MATH5735 Modules and Representation Theory Lecture …web.maths.unsw.edu.au/~danielch/modules12/beeren_notes.pdf · 2013-09-06 · Joel Beeren Modules Lecture Notes (2) Ais naturally

MATH5735 Modules and Representation Theory

Lecture Notes

Joel Beeren

Semester 1, 2012

Contents

1 Why study modules? 41.1 Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 How do you study modules? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Rings and Algebras 42.1 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 Some examples of Algebras 63.1 Free Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Weyl Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4 Module Basics 94.1 Submodules and Quotient Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

5 Module Homomorphisms 115.1 Isomorphism Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125.2 Universal property of quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

6 Direct Sums and Products 146.1 Canonical Injections and Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146.2 Universal Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156.3 Decomposability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

7 Free Modules and Generators 167.1 Free Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167.2 Generating Submodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

8 Modules over a PID 188.1 Structure theorem for modules over a PID . . . . . . . . . . . . . . . . . . . . . . . . . 198.2 Elementary Row and Column Operations . . . . . . . . . . . . . . . . . . . . . . . . . 20

9 Proof of Structure Theorem 20

10 Applications of the Structure Theorem 2210.1 Endomorphism Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2310.2 Jordan Canonical Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1

Joel Beeren Modules Lecture Notes

11 Exact Sequences and Splitting 2411.1 Exact Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2411.2 Splitting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2511.3 Idempotents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

12 Chain Conditions 2712.1 Submodules and Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2712.2 Finite Generation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

13 Noetherian Rings 2913.1 Almost Normal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2913.2 Hilbert’s Basis Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

14 Composition Series 3114.1 Simple Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3114.2 Composition series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3114.3 Jordan Holder Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

15 Semisimple Modules 33

16 Semisimple rings and Group Representations 3516.1 Group Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

17 Maschke’s Theorem 3817.1 Reynold’s Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3817.2 HomR (M,N) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

18 Wedderburn’s Theorem 4118.1 Schur’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4118.2 Structure Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4218.3 Module Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4318.4 Finite dimensional Semisimple Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 44

19 One-dimensional Representations 4419.1 Abelianisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4419.2 One-dimensional representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

20 Centres of Group Algebras 4720.1 Centres of Semisimple Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4720.2 Centres of group algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4820.3 Irreducible representations of dihedral groups . . . . . . . . . . . . . . . . . . . . . . . 49

21 Categories and Functors 5021.1 Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5121.2 Hom functor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

22 Tensor Products I 5322.1 Construction of Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5322.2 Universal property of tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . 5322.3 Functoriality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5422.4 Bimodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

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23 Tensor Products II 5523.1 Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5523.2 Distributive Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5623.3 Case when R is commutative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5723.4 Tensor Products of Group Representations . . . . . . . . . . . . . . . . . . . . . . . . . 58

24 Characters 5824.1 Linear Algebra Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5824.2 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5924.3 Character Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6024.4 Dual Modules and Contragredient Representation . . . . . . . . . . . . . . . . . . . . . 6024.5 Hom as tensor product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

25 Orthogonality 6125.1 Characters of ⊕,⊗ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6125.2 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6225.3 Decomposing kG-modules into a direct sum of simples . . . . . . . . . . . . . . . . . . 63

26 Example from Harmonic Motion 64

27 Adjoint Associativity and Induction 6427.1 Adjoint Associativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6427.2 Induced Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

28 Annihilators and the Jacobson Radical 6628.1 Restriction Functor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6628.2 Annihilators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6728.3 Jacobson Radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

29 Nakayama Lemma and the Wedderburn-Artin Theorem 6929.1 NAK Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6929.2 Properties of the Jacobson Radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7029.3 Wedderburn-Artin Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

30 Radicals and Artinian Rings 7130.1 Nilpotence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7130.2 DCC implies ACC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7230.3 Computing Some Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

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1 Why study modules?

Modules appear all over mathematics but it is good to keep the following setup in mind. This ariseswhen we have symmetry in a linear context.

1.1 Setup

BIG EXAMPLE HERE (seems unnecessary)

1.2 How do you study modules?

We take our inspiration from linear algebra and study vector spaces. The key theorem of vectorspacesis that any finitely spanned vector space V over the field k has form

V ∼= k ⊕ k ⊕ · · · ⊕ k.

We seek a similar result for modules – that is, can you write some “finite” modules as a direct sum ofones which are easy to understand? If not, how do you analyse them?

2 Rings and Algebras

2.1 Rings

Definition 2.1. A ring R is

(1) An abelian group with group addition denoted + and group zero 0; equipped with

(2) an associative multiplication map µ : R×R→ R, (r, r′) 7→ rr′ satisfying

(a) (distributive law): for r, r′, s ∈ R we have

(r + r′)s = rs+ r′s, and

s(r + r′) = sr + sr′.

(b) (multiplicative identity) there is an element 1R = 1 ∈ R such that

1r = r1 = r ∀r ∈ R.

We say R is commutative if multiplication is commutative (that is, rs = sr for all r, s ∈ R).The group of units is

R∗ = {r ∈ R | ∃s ∈ R.sr = 1 = rs} .

If R∗ = R \ {0} we say R is a division ring, and further if R is commutative then it is a field.

Example 2.1. R,C,Q,Q(i) are fields

Example 2.2. Z,Z(i) are rings

Example 2.3. For k a field, R a ring, R[x1, . . . , xn] is the polynomial ring in n variables over R, andk(x1, . . . , xn) is the field of rational functions in n variables.

Furthermore, Mn(R) is the ring of n×n matrices over R, and if V is a vector space then Endk (V ),the set of linear maps V → V is a ring with pointwise addition and composition for multiplication.

Definition 2.2. Let R be a ring. A subgroup S of the underlying additive group of R is

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(1) a subring if

• 1R ∈ S; and

• for s, s′ ∈ S, we have ss′ ∈ S.

(2) an (two-sided) ideal if for all r ∈ R, s ∈ S, we have sr, rs ∈ S.

Example 2.4. Let R be a ring. We have the opposite ring Rop where

Rop = {r◦ | r ∈ R}

which has the same addition as in R but r◦s◦ = (sr)◦.

Exercise. Check the ring axioms for Rop.

Definition 2.3. Let R be a ring. The centre of R is

Z(R) = {z ∈ R | zr = rz ∀r ∈ R} .

Exercise. Check that Z(R) is a subring of R.

Fact 2.1. Let I CR be an ideal of R. Then there is a quotient ring R/I with

• addition: (r + I) + (r′ + I) = (r + r′) + I; and

• multiplication: (r + I)(r′ + I) = rr′ + I.

Definition 2.4. A map of rings ϕ : R→ S is a ring homomorphism if (for all r, r′ ∈ R)

(1) ϕ(r + r′) = ϕ(r) + ϕ(r′)

(2) ϕ(rr′) = ϕ(r)ϕ(r′)

(3) ϕ(1R) = 1S

Fact 2.2. Let ϕ : R→ S be a ring homomorphism. Then

• im(ϕ) ⊆ S is a subring

• kerϕCR

2.2 Algebras

Fix a commutative ring R.

Definition 2.5. An R-algebra (also called an algebra over R) is a ring A equipped with a ringhomomorphism ι : R→ Z(A).

An R-subalgebra is a subring B of A with Z(R) ⊆ Z(B) (so B is naturally a subalgebra).

Example 2.5. The following are R-algebras:

(1) R[x1, . . . , xn] – here we take ι : R→ R[x1, . . . , xn] as the map r 7→ r + 0x1 + · · ·+ 0xn

(2) Mn(R) – take ι : R→Mn(R) by r 7→ rIn.

Proposition 2.1. Let A 6= 0 be an algebra over a field k.

(1) The unit map ι : k → A is injective, so we will often consider k as a subring of A by identifyingit with ι(k)

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(2) A is naturally a vector space over k.

Proof.

(1) k is a field, so ker ι = k or 0. We know that ι(1k) = 1A so ker ι = 0 and hence ι is injective.

(2) We set vector addition to be the same as ring addition, and scalar multiplication the same asring multiplication by elements of k identified with ι(k). The vector space axioms then followfrom the ring axioms.

Definition 2.6. A ring homomorphism ϕ : A→ B of k-algebras is a k-algebra homomorphism ifit is k-linear.

Definition 2.7. Let G be a group and k a field. Define the vector space

kG :=⊕σ∈G

kσ

of all formal (finite) linear combinations of elements of G. Then kG is a k-algebra called the groupalgebra of G with ring multiplication induced by group multiplication in the following sense:(∑

αgigi

)(∑βgjgj

)=∑g∈G

∑gigj=g

αgiβgjg

.

Exercise. Check the k-algebra axioms for kG.

Exercise. For the cyclic group G = {1, σ} and field k with char k 6= 2, show that there is a k-algebraisomorphism

kG −→ k × k1 7→ (1, 1)

σ 7→ (1,−1)

Exercise. Generalise the previous to larger cyclic groups

Exercise. Find a noncommutative ring R with R ∼= Rop.

3 Some examples of Algebras

Fix a commutative ring R.

Definition 3.1. Let A1, A2 be R-algebras with unit maps ιj : R → Aj for j = 1, 2. An R-algebrahomomorphism (resp. isomorphism) ϕ : A1 → A2 is a ring homomorphism (resp. isomorphism) suchthat the following diagram commutes

R

ι1~~}}}}

}}}

ι2

AAA

AAAA

A1 ϕ// A2

(that is, ι2 = ϕ ◦ ι1).

Example 3.1. C is both an R-algebra and a C-algebra (with ι defined as inclusion). Now, theconjugation map

ϕ : C→ C; z 7→ z

is an R-algebra isomorphism but not a C-algebra isomorphism (as ι2 = ϕ ◦ ι1 ⇔ z = z for all z ∈ R).

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3.1 Free Algebra

Let {xi}i∈I be a set of noncommuting variables that commute with scalars in R. Let R〈xi〉i∈I denotethe set of noncommutative polynomials in the xi’s with coefficients in R. That is, expressions of theform ∑

n≥0

∑i1,...,in∈I

ri1ri2 . . . rinxi1xi2 . . . xin , rij ∈ R ∀j

where all but finitely many ri1 , . . . , rin are zero.If I = {1, 2, . . . , n}, we write R〈xi〉i∈I = R〈x1, x2, . . . , xn〉.

Proposition 3.1. R〈xi〉i∈I is an R-algebra when endowed with

(1) ring addition: just add coefficients

(2) ring multiplication: induced by concatenation of monomials

(3) unit map: R→ R〈xi〉i∈I with r 7→ r, the constant polynomial.

This is called the free R-algebra on {xi}i∈I .

Proof. Elementary but long and tedious.

Example 3.2. A = Z〈x, y〉. See

(2x+ xy)(3yx− y2x) = 6xyx− 2xy2x+ 2xy2x− xy3x

= 6xyx+ xy2x− xy3x.

Exercise. For the dihedral group Dn, k a field, show that

kDn∼=

k〈x, y〉〈x2 − 1, yn − 1, yx− xyn−1〉

.

3.2 Weyl Algebra

Let k be a field, V a vector space. Recall that Endk (V ) is the ring of linear maps T : V → V . ThenEndk (V ) is also a k-algebra. How? Note that α idV , the scalar multiplication by α map is an elementof Endk (V ), so we have the unit map

ι : k → Endk (V ) ; α 7→ α idV .

We need to check that α idV ∈ Z(Endk (V )) but this is obvious.

Definition 3.2. A differental operator with polynomial coefficients is a C-linear map of theform

D =n∑i=0

pi(x)∂i : C[x] −→ C[x]

f(x) 7→n∑i=0

pi(x)dif

dxi

where pi(x) ∈ C[x] for all i.

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We denote the set of these operators by A1 (note thatD is an element of the C-algebra EndC (C[x])).Furthermore, we see that

∂x = x∂ + 1

Why? For p(x) ∈ C[x], we have

(∂x)(p(x)) =d

dx(xp(x))

= xdp

dx+ p(x)

= (x∂ + 1)(p(x)).

Proposition 3.2. A1 is a C-subalgebra of EndC (C[x]). It is called the (first) Weyl algebra.

Proof. Note that C ⊂ A1, so we check closure axioms. A1 is clearly closed under addition, and contains0 and 1. Furthermore, any D ∈ A1 has the form

N∑i,j=0

αijxi∂j , αij ∈ C

so by the distributive law, it suffices to show (exercise)

xi1∂j1xi2∂j2 ∈ A1.

However, using the Weyl relation we can push all of the ∂’s to the right of all x’s to see that this holdstrue.

Proposition 3.3. There is a C-algebra isomorphism

ϕ :C〈x, y〉

〈yx− xy − 1〉−→ A1 = C〈x, ∂〉

x+ 〈yx− xy − 1〉 7→ x

y + 〈yx− xy − 1〉 7→ ∂.

Proof. Certainly there is a “substitution homomorphism” ϕ : C〈x, y〉 → A1 which maps x 7→ x andy 7→ ∂; and more generally p(x, y) 7→ p(x, ∂). By the Weyl relation, ker ϕ 3 yx − xy − 1 and soker ϕ ⊇ 〈yx− xy − 1〉 = I.

By the universal property of quotients, there is an induced ring homomorphism

ϕ :C〈x, y〉I

→ A1.

This is clearly surjective. We now check that kerϕ = 0. By the proof of the previous proposition, anynonzero element in A1 can be written in the form

D =N∑j=n

pj(x)yj

with say pn(x) 6= 0. Then

[ϕ(D)](xn) =

N∑j=n

pj(x)∂j

(xn)

= pn(x)dnxn

dxn+ . . .

= n!pn(x) 6= 0.

So ϕ(D) 6= 0 and therefore kerϕ = 0.

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Exercise. The quaternions can be represented as

H :=R〈i, j〉

〈i2 − 1, j2 − 1, ij + ji〉.

Show that there is an R-algebra homomorphism H→M2(C) such that

i 7→(i 00 −i

)j 7→

(0 1−1 0

).

Hence or otherwise show that H is a division ring.

4 Module Basics

Fix a ring R.

Definition 4.1. A (right) R-module is an additive group M equipped with a scalar multiplicationmap M ×R→M : (m, r) 7→ mr such that the following axioms hold (for all m,m′ ∈M , r, r′ ∈ R)

(1) m1 = m

(2) (associativity) (mr)r′ = m(rr′)

(3) (distributivity) (m+m′)r = mr +m′r and m(r + r′) = mr +mr′

Remark. We can similarly define left R-modules.

Example 4.1. For a field k, a right k-module is just a vector space over k with scalars on the right.Here module axioms are equivalent to vector space axioms.

Example 4.2. R itself is a left and right R-module with module addition and scalar multiplicationequal to ring addition and multiplication respectively. Note that the module axioms are a consequenceof the ring axioms in this case.

Exercise. For any R-module M and m ∈M we have

(1) m0 = 0

(2) m(−1) = −m.

The following tells us that Z-modules are “just” abelian groups.

Proposition 4.1. Let M be an abelian group. There is a unique Z-module structure on it extendingthe additive group structure.

Proof. (Sketch) We show the Z-module structure is unique by showing how the module axioms com-pletely determine scalar multiplication by n ∈ Z. If n > 0 then axioms (1) and (3) imply that form ∈M ,

mn = m(1 + 1 + · · ·+ 1)

= m1 +m1 + · · ·+m1

= m+m+ · · ·+m.

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For n = 0, we have m0 = 0. Also, if n < 0 then

mn = m((−n)(−1))

= (m(−n))(−1)

= −(m+ · · ·+m)

= −m−m · · · −m.

Exercise. Check that this actually gives a Z-module structure.

Exercise. Explain and prove the following statement: “A right R-module is just a left Rop-module”.In particular, if R is commutative, so there is a natural isomorphism R ↔ Rop then left and rightmodules are the same thing.

Example 4.3. Let S = Mn(R), and let Rn be the set of n-tuples (i.e. row vectors) with entries inR. Then Rn is a right S-module with module addition and scalar multiplication defined by matrixoperations.

Exercise. Check the module axioms.

Remark. Similarly, with Rn being column vectors, then Rn is a left S-module.

We can “change scalars” with

Proposition 4.2. Let ϕ : S → R be a ring homomorphism, M a right R-module. Then M is naturallya right S module with the same additive structure but multiplication defined as

ms := mϕ(s).

We denote the S-module by MS to distinguish it from the original.

Proof. This is an exercise in checking axioms. For example, let m ∈M , s, s′ ∈ S and see

m(s+ s′) = mϕ(s+ s′) = m(ϕ(s) + ϕ(s′)) = mϕ(s) +mϕ(s′) = ms+ms′.

Corollary 4.1. For R commutative, any R-algebra A is also a left and right R-module.

Proof. Follows from proposition 4.2 and example 4.2.

4.1 Submodules and Quotient Modules

Proposition 4.3. Let M be an R-module. An R-submodule is an additive subgroup N of M whichis closed under scalar multiplication. In this case, N is an R-module and we write N ≤ M . A rightideal of R is a submodule of the right module RR, whereas a left ideal of R is a submodule of theleft module RR.

Exercise. Let S = M2(R). Then(R 0R 0

)is a left ideal but not a right ideal.

Remark. An ideal is a subset which is both a left and right ideal.

Exercise. The intersection of submodules is a submodule.

