ker(φ).

Example. Suppose

I C R

is an ideal, and 1

∈ I

. Then for any

r ∈ R

, the

axioms entail 1

· r ∈ I. But 1

· r = r. So if 1

∈ I, then I = R.

In other words, every proper ideal does not contain 1. In particular, every

proper ideal is not a subring, since a subring must contain 1.

We are starting to diverge from groups. In groups, a normal subgroup is a

subgroup, but here an ideal is not a subring.

Example. We can generalize the above a bit. Suppose

I C R

and

u ∈ I

is a

unit, i.e. there is some

v ∈ R

such that

u · v

= 1

. Then by strong closure,

= u · v ∈ I. So I = R.

Hence proper ideals are not allowed to contain any unit at all, not just 1

Example. Consider the ring

of integers. Then every ideal of

is of the form

nZ = {··· , −2n, −n, 0, n, 2n, ···} ⊆ Z.

It is easy to see this is indeed an ideal.

To show these are all the ideals, let

I C Z

. If

{

}

, then

= 0

. Otherwise,

let

n ∈ N

be the smallest positive element of

. We want to show in fact

Certainly nZ ⊆ I by strong closure.

Now let m ∈ I. By the Euclidean algorithm, we can write

m = q · n + r

with 0

≤ r < n

. Now

n, m ∈ I

. So by strong closure,

m, q · n ∈ I

. So

m −q ·n ∈ I

. As

is the smallest positive element of

, and

r < n

, we must

have r = 0. So m = q · n ∈ nZ. So I ⊆ nZ. So I = nZ.

The key to proving this was that we can perform the Euclidean algorithm on

. Thus, for any ring

in which we can “do Euclidean algorithm”, every ideal

is of the form

{a · r

r ∈ R}

for some

a ∈ R

. We will make this notion

precise later.

Definition (Generator of ideal). For an element a ∈ R, we write

(a) = aR = {a · r : r ∈ R}C R.

This is the ideal generated by a.

In general, let a

, a

, ··· , a

∈ R, we write

, a

, ··· , a

) = {a

+ ··· + a

: r

, ··· , r

∈ R}.

This is the ideal generated by a

, ··· , a

We can also have ideals generated by infinitely many objects, but we have to

be careful, since we cannot have infinite sums.

Definition (Generator of ideal). For

A ⊆ R

a subset, the ideal generated by

(A) =

(

a∈A

· a : r

∈ R, only finitely-many non-zero

)

These ideals are rather nice ideals, since they are easy to describe, and often

have some nice properties.

Definition (Principal ideal). An ideal

is a principal ideal if

= (

) for some

a ∈ R.

So what we have just shown for

is that all ideals are principal. Not all

rings are like this. These are special types of rings, which we will study more in

depth later.

Example. Consider the following subset:

{f ∈ R[X] : the constant coefficient of f is 0}.

This is an ideal, as we can check manually (alternatively, it is the kernel of the

“evaluate at 0” homomorphism). It turns out this is a principal ideal. In fact, it

is (X).

We have said ideals are like normal subgroups. The key idea is that we can

divide by ideals.

Definition (Quotient ring). Let

I C R

. The quotient ring

R/I

consists of the

(additive) cosets

with the zero and one as 0

and 1

, and operations

+ I) + (r

+ I) = (r

+ r

) + I

+ I) · (r

+ I) = r

+ I.

Proposition. The quotient ring is a ring, and the function

R → R/I

r 7→ r + I

is a ring homomorphism.

This is true, because we defined ideals to be those things that can be

quotiented by. So we just have to check we made the right definition.

Just as we could have come up with the definition of a normal subgroup by

requiring operations on the cosets to be well-defined, we could have come up

with the definition of an ideal by requiring the multiplication of cosets to be

well-defined, and we would end up with the strong closure property.

Proof.

We know the group (

R/I,

R/I

) is well-defined, since

is a (normal)

subgroup of R. So we only have to check multiplication is well-defined.

Suppose

and

. Then

− r

∈ I

and

− r

= a

∈ I. So

= (r

+ a

)(r

+ a

) = r

+ r

+ a

By the strong closure property, the last three objects are in

. So

+ I.

It is easy to check that 0

and 1

are indeed the zero and one, and

the function given is clearly a homomorphism.

Example. We have the ideals

nZ C Z

. So we have the quotient rings

Z/nZ

The elements are of the form m + nZ, so they are just

0 + nZ, 1 + nZ, 2 + nZ, ··· , (n − 1) + nZ.

Addition and multiplication are just what we are used to — addition and

multiplication modulo n.

Note that it is easier to come up with ideals than normal subgroups — we

can just pick up random elements, and then take the ideal generated by them.

Example. Consider (

)

C C

[

]. What is

[

]

(

)? Elements are represented

+ a

X + a

+ ··· + a

+ (X).

But everything but the first term is in (

). So every such thing is equivalent to

+ (

). It is not hard to convince yourself that this representation is unique.

So in fact C[X]/(X)

∼

C, with the bijection a

+ (X) ↔ a

If we want to prove things like this, we have to convince ourselves this

representation is unique. We can do that by hand here, but in general, we want

to be able to do this properly.

Proposition (Euclidean algorithm for polynomials). Let

be a field and

f, g ∈ F[X]. Then there is some r, q ∈ F[X] such that

f = gq + r,

with deg r < deg g.

This is like the usual Euclidean algorithm, except that instead of the absolute

value, we use the degree to measure how “big” the polynomial is.

Proof. Let deg(f) = n. So

f =

i=0

and a

6= 0. Similarly, if deg g = m, then

g =

i=0

with b

6= 0. If n < m, we let q = 0 and r = f, and done.

Otherwise, suppose n ≥ m, and proceed by induction on n.

We let

= f − a

−1

n−m

This is possible since

= 0, and

is a field. Then by construction, the

coefficients of X

cancel out. So deg(f

) < n.

If n = m, then deg(f

) < n = m. So we can write

f = (a

−1

n−m

)g + f

and

deg

(

)

< deg

(

). So done. Otherwise, if

n > m

, then as

deg

(

)

< n

, by

induction, we can find r

, q

such that

= gq

+ r

and deg(r

) < deg g = m. Then

f = a

−1

n−m

g + q

g + r

= (a

−1

n−m

+ q

)g + r

So done.

Now that we have a Euclidean algorithm for polynomials, we should be able

to show that every ideal of

[

] is generated by one polynomial. We will not

prove it specifically here, but later show that in general, in every ring where the

Euclidean algorithm is possible, all ideals are principal.

We now look at some applications of the Euclidean algorithm.

Example. Consider

[

], and consider the principal ideal (

+ 1)

C R

[

We let R = R[X]/(X

+ 1).

Elements of R are polynomials

+ a

X + a

+ ··· + a

| {z }

+(X

+ 1).

By the Euclidean algorithm, we have

f = q(X

+ 1) + r,

with

deg

(

)

2, i.e.

. Thus

+ (

+ 1) =

+ (

+ 1). So every

element of R[X]/(X

+ 1) is representable as a + bX for some a, b ∈ R.

Is this representation unique? If

+ (

+ 1) =

+ (

+ 1),

then the difference (

a −a

) + (

b −b

)

X ∈

(

+ 1). So it is (

+ 1)

for some

This is possible only if

= 0, since for non-zero

, we know (

+ 1)

has degree

at least 2. So we must have (

a − a

) + (

b − b

)

= 0. So

. So

the representation is unique.

What we’ve got is that every element in

is of the form

, and

+ 1 = 0, i.e.

−

1. This sounds like the complex numbers, just that we

are calling it X instead of i.

To show this formally, we define the function

φ : R[X]/(X

+ 1) → C

a + bX + (X

+ 1) 7→ a + bi.

This is well-defined and a bijection. It is also clearly additive. So to prove this

is an isomorphism, we have to show it is multiplicative. We check this manually.

We have

φ((a + bX + (X

+ 1))(c + dX + (X

+ 1)))

= φ(ac + (ad + bc)X + bdX

+ (X

+ 1))

= φ((ac − bd) + (ad + bc)X + (X

+ 1))

= (ac − bd) + (ad + bc)i

= (a + bi)(c + di)

= φ(a + bX + (X

+ 1))φ(c + dX + (X

+ 1)).

So this is indeed an isomorphism.

This is pretty tedious. Fortunately, we have some helpful results we can use,

namely the isomorphism theorems. These are exactly analogous to those for

groups.

Theorem (First isomorphism theorem). Let

R → S

be a ring homomorphism.

Then ker(φ) C R, and

ker(φ)

∼

im(φ) ≤ S.

Proof. We have already seen ker(φ) C R. Now define

Φ : R/ ker(φ) → im(φ)

r + ker(φ) 7→ φ(r).

This well-defined, since if

ker

(

) =

ker

(

), then

r − r

∈ ker

(

). So

φ(r − r

) = 0. So φ(r) = φ(r

We don’t have to check this is bijective and additive, since that comes for

free from the (proof of the) isomorphism theorem of groups. So we just have to

check it is multiplicative. To show Φ is multiplicative, we have

Φ((r + ker(φ))(t + ker(φ))) = Φ(rt + ker(φ))

= φ(rt)

= φ(r)φ(t)

= Φ(r + ker(φ))Φ(t + ker(φ)).

This is more-or-less the same proof as the one for groups, just that we had a

few more things to check.

Since there is the first isomorphism theorem, we, obviously, have more

coming.

Theorem (Second isomorphism theorem). Let

R ≤ S

and

J CS

. Then

J ∩RCR

and

R + J

= {r + J : r ∈ R} ≤

is a subring, and

R ∩ J

∼

R + J

Proof. Define the function

φ : R → S/J

r 7→ r + J.

Since this is the quotient map, it is a ring homomorphism. The kernel is

ker(φ) = {r ∈ R : r + J = 0, i.e. r ∈ J} = R ∩ J.

Then the image is

im(φ) = {r + J : r ∈ R} =

R + J

Then by the first isomorphism theorem, we know

R ∩ J C R

, and

R+J

≤ S

, and

R ∩ J

∼

R + J

Before we get to the third isomorphism theorem, recall we had the subgroup

correspondence for groups. Analogously, for I C R,

{subrings of R/I} ←→ {subrings of R which contain I}

L ≤

−→ {x ∈ R : x + I ∈ L}

≤

←− I C S ≤ R.

This is exactly the same formula as for groups.

For groups, we had a correspondence for normal subgroups. Here, we have a

correspondence between ideals

{ideals of R/I} ←→ {ideals of R which contain I}

It is important to note here that quotienting in groups and rings have different

purposes. In groups, we take quotients so that we have simpler groups to work

with. In rings, we often take quotients to get more interesting rings. For example,

[

] is quite boring, but

[

]

(

+ 1)

∼

is more interesting. Thus this ideal

correspondence allows us to occasionally get interesting ideals from boring ones.

Theorem (Third isomorphism theorem). Let

I C R

and

J C R

, and

I ⊆ J

Then J/I C R/I and











∼

Proof. We define the map

φ : R/I → R/J

r + I 7→ r + J.

This is well-defined and surjective by the groups case. Also it is a ring homo-

morphism since multiplication in

R/I

and

R/J

are “the same”. The kernel

ker(φ) = {r + I : r + J = 0, i.e. r ∈ J} =

So the result follows from the first isomorphism theorem.

Note that for any ring

, there is a unique ring homomorphism

Z → R

, given

ι : Z → R

n ≥ 0 7→ 1

+ 1

+ ··· + 1

| {z }

n times

n ≤ 0 7→ −(1

+ 1

+ ··· + 1

| {z }

−n times

)

Any homomorphism

Z → R

must be given by this formula, since it must send the

unit to the unit, and we can show this is indeed a homomorphism by distributivity.

So the ring homomorphism is unique. In fancy language, we say

is the initial

object in (the category of) rings.

We then know ker(ι) C Z. Thus ker(ι) = nZ for some n.

Definition (Characteristic of ring). Let

be a ring, and

Z → R

be the

unique such map. The characteristic of

is the unique non-negative

such

that ker(ι) = nZ.

Example. The rings

Z, Q, R, C

all have characteristic 0. The ring

Z/nZ

has

characteristic n. In particular, all natural numbers can be characteristics.

The notion of the characteristic will not be too useful in this course. How-

ever, fields of non-zero characteristic often provide interesting examples and

counterexamples to some later theory.

2.3

Integral domains, field of factions, maximal and prime

ideals

Many rings can be completely nothing like

. For example, in

, we know that if

a, b 6

= 0, then

ab 6

= 0. However, in, say,

, we get 2

= 0, but 2

3 = 0. Also,

has some nice properties such as every ideal is principal, and every integer

has an (essentially) unique factorization. We will now classify rings according to

which properties they have.

We start with the most fundamental property that the product of two non-

zero elements are non-zero. We will almost exclusively work with rings that

satisfy this property.

Definition (Integral domain). A non-zero ring

is an integral domain if for all

a, b ∈ R, if a · b = 0

, then a = 0

or b = 0

An element that violates this property is known as a zero divisor.

Definition (Zero divisor). An element

x ∈ R

is a zero divisor if

x 6

= 0 and

there is a y 6= 0 such that x · y = 0 ∈ R.

In other words, a ring is an integral domain if it has no zero divisors.

Example. All fields are integral domains, since if

a · b

= 0, and

b 6

= 0, then

a = a · (b · b

−1

) = 0. Similarly, if a 6= 0, then b = 0.

Example. A subring of an integral domain is an integral domain, since a zero

divisor in the small ring would also be a zero divisor in the big ring.

Example. Immediately, we know

Z, Q, R, C

are integral domains, since

is a

field, and the others are subrings of it. Also,

[

]

≤ C

is also an integral domain.

These are the nice rings we like in number theory, since there we can sensibly

talk about things like factorization.

It turns out there are no interesting finite integral domains.

Lemma. Let

be a finite ring which is an integral domain. Then

is a field.

Proof. Let a ∈ R be non-zero, and consider the ring homomorphism

a · − : R → R

b 7→ a · b

We want to show this is injective. For this, it suffices to show the kernel is trivial.

r ∈ ker

(

a · −

), then

a · r

= 0. So

= 0 since

is an integral domain. So the

kernel is trivial.

Since

is finite,

a · −

must also be surjective. In particular, there is an

element

b ∈ R

such that

a · b

= 1

. So

has an inverse. Since

was arbitrary,

R is a field.

So far, we know fields are integral domains, and subrings of integral domains

are integral domains. We have another good source of integral domain as follows:

Lemma. Let R be an integral domain. Then R[X] is also an integral domain.

Proof.

We need to show that the product of two non-zero elements is non-zero.

Let f, g ∈ R[X] be non-zero, say

f = a

+ a

X + ··· + a

∈ R[X]

g = b

+ b

X + ··· + b

∈ R[X],

with

, b

= 0. Then the coefficient of

n+m

. This is non-

zero since

is an integral domain. So

is non-zero. So

[

] is an integral

domain.

