TIFR Notes on Semisimple Rings, Central Simple Algebras, Etc. · 2 Semi-Simple Modules and Rings 13 3 Central Simple Algebras 25 4 Brauer Groups of some Fields 39 5 Representations

TIFR Notes on Semisimple Rings,

Central Simple Algebras, Etc.

August 3, 2007

2

Contents

1 Tensor Product 3

2 Semi-Simple Modules and Rings 13

3 Central Simple Algebras 25

4 Brauer Groups of some Fields 39

5 Representations of Finite Groups 47

1

2 CONTENTS

Chapter 1

Tensor Product

§ 1. Tensor product of modules and linear

maps.

Let M,N and P be abelian groups. A map f : M × N → P is calledbi-additive if

f(x1 + x2, y) = f(x1, y) + f(x2, y)

andf(x, y1 + y2) = f(x, y1) + f(x, y2)

for x, x1, x2 ∈ M and y, y1, y2 ∈ N . In other words, f is bi-additive if themaps x 7→ f(x, y) and y 7→ f(x, y) are homomorphisms. It follows that

f(x, 0) = f(0, y) = 0

andf(x,−y) = f(−x, y) = −f(x, y)

for x ∈M and y ∈ N .

Let A be a ring, M a right A-module, N a left A-module and P an abeliangroup. Then a map f : M ×N → P is called balanced (over A, to be precise)if it is bi-additive and satisfies

f(xa, y) = f(x, ay)

for x ∈M, y ∈ N and a ∈ A.

3

4 CHAPTER 1. TENSOR PRODUCT

A tensor product of M and N is a pair (T, φ), where T is an abelian groupand φ is a balanced map from M × N into T , which satisfies the followingcondition:

given any abelian group P and a balanced map f : M × N → P , thereexists a unique homomorphism of groups f : T → P such that the diagram

M ×N

φ

f

""E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

Tf

// P

is commutative.

It is easily seen that if (T ′, φ′) is another tensor product of M and N ,then there exists an isomorphism of groups α : T → T ′ such that φ′ = α φ.In this sense the tensor product is unique (provided that it exists). We nowproceed to show that it does exist.

Let F denote the free abelian group with the set M ×N as a basis. LetH denote the subgroup of F generated by the elements of the form

(x1 + x2, y) − (x1, y) − (x2, y),

(x, y1 + y2) − (x, y1) − (x, y2),

(xa, y) − (x, ay),

for x, x1, x2 ∈ M ; y, y1, y2 ∈ N ; a ∈ A. The natural homomorphism : F →F/H, restricted to M × N , gives a balanced map φ : M × N → F/H,and φ(M × N) generates F/H. Now let P be an abelian group and letf : M × N → P be a balanced map. Then f extends to a homomorphismf : F → P . Since f is balanced f vanishes on H and therefore passes downto a homomorphism f : F/H → P . Evidently the diagram

M ×N

φ

f

!!D

D

D

D

D

D

D

D

D

D

D

D

D

D

D

D

D

D

F/Hef

// P

5

is commutative. Further, f is unique since φ(M × N) generates F/H.Thus (F/H, φ) is a tensor product of M and N .

We shall write M⊗AN for F/H and x⊗y for φ((x, y)), where x ∈ Mand y ∈ N . Since (x, y) 7→ x⊗y is balanced, it follows that every elementof M⊗AN can be written in the form

∑1≤i≤n xi⊗yi for some xi ∈ M and

yi ∈ N .

Remark. M⊗AN can be 0 even if neither M nor N is 0. For instance,Q⊗ZZ/(n) = 0 for n 6= 0.

Let M,M ′ be two right A-modules and N,N ′ two left A-modules. Iff : M →M ′ and g : N → N ′ are A-linear maps, then the composite map

M ×N(f,g)−−−→ M ′ ×N ′ −−−→ M ′⊗AN

′

is balanced. Thus there exists a unique homomorphism of groups f⊗g :M⊗AN →M ′⊗AN

′ such that the diagram

M ×N(f,g)−−−→ M ′ ×N ′

yy

M⊗AN −−−→f⊗g

M ′⊗AN′

is commutative. The map f⊗g is called the tensor product of f and g; it ischaracterized by

(f⊗g)(x⊗y) = f(x)⊗g(y).The map

HomA(M,M ′) × HomA(N,N ′) −→ HomZ(M⊗AN,M′⊗AN

′)

given by (f, g) 7→ f⊗g is bi-additive.

Remarks.

1. 1M⊗1N = 1M⊗N .

2. if Mf−→ M ′ f ′

−→ M ′′ (resp. Ng−→ N ′ g′−→ N ′′) are A-linear maps,

where the M ’s (resp. N ’s) are right (resp. left) A-modules, then

(f ′ f)⊗(g′ g) = (f ′⊗g′) (f⊗g).


§ 2. Structure of module on tensor product

We recall that given two rings A and B, an abelian group M is said to bean A-B-bimodule if M is a module over A and B, and if the endomorphismof M effected by any element of A commutes with the endomorphism of Meffected by any element of B. In other words the multiplication in M byany element of B is A-linear. If M is an A-B-bimodule on which A operatesfrom the left and B operates from the right, then we shall say that M is anA-B-bimodule of the type AMB. Similar explanations for A,BM and MA,B.

Examples.

1. Any two-sided ideal of A is an A-A-bimodule in an obvious manner.

2. If h : A→ B is a homomorphism of rings, then B can be considered asan A-B-bimodule of the type ABB or BBA.

3. If A is a commutative ring, then any A-module can be considered as anA-A-bimodule by taking the given A-module structure on M twice.

Let N be a left A-module and let M be an A-B-bimodule of the type

BMA. We know that the multiplication θb in M by an element b of B isA-linear. This gives rise to an endomorphism θb⊗1 of the group M⊗AN .From the properties of tensor product of linear maps mentioned at the endof §1, it follows that b 7→ θb⊗1 is a homomorphism of rings from B intoEndZ(M⊗AN). ThusM⊗AN acquires the structure of a leftB-module whichis characterized by the formula

b (x⊗y) = (bx)⊗y

for b ∈ B, x ∈ M and y ∈ N . If BM′A is another bimodule, f : M → M ′

linear over A and B, and g : N → N ′ an A-linear map of left A-modulesthen f⊗g is B-linear.

A similar discussion applies to the case where M is an A-B-bimodule ofthe type MA,B or N is an A-B-bimodule (of course A operating on N fromthe left).

Proposition 2.1. If M (resp. N) is a right (resp. left) A-module, then

M⊗AA ∼= M as right A-modules (resp. A⊗AN ∼= N as left A-modules).

7

Proof. The map M × A → M given by (x, a) 7→ xa is balanced and givesrise to a homomorphism φ : M⊗AA→M with φ(x⊗a) = xa. Clearly φ is A-linear. Define an A-linear map ψ : M →M⊗AA by ψ(x) = x⊗1. It is easilychecked that φ and ψ are inverses of each other. So φ is an isomorphism.Similarly A⊗AN is isomorphic to N as left A-modules.

Proposition 2.2. Let A and B be two rings. Let M be a right A-module,

N an A-B-bimodule of the type ANB, and P a left B-module. Then there is

an isomorphism of groups

M⊗A(N⊗BP ) −→∼ (M⊗AN)⊗BP

which maps x⊗(y⊗z) to (x⊗y)⊗z.Proof. For each x ∈M , define a map

αx : N × P → (M⊗AN)⊗BP

by αx((y, z)) = (x⊗y)⊗z. The map αx is balanced (over B) and gives rise toa homomorphism αx : N⊗BP → (M⊗AN)⊗BP . Now define a map

β : M × (N⊗BP ) → (M⊗AN)⊗BP

by β((x,w)) = αx(w) for x ∈ M and w ∈ N⊗BP . Then β is balanced (overA) and gives rise to a homomorphism

β : M⊗A(N⊗BP ) → (M⊗AN)⊗BP

such that β(x⊗(y⊗z)) = (x⊗y)⊗z. Similarly we have a homomorphism (say

γ) in the other direction such that γ((x⊗y)⊗z) = x⊗(y⊗z). The maps β

and γ are inverses of each other and hence β is an isomorphism.