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Proposition 4.4. Let N be a submodule of M . The quotient abelian group M/N is naturally anR-module with scalar multiplication

(m+N)r := mr +N.

Proof. First note that multiplication is well defined for if n ∈ N we have

(m+ n+N)r = (m+ n)r +N

= mr + nr +N

= mr +N

since nr ∈ N . It remains to check the module axioms.

Exercise. Let R be the Weyl algebra A1 = C〈x, ∂〉. M = C[x] is a left R-module with the usualaddition and scalar multiplication defined as(∑

i

pi(x)∂i

)p(x) :=

∑i

pi(x)dip(x)

dxi.

5 Module Homomorphisms

Fix a ring R.

Definition 5.1. Let M,N be R-modules. A function ϕ : M → N is an R-module homomorphismif it is a homomorphism of abelian groups such that (for all m ∈M, r ∈ R)

ϕ(mr) = ϕ(m)r.

In this case we also say ϕ is R-linear. We denote the set of these as HomR (M,N).

Example 5.1. k-module homomorphisms are just k-linear maps.

Proposition 5.1. HomR (M,N) is an abelian group endowed with group addition

(ϕ1 + ϕ2)(m) = ϕ1(m) + ϕ2(m).

Proof. Note that addition is well defined as ϕ1, ϕ2 are R-linear. Indeed it is additive and for m ∈M, r ∈ R we have

(ϕ1 + ϕ2)(mr) = ϕ1(mr) + ϕ2(mr)

= ϕ1(m)r + ϕ2(m)r

= (ϕ1(m) + ϕ2(m))r

= (ϕ1 + ϕ2)(m)r.

Exercise. Check group axioms.

Proposition 5.2. For any R-module M , there is the following isomorphism of abelian groups

Φ : M −→ HomR (RR,M)

m 7→ (λm : r 7→ mr)

with inverseΨ : ϕ 7→ ϕ(1).

Page 11 of 73


Proof. Check λm is R-linear so Φ is well defined. The distributive law implies that

λm(r + r′) = m(r + r′) = mr +mr′ = λm(r) + λm(r′).

Associativity gives usλm(rr′) = m(rr′) = (mr)r′ = λm(r)r′.

Φ is additive by the other distributive law.It now suffices to check that Ψ is the inverse of Φ. We observe that

(ΨΦ)(m) = Ψ(λm) = λm(1) = m1 = m

[(ΦΨ)(ϕ)](r) = [Φ(ϕ(1))](r) = ϕ(1)r = ϕ(r)

so ΦΨ = id.

Proposition 5.3. As for vector spaces, the composition of R-linear maps is R-linear.

Proof. Exercise.

Proposition 5.4. Let ϕ : M → N be a homomorphism of right R-modules. Then

(1) kerϕ ≤M

(2) im(ϕ) ≤ N

Proof. (1) Exercise.

(2) Any element in im(ϕ) has the form ϕ(m) for m ∈M . Then for r ∈ R, see

ϕ(m)r = ϕ(mr) ∈ im(ϕ),

so im(ϕ) is closed under scalar multiplication. We know im(ϕ) is a subgroup so it must be asubmodule.

Proposition 5.5. Let N ≤M . Then there are module homomorphisms

(1) The inclusion map ι : N ↪→M

(2) the projection map π : M →M/N defined by m 7→ m+N .

Proof. Exercise.

5.1 Isomorphism Theorems

Theorem 5.1 (First Isomorphism Theorem). Let ϕ : M → N be an R-module homomorphism. Thenthe following is a well-defined isomorphism of R-modules:

ϕ :M

kerϕ↔ im(ϕ)

m+ kerϕ 7→ ϕ(m)

Page 12 of 73


Proof. Group theory implies that ϕ is a well defined group isomorphism. We need only check ϕ isR-linear. But observe

ϕ((m+ kerϕ)r) = ϕ(mr + kerϕ)

= ϕ(mr)

= ϕ(m)r

= ϕ(m+ kerϕ)r.

We can similarly conclude

Theorem 5.2. (1) Let N,N ′ ≤M . Then N +N ′ ≤M and

N +N ′

N∼=

N ′

N ∩N ′

(2) Given M ′′ ≤M ′ ≤M thenM/M ′′

M ′/M ′′∼=M

M ′.

Example 5.2. Let R be a ring, and S = M2(R). Recall that(R R

) ∼= R2 is a right S-module. Letm :

(1 0

)∈ R2 so by proposition (5.2) we get the homomorphism λm : s 7→

(1 0

)s. It is surjective,

so im(λm) = M , and

kerλm =

(0 0R R

).

The first isomorphism theorem gives

M2(R)

kerλm∼=(R R

).

5.2 Universal property of quotients

Theorem 5.3. Let M ′ ≤M and N another R-module. We have a bijection

Ψ : HomR

(M/M ′, N

)−→

{ϕ ∈ HomR (M,N) | ϕ(M ′) = 0

}ϕ 7→ ϕ ◦ π.

Proof. Note that Ψ is well defined. It is injective as π is surjective, so we check Ψ is surjective. Letϕ ∈ HomR (M,N) with ϕ(M ′) = 0. Then we get the well-defined map

ϕ : M/M ′ −→ N

m+m′ 7→ ϕ(m+M ′) = ϕ(m).

Exercise. Check that ϕ is R-linear.

We note that[Ψ(ϕ)](m) = ϕ(m+M ′) = ϕ(m+M ′) = ϕ(m)

so Ψ(ϕ) = ϕ and thus Ψ is also surjective.

Example 5.3. What is HomZ (Z/2Z,Z/3Z)?The universal property of quotients tells us that these are the homomorphisms

ϕ : Z→ Z/3Z

such that ϕ(2Z) = 0. But proposition (5.2) tells us that HomZ (Z,Z/3Z) = Z/3Z, so we need a ∈ Z/3Zsuch that a(2Z) = 0. It follows that a = 0 as 2 ∈ (Z/3Z)∗.

Page 13 of 73


6 Direct Sums and Products

Let R be a ring, and let {Mi}i∈I be a set of (right) R-modules. We recall∏i∈I

Mi = {(mi)i∈I | mi ∈Mi} .

We also consider the following subset

⊕i∈I

Mi =

{(mi)i∈I ∈

∏i∈I

Mi | all but finitely many mi are zero

}.

Remark. If I = {1, 2, . . . , n} then∏

Mi =⊕

Mi = M1 × · · · ×Mn, and the elements will be eitherwritten as m1

...mn

or(m1 . . . mn

).

As usual, Mn denotes M ×M × · · · ×M .

Proposition 6.1.∏i∈I

Mi is an R-module when endowed with coordinate wise addition and scalar

multiplication. This module is called the direct product.⊕i∈IMi is a submodule of

∏Mi and is called the direct sum.

Proof. Check axioms.

Exercise. 0r = 0 in any module.

6.1 Canonical Injections and Projections

Proposition 6.2. Let {Mi}i∈I be a set of (right) R-modules. Then we have the following R-modulehomomorphisms for each j ∈ I

(1) Canonical Injection:

ιj : Mj −→⊕i∈I

Mi

m 7→ (0, . . . ,m, 0, . . . , 0)

(2) Canonical Projection:

πj :∏i∈I

Mi −→Mj

(mi)i∈I 7→ mj

Proof. Check R-linearity.

Page 14 of 73


6.2 Universal Property

Let {Mi}i∈I be as above, and N another R-module. We have the following inverse isomorphisms ofabelian groups

(1)

HomR

(⊕Mi, N

)Ψ↔Φ

∏i∈I

HomR (Mi, N)

f 7→ (f ◦ ιi)i∈I(f : (mi) 7→

∑fimi

)←[ (fi)i∈I .

(2)

HomR

(N,⊕

Mi

)Ψ↔Φ

∏i∈I

HomR (N,Mi)

f 7→ (π ◦ f)i∈I

(f : n 7→ (fi(n))i∈I)←[ (fi)i∈I .

Proof. We do (1) only as (2) is similar. First note that Φ is well defined, f ◦ ιi is R-linear as it is thecomposition of R-linear maps.

Exercise. One sees easily that Φ is additive.

It suffices to show that Ψ is an inverse. But observe that

[(ΨΦ)(f)]((mi)i∈I) = [Ψ((f ◦ ιi)i∈I)](mi)i∈I

=∑i∈I

(f ◦ ιi)(mi)

= f(∑

ιi(mi))

= f((mi)i∈I).

So ΨΦ = id. Furthermore, we have

[(ΦΨ)((fi)i∈I)]j(m) = [(Ψ((fi)i∈I ◦ ιi))]j(m)

= Ψ((fi)i∈I) ◦ ιj(m)

= fj(m).

So ΦΨ = id and we are done.

Remark. If I = {1, 2, . . . , n} we usually interpret the map Ψ by

Ψ((f1, . . . , fn)) :

m1...mn

7→ n∑i=1

fi(mi).

That is, Ψ((f1, . . . , fn)) is just left multiplication by (f1, . . . , fn) if we write the elements of⊕Mi as

column vectors.

Page 15 of 73


Example 6.1. Let R be a ring, S = M2(R), and recall that R2 =(R R

)is a right S-module. We

also have two S-module maps given by left multiplication by(1 0

); f1 : S → R2(

0 1)

; f2 : S → R2.

We know thatHomS

(S,R2 ×R2

)= HomS

(S,R2

)×HomS

(S,R2

).

We have the S-module map (f1

f2

):

(a bc d

)7→((a b

)(c d

)) .We see that S ∼= R2 ×R2 as right S-modules.

6.3 Decomposability

Definition 6.1. We say that a nonzero R-module is decomposable if it is isomorphic to the directsum of two nonzero submodules. Otherwise it is indecomposable.

Example 6.2. Let R = k[x]. For n ≥ 1, 〈xn〉 is an ideal, and therefore M = R/〈xn〉 is a modulewhich is indecomposable.

Why? From the internal characterisation of direct product groups, we need only show that anytwo submodules intersect nontrivially. We show indeed that any nonzero N ≤M contains xn−1 +〈xn〉.We may suppose that

N 3 αjxj + · · ·+ αn−1xn−1 + 〈xn〉, αj 6= 0 =: n.

Then nα−1j xn−1−j = xn−1 + 〈xn〉.

7 Free Modules and Generators

7.1 Free Modules

Fix a ring R.

Definition 7.1. A free module is one which is isomorphic to a direct sum of copies of R. That is,M is a free module if

M ∼=⊕i∈I

R (=: RI if I = {1, 2, . . . , n} .)

Proposition 7.1. Let R be a commutative ring, and G a group. Let

RG =⊕g∈G

gR

be the free right (or left) R-module with “basis” the elements of G. The R-module structure on RGextends to an R-algebra structure with multiplication induced by group multiplication, that is,∑

g∈Ggrg

(∑h∈G

hsh

):=∑l∈G

l

∑gh=l

rgsh

,

and unit mapι : R→ RG; r 7→ 1r.

This is called the group algebra.

Proof. Exercise.

Page 16 of 73


7.2 Generating Submodules

Let M be a module, I an index set. Recall there is a group isomorphism (from the universal property)

HomR

(⊕i∈I

R,M

)∼=∏i∈I

HomR (R,M)prop5.2∼=

∏i∈I

M.

We then ask, what is the homomorphism in HomR

(RI ,M

)corresponding to (mi)i∈I ∈

∏i∈IM?

The answer is called the universal property for free modules, and is

(mi)i∈I : (ri)i∈I 7→∑i∈I

miri.

Such an expression of element of M is called a (right) R-linear combination of the mi’s.

Example 7.1. We see

HomR (Rm, Rn) ∼=m∏i=1

Rn = (Rn)m = Mnm(R).

Homomorphisms corresponding to n×m matrices are given by left multiplication.

Example 7.2. Let R be a commutative ring, and G a group. Let H be a subgroup of G.

Exercise. Show RH is an R-subalgebra of RG.

Hence RG is also a (right) RH-module. In fact, it is a free RH-module.Why? For each left coset C of H in G, we pick a representative gC so C = gCH. The universal

property of free modules gives a homomorphism

(gC)C∈G/H :⊕

C∈G/H

(RH) −→ RG

(aC)C∈G/H 7→∑

C∈G/H

gCaC .

Exercise. This is clearly bijective so gives an isomorphism.

Proposition 7.2. Let M be a (right) R-module, and L a subset. The submodule generated by L

is the set of all R-linear combinations of elements of L. It is a submodule of M , and is denoted∑l∈L

lR.

Proof.∑

lR is a submodule as it is the image of the R-linear map

(l)l∈L :⊕l∈L

R −→M

given by the universal property of free modules.

Definition 7.2. Let M be an R-module. A set of generators for M is a subset L such that thesubmodule generated by L is M itself. We also say L generates M .

We say M is finitely generated if we can take L to be finite, and say M is finite if |L| = 1.

Example 7.3. Any module M is generated by the subset M .

Example 7.4. The right (or left) R-module R is cyclic, generated by 1R.

Page 17 of 73


Corollary 7.1. Any module is a quotient of a free module by the first isomorphism theorem and thesurjectivity of

(m)m∈M :⊕m∈M

R −→M.

Example 7.5. Let R = C〈x, ∂〉 and suppose V is a C-vector space of functions on which polynomialdifferential operators act (e.g. V = C[x] or meromorphic functions on C). Suppose D ∈ R is adifferential operator and consider the differential equation Df = 0 with solution f ∈ V .

Recall V is a left R-module and so right multiplication by f ∈ V gives an R-module homomorphism

ϕ : R −→ V

D 7→ Df.

Since we assumped that Df = 0, we know that D ∈ kerϕ. Hence RD ⊆ kerϕ. The universal propertyof quotients gives an R-module homomorphism

ϕ : R/RD −→ V

R/RD is a cyclid module generated by 1 +RD. You can reverse this to see that the solutions to thedifferential equation Df = 0 are given by R-linear maps R/RD −→ V .

Example 7.6. Let R = k〈x, y〉 be the ring of polynomials in noncommutative variables x, y over afield k. We know R is cyclic and therefore finitely generated, but the right ideal

∞∑i=0

xiyR

is not finitely generated (exercise).

8 Modules over a PID

Fix a PID R. Recall a PID is a commutative domain where every ideal is generated by a singleelement. Examples include Z, Z[i], k[x] for k a field.

Fact 8.1. Every PID is a UFD and we can define a gcd. In fact, for a, b ∈ R, then 〈gcd(a, b)〉 = 〈a, b〉.

Lemma 8.1. Any submodule M of Rn is finitely generated.

Proof. By induction on n, with n = 0 automatic.Suppose that n > 0 and let π : Rn → R be the projection onto the first factor (which is a linear

map). Let ι : M → Rn be inclusion. Consider the linear map π|M = π ◦ ι : M → Rn → R.im(π|M ) C R so is generated by m0 ∈ R as R is a PID. We pick m0 ∈ M such that π(m0) = m0.

Now, kerπ|M = kerπ∩M ≤ kerπ ∼= Rn−1. Induction implies that kerπ|M is generated by m1, . . . ,ms.It suffices to show that m0,m1, . . . ,ms generate M .

Consider m ∈M . Then im(π|M ) 3 πm = m0r for some r ∈ R. Hence

π(m−m0r) = π(m)− π(m0)r = π(m)− π(m) = 0,

so kerπ|M = kerπ ∩ M 3 m − m0r. But any element of kerπ ∩ M can be written as∑rimi so

m = m0r +∑rimi and so m0, . . . ,ms generate M .

Page 18 of 73


8.1 Structure theorem for modules over a PID

Theorem 8.1. Let R be a PID and M a finitely generated R-module. Then

M ∼= Rr ⊕ R

〈a1〉⊕ R

〈a2〉⊕ · · · ⊕ R

〈as〉

for some ai ∈ R with a1 | a2 | · · · | as.

The proof of the theorem occupies the next two sections.Recall that given a finite set of generators m1, . . . ,mn for M we get a surjective homomorphism

π =(m1 m2 . . . ms

): Rn →M.

Lemma 8.1 implies that kerπ is also finitely generated so picking a set of generators for it gives asurjective homomorphism Rm → kerπ. The composite map

(Rm → kerπ ↪→ Rn) ∈ HomR (Rm, Rn)

is given by an n×m matrix over R (by example 7.1). Call this matrix Φ.

Remark. (1) Given a module isomorphism Ψ : Rn → Rn, we get a new surjective isomorphismπΨ : Rn → Rn → M which corresponds to a change of basis/generators for M . This has theeffect of changing

Φ 7→ Ψ−1Φ.

(2) Similarly, changing basis in Rm, we can change Φ to Φ ◦ Ψ for some module isomorphismΨ : Rm → Rm.

(3) The first isomorphism theorem implies that M ∼= Rn/ im(Φ) = kerπ.

Proposition 8.1. (1) Let S be a ring, N an S-module. The set AutS (N) of module automorphismsψ : N → N is a subgroup of the permutation group on N .

(2) Let GLn(R) = {ψ ∈Mn(R) | detψ ∈ R∗}. Then GLn(R) is a subgroup of AutR (Rn).

Proof. (1) Easy exercise in checking axioms.

(2) Follows from Cramer’s rule about the solution to differential equations.