So, for instance, Z[X] is an integral domain.

We can also iterate this.

Notation. Write

[

X, Y

] for (

[

])[

], the polynomial ring of

in two vari-

ables. In general, write R[X

, ··· , X

] = (···((R[X

])[X

]) ···)[X

Then if R is an integral domain, so is R[X

, ··· , X

We now mimic the familiar construction of

from

. For any integral

domain

, we want to construct a field

that consists of “fractions” of elements

. Recall that a subring of any field is an integral domain. This says the

converse — every integral domain is the subring of some field.

Definition (Field of fractions). Let

be an integral domain. A field of fractions

F of R is a field with the following properties

(i) R ≤ F

(ii)

Every element of

may be written as

a ·b

−1

for

a, b ∈ R

, where

−1

means

the multiplicative inverse to b 6= 0 in F .

For example, Q is the field of fractions of Z.

Theorem. Every integral domain has a field of fractions.

Proof.

The construction is exactly how we construct the rationals from the

integers — as equivalence classes of pairs of integers. We let

S = {(a, b) ∈ R × R : b 6= 0}.

We think of (a, b) ∈ S as

. We define the equivalence relation ∼ on S by

(a, b) ∼ (c, d) ⇔ ad = bc.

We need to show this is indeed a equivalence relation. Symmetry and reflexivity

are obvious. To show transitivity, suppose

(a, b) ∼ (c, d), (c, d) ∼ (e, f),

i.e.

ad = bc, cf = de.

We multiply the first equation by f and the second by b, to obtain

adf = bcf, bcf = bed.

Rearranging, we get

d(af − be) = 0.

Since

is in the denominator,

d 6

= 0. Since

is an integral domain, we must

have

af −be

= 0, i.e.

. So (

a, b

)

∼

(

e, f

). This is where being an integral

domain is important.

Now let

F = S/∼

be the set of equivalence classes. We now want to check this is indeed the field

of fractions. We first want to show it is a field. We write

= [(

a, b

)]

∈ F

, and

define the operations by

ad + bc

These are well-defined, and make (

, ·,

) into a ring. There are many

things to check, but those are straightforward, and we will not waste time doing

that here.

Finally, we need to show every non-zero element has an inverse. Let

= 0

i.e.

, or a ·1 6= b · 0 ∈ R, i.e. a 6= 0. Then

∈ F is defined, and

= 1

has a multiplicative inverse. So F is a field.

We now need to construct a subring of

that is isomorphic to

. To do so,

we need to define an injective isomorphism φ : R → F . This is given by

φ : R → F

r 7→

This is a ring homomorphism, as one can check easily. The kernel is the set of

all

r ∈ R

such that

= 0, i.e.

= 0. So the kernel is trivial, and

is injective.

Then by the first isomorphism theorem, R

∼

im(φ) ⊆ F .

Finally, we need to show everything is a quotient of two things in

. We

have





−1

as required.

This gives us a very useful tool. Since this gives us a field from an integral

domain, this allows us to use field techniques to study integral domains. Moreover,

we can use this to construct new interesting fields from integral domains.

Example. Consider the integral domain

[

]. Its field of fractions is the field

of all rational functions

p(X)

q(X)

, where p, q ∈ C[X].

To some people, it is a shame to think of rings as having elements. Instead,

we should think of a ring as a god-like object, and the only things we should

ever mention are its ideals. We should also not think of the ideals as containing

elements, but just some abstract objects, and all we know is how ideals relate to

one another, e.g. if one contains the other.

Under this philosophy, we can think of a field as follows:

Lemma. A (non-zero) ring

is a field if and only if its only ideals are

{

}

and

Note that we don’t need elements to define the ideals

{

}

and

{

}

can be

defined as the ideal that all other ideals contain, and

is the ideal that contains

all other ideals. Alternatively, we can reword this as “R is a field if and only if

it has only two ideals” to avoid mentioning explicit ideals.

Proof.

(

⇒

) Let

I C R

and

be a field. Suppose

x 6

= 0

∈ I

. Then as

is a unit,

I = R.

(

⇐

) Suppose

x 6

= 0

∈ R

. Then (

) is an ideal of

. It is not

{

}

since it

contains

. So (

) =

. In other words 1

∈

(

). But (

) is defined to be

{x · y

y ∈ R}

. So there is some

u ∈ R

such that

x · u

= 1

. So

is a unit.

Since x was arbitrary, R is a field.

This is another reason why fields are special. They have the simplest possible

ideal structure.

This motivates the following definition:

Definition (Maximal ideal). An ideal

of a ring

is maximal if

I 6

and

for any ideal J with I ≤ J ≤ R, either J = I or J = R.

The relation with what we’ve done above is quite simple. There is an easy

way to recognize if an ideal is maximal.

Lemma. An ideal I C R is maximal if and only if R/I is a field.

Proof. R/I

is a field if and only if

{

}

and

R/I

are the only ideals of

R/I

. By

the ideal correspondence, this is equivalent to saying

and

are the only ideals

of R which contains I, i.e. I is maximal. So done.

This is a nice result. This makes a correspondence between properties of

ideals I and properties of the quotient R/I. Here is another one:

Definition (Prime ideal). An ideal

of a ring

is prime if

I 6

and whenever

a, b ∈ R are such that a · b ∈ I, then a ∈ I or b ∈ I.

This is like the opposite of the property of being an ideal — being an ideal

means if we have something in the ideal and something outside, the product is

always in the ideal. This does the backwards. If the product of two random

things is in the ideal, then one of them must be from the ideal.

Example. A non-zero ideal nZ C Z is prime if and only if n is a prime.

To show this, first suppose

is a prime, and

a · b ∈ pZ

. So

p | a · b

. So

p | a or p | b, i.e. a ∈ pZ or b ∈ pZ.

For the other direction, suppose

is a composite number (

p, q 6

= 1).

Then n ∈ nZ but p 6∈ nZ and q 6∈ nZ, since 0 < p, q < n.

So instead of talking about prime numbers, we can talk about prime ideals

instead, because ideals are better than elements.

We prove a result similar to the above:

Lemma. An ideal I C R is prime if and only if R/I is an integral domain.

Proof.

Let

be prime. Let

I, b

I ∈ R/I

, and suppose (

)(

) = 0

R/I

By definition, (

)(

) =

. So we must have

ab ∈ I

. As

is prime,

either

a ∈ I

b ∈ I

. So

= 0

R/I

= 0

R/I

. So

R/I

is an integral

domain.

Conversely, suppose

R/I

is an integral domain. Let

a, b ∈ R

be such that

ab ∈ I

. Then (

)(

) =

= 0

R/I

∈ R/I

. Since

R/I

is an integral

domain, either

= 0

R/I

= 0

R/i

, i.e.

a ∈ I

b ∈ I

. So

is a prime

ideal.

Prime ideals and maximal ideals are the main types of ideals we care about.

Note that every field is an integral domain. So we immediately have the following

result:

Proposition. Every maximal ideal is a prime ideal.

Proof. I C R

is maximal implies

R/I

is a field implies

R/I

is an integral domain

implies I is prime.

The converse is not true. For example,

{

} ⊆ Z

is prime but not maximal.

Less stupidly, (

)

∈ Z

[

X, Y

] is prime but not maximal (since

[

X, Y

]

(

)

∼

[

]). We can provide a more explicit proof of this, which is essentially the

same.

Alternative proof.

Let

be a maximal ideal, and suppose

a, b 6∈ I

but

ab ∈ I

Then by maximality,

+ (

) =

+ (

) =

= (1). So we can find some

p, q ∈ R

and n, m ∈ I such that n + ap = m + bq = 1. Then

1 = (n + ap)(m + bq) = nm + apm + bqn + abpq ∈ I,

since n, m, ab ∈ I. This is a contradiction.

Lemma. Let

be an integral domain. Then its characteristic is either 0 or a

prime number.

Proof.

Consider the unique map

Z → R

, and

ker

(

) =

. Then

is the

characteristic of R by definition.

By the first isomorphism theorem,

Z/nZ

(

)

≤ R

. So

Z/nZ

is an

integral domain. So nZ C Z is a prime. So n = 0 or a prime number.

2.4 Factorization in integral domains

We now move on to tackle the problem of factorization in rings. For sanity,

we suppose throughout the section that

is an integral domain. We start by

making loads of definitions.

Definition (Unit). An element

a ∈ R

is a unit if there is a

b ∈ R

such that

ab = 1

. Equivalently, if the ideal (a) = R.

Definition (Division). For elements

a, b ∈ R

, we say

divides

, written

a | b

if there is a c ∈ R such that b = ac. Equivalently, if (b) ⊆ (a).

Definition (Associates). We say

a, b ∈ R

are associates if

for some unit

c. Equivalently, if (a) = (b). Equivalently, if a | b and b | a.

In the integers, this can only happen if

and

differ by a sign, but in more

interesting rings, more interesting things can happen.

When considering division in rings, we often consider two associates to be

“the same”. For example, in Z, we can factorize 6 as

6 = 2 · 3 = (−2) · (−3),

but this does not violate unique factorization, since 2 and

−

2 are associates (and

so are 3 and −3), and we consider these two factorizations to be “the same”.

Definition (Irreducible). We say

a ∈ R

is irreducible if

a 6

= 0,

is not a unit,

and if a = xy, then x or y is a unit.

For integers, being irreducible is the same as being a prime number. However,

“prime” means something different in general rings.

Definition (Prime). We say

a ∈ R

is prime if

is non-zero, not a unit, and

whenever a | xy, either a | x or a | y.

It is important to note all these properties depend on the ring, not just the

element itself.

Example. 2 ∈ Z is a prime, but 2 ∈ Q is not (since it is a unit).

Similarly, the polynomial 2

X ∈ Q

[

] is irreducible (since 2 is a unit), but

2X ∈ Z[X] not irreducible.

We have two things called prime, so they had better be related.

Lemma. A principal ideal (

) is a prime ideal in

if and only if

= 0 or

prime.

Proof.

(

⇒

) Let (

) be a prime ideal. If

= 0, then done. Otherwise, as prime

ideals are proper, i.e. not the whole ring,

is not a unit. Now suppose

r | a · b

Then

a · b ∈

(

). But (

) is prime. So

a ∈

(

) or

b ∈

(

). So

r | a

r | b

. So

prime.

(

⇐

) If

= 0, then (0) =

{

} C R

, which is prime since

is an integral

domain. Otherwise, let

r 6

= 0 be prime. Suppose

a · b ∈

(

). This means

r | a · b

So r | a or r | b. So a ∈ (r) and b ∈ (r). So (r) is prime.

Note that in

, prime numbers exactly match the irreducibles, but prime

numbers are also prime (surprise!). In general, it is not true that irreducibles

are the same as primes. However, one direction is always true.

Lemma. Let r ∈ R be prime. Then it is irreducible.

Proof.

Let

r ∈ R

be prime, and suppose

. Since

r | r

, and

prime, we must have

r | a

r | b

. wlog,

r | a

. So

for some

c ∈ R

. So

rcb

. Since we are in an integral domain, we must have 1 =

. So

a unit.

We now do a long interesting example.

Example. Let

R = Z[

√

−5] = {a + b

√

−5 : a, b ∈ Z} ≤ C.

By definition, it is a subring of a field. So it is an integral domain. What are

the units of the ring? There is a nice trick we can use, when things are lying

inside C. Consider the function

N : R → Z

≥0

given by

N(a + b

√

−5) 7→ a

+ 5b

It is convenient to think of this as

z 7→ z¯z

|z|

. This satisfies

(

z · w

) =

(

)

(

). This is a desirable thing to have for a ring, since it immediately

implies all units have norm 1 — if

r ·s

= 1, then 1 =

(1) =

(

) =

(

)

(

So N (r) = N(s) = 1.

So to find the units, we need to solve

+ 5

= 1, for

and

units. The

only solutions are

1. So only

∈ R

can be units, and these obviously are

units. So these are all the units.

Next, we claim 2

∈ R

is irreducible. We again use the norm. Suppose 2 =

Then 4 =

(2) =

(

)

(

). Now note that nothing has norm 2.

+ 5

can

never be 2 for integers

a, b ∈ Z

. So we must have, wlog,

(

) = 4

, N

(

) = 1.

must be a unit. Similarly, we see that 3

1 +

√

−5,

−

√

−5

are irreducible

(since there is also no element of norm 3).

We have four irreducible elements in this ring. Are they prime? No! Note

that

(1 +

√

−5)(1 −

√

−5) = 6 = 2 · 3.

We now claim 2 does not divide 1 +

√

−5 or 1 −

√

−5. So 2 is not prime.

To show this, suppose 2

1 +

√

−5

. Then

(2)

| N

(1 +

√

−5

). But

(2) = 4

and

(1 +

√

−5

) = 6, and 4

6. Similarly,

−

√

−5

) = 6 as well. So

2 - 1 ±

√

−5.

There are several life lessons here. First is that primes and irreducibles are

not the same thing in general. We’ve always thought they were the same because

we’ve been living in the fantasy land of the integers. But we need to grow up.

The second one is that factorization into irreducibles is not necessarily unique,

since 2 · 3 = (1 +

√

−5)(1 −

√

−5) are two factorizations into irreducibles.

However, there is one situation when unique factorizations holds. This is

when we have a Euclidean algorithm available.

Definition (Euclidean domain). An integral domain

is a Euclidean domain

(ED) if there is a Euclidean function φ : R \ {0} → Z

≥0

such that

(i) φ(a · b) ≥ φ(b) for all a, b 6= 0

(ii) If a, b ∈ R, with b 6= 0, then there are q, r ∈ R such that

a = b · q + r,

and either r = 0 or φ(r) < φ(b).

What are examples? Every time in this course where we said “Euclidean

algorithm”, we have an example.

Example. Z is a Euclidean domain with φ(n) = |n|.

Example. For any field F, F[X] is a Euclidean domain with

φ(f) = deg(f ).

Example. The Gaussian integers

[

]

≤ C

is a Euclidean domain with

φ(z) = N(z) = |z|

. We now check this:

(i) We have φ(zw) = φ(z)φ(w) ≥ φ(z), since φ(w) is a positive integer.

(ii) Given a, b ∈ Z[i], b 6= 0. We consider the complex number

∈ C.

Consider the following complex plane, where the red dots are points in

Z[i].

By looking at the picture, we know that there is some

q ∈ Z

[

] such that



− q



< 1. So we can write

= q + c

with |c| < 1. Then we have

a = b · q + b · c

|{z}

We know

a − bq ∈ Z

[

], and

(

) =

(

) =

(

)

(

)

< N

(

) =

(

So done.

This is not just true for the Gaussian integers. All we really needed was that

R ≤ C

, and for any

x ∈ C

, there is some point in

that is not more than 1 away

from x. If we draw some more pictures, we will see this is not true for Z[

√

−5].

Before we move on to prove unique factorization, we first derive something

we’ve previously mentioned. Recall we showed that every ideal in

is principal,

and we proved this by the Euclidean algorithm. So we might expect this to be

true in an arbitrary Euclidean domain.