The above discussion about the structure of module on tensor productcan be applied in particular to the following two situations:

(1) Change of rings. Let h : A → B be a homomorphism of rings andlet N be a left A-module. Then B⊗AN is a left B-module in a canonicalmanner (consider B as an A-B-bimodule of the type BBA). We say that theB-module B⊗AN is obtained by a change of base ring (via h). Similarly ifM is a right A-module, then M⊗AB is canonically a right B-module.

The first proposition above says that the module obtained by a change ofrings via the identity map of the base ring is the same as the original module.


The second proposition (combined with the first) gives the transitivity ofchange of rings: if h′ : B → C is another homomorphism of rings and Nis an A-module, say a left module, then the C-module obtained by changeof rings through h′ h : A → C is canonically isomorphic to the C-moduleobtained via h′ from the B-module B⊗AN ontained via h from N . In fact

C⊗B(B⊗AN) ∼= (C⊗BB)⊗AN) ∼= C⊗AN.

(2) Tensor product over commutative rings. Let A be a commutative

ring and letM andN be A-modules. Then bothM andN are A-A-bimodulesand M⊗AN can be made into an A-module either through M or through N(these two module structures on M⊗AN are identical). Moreover; the mapM×N →M⊗AN is A-bilinear in the sense of the following definition: GivenA-modules M,N and P , a map f : M × N → P is called A-bilinear if it isbi-additive and satisfies

f(ax, y) = af(x, y) = f(x, ay)

for x ∈M, y ∈ N, a ∈ A.

A bilinear map is balanced.

If f : M × N → P is A-bilinear, then the map f : M⊗AN → P isA-linear.

Proposition 2.3. If M and N are modules over a commutative ring A, then

M⊗AN and N⊗AM are isomorphic A-modules.

Proof. The map α : M × N → N⊗AM given by α((x, y)) = y⊗x is A-bilinear and gives rise to an A-linear map α : M⊗AN → N⊗AM such thatα(x⊗y) = y⊗x. Similarly we have an A-linear map β : N⊗AM → M⊗AN

such that β(y⊗x) = x⊗y. Clearly α and β are inverses of each other andhence they are isomorphisms.

§ 3. Tensor product of direct sums.

Porposition 3.1 Let (Mi)i∈I and (Nj)j∈J be families of right and left A-

modules respectively. There is an isomorphism(⊕

i∈I

Mi

)⊗A

(⊕

j∈J

Nj

)−→

⊕

(i,j)∈I×J

(Mi

⊗ANj

)

9

which maps(∑

i xi

)⊗(∑

j yj

)to∑

i,j xi⊗yj.

Proof. The map

α :

(⊕

i∈I

Mi

)×(⊕

j∈J

Nj

)−→

⊕

(i,j)∈I×J

(Mi

⊗ANj

)

given by(∑

i xi,∑

j yj

) α7−→ ∑i,j xi⊗yj is balanced and gives rise to a ho-

momorphism

α :

(⊕

i∈I

Mi

)⊗A

(⊕

j∈J

Nj

)−→

⊕

(i,j)∈I×J

(Mi

⊗ANj

)

such that(∑

i xi

)⊗(∑

j yj

)eα7−→ ∑

i,j xi⊗yj. On the other hand, the inclu-sions Mi → ⊕i∈IMi and Nj → ⊕j∈JNj give rise to homomorphisms

βij : Mi⊗ANj →(⊕

i∈I

Mi

)⊗A

(⊕

j∈J

Nj

).

There exists a unique homomorphism

β :⊕

(i,j)∈I×J

(Mi

⊗ANj

)→(⊕

i∈I

Mi

)⊗A

(⊕

j∈J

Nj

)

which extends all the βij’s. Clearly α and β are inverses of each other andhence they are isomorphisms.

Consequences.

1. Let F be a free left A-module with a basis (ej)j∈J say, and let M be aright A-module. Then an element of M⊗AF can be uniquely writtenin the form

∑j xj⊗ej with xj ∈M and xj = 0 for almost all j ∈ J .

2. If F is a free left A-module and f : M →M ′ a monomorphism of rightA-modules, then the induced map f⊗1 : M⊗AF → M ′⊗AF is also amonomorphism.

3. Let h : A → B be a homomorphism of rings and let F be a free leftA-module with a basis (ej)j∈J . Then B⊗AF is a free left B-modulewith a basis (1⊗ej)j∈J .


4. Let A be a ring which admits a homomorphism into a division ringand let F be a free A-module. Then any two bases of F have thesame cardinality. This holds in particular if A is commutative, and thecommon cardinality of any two bases of F is called the rank of F .

5. Let A be a commutative ring and let F ′ and F be free A-modules withbases (e′i)i∈I and (ej)j∈J respectively. Then F ′⊗AF is a free A-modulewith (e′i⊗ej)(i,j)∈I×J as a basis. In particular it follows that

rank of F ′⊗AF = (rank of F ′) × (rank of F ).

Remark. It is not true in general that M⊗A

(Π

j∈JNj

)is isomorphic to

Πj∈J

(M⊗ANj

).

Exercise. Let K be a field, V and W vector spaces over K, V1 a subspace ofV , and W1 a subspace of W . Show that

(V1⊗KW

)∩(V⊗KW1

)= V1⊗KW1,

where the vector spaces involved are identified to subspaces of V⊗KW bymeans of obvious canonical maps.

§ 4. Tensor product of algebras.

Let K be a commutative ring. If A is a K-algebra, then the multiplicationin A is a K-bilinear map of A × A into A. Thus we have a K-linear mapmA : A⊗KA → A such that mA(a⊗a′) = aa′. If B is another K-algebra, wehave a sequence of maps

(A⊗KB

)×(A⊗kB

) φ−→(A⊗KB

)⊗K

(A⊗KB

) τ−→(A⊗KA

)⊗K

(B⊗KB

) ma⊗mb−→ A⊗KB ,

where the map φ is obvious and τ is the unique isomorphism of K-modulessuch that

τ((a⊗b) ⊗ (a′⊗b′)

)= (a⊗a′) ⊗ (b⊗b′).

The composite of these three maps is aK-bilinear map of(A⊗KB

)×(A⊗kB

)

into A⊗KB. This defines a multiplication in the K-module A⊗KB, whichis easily seen to be associative. Also 1⊗1 acts as identity. Thus A⊗KBbecomes a K-algebra and is called the tensor product of the K-algebras Aand B.

11

If A,B and C are K-algebras, then we have isomorphisms of K-algebras:

1. K⊗KA −→∼ A,

2. A⊗KB −→∼ B⊗KA,

3. A⊗K

(B⊗KC

)−→∼

(A⊗KB

)⊗KC.

(The maps are obvious and so are the proofs.)

There are two natural homomorphisms of K-algebras iA : A → A⊗KBand iB : B → A⊗KB given by iA(a) = a⊗1 and iB(b) = 1⊗b. Thesehomomorphisms have the property that iA(a) commutes with iB(b) for alla ∈ A and b ∈ B.

Remark. In general the maps iA and iB are not injective. But if K is a fieldand the algebras A and B are non-zero, then they are injective and they canbe used to identify A and B with their images A⊗1 and 1⊗B respectively inA⊗B.

Proposition 4.1. Let A,B and C be algebras over a commutative ring K,

and let f1 : A → C and f2 : B → C be homomorphisms of K-algebras such

that f1(a)f2(b) = f2(b)f1(a) for a ∈ A, b ∈ B. Then there exists a unique

homomorphism of K-algebras f : A⊗KB → C such that f1 = f iA and

f2 = f iB.

Proof. The map A×B → C given by (a, b) 7→ f1(a)f2(b) is K-bilinear andgives rise to a K-linear map f : A⊗KB → C satisfying f(a⊗b) = f1(a)f2(b).We now check that f preserves multiplication, it suffices to consider elementsof A⊗KB of the type a⊗b. We have then

f((a⊗b)(a′⊗b′)

)= f

(aa′⊗bb′

)= f1(aa

′)f2(bb′)

= f1(a)f1(a′)f2(b)f2(b

′) = f1(a)f2(b)f1(a′)f2(b

′)

= f(a⊗b

)f(a′⊗b′

).