This reduces theorem (8.1) to

Theorem 8.2. Given any n×m matrix Φ with entries in R, there are Ψl ∈ GLn(R) and Ψr ∈ GLm(R)such that

ΨlΦΨr =

a1 0 . . . 0

0. . .

...... as

...

0 . . . . . . 0...

......

0 . . . . . . 0

Page 19 of 73


Proof that theorem (8.2) implies theorem (8.1). Remark (3) implies that M ∼= Rn/ im(Φ). By remark(1),(3) as well as proposition 8.1 and theorem 8.2, we can assume that

Φ = diag {a1, . . . , as, 0, . . . , 0} .

So

im(Φ) =

a1

0...0

R⊕

0a2

0...0

R⊕ · · · ⊕

0...0as0...0

R.

Exercise. The universal property of free modules and direct products show that we have a surjectiveR-module homomorphism

Rn −→ R/〈a1〉 ⊕ · · · ⊕R/〈as〉 ⊕Rn−s

(r1, . . . , rn) 7→ (r1 + 〈a1〉, . . . , rs + 〈as〉, rs+1, . . . , rn).

Furthermore, the kernel of this map is the same as the image of Φ above, so the first isomorphismtheorem completes the proof.

8.2 Elementary Row and Column Operations

One can perform elementary row and column operations (ERO,ECOs) by left (resp. right) multipli-cation by elements of GLn(R)

(1) row (resp. column) swaps have determinant −1

(2) adding multiples of one row (resp. column) to another has determinant 1

(3) scalar multiplication of a row (resp. column) by µ has determinant µ.

9 Proof of Structure Theorem

Fix a PID R. We need to prove theorem (8.2).

We may as well assume that Φ 6= 0. Then we are reduced to proving

Theorem 9.1. There are Ψl ∈ GLn(R), Ψr ∈ GLm(R) such that the (1, 1) entry Φ11 of Φ = ΨlΦΨr

divides every other entry of Φ.

Proof that theorem (9.1) implies theorem (8.2). Note (exercise) that Φ11 6= 0. Then by factoring outΦ11 and using EROs and ECOs we can assume

Φ = Φ11

(1 00 Φ1

).

Now we use induction on n (or m) and the matrix Φ1 to get the result.

Let us now prove theorem (9.1). Using EROs and ECOs we can assume that Φ11 6= 0. Useinduction on the number of prime factors of Φ11.

If Φ11 is a unit, then theorem (9.1) holds. We need two lemmas.

Page 20 of 73


Lemma 9.1. Let a 6= 0, b ∈ R and d = gcd(a, b). Then there is some Ψ ∈ GL2(R) with

Ψ

(ab

)=

(dc

)with c ∈ 〈a, b〉 = 〈d〉.

Proof. Note that dR = aR+ bR so ∃f, g ∈ R such that d = af + bg. Then

Ψ =

(f g

− bd

ad

)works as det Ψ = 1.

Lemma 9.2. Let 0 6= a, b ∈ R and d = gcd(a, b). Then there are Ψl,Ψr ∈ GL2(R) such that

Ψl

(a 00 b

)Ψr =

(d c1

c2 c3

)with ci ∈ 〈d〉.

Proof. As before we have f, g ∈ R such that d = af + bg. Then

Ψl =

(1 g0 1

)Ψr =

(f −11 0

)work.

We now continue the proof of theorem (9.1). Note first that we can assume Φ11 divides all entriesin its column. Indeed, suppose not and without loss of generality we may assume Φ11 - Φ21. Then bylemma (9.1) there is a Φ ∈ GL2(R) such that

(Φ 00 In−2

)Φ11 . . .Φ21

...

=

(d ∗∗ ∗

)

where d = gcd(Φ11,Φ21). Then d has fewer prime factors than Φ11, and so we are done by induction.Similarly using the transpose of lemma (9.1) we can assume Φ11 divides all entries in the first row.

Applying EROs and ECOs we can assume

Φ =

(Φ11 00 Φ

).

If Φ11 does not divide all entries of Φ, then we can use EROs and ECOs to ensure that Φ11 - Φ22.Lemma (9.2) implies that there exists Ψl,Ψr ∈ GL2(R) such that(

Ψl 00 In−2

)Φ

(Ψr 00 In−2

)=

(d∗

)where d = gcd(Φ11,Φ22) has fewer prime factors than Φ11. This proves theorem (9.1) and hencetheorem (8.2) and the structure theorem for finitely generated modules over a PID.

Page 21 of 73


Example 9.1. Let K be the Z submodule of Z3 generated by 3

2−2

,

586

,

(45

)Write M = Z3/K as a direct sum of cyclic modules.

M = Z3/ im(Φ) where

Φ =

3 5 42 8 5−2 6 2

.

EROs and ECOs reduce this to 1 0 00 7 00 0 0

so M ∼= Z/7Z⊕ Z.

Example 9.2. Let M = Z2/(Z(m1

)+Z

(n2

)) for some m,n ∈ Z. Show (assuming M is finite) that

|M | = |2m− n|.

We have M ∼= Z2/ im(Φ) where

Φ =

(m n1 2

).

Use theorem (8.2) to find Ψl,Ψr ∈ GL2(Z) with

ΨlΦΨr =

(a 00 b

)for a, b ∈ Z.

Then M ∼= Z/aZ⊕ Z/bZ and |M | = |ab| = | det(ΨlΦΨr)| = |2m− n|.

Proposition 9.1. GLn(R) = AutR (Rn).

Proof. We say in section 8 that GLn(R) ≤ AutR (Rn). Suppose that Φ ∈ HomR (Rn, Rn) and det Φ =∆ 6∈ R∗. Then there are Ψl,Ψr ∈ GLn(R) such that ΨlΦΨr is diagonal and ∆ = ua1a2 . . . an for someu ∈ R∗.

If ∆ = 0, then Φ is not injective. If ∆ 6= 0 but is not a unit then

Rn

im(Φ)∼=

R

a1R⊕ · · · ⊕ R

anR.

At least one of the ai’s is not a unit so R/aiR 6= 0. So im(Φ) 6= Rn and thus Φ is not surjective.Therefore Φ ∈ AutR (Rn).

10 Applications of the Structure Theorem

Theorem 10.1 (Alternate structure theorem). Let R be a PID and M a finitely generated R-module.Then M is a direct sum of cyclic modules of the form R or R/〈pn〉 where p ∈ R is a prime, and n ∈ N.

Page 22 of 73


Proof. From the usual structure theorem, we know M is a direct sum of cyclic modules, so we maysuppose that M = R/〈a〉 for some a ∈ R \ {0}. Prime factorise a = pn1

1 pn12 . . . pnr

r and use the ChineseRemainder theorem to get a ring isomorphism

R

〈a〉∼=

R

〈pn11 〉× · · · × R

〈pnrr 〉

.

This is clearly R-linear as well, so we are done.

10.1 Endomorphism Ring

Let R be a ring, and M an R-module. Let

EndR (M) = {ϕ : M →M | ϕ is R-linear} = HomR (M,M) .

Proposition 10.1. The abelian group EndR (M) is a ring when endowed with ring multiplicationequal to composition of homomorphisms. It is called the endomorphism ring of M and its elementsare called endomorphisms of M .

Proof. Exercise in checking axioms.

Definition 10.1. Let R,S be rings. An (R,S)-bimodule is a left R-module M which is also a rightS-module with the same additive structure such that we have the associative law

(rm)s = r(ms).

Example 10.1. Let R be a commutative ring, and M a right R-module. It is also a left R-modulewith the same addition and left multiplication defined as

r ·m := mr.

This makes M an (R,R)-bimodule.

Exercise. Check axioms.

Example 10.2. R itself is a left and right R-module and hence an (R,R)-bimodule.

Proposition 10.2. Let R be a ring, and M a right R-module. Then M is an (EndR (M) , R)-bimodulewith left multiplication defined by

ϕ ·m := ϕ(m).


We now have an alternative view of bimodules:

Proposition 10.3. Let R,S be rings, and M a right R-module, which by proposition (10.1) is an(EndR (M) , R)-bimodule.

(1) Given a ring homomorphismϕ : S → EndR (M) ,

the left EndR (M)-module structure on M “restricts” to a left S-module structure by proposition(4.2). This makes M an (S,R)-bimodule.

(2) Any (S,R)-bimodule arises in this fashion.

Proof. (1) Exercise in checking the associative law for bimodules.

(2) Let M be an (S,R)-bimodule. We construct a ring homomorphism ϕ : S → EndR (M) byϕ(s) =left multiplication by s on M . Note ϕ(s) is R-linear.

Exercise. Check ϕ is a ring homomorphism.

Page 23 of 73


10.2 Jordan Canonical Forms

Suppose k is an algebraically closed field, so the primes in k[x] have the form β(x−α) with α ∈ k, β ∈k∗.

Let X ∈Mn(k) = Endk (kn). We have the substitution homomorphism

Φ : k[x] −→Mn(k)

p(x) 7→ p(X).

Hence proposition (10.2)(1) implies that we get a (k[x], k)-bimodule structure on kn.Thus by the alternate structure theorem we have an isomorphism of left k[x]-modules (also of

bimodules)

kn ∼=k[x]

〈(x− α1)n1〉⊕ · · · ⊕ k[x]

〈(x− αr)nr〉.

X acts on kn by left multiplication by x. Therefore the direct sum decomposition above implies thatX is similar to a block diagonal matrix, with block sizes n1 × n1, . . . , nr × nr.

The ith block with respect to the basis 1, x− αi, (x− αi)2, . . . , (x− αi)ni−1 is in Jordan canonicalform.

11 Exact Sequences and Splitting

11.1 Exact Sequences

Fix a ring R.

Definition 11.1. A complex of R-modules is a sequence of R-modules and R-module homomor-phisms of the form

M• : · · · →Mi−1di−1−−−→Mi

di−→Mi+1 → . . .

such that d2 := didi−1 = 0 for all i.

Remark. d2 = 0 if and only if di(di−1(Mi−1)) = 0 if and only if ker di ⊆ im(di−1).

Definition 11.2. A complex M• : · · · →Mi−1 →Mi → . . . is exact at Mi if ker di = im(di−1).We say M• is exact if it is exact at all Mi.

Example 11.1. Let R = Z. We have the complex of Z-modules

· · · → Z4Z

ι−→ Z4Z

ι−→ . . .

where each stage is the map n+ 4Z 7→ 2n+ 4Z. This is exact.

Example 11.2. (1) 0→Mf−→ N is an exact sequence of R-modules if and only if f is an R-linear

injection (as exactness implies ker f = 0).

(2) Mg−→ N → 0 is exact if and only if im(g) = N .

(3) Given a submodule M ≤ N we get an exact sequence

0→M ↪→ Nπ−→ N/M → 0

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Corollary 11.1. Consider the sequence of R-modules and R-module homomorphisms

M• : 0→M ′f−→M

g−→M ′′ → 0.

Then M is exact if and only if

(1) f is injective

(2) g is surjective

(3) ker g = im(f).

In this case, M ′′ ∼= M/ im(f). We say M• is a short exact sequence (SES).

11.2 Splitting

First some motivation. Recall that given R-modules M ′,M ′′ we have the canonical injection

ι : M ′ →M ′ ⊕M ′′

and projectionπ : M ′ ⊕M ′′ →M ′.

Notice that πι = id.

Proposition 11.1. Consider the R-module homomorphisms f : M → N and g : N → M such thatgf = idM .

(1) Then f is injective, g is surjective and we say f is a split injection and g is a split surjection.

(2) We have a direct sum decomposition(f idker g

): M ⊕ ker g

∼−→ N,

and we say M is a direct summand of N .

(3) Given either f or g, we say the other map splits it (e.e. f splits g and g splits f).

Proof. (1) Easy exercise in set theory.

(2) We construct an inverse as follows. First note that

idN −fg : N → N

has image in ker g. This is because

g(idN −fg) = g − gfg = g − g = 0.

It follows that we can assume idN −fg : N → ker g. We claim that the inverse to Φ =(f idker g

)is

Ψ =

(g

idN −fg

).

To see this, we observe that

ΦΨ =(f idker g

)( gidN −fg

)= fg + id |ker g(idN −fg)

= fg + idN −fg= idN .

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Similarly, we have

ΨΦ =

(g

idN −fg

)(f idker g

)=

(gf g idker g

idN f − fgf (id−fg)|ker g

)= idM⊕ker g .

Corollary 11.2. Consider a short exact sequence of R-modules

0→M ′f−→M

g−→M ′′ → 0.

(1) f is a split injection if and only if g is a split surjection

(2) in this case, M ∼= M ′ ⊕M ′′ and we say the short exact sequence is split.

Proof. Exercise.

Example 11.3.

0→ Z3Z

2−→ Z6Z

g−→ Z2Z→ 0

with 2 : n+ 3Z 7→ 2n+ 6Z, and g : n+ 6Z 7→ n+ 2Z. This is a split short exact sequence, where g issplit by the “multiplication by 3” map, that is, h : n+ 2Z 7→ 3n+ 2Z.

Exercise. Show that gh =multiplication by 3 which is the identity on Z/2Z.

Example 11.4.

0→ Z2Z

2−→ Z4Z

π−→ Z2Z→ 0

is exact but not split as Z/4Z 6∼= Z/2Z⊕ Z/2Z.

11.3 Idempotents

Definition 11.3. Let R be a ring, and e ∈ R. e is an idempotent if e2 = e.

Exercise. Let M be an (R,S)-bimodule. Show that left multiplication by e

λe : M → eM

is a split surjection of right S-modules. It is split by the inclusion map eM ↪→M .Show that M ∼= eM ⊕ (1− e)M .

Exercise. Consider a finite set of idempotents {e1, . . . , en} ⊆ R. We say they are complete and

orthogonal if∑i

ei = 1 and eiej = 0 if i 6= j.

Show thatM ∼= e1M ⊕ · · · ⊕ enM.

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12 Chain Conditions

Let R be a ring, and M a (right) R-module.

Definition 12.1. M is

(1) noetherian if it satisfies the ascending chain condition (ACC): for any chain of submodules

M1 ≤M2 ≤ · · · ≤M

stabilises in the sense that Mn = Mn+1 = . . . for all n > N .

(2) artinian if it satisfies the descending chain condition (DCC): for any chain of submodules

M ≥M1 ≤M2 ≥ . . .

stabilises.

Example 12.1. Let k be a field, and V a finite dimensional vector space. Then V is both noetherianand artinian as any strictly increasing or decreasing chain of subspaces has strictly increasing/decresingdimension.

Example 12.2. Let R be a PID. Then M = RR is noetherian but not artinian.Why? Pick a nonzero, non-unit r ∈ R. We obtain a strictly decreasing chain of submodules

R 〈r〉〈r2〉 . . .

so RR is not artinian. However, consider the strictly increasing chain of ideals

〈r1〉 ≤ 〈r2〉 ≤ 〈r3〉 ≤ . . .

so r2 | r1, rn | rn−1. Since the number of prime factors will stabilise, M is noetherian.

12.1 Submodules and Quotients

Proposition 12.1. Let R be a ring and N a submodule of an R-module M . Then M is noetherian(resp. artinian) if and only if N and M/N are.

Proof. We only prove the noetherian result. First, suppose that M satisfies the ACC, and thereforeN satisfies the ACC trivially. Then let π : M →M/N be the projection map, and consider the chainof submodules

M1 ≤M2 ≤ . . .Exercise. We get an ascending chain of submodules of M

π−1(M1) ≤ π−1(M2) ≤ . . .

ACC on M implies that for n large enough, this chain stabilises. Now π is surjective so ππ−1(Mi) =Mi, and so the first chain stabilises in M/N and thus the ACC holds for M/N as well.

Conversely, suppose that the ACC holds for N and M/N . Consider a chain of submodules of M

M1 ≤M2 ≤ . . .

We then get induced chains on N and M/N :

M1 ∩N ≤M2 ∩N ≤ . . .π(M)1 ≤ π(M2) ≤ . . .

ACC on N and N/M means that these chains stabilise and we may assume that n is large enoughthat Mn ∩N = Mn+1 ∩N = . . . and π(Mn) = π(Mn+1) = . . . . It now suffices to prove

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Claim 12.1. Suppose given L1 ≤ L2 ≤ . . .M with L1∩N = L2∩N and π(L1) = π(L2) Then L1 = L2.

Proof. It suffices to show that L2 ≤ L1. Pick m2 ∈ L2 and we know

π(m1) = m2 +N ∈ π(L2) = π(L1)

so m2 +N = m1 +N for some m1 ∈ L1. Therefore m2 −m1 ∈ N ∩ L1 ⊆ L1.It follows that m2 = m1 + (m2 −m1) ∈ L1.

Remark. Proposition (12.1) is equivalent to the following statement.For any short exact sequence of R-modules

0→M ′ →M →M ′′ → 0

then M is noetherian (resp. artinian) if and only if M ′ and M ′′ are.

Corollary 12.1. (1) If M1,M2 are noetherian (resp. artinian) then so is M1 ⊕M2.

(2) If M1,M2 are noetherian (resp. artinian) submodules of M , then M1 +M2 is noetherian (resp.artinian).

Proof. (1) We have a split exact sequence

0→M1ι1−→M1 ⊕M2

π2−→M2 → 0.