Definition (Principal ideal domain). A ring

is a principal ideal domain (PID)

if it is an integral domain, and every ideal is a principal ideal, i.e. for all

I C R

there is some a such that I = (a).

Example. Z is a principal ideal domain.

Proposition. Let

be a Euclidean domain. Then

is a principal ideal domain.

We have already proved this, just that we did it for a particular Euclidean

domain Z. Nonetheless, we shall do it again.

Proof.

Let

have a Euclidean function

R \ {

} → Z

≥0

. We let

I C R

be a

non-zero ideal, and let

b ∈ I \ {

}

be an element with

(

) minimal. Then for

any a ∈ I, we write

a = bq + r,

with

= 0 or

(

)

< φ

(

). However, any such

must be in

since

a −bq ∈ I

So we cannot have

(

)

< φ

(

). So we must have

= 0. So

. So

a ∈

(

Since this is true for all

a ∈ I

, we must have

I ⊆

(

). On the other hand, since

b ∈ I, we must have (b) ⊆ I. So we must have I = (b).

This is exactly, word by word, the same proof as we gave for the integers,

except we replaced the absolute value with φ.

Example.

is a Euclidean domain, and hence a principal ideal domain. Also,

for any field F, F[X] is a Euclidean domain, hence a principal ideal domain.

Also, Z[i] is a Euclidean domain, and hence a principal ideal domain.

What is a non-example of principal ideal domains? In

[

], the ideal

, X

)

C Z

[

] is not a principal ideal. Suppose it were. Then (2

, X

) = (

). Since

∈

, X

) = (

), we know 2

∈

(

) , i.e. 2 =

f · g

for some

. So

has degree

zero, and hence constant. So f = ±1 or ±2.

1, since

1 are units, then (

) =

[

]. But (2

, X

)

[

], since,

say, 1

6∈

, X

). If

2, then since

X ∈

, X

) = (

), we must have

| X

but this is clearly false. So (2, X) cannot be a principal ideal.

Example. Let

A ∈ M

n×n

(

) be an

n × n

matrix over a field

. We consider

the following set

I = {f ∈ F[X] : f(A) = 0}.

This is an ideal — if

f, g ∈ I

, then (

)(

) =

(

) +

(

) = 0. Similarly, if

f ∈ I and h ∈ F[X], then (f g)(A) = f(A)g(A) = 0.

But we know

[

] is a principal ideal domain. So there must be some

m ∈ F[X] such that I = (m) for some m.

Suppose

f ∈ F

[

] such that

(

) = 0, i.e.

f ∈ I

. Then

m | f

. So

a polynomial that divides all polynomials that kill

, i.e.

is the minimal

polynomial of A.

We have just proved that all matrices have minimal polynomials, and that

the minimal polynomial divides all other polynomials that kill

. Also, the

minimal polynomial is unique up to multiplication of units.

Let’s get further into number theory-like things. For a general ring, we

cannot factorize things into irreducibles uniquely. However, in some rings, this

is possible.

Definition (Unique factorization domain). An integral domain

is a unique

factorization domain (UFD) if

(i) Every non-unit may be written as a product of irreducibles;

(ii)

···p

···q

with

, q

irreducibles, then

, and they can

be reordered such that p

is an associate of q

This is a really nice property, and here we can do things we are familiar with

in number theory. So how do we know if something is a unique factorization

domain?

Our goal is to show that all principal ideal domains are unique factorization

domains. To do so, we are going to prove several lemmas that give us some

really nice properties of principal ideal domains.

Recall we saw that every prime is an irreducible, but in

[

√

−5

], there are

some irreducibles that are not prime. However, this cannot happen in principal

ideal domains.

Lemma. Let

be a principal ideal domain. If

p ∈ R

is irreducible, then it is

prime.

Note that this is also true for general unique factorization domains, which

we can prove directly by unique factorization.

Proof.

Let

p ∈ R

be irreducible, and suppose

p | a · b

. Also, suppose

p - a

. We

need to show p | b.

Consider the ideal (

p, a

)

C R

. Since

is a principal ideal domain, there is

some d ∈ R such that (p, a) = (d). So d | p and d | a.

Since

d | p

, there is some

such that

. As

is irreducible, either

or d is a unit.

is a unit, then

−1

, and this divides

. So

−1

for some

This is a contradiction, since p - a.

Therefore

is a unit. So (

p, a

) = (

) =

. In particular, 1

∈

(

p, a

). So

suppose 1

, for some

r, s ∈ R

. We now take the whole thing and

multiply by b. Then we get

b = rpb + sab.

We observe that

is divisible by

, and so is

. So

is divisible by

. So

done.

This is similar to the argument for integers. For integers, we would say if

p - a

then

and

are coprime. Therefore there are some

r, s

such that 1 =

Then we continue the proof as above. Hence what we did in the middle is to do

something similar to showing p and a are “coprime”.

Another nice property of principal ideal domains is the following:

Lemma. Let

be a principal ideal domain. Let

⊆ I

⊆ ···

be a chain

of ideals. Then there is some N ∈ N such that I

= I

n+1

for some n ≥ N.

So in a principal ideal domain, we cannot have an infinite chain of bigger

and bigger ideals.

Definition (Ascending chain condition). A ring satisfies the ascending chain

condition (ACC) if there is no infinite strictly increasing chain of ideals.

Definition (Noetherian ring). A ring that satisfies the ascending chain condition

is known as a Noetherian ring.

So we are proving that every principal ideal domain is Noetherian.

Proof.

The obvious thing to do when we have an infinite chain of ideals is to

take the union of them. We let

I =

∞

[

n≥1

which is again an ideal. Since

is a principal ideal domain,

= (

) for some

a ∈ R. We know a ∈ I =

∞

n=0

. So a ∈ I

for some N . Then we have

(a) ⊆ I

⊆ I = (a)

So we must have I

= I. So I

= I

= I for all n ≥ N.

Notice it is not important that

is generated by one element. If, for some

reason, we know

is generated by finitely many elements, then the same argument

work. So if every ideal is finitely generated, then the ring must be Noetherian.

It turns out this is an if-and-only-if — if you are Noetherian, then every ideal is

finitely generated. We will prove this later on in the course.

Finally, we have done the setup, and we can prove the proposition promised.

Proposition. Let

be a principal ideal domain. Then

is a unique factoriza-

tion domain.

Proof. We first need to show any (non-unit) r ∈ R is a product of irreducibles.

Suppose

r ∈ R

cannot be factored as a product of irreducibles. Then it is

certainly not irreducible. So we can write

, with

, s

both non-units.

Since

cannot be factored as a product of irreducibles, wlog

cannot be

factored as a product of irreducibles (if both can, then

would be a product of

irreducibles). So we can write

, with

, s

not units. Again, wlog

cannot be factored as a product of irreducibles. We continue this way.

By assumption, the process does not end, and then we have the following

chain of ideals:

(r) ⊆ (r

) ⊆ (r

) ⊆ ··· ⊆ (r

) ⊆ ···

But then we have an ascending chain of ideals. By the ascending chain condition,

these are all eventually equal, i.e. there is some

such that (

) = (

n+1

) =

(

n+2

) =

···

. In particular, since (

) = (

n+1

), and

n+1

, then

n+1

is a unit. But this is a contradiction, since

n+1

is not a unit. So

must be a

product of irreducibles.

To show uniqueness, we let

···p

···q

, with

, q

irreducible.

So in particular

| q

···q

. Since

is irreducible, it is prime. So

divides

some

. We reorder and suppose

| q

. So

·a

for some

. But since

is irreducible,

must be a unit. So

, q

are associates. Since

is a principal

ideal domain, hence integral domain, we can cancel p

to obtain

···p

= (aq

···q

We now rename aq

as q

, so that we in fact have

···p

= q

···q

We can then continue to show that

and

are associates for all

. This also

shows that

, or else if

, saw, then

k+1

···p

= 1, which is a

contradiction.

We can now use this to define other familiar notions from number theory.

Definition (Greatest common divisor).

is a greatest common divisor (gcd) of

, a

, ··· , a

d | a

for all

, and if any other

satisfies

| a

for all

, then

| d.

Note that the gcd of a set of numbers, if exists, is not unique. It is only

well-defined up to a unit.

This is a definition that says what it means to be a greatest common divisor.

However, it does not always have to exist.

Lemma. Let

be a unique factorization domain. Then greatest common

divisors exists, and is unique up to associates.

Proof.

We construct the greatest common divisor using the good-old way of

prime factorization.

We let

, p

, ··· , p

be a list of all irreducible factors of

, such that no

two of these are associates of each other. We now write

= u

j=1

where n

∈ N and u

are units. We let

= min

and choose

d =

j=1

As, by definition, m

≤ n

for all i, we know d | a

for all i.

Finally, if d

| a

for all i, then we let

= v

j=1

Then we must have

≤ n

for all

i, j

. So we must have

≤ m

for all

. So

| d.

Uniqueness is immediate since any two greatest common divisors have to

divide each other.

2.5 Factorization in polynomial rings

Since polynomial rings are a bit more special than general integral domains, we

can say a bit more about them.

Recall that for

a field, we know

[

] is a Euclidean domain, hence a

principal ideal domain, hence a unique factorization domain. Therefore we know

(i) If I C F [X], then I = (f ) for some f ∈ F [X].

(ii) If f ∈ F [X], then f is irreducible if and only if f is prime.

(iii)

Let

be irreducible, and suppose (

)

⊆ J ⊆ F

[

]. Then

= (

) for some

. Since (

)

⊆

(

), we must have

for some

. But

is irreducible.

So either

is a unit. If

is a unit, then (

) =

[

]. If

is a unit,

then (

) = (

). So (

) is a maximal ideal. Note that this argument is valid

for any PID, not just polynomial rings.

(iv)

Let (

) be a prime ideal. Then

is prime. So

is irreducible. So (

) is

maximal. But we also know in complete generality that maximal ideals are

prime. So in

[

], prime ideals are the same as maximal ideals. Again,

this is true for all PIDs in general.

(v) Thus f is irreducible if and only if F [X]/(f ) is a field.

To use the last item, we can first show that

[

]

(

) is a field, and then use this

to deduce that

is irreducible. But we can also do something more interesting

— find an irreducible f , and then generate an interesting field F [X]/(f).

So we want to understand reducibility, i.e. we want to know whether we can

factorize a polynomial

. Firstly, we want to get rid of the trivial case where we

just factor out a scalar, e.g. 2

+ 2 = 2(

+ 1)

∈ Z

[

] is a boring factorization.

Definition (Content). Let

be a UFD and

···

∈ R

[

The content c(f ) of f is

c(f) = gcd(a

, a

, ··· , a

) ∈ R.

Again, since the gcd is only defined up to a unit, so is the content.

Definition (Primitive polynomial). A polynomial is primitive if

(

) is a unit,

i.e. the a

are coprime.

Note that this is the best we can do. We cannot ask for

(

) to be exactly 1,

since the gcd is only well-defined up to a unit.

We now want to prove the following important lemma:

Lemma (Gauss’ lemma). Let

be a UFD, and

f ∈ R

[

] be a primitive

polynomial. Then

is reducible in

[

] if and only if

is reducible

[

], where

F is the field of fractions of R.

We can’t do this right away. We first need some preparation. Before that,

we do some examples.

Example. Consider

+ 1

∈ Z

[

]. This has content 1 so is primitive. We

show it is not reducible in Z[X], and hence not reducible in Q[X].

Suppose

is reducible in

[

]. Then by Gauss’ lemma, this is reducible in

Z[X]. So we can write

+ X + 1 = gh,

for some polynomials

g, h ∈ Z

[

], with

g, h

not units. But if

and

are not

units, then they cannot be constant, since the coefficients of

+ 1 are all

1 or 0. So they have degree at least 1. Since the degrees add up to 3, we wlog

suppose g has degree 1 and h has degree 2. So suppose

g = b

+ b

X, h = c

+ c

X + c

Multiplying out and equating coefficients, we get

= 1

and

must be

1. So

is either 1 +

− X, −

1 +

−

− X

, and

hence has

1 as a root. But this is a contradiction, since

1 is not a root of

+ X + 1. So f is not reducible in Q. In particular, f has no root in Q.

We see the advantage of using Gauss’ lemma — if we worked in

instead,

we could have gotten to the step

= 1, and then we can do nothing, since

and c

can be many things if we live in Q.

Now we start working towards proving this.

Lemma. Let R be a UFD. If f, g ∈ R[X] are primitive, then so is f g.

Proof. We let

f = a

+ a

X + ··· + a

g = b

+ b

X + ··· + b

where

, b

= 0, and

f, g

are primitive. We want to show that the content of

fg is a unit.

Now suppose

is not primitive. Then

(

) is not a unit. Since

is a

UFD, we can find an irreducible p which divides c(fg).

By assumption,

(

) and

(

) are units. So

p - c

(

) and

p - c

(

). So suppose

p | a

, . . . ,

p | a

k−1

but

p - a

. Note it is possible that

= 0. Similarly,

suppose p | b

, p | b

, ··· , p | b

`−1

, p - b

We look at the coefficient of X

k+`

in f g. It is given by

i+j=k+`

= a

k+`

+ ··· + a

k+1

`−1

+ a

k−1

`+1

+ ··· + a

`+k

By assumption, this is divisible by p. So

p |

i+j=k+`

However, the terms

k+`

···

k+1

`−1

, is divisible by

, as

p | b

for

j < `

Similarly,

k−1

`+1

···

`+k

is divisible by

. So we must have

p | a

is irreducible, and hence prime, we must have

p | a

p | b

. This is a

contradiction. So c(fg) must be a unit.

Corollary. Let

be a UFD. Then for

f, g ∈ R

[

], we have that

(

) is an

associate of c(f)c(g).

Again, we cannot say they are equal, since content is only well-defined up to

a unit.

Proof.

We can write

(

)

and

(

)

, with

and

primitive. Then

fg = c(f)c(g)f

Since

is primitive, so

(

)

(

) is a gcd of the coefficients of

, and so is

c(fg), by definition. So they are associates.

Finally, we can prove Gauss’ lemma.

Lemma (Gauss’ lemma). Let

be a UFD, and

f ∈ R

[

] be a primitive

polynomial. Then

is reducible in

[

] if and only if

is reducible

[

], where

F is the field of fractions of R.

Proof.

We will show that a primitive

f ∈ R

[

] is reducible in

[

] if and only

if f is reducible in F [X].

One direction is almost immediately obvious. Let

be a product in

[

] with

g, h

not units. As

is primitive, so are

and

. So both have degree

> 0. So g, h are not units in F [X]. So f is reducible in F [X].

The other direction is less obvious. We let

[

], with

g, h

not units.

and

have degree

0, since

is a field. So we can clear denominators

by finding

a, b ∈ R

such that (

)

(

)

∈ R

[

] (e.g. let

be the product of

denominators of coefficients of g). Then we get

abf = (ag)(bh),

and this is a factorization in

[

]. Here we have to be careful — (

) is one

thing that lives in

[

], and is not necessarily a product in

[

], since

might

not be in R[X]. So we should just treat it as a single symbol.