Thus f is a homomorphism of K-algebras. The uniqueness of f is clear.

The commutativity between f1(a) and f2(b) is clearly necessary for theexistence of f .

Change of rings. Let h : K → K ′ be a homomorphism of rings, whereK ′ is commutative, and let A be a K-algebra. Then the homomorphism


iK′ : K ′ → K ′⊗KA maps K ′ into the center of K ′⊗KA, and the latterbecomes a K ′-algebra. We say that the K ′-algebra K ′⊗KA is obtained by achange of base ring (via h). Note that the structure of K ′-module carried byK ′⊗KA as a K ′-algebra is the same as the one it acquires according to themethod described in §2.

As in the case of modules, the transitivity of change of rings holds foralgebras.

Exercise. Let A be an algebra over a commutative ring K. Show that

1. Mm(K)⊗KMn(K) ∼= Mmn(K),

2. A⊗KMn(K) ∼= Mn(A) .

(Here ∼= stands for isomorphism of K-algebras.)

Chapter 2

Semi-Simple Modules andRings

In what follows, a “ring” will always mean a ring with identity and a “sub-ring” will mean a subring which contains the identity of the over-ring. Ho-momorphisms of rings will be assumed to map identity elements into identityelements. All modules will be unitary and unless otherwise stated, the word“module” will mean a left module.

§ 1. Semi-simple Modules.

Let A be a ring. An A-module M is called simple if M 6= 0 and it has nosubmodules other than M and 0. An A-module M is called semi-simple if itis the sum of a family of simple A-submodules.

Note that, by definition, the module 0 is semi-simple (being the directsum of an empty family of simple submodules).

Clearly the direct sum of a family of semi-simple modules is semi-simple.

Lemma 1.1 (Schur). Let A be a ring and let f : M → N be an A-linear

map.

1. If M is simple, then either f = 0 or f is injective.

2. If N is simple then either f = 0 or f is surjective.

3. If both M and N are simple, then either f = 0 or f is an isomorphism.

13

14 CHAPTER 2. SEMI-SIMPLE MODULES AND RINGS

Proof.

1. Since M is simple, the submodule ker f of M is either M or 0.

2. Since N is simple, the submodule f(M) of N is either 0 or N .

3. Follows from 1 and 2.

Corollary. The endomorphism ring of a simple module is a division ring.

Remark. A homomorphic image of a semi-simple module is semi-simple.In fact, letM =

∑i∈I Si be a semi-simple A-module with Si simple and let

f : M → N be a surjective A-linear map. By Schur’s lemma, f |Si: Si → N

is either 0 or injective. Hence f(Si) is either 0 or is isomorphic to Si andtherefore simple. Since N =

∑i∈I f(Si), it follows that N is semi-simple.

Theorem 1.1. Let A be a ring and M an A-module. The following condi-

tions are equivalent.

1. M is semi-simple.

2. M is a direct sum of simple modules.

3. Every submodule of M is a direct summand of M .

We first state two propositions and using them prove the above theorem.

Proposition 1.1. Let M(6= 0) be a finitely generated A-module. Then Mcontains a maximal A-submodule.

Proposition 1.2. Let M be an A-module, N an A-submodule of M and

(Si)i∈I a family of simple A-submodules of M such that M = N +∑

i∈I Si.

Then there exists a subset J of I such that M is the direct sum of the sub-

modules N and Sj with j ∈ J .

Proof of theorem 1.1. 1. ⇒ 2: Let M =∑

i∈I Si with Si simple. ApplyProposition 1.2 with N = 0.

2. ⇒ 3: Let M = ⊕i∈ISi with Si simple. If N is any submodule of M ,we have M = N +

∑i∈I Si. By Proposition 1.2, there exists a subset J of I

such that M = N⊕(∑

j∈J Sj), which shows that N is a direct summand ofM .

15

3. ⇒ 1: We first remark that the property 3 holds for any submodule Nof M . For let P be a submodule of N . There exists a submodule Q of Msuch that M = P⊕Q. It is easily seen that N = P⊕(Q ∩N). Next we notethat a module M(6= 0) satisfying the condition 3 always contains a simplesubmodule. In fact let N(6= 0) be a finitely generated submodule of M (forinstance N can be taken to be Ax for some non-zero element x of M). ByProposition 1.1, N has a maximal submodule P . Since N has the property3, there exists a submodule S of N such that N = P⊕S. In particular S isisomorphic to N/P and therefore simple.

To prove 3 ⇒ 1, we can assume that M 6= 0. Let N be the sum of allsimple submodules of M . We claim that N = M . For, otherwise we haveM = N⊕P for a non-zero submodule P of M . Since P has the property 3,it has a simple submodule. This contradicts the definition of N .

Proof of Proposition 1.1. Let (Mi)i∈I be a family of proper submodulesof M , totally ordered by inclusion. The union N =

∑i∈I Mi is again an

A-submodule of M . Further N 6= M . For, let x1, x2, . . . , xn be a systemof generators of M . If N = M , then xk ∈Mik , k = 1, 2, . . . , n, which implies(since (Mi) is a totally ordered family) that x1, x2, . . . , xn ∈ Mi for some i.This gives M = Mi. Contradiction. Thus by Zorn’s lemma, the set of propersubmodules of M has a maximal element, i.e., a maximal submodule of M .

Proof of Proposition 1.2. Let S be the set of subsets K of I ordered byinclusion such that the sum N +

∑k∈K Sk is a direct sum of the submodules

N and (Sk)k∈K . Since this property is satisfied for K = ∅, the set S is notempty. Let (Kγ)γ∈Γ be a totally ordered family of elements of S. We claimthatK =

⋃γ∈ΓKγ is an element of S. For, let x+x1+· · ·+xn = 0 with x ∈ N

and xr ∈ Skr, kr ∈ K (r = 1, 2, . . . , n). We have kr ∈ Kγ(r = 1, 2, . . . , n) for

a suitable γ ∈ Γ. Therefore x = x1 = · · · = xn = 0. Thus N +∑

k∈K Sk is adirect sum of the submodules N and (Sk)k∈K , i.e., K ∈ S. By Zorn’s lemma,there exists a maximal element J ∈ S. Let M ′ = N +

∑j∈J Sj. We assert

that M = M ′. For, suppose M 6= M ′. Then there exists an i ∈ I such thatSi * M ′ so that Si∩M ′ = 0 (because Si is simple). This shows that the sumN +

∑j∈J Sj + Si is direct. Thus J ∪ i ∈ S, contradicting the maximality

of J .

Corollary. A submodule of a semi-simple module is semi-simple.


Remarks.

1. If M =∑

i∈I Si with Si simple and S is a simple submodule of M ,then S is isomorphic to Si for some i ∈ I. For, by 3 of Theorem1.1, Sis a direct summand of M and hence there exists a surjective A-linearmap p : M → S. There exists an i ∈ I such that p(Si) 6= 0 (otherwisep = 0). By Schur’s lemma p : Si → S is an isomorphism.

2. If M = S1⊕ · · ·⊕Sn = S ′1⊕ · · · ⊕S ′

n with Si and S ′j simple, then n = m

and there exists a permutation σ of 1, 2, . . . , n such that Si∼= S ′

σ(i)

for i = 1, 2, . . . , n. This follows, for example, from the Jordan-Holdertheorem for modules.

Exercises.

1. Let M be a semi-simple module. Then M has finite length n if andonly if M is a direct sum of n simple modules.

2. Show that a vector space is semi-simple.

3. Show that if n > 1, the Z-module Z/(n) is semi-simple if and only if nis a product of distinct primes.

§ 2. Isotypical Components.

Let S be a simple A-module. An A-module M is said to be isotypical of

type S, if M is the sum of a family of simple submodules each of which isisomorphic to S.