(2) We have a surjective homomorphism(ι ι

): M1 ⊕M2 →M1 +M2

(m1,m2) 7→ m1 +m2,

so (1) and proposition (12.1) give the result.

12.2 Finite Generation

Proposition 12.2. An R-module M is noetherian if and only if every submodule is finitely generated.

Proof. Suppose M is infinitely generated. If suffices to construct a strictly ascending chain of finitelygenerated submodules inductively: set M0 = 0 and

Mn =n∑i=1

fiR

where fn+1 is any element of M \Mn (this is possible because M is infinitely generated but Mn isfinitely generated). Putting

Mn+1 = Mn + fn+1R.

Conversely, consider an ascending chain of submodules

M1 ≤M2 ≤ . . .

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Exercise.M ′ :=

⋃i≥1

Mi

is a submodule.

Therefore it is finitely generated by say m1, . . . ,mr. Hence we can pick n large enough so Mn 3m1, . . . ,mr. Thus

M ′ =

r∑i=1

miR = Mn = Mn+1 = . . .

so the ACC holds.

13 Noetherian Rings

Definition 13.1. A ring R is (right) noetherian (resp. artinian) if the module RR is noetherian (resp.artinian).

Example 13.1. Let k be a field. Then any finite dimensional k-algebra A is right and left notherianand artinian.

Furthermore, any PID is noetherian, but k〈x, y〉 is not noetherian or artinian.

Proposition 13.1. Let R be a right noetherian ring.

(1) An M -module M is noetherian if and only if M is finitely generated.

(2) Every submodule of a finitely generated R-module is itself finitely generated.

Proof. (1)⇒ (2) by proposition (12.1), so it remains to prove (1). The forward direction follows fromproposition (12.2). Conversely, if M is finitely generated, there exists elements m1, . . . ,mn ∈M withM = m1R+ · · ·+mnR. Each miR is noetherian as it is a quotient of RR, and hence M is noetherianby corollary (12.1).

Exercise. If R is right noetherian, so is R/I for every I CR.

13.1 Almost Normal Extensions

Definition 13.2. Let S and R be rings such that R is a subring of S. We say S is an almost normalextension of R with almost normal generator a if

(1) R+ aR = R+Ra, so (exercise: induction)

R+ aR+ arR+ · · ·+ adR = R+Ra+ · · ·+Rad

that is, any “right” polynomial in a (with coefficients in R) of degree ≤ d is also a left polynomialin a with degree ≤ d.

(2) S =⋃d≥0

R+Ra+ · · ·+Rad

We then write S = R〈a〉.

Example 13.2. Let R be a ring. Then R[x1, . . . , xn] = (R[x1, . . . , xn−1]) [xn] is an almost normalextension of R[x1, . . . , xn−1] with almost normal generator xn.

Example 13.3. Consider the Weyl algebra C〈x, ∂〉 = C[x]〈∂〉 – this is an almost normal extension ofC[x] with almost normal generator ∂.

Why? We know (by the Weyl relation) ∂x = x∂ + 1 and so C[x] + ∂C[x] = C[x]∂ + C[x].

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13.2 Hilbert’s Basis Theorem

Theorem 13.1 (Hilbert). Let R be a right noetherian ring and let S = R〈a〉 be an almost normalextension with almost normal generator a. The S is right noetherian.

Proof. By proposition (12.2), it suffices to show that any right ideal I C S is finitely generated. Weconsider the set of “degree j leading coefficients”

Ij :=

{rj ∈ R |

j∑i=0

riai ∈ I

}⊆ R

for each j ≥ 0.Note that Ij ⊆ Ij+1 since I is closed under right multiplication by a.

Claim 13.1. Ij is also a right ideal of R.

Proof. Let rj ∈ Ij , so we have some

j∑i=0

riai ∈ I. Note that Ij is an additive subgroup because I is.

Now, let r ∈ R. Definition 13.2(1) implies that raj = ajr′+ lower order terms in a. Then

I 3j∑i=0

ririair′ = rja

jr′ + lower order terms

= rjraj +O(aj−1)

and so rjr ∈ Ij .

ACC on RR implies that for d large enough, we have Id = Id+1 = . . . . Note

I≤d = (R+Ra+ · · ·+Rad) ∩ I = (R+ aR+ · · ·+ adR) ∩ I

is a finitely generated R-module since it is a submodule of the noetherian module R+· · ·+Rad. Thus wecan find f1, . . . , fs ∈ I≤d with f1R+ · · ·+fsR = I≤d. It suffices to show that I ′ := f1S+ · · ·+fsS ⊇ I,since I ⊃ I ′.

Define

I ′j :=

{rj ∈ R |

j∑i=0

riai ∈ I ′

}⊂ Ij .

But

I ′ ∩ (R+ · · ·+ adR) ≥ (f1R+ · · ·+ fsR) ∩ (R+ aR+ · · ·+ adR) = I ∩ (R+ aR+ · · ·+ adR) (13.1)

and so I ′d = Id = Id+1 = I ′d+1 so I ′t = It for t ≥ d. We now show

r = rtat + rt−1a

t−1 + · · ·+ r0 ∈ I

is in I ′ by induction on t.

If t ≤ d, then this holds by (13.1). Otherwise, we know I ′t = It so there exists

t∑i=0

r′iai ∈ I ′ with

r′t = rt. Then r − r′ ∈ I ∩ (R+ · · ·+Rat−1) so we are done by induction.

Corollary 13.1. If R a noetherian ring, so is R[x1, . . . , xn]. Furthermore, the Weyl algebra is noethe-rian.

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14 Composition Series

14.1 Simple Modules

Let R be a ring.

Definition 14.1. An R-module M 6= 0 is called simple or irreducible if the only submodules of Mare 0 and M itself.

A right ideal I CR is maximal if the only ideal strictly containing I is R. We say I is minimalif the only ideals contained in I are 0 and I.

Proposition 14.1. A right module M is simple if and only if M ∼= R/I for some maximal I CR.

Proof. Suppose M is simple, and pick m ∈M \ {0}. Consider the homomorphism λm : R→M whichmaps r 7→ mr. Note that λm is surjective as im(λm) is a nonzero submodule of the simple module M .The first isomorphism theorem implies that M ∼= R/I for some I = kerλm. Now, I is maximal, for ifnot we can find an ideal I ′ with I � I ′ � R. This gives a nontrivial submodule of M correspondingto I ′/I.

The converse is clear by reversing the above argument.

Example 14.1. Let R be a PID. Simple modules correspond to R/〈p〉 for some p ∈ R prime.

Example 14.2. Let R be the Weyl algebra C〈x, ∂〉, and M the left R-module C[x]. Then M issimple for given a submodule N containing some nonzero element, say p(x) = pnx

n + · · · + p0 6= 0,then N 3 1

pnn!ddxn p(x) = 1, so N = M .

14.2 Composition series

Definition 14.2. Let M be an R-module. A composition series is a chain of submodules

0 < M1 < M2 < · · · < Mn = M

such that Mi+1/Mi is simple for all i.In this case, we call Mi+1/Mi the composition factors of M and say M has finite length.

Example 14.3. Let R = k[x], and M = k[x]/〈x2〉. Then M has composition series

0 <〈x〉〈x2〉

< M.

Thusk[x]/〈x2〉〈x〉/〈x2〉

∼=k[x]

〈x〉is simple. Also, we have the isomorphism

k[x]

〈x〉∼=〈x〉〈x2〉

that maps 1 7→ x+ 〈x2〉. Therefore the composition series has a repeated composition factor, k[x]/〈x〉.

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14.3 Jordan Holder Theorem

Theorem 14.1. Let M be an R-module with two composition series

0 < M1 < M2 < · · · < Mn = M (14.1)

and0 < M ′1 < M ′2 < · · · < M ′m = M. (14.2)

Then n = m (called the length of M). Also, there exists a permutation σ of {1, . . . , n} such that

Mi

Mi−1

∼=M ′σ(i)

M ′σ(i−1)

,

that is, the composition factors are equal up to isomorphism.

Proof. By induction on n with the case n = 0 clear.Suppose that n ≥ 1. Note that we have the following composition series for M/M1:

0 =M1

M1<M2

M1< · · · < M

M1(14.3)

which has the same composition factors as (14.1) except one copy of M1 is removed.We construct another composition series for M/M1 as follows. Note that M1 is simple, so for any

i, we have M ′i ∩M1 is either 0 or all of M1, in which case M ′i ⊇M1. Hence we can find a j such thatM ′j ∩M1 = 0 but M ′j+1 ∩M1 = M1. Note (M1 +M ′j)/M

′j is a nonzero submodule of M ′j+1/M

′j and so

must equal it.Hence

M1∼=

M1

M1 ∩M ′j

∼=M1 +M ′j

M ′j

∼=M ′j+1

M ′j.

It suffices by induction to now prove

Claim 14.1. We have the following composition series for M/M1

0 <M ′1 +M1

M1<M ′2 +M1

M1< · · · <

M ′j +M1

M1=M ′j+1

M1< · · · < M

M1(14.4)

and this has the same composition factors as (14.2) except one copy of M ′j+1/M′j∼= M1 is removed.

Proof. For i > j, we haveM ′i+1/M1

M ′i/M1

∼=M ′i+1

M ′isimple.

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For i ≤ j, we have composition factors

(M ′i+1 +M1)/M1

(M ′i +M1)/M1

∼=M ′i+1 +M1

M ′1 +M1

∼=M ′i+1 + (M ′i +M1)

M ′i +M1

∼=M ′i+1

M ′i+1 ∩ (M ′i +M1)

∼=M ′i+1

M ′i.

If m ∈M ′i+1 and m = mi +m1 with mi ∈M ′i and m1 ∈M1, then

m−mi = m1 ∈M ′i+1 ∩M1 = 0

and so m1 = 0, m ∈M ′i and we are done.

Exercise. If M1, . . . ,Mn are simple, find a composition series for M1 ⊕ · · · ⊕Mn.

Remark. Any simple module is noetherian and artinian.

Exercise. Show M has finite length if and only if M is noetherian and artinian.

15 Semisimple Modules

Let R be a ring.

Proposition 15.1. A (right) R-module M is semisimple if any of the following equivalent conditionshold

(1) Any submodule M ′ ≤M is a direct summand

(2) Any surjective homomorphism M →M ′′ is split

(3) Any short exact sequence of the form

0→M ′ →M →M ′′ → 0

splits.

Proof. Easy exercise.

Example 15.1. Let R = k[x], with k a field. Then k[x]/〈x2〉 is not semisimple because

0→ 〈x〉〈x2〉

↪→M → k[x]

〈x2〉→ 0

is not split. However,

M =k[x]

〈x(x− 1)〉∼=k[x]

〈x〉⊕ k[x]

〈x− 1〉is semisimple (as the direct summands are semisimple).

Page 33 of 73


Lemma 15.1. Let M be a semisimple module, and N ≤M . Then N is semisimple.

Proof. Let N ′ ≤ N ≤M . As M is semisimple, the map N ′ ↪→M is split by π : M → N ′. Restrict π toN , and hence π|N : N → N ′ also splits N ′ ↪→ N , so by proposition (15.1)(1) N is also semisimple.

Lemma 15.2. Let M 6= 0 be a cyclic module. Then there exists a quotient M/N which is simple.

Proof. M cyclic implies that M = mR for some m ∈ M . We thus find a surjective homomorphismR

m−→ M . The first isomorphism theorem then implies that we can assume M = R/I for some rightideal I. Zorn’s lemma tells us that we can find a right ideal J such that

(1) J ⊇ I;

(2) 1 6= J ; and

(3) J is maximal with respect to these properties.

Call this maximal J Jmax. Indeed, let S be the set of such J , then given an increasing chain of rightideals in S

I ⊆ J1 ⊆ J2 ⊆ . . .

we haveJ∞ =

⋃i∈N

Ji

is an upper bound in S since

Exercise. J∞ is a right ideal which satisfies the properties above.

The maximality of Jmax implies that any right ideal strictly containing Jmax contains 1 andtherefore is R.

HenceR/I

Jmax/I∼=

R

Jmax

is the desired simple quotient.

Lemma 15.3. Let N ≤M . Then there exists a submodule N ′ ≤M such that

(a) N ′ is a direct sum of simple modules

(b) N ∩N ′ = 0

(c) for any simple submodule Ns ≤M we have

(N +N ′) ∩Ns 6= 0.

Proof. Let {Ns}s∈S be the set of simple submodules of M and S ⊂ P(S) be the set of J ⊆ S suchthat

(1) NJ :=∑j∈J

Nj is a direct sum in the sense that the natural map

⊕j∈J

Nj −→∑

Nj

is an isomorphism.

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(2) N ∩NJ = 0.

By Zorn’s lemma

Exercise. S has a maximal element, say Jmax.

We wish to show N ′ = NJmax satisfies the lemma. Note that (1) and (2) give (a) and (b). We needonly check (c) so suppose instead that there exists a simple module Ns with

(N +N ′) ∩Ns = 0. (15.1)

We will show that J ′ = Jmax ∪ {S} contradicts the maximality of Jmax. Note that Ns ∩ N ′ =NS ∩NJmax = 0, so

NJ ′ = NS ⊕N ′

so (1) holds. We show (2) holds for J ′, that is, N ∩NJ ′ = 0.Suppose instead that (NS + N ′) ∩ N 3 n = nS + n′, Since (c) is assumed to be false, n − n′ =

ns ∈ (N + N ′) ∩ NS so n − n′ = ns = 0. This means that n = n′ ∈ N ∩ N ′ = 0 by (b). This is acontradiction, so (NS +N ′) ∩N = 0.

Theorem 15.1. The following are equivalent for an R-module M

(1) M is semisimple

(2) M ∼=⊕Mj with Mj simple for all j.

Proof. Suppose that M is semisimple. By lemma (15.3) with N = 0 we get a direct sum of simplesN ′ ≤M such that

N ′ ∩NS = 0

for NS simple. It suffices to show that M = N ′. Proposition (15.1) implies that M = N ′⊕N ′′ for someN ′′ ≤M . Pick n′′ ∈ N ′′ \ {0}, so the submodule n′′R is also semisimple by lemma (15.1). By lemma(15.2), n′′R has a simple quotient N ′′′. The semisimplicity of n′′R implies that it has a submodule NS

isomorphic to N ′′′. But NS ∩N ′ ⊆ N ′′ ∩N ′ = 0, so (c) in lemma (15.3) is violated.

Conversely, Let N ≤ M , N ′ be as in lemma (15.3). It suffices to show M = N ′ +N for we knowN ∩N ′ = 0.

SupposeM 6= N+N ′, so there existsMj ≤ N+N ′. ButMj is simple, and therefore (N+N ′)∩Mj =0. This contradicts (c) in lemma (15.3), so N +N ′ = M .

Exercise. Let R = Z[x]. Show that R/〈6x〉 is not semisimple, but R/〈6, x〉 is.

16 Semisimple rings and Group Representations

Proposition 16.1. A ring R is semisimple if either of the following equivalent conditions hold

(1) Any short exact sequence of right R-modules

0→M ′ →M →M ′′ → 0

splits

(2) RR is semisimple.

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Proof. It is clear that (1) ⇒ (2) on setting M = RR. For (2) ⇒ (1), (2) and theorem (15.1) implythat any free module is a direct sum of simple modules and so is semisimple. Any M is a quotient ofa free module F , and the semisimplicity of F implies that M is isomorphic to a summand of F so issemisimple by lemma (15.1).

Example 16.1. Let D be a division ring. Then D is right and left semisimple since DD and DD aresimple and therefore semisimple.

More generally,

Proposition 16.2. Let D be a division ring and R = Mn(D). Then

(1) Dn is a simple right R-module

(2) Mn(D) is semisimple.

Proof. (1) Consider N ≤ Dn which is nonzero. It suffices to show N = Dn.

Pick (α1, . . . , αn) ∈ N with αi 6= 0. Let ϕ ∈ Mn(D) with α−1i in the (1, i)th entry and zeroes

elsewhere. Then N 3 (α1, . . . , αn)ϕ = (1, 0, . . . , 0) and so

N 3 (1, 0, . . . , 0)

(β1 β2 . . . βn∗ ∗ ∗ ∗

)= (β1, . . . , βn) ∈ Dn.

So N = Dn and Dn must be simple.

(2) R = Mn(D) is right semisimple because RR has the following direct sum decomposition intosimple modules (using (1))

Mn(D) =

D . . . D...

. . ....

D . . . D

=

Dn

0...0

⊕ · · · ⊕

0...0Dn

∼= Dn ⊕ · · · ⊕Dn.

Similarly R is left semisimple.

Proposition 16.3. Let R1, . . . , Rn be rings, and R = R1 × · · · ×Rn their product. Then

(1) Given Ri-modules Mi for each i, we obtain an R-module structure on

M = M1 × · · · ×Mn

with scalar multoplication defined by

(m1, . . . ,mn)(r1, . . . , rn) := (m1r1, . . . ,mnrn).

(2) Conversely, every R-module has this form (the Mi are called the components of M).

Proof. (1) Check module axioms.

(2) We give a sketch proof in the case n = 2. Let R = R1×R2, and consider the nonunital subrings(1 0

)R = R1 × {0} ∼= R1 and

(0 1

)R ∼= R2. Let M be an R-module, and M1 = M

(1 0

),

M2 = M(0 1

).