We now write

(ag) = c(ag)g

(bh) = c(bh)h

where g

, h

are primitive. So we have

ab = c(abf) = c((ag)(bh)) = u · c(ag)c(bh),

where u ∈ R is a unit, by the previous corollary. But also we have

abf = c(ag)c(gh)g

= u

−1

abg

So cancelling ab gives

f = u

−1

∈ R[X].

So f is reducible in R[X].

If this looks fancy and magical, you can try to do this explicitly in the case

where R = Z and F = Q. Then you will probably get enlightened.

We will do another proof performed in a similar manner.

Proposition. Let

be a UFD, and

be its field of fractions. Let

g ∈ R

[

] be

primitive. We let

J = (g) C R[X], I = (g) C F [X].

Then

J = I ∩ R[X].

In other words, if

f ∈ R

[

] and we can write it as

, with

h ∈ F

[

], then

in fact h ∈ R[X].

Proof.

The strategy is the same — we clear denominators in the equation

and then use contents to get that down in R[X].

We certainly have J ⊆ I ∩ R[X]. Now let f ∈ I ∩ R[X]. So we can write

f = gh,

with h ∈ F [X]. So we can choose b ∈ R such that bh ∈ R[X]. Then we know

bf = g(bh) ∈ R[X].

We let

(bh) = c(bh)h

for h

∈ R[X] primitive. Thus

bf = c(bh)gh

Since

is primitive, so is

. So

(

) =

(

) for

a unit. But

is really a

product in R[X]. So we have

c(bf) = c(b)c(f) = bc(f).

So we have

bf = ubc(f)gh

Cancelling b gives

f = g(uc(f)h

So g | f in R[X]. So f ∈ J.

From this we can get ourselves a large class of UFDs.

Theorem. If R is a UFD, then R[X] is a UFD.

In particular, if R is a UFD, then R[X

, ··· , X

] is also a UFD.

Proof.

We know

[

] has a notion of degree. So we will combine this with the

fact that R is a UFD.

Let

f ∈ R

[

]. We can write

(

)

, with

primitive. Firstly, as

is a

UFD, we may factor

c(f) = p

···p

for

∈ R

irreducible (and also irreducible in

[

]). Now we want to deal with

If f

is not irreducible, then we can write

= f

with

, f

both not units. Since

is primitive,

, f

also cannot be constants.

So we must have

deg f

, deg f

0. Also, since

deg f

, we must

have

deg f

, deg f

< deg f

. If

, f

are irreducible, then done. Otherwise, keep

on going. We will eventually stop since the degrees have to keep on decreasing.

So we can write it as

= q

···q

with q

irreducible. So we can write

f = p

···p

···q

a product of irreducibles.

For uniqueness, we first deal with the p’s. We note that

c(f) = p

···p

is a unique factorization of the content, up to reordering and associates, as

a UFD. So cancelling the content, we only have to show that primitives can be

factored uniquely.

Suppose we have two factorizations

= q

···q

= r

···r

Note that each

and each

is a factor of the primitive polynomial

, so are

also primitive. Now we do (maybe) the unexpected thing. We let

be the

field of fractions of

, and consider

, r

∈ F

[

]. Since

is a field,

[

] is

a Euclidean domain, hence principal ideal domain, hence unique factorization

domain.

By Gauss’ lemma, since the

and

are irreducible in

[

], they are also

irreducible in

[

]. As

[

] is a UFD, we find that

, and after reordering,

and q

are associates, say

= u

with

∈ F

[

] a unit. What we want to say is that

is a unit times

[

Firstly, note that u

∈ F as it is a unit. Clearing denominators, we can write

= b

∈ R[X].

Taking contents, since

, q

are primitives, we know

and

are associates, say

= v

with

∈ R

a unit. Cancelling

on both sides, we know

as required.

The key idea is to use Gauss’ lemma to say the reducibility in

[

] is the

same as reducibility in

[

], as long as we are primitive. The first part about

contents is just to turn everything into primitives.

Note that the last part of the proof is just our previous proposition. We

could have applied it, but we decide to spell it out in full for clarity.

Example. We know

[

] is a UFD, and if

is a UFD, then

[

, ··· , X

] is

also a UFD.

This is a useful thing to know. In particular, it gives us examples of UFDs

that are not PIDs. However, in such rings, we would also like to have an easy to

determine whether something is reducible. Fortunately, we have the following

criterion:

Proposition (Eisenstein’s criterion). Let R be a UFD, and let

f = a

+ a

X + ··· + a

∈ R[X]

be primitive with a

6= 0. Let p ∈ R be irreducible (hence prime) be such that

(i) p - a

;

(ii) p | a

for all 0 ≤ i < n;

(iii) p

- a

Then

is irreducible in

[

], and hence in

[

] (where

is the field of fractions

of R).

It is important that we work in

[

] all the time, until the end where we

apply Gauss’ lemma. Otherwise, we cannot possibly apply Eisenstein’s criterion

since there are no primes in F .

Proof. Suppose we have a factorization f = gh with

g = r

+ r

X + ··· + r

h = s

+ s

X + ··· + s

for r

, s

6= 0.

We know

. Since

p - a

, so

p - r

and

p - s

. We can also look at

bottom coefficients. We know

. We know

p | a

and

- a

. So

divides exactly one of r

and s

. wlog, p | r

and p - s

Now let j be such that

p | r

, p | r

, ··· , p | r

j−1

, p - r

We now look at a

. This is, by definition,

= r

+ r

j−1

+ ··· + r

j−1

+ r

We know r

, ··· , r

j−1

are all divisible by p. So

p | r

+ r

j−1

+ ··· + r

j−1

Also, since

p - r

and

p - s

, we know

p - r

, using the fact that

is prime. So

p - a

. So we must have j = n.

We also know that

j ≤ k ≤ n

. So we must have

. So

deg g

Hence

n − h

= 0. So

is a constant. But we also know

is primitive. So

must be a unit. So this is not a proper factorization.

Example. Consider the polynomial

− p ∈ Z

[

] for

a prime. Apply

Eisenstein’s criterion with

, and observe all the conditions hold. This is

certainly primitive, since this is monic. So

− p

is irreducible in

[

], hence

[

]. In particular,

− p

has no rational roots, i.e.

√

is irrational (for

n > 1).

Example. Consider a polynomial

f = X

p−1

+ X

p−2

+ ··· + X

+ X + 1 ∈ Z[X],

where

is a prime number. If we look at this, we notice Eisenstein’s criteria

does not apply. What should we do? We observe that

f =

− 1

X − 1

So it might be a good idea to let Y = X − 1. Then we get a new polynomial

f =

f(Y ) =

(Y + 1)

− 1

= Y

p−1





p−2





p−3

+ ··· +



p − 1



When we look at it hard enough, we notice Eisenstein’s criteria can be applied —

we know

p |





for 1

≤ i ≤ p −

1, but



p−1



. So

is irreducible in

[

Now if we had a factorization

f(X) = g(X)h(X) ∈ Z[X],

then we get

f(Y ) = g(Y + 1)h(Y + 1)

in Z[Y ]. So f is irreducible.

Hence none of the roots of

are rational (but we already know that — they

are not even real!).

2.6 Gaussian integers

We’ve mentioned the Gaussian integers already.

Definition (Gaussian integers). The Gaussian integers is the subring

Z[i] = {a + bi : a, b ∈ Z} ≤ C.

We have already shown that the norm

(

) =

is a Euclidean

function for

[

]. So

[

] is a Euclidean domain, hence principal ideal domain,

hence a unique factorization domain.

Since the units must have norm 1, they are precisely

, ±i

. What does

factorization in Z[i] look like? What are the primes? We know we are going to

get new primes, i.e. primes that are not integers, while we will lose some other

primes. For example, we have

2 = (1 + i)(1 − i).

So 2 is not irreducible, hence not prime. However, 3 is a prime. We have

(3) = 9. So if 3 =

, with

u, v

not units, then 9 =

(

)

(

), and neither

(

) nor

(

) are 1. So

(

) =

(

) = 3. However, 3 =

has no

solutions with

a, b ∈ Z

. So there is nothing of norm 3. So 3 is irreducible, hence

a prime.

Also, 5 is not prime, since

5 = (1 + 2i)(1 − 2i).

How can we understand which primes stay as primes in the Gaussian integers?

Proposition. A prime number

p ∈ Z

is prime in

[

] if and only if

p 6

for a, b ∈ Z \ {0}.

The proof is exactly what we have done so far.

Proof. If p = a

+ b

, then p = (a + ib)(a − ib). So p is not irreducible.

Now suppose

, with

u, v

not units. Taking norms, we get

(

)

(

). So if

and

are not units, then

(

) =

(

) =

. Writing

u = a + ib, then this says a

+ b

= p.

So what we have to do is to understand when a prime

can be written as a

sum of two squares. We will need the following helpful lemma:

Lemma. Let

be a prime number. Let

Z/pZ

be the field with

elements.

Let

}

be the group of invertible elements under multiplication. Then

∼

p−1

Proof.

Certainly

has order

p −

1, and is abelian. We know from the classifi-

cation of finite abelian groups that if

is not cyclic, then it must contain a

subgroup

×C

for

m >

1 (we can write it as

×C

×···

, and that

| d

So C

has a subgroup isomorphic to C

We consider the polynomial

−

∈ F

[

], which is a UFD. At best, this

factors into

linear factors. So

−

1 has at most

distinct roots. But if

× C

≤ F

, then we can find

elements of order dividing

. So there

are

elements of

which are roots of

−

1. This is a contradiction. So

is cyclic.

This is a funny proof, since we have not found any element that has order

p − 1.

Proposition. The primes in Z[i] are, up to associates,

(i) Prime numbers p ∈ Z ≤ Z[i] such that p ≡ 3 (mod 4).

(ii)

Gaussian integers

z ∈ Z

[

] with

(

) =

z¯z

for some prime

such that

p = 2 or p ≡ 1 (mod 4).

Proof.

We first show these are primes. If

p ≡

3 (

mod

4), then

p 6

, since

a square number mod 4 is always 0 or 1. So these are primes in Z[i].

On the other hand, if

(

) =

, and

, then

(

)

(

) =

. So

(

)

is 1 or

(

) is 1. So

is a unit. Note that we did not use the condition

that p 6≡ 3 (mod 4). This is not needed, since N(z) is always a sum of squares,

and hence N (z) cannot be a prime that is 3 mod 4.

Now let

z ∈ Z

[

] be irreducible, hence prime. Then

¯z

is also irreducible. So

(

) =

z¯z

is a factorization of

(

) into irreducibles. Let

p ∈ Z

be an ordinary

prime number dividing N (z), which exists since N (z) 6= 1.

Now if

p ≡

3 (

mod

4), then

itself is prime in

[

] by the first part of the

proof. So

p | N

(

) =

z¯z

. So

p | z

p | ¯z

. Note that if

p | ¯z

, then

p | z

by taking

complex conjugates. So we get

p | z

. Since both

and

are irreducible, they

must be equal up to associates.

Otherwise, we get

= 2 or

p ≡

1 (

mod

4). If

p ≡

1 (

mod

4), then

p −

1 = 4

for some

k ∈ Z

. As

∼

p−1

, there is a unique element of order 2 (this

is true for any cyclic group of order 4

— think

). This must be [

−

∈ F

Now let a ∈ F

be an element of order 4. Then a

has order 2. So [a

] = [−1].

This is a complicated way of saying we can find an

such that

p | a

+ 1.

Thus

p |

(

)(

a − i

). In the case where

= 2, we know by checking directly

that 2 = (1 + i)(1 − i).

In either case, we deduce that

(or 2) is not prime (hence irreducible),

since it clearly does not divide

a ± i

(or 1

± i

). So we can write

, for

, z

∈ Z[i] not units. Now we get

= N(p) = N(z

)N(z

As the

are not units, we know

(

) =

(

) =

. By definition, this means

p = z

¯z

= z

¯z

. But also p = z

. So we must have ¯z

= z

Finally, we have

¯z

| N

(

) =

z¯z

. All these

are irreducible. So

must be an associate of z

(or maybe ¯z

). So in particular N (z) = p.

Corollary. An integer

n ∈ Z

≥0

may be written as

(as the sum of two

squares) if and only if “when we write

···p

as a product as distinct

primes, then p

≡ 3 (mod 4) implies n

is even”.

We have proved this in the case when n is a prime.

Proof. If n = x

+ y

, then we have

n = (x + iy)(x − iy) = N(x + iy).

Let

. So we can write

···α

as a product of irreducibles in

[

By the proposition, each

is either

(a genuine prime number with

p ≡

(

mod

4)), or

(

) =

is a prime number which is either 2 or

≡

1 (

mod

4). We

now take the norm to obtain

N = x

+ y

= N(z) = N(α

)N(α

) ···N(α

Now each

(

) is either

with

p ≡

3 (

mod

4), or is just

for

= 2 or

p ≡

(

mod

4). So if

is the largest power of

divides

, we find that

must be

even if p ≡ 3 (mod 4).

Conversely, let

···p

be a product of distinct primes. Now for

each p

, either p

≡ 3 (mod 4), and n

is even, in which case

= (p

)

= N(p

);

= 2 or

≡

1 (

mod

4), in which case, the above proof shows that

(

)

for some α

. So p

= N(α

Since the norm is multiplicative, we can write

as the norm of some

z ∈ Z

[

n = N(z) = N(x + iy) = x

+ y

as required.

Example. We know 65 = 5

13. Since 5

≡

1 (

mod

4), it is a sum of squares.

Moreover, the proof tells us how to find 65 as the sum of squares. We have to

factor 5 and 13 in Z[i]. We have

5 = (2 + i)(2 − i)

13 = (2 + 3i)(2 − 3i).

So we know

65 = N(2 + i)N(2 + 3i) = N ((2 + i)(2 + 3i)) = N(1 + 8i) = 1

+ 8

But there is a choice here. We had to pick which factor is

and which is

¯α

. So

we can also write

65 = N((2 + i)(2 − 3i)) = N(7 − 4i) = 7

+ 4

So not only are we able to write them as sum of squares, but this also gives us

many ways of writing 65 as a sum of squares.

2.7 Algebraic integers

We generalize the idea of Gaussian integers to algebraic integers.

Definition (Algebraic integer). An

α ∈ C

is called an algebraic integer if it is

a root of a monic polynomial in

[

], i.e. there is a monic

f ∈ Z

[

] such that

f(α) = 0.

We can immediately check that this is a sensible definition — not all complex

numbers are algebraic integers, since there are only countably many polynomials

with integer coefficients, hence only countably many algebraic integers, but there

are uncountably many complex numbers.

Notation. For

an algebraic integer, we write

[

]

≤ C

for the smallest subring

containing α.