Let M = ⊕i∈ISi with Si simple. By collecting together isomorphic Si’swe can write M = ⊕γ∈ΓMγ, where Mγ is the direct sum of submodulesisomorphic to simple module Sγ and such that Sγ is not isomorphic to Sγ′

for γ 6= γ′. If S is a simple submodule of M , then there exists γ ∈ Γ suchthat S ∼= Sγ. We assert that S ⊆ Mγ. In fact, consider the natural mapp : M → M/Mγ . If p restricted to S were not 0, then by Schur’s lemmaM/Mγ = ⊕γ′∈Γ\γMγ′ would contain a submodule isomorphic to S, which isa contradiction. So, p(S) = 0, i.e., S ⊆ Mγ . This shows that Mγ is the sumof all submodules of M which are isomorphic to Sγ . The submodules Mγ arecalled isotypical components of M .

17

Proposition 2.1. The isotypical components of a semi-simple module Mare left invariant under every endomorphism of M . Every sub-module of M ,

invariant under all endomorphisms of M is a sum of isotypical components

of M .

Proof. Let f be an endomorphism of M and let Mγ be an isotypical com-ponent of type Sγ. If S is any simple submodule of Mγ, then by Schur’slemma f(S) is either 0 or is isomorphic to S and therefore to Sγ. In any casef(S) ⊆Mγ. This shows that f(Mγ) ⊆Mγ.

Let N be a submodule of M invariant under all endomorphisms of M .It is enough to show that if N contains a simple submodule S of M , thenN contains every simple sub-module S ′ isomorphic to S. Let f be an en-domorphism of M such that f(S) = S ′. Since f(N) ⊆ N , it follows thatS ′ ⊆ N .

Corollary. Let M be a semi-simple module with isotypical components

Mγγ∈Γ. Then EndA(M) ∼=∏

γ∈Γ EndA(Mγ).

Proof. Let f ∈ EndA(M). By the above proposition, f induces an endo-morphism fγ of Mγ. The mapping f 7→ (fγ)γ∈Γ is clearly an isomorphism ofEndA(M) onto

∏γ∈Γ EndA(Mγ).

Proposition 2.2. Let M be an A-module and let M = M1⊕ · · ·⊕Mn, where

each Mi is isomorphic to an A-module N . Then EndA(M) is isomorphic to

the ring of n× n matrices with entries in EndA(N).

Proof. Let φi : Mi → N be an isomorphism. Let pi : M → Mi denote theith projection for 1 ≤ i ≤ n. If f ∈ EndA(M), define an endomorphism fij

of N by settingfij = φi pi f φ−1

j .

The mapping f 7→ (fij) is clearly an additive homomorphism from EndA(M)into Mn(EndA(N)). For f, g ∈ EndA(M), we have

∑1≤k≤n(gik fkj) =

∑k(φi pi g φ−1

k ) (φk pk f φ−1j )

=∑

k(φi pi g pk f φ−1j )

= φi pi g (∑

k pk) f φ−1j

= φi pi g f φ−1j (∵

∑k pk = id)

= (g f)ij.

This shows that f 7→ (fij) is a homomorphism of rings. It is easily seenthat this homomorphism is an isomorphism.


Corollary. Let M be a semi-simple A-module of finite length n, isotypical

of type S. Let D denote the (division) ring of endomorphisms of S. Then

EndA(M) is isomorphic to Mn(D).

§ 3. Semi-simple Rings.

A ring A is said to be semi-simple if it is semi-simple as a left module overitself.

Exercise. A finite direct product of rings is semi-simple if and only if eachfactor is semi-simple.

Proposition 3.1. A ring A is semi-simple if and only if every A-module is

semi-simple.

Proof. We need only show the “only if” part. So suppose A is semi-simple.Then any free A-module is semi-simple since it is direct sum of a familyof modules isomorphic to A. Since any A-module is a quotient of a freeA-module, we are through.

Proposition 3.2. Let A be a semi-simple ring. Then A is the direct sum of

a finite number of simple left ideals. Every simple A-module is isomorphic

to one of these left ideals. In particular, there are only finitely many non-

isomorphic simple A-modules.

Proof. Let A = ⊕i∈ISi with Si simple left ideals of A. Let 1 = x1 + · · ·+xn

with xk ∈ Sik , 1 ≤ k ≤ n. Then clearly A = Si1⊕ · · ·⊕Sin . Let now S be asimple A-module and let x be a non-zero element of S. The map a 7→ ax is asurjective A-linear map A→ S with kernel N (say). Then A = N⊕S ′, whereS ′ is a left ideal of A isomorphic to S. Since S is simple we are through, inview of Remark 1 following the proof of Proposition 1.2.

Theorem 3.1 (Wedderburn). A semi-simple ring is a finite direct product

of matrix rings over division rings.

Proof. Let A be a semi-simple ring. By Proposition 1.6, A = S1⊕ · · ·⊕Sn

where Si are simple left ideals, which shows that A is of finite length asan A-module. Let M1, . . . ,Mr be the isotypical components of A. Thenby Corollary 1.3, EndA(A) ∼=

∏1≤i≤r EndA(Mi). Since EndA(A) ∼= Ao (the

opposite ring of A) and by Corollary 1.4, EndA(Mi) is isomorphic to a matrix

19

ring over a division ring, it follows that Ao is isomorphic to a finite directproduct of matrix rings over division rings. Since for any ring B, (Mn(B))o ∼=Mn(Bo) (the map α 7→ αt does it), it follows that A is a finite direct productof matrix rings over division rings.

§ 4. Simple Rings.

A ring A is said to be simple if it is a non-zero semi-simple ring with notwo-sided ideals other than 0 and A.

Proposition 4.1. Let A be a semi-simple ring. Then A is simple if and

only if all simple A-modules are isomorphic.

Proof. Let A be simple. An isotypical component of A is a two-sided idealsince the right multiplication by any element of A being an A-linear mapleaves this component invariant. Thus it must be A. In view of Proposition1.6, this means that any two simple modules are isomorphic.

Conversely, suppose all simple A-modules are isomorphic and let a be anon-zero two-sided ideal ofA. Since a is a two-sided ideal, it is invariant underevery endomorphism of the left A-module A. By Proposition 1.3, it followsthat a is a sum of isotypical components of A. But since by assumption allsimple A-modules are isomorphic, there is only one isotypical component,i.e., A. Thus a = A.

Theorem 4.1 (Wedderburn). A ring is simple if and only if it is iso-

morphic to Mn(D) for some division ring D. The integer n and (up to

isomorphism) the division ring D are uniquely determined.

Proof. Let A be a simple ring. In view of Theorem 1.2, A is a finite directproduct of matrix rings over division rings. But, since A has no two-sidedideals other than 0 and A, the number of factors in such a decompositionmust be one. This proves that A ∼= Mn(D) for some division ring D andsome integer n.

We show now conversely that if D is a division ring then A = Mn(D)is simple. Let Eij be the matrix whose (i, j)th entry is one and all otherentries are zero. Let Sj =

∑1≤j≤nDEij. It is clear that Sj is a left ideal

of A, 1 ≤ j ≤ n and A = S1⊕ · · ·Sn. We assert that the Sj are simple.For, if x =

∑i λiEij ∈ Sj with λk 6= 0, then Eij = λ−1

k Eikx, ∀i. Hence the


only non-zero sub-module of Sj is Sj, i.e., Sj is simple. Further the rightmultiplication by Ejj maps Sj isomorphically onto Sj, showing that A is asimple ring of length n.

Now we show that if Mn(D) ∼= Mn′(D′), then n = n′ and D ∼= D′.Comparing the length of Mn(D) and Mn′(D′), we have n = n′.

To show that D ∼= D′, it suffices to prove that if A = Mn(D), thenEndA(S1) ∼= Do. For any λ ∈ D, the right multiplication RλE11

by λE11

clearly induces an A-endomorphism of S1. Thus we have a mapping φ :D → EndA(S1). This is easily seen to be an anti-homomorphism of rings.Since D is a division ring, this map is injective. Let f ∈ EndA(S1). Letf(E11) =

∑i λ1iEi1. Since f is A-linear,∑

i

λ1iEi1 = f(E11) = f(E211) = E11f(E11) = λ11E11.

The endomorphism ψ = f − λ11E11 is not an isomorphism since ψ(E11) = 0.Therefore by Schur’s lemma, f = λ11E11. This shows that φ is an anti-isomorphism.