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Exercise. Note M1 is a module over R1 =(1 0

)R “since”

M1R1 = M(1 0

) (1 0

)R

= M(1 0

)R

= MR(1 0

)= M

(1 0

)= M1.

Similarly, M2 is an R2-module.

it now suffices to show

ϕ : M1 ×M2 −→M

(m1,m2) 7→ m1 +m2

is an R-module isomorphism by showing

Exercise. (a) M1 + M2 = M since given any m ∈ M we have m = m(1 1

)= m

(1 0

)+

m(0 1

)(b) M1 ∩M2 = 0 since given m ∈M1 ∩M2, m ∈M1 ⇒ m1R1 = m

(1 0

)and m ∈M2 ⇒ m =

m1R2 = m(0 1

). It follows that m

(0 1

) (1 0

)= m = 0.

(c) Show ϕ is actually R-linear.

Corollary 16.1. Any finite product R = R1 × · · · ×Rn of semisimple rings is also semisimple.

Proof. Each Ri ∼=⊕ir

j=1Mij for simple Ri-modules Mij .

Exercise. Note that these sums are finite as 1 must have a nonzero component in each Mij .

So

RR =

i1⊕j=1

M1j × 0× · · · × 0

⊕ · · · ⊕ in⊕j=1

0× · · · × 0×Mnj

is a decomposition of RR into a direct sum of simple modules.

16.1 Group Representations

Let G be a group, and R a commutative ring.

Definition 16.1. An (R-linear) representation of G is a group homomorphism

ρ : G −→ AutR (V )

for some R-module V .

Remark. (1) AutR (V ) is a subgroup of the permutation group of V so the group representationabove gives a G-set such that each g ∈ G acts linearly on V .

Exercise. Given any G-set such that every g ∈ G acts linearly on V , we obtain a grouprepresentation.

Such G-sets are called G-modules.

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(2) (a) Given any left RG-module M , we may consider it an (RG,R)-bimodule with right actionof R equal to the left action. This is well defined as R ⊆ Z(RG).

(b) Proposition (10.2) implies that such an (RG,R)-bimodule structure is equivalent to givingthe right R-module M and a ring homomorphism

Φ : RG −→ EndR (M)

(c) Given such an R-algebra homomorphism we obtain by restriction a group representation

ρ : G→ AutR (M) .

Conversely, any such group representation extends linearly to an R-algebra homomorphismRG→ EndR (M).

Exercise. Let G = 〈σ〉 be a cyclic group of order n, and A = CG, M = C. We have the grouprepresentation

ρ : G→ AutA (C) = C∗

defined by σ 7→ ω, some nth root of unity.This gives by remark (2) a CG-module with left action(

n−1∑i=0

αiσi

)α :=

n−1∑i=0

αiωiα.

17 Maschke’s Theorem

17.1 Reynold’s Operator

Let G be a finite group of order n, and R a commutative ring such that n ∈ R∗. For example, it R isa field then this condition just means charR - n.

Let M be a left RG-module.

Proposition 17.1. The fixed submodule of M is

MG := {m ∈M | gm = m ∀g ∈ G} .

This is an R-submodule of M .


Note that M can be viewed as an (RG,R)-bimodule with right R-action the same as the leftaction. Consider the element

e =1

|G|∑g∈G

g ∈ RG

which is well defined since |G| ∈ R∗.Now consider the right R-linear map induced by left multoplication by e,

(·)\ : M −→M.

This is called the Reynold’s operator.

Lemma 17.1. (1) im((·)\) ⊆MG

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(2) The induced map (·)\ : M → MG splits the inclusion map ι : MG ↪→ M (that is, (·)\ projectsonto MG).

Proof. (1) Let m ∈M , g ∈ G. Then

gm\ = g1

|G|∑h∈G

hm

=1

|G|∑h∈G

ghm

=1

|G|∑h′∈G

h′m

= m\.

So m\ ∈MG and (1) holds.

(2) We need to check (·)\ ◦ ι : MG ↪→M →MG is the identity on MG. Let m ∈MG, then

(ιm)\ = m\

=1

|G|∑g∈G

gm

=1

|G||G|m since m ∈MG

= m.

So (·)\ ◦ ι = idMG and \ splits ι.

17.2 HomR (M,N)

Let R be a commutative ring, G a finite group and M,N R-modules.

Proposition 17.2. The abelian group HomR (M,N) has the structure of an R-module with scalarmultiplication defined by

(rf)(m) := rf(m) = f(rm)

for r ∈ R, f ∈ HomR (M,N) and m ∈M .


Proposition 17.3. Suppose now that M,N are RG-modules, corresponding to R-linear group rep-resentations ρM : G→ AutR (M) and ρN : G→ AutR (N).

Then H := HomR (MR, NR) is a left RG-module with corresponding group representation

ρ : G −→ AutR (HR)

g 7→((M → N) 3 f 7→ ρN (g) ◦ f ◦ ρM (g−1) ∈ (M →M → N → N)

).

Proof. It suffices to show ρ is a well defined group homomorphism.

(1) Note ρN (g) ◦ f ◦ ρM (g−1) ∈ H since it is a composition of R-linear maps

(2) Also ρ(g) : f 7→ ρN (g) ◦ f ◦ ρM (g−1) is in AutR (H) since it is clearly invertible (with inverseρ(g−1)) and it is clearly (exercise) R-linear.

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(3) ρ is a group homomorphism, for given g, h ∈ G we have for all f ∈ H,

[ρ(gh)](f) = ρN (gh) ◦ f ◦ ρM (h−1g−1)

= ρN (g) ◦ ρN (h) ◦ f ◦ ρM (h−1) ◦ ρM (g−1)

= ρ(g)(ρ(h)f)

= (ρ(g)ρ(h))(f).

Lemma 17.2. Let M,N be RG-modules. A function f : M → N is RG-linear if and only if it isR-linear and for all g ∈ G, m ∈M ,

f(gm) = gf(m).

Proof. The forward direction is clear. For the converse, it suffices to check

f

∑g∈G

rgg

m

=

∑g∈G

rgg

f(m).

But

f

∑g∈G

rgg

m

= f

(∑g∈g

rggm

)

=∑g∈G

rgf(gm)

=

∑g∈G

rgg

f(m).

Proposition 17.4. Let M,N be RG-modules. Then

HomR (M,N)G = HomRG (M,N) .

Proof. Let f ∈ HomR (M,N) and g ∈ G. Then

gf = f ⇔ [ρ(g)]f = f

⇔ f ◦ πM (g)ρN (g) ◦ f⇔ ∀m ∈M, f(gm) = gf(m).

Lemma (17.2) then implies that f ∈ HomR (M,N)G ⇔ f ∈ HomRG (M,N).

Theorem 17.1 (Maschke). Let G be a finite group, k a field such that char k - |G|. Then kG is left(and right) semisimple.

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Proof. Consider the inclusion of kG-modules

ι : N →M.

It suffices to show that ι splits.Now k is a field and therefore is semisimple, so there is a k-module splitting π : M → N ∈

Homk (M,N). Proposition (17.3) and lemma (17.1) imply that π\ ∈ Homk (M,N)G = HomkG (M,N).It now suffices to show π\ splits ι, that is,

π\ ◦ ι = idN .

Let n ∈ N , then

π\(ι(n)) = π\(n)

=

1

|G|∑g∈G

g

π

(n)

=1

|G|∑g∈G

gπ(g−1n)

=1

|G|∑g∈G

gg−1n

=1

|G|∑g∈G

n

= n.

So π\ ◦ ι = idN and therefore π\ splits ι and hence kG is left semisimple.

Exercise. Let G = 〈σ〉 ∼= Z/2Z act on the vector space V of real valued functions on R by

(σf)(x) := f(−x).

V is an RG-module. Explore the Reynold’s operator and the fixed submodule in this case.

18 Wedderburn’s Theorem

Let R be a ring.

18.1 Schur’s Lemma

Lemma 18.1 (Schur). Let M,N be simple R-modules.

(1) Any R-module homomorphism ϕ : M → N is either 0 or invertible

(2) EndR (M) is a division ring

Proof. It is clear that (1) implies (2) as every nonzero element of EndR (M) is invertible by (1). So itremains to prove (1).

Suppose ϕ : M → N is not the zero map. Then im(ϕ) ≤ N is nonzero, so we must have im(ϕ) = Nas N is simple. Also, kerϕ ≤ M and kerϕ 6= M so kerϕ = 0 and thus ϕ is injective. So ϕ is anisomorphism.

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18.2 Structure Theory

Lemma 18.2. The isomorphism of abelian groups

Φ : E −→ EndR (R) = HomR (RR, RR)

r 7→ (λr : s 7→ rs)

is an isomorphism of rings.

Proof. Just note that λ1 = idR = 1EndR(R) and for r, s ∈ R, we have

Φ(rs) = λrs = λrλs = Φ(r)Φ(s).

Theorem 18.1 (Wedderburn). Let R be a right semisimple ring. Then

R ∼= Mn1(D1)× · · · ×Mnr(Dr)

where the Di are division rings, and r, ni ∈ Z.

Proof. Theorem (15.1) implies that

RR ∼= V n11 ⊕ V n2

2 ⊕ · · · ⊕ V nrr

for some simple nonisomorphic modules Vi.

Exercise. The sum is finite because the image of 1 in each component must be nonzero.

Schur’s lemma implies that

HomR (Vi, Vj) =

{0 i 6= j

Di i = j

where Di is a division ring.Now, lemma (18.2) implies that we have the following ring isomorphisms

R ∼= EndR (RR)

= EndR (V n11 ⊕ · · · ⊕ V nr

r )

∼=r⊕

i,j=1

HomR

(V nii , V

nj

j

)by the universal property of ⊕

= Mn1(D1)× · · · ×Mnr(Dr).

Remark. The Mni(Di) are called the simple or Wedderburn components of R. They are welldefined by the following addendum:

Addendum 18.1. The numbers ni and division rings Di are uniquely determined by R up to iso-morphism and permutation.

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Proof. Continuing the above notation, suppose also that

R ∼= Mn′1(D′1)× · · · ×Mn′s(D

′s)

for some division rings D′i. Then (abusing notation)

RR =(D′

n′11

)n′1 ⊕ · · · ⊕ (D′n′ss )n′s .The Jordan-Holder theorem implies that (on reindexing) we may suppose ni = n′i and Vi ∼= (D′i)

n′i

(and r = s).To show D′i

∼= Di, we use

Lemma 18.3. For R = Mn(D) with D a division ring, we have

EndR (Dn) = D.

Proof. Note that EndR (Dn) is a subring of EndD (Dn) = Mn(D).

Exercise. Show it is the subring of scalar matrices.

Corollary 18.1. A right semisimple ring is left semisimple and conversely a left semisimple ring isright semisimple.

Proof. If R is right semisimple, then R ∼= Mn1(D1)×· · ·×Mnr(Dr) where Di are division rings. Theseare left semisimple.

For the converse, we can apply the left hand version of Wedderburn’s theorem, or note

Exercise. R is left semisimple if and only if Rop is right semisimple.

So left semisimplicity implies right semisimplicity.

Scholium 1. Any semisimple ring is noetherian and artinian.

18.3 Module Theory

Corollary 18.2. Let R ∼= Mn1(D1)× · · · ×Mnr(Dr) where the Di are all division rings.

(1) The only simple modules up to isomorphism are Dn11 , . . . , Dn1

r

(2) Any R-module is a direct sum of these simples.

Proof. (1)⇒ (2) since R is semisimple, so it remains to prove (1).Let M be a simple right R-module. Proposition (14.1) implies that we know M ∼= R/I for some

maximal ideal I.Now M is a composition factor for R since any composition series for I gives one for R. But

RR ∼=⊕

(Dnii )ni so the composition factors are all of the form Dni

i .

Definition 18.1. For R ∼= Mn1(D1) × · · · ×Mnr(Dr) with each Di a division ring, the componentsof any R-module are called isotypic components.

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18.4 Finite dimensional Semisimple Algebras

Let k be an algebraically closed field, and R a finite dimensional k-algebra.

Lemma 18.4. If R is a division ring, then R = k.

Proof. Exercise (or note for r ∈ R, k[r] ⊂ R is a finite field extension of k).

Corollary 18.3. Let R be a semisimple k-algebra whose simple modules have dimension n1, . . . , nr.Then

dim kR = n21 + · · ·+ n2

r .

Proof. By Wedderburn’s theorem and lemma (18.4) we know R ∼= Mn1(k)×· · ·×Mnr(k) so the simplesare kn1 , . . . , knr . Then

dim kR = dim∏

Mni(k) =∑

n2i .

Example 18.1. Let G = S3, the symmetric group on three symbols. Maschke’s theorem implies that

CG is semisimple, so CG ∼=r∏i=1

Mni(C), where

n21 + · · ·+ n2

r = dimCCG = |G| = 6.

The only possibilities are

6 = 12 + · · ·+ 12 or

6 = 12 + 12 + 22.

But CG 6∼= C× C× · · · × C, as C6 is commutative but CG is not. So

CG ∼= C× C×M2(C).

It follows that there are exactly three simple CG-modules up to isomorphism.

19 One-dimensional Representations

19.1 Abelianisation

Let G be a group.

Definition 19.1. Let g, h ∈ G. The commutator of g and h is

[g, h] = g−1h−1gh.

The subbgroup of G generated by these is called the commutator subgroup and is denoted [G,G].

Lemma 19.1. (1) C := [G,G] EG

(2) G/C is an abelian group called the abelianisation of G (denoted Gab)

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(3) Let H be any abelian group and ϕ : G→ H any group homomorphsim. Then

kerϕ ⊇ C = [G,G]

so there is a commutative diagram of the form

Gϕ //

π

!!DDD

DDDD

D H

G/C

ϕ

==zz

zz

(that is, ϕ = ϕ ◦ π).

Proof. (1) Let f, g, h ∈ G. Then

f−1[g, h]f = f−1g−1h−1ghf

= (f−1g−1f)(f−1h−1f)(f−1gf)(f−1hf)

= [f−1gf, f−1hf ]

∈ [G,G]

so f−1[g, h]f ∈ [G,G] and hence [G,G] EG.

(2) We have

(gC)(hC) = gh(h−1g−1hg)C

= hgC

= hCgC

so G/C is abelian.

(3) It suffices to show ϕ([g, h]) = 1H . But

ϕ(g−1h−1gh) = ϕ(g)−1ϕ(h)−1ϕ(g)ϕ(h) = 1

since H is abelian.

19.2 One-dimensional representations

Let k be a field.

Lemma 19.2. The isomorphism classes of one-dimensional kG-modules are in one-to-one correspon-dence with group representations of the form

ρ : G→ Autk (k) = k∗ = GL1(k)

that is, one-dimensional representations.

Proof. Suppose we have a one-dimensional kG-module V . It corresponds to a group representation ofthe form

ρV : G→ Autk (V ) ∼= k∗

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with the isomorphism given by the map α ∈ k∗ 7→ λα, the multiplication by α map. This one-dimensional representation depends only on the isomorphism class of V , for if ϕ : V → W is akG-module isomorphism then for any v ∈ V, g ∈ G we have

ϕ(gv) = gϕ(v).

Thereforeϕ(ρV (g)v) = ρW (g)ϕ(v) (19.1)

and hence V ∼= W ⇒ ρV = ρW .Conversely, if ρV = ρW then by (19.1) any vector space isomorphism is an isomorphism of kG-

modules. We also know any one-dimensional representation arises from the above since there is akG-module structure on k associated to any one-dimensional representation

ρ : G→ k∗ = Autk (k) .

Exercise. Extend this to n-dimensional kG-modules as follows.An n-dimensional representation is a group representation of the form

ρ : G −→ Autk (kn) ∼= GLn(k).

We put an equivalence relation on the representations by ρ1 ∼ ρ2 if ρ1 = cA ◦ ρ2 : Gρ2−→ GLn(k) →

GLn(k) where cA is conjugation by some matrix A ∈ GLn(k).Equivalence classes of n-dimensional representations correspond to the isomorphism classes of n-

dimensional kG-modules.

Corollary 19.1. The isomorphism classes of one-dimensional modules correspond to the elements ofthe abelian group HomZ (Gab, k

∗).

Proof. Use lemmas (19.1)(3) and (19.2).

Remark. (1) If G is finite and you know to write

G ∼= Z/a1Z⊕ · · · ⊕ Z/arZ

then we know

HomZ (G, k∗) = HomZ

(r⊕i=1

Z/aiZ, k∗)

=

r∏i=1

HomZ (Z/aiZ, k∗)

=

r∏i=1

µai,k

where µai,k is the subgroup of aith roots of unity in k. The last equality follows by the universalproperty of quotients and the fact

HomZ (Z, k∗) = k∗

so then HomZ (Z/aiZ, k∗) are those α ∈ k∗ with αai = 1.

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(2) The zero in HomZ (Gab, k∗) is the trivial representation ρ : G 7→ 1k.

Corollary 19.2. Let k be an algebraically closed field with char k - |G|, and G a finite group. Then

kG ∼=|Gab|∏i=1

k ×s∏j=1

Mnj (k)

where all nj ≥ 2.