This can also be defined for arbitrary complex numbers, but it is less inter-

esting.

We can also construct

[

] by taking it as the image of the map

[

]

→ C

given by g 7→ g(α). So we can also write

Z[α] =

Z[X]

, I = ker φ.

Note that

is non-empty, since, say,

f ∈ I

, by definition of an algebraic integer.

Proposition. Let α ∈ C be an algebraic integer. Then the ideal

I = ker(φ : Z[X] → C, f 7→ f(α))

is principal, and equal to (f

) for some irreducible monic f

This is a non-trivial theorem, since

[

] is not a principal ideal domain. So

there is no immediate guarantee that I is generated by one polynomial.

Definition (Minimal polynomial). Let

α ∈ C

be an algebraic integer. Then

the minimal polynomial is a polynomial

is the irreducible monic such that

I = ker(φ) = (f

Proof.

By definition, there is a monic

f ∈ Z

[

] such that

(

) = 0. So

f ∈ I

I 6

= 0. Now let

∈ I

be such a polynomial of minimal degree. We may

suppose that

is primitive. We want to show that

= (

), and that

irreducible.

Let

h ∈ I

. We pretend we are living in

[

]. Then we have the Euclidean

algorithm. So we can write

h = f

q + r,

with

= 0 or

deg r < deg f

. This was done over

[

], not

[

]. We now clear

denominators. We multiply by some a ∈ Z to get

ah = f

(aq) + (ar),

where now (

)

(

)

∈ Z

[

]. We now evaluate these polynomials at

. Then

we have

ah(α) = f

(α)aq(α) + ar(α).

We know

(

) =

(

) = 0, since

and

are both in

. So

(

) = 0. So

(

)

∈ I

. As

∈ I

has minimal degree, we cannot have

deg

(

) =

deg

(

)

deg(f

). So we must have r = 0.

Hence we know

ah = f

· (aq)

is a factorization in

[

]. This is almost right, but we want to factor

, not

Again, taking contents of everything, we get

ac(h) = c(ah) = c(f

(aq)) = c(aq),

is primitive. In particular,

a | c

(

). This, by definition of content, means

(

) can be written as

a¯q

, where

¯q ∈ Z

[

]. Cancelling, we get

¯q ∈ Z

[

]. So

we know

h = f

q ∈ (f

So we know I = (f

To show f

is irreducible, note that

Z[X]

)

∼

Z[X]

ker φ

∼

im(φ) = Z[α] ≤ C.

Since

is an integral domain, so is

(

). So we know

[

]

(

) is an integral

domain. So (f

) is prime. So f

is prime, hence irreducible.

If this final line looks magical, we can unravel this proof as follows: suppose

for some non-units

. Then since

(

) = 0, we know

(

)

(

) = 0.

Since

(

)

, q

(

)

∈ C

, which is an integral domain, we must have, say,

(

) = 0.

But then deg p < deg f

, so p 6∈ I = (f

). Contradiction.

Example.

(i) We know α = i is an algebraic integer with f

= X

+ 1.

(ii) Also, α =

√

2 is an algebraic integer with f

= X

− 2.

(iii)

More interestingly,

(1 +

√

−3

) is an algebraic integer with

− X − 1.

(iv)

The polynomial

− X

d ∈ Z

[

] with

d ∈ Z

≥0

has precisely one real

root

, which is an algebraic integer. It is a theorem, which will be proved

in IID Galois Theory, that this

cannot be constructed from integers

via +

, −, ×, ÷,

√

. It is also a theorem, found in IID Galois Theory, that

degree 5 polynomials are the smallest degree for which this can happen (the

prove involves writing down formulas analogous to the quadratic formula

for degree 3 and 4 polynomials).

Lemma. Let α ∈ Q be an algebraic integer. Then α ∈ Z.

Proof.

Let

∈ Z

[

] be the minimal polynomial, which is irreducible. In

[

the polynomial

X − α

must divide

. However, by Gauss’ lemma, we know

f ∈ Q

[

] is irreducible. So we must have

X − α ∈ Z

[

]. So

is an

integer.

It turns out the collection of all algebraic integers form a subring of

. This

is not at all obvious — given

f, g ∈ Z

[

] monic such that

(

) =

(

) = 0, there

is no easy way to find a new monic

such that

(

) = 0. We will prove this

much later on in the course.

2.8 Noetherian rings

We now revisit the idea of Noetherian rings, something we have briefly mentioned

when proving that PIDs are UFDs.

Definition (Noetherian ring). A ring is Noetherian if for any chain of ideals

⊆ I

⊆ ··· ,

there is some N such that I

= I

N+1

= I

N+2

= ···.

This condition is known as the ascending chain condition.

Example. Every finite ring is Noetherian. This is since there are only finitely

many possible ideals.

Example. Every field is Noetherian. This is since there are only two possible

ideals.

Example. Every principal ideal domain (e.g.

) is Noetherian. This is easy to

check directly, but the next proposition will make this utterly trivial.

Most rings we love and know are indeed Noetherian. However, we can

explicitly construct some non-Noetherian ideals.

Example. The ring

[

, X

, ···

] is not Noetherian. This has the chain of

strictly increasing ideals

) ⊆ (X

, X

) ⊆ (X

, X

) ⊆ ··· .

We have the following proposition that makes Noetherian rings much more

concrete, and makes it obvious why PIDs are Noetherian.

Definition (Finitely generated ideal). An ideal

is finitely generated if it can

be written as I = (r

, ··· , r

) for some r

, ··· , r

∈ R.

Proposition. A ring is Noetherian if and only if every ideal is finitely generated.

Every PID trivially satisfies this condition. So we know every PID is Noethe-

rian.

Proof. We start with the easier direction — from concrete to abstract.

Suppose every ideal of

is finitely generated. Given the chain

⊆ I

⊆ ···

consider the ideal

I = I

∪ I

∪ ··· .

This is obviously an ideal, and you will check this manually in example sheet 2.

We know I is finitely generated, say I = (r

, ··· , r

), with r

∈ I

. Let

K = max

i=1,··· ,n

Then r

, ··· , r

∈ I

. So I

= I. So I

= I

K+1

= I

K+2

= ···.

To prove the other direction, suppose there is an ideal

I C R

that is not

finitely generated. We pick

∈ I

. Since

is not finitely generated, we know

) 6= I. So we can find some r

∈ I \ (r

Again (

, r

)

. So we can find

∈ I \

(

, r

). We continue on, and then

can find an infinite strictly ascending chain

) ⊆ (r

, r

) ⊆ (r

, r

) ⊆ ··· .

So R is not Noetherian.

When we have developed some properties or notions, a natural thing to ask

is whether it passes on to subrings and quotients.

is Noetherian, does every subring of

have to be Noetherian? The

answer is no. For example, since

[

, X

, ···

] is an integral domain, we can

take its field of fractions, which is a field, hence Noetherian, but Z[X

, X

, ···]

is a subring of its field of fractions.

How about quotients?

Proposition. Let

be a Noetherian ring and

be an ideal of

. Then

R/I

Noetherian.

Proof.

Whenever we see quotients, we should think of them as the image of a

homomorphism. Consider the quotient map

π : R → R/I

x 7→ x + I.

We can prove this result by finitely generated or ascending chain condition. We

go for the former. Let

J C R/I

be an ideal. We want to show that

is finitely

generated. Consider the inverse image

−1

(

). This is an ideal of

, and is

hence finitely generated, since

is Noetherian. So

−1

(

) = (

, ··· , r

) for

some r

, ··· , r

∈ R. Then J is generated by π(r

), ··· , π(r

). So done.

This gives us many examples of Noetherian rings. But there is one important

case we have not tackled yet — polynomial rings. We know

[

] is not a PID,

since (2

, X

) is not principal. However, this is finitely generated. So we are not

dead. We might try to construct some non-finitely generated ideal, but we are

bound to fail. This is since

[

] is a Noetherian ring. This is a special case of

the following powerful theorem:

Theorem (Hilbert basis theorem). Let

be a Noetherian ring. Then so is

R[X].

Since Z is Noetherian, we know Z[X] also is. Hence so is Z[X, Y ] etc.

The Hilbert basis theorem was, surprisingly, proven by Hilbert himself. Before

that, there were many mathematicians studying something known as invariant

theory. The idea is that we have some interesting objects, and we want to look

at their symmetries. Often, there are infinitely many possible such symmetries,

and one interesting question to ask is whether there is a finite set of symmetries

that generate all possible symmetries.

This sounds like an interesting problem, so people devoted much time, writing

down funny proofs, showing that the symmetries are finitely generated. However,

the collection of such symmetries are often just ideals of some funny ring. So

Hilbert came along and proved the Hilbert basis theorem, and showed once and

for all that those rings are Noetherian, and hence the symmetries are finitely

generated.

Proof.

The proof is not too hard, but we will need to use both the ascending

chain condition and the fact that all ideals are finitely-generated.

Let

I C R

[

] be an ideal. We want to show it is finitely generated. Since we

know R is Noetherian, we want to generate some ideals of R from I.

How can we do this? We can do the silly thing of taking all constants of

i.e.

I ∩ R

. But we can do better. We can consider all linear polynomials, and

take their leading coefficients. Thinking for a while, this is indeed an ideal.

In general, for n = 0, 1, 2, ···, we let

= {r ∈ R : there is some f ∈ I such that f = rX

+ ···} ∪ {0}.

Then it is easy to see, using the strong closure property, that each ideal

is an

ideal of

. Moreover, they form a chain, since if

f ∈ I

, then

Xf ∈ I

, by strong

closure. So I

⊆ I

n+1

for all n.

By the ascending chain condition of

, we know there is some

such that

N+1

···

. Now for each 0

≤ n ≤ N

, since

is Noetherian, we can write

= (r

(n)

, r

(n)

, ··· , r

(n)

k(n)

Now for each r

(n)

, we choose some f

(n)

∈ I with f

(n)

= r

(n)

+ ···.

We now claim the polynomials

(n)

for 0

≤ n ≤ N

and 1

≤ i ≤ k

(

) generate

Suppose not. We pick g ∈ I of minimal degree not generated by the f

(n)

There are two possible cases. If deg g = n ≤ N, suppose

g = rX

+ ··· .

We know r ∈ I

. So we can write

r =

(n)

for some λ

∈ R, since that’s what generating an ideal means. Then we know

(n)

= rX

+ ··· ∈ I.

But if

is not in the span of the

(j)

, then so isn’t

g −

(n)

. But this has

a lower degree than g. This is a contradiction.

Now suppose

deg g

n > N

. This might look scary, but it is not, since

= I

. So we write the same proof. We write

g = rX

+ ··· .

But we know r ∈ I

= I

. So we know

r =

(N)

Then we know

n−N

(n)

= rX

+ ··· ∈ I.

Hence

g − X

n−N

(N)

has smaller degree than

, but is not in the span of

(j)

As an aside, let

E ⊆ F

[

, X

, ··· , X

] be any set of polynomials. We view

this as a set of equations

= 0 for each

f ∈ E

. The claim is that to solve the

potentially infinite set of equations

, we actually only have to solve finitely

many equations.

Consider the ideal (

)

C F

[

, ··· , X

]. By the Hilbert basis theorem, there

is a finite list f

, ··· , f

such that

, ··· , f

) = (E).

We want to show that we only have to solve

(

) = 0 for these

. Given

(α

, ··· , α

) ∈ F

, consider the homomorphism

: F [X

, ··· , X

] → F

7→ α

Then we know (

, ··· , α

)

∈ F

is a solution to the equations

if and only

if (

)

⊆ ker

(

). By our choice of

, this is true if and only if (

, ··· , f

)

⊆

ker

(

). By inspection, this is true if and only if (

, ··· , α

) is a solution to

all of

, ··· , f

. So solving

is the same as solving

, ··· , f

. This is useful

in, say, algebraic geometry.

3 Modules

Finally, we are going to look at modules. Recall that to define a vector space,

we first pick some base field

. We then defined a vector space to be an abelian

group

with an action of

(i.e. scalar multiplication) that is compatible

with the multiplicative and additive structure of F.

In the definition, we did not at all mention division in

. So in fact we can

make the same definition, but allow

to be a ring instead of a field. We call

these modules. Unfortunately, most results we prove about vector spaces do use

the fact that

is a field. So many linear algebra results do not apply to modules,

and modules have much richer structures.

3.1 Definitions and examples

Definition (Module). Let

be a commutative ring. We say a quadruple

(M, +, 0

, ·) is an R-module if

(i) (M, +, 0

) is an abelian group

(ii) The operation · : R × M → M satisfies

(a) (r

+ r

) · m = (r

· m) + (r

· m);

(b) r · (m

+ m

) = (r · m

) + (r · m

);

· (r

· m) = (r

· r

) · m; and

(d) 1

· m = m.

Note that there are two different additions going on — addition in the ring

and addition in the module, and similarly two notions of multiplication. However,

it is easy to distinguish them since they operate on different things. If needed,

we can make them explicit by writing, say, +

and +

We can imagine modules as rings acting on abelian groups, just as groups

can act on sets. Hence we might say “

acts on

” to mean

is an

-module.

Example. Let

be a field. An

-module is precisely the same as a vector space

over F (the axioms are the same).

Example. For any ring R, we have the R-module R

= R × R × ··· × R via

r · (r

, ··· , r

) = (rr

, ··· , rr

using the ring multiplication. This is the same as the definition of the vector

space F

for fields F.

Example. Let I C R be an ideal. Then it is an R-module via

r ·

a = r ·

a, r

= r

Also, R/I is an R-module via

r ·

R/I

(a + I) = (r ·

a) + I,

Example. A

-module is precisely the same as an abelian group. For

abelian group, we have

Z × A → A

(n, a) 7→ a + ··· + a

| {z }

n times

where we adopt the notation

a + ··· + a

| {z }

−n times

= (−a) + ··· + (−a)

| {z }

n times

and adding something to itself 0 times is just 0.

This definition is essentially forced upon us, since by the axioms of a module,

we must have (1, a) 7→ a. Then we must send, say, (2, a) = (1 + 1, a) 7→ a + a.

Example. Let

be a field and

a vector space over

, and

V → V

be a

linear map. Then V is an F[X]-module via

F[X] × V → V

(f, v ) 7→ f (α)(v).

This is a module.

Note that we cannot just say that

is an

[

]-module. We have to specify

the α as well. Picking a different α will give a different F[X]-module structure.

Example. Let

R → S

be a homomorphism of rings. Then any

-module

may be considered as an R-module via

R × M → M

(r, m) 7→ φ(r) ·

Definition (Submodule). Let

be an

-module. A subset

N ⊆ M

is an

-submodule if it is a subgroup of (

), and if

n ∈ N

and

r ∈ R

, then

rn ∈ N. We write N ≤ M.

Example. We know

itself is an

-module. Then a subset of

is a submodule

if and only if it is an ideal.

Example. A subset of an

-module

, where

is a field, is an

-submodule if

and only if it is a vector subspace of V .