Corollary 1. Let D be a division ring and let M and N be two Mn(D)-modules which have the same dimension as vector spaces over D. Then Mand N are isomorphic as Mn(D) modules.

Proof. Since Mn(D) is simple, we have M ∼= S⊕ · · ·⊕S (m times) andN ∼= S⊕ · · ·⊕S (p times), where S is the unique simple MN(D)-module.Since dimD(M) = dimD(N) it follows that m = p. Thus M and N areisomorphic as MN(D)-modules.

A ring A is said to be right-semisimple if A, considered as a right A-module, is semi-simple; or in other words, if the opposite ring Ao is semi-simple.

Corollary 2. A ring A is semi-simple if and only if A is right-semisimple.

Proof. Suppose A is semi-simple. Then , by Theorem 1.2,

A =r∏

i=1

Mni(Di)

where Di are division rings. Hence Ao =∏

i Mni(Di)

o =∏

i Mni(Do

i ) whichshows that Ao is semi-simple, i.e., A is right-semisimple. The other implica-tion is similarly proved.

21

Corollary 3. If A =∏r

i=1 Mni(Di), Di division rings, then the integers ni, r

and Di (upto isomorphism) are unique.

Proof. We first remark that if A is the direct product of simple rings Ai, 1 ≤i ≤ r, then Ai are unique. In fact Ai, being a two-sided ideal, is a sum ofisotypical components of A. However since Ai is simple and the simple leftideals of Ai are precisely the simple left ideals of A contained in Ai, it followsthat Ai must be an isotypical component of A. This proves the uniquenessof the Ai, 1 ≤ i ≤ r. Now, the corollary follows from the theorem above.

§ 5. Radical and Semisimplicity.

Let A be a ring. The radical of A denoted by r(A) is defined to be theintersection of all maximal left ideals of A.

Proposition 5.1. Let A be a ring. Then r(A) is the intersection of the

annihilators of all simple (left) A-modules. In particular r(A) is a two sided

ideal of A.

Proof. We have

⋂

Msimple

ann M =⋂

M

⋂

x∈M\0

ann x =⋂

a

where the intersection is taken over those maximal left ideals a such that a isthe annihilator of a nonzero element of some simple A-module. However, anymaximal left ideal a has this property, since A/a is simple and a = ann 1,where 1 is the class of 1. This proves the first part of the proposition. Sincethe annihilator of any module is a two-sided ideal, the second assertion followsfrom the first.

Proposition 5.2. Let A be any ring and r(A) its radical. Then the radical

of A/r(A) is zero.

Proof. Let η : A → A/r(A) be the canonical homomorphism. Then theassignment a 7→ η(a) induces a bijection between the maximal left ideals ofA and those of A/r(A). It follows that

r(A/r(A)) =⋂

a

η(a)


where a runs over the set of maximal left ideals of A. But⋂

a

η(a) = η(r(A)) = 0

and the Proposition is proved.

Proposition 5.3. Let x ∈ A. Then x ∈ r(A) ⇐⇒ 1 − yx is invertible for

every y ∈ A.

Proof. We first prove ⇒. Let x ∈ r(A). Since the subset 1 − yx : y ∈ Ais a semigroup under multiplication, it is enough to show that 1 − yx has aleft inverse for any y ∈ A. In fact, if 1− yx does not have a left inverse, thenA(1 − yx) is a proper left ideal. If a is a maximal left ideal with 1 − yx ∈ a,it follows, since yx ∈ a, that 1 ∈ a, a contradiction.

We now prove ⇐. Let x ∈ A with 1 − yx invertible for every y ∈ A. Leta be any maximal left ideal. Suppose x /∈ a. Then, we have a + Ax = Aso that 1 − yx ∈ a for some y ∈ A. In particular 1 − yx is not invertible, acontradiction. Hence x ∈ a. Since a is arbitrary, it follows that x ∈ r(A).

Corollary 1. The radical r(A) is the largest two-sided ideal a such that 1−xis invertible for every x ∈ a.

The right radical of A is, by definition, the intersection of all maximalright ideals of A.

In view of the symmetric nature of Corollary 1.8, it holds also for theright radical. Hence we have

Corollary 2. The right radical of A coincides with the radical of A.

Nakayama’s lemma. If M is a finitely generated A-module such that

r(A)M = M , then M = 0.

Proof. Let if possible r(A)M = M and M 6= 0. Let x1, . . . , xn, n > 0 bea minimal set of generators of M . Since M = r(A)M , it is easily seen thatwe can write x1 =

∑1≤i≤n aixi with ai ∈ r(A), i.e., (1−a1)x1 =

∑2≤i≤n aixi.

Since 1 − a1 is invertible, it follows that x1 =∑

2≤i≤n bixi with bi ∈ A.This shows that x2, . . . , xn is a set of generators of M , contradicting theminimality of x1, . . . , xn.

An element a ∈ A is said to be nilpotent if an = 0 for some integer n > 0.An ideal a of A is said to be a nil ideal if every element of a is nilpotent.

23

An ideal a is said to be nilpotent if an = 0 for some n > 0. Clearly everynilpotent ideal is nil.

Remark. Any nil ideal a is contained in r(A). In fact, let a ∈ a. Then, forany y ∈ A, ya ∈ a and hence nilpotent. Let (ya)n = 0. Then 1 − ya has aninverse, viz. 1 + ya+ · · ·+ (ya)n−1. Thus, by Proposition 5.3, it follows thata ∈ r(A).

Proposition 5.4. Let A be an Artinian ring. Then r(A) is nilpotent. In

fact r(A) is the largest nilpotent (left or right or two-sided) ideal of A.

Proof. Since A is artinian, the descending chain of ideals

r(A) ⊃ r(A)2 ⊃ · · · ⊃ r(A)n ⊃ · · ·

terminates. Let r(A)m = r(A)m+1 = · · · . Write a = r(A)m. Then a2 = a.Suppose a 6= 0 and let b be a left ideal contained in a, minimal with theproperty that ab 6= 0. Since a(ab) = a2b = ab 6= 0 and ab ⊂ b, we have

ab ⊂ b (1)

Now let b ∈ b such that ab 6= 0. Then a(Ab) 6= 0 and Ab ⊂ b. By theminimality of b we have Ab = b which shows that b is finitely generated.From equation 1.1, it follows, by Nakayama’s lemma, that b = 0 (sinceab = b ⇒ r(A)b = b).

The second part of the proposition is clear in view of the remark precedingthe proposition.

Corollary. In an artinian ring any nil ideal is nilpotent.

Proof. As we have seen, such an ideal is contained in r(A) and hence isnilpotent since r(A) is nilpotent.

Lemma. In an artinian ring the radical is the intersection of a finite number

of maximal left ideals.

Proof. Let F be the family of finite intersections of maximal left ideals ofA. Since A is artinian there exists a minimal element, say a = ∩n

i=1bi of F.Since for any maximal ideal b we must have a ∩ b = a, we have a = r(A).

Theorem 5.1. Let A be a ring. Then the following statements are equiva-

lent:


1. A is semi-simple.

2. A is artinian with r(A) = 0.

3. A is artinian and has no non-zero nilpotent left ideal.

Proof. The statements 2. and 3. are equivalent in view of Proposition 5.4.We shall prove that 1. and 2. are equivalent.

1. ⇒ 2. : Since A is semi-simple, by Proposition 3.2, A is a finite directsum of simple left ideals Si. This shows that A is of finite length as an A-module and hence artinian. We also have 0 = ann (A) = ∩iann (Si) ⊃ r(A).

2. ⇒ 1.: Since A is artinian, by the lemma above, there exists maximalleft ideals a1, . . . , an such that

n⋂

i=1

ai = r(A) = 0.

Thus the mapping x 7→ (x mod ai) is anA-monomorphism ofA intoA/a1⊕ · · ·⊕A/an.Since A/a1⊕ · · ·⊕A/an is semi-simple, it follows that A is semi-simple.

Corollary. A ring A is semi-simple if and only if A is artinian and has no

two sided ideals other than 0 and A.