Proof. Follows from section 18 and the fact that the number of one-dimensional representations is

|HomZ (Gab, k∗)| = |Gab|

Example 19.1. Let G = A4, the alternating group on four symbols. Then |A4| = 12. Let

H := {(12)(34), (13)(24), (14)(23)} ⊂ A4

Exercise. This is a normal subgroup of A4 by computation or Sylow’s theorem.

In fact

Exercise. Show H = [G,G]

and so Gab = G/H ∼= Z/3Z. It follows that |G/H| = 3. Hence the one-dimensional representationsover C are the elements of HomZ (Gab,C∗), and there are |Gab| = 3 of these. They are

ρ0 : G −→ 1 ρ1 : G −→ C∗ ρ2 : G −→ C∗

H 7→ 1 H 7→ 1

(123) 7→ e2πi/3 (123) 7→ e4πi/3

(123)2 7→ e4πi/3 (123)2 7→ e2πi/3

Since dimCCG = |G| = 12, corollary (19.2) implies that

CG ∼= C× C× C×M3(C).

Therefore there are 4 simple CG-modules up to isomorphism – three with dimension 1 and one withdimension 3.

Exercise. Figure out the three-dimensional representation.

20 Centres of Group Algebras

20.1 Centres of Semisimple Rings

Proposition 20.1. Let R = R1 × · · · ×Rr. Then

Z(R) = Z(R1)× · · · × Z(Rr).

Proof. z = (z1, . . . , zr) ∈ Z(R) if and only if for any (r1, . . . , rr) ∈ R we have (z1, . . . , zr)(r1, . . . , rr) =(r1, . . . , rr)(z1, . . . , zr). Therefore we must have ziri = rizi for all i, and so zi ∈ Z(Ri).

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Proposition 20.2. Let R be a ring. Then Z(Mn(R)) = Z(R).[Note: the right hand side of the equality is technically Z(R)1Mn(R).]

Proof. Easy exercise – check commutativity with row and column swaps.

Corollary 20.1. Let k be an algebraically closed field and A a finite dimensional semisimple k-algebra.Then

dimk Z(A) = #(Wedderburn components of A) = #(isomorphism classes of simple kG-modules).

Proof. Wedderburn’s theorem implies that A ∼=r∏i=1

Mni(k). Then propositions (20.1) and (20.2) imply

that

Z(A) =r∏i=1

Z(k)

=r∏i=1

k

so dimk Z(A) = #(Wedderburn components of A).

20.2 Centres of group algebras

Let R be a commutative ring, and G a group.

Lemma 20.1.Z(RG) = {z ∈ RG | zg = gz ∀g ∈ G} .

Proof. We see ⊆ so we check the other inclusion, so assume z ∈ RG commutes with every g ∈ G.Consider an arbitrary element of RG,

∑rgg. Then

z∑g∈G

rgg =∑g∈G

zrgg

=∑g∈G

rgzg

=∑g∈G

rggz

=

∑g∈G

rgg

z.

Let C ⊂ G be a conjugacy class and ΣC =∑g∈C

g.

Proposition 20.3. Z(RG) =⊕

C RΣC where C ranges over all conjugacy classes of G.

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Proof. By lemma (20.1),

z =∑

rgg ∈ Z(RG)⇔ zh = hz ∀h ∈ G

⇔∑g∈G

rgg =∑g∈G

rgh−1gh ∀h ∈ G

⇔∑g∈G

rgg =∑l∈G

rll where l = h−1gh

⇔ ∀g, h ∈ G we have rg = rhgh−1

⇔ for each conjugacy class, there is a scalar rC ∈ Rsuch that rg = rC for all g ∈ C

⇔ z =∑C

rcΣC .

Corollary 20.2. Let k be an algebraically closed field and G a finite group with char k - G. Then

#(simple kG-modules up to isomorphism) = #(conjugacy classes of G).

Proof. Maschke’s theorem implies that kG is semisimple, so corollary (20.1) implies that dimk Z(kG)is equal to the number of simple kG-modules. But proposition (20.3) implies that dimk Z(kG) is equalto the number of conjugacy classes of G, so we are done.

20.3 Irreducible representations of dihedral groups

LetG = Dn = 〈σ, τ | σn = 1 = τ2, τ−1στ = σ−1〉

be the dihedral group of order 2n with n ≥ 3 odd. We will work over k = C.First step is to find Gab. See

[σ, τ ] = σ−1τ−1στ = σ−2

which generates 〈σ〉 since n is odd. Hence 〈σ〉 ⊆ [G,G], but G/〈σ〉 ∼= Z/2Z is abelian, so 〈σ〉 = [G,G]and Gab = Z/2Z.

The one-dimensional complex representations correspond to elements of

HomZ (Gab,C∗) ∼= {±1} .

Thus the representations are

ρ0 : G −→ 1 ρ−1 : G −→ C∗

〈σ〉 7→ 1 〈σ〉 7→ 1

〈τ〉 7→ 1 〈τ〉 7→ −1.

We now find the conjugacy classes. στσ−1 = τσ−2 shows (exercise) that all reflectionsτ, τσ, . . . , τσn−1 are conjugate.

The other conjugacy classes are{σ, σ−1

},{σ2, σ−2

}, . . . ,

{σ(n−1)/2, σ(n+1)/2

}.

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There are (n− 1)/2 + 2 = (n+ 3)/2 conjugacy classes. It then follows that there are (n+ 3)/2 simpleCG-modules. Furthermore, we have

dimCCDn = 2n = 2 +

(n− 1

2

)· 4 = 12 + 12 + 22 + 22 + · · ·+ 22︸︷︷︸

(n−1)/2 terms

.

Thus corollary (20.2) implies that

CDn∼= C× C×

n−12∏i=1

M2(C).

There are two one-dimensional simple CDn-modules, all of the others are dimension 2. Thesetwo-dimensional simples are given by representations as follows for j = 1, . . . , n−1

2

πj : G −→ GL2(C)

σ 7→(e2πij/n 0

0 e−2πij/n

)τ 7→

(0 11 0

).

Note that these are simple because they are not direct sums of one-dimensionals. Also, they arenonisomorphic because the eigenvalues of ρj(σ) differ (see exercise in section 19).

Exercise. Do this for n even.

21 Categories and Functors

Definition 21.1. A category C consists of the following data:

(1) A class Obj C of objects

(2) A collection of sets HomC (X,Y ), one for each ordered pair X,Y ∈ Obj C whose elements arecalled morphisms (from X → Y ), denoted by ϕ : X → Y .

(3) A collection of mappings, one for each ordered triple X,Y, Z ∈ Obj C , with

HomC (X,Y )×HomC (Y,Z) −→ HomC (X,Z)

(ϕ,ψ) 7→ ψϕ = ψ ◦ ϕ

called the composition of ϕ,ψ.

These data also satisfy the following conditions:

(a) The sets HomC (X,Y ) are piecewise distinct

(b) The composition of morphisms is associative: (ξψ)ϕ = ξ(ψϕ) for any ϕ : X → Y, ψ : Y →Z, ξ : Z →W .

(c) For any X ∈ Obj C , ∃ an identity morphism idX : X → X such that (idX)ϕ = ϕ and ψ(idX) = ψwhenever these compositions are defined.

Example 21.1. Grp is the category of groups with

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• Obj Grp: the class of all groups

• HomGrp (G,H): group homomorphisms from G to H

• composition: composition of group homomorphisms.

Example 21.2. Ring, the category of rings

Example 21.3. Let R be a ring. We then have two induced categories

• R-mod: the category of left R-modules

• mod-R: the category of right R-modules

Example 21.4. Let R be a commutative ring. We then have the category R-alg of R-algebras.

Example 21.5. Top, the category of topological spaces.

21.1 Functors

Definition 21.2. A covariant functor F from category C to category D (notated F : C → D)consists of the following data:

(1) A mapping from Obj C to Obj D denoted by X 7→ FX

(2) For each X,Y ∈ Obj C , a function

F : HomC (X,Y ) −→ HomD (FX,FY ) .

These data also satisfy the following conditions:

(a) F (ψϕ) = FψFϕ for any ϕ,ψ ∈ Mor(C ) defined

(b) F (idX) = idFX for all X ∈ Obj C

If F is a functor, we also say F is functorial.

Example 21.6 (Group algebra functor). Fix R a commutative ring. We have the following covariantfunctor

F : Grp −→ R− alg

defined by the mappings

• Objects: for G ∈ Grp, have FG = RG, the group algebra over R

• Morphisms: for ϕ : G→ H we have

Fϕ : RG −→ RH∑rgg 7→

∑rgϕ(g).

We need Fϕ to be an R-algebra map, that is

F : HomGrp (G,H) −→ HomR−alg (RG,RH)

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It is clear that ϕ is R-linear so we check that products are preserved:

Fϕ

((∑g

rgg)(∑h

shh)

)= Fϕ

∑k∈G

∑gh=k

(rgsh)k

=∑k∈G

∑gh=k

rgsh

ϕ(k)

=∑k∈G

∑gh=k

rgsh

ϕ(g)ϕ(h)

=

(∑g

rgϕ(g)

)(∑h

shϕ(h)

)= Fϕ(

∑rgg)Fϕ(

∑shh).

Exercise. Check F is a functor.

Definition 21.3. A contravariant functor F : C → D is defined similarly to definition (21.2)except (2) is replaced with

(2′) For each X,Y ∈ Obj C , a function

F : HomC (X,Y ) −→ HomD (FY, FX)

and (a) is replaced with

(a′) F (ϕψ) = F (ψ)F (ϕ) for any ϕ,ψ ∈ Mor(C ) defined.

Remark. Let R,S be rings. Note that R-mod (and mod-R) have the additional property that forM,N ∈ ObjR-mod,

HomR−mod (M,N) = HomR (M,N)

are abelian groups, so a functor F : R-mod −→ S-mod is additive if all of the morphism maps aregroup homomorphisms.

21.2 Hom functor

Let R be a ring, and M,M ′, N ∈mod-R. We define a contravariant additive functor

Fn = HomR (−, N) : mod−R −→mod− Z

by

• Fn(M) = HomR (M,N), and

•

FN HomR

(M,M ′

)−→ HomZ

(HomR

(M ′, N

),HomR (M,N)

)ϕ 7→

((M ′

f−→ N) 7→ (Mϕ−→M ′

f−→ N)).

Proposition 21.1. Fn = HomR (−, N) is indeed a contravariant additive functor. We also sayHomR (M,N) is functorial in M .

Similarly, HomR (M,N) is covariantly functorial in N .

Proof. Mainly exercise.

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22 Tensor Products I

22.1 Construction of Tensor Products

Let R be a ring, M ∈mod-R and N ∈ R-mod. We wish to construct an abelian group M ⊗R N .

We start by considering the free Z-module Z(M × N) with free generators (m,n) as m,n varyover M,N . Consider the subgroup K generated by elements of the form below, where m,m′ ∈ M ,n, n′ ∈ N and r ∈ R

(a) (m+m′, n)− (m,n)− (m′, n)

(b) (m,n+ n′)− (m,n)− (m,n′)

(c) (mr, n)− (m, rn)

Definition 22.1. The tensor product of M and N is the abelian group M ⊗R N = Z(M ×N)/K.We denote the image of (m,n) in M⊗RN by m⊗n and call it an elementary tensor. (a),(b),(c)

above give the following relations

(1) (m+m′)⊗ n = m⊗ n+m′ ⊗ n

(2) m× (n+ n′) = m⊗ n+m⊗ n′

(3) mr ⊗ n = m⊗ rn.

Remark.0M ⊗ n = 0⊗ n+ 0⊗ n =⇒ 0⊗ n = 0.

Example 22.1. Z/2Z⊗Z Z/3Z = 0.We see this as Z/2Z⊗Z Z/3Z is generated by elementary tensors

a⊗ b = a⊗ 4b = 2a⊗ 2b = 0⊗ 2b = 0.

22.2 Universal property of tensor products

Definition 22.2. Let M ∈mod-R, N ∈ R-mod and A an abelian group. A function ϕ : M×N → Ais mid-linear if for m,m′ ∈M , n, n′ ∈ N and r ∈ R we have

(1) ϕ(m+m′, n) = ϕ(m,n) + ϕ(m′, n)

(2) ϕ(m,n+ n′) = ϕ(m,n) + ϕ(m,n′)

(3) ϕ(mr, n) = ϕ(m, rn).

Example 22.2.

ι⊗ : M ×N −→M ⊗R N(m,n) 7→ m⊗ n

is mid-linear by definition (22.1).

Proposition 22.1 (Universal property). With the above notation, there are inverse bijections

ϕ 7−→ (ϕ : m⊗ n 7→ ϕ(m,n))

{mid-linear ϕ : M ×N → A} Φ←→Ψ{abelian group homs ϕ : M ⊗R N → A}

ϕ = ϕ ◦ ι⊗ ←− [ ϕ.

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Proof. Note first that by the universal property of free modules we have a group homomorphism≈ϕ : Z(M × N) → A such that (m,n) 7→ ϕ(m,n). By the universal property of quotients, we need≈ϕ(K) = 0 where K is as in definition (22.1). But this follows precisely because ϕ is mid-linear. So Φis well defined.

Exercise. Check Ψ is well defined, that is, ϕ ◦ ι⊗ is indeed mid-linear.

We now check ΨΦ = id.

[(ΨΦ)ϕ] (m,n) = [(Φϕ) ◦ ι⊗] (m,n)

= (Φϕ)(m⊗ n)

= ϕ(m,n)

so ΨΦ(ϕ) = ϕ and ΨΦ = id.

Exercise. Check ΦΨ = id.

22.3 Functoriality

Let R be a ring, M,M ′ ∈ mod-R, N ∈ R-mod. Consider an R-linear map ϕ : M → M ′ and theinduced homomorphism

ϕ⊗R N : M ⊗R N −→M ′ ⊗R Nm⊗ n 7−→ ϕ(m)⊗ n

Note that this is a homomorphism by the universal property as we can check the mid-linearity of

ϕ : M ×N −→M ′ ⊗R N(m,n) 7−→ ϕ(m)⊗ n.

Proposition 22.2. M ⊗R N is covariantly functorial and additive in both M and N .

Proof. We have 1M ⊗R N = 1M⊗RN . Furthermore, given Mϕ−→M ′

ϕ′−→M ′′, we have

(ϕ′ϕ)⊗R N : M ⊗R N →M ′′ ⊗R N = (ϕ′ ⊗R N)(ϕ⊗R N) : M ⊗R N →M ′ ⊗R N →M ′′ ⊗R N.

Exercise. Check additivity, and the case for N .

22.4 Bimodules

Proposition 22.3. Let R,S, T ∈ Ring, M ∈mod-R, N ∈ R-mod.

(1) If M is an (S,R)-bimodule, then M⊗RN is a left S-module with multiplication s(m⊗n) = sm⊗n

(2) If M is an (R, T )-bimodule, then M ⊗R N is a right T -module, with multiplication (m⊗ n)t =m⊗ nt.

If both (1) and (2) occur, then M ⊗R N is an (S, T )-bimodule.

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Proof. Mainly exercise in checking module axioms. Alternatively, (1) and (2) can be proven along thefollowing lines.

Note λs : M → M given by left multiplication by s is a homomorphism MR → MR. By functori-ality, we have the induced homomorphism of abelian groups

λs ⊗N : M ⊗R N −→M ⊗R N

giving left multiplication by s on M ⊗R N . Functoriality and additivity imply the module axioms,and we are done.

23 Tensor Products II

23.1 Identity

Let R be a ring, and recall R is an (R,R)-bimodule.

Proposition 23.1. Let M ∈ R-mod.

(1) Then there are inverse left R-mod isomorphisms

m 7−→ 1⊗m

MΦ←→Ψ

R⊗RM

rm←− [ r ⊗m

(2) If M is an (R,S)-bimodule for some ring S then Φ,Ψ are right S-linear.

Proof. We check Ψ is well defined by the universal property for tensor products, that is, check

Ψ : R×M −→M

(r,m) 7−→ rm

is mid-linear. Ψ is clearly additive in r and m by the distributive law, and for r′ ∈ R we have

(rr′,m)Ψ7−→ (rr′)m

Ψ←− [ (r, r′m).

Now we check Ψ,Φ are inverses: for r ∈ R, m ∈M

ΦΨ(r ⊗m) = Φ(rm) = 1⊗ rm = r ⊗m

so ΦΨ = id. Similarly ΨΦ = id.

Exercise. Check R-linearity and S-linearity in case (2).

Remark. Similarly, N ⊗R R ∼= N for N ∈mod-R.

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23.2 Distributive Law

Let {Mi}i∈I be a set of right R-modules, and N ∈ R-mod. Recall that we have the canonical injection

ιj : Mj −→⊕i∈I

Mi

m 7−→ (0, . . . , 0,m, 0, . . . ).

Functoriality implies that we have the R-linear map

ιj ⊗N : Mj ⊗N −→

(⊕i∈I

Mi

)⊗N

for each j ∈ I. Furthermore, the universal property of ⊕ gives the map

Φ :⊕i∈I

(Mi ⊗N) −→

(⊕i∈I

Mi

)⊗N.