Definition (Quotient module). Let

N ≤ M

be an

-submodule. The quotient

module M/N is the set of N-cosets in (M, +, 0

), with the R-action given by

r · (m + N) = (r · m) + N.

It is easy to check this is well-defined and is indeed a module.

Note that modules are different from rings and groups. In groups, we had

subgroups, and we have some really nice ones called normal subgroups. We are

only allowed to quotient by normal subgroups. In rings, we have subrings and

ideals, which are unrelated objects, and we only quotient by ideals. In modules,

we only have submodules, and we can quotient by arbitrary submodules.

Definition (

-module homomorphism and isomorphism). A function

M →

between

-modules is an

-module homomorphism if it is a homomorphism

of abelian groups, and satisfies

f(r · m) = r · f(m)

for all r ∈ R and m ∈ M.

An isomorphism is a bijective homomorphism, and two

-modules are

isomorphic if there is an isomorphism between them.

Note that on the left, the multiplication is the action in

, while on the

right, it is the action in N.

Example. If

is a field and

V, W

are

-modules (i.e. vector spaces over

then an F-module homomorphism is precisely an F-linear map.

Theorem (First isomorphism theorem). Let

M → N

be an

-module

homomorphism. Then

ker f = {m ∈ M : f(m) = 0} ≤ M

is an R-submodule of M. Similarly,

im f = {f(m) : m ∈ M } ≤ N

is an R-submodule of N. Then

ker f

∼

im f.

We will not prove this again. The proof is exactly the same.

Theorem (Second isomorphism theorem). Let A, B ≤ M. Then

A + B = {m ∈ M : m = a + b for some a ∈ A, b ∈ B} ≤ M,

and

A ∩ B ≤ M.

We then have

A + B

∼

A ∩ B

Theorem (Third isomorphism theorem). Let N ≤ L ≤ M. Then we have

∼











Also, we have a correspondence

{submodules of M/N} ←→ {submodules of M which contain N }

It is an exercise to see what these mean in the cases where

is a field, and

modules are vector spaces.

We now have something new. We have a new concept that was not present

in rings and groups.

Definition (Annihilator). Let

be an

-module, and

m ∈ M

. The annihilator

of m is

Ann(m) = {r ∈ R : r · m = 0}.

For any set S ⊆ M, we define

Ann(S) = {r ∈ R : r · m = 0 for all m ∈ S} =

m∈S

Ann(m).

In particular, for the module M itself, we have

Ann(M) = {r ∈ R : r · m = 0 for all m ∈ M} =

m∈M

Ann(m).

Note that the annihilator is a subset of

. Moreover it is an ideal — if

r ·m

= 0 and

s ·m

= 0, then (

)

·m

r ·m

s ·m

= 0. So

s ∈ Ann

(

Moreover, if r · m = 0, then also (sr) · m = s · (r · m) = 0. So sr ∈ Ann(m).

What is this good for? We first note that any

m ∈ M

generates a submodule

Rm as follows:

Definition (Submodule generated by element). Let

be an

-module, and

m ∈ M . The submodule generated by m is

Rm = {r · m ∈ M : r ∈ R}.

We consider the R-module homomorphism

φ : R → M

r 7→ rm.

This is clearly a homomorphism. Then we have

Rm = im(φ),

Ann(m) = ker(φ).

The conclusion is that

∼

R/ Ann(m).

As we mentioned, rings acting on modules is like groups acting on sets. We can

think of this as the analogue of the orbit-stabilizer theorem.

In general, we can generate a submodule with many elements.

Definition (Finitely generated module). An

-module

is finitely generated

if there is a finite list of elements m

, ··· , m

such that

M = Rm

+ Rm

+ ··· + Rm

= {r

+ r

+ ··· + r

: r

∈ R}.

This is in some sense analogous to the idea of a vector space being finite-

dimensional. However, it behaves much more differently.

While this definition is rather concrete, it is often not the most helpful

characterization of finitely-generated modules. Instead, we use the following

lemma:

Lemma. An

-module

is finitely-generated if and only if there is a surjective

R-module homomorphism f : R

 M for some finite k.

Proof. If

M = Rm

+ Rm

+ ··· + Rm

we define f : R

→ M by

, ··· , r

) 7→ r

+ ··· + r

It is clear that this is an

-module homomorphism. This is by definition

surjective. So done.

Conversely, given a surjection f : R

 M, we let

= f(0, 0, ··· , 0, 1, 0, ··· , 0),

where the 1 appears in the ith position. We now claim that

M = Rm

+ Rm

+ ··· + Rm

So let m ∈ M. As f is surjective, we know

m = f(r

, r

, ··· , r

)

for some r

. We then have

f(r

, r

, ··· , r

)

= f((r

, 0, ··· , 0) + (0, r

, 0, ··· , 0) + ··· + (0, 0, ··· , 0, r

))

= f(r

, 0, ··· , 0) + f(0, r

, 0, ··· , 0) + ··· + f(0, 0, ··· , 0, r

)

= r

f(1, 0, ··· , 0) + r

f(0, 1, 0, ··· , 0) + ··· + r

f(0, 0, ··· , 0, 1)

= r

+ r

+ ··· + r

So the m

generate M .

This view is a convenient way of thinking about finitely-generated modules.

For example, we can immediately prove the following corollary:

Corollary. Let

N ≤ M

and

be finitely-generated. Then

M/N

is also finitely

generated.

Proof.

Since

is finitely generated, we have some surjection

 M

Moreover, we have the surjective quotient map

M  M/N

. Then we get the

following composition

M M/N,

f q

which is a surjection, since it is a composition of surjections. So

M/N

is finitely

generated.

It is very tempting to believe that if a module is finitely generated, then its

submodules are also finitely generated. It would be very wrong to think so.

Example. A submodule of a finitely-generated module need not be finitely

generated.

We let

[

, X

, ···

]. We consider the

-module

, which is

finitely generated (by 1). A submodule of the ring is the same as an ideal.

Moreover, an ideal is finitely generated as an ideal if and only if it is finitely

generated as a module. We pick the submodule

I = (X

, X

, ···),

which we have already shown to be not finitely-generated. So done.

Example. For a complex number

, the ring

[

] (i.e. the smallest subring

containing

) is a finitely-generated as a

-module if and only if

is an

algebraic integer.

Proof is left as an exercise for the reader on the last example sheet. This allows

us to prove that algebraic integers are closed under addition and multiplication,

since it is easier to argue about whether Z[α] is finitely generated.

3.2 Direct sums and free modules

We’ve been secretly using the direct sum in many examples, but we shall define

it properly now.

Definition (Direct sum of modules). Let

, M

, ··· , M

-modules. The

direct sum is the R-module

⊕ M

⊕ ··· ⊕ M

which is the set M

× M

× ··· × M

, with addition given by

, ··· , m

) + (m

, ··· , m

) = (m

+ m

, ··· , m

+ m

and the R-action given by

r · (m

, ··· , m

) = (rm

, ··· , rm

We’ve been using one example of the direct sum already, namely

= R ⊕ R ⊕ ··· ⊕ R

| {z }

n times

Recall we said modules are like vector spaces. So we can try to define things like

basis and linear independence. However, we will fail massively, since we really

can’t prove much about them. Still, we can define them.

Definition (Linear independence). Let

, ··· , m

∈ M

. Then

, ··· , m

}

is linearly independent if

i=1

= 0

implies r

= r

= ··· = r

= 0.

Lots of modules will not have a basis in the sense we are used to. The next

best thing would be the following:

Definition (Freely generate). A subset S ⊆ M generates M freely if

(i) S generates M

(ii)

Any set function

S → N

to an

-module

extends to an

-module

map θ : M → N.

Note that if

, θ

are two such extensions, we can consider

−θ

M → N

Then

− θ

sends everything in

to 0. So

S ⊆ ker

(

− θ

)

≤ M

. So the

submodule generated by S lies in ker(θ

− θ

) too. But this is by definition M.

M ≤ ker

(

− θ

)

≤ M

, i.e. equality holds. So

− θ

= 0. So

. So

any such extension is unique.

Thus, what this definition tells us is that giving a map from

exactly the same thing as giving a function from S to N .

Definition (Free module and basis). An

-module is free if it is freely generated

by some subset S ⊆ M, and S is called a basis.

We will soon prove that if

is a field, then every module is free. However, if

R is not a field, then there are non-free modules.

Example. The

-module

is not freely generated. Suppose

were

generated by some

S ⊆ Z/

. Then this can only possibly be

{

}

. Then

this implies there is a homomorphism

Z → Z

sending 1 to 1. But it does

not send 0 = 1 + 1 to 1 + 1, since homomorphisms send 0 to 0. So

is not

freely generated.

We now want to formulate free modules in a way more similar to what we

do in linear algebra.

Proposition. For a subset

, ··· , m

} ⊆ M

, the following are equivalent:

(i) S generates M freely.

(ii) S generates M and the set S is independent.

(iii) Every element of M is uniquely expressible as

+ r

+ ··· + r

for some r

∈ R.

Proof.

The fact that (ii) and (iii) are equivalent is something we would expect

from what we know from linear algebra, and in fact the proof is the same. So

we only show that (i) and (ii) are equivalent.

Let S generate M freely. If S is not independent, then we can write

+ ··· + r

= 0,

with

∈ R

and, say,

non-zero. We define the set function

S → R

sending

7→

and

7→

0 for all

i 6

= 1. As

generates

freely, this

extends to an R-module homomorphism θ : M → R.

By definition of a homomorphism, we can compute

0 = θ(0)

= θ(r

+ r

+ ··· + r

)

= r

θ(m

) + r

θ(m

) + ··· + r

θ(m

)

= r

This is a contradiction. So S must be independent.

To prove the other direction, suppose every element can be uniquely written

···

. Given any set function

S → N

, we define

M → N

θ(r

+ ··· + r

) = r

ψ(m

) + ··· + r

ψ(m

This is well-defined by uniqueness, and is clearly a homomorphism. So it follows

that S generates M freely.

Example. The set

{

} ∈ Z

generates

. However, they do not generate

freely, since

3 · 2 + (−2) · 3 = 0.

Recall from linear algebra that if a set

spans a vector space

, and it is not

independent, then we can just pick some useless vectors and throw them away

in order to get a basis. However, this is no longer the case in modules. Neither

2 nor 3 generate Z.

Definition (Relations). If

is a finitely-generated

-module, we have shown

that there is a surjective

-module homomorphism

→ M

. We call

ker

(

)

the relation module for those generators.

Definition (Finitely presented module). A finitely-generated module is finitely

presented if we have a surjective homomorphism

→ M

and

ker φ

is finitely

generated.

Being finitely presented means I can tell you everything about the module

with a finite amount of paper. More precisely, if

, ··· , m

}

generate

and

, n

, ··· , n

} generate ker(φ), then each

= (r

, ···r

)

corresponds to the relation

+ r

+ ··· + r

= 0

. So

is the module generated by writing down

-linear combinations

, ··· , m

, and saying two elements are the same if they are related to one

another by these relations. Since there are only finitely many generators and

finitely many such relations, we can specify the module with a finite amount of

information.

A natural question we might ask is if

n 6

, then are

and

the same?

In vector spaces, they obviously must be different, since basis and dimension are

well-defined concepts.

Proposition (Invariance of dimension/rank). Let

be a non-zero ring. If

∼

as R-modules, then n = m.

We know this is true if

is a field. We now want to reduce this to the case

where R is a ring.

is an integral domain, then we can produce a field by taking the field of

fractions, and this might be a good starting point. However, we want to do this

for general rings. So we need some more magic.

We will need the following construction:

Let I C R be an ideal, and let M be an R-module. We define

IM = {am ∈ M : a ∈ I, m ∈ M} ≤ M.

So we can take the quotient module M/IM, which is an R-module again.

Now if b ∈ I, then its action on M/IM is

b(m + IM) = bm + IM = IM.

So everything in

kills everything in

M/IM

. So we can consider

M/IM

as an

R/I module by

(r + I) · (m + IM) = r · m + IM.

So we have proved that

Proposition. If

I C R

is an ideal and

is an

-module, then

M/IM

is an

R/I module in a natural way.

We next need to use the following general fact:

Proposition. Every non-zero ring has a maximal ideal.

This is a rather strong statement, since it talks about “all rings”, and we can

have weird rings. We need to use a more subtle argument, namely via Zorn’s

lemma. You probably haven’t seen it before, in which case you might want to

skip the proof and just take the lecturer’s word on it.

Proof.

We observe that an ideal

I C R

is proper if and only if 1

6∈ I

. So every

increasing union of proper ideals is proper. Then by Zorn’s lemma, there is a

maximal ideal (Zorn’s lemma says if an arbitrary union of increasing things is

still a thing, then there is a maximal such thing, roughly).

With these two notions, we get

Proposition (Invariance of dimension/rank). Let

be a non-zero ring. If

∼

as R-modules, then n = m.

Proof.

Let

be a maximal ideal of

. Suppose we have

∼

. Then we

must have

∼

as R/I modules.

But staring at it long enough, we figure that

∼





and similarly for

. Since

R/I

is a field, the result follows by linear algebra.

The point of this proposition is not the result itself (which is not too inter-

esting), but the general constructions used behind the proof.

3.3 Matrices over Euclidean domains

This is the part of the course where we deliver all our promises about proving

the classification of finite abelian groups and Jordan normal forms.

Until further notice, we will assume R is a Euclidean domain, and we write

R \{

} → Z

≥0

for its Euclidean function. We know that in such a Euclidean

domain, the greatest common divisor gcd(a, b) exists for all a, b ∈ R.

We will consider some matrices with entries in R.

Definition (Elementary row operations). Elementary row operations on an

m × n matrix A with entries in R are operations of the form

(i)

Add

c ∈ R

times the

th row to the

th row. This may be done by

multiplying by the following matrix on the left:







1 c







where c appears in the ith column of the jth row.

(ii)

Swap the

th and

th rows. This can be done by left-multiplication of the

matrix







0 1

1 0







Again, the rows and columns we have messed with are the

th and

rows and columns.

(iii)

We multiply the

th row by a unit

c ∈ R

. We do this via the following

matrix:













Notice that if

is a field, then we can multiply any row by any non-zero

number, since they are all units.

We also have elementary column operations defined in a similar fashion, corre-

sponding to right multiplication of the matrices. Notice all these matrices are

invertible.

Definition (Equivalent matrices). Two matrices are equivalent if we can get from

one to the other via a sequence of such elementary row and column operations.

Note that if A and B are equivalent, then we can write

B = QAT

−1

for some invertible matrices Q and T

−1

The aim of the game is to find, for each matrix, a matrix equivalent to it

that is as simple as possible. Recall from IB Linear Algebra that if

is a field,

then we can put any matrix into the form



0 0



via elementary row and column operations. This is no longer true when working

with rings. For example, over Z, we cannot put the matrix



2 0

0 0



into that form, since no operation can turn the 2 into a 1. What we get is the

following result:

Theorem (Smith normal form). An

m × n

matrix over a Euclidean domain

is equivalent to a diagonal matrix













with the d

all non-zero and

| d

| ··· | d

Note that the divisibility criterion is similar to the classification of finitely-

generated abelian groups. In fact, we will derive that as a consequence of the

Smith normal form.