Proof. If A is simple, then A is semi-simple in particular, and thereforeartinian by above theorem.

Conversely, if A is artinian and has no two sided ideals other than 0 andA, then r(A) = 0 and it follows from the above theorem that A is semi-simpleand hence simple.

Corollary. If A is an artinian ring, then A/r(A) is semi-simple.

Proof. By Proposition 5.1, r(A/r(A)) = 0. Since A is artinian, so is A/r(A).Now the Corollary follows from the theorem above.

Chapter 3

Central Simple Algebras andthe Brauer Group

In this chapter, K will denote a (commutative) field and ⊗ will mean ⊗K .By a K-algebra, we shall mean an associative algebra over K. Unless oth-erwise stated, the K-vector space dimension of every K-algebra A, denotedby [A : K], will be assumed to be finite. Let A be a K-algebra. The naturalhomomorphism K → A (given by k 7→ k1) is a K-algebra monomorphismand we shall often identify K with its image in A under this monomorphism.

§ 6. Central Simple Algebras.

We say that a K-algebra A is central if center A = K, and central simple ifit is both central and simple. We shall call a ring A quasi-simple if it has notwo sided ideals other than 0 and A.

Theorem 6.1. Let A be a central simple K-algebra. If B is a (not nec-

essarily finite dimensional) K-algebra which is quasi-simple, then A⊗B is

quasi-simple.

To prove the theorem, we need the following

Lemma. If B is a quasi-simple ring then for any integer n ≥ 1, the matrix

ring Mn(B) is quasi-simple.

Proof. Let a ∈ Mn(B) be a two sided ideal 6= 0. Let a′ be the subset ofB consisting of all entries of elements of a. We assert that a′ is a two sided

25

26 CHAPTER 3. CENTRAL SIMPLE ALGEBRAS

ideal of B. For, let∑bijEij be an element of a, where bij ∈ B and Eij

denotes the n×n matrix whose (i, j)th entry is 1 and all others are zero. For1 ≤ k, k′, l, l′ ≤ n and b, c ∈ B, we have bbk′l′cEkl = (bEkk′)(

∑bijEij)(cEl′l)

is in a. This shows in particular that a′ is stable under multiplication on theleft and right by elements of B. Also taking b = c = 1 we see that if a ∈ a′,then aEkl ∈ a so that (a1 + a2)Ekl ∈ a and hence a1 + a2 ∈ a′. This showsthat a′ is a two sided ideal of B. Therefore a′ = B, i.e., 1 ∈ a′. So, we haveEkl ∈ a for all k, l. This shows that a = Mn(B).

Proof of the theorem: Since A is simple, by Theorem 4.1, there exists adivision ring D such that A ∼= Mn(D). It is clear that D is a central divisionalgebra over K. We have A⊗B ∼= Mn(D)⊗B ∼= Mn(D⊗B). If D⊗B isquasi-simple the lemma above would imply that the same is true of A⊗B,and the theorem would be proved. So, it is enough to prove the theorem inthe case when A is a central division algebra.

Let then D be a central division algebra and let a be any nonzero two-sided ideal of D⊗B. Let (eα)α∈I be a basis of B over K. Clearly, anyelement a ∈ a can be written uniquely in the form a =

∑α∈I aα⊗eα with

aα ∈ D, aα = 0 for almost all α. Let us write J(a) = α ∈ I : aα 6= 0. Thenfor each a ∈ a, J(a) is a finite subset of I. Let c =

∑cα⊗eα be a nonzero

element of a such that J(c) is minimal in the set J(a) : a ∈ a, a 6= 0.Multiplying c by an element of D, we can clearly assume that at least onecα, say cβ, is equal to 1. Since a is a two-sided ideal, we have, for any d ∈ D,

c′ = (d⊗1)c− c(d⊗1) =∑

(dcα − cαd)⊗eα ∈ a.

Since cβ = 1, dcβ = cβd and thus J(c′) $ J(c). Hence, by the minimality ofJ(c), it follows that c′ = 0, i.e., dcα = cαd for all α. Since d is arbitrary andD is central, it follows that cα ∈ K. In other words c ∈ a ∩ (1⊗B). Thusa ∩ (1⊗B) is a nonzero two-sided ideal of 1⊗B. Since B is quasi-simple, itfollows that a ∩ (1⊗B) = 1⊗B. In particular, 1⊗1 ∈ a, i.e., a = A⊗B, andthe theorem is proved.

If A is any ring and E a non-empty subset of A, the commutant of E,denoted by E ′, is by definition the set a ∈ A : ae = ea ∀ e ∋ E. ClearlyE ′ is a subring of A.

If A and B are K-algebras and C ⊂ A and D ⊂ B are K-subalgebras,then it is clear that the induced K-algebra homomorphism C⊗D → A⊗B

27

is a monomorphism. We shall identify C⊗D under this mapping with aK-subalgebra of A⊗B.

Proposition 6.1. Let A and B be K-algebras (not necessarily finite dimensional).If C ⊂ A,D ⊂ B are K-subalgebras and C ′ (resp. D′, (C⊗D)′) is the com-

mutant of C (resp. D,C⊗D) in A (resp. B,A⊗B), then we have

(C⊗D)′ = C ′⊗D′.

Proof. It is trivially seen that C ′⊗D′ ⊂ (C⊗D)′. On the other hand, let(eα) be a K-basis of B and let λ =

∑aα⊗eα ∈ (C⊗D)′ with aα ∈ A. For

any c ∈ C, we have (c⊗1)λ = λ(c⊗1), i.e.,∑caα⊗eα =

∑aαc⊗eα, which

implies that caα = aαc for every α. Since c ∈ C is arbitrary, it follows thataα ∈ C ′. In other words, λ ∈ C ′⊗B. Similarly, it follows that λ ∈ A⊗D′,so that λ ∈ (C ′⊗B) ∩ (A⊗D′). Thus (C⊗D)′ ⊂ C ′⊗D′, and this proves theproposition.

Corollary 1. If A and B are K-algebras, we have

center (A⊗B) = center A ⊗ center B.

Proof. Note that with our notation, center A = A′, center B = B′ andcenter (A⊗B) = (A⊗B)′.

Corollary 2. If A and B are central simple algebras over K, then A⊗B is

central simple over K.

Proof. By Theorem 6.1, A⊗B is quasi-simple. On the other hand, sinceA⊗B is clearly a finite dimensional algebra over K, it follows that A⊗B issimple. By Corollary 1, center (A⊗B) = center A ⊗ center B = K⊗K = K.This proves the corollary.

Corollary 3. If A is a central simple K-algebra and L any field extension

of K, then L⊗KA is a central simple algebra over L.

Proof. Since L⊗A is a finite dimensional algebra over L and since by The-orem 6.1, it is quasi-simple, it follows that L⊗A is a simple L-algebra. Onthe other hand, by corollary 1, it follows that center (L⊗A) = L⊗K = L,and the corollary is proved.


Corollary 4. If A is a central simple algebra over K, then [A : K] is a

perfect square.

Proof. Let L be the algebraic closure of K. By corollary 3, L⊗A iscentral simple over L. Thus L⊗A ∼= Mn(D) for some finite division algebraD over L. However , since L is algebraically closed, the only finite divisionalgebra over L is L itself. Thus L⊗A ∼= Mn(L), so that [A : K] = [L⊗A :L] = n2.

Let A be a central simple K-algebra and A be the opposite ring of A.Clearly, A is again a central simple K-algebra. For any a ∈ A, let La denotethe K-linear endomorphism of A given by left multiplication by a. Similarly,let Ra denote the right multiplication by a. The mappings φ : A→ EndK(A)and ψ : A → EndK(A) given respectively by φ(a) = La and ψ(a) = Ra

are K-algebra homomorphisms. Since every element of φ(A) commutes withevery element of ψ(A), we have an induced K-algebra homomorphism

θ : A⊗A → EndK(A)

defined by θ(a⊗b) = φ(a)ψ(b) = LaRb.

Theorem 6.2. The map θ : A⊗A → EndK(A) is an isomorphism of K-

algebras.