Proposition 23.2. (1) Φ above is an isomorphism.

(2) If all Mi are (S,R)-bimodules so is⊕Mi and Φ is left S-linear

(3) If N is an (R, T )-bimodule then Φ is right T -linear.

Proof. Mainly exercise.Recall we have the canonical projection

πj :∏i∈I

Mi −→Mj

(mi)i∈I 7−→ mj .

The universal property of the direct product gives map and functoriality of −⊗N give(∏i∈I

Mi

)⊗N −→

∏i∈I

(Mi ⊗N).

Exercise. Show that this restricts to a map(⊕i∈I

Mi

)⊗R N −→

⊕i∈I

(Mi ⊗R N)

which is inverse to Φ.

Exercise. Check S, T -lienarity in cases (2) and (3).

Remark. For M ∈mod-R, {Ni}i∈I ⊆ R-mod, we have similarly that

M ⊗R

(⊕i∈I

Ni

)∼−→⊕i∈I

(M ⊗R Ni).

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Example 23.1. Complexification of a real vector space V =⊕n

i=1Rei. Its complexification is theC-module C⊗R V .

Note that C is a (C,R)-bimodule. Now

C⊗R V = C⊗R

(n⊕i=1

Rei

)

=n⊕i=1

(C⊗R Rei)

=n⊕i=1

C(1⊗ ei)

as C⊗R Rei ∼= C by proposition (23.1).

23.3 Case when R is commutative

Now let R be a commutative ring, and M,N left R-modules. R commutative implies that we canthink of M,N as (R,R)-bimodules with left multiplication equal to right multiplication. Hence wecan define the (R,R)-bimodule M ⊗R N .

However, for r ∈ R, m ∈M, n ∈ N we have

r(m⊗ n) = (rm)⊗ n = (mr)⊗ n = m⊗ (rn) = m⊗ (nr) = (m⊗ n)r

so we always have that left multiplication by r is equal to right multiplication. Thus we get no moreinformation than a left module.

Remark. An upshot of the discussion above is that tensor products of left/right R-modules areleft/right R-modules.

Corollary 23.1. Rn ⊗R Rm ∼= Rnm.More precisely, (

n⊕i=1

Rei

)⊗R

m⊕j=1

Rfj

∼= n⊕i=1

m⊕j=1

R(ei ⊗ fj)

for ei, fj free generators.

Now, consider R-linear endomorphisms ϕ : Rn → Rn and ψ : Rm → Rm given by matrices(ϕij) ∈Mn(R) and (ψij) ∈Mm(R). Functoriality of ⊗ gives a new R-linear endomorphism

ϕ⊗ ψ : Rn ⊗R Rm −→ Rn ⊗R Rm

a⊗ b 7−→ ϕ(a)⊗ ψ(b).

We can represent this with some matrix ((ϕ⊗ ψ)ij) ∈Mnm(R).

Proposition 23.3.tr ((ϕ⊗ ψ)ij) = tr(ϕij) tr(ψij).

Proof. Use the free basis from corollary (23.1), that is, {ei}ni=1 and {fj}mj=1. Then observe

(ϕ⊗ ψ)(ei ⊗ fj) = ϕ(ei)⊗ ψ(fj)

=

(n∑u=1

ϕuieu

)⊗

(m∑v=1

ψvjfv

).

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The coefficient of ei ⊗ fj here is ϕiiψjj . Therefore

tr((ϕ⊗ ψ)ij) =n∑i=1

m∑j=1

ϕiiψjj

=n∑i=1

ϕii

m∑j=1

ψjj

= tr(ϕij) tr(ψij).

23.4 Tensor Products of Group Representations

Let R be a commutative ring, and G a group.

Proposition 23.4. Given two group representations ρV : G → AutR (V ) and ρW : G → AutR (W ),we get a new group representation

ρV⊗RW : G −→ AutR (V ⊗RW )

defined byρV⊗RW (g) = ρV (g)⊗R ρW (g).

This is called the tensor product representation and the RG-module V ⊗RW is called the tensorproduct.

Proof. For g, h ∈ G we check ρV⊗RW is multiplicative.

ρV⊗RW (gh) = ρV (gh)⊗ ρW (gh)

= ρV (g)ρV (h)⊗ ρW (g)ρW (h)ex= (ρV (g)⊗ ρW (g)) ◦ (ρV (h)⊗ ρW (h))

= ρV⊗RW (g) ◦ ρV⊗RW (h).

[Note: the exercise follows from functoriality].

Exercise. Show that the set of one-dimensional representations of G over k form a group under ⊗which is essentially equal to the group

G = HomZ (Gab, k∗) .

24 Characters

24.1 Linear Algebra Recap

Fix a field k.

Lemma 24.1. Let T ∈Mn(k) and U ∈ GLn(k). Then

tr(T ) = tr(U−1TU).

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Proof.det(U−1TU − λI) = det(U−1(T − λI)U) = det(T − λI).

Corollary 24.1. Let V be a finite dimensional k-space, and ϕ : V → V a linear map. Then the traceof ϕ, tr(ϕ) := tr(T ) is independent of the matrix T representing it (with respect to any basis for V inboth the domain and codomain).

Addendum 24.1. Given a linear isomorphism ψ : V → V then

tr(ψ−1ϕψ) = tr(ϕ)

because both ψ−1ϕψ and ϕ can be represented by the same matrix.

Lemma 24.2. Let V be a finite dimensional k-space, and suppose V = V1⊕V2. The linear projectiononto V1, that is, ϕ : V1 ⊕ V2 7→ V1 ↪→ V1 ⊕ V2 has trace trϕ = dimV1.

Proof. Exercise – show that you can find a matrix representing it of the form(IdimV1 0

0 0

)

24.2 Characters

Let G be a group and V a finite dimensional kG-module with corresponding group representationρV : G→ Autk (V ). Note we have ρV (g) : V → V is linear for all g ∈ G.

Definition 24.1. The character of V (or ρV ) is the function

χV : G −→ k

g 7−→ tr(ρV (g)).

We say χV is irreducible if V is an irreducible (that is, simple) kG-module.

Fact 24.1.χV (1) = tr(ρV (1)) = tr(idV ) = dimV.

Characters are invariants of kG-modules in the following sense:

Proposition 24.1. Let ϕ : V∼

−→W be an isomorphism of finite dimensional kG-modules. ThenχV = χW .

Proof. The kG-linearity of ϕ implies that for all g ∈ G, v ∈ V we have ϕ(gv) = g(ϕ(v)). So ϕ(ρV (g)) =ρW (g)ϕ and therefore ρV (g) = ϕ−1ρW (g)ϕ. It follows by the addendum that χV = χW .

Definition 24.2. A function χ : G → k is a class function if it is constant on conjugacy classes.The set of these will be denoted Fk(g). Hence given a conjugacy class C ⊆ G we can define

χ(C) = χ(g) for g ∈ C.

Corollary 24.2. Characters are class functions.

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Proof. Given a character χV , for g, h ∈ G we have

ρV (h−1gh) = ρV (h−1)ρV (g)ρV (h).

The addendum implies that χV (h−1gh) = χV (g).

Example 24.1. LetG = S3 = D3 = 〈σ, τ | σ3 = τ2 = 1, τσ = σ−1τ〉.

Consider the irreducible kG-module V given by the representation

ρV : G −→ GL2(k)

σ 7−→(e2πi/3 0

0 e−2πi/3

)τ 7−→

(0 11 0

)See

χV (1) = dimV = 2

χV (σ) = χV (σ2) = 2 cos2π

3= 1

χV (τ) = χV (στ) = χV (σ2τ) = 0.

24.3 Character Tables

Definition 24.3. Suppose now G is finite, and k is algebraically closed with char k - |G|. LetC1, . . . , Cr be the conjugacy classes of G, and χ1, . . . , χr the irreducible characters.

The character table for G is the r × r-matrix

X = (χi(Cj))ri,j=1

Example 24.2. Let G = S3 = D3.

Conjugacy Classes

Irreducible characters 1{σ, σ2

} {τ, στ, σ2τ

}trivial = χ1 1 1 1

det = χ2 1 1 -1

χV 2 1 0

Table 1: Character table for D3 = S3

24.4 Dual Modules and Contragredient Representation

Recall we have the trivial representation

ρ0 : G −→ 1 ↪→ k∗

which makes k a left kG-module.

Definition 24.4. Given a kG-module V , the dual module is the kG-module V ∗ = Homk (V, k).

Exercise. Let V = kn, and consider the inner product 〈v,v′〉 := vTv′ ∈ k.

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(1) Show that the map Φ : V → V defined by v 7→ 〈v,−〉 is a linear isomorphism

(2) Let Ψ : V → V be a linear map (so it is a matrix). Then show that

〈v,Ψv′〉 = 〈ΨTv,v′〉.Hence the natural induced map

Ψ∗ : V ∗ −→ V ∗

(f : V → k) 7−→ (f ◦Ψ : V → V → k)

is represented by the matrix ΨT .

Proposition 24.2. Let V be a finite dimensional left kG-module, and let g ∈ G. Then

χV ∗(g) = χV (g−1).

Proof. By the definition of V ∗, we have ρV ∗(g) = ρV (g−1)∗ = ρV (g−1)T . Taking traces gives

χV ∗(g) = χV (g−1).

24.5 Hom as tensor product

Proposition 24.3. Let V,W be finite dimensional left kG-modules. Then there is an isomorphismof kG-modules

Φ : W ⊗k V ∗ −→ Homk (V,W )

w ⊗ f 7−→ (v 7→ f(v)w).

Proof. Exercise. Check Φ is an isomorphism of vector spaces and now note that with the abovenotation, for g ∈ G

[Φ(g(w ⊗ f))](v) = Φ(gw ⊗ (f ◦ ρV (g−1))v

= f(g−1v)gw

= g(f(g−1v)w)

= [gΦ(w ⊗ f)](v).

25 Orthogonality

25.1 Characters of ⊕,⊗

Let k be a field, G a group and V,W finite dimensional left kG-modules.

Lemma 25.1. For any g ∈ G,

(1) χV⊕W = χV (g) + χW (g)

(2) χV⊗W = χV (g)χW (g)

Proof. (1) Just take the trace of

ρV⊕W (g) =

(ρV (g) 0

0 ρW (g)

)and the result follows.

(2) Follows from proposition (23.2) [since ρV⊗W = ρV (g)⊗ ρW (g)].

Page 61 of 73


25.2 Orthogonality

For the rest of this section, assume k is an algebraically closed field and hk - |G| <∞ (so kG is finitedimensional and semisimple by Maschke’s theorem).

For V,W left kG-modules, recall Homk (V,W ) is also a kG-module. Hence given a ∈ kG we havethe linear map λa : Homk (V,W )→ Homk (V,W ) defined by left multiplication by a.

From section 17 we know if e = 1|G|

∑g∈G

g then the Reynold’s operator λe is a projection onto

Homk (V,W )G = HomkG (V,W ).

Theorem 25.1 (First orthogonality relation). Let V,W be simple left kG-modules, and χV , χW theircharacters. Then

1

|G|∑g∈G

χV (g−1)χW (g)(∗)= dimk

(Homk (V,W )G

)Schur

=

{1 V ∼= W

0 V 6∼= W.

Proof. Schur’s lemma implies that it suffices to prove (∗). We have

dimk

(Homk (V,W )G

)= tr (λe : Homk (V,W )→ Homk (V,W )) [lemma 24.2]

=1

|G|∑g∈G

tr (λg : Homk (V,W )→ Homk (V,W ))

=1

|G|∑g∈G

tr (λg : V ∗ ⊗W → V ∗ ⊗W ) [prop 24.3]

=1

|G|∑g∈G

χV ∗⊗W (g)

=1

|G|∑g∈G

χV ∗(g)χW (g) [lemma 25.1]

=1

|G|∑g∈G

χV (g−1χW (g) [prop 24.2]

Theorem 25.1 suggests defining the following inner product (that is, mid-k-linear map) on Fk(G)×Fk(G):

〈χ, χ′〉 :=1

|G|∑g∈G

χ(g−1)χ′(g).

Let χ1, . . . , χr be the irreducible characters of G, and C1, . . . , Cr the conjugacy classes. Theorem25.1 can be restated as

Corollary 25.1. {χ1, . . . , χr} is an orthonormal basis for the r-dimensional space of class functionsFk(G).

There is another orthogonality relation.

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Recall that X = (χi(Cj))ri,j=1 is the character table. Let

Γ =

|C1| 0 . . . 0

... |C2|...

.... . .

...0 . . . 0 |Cr|

Y = (χi(C

−1j ))ri,j=1.

Then X,Γ, Y ∈Mr(k).

Exercise. (1) If k = C, Y = XT

(2) Theorem 25.1 can be rewritten asXΓY = |G|Ir

(3) ΓY X = |G|Ir

Then we have

Theorem 25.2 (Second orthogonality relation).

1

|G|

r∑l=1

χl(C−1i )χl(Cj) = δij .

25.3 Decomposing kG-modules into a direct sum of simples

Theorem 25.3. Let k be an algebraically closed field with char k = 0. Let V be a finite dimensionalkG-module. Then χV determines the isomorphism class of V as follows:

(1) χV =∑i

niχI where the Fourier coefficients are given by

ni = 〈χV , χi〉

(2) V ∼=⊕r

i=1 Vnii where Vi is the simple kG-module with character χi.

Proof. Easy. We know V ∼=⊕V nii for some ni ∈ N, but lemma (25.1)(1) implies that χV =

∑niχi.

Part (1) (and hence (2)) follows from orthogonality.

Proposition 25.1. Let k be an algebraically closed field, char k = 0 and V a finite dimensionalkG-module. Then V is simple if and only if

〈χV , χV 〉 = 1.

Proof. Write χV =∑niχi, then

〈χV , χV 〉 =∑

n2i .

This is 1 if and only if all ni = 0 except one.

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g ∈ Ci 1 (12)(34) (123) (132)|Ci| 1 3 4 4

χ0 1 1 1 1

χ1 1 1 ω ω2

χ2 1 1 ω2 ω

χ3 3 = dimVi a b c

Table 2: Character table for A4

Example 25.1. G = A4 ≤ S3.Let ω = e2πi/3. We saw before what all of the one-dimensional representations of A4 are, these

give the first three rows of the character table below.Using orthogonality, we can deduce the values for the three-dimensional character χ3. We know

〈χ0, χ3〉 = 0 =⇒ 3 + 3a+ 4b+ 4c = 0

〈χ1, χ3〉 = 0 =⇒ 3 + 3a+ 4ω2b+ 4ωc = 0

〈χ2, χ3〉 = 0 =⇒ 3 + 3a+ 4ωb+ 4ω2c = 0

From these, we deduce a = −1, b = c = 0.

26 Example from Harmonic Motion

OMITTED

27 Adjoint Associativity and Induction

To keep track of scalars acting on modules we let RM denote a left R-module, NS a right S-moduleand RLS an (R,S)-bimodule.

27.1 Adjoint Associativity

Let R,S, T be rings, M,N modules with scalars given in the above notation.

Proposition 27.1. (1) The abelian group HomR (RMS ,RN) is a left S-module with scalar multi-plication given by

(sϕ)(m) := ϕ(ms)

(2) Similarly HomR (RM,RNT ) is a right T -module with scalar multiplication

(ϕt)(m) := ϕ(m)t

(3) HomR (RMS ,RNT ) is an (S, T )-bimodule.

Proof. Exercise: this can be proved using the functoriality of HomR (−,−) of just checking moduleaxioms directly.

For a subring S of a ring R, recall we have an (R,S)-bimodule RRS .

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Proposition 27.2. The isomorphism of abelian groups from section 5

Φ : S HomR (RRS ,RM) −→M

ϕ 7−→ ϕ(1)

is S-linear.

Proof. Just check S-linearity. We know Φ is additive so consider s ∈ S, so we have

Φ(sϕ) = (sϕ)(1)

= ϕ(s)

= sϕ(1)

= sΦ(ϕ).

Theorem 27.1 (Adjoint Associativity). Let R,S, T be rings and consider modules RBS, SM and RN .

(1) The following is an isomorphism of groups

Φ : HomR (RBS ⊗S SM,RN) −→ HomS (SM, S HomR (RBS ,RN))

ϕ 7−→ (m 7→ ϕ(−⊗m))

(2) If M is an (S, T )-bimodule, this map is left T -linear. If N is an (R, T )-bimodule, then this mapis right T -linear.

Proof. Purely formal but long and tedious and mainly left as an exercise.

Exercise. • Check Φ is well defined, that is

– ϕ(−⊗m) is R-linear

– m 7→ ϕ(−⊗m) is S-linear

• Check Φ is aditive

• Use the universal property to construct an inverse Ψ to Φ as follows:

Suppose given S-linear map ψ : M → HomR (B,N). We get a mid-S-linear map

ϕ : B ⊗M −→ N

(b,m) 7−→ [ψ(m)](b).

Indeed, this is clearly additive in m and b, and

ϕ(bs,m) = [ψ(m)](bs)

= [s(ψ(m))](b)

= [ϕ(sm)](b)

= ϕ(b, sm).

• Check Ψ = Φ−1

• Check linearity in the case (2).