Definition (Invariant factors). The

obtained in the Smith normal form are

called the invariant factors of A.

We first exhibit the algorithm of producing the Smith normal form with an

algorithm in Z.

Example. We start with the matrix





3 7 4

1 −1 2

3 5 1





We want to move the 1 to the top-left corner. So we swap the first and second

rows to obtain.





1 −1 2

3 7 4

3 5 1





We then try to eliminate the other entries in the first row by column operations.

We add multiples of the first column to the second and third to obtain





1 0 0

3 10 −2

3 8 −5





We similarly clear the first column to get





1 0 0

0 10 −2

0 8 −5





We are left with a 2 × 2 matrix to fiddle with.

We swap the second and third columns so that 2 is in the 2

2 entry, and

secretly change sign to get





1 0 0

0 2 10

0 5 8





We notice that (2

5) = 1. So we can use linear combinations to introduce a 1 at

the bottom





1 0 0

0 2 10

0 1 −12





Swapping rows, we get





1 0 0

0 1 −12

0 2 10





We then clear the remaining rows and columns to get





1 0 0

0 1 0

0 0 34





Proof.

Throughout the process, we will keep calling our matrix

, even though

it keeps changing in each step, so that we don’t have to invent hundreds of names

for these matrices.

= 0, then done! So suppose

A 6

= 0. So some entry is not zero, say,

= 0. Swapping the

th and first row, then

th and first column, we arrange

that

= 0. We now try to reduce

as much as possible. We have the

following two possible moves:

(i)

If there is an

not divisible by

, then we can use the Euclidean

algorithm to write

= qA

+ r.

By assumption,

r 6

= 0. So

(

)

< φ

(

) (where

is the Euclidean

function).

So we subtract

copies of the first column from the

th column. Then

in position (1

, j

), we now have

. We swap the first and

th column such

that

is in position (1

1), and we have strictly reduced the value of

the first entry.

(ii)

If there is an

not divisible by

, we do the same thing, and this

again reduces φ(A

We keep performing these until no move is possible. Since the value of

(

)

strictly decreases every move, we stop after finitely many applications. Then we

know that we must have

dividing all

and

. Now we can just subtract

appropriate multiples of the first column from others so that

= 0 for

j 6

= 1.

We do the same thing with rows so that the first row is cleared. Then we have a

matrix of the form

A =







d 0 ··· 0

. C







We would like to say “do the same thing with

”, but then this would get us a

regular diagonal matrix, not necessarily in Smith normal form. So we need some

preparation.

(iii) Suppose there is an entry of C not divisible by d, say A

with i, j > 1.

A =







d 0 ··· 0 ··· 0

0 A







We suppose

= qd + r,

with

r 6

= 0 and

(

)

< φ

(

). We add column 1 to column

, and subtract

times row 1 from row

. Now we get

in the (

i, j

)th entry, and we want

to send it back to the (1

1) position. We swap row

with row 1, swap

column j with row 1, so that r is in the (1, 1)th entry, and φ(r) < φ(d).

Now we have messed up the first row and column. So we go back and do

(i) and (ii) again until the first row and columns are cleared. Then we get

A =







0 ··· 0

0 C







where

φ(d

) ≤ φ(r) < φ(d).

As this strictly decreases the value of

(

), we can only repeat this finitely

many times. When we stop, we will end up with a matrix

A =







d 0 ··· 0

. C







and

divides every entry of

. Now we apply the entire process to

. When

we do this process, notice all allowed operations don’t change the fact that

divides every entry of C.

So applying this recursively, we obtain a diagonal matrix with the claimed

divisibility property.

Note that if we didn’t have to care about the divisibility property, we can

just do (i) and (ii), and we can get a diagonal matrix. The magic to get to the

Smith normal form is (iii).

Recall that the

are called the invariant factors. So it would be nice if we

can prove that the

are indeed invariant. It is not clear from the algorithm

that we will always end up with the same

. Indeed, we can multiply a whole

row by

−

1 and get different invariant factors. However, it turns out that these

are unique up to multiplication by units.

To study the uniqueness of the invariant factors of a matrix

, we relate

them to other invariants, which involves minors.

Definition (Minor). A

k ×k

minor of a matrix

is the determinant of a

k ×k

sub-matrix of

(i.e. a matrix formed by removing all but

rows and all but

columns).

Any given matrix has many minors, since we get to decide which rows and

columns we can throw away. The idea is to consider the ideal generated by all

the minors of matrix.

Definition (Fitting ideal). For a matrix

, the

th Fitting ideal

Fit

(

)

C R

the ideal generated by the set of all k × k minors of A.

A key property is that equivalent matrices have the same Fitting ideal, even

if they might have very different minors.

Lemma. Let A and B be equivalent matrices. Then

Fit

(A) = Fit

(B)

for all k.

Proof.

It suffices to show that changing

by a row or column operation does

not change the Fitting ideal. Since taking the transpose does not change the

determinant, i.e.

Fit

(

) =

Fit

(

), it suffices to consider the row operations.

The most difficult one is taking linear combinations. Let

be the result of

adding

times the

th row to the

th row, and fix

k ×k

minor of

. Suppose

the resultant matrix is C

. We then want to show that det C

∈ Fit

(A).

If the

th row is outside of

, then the minor

det C

is unchanged. If both

the

th and

th rows are in

, then the submatrix

changes by a row operation,

which does not affect the determinant. These are the boring cases.

Suppose the

th row is in

and the

th row is not. Suppose the

th row is

, ··· , f

. Then C is changed to C

, with the jth row being

+ cf

, C

+ cf

, ··· , C

+ cf

We compute det C

by expanding along this row. Then we get

det C

= det C + c det D,

where

is the matrix obtained by replacing the

th row of

with (

, ··· , f

The point is that

det C

is definitely a minor of

, and

det D

is still a minor of

, just another one. Since ideals are closed under addition and multiplications,

we know

det(C

) ∈ Fit

(A).

The other operations are much simpler. They just follow by standard properties

of the effect of swapping rows or multiplying rows on determinants. So after any

row operation, the resultant submatrix C

satisfies

det(C

) ∈ Fit

(A).

Since this is true for all minors, we must have

Fit

(B) ⊆ Fit

(A).

But row operations are invertible. So we must have

Fit

(A) ⊆ Fit

(B)

as well. So they must be equal. So done.

We now notice that if we have a matrix in Smith normal form, say

B =













then we can immediately read off

Fit

(B) = (d

···d

This is clear once we notice that the only possible contributing minors are from

the diagonal submatrices, and the minor from the top left square submatrix

divides all other diagonal ones. So we have

Corollary. If A has Smith normal form

B =













then

Fit

(A) = (d

···d

So d

is unique up to associates.

This is since we can find

by dividing the generator of

Fit

(

) by the

generator of Fit

k−1

(A).

Example. Consider the matrix in Z:

A =



2 0

0 3



This is diagonal, but not in Smith normal form. We can potentially apply the

algorithm, but that would be messy. We notice that

Fit

(A) = (2, 3) = (1).

So we know d

= ±1. We can then look at the second Fitting ideal

Fit

(A) = (6).

So d

= ±6. So we must have d

= ±6. So the Smith normal form is



1 0

0 6



That was much easier.

We are now going to use Smith normal forms to do things. We will need

some preparation, in the form of the following lemma:

Lemma. Let

be a principal ideal domain. Then any submodule of

generated by at most m elements.

This is obvious for vector spaces, but is slightly more difficult here.

Proof. Let N ≤ R

be a submodule. Consider the ideal

I = {r ∈ R : (r, r

, ··· , r

) ∈ N for some r

, ··· , r

∈ R}.

It is clear this is an ideal. Since

is a principle ideal domain, we must have

I = (a) for some a ∈ R. We now choose an

n = (a, a

, ··· , a

) ∈ N.

Then for any vector (

, r

, ··· , r

)

∈ N

, we know that

∈ I

. So

a | r

. So we

can write

= ra.

Then we can form

, r

, ··· , r

) − r(a, a

, ··· , a

) = (0, r

− ra

, ··· , r

− ra

) ∈ N.

This lies in

N ∩

(

{

} × R

m−1

)

≤ R

m−1

. Thus everything in

can

be written as a multiple of

plus something in

. But by induction, since

≤ R

m−1

, we know

is generated by at most

m −

1 elements. So there are

, ··· , n

∈ N

generating N

. So n, n

, ··· , n

generate N .

If we have a submodule of

, then it has at most

generators. However,

these might generate the submodule in a terrible way. The next theorem tells us

there is a nice way of finding generators.

Theorem. Let

be a Euclidean domain, and let

N ≤ R

be a submod-

ule. Then there exists a basis

, ··· , v

such that

is generated by

, d

, ··· , d

for some 0 ≤ r ≤ m and some d

∈ R such that

| d

| ··· | d

This is not hard, given what we’ve developed so far.

Proof.

By the previous lemma,

is generated by some elements

, ··· , x

with

n ≤ m

. Each

is an element of

. So we can think of it as a column vector

of length m, and we can form a matrix

A =





↑ ↑ ↑

··· x

↓ ↓ ↓





We’ve got an

m ×n

matrix. So we can put it in Smith normal form! Since there

are fewer columns than there are rows, this is of the form













Recall that we got to the Smith normal form by row and column operations.

Performing row operations is just changing the basis of

, while each column

operation changes the generators of N.

So what this tells us is that there is a new basis

, ··· , v

such

that

is generated by

, ··· , d

. By definition of Smith normal form, the

divisibility condition holds.

Corollary. Let

be a Euclidean domain. A submodule of

is free of rank

at most

. In other words, the submodule of a free module is free, and of a

smaller (or equal) rank.

Proof.

Let

N ≤ R

be a submodule. By the above, there is a basis

, ··· , v

such that

is generated by

, ··· , d

for

r ≤ m

. So it is certainly

generated by at most

elements. So we only have to show that

, ··· , d

are

independent. But if they were linearly dependent, then so would be

, ··· , v

But

, ··· , v

are a basis, hence independent. So

, ··· , d

generate

freely. So

∼

Note that this is not true for all rings. For example, (2

, X

)

C Z

[

] is a

submodule of Z[X], but is not isomorphic to Z[X].

Theorem (Classification of finitely-generated modules over a Euclidean domain).

Let R be a Euclidean domain, and M be a finitely generated R-module. Then

∼

)

⊕

)

⊕ ··· ⊕

)

⊕ R ⊕ R ⊕ ··· ⊕ R

for some d

6= 0, and

| d

| ··· | d

This is either a good or bad thing. If you are pessimistic, this says the world

of finitely generated modules is boring, since there are only these modules we

already know about. If you are optimistic, this tells you all finitely-generated

modules are of this simple form, so we can prove things about them assuming

they look like this.

Proof.

Since

is finitely-generated, there is a surjection

→ M

. So by

the first isomorphism, we have

∼

ker φ

Since

ker φ

is a submodule of

, by the previous theorem, there is a basis

, ··· , v

such that

ker φ

is generated by

, ··· , d

for 0

≤ r ≤ m

and d

| d

| ··· | d

. So we know

∼

((d

, 0, ··· , 0), (0, d

, 0, ··· , 0), ··· , (0, ··· , 0, d

, 0, ··· , 0))

This is just

)

⊕

)

⊕ ··· ⊕

)

⊕ R ⊕ ··· ⊕ R,

with m − r copies of R.

This is particularly useful in the case where

, where

-modules are

abelian groups.

Example. Let A be the abelian group generated by a, b, c with relations

2a + 3b + c = 0,

a + 2b = 0,

5a + 6b + 7c = 0.

In other words, we have

A =

((2, 3, 1), (1, 2, 0), (5, 6, 7))

We would like to get a better description of

. It is not even obvious if this

module is the zero module or not.

To work out a good description, We consider the matrix

X =





2 1 5

3 2 6

1 0 7





To figure out the Smith normal form, we find the fitting ideals. We have

Fit

(X) = (1, ···) = (1).

So d

= 1.

We have to work out the second fitting ideal. In principle, we have to check

all the minors, but we immediately notice



2 1

3 2



= 1.

So Fit

(X) = (1), and d

= 1. Finally, we find

Fit

(X) =









2 1 5

3 2 6

1 0 7









= (3).

So d

= 3. So we know

∼

(1)

⊕

(1)

⊕

(3)

∼

(3)

∼

If you don’t feel like computing determinants, doing row and column reduction

is often as quick and straightforward.

We re-state the previous theorem in the specific case where

, since this

is particularly useful.

Corollary (Classification of finitely-generated abelian groups). Any finitely-

generated abelian group is isomorphic to

× ··· × C

× C

∞

× ··· × C

∞

where C

∞

∼

Z is the infinite cyclic group, with

| d

| ··· | d

Proof.

Let

, and apply the classification of finitely generated

-modules.

Note that if the group is finite, then there cannot be any

∞

factors. So it

is just a product of finite cyclic groups.

Corollary. If A is a finite abelian group, then

∼

× ··· × C

with

| d

| ··· | d

This is the result we stated at the beginning of the course.

Recall that we were also to decompose a finite abelian group into products of

the form

, where

is a prime, and we said it was just the Chinese remainder

theorem. This is again in general true, but we, again, need the Chinese remainder

theorem.

Lemma (Chinese remainder theorem). Let

be a Euclidean domain, and

a, b ∈ R be such that gcd(a, b) = 1. Then

(ab)

∼

(a)

(b)

as R-modules.

The proof is just that of the Chinese remainder theorem written in ring

language.

Proof. Consider the R-module homomorphism

φ :

(a)

(b)

→

(ab)

+ (a), r

+ (b)) 7→ br

+ ar

+ (ab).

To show this is well-defined, suppose

+ (a), r

+ (b)) = (r

+ (a), r

+ (b)).

Then

= r

+ xa

= r

+ yb.

+ ar

+ (ab) = br

+ xab + ar

+ yab + (ab) = br

+ ar

+ (ab).

So this is indeed well-defined. It is clear that this is a module map, by inspection.

We now have to show it is surjective and injective. So far, we have not used

the hypothesis, that

gcd

(

a, b

) = 1. As we know

gcd

(

a, b

) = 1, by the Euclidean

algorithm, we can write

1 = ax + by

for some x, y ∈ R. So we have

φ(y + (a), x + (b)) = by + ax + (ab) = 1 + (ab).

So 1 ∈ im φ. Since this is an R-module map, we get

φ(r(y + (a), x + (b))) = r · (1 + (ab)) = r + (ab).

The key fact is that

(

) as an

-module is generated by 1. Thus we know

is surjective.

Finally, we have to show it is injective, i.e. that the kernel is trivial. Suppose

φ(r

+ (a), r

+ (b)) = 0 + (ab).

Then

+ ar

∈ (ab).