Proof. Since A is central simple, it follows from Corollary 2 to Proposition6.1 that A⊗A is simple. Since ker θ is a two-sided ideal of A⊗A andker θ 6= A⊗A, it follows that ker θ = (0). Thus, θ is a monomorphism.However, since dim(A⊗A) = (dimA)2 = dim EndK(A), it follows that θ issurjective. This proves the theorem.

Corollary. If A is a central simple K-algebra with [A : K] = m then

A⊗A ∼= Mm(K).

Proof. Note that EndK(A) ∼= Mm(K).

§ 7. The Brauer Group.

We say that two central simple algebras are equivalent (or Brauer equivalent),and write A ∼ B, if there exist matrix algebras Mm(K) and Mn(K) such thatA⊗Mm(K) ∼= B⊗Mn(K), i.e., Mm(A) ∼= Mn(B). It is trivially checked that

29

this defines an equivalence relation. It is also clear that isomorphic algebrasare equivalent.

Proposition 7.1. Let A,B be central simple K-algebras and let DA (resp.DB) denote the division algebra of A ( resp. B). Then A ∼ B if and only if

DA∼= DB.

Proof. Let A ∼= Mk(DA) and let B ∼= Ml(DB). If A ∼ B, then for somem,n,Mm(A) ∼= Mn(B), i.e., Mmk(DA) ∼= Mnl(DB). This implies by Theorem4.1 that DA

∼= DB. Conversely, suppose DA∼= DB. Then

Ml(A) ∼= Ml(Mk(DA)) ∼= Mlk(DA) ∼= Mlk(DB) ∼= Mk(Ml(DB)) ∼= Mk(B).

This proves the proposition.

The set of equivalence classes of central simple K-algebras is denoted byBr(K). For any central simple K-algebra A, we shall denote its equivalenceclass by [A]. If A and B are central simple K-algebras, then it follows byCorollary 2 to Proposition 6.1 that A⊗B is again central simple. We definea binary composition in Br(K) by setting [A] · [B] = [A⊗B]. We check thatthis is well-defined. In fact, let A ∼ A′ and B ∼ B′ and let

A ⊗ Mm(K) ∼= A′ ⊗ Mm′(K), B ⊗ Mn(K) ∼= B′ ⊗ Mn′(K).

We then have

A ⊗ B ⊗ Mmn(K) ∼= A ⊗ Mm(K) ⊗ B ⊗ Mn(K)∼= A′ ⊗ Mm′(K) ⊗ B′ ⊗ Mn′(K)∼= A′ ⊗ B′ ⊗ Mm′n′(K),

showing that A⊗B ∼ A′⊗B′. In view of the commutativity and associativityof tensor product, it follows that Br(K) is a commutative semigroup underthis composition.

Proposition 7.2. With the above composition Br(K) is an abelian group.

The identity of this group is [K], the class of K and consists of all matrix

algebras over K. The inverse of [A] is [A].

Proof. Since A⊗K ∼= A, it is clear that [K] is the identity. By the Corollaryto Theorem 6.2, it follows that for any central simple algebra A, [A⊗A] =[K], showing that [A] is invertible and has [A] for its inverse.

The group Br(K) defined above is called the Brauer group of K.


§ 8. The Skolem-Noether Theorem.

Theorem 8.1 (Skolem-Noether). Let A be a central simple K-algebra and

let B be a simple K-algebra. If f, g : B → A are K-algebra monomorphisms,

then there exists an invertible element u ∈ A such that for any b ∈ B, we

have g(b) = uf(b)u−1.

Proof. Suppose first that A is a matrix algebra Mn(K). Then to say thatwe are given two monomorphisms f, g : B → Mn(K) simply means thatwe are given two B-module structures on Kn. It follows by Corollary 1 toTheorem 4.1 that these modules are isomorphic. This means that there existsan invertible element u ∈ Mn(K) such that for any b ∈ B, the diagram

Kn u−−−→ Kn

f(b)

yyg(b)

Kn u−−−→ Kn

is commutative. But this is precisely the statement of the theorem in thiscase.

Let now A be any central simple algebra and let A be its opposite. Wehave K-algebra monomorphisms f⊗1, g⊗1 : B⊗A → A⊗A (1 denotingthe identity map of A). Since A⊗A ∼= Mn(K) for some n, we get aninvertible element u ∈ A⊗A such that

(∗) (g ⊗ 1) (b ⊗ a) = u (f ⊗ 1)(b ⊗ a) u−1

for every b ∈ B, a ∈ A. Putting b = 1, we find that

1 ⊗ a = (g ⊗ 1) (1 ⊗ a) = u (1 ⊗ a) u−1.

In other words, u commutes with 1⊗a; since a ∈ A is arbitrary, it followsthat u ∈ (1⊗A)′. But by Proposition 6.1, (1⊗A)′ = A⊗1. Thus u =u1⊗1 ∈ A⊗1 with u1 an invertible element of A. It is clear by putting a = 1in (∗) that u1 satisfies the required property, and that finishes the proof ofthe above theorem.

Corollary. Every K-algebra automorphism of a central simple K-algebra is

an inner automorphism.

Proof. In the Skolem-Noether theorem, take B = A, f = 1 and g = thegiven automorphism.

31

Proposition 8.1. Let A be a central simple algebra, B a simple sub-algebra

and B′ the commutant of B in A. Then B′ is simple, the commutant B′′ of

B′ is B and we have [B : K][B′ : K] = [A : K].

Proof. Let EndK(B) be the K-algebra of endomorphisms of the K-vectorspace B. Since EndK(B) is a matrix algebra over K, it is a central simplealgebra over K. Since A is central simple, it follows that A⊗EndK(B) is alsocentral simple. The inclusion of B in A induces a K-algebra monomorphismf : B → A⊗EndK(B). On the other hand, B can be embedded in EndK(B)under the map b 7→ Lb where Lb is the left multiplication by b. This inducesa K-algebra monomorphism g : B → A⊗EndK(B) such that g = (Int u) f(where Int u is the inner automorphism of A⊗EndK(B) given by u). ThusInt u maps f(B) isomorphically onto g(B) and hence the commutant f(B)′

onto g(B)′. But, by Proposition 6.1, it is clear that f(B)′ = B′⊗EndK(B)and g(B)′ = A⊗B. Thus B′⊗EndK(B) ∼= A⊗B. Since B is simple, itfollows that A⊗B is simple and hence B′⊗EndK(B) is also simple, whichimplies that B′ is simple. Equating dimensions of B′⊗EndK(B) and A⊗B,we get

[B′ : K][B : K]2 = [A : K][B : K]

which gives [B′ : K][B : K] = [A : K]. Applying this formula to the simplesubalgebra B′, we get

[B′′ : K][B′ : K] = [A : K]

so that we get [B′′ : K] = [B : K]. Since B ⊂ B′′, it follows that B = B′′,and this completes the proof of the proposition.

Corollary 1. If B is a central simple subalgebra of a central simple algebra

A, then B′ is also central simple and the inclusions B → A,B′ → A induce

an isomorphism B⊗B′ −→∼ A.

Proof. Clearly, the inclusions B → A,B′ → A induce a K-algebra homo-morphism φ : B⊗B′ → A. Since B is central simple, it follows that B⊗B′ issimple and hence φ is a monomorphism. Now, we have

[B⊗B′ : K] = [B : K][B′ : K] = [A : K]

by the above Proposition. Hence φ is surjective. That the center of B′ is Kis now clear.


Corollary 2. Let A be a central simple algebra over a field K and let L be a

commutative subfield of A containing K. Then the following conditions are

equivalent :

1. L is a maximal commutative subring of A.

2. L coincides with its commutant.

3. [A : K] = [L : K]2.

Proof. 1. ⇒ 2. : Let L′ be the commutant of L. Since L is commutative,it follows that L ⊂ L′. Let x ∈ L′. Then, the subring of A generated by Land x is commutative. By the maximality of L, it follows that x ∈ L. Sincex ∈ L′ is arbitrary, it follows that L′ ⊂ L, i.e., L′ = L.

2. ⇒ 1. : Let L1 ⊃ L, be a commutative subring of A. Since L1 iscommutative, it follows that L1 ⊂ L′ = L. Thus L1 = L and L is a maximalcommutative subring of A.