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27.2 Induced Modules

Let k be a field, G a group and H a subgroup of G. kH is a subring of kG, and we also get thebimodule kGkGkH .

Definition 27.1. Let V be a left kH-module. The induced module is the kG-module

kGkGkH ⊗kH V =: IndGH (V ) .

Corollary 27.1 (Frobenius reciprocity). With this notation,

HomkG (kG⊗kH V, kGW ) = HomkH (V, kHW ) .

Proof. Use theorem 27.1 and proposition 27.2.

Example 27.1. Consider the dihedral group

G = Dn = 〈σ, τ | σn = 1 = τ2, τσ = σ−1τ〉.

Let H = 〈σ〉.Suppose we are given a one-dimensional CH-module V = Cv. Hence σv = ωv for some nth root

of unity ω. What is IndGH (V ) = CG⊗CH V ? Observe

CG⊗CH V = (CG⊗CH V )⊕ (τCH ⊗CH V )

= (1⊗ V )⊕ (τ ⊗ V )

as C-spaces. Therefore W = IndGH (V ) is two-dimensional with C-basis {1⊗ v, τ ⊗ v}.

What is ρW : G −→ GL2(C) with respect to this basis?We note that τ swaps 1⊗ v with τ ⊗ v so

ρW (τ) =

(0 11 0

).

Furthermore, we have

σ(1⊗ v) = σ ⊗ v = 1⊗ σv = 1⊗ ωv = ω(1⊗ v)

σ(τ ⊗ v) = (στ ⊗ v) = τσ−1 ⊗ v = τ ⊗ ω−1v = ω−1(τ ⊗ v)

so

ρW (σ) =

(ω 00 ω−1

).

If ω 6= ω−1 these give the two-dimensional simple CG-modules.

28 Annihilators and the Jacobson Radical

28.1 Restriction Functor

Let R be a ring, I CR and consider the canonical quotient map π : R −→ R/I.Recall from section 5 (using the universal property of quotients) that we have a “restriction”

functor

ρ : mod−R/I −→mod−RM 7→MR

where the R-module structure on M is given by

mr := mπ(r) = m(r + I).

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Exercise. This is an additive functor.

Proposition 28.1. The restriction functor ρ identifies the class of (right) R/I-modules with the classof right R-modules such that MI = 0.

Proof. Suppose M ∈mod-R/I. Then it is clear MRI = 0.Conversely, suppose MR ∈mod-R such that MRI = 0. We get an R/I-module on the underlying

abelian group M of MR by defining m(r + I) := mr, which is well defined because MRI = 0.

28.2 Annihilators

Definition 28.1. Let R be a ring, M ∈mod-R, I ER, N ⊆M . We say I annihilates N if NI = 0.[Note that NI = {

∑imiri | mi ∈ N, ri ∈ I}].

The right annihilator of N in R is

r annR (N) = annR (N) = {r ∈ R | nr = 0 ∀n ∈ N}

that is, the maximal subset of R which annihilates N .

Proposition 28.2. With the above notation,

(1) annR (N) is a right ideal of R

(2) If N is a submodule then annR (N) is an ideal.

Proof. Just need to check closure axioms. Let n ∈ N , r′, r′′ ∈ annR (N) , r ∈ R.

(1) • n0 = 0 =⇒ 0 ∈ annR (N)

• n(r′ + r′′) = 0 =⇒ r′ + r′′ ∈ annR (N)

• n(−r′) = −nr′ = 0 =⇒ −r′ ∈ annR (N)

• n(r′r) = 0r = 0 =⇒ r′r ∈ annR (N)

(2) Since nr ∈ N we have n(rr′) = (nr)r′ = 0 so rr′ ∈ annR (N).

Example 28.1. Consider the product of rings R = R1 × · · · ×Rn. Let I = 0×R2 × · · · ×Rn be thekernel of the projection map π1 : R→ R1 (and therefore an ideal of R).

Recall every R-module has the form M1 × · · · × Mn for Ri-modules Mi. Those of the formM1 × 0× · · · × 0 are annihilated by I.

Example 28.2. Let NR, N′R ≤ MR. Suppose I annihilates N and I ′ annihilates N ′. Then I ∩ I ′

annihilates N +N ′.Indeed, see that given any r ∈ I ∩ I ′, n ∈ N, n′ ∈ N ′ we have

(n+ n′)r = nr + nr′ = 0.

Lemma 28.1. Let M ∈mod-R.

(1) Given any m ∈M we have mR ∼= R/ annR (m).

(2) Suppose M is simple. Then for any m ∈M \{0}, annR (M) is a maximal right ideal. Conversely,any maximal right ideal I CR is the annihilator of 1 + I ∈ R/I in the simple module R/I.

Proof. (1)⇒ (2) exercise.

For (1) use the first isomorphism theorem on RλM−−→M .

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28.3 Jacobson Radical

Theorem 28.1. Let R be a ring.

(1) The following subsets of R are equal:

J1 =⋂S

annR (S) for S ranging over all simple R-modules

J2 =⋂I

I for I ranging over all maximal right ideals of R

(2) The common subset J1 = J2 in (1) is an ideal, called the Jacobson radical of R and is denotedJ(R) (or sometimes radR).

(3) Via the restriction functor ρ : mod-R/J(R) −→mod-R, the semisimple R-modules are preciselythe semisimple R/J(R) modules.

Proof. Proposition (28.2)(2) implies (2). Furthermore, proposition (28.1) implies (3). It remains toprove (1). But we have

J1 =⋃S

annR (S)

=⋃S

⋃s∈S

annR (s)

=⋃

S,s∈S\{0}

annR (s)

=⋃

I max

I

= J2.

Definition 28.2. Say R is semiprimitive if J(R) = 0.

Example 28.3. Z is seimprimitive as

J(Z) = 〈2〉 ∩ 〈3〉 ∩ · · · = 0.

Definition 28.3. Let R be a ring and r ∈ R. We say r is nilpotent if rn = 0 for some n > 0.Similarly, S ⊆ R is nilpotent if Sn = 0 for some n.

Proposition 28.3. Let R be a ring, I ⊆ R be a nilpotent right ideal. Then I ⊆ J(R).

Proof. It suffices to show I ⊆ annR (S) for any simple R-module S.Note that SI is a submodule of S so by simplicity is either 0 or S. Suppose by way of contradiction

that SI = S. Pick n large enough so In = 0, then

S = SI = SI2 = · · · = SIn = 0

which is a contradiction. So SI = 0 and therefore I ⊆ annR (S) as required.

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Example 28.4. Let G be a finite group, k a field with char k | G. Consider Σ =∑

g∈G g. Note thatfor any h ∈ G we have

hΣ = Σ = Σh.

Therefore, kΣ is an ideal of kG. Also, (kΣ)2 = kΣ2 = k|G|Σ = 0, so kΣ is nilpotent, and J(kG) ⊇ kΣ.

Exercise. Consider rings R,S and (R,S)-bimodule B. We have the ring(R B0 S

)with entrywise addition and matrix multiplication (which is well defined as B is an (R,S)-bimodule).

Then (0 B0 0

)is a nilpotent ideal.

29 Nakayama Lemma and the Wedderburn-Artin Theorem

Fix a ring R.

29.1 NAK Lemma

Lemma 29.1. Let I be a right ideal of R. Then every element of 1 + I has a right inverse in R if andonly if 1 + I ⊂ R∗.

Proof. The reverse direction is clear as two-sided inverses are trivially right inverses as well.We prove the forward direction. Suppose r ∈ I, so 1 + r has right inverse 1 + s ∈ R. Then

1 = (1 + r)(1 + s) = 1 + r + s+ rs

so s = −r − rs ∈ I. Therefore 1 + s has right inverse, say 1 + r′. Byt then

(1 + r) = (1 + r)(1 + s)(1 + r′) = (1 + r′).

Hence (1 + s) is also a left inverse of 1 + r′ = 1 + r, and so 1 + r ∈ R∗.

Lemma 29.2 (Nakayama-Azumaya-Krull (NAK)). Let M be a finitely generated R-module and I aright ideal of R such that 1 + I ⊆ R∗. If MI = M then M = 0 – that is, MI (M unless M = 0.

Proof. By induction on the number of generators n with n = 0 being what we wish to show.Suppose M = m1R+ · · ·+mnR. If MI = M 3 mn = m1r1 + · · ·+mnrn with ri ∈ I. So

mn(1− rn) = m1r1 + · · ·+mn−1rn−1.

But −rn ∈ I so 1− rn ∈ R∗ by the assumption. Therefore

mn = m1r1(1− rn)−1 + · · ·+mn−1rn−1(1− rn)−1

so M is generated by m1, . . . ,mn−1 and we are done by induction.

Corollary 29.1. Let I be a right ideal of R such that 1 + I ∈ R∗. For any simple right R-module Mwe have MI = 0. In particular, I ⊆ J(R).

Proof. Simplicity of M implies that MI = 0 or M . Nakayama’s lemma then tells us that MI = 0.

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29.2 Properties of the Jacobson Radical

Proposition 29.1. 1 + J(R) ⊆ R∗.

Proof. Let r ∈ J(R), and suppose by way of contradiction that 1 + r is not invertible. Then lemma29.1 implies that it doesn’t have a right inverse, and so (1 + r)R ( R.

Zorn’s lemma implies that there is a maximal right ideal I which contains (1 + r)R. But nowI ⊇ J(R) so I 3 r, 1 + r =⇒ I 3 1 which contradicts the maximality of I.

Theorem 29.1. The Jacobson radical is equal to any of the subsets listed below

Ja(R) =⋂

S left simple

annR (S)

Jb(R) =⋂

I left maximal

I

Jc(R) = the largest right (or left or two-sided) ideal such that 1 + I ⊆ R∗.

Proof. Ja(R) = Jb(R) by the same proof as in theorem (28.1). Corollary (29.1) and proposition (29.1)imply that J(R) is a maximal right (or left) ideal satisfying 1 + I ⊆ R∗. It is two-sided so the threesubsets in (3) are all equal to J(R).

Left-right symmetry in (3) gives Ja(R) = J(R).

29.3 Wedderburn-Artin Theorem

Theorem 29.2 (Wedderburn-Artin). A ring R is semisimple if and only if it is semiprimitive and is(right) artinian.

Proof. Suppose that R is semisimple, so RR =⊕n

j=1 Ij where Ij are simple right R-modules (that is,minimal right ideals). Since Ij is artinian, so is RR.

Note Mj :=∑i 6=j

Ii is a maximal right ideal as R/Mj∼= Ij which is simple. It follows that

J(R) ⊆n⋂j=1

Mj = 0.

For the converse, we note that the descending chain condition and semiprimitivity of R imply thefollowing facts

(a) Any right ideal I contains a minimal right ideal I ′

(b) Any minimal right ideal I of R is a direct summand of RR

Why? J(R) =⋂L

L where L ranges over all maximal right ideals. Thus there is a maximal right

ideal L such that I ∩L ( I. But I is simple, so I ∩L = 0. Also, L is maximal and so I +L = Rand therefore R = I ⊕ L.

We wish to show RR is a direct sum of simple modules by constructing two sequences of idealsIn, I

′n such that

(i) RR ∼= In ⊕ I ′n

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(ii) In is the internal direct sum of minimal right ideals

(iii) I ′n ) I ′n+1.

Given these sequences, the descending chain condition implies that eventually I ′n = 0 for large enoughn, so RR = In is a direct sum of simples. To generate the sequences, we start with I0 = 0 so I ′0 = R.Assume that In, I

′n are defined. Fact (a) implies that we can find a minimal right ideal I ′′n contained

in I ′n. Fact (b) then implies that the inclusion map I ′′n ↪→ RR splits, and so the inclusion I ′′n ↪→ I ′n alsosplits. Hence

I ′n = I ′′n ⊕ I ′n+1

for some ideal I ′n+1. Define In+1 = In ⊕ I ′′n and we are done.

Definition 29.1. A ring R is simple if it has no nontrivial ideals (that is, the only two-sided idealsare 0 and R).

Proposition 29.2. A ring R is simple (right) artinian if and only if R ∼= Mn(D) for some divisionring D.

Proof. Suppose R is a simple artinian ring. Then simplicity implies that J(R) = 0 so the Wedderburn-Artin theorem implies that R is semisimple.

The wedderburn theorem then implies that R ∼=n∏i=1

Mni(Di) for division rings Di. Simplicity then

implies that n = 1 (else Mn1(D1)× 0× · · · × 0 is a nontrivial ideal).

Proof of the converse is an exercise in the computation of ideals.

30 Radicals and Artinian Rings

30.1 Nilpotence

Theorem 30.1. Let R be a (right) artinian ring. Then J(R) is the unique largest nilpotent right idealof R.

Proof. Proposition (28.3) implies that any nilpotent right ideal is contained in J(R) so it suffices toshow J(R) is nilpotent. The DCC implies that for n large enough, we have

J(R)n = J(R)n+1 = . . .

Suppose J(R) is not nilpotent, so J(R)n+1 6= 0 (that is, J(R)J(R)n 6= 0). Hence DCC implies thatwe can find a right ideal I satisfying

IJ(R)n 6= 0 (30.1)

such that I is minimal amongst right ideals satisfying (30.1). Pick x ∈ I such that xJ(R)n 6= 0. ThenxR ⊆ I and xRJ(R)n 6= 0 so xR = I.

Also, 0 6= IJ(R)n ≤ I and

(IJ(R)n)J(R)n = IJ(R)2n = IJ(R) 6= 0

(this comes from the fact that the minimality of I implies that I = IJ(R)n). Now, the Nakayamalemma and the fact that I is finitely generated (by x) imply that I = 0, a contradiction.

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Lemma 30.1. Let R be a ring, and I an ideal contained in J(R). Then

J(R/I) = J(R)/I.

Proof. This is clear from the definition of J(R/I) and J(R) as an intersection of maximal right (orleft) ideals.

Corollary 30.1. Let R be a (right) artinial ring. Then the Jacobson radical is the only nilpotent idealI with R/I semisimple.

Proof. We know J(R) is nilpotent. Also, lemma (30.1) implies that J(R/J(R)) = J(R)/J(R) = 0,so R/J(R) is semiprimitive. Furthermore, as R is artinian, so is R/J(R). The Wedderburn-Artintheorem then implies that R/J(R) is semisimple.

Suppose conversely that I C R is nilpotent, and R/I is semisimple. We know I ⊆ J(R). Also,the nilpotency of J(R) implies that J(R)/I CR/I is nilpotent. But R/I is semisimple, and thereforehas no nonzero nilpotent ideals (as any such is contained in J(R/I)). Therefore J(R)/I = 0 and thusJ(R) = I.

30.2 DCC implies ACC

Theorem 30.2 (Hopkins-Levitzki). Let R be a right artinian ring. Then R is also right noetherian.

Proof. Theorem (30.1) implies that we have a sequence of ideals

R > J(R) > J(R)2 > · · · > J(R)n = 0.

It suffices to show each quotient Mi = J(R)i/J(R)i+1 is a noetherian right R-module.Note that J(R) annihilates Mi so Mi is an R/J(R)-module. Now, the Wedderburn-Artin theorem

implies that R/J(R) is semisimple. So Mi is a direct sum of simple R/J(R)-modules (which correspondto simple R-modules). But Mi is also an artinian module, so this is a (exercise) finite direct sum.Now, simple modules are noetherian so Mi is as well.

30.3 Computing Some Radicals

Example 30.1. Let R1, R2 be simple artinian rings and consider the (R1, R2)-bimodule R1BR2 suchthat BR2 is artinian.

Exercise. Then

R =

(R1 B0 R2

)is artinian because it is right artinian over

(R1 00 R2

).

Then J =

(0 B0 0

)is a nilpotent ideal and

R/J =

(R1 00 R2

)∼= R1 ×R2

which is semisimple.Corollary (30.1) implies that J = J(R).

Page 72 of 73


A consequence of this is that the subring R =

(M2(k) M2(k)

0 M2(k)

)of M4(k) has Jacobson radical(

0 M2(k)0 0

).

Definition 30.1. Let R be a ring, and x ∈ R. We say x is normal if xR = Rx.In this case, xR is a two-sided ideal.

Remark. Let x ∈ R be normal and nilpotent. Then xR is a nilpotent ideal.Why? Suppose xn = 0, then

(xR)n = xRxR . . . xR = x2R2xRxR . . . xR = xnR = 0.

Example 30.2. Let k be a field, ζ ∈ k be a primitive nth root of unity. We let

R =k〈x, y〉

〈yx− ζxy, xn, yn − α〉

for some α ∈ k such that yn − α ∈ k[y] is irreducible. We then want to find J(R).

R is finite dimensional over k with spanning set{xiyj

}n−1

i,j=0, so R is artinian.

Now x ∈ R is nilpotent and is normal because by using yx− ζxy = 0 we camn write any p(x, y)xas xq(x, y) for some q(x, y) ∈ R. Thus Rx ⊆ xR. A similar argument gives xR ⊆ Rx and so x isnormal.

The remark then implies that xR ⊆ J(R).

Claim 30.1. xR = J(R).

This follows because

R/xR =k〈x, y〉

〈yx− ζxy, xn, yn − α, x〉

=k[y]

〈yn − α〉

which is a field. This is semisimple so corollary (30.1) implies xR = J(R).

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