So we can write

+ ar

= abx

for some

x ∈ R

. Since

a | ar

and

a | abx

, we know

a | br

. Since

and

are

coprime, unique factorization implies a | r

. Similarly, we know b | r

+ (a), r

+ (b)) = (0 + (a), 0 + (b)).

So the kernel is trivial.

Theorem (Prime decomposition theorem). Let

be a Euclidean domain, and

M be a finitely-generated R-module. Then

∼

⊕ N

⊕ ··· ⊕ N

where each N

is either R or is R/(p

) for some prime p ∈ R and some n ≥ 1.

Proof. We already know

∼

)

⊕ ··· ⊕

)

⊕ R ⊕ ··· ⊕ R.

So it suffices to show that each R/(d

) can be written in that form. We let

d = p

···p

with

distinct primes. So each

is coprime to each other. So by the lemma

iterated, we have

)

∼

)

⊕ ··· ⊕

)

3.4 Modules over F[X] and normal forms for matrices

That was one promise delivered. We next want to consider the Jordan normal

form. This is less straightforward, since considering

directly as an

module

would not be too helpful (since that would just be pure linear algebra). Instead,

we use the following trick:

For a field

, the polynomial ring

[

] is a Euclidean domain, so the results

of the last few sections apply. If

is a vector space on

, and

V → V

is a

linear map, then we can make V into an F[X]-module via

F[X] × V → V

(f, v) 7→ (f(α))(v).

We write V

for this F[X]-module.

Lemma. If

is a finite-dimensional vector space, then

is a finitely-generated

F[X]-module.

Proof.

If v

, ··· ,

generate

as an

-module, i.e. they span

as a vector

space over

, then they also generate

as an

[

]-module, since

F ≤ F

[

Example. Suppose

∼

[

]

(

) as

[

]-modules. Then in particular

they are isomorphic as

-modules (since being a map of

-modules has fewer

requirements than being a map of F[X]-modules).

Under this bijection, the elements 1

, X, X

, ··· , X

r−1

∈ F

[

]

(

) form a

vector space basis for

. Viewing

[

]

(

) as an

-vector space, the action

of X has the matrix







0 0 ··· 0 0

1 0 ··· 0 0

0 1 ··· 0 0

0 0 ··· 1 0







We also know that in V

, the action of X is by definition the linear map α. So

under this basis, α also has matrix







0 0 ··· 0 0

1 0 ··· 0 0

0 1 ··· 0 0

0 0 ··· 1 0







Example. Suppose

∼

F[X]

((X − λ)

)

for some λ ∈ F. Consider the new linear map

β = α − λ · id : V → V.

Then

∼

[

]

(

), for

X − λ

. So there is a basis for

so that

looks







0 0 ··· 0 0

1 0 ··· 0 0

0 1 ··· 0 0

0 0 ··· 1 0







So we know α has matrix







λ 0 ··· 0 0

1 λ ··· 0 0

0 1 ··· 0 0

0 0 ··· 1 λ







So it is a Jordan block (except the Jordan blocks are the other way round, with

zeroes below the diagonal).

Example. Suppose V

∼

F[X]/(f) for some polynomial f, for

f = a

+ a

X + ··· + a

r−1

+ X

This has a basis 1, X, X

, ··· , X

r−1

as well, in which α is

c(f) =







0 0 ··· 0 −a

1 0 ··· 0 −a

0 1 ··· 0 −a

0 0 ··· 1 −a

r−1







We call this the companion matrix for the monic polynomial f .

These are different things that can possibly happen. Since we have already

classified all finitely generated

[

] modules, this allows us to put matrices in a

rather nice form.

Theorem (Rational canonical form). Let

V → V

be a linear endomorphism

of a finite-dimensional vector space over

, and

be the associated

[

]-module.

Then

∼

F[X]

)

⊕

F[X]

)

⊕ ··· ⊕

F[X]

)

with

| f

| ··· | f

. Thus there is a basis for

in which the matrix for

the block diagonal







c(f

) 0 ··· 0

0 c(f

) ··· 0

0 0 ··· c(f

)







This is the sort of theorem whose statement is longer than the proof.

Proof.

We already know that

is a finitely-generated

[

]-module. By the

structure theorem of F[X]-modules, we know

∼

F[X]

)

⊕

F[X]

)

⊕ ··· ⊕

F[X]

)

⊕ 0.

We know there are no copies of

[

], since

is finite-dimensional over

, but

[

] is not. The divisibility criterion also follows from the structure

theorem. Then the form of the matrix is immediate.

This is really a canonical form. The Jordan normal form is not canonical,

since we can move the blocks around. The structure theorem determines the

factors

up to units, and once we require them to be monic, there is no choice

left.

In terms of matrices, this says that if

is represented by a matrix

A ∈ M

n,n

(

)

in some basis, then A is conjugate to a matrix of the form above.

From the rational canonical form, we can immediately read off the minimal

polynomial as

. This is since if we view

as the decomposition above, we

find that

(

) kills everything in

F[X]

)

. It also kills the other factors since

| f

for all

. So

(

) = 0. We also know no smaller polynomial kills

, since it

does not kill

F[X]

)

Similarly, we find that the characteristic polynomial of α is f

···f

Recall we had a different way of decomposing a module over a Euclidean

domain, namely the prime decomposition, and this gives us the Jordan normal

form.

Before we can use that, we need to know what the primes are. This is why

we need to work over C.

Lemma. The prime elements of

[

] are the

X − λ

for

λ ∈ C

(up to multipli-

cation by units).

Proof.

Let

f ∈ C

[

]. If

is constant, then it is either a unit or 0. Otherwise, by

the fundamental theorem of algebra, it has a root

. So it is divisible by

X − λ

So if

is irreducible, it must have degree 1. And clearly everything of degree 1

is prime.

Applying the prime decomposition theorem to

[

]-modules gives us the

Jordan normal form.

Theorem (Jordan normal form). Let

V → V

be an endomorphism of a

vector space V over C, and V

be the associated C[X]-module. Then

∼

C[X]

((X − λ

)

⊕

C[X]

((X − λ

)

⊕ ··· ⊕

C[X]

((X − λ

)

where

∈ C

do not have to be distinct. So there is a basis of

in which

has

matrix







(λ

) 0

(λ

)

0 J

(λ

)







where

(λ) =







λ 0 ··· 0

1 λ ··· 0

0 ··· 1 λ







is an m × m matrix.

Proof.

Apply the prime decomposition theorem to

. Then all primes are of

the form

X −λ

. We then use our second example at the beginning of the chapter

to get the form of the matrix.

The blocks

(

) are called the Jordan

-blocks. It turns out that the Jordan

blocks are unique up to reordering, but it does not immediately follow from what

we have so far, and we will not prove it. It is done in the IB Linear Algebra

course.

We can also read off the minimal polynomial and characteristic polynomial

of α. The minimal polynomial is

(X − λ)

where

is the size of the largest

-block. The characteristic polynomial of

(X − λ)

where b

is the sum of the sizes of the λ-blocks. Alternatively, it is

i=1

(X − λ

)

From the Jordan normal form, we can also read off another invariant, namely

the size of the λ-space of α, namely the number of λ-blocks.

We can also use the idea of viewing

as an

[

] module to prove Cayley-

Hamilton theorem. In fact, we don’t need F to be a field.

Theorem (Cayley-Hamilton theorem). Let

be a finitely-generated

-module,

where

is some commutative ring. Let

M → M

be an

-module homomor-

phism. Let

be a matrix representation of

under some choice of generators,

and let p(t) = det(tI − A). Then p(α) = 0.

Proof. We consider M as an R[X]-module with action given by

(f(X))(m) = f(α)m.

Suppose e

, ··· , e

span M , and that for all i, we have

α(e

) =

j=1

Then

j=1

(Xδ

− a

= 0.

We write C for the matrix with entries

= Xδ

− a

∈ F[X].

We now use the fact that

adj(C)C = det(C)I,

which we proved in IB Linear Algebra (and the proof did not assume that the

underlying ring is a field). Expanding this out, we get the following equation (in

F[X]).

(X)I = det(XI − A)I = (adj(XI − A))(XI − A).

Writing this in components, and multiplying by e

, we have

(X)δ

j=1

(adj(XI − A)

)(Xδ

− a

Then for each i, we sum over k to obtain

k=1

(X)δ

j,k=1

(adj(XI − A)

)(Xδ

− a

= 0,

by our choice of

. But the left hand side is just

(

)

. So

(

) acts

trivially on all of the generators

. So it in fact acts trivially. So

(

) is the

zero map (since acting by X is the same as acting by α, by construction).

Note that if we want to prove this just for matrices, we don’t really need the

theory of rings and modules. It just provides a convenient language to write the

proof in.

3.5 Conjugacy of matrices*

We are now going to do some fun computations of conjugacy classes of matrices,

using what we have got so far.

Lemma. Let

α, β

V → V

be two linear maps. Then

∼

[

]-modules

if and only if

and

are conjugate as linear maps, i.e. there is some

V → V

such that α = γ

−1

βγ.

This is not a deep theorem. This is in some sense just some tautology. All

we have to do is to unwrap what these statements say.

Proof.

Let

→ V

be an

[

]-module isomorphism. Then for v

∈ V

, we

notice that β(v) is just X · v in V

, and α(v) is just X · v in V

. So we get

β ◦ γ(v) = X · (γ(v)) = γ(X · v) = γ ◦ α(v),

using the definition of an F[X]-module homomorphism.

So we know

βγ = γα.

α = γ

−1

βγ.

Conversely, let

V → V

be a linear isomorphism such that

−1

βγ

. We

now claim that

→ V

is an

[

]-module isomorphism. We just have to

check that

γ(f · v ) = γ(f(α)(v))

= γ(a

+ a

α + ··· + a

)(v)

= γ(a

v) + γ(a

α(v)) + γ(a

(v)) + ··· + γ(a

(v))

= (a

+ a

β + a

+ ··· + a

)(γ(v))

= f · γ(v).

So classifying linear maps up to conjugation is the same as classifying modules.

We can reinterpret this a little bit, using our classification of finitely-generated

modules.

Corollary. There is a bijection between conjugacy classes of

n × n

matrices

over

and sequences of monic polynomials

, ··· , d

such that

| d

| ··· | d

and deg(d

···d

) = n.

Example. Let’s classify conjugacy classes in

(

), i.e. we need to classify

F[X]-modules of the form

F[X]

)

⊕

F[X]

)

⊕ ··· ⊕

F[X]

)

which are two-dimensional as

-modules. As we must have

deg

(

···d

) = 2,

we either have a quadratic thing or two linear things, i.e. either

(i) r = 1 and deg(d

) = 2,

(ii) r

= 2 and

deg

(

) =

deg

(

) = 1. In this case, since we have

| d

, and

they are both monic linear, we must have d

= d

= X − λ for some λ.

In the first case, the module is

F[X]

)

where, say,

= X

+ a

X + a

In the second case, we get

F[X]

(X − λ)

⊕

F[X]

(X − λ)

What does this say? In the first case, we use the basis 1

, X

, and the linear map

has matrix



0 −a

1 −a



In the second case, this is



λ 0

0 λ



Do these cases overlap? Suppose the two of them are conjugate. Then they have

the same determinant and same trace. So we know

−a

= 2λ

= λ

So in fact our polynomial is

+ a

X + a

= X

− 2λ + λ

= (X − λ)

This is just the polynomial of a Jordan block. So the matrix



0 −a

1 −a



is conjugate to the Jordan block



λ 0

1 λ



but this is not conjugate to

λI

, e.g. by looking at eigenspaces. So these cases

are disjoint.

Note that we have done more work that we really needed, since

λI

is invariant

under conjugation.

But the first case is not too satisfactory. We can further classify it as follows.

If X

+ a

X + a

is reducible, then it is

(X − λ)(X − µ)

for some µ, λ ∈ F. If λ = µ, then the matrix is conjugate to



λ 0

1 λ



Otherwise, it is conjugate to



λ 0

0 µ



In the case where

is irreducible, there is nothing we can do

in general. However, we can look at some special scenarios and see if there is

anything we can do.

Example. Consider

(

3). We want to classify its conjugacy classes. By

the general theory, we know everything is conjugate to



λ 0

0 µ





λ 0

1 λ





0 −a

1 −a



with

+ a

X + a

irreducible. So we need to figure out what the irreducibles are.

A reasonable strategy is to guess. Given any quadratic, it is easy to see if it

is irreducible, since we can try to see if it has any roots, and there are just three

things to try. However, we can be a bit slightly more clever. We first count how

many irreducibles we are expecting, and then find that many of them.

There are 9 monic quadratic polynomials in total, since

, a

∈ Z/

3. The

reducibles are (

X − λ

)

or (

X − λ

)(

X − µ

) with

λ 6

. There are three of each

kind. So we have 6 reducible polynomials, and so 3 irreducible ones.

We can then check that

+ 1, X

+ X + 2, X

+ 2X + 2

are the irreducible polynomials. So every matrix in

(

3) is either congruent



0 −1

1 0





0 −2

1 −1





0 −2

1 −2





λ 0

0 µ





λ 0

1 λ



where

λ, µ ∈

(

(since the matrix has to be invertible). The number of

conjugacy classes of each type are 1

2. So there are 8 conjugacy classes.

The first three classes have elements of order 4, 8, 8 respectively, by trying. We

notice that the identity matrix has order 1, and



λ 0

0 µ



has order 2 otherwise. Finally, for the last type, we have

ord



1 0

1 1



= 3, ord



2 0

1 2



= 6

Note that we also have

|GL

(Z/3)| = 48 = 2

· 3.

Since there is no element of order 16, the Sylow 2-subgroup of

(

3) is not

cyclic.

To construct the Sylow 2-subgroup, we might start with an element of order

8, say

B =



0 1

1 2



To make a subgroup of order 6, a sensible guess would be to take an element of

order 2, but that doesn’t work, since

will give you the element of order 2.

Instead, we pick

A =



0 2

1 0



We notice

−1

BA =



0 1

2 0



0 1

1 2



0 2

1 0





1 2

0 2



0 2

1 0





2 2

2 0



= B

So this is a bit like the dihedral group.

We know that

hBi C hA, Bi.

Also, we know

|hBi|

= 8. So if we can show that

hBi

has index 2 in

hA, Bi

, then

this is the Sylow 2-subgroup. By the second isomorphism theorem, something

we have never used in our life, we know

hA, Bi

hBi

∼

hAi

hAi ∩ hBi

We can list things out, and then find

hAi ∩ hBi =



2 0

0 2



∼

We also know hAi

∼

. So we know

|hA, Bi|

|hBi|

= 2.

So |hA, Bi| = 16. So this is the Sylow 2-subgroup. in fact, it is

hA, B | A

= B

= e, A

−1

BA = B

We call this the semi-dihedral group of order 16, because it is a bit like a dihedral

group.

Note that finding this subgroup was purely guesswork. There is no method

to know that A and B are the right choices.