2.⇒ 3. : By the above Proposition we have

[A : K] = [L : K][L′ : K] = [L : K]2.

3. ⇒ 2. : Suppose L′ is the commutant of L. We have L′ ⊃ L; onthe other hand, [L : K]2 = [A : K] = [L : K][L′ : K] which implies that[L : K] = [L′ : K]. Hence L′ = L.

Corollary 3. Let D be a central division algebra over K. If L is a maximal

commutative subfield of D containing K, then [D : K] = [L : K]2.

Proof. Note that any commutative subring of D containing K is a subfield.And now Corollary 2 applies.

§ 9. Splitting Fields.

Let L be a field extension of K. The assignment A 7→ L⊗KA clearly inducesa homomorphism Br(K) → Br(L). Let A be a central simpleK-algebra. Wesay that L is a splitting field for A or that L splits A if L⊗A is L-isomorphicto Mn(L) for some n. This simply means that [A] is in the kernel of thehomomorphism Br(K) → Br(L). For example, if L ⊃ K is an algebraicallyclosed field, then L splits any central simple K-algebra A.

33

Theorem 9.1. Let L/K be a finite field extension. For any central simple

K-algebra A, the following conditions are equivalent :

1. L is a splitting field for A.

2. L is a maximal commutative subring of some central simple K-algebra

equivalent to A.

Proof. 1.⇒ 2. : Let φ : L⊗A −→∼ EndL(V ) be an L-isomorphism, where Vis a finite dimensional L-vector space. Since L is finite dimensional over K,V is finite dimensional over K and EndK(V ) is a central simple K-algebracontaining EndL(V ). Let C be the commutant of φ(1⊗A) in EndK(V ). Sincethe commutant of φ(L⊗A) in EndK(V ) is clearly L and 1⊗A ⊂ L⊗A, itfollows that C ⊃ L. Since φ(1⊗A) is central simple it follows from Corollary1 to Proposition 8.1 that C is central simple and that A⊗C ∼= EndK(V ), i.e.,A ∼ C. If we set B = C, then clearly B ⊃ L and we would be throughprovided we show that L is a maximal commutative subring of B. In view ofCorollary 2 to Proposition 8.1, it is enough to show that [B : K] = [L : K]2.But we have [B : K] = [C : K] and

[A : K][C : K] = [EndK(V ) : K] = [EndL(V ) : K][L : K]

= [L⊗KA : K][L : K] = [L : K]2[A : K]

which shows that [C : K] = [L : K]2.

2. ⇒ 1. : It is enough to show that if L is a maximal commutativesubring of A, then L splits A. We have, by Theorem 6.2, an isomorphismA⊗A −→∼ EndK(A). Since L ⊂ A and L is commutative, it follows thatL ⊂ A. The commutant of 1⊗L in A⊗A is, by Proposition 6.1, A⊗L.On the other hand, the commutant of L in EndK(A) is EndL(A). ThusA⊗L ∼= EndL(A) = Mn(L) with n = [A : L], and this finishes the proof.

Corollary. Any maximal commutative subfield of a central division algebra

D is the splitting field for D.

§ 10. Existence of Galois Splitting Fields.

Lemma.Let D 6= K be a central division algebra over K. Then D contains

a separable algebraic extension of K containing K properly.


Proof. Suppose that the lemma is false. If there exists some element α ∈ Dwhich is not purely inseparable over K, then K(α) contains a separable sub-extension of K other than K, and this would prove the lemma. Supposetherefore that every element of D is purely inseparable over K. Thus everyelement x ∈ D satisfies xpe ∈ K for some positive integer e depending onx, where p is the characteristic of K. Since D is finite dimensional over Kthere exists an integer e such that xpe ∈ K for all x ∈ D. Let 1 = e1, . . . , en

be a K-basis of D, and let x =∑

i xiei ∈ D, with xi ∈ K. Then xpe

=∑j Pj(x1, . . . , xn)ej, where the Pj are polynomials in xi whose coefficients

can be expressed in the terms of the structure constants of D. We have, byour hypothesis, Pj(x1, . . . , xn) = 0 for j 6= 1 and for all systems of values ofxi in K. Since we can suppose that K is infinite (since if K is finite, everyfinite extension of K is separable) we get that Pj(j 6= 1) vanish identically.This implies that the same condition xpe ∈ K also holds if we extend the basefield. In particular this should be true if we extend K to its algebraic closureL. But then L⊗D is a matrix algebra and there are therefore idempotents ofL⊗D which certainly do not satisfy our condition. This is a contradiction,and the lemma is proved.

Theorem 10.1. Every central division algebra D over a field K contains a

maximal commutative subfield which is separable over K.

Proof. Let L be a subfield of D which is a maximal separable extension ofK. We assert that L is a maximal commutative subfield of D. For, if not, letL′ 6= L be the commutant of L. Then L′ is a division algebra of center L. Bythe above lemma there exists a proper separable extension of L contained inL′. But this is a contradiction to our assumption on L. This shows that Lis maximal commutative subfield, and the theorem is proved.

Corollary 1. Let K be a field and L ⊃ K be a separably algebraically closed

field (i.e., a field which has not proper separable algebraic extensions). Then

L splits any central simple algebra over K.

Proof. Let A be any central simple algebra over K. Then L⊗A ∼= Mn(D)where D is a (finite) central division algebra over L. If D 6= L, D mustcontain, by the above Theorem, a proper finite separable extension of L. Thisis however impossible, since L is, by our assumption, separably algebraicallyclosed. Thus D = L and L splits A.

Corollary 2. Every central simple algebra A over a field K admits of split-

35

ting field which is a (finite) Galois extension of K.

Proof. Let D be the division algebra of A. By the above theorem, Dcontains a maximal commutative subfield L which is separable over K. Thefield L splits D by Corollary to Theorem 9.1, and hence splits A also. Now,let L′ be the normal closure of L. Clearly L′ is finite and Galois extension ofK which splits A.

§ 11. Reduced Norm.

Let A be a central simple algebra over a field K and let L be a splitting fieldfor A. Choose an L-isomorphism

φ : L ⊗ A −→∼ Mn(L).

For any x ∈ A the element detφ(1⊗x) is independent of the isomorphism φ.In fact, if φ′ : L⊗A −→∼ Mn(L) is another L-isomorphism, then by Skolem-Noether theorem, there exists an invertible element u ∈ Mn(L) such thatφ′(z) = uφ(z)u−1 for every z ∈ L⊗A so that

detφ′(1⊗x) = det(u φ(1⊗x) u−1) = detφ(1⊗x).

We will call detφ(1⊗x) the reduced norm of x (w.r.t. L) and denote it byNL

rd(x). We shall now show that this element belongs toK and is independentof L. We begin by making a few general remarks.

Let L,L′ be splitting fields forA and let θ : L→ L′ be aK-monomorphism.Let φ : L⊗A −→∼ Mn(L) be an L-isomorphism. Then there exists an L′-isomorphism

φ′ : L′⊗A = L′⊗LL⊗A −→∼ L′⊗LMn(L) = Mn(L′)

such that the diagram

L⊗A φ−−−→ Mn(L)

θ⊗1

yyMn(θ)

L′⊗A φ′

−−−→ Mn(L′)

is commutative. It follows from this that we have for any x ∈ A

(∗) NL′

rd (x) = θ(NLrd(x)).


Let now L be a Galois splitting field for A. Then, for any element σ ∈G(L/K), we have by (∗) that

NLrd(x) = σ(NL

rd(x)).

This shows that NLrd(x) belongs to K. Let now L′ be any splitting field of

A. There exists a field extension L′′ of K and K-monomorphisms θ : L →L′′, θ′ : L′ → L′′. By (∗), we have

θ′(NL′

rd (x)) = NL′′

rd (x) = θ(NLrd(x)) = NL

rd(x)

the last equality being a consequence of the fact that L is Galois over K.This shows that NL′

rd (x) is independent of L′, and we call it the reduced norm

of x. and denote it by Nrd(x). The map x 7→ Nrd(x) is a multiplicative mapof A into K and induces a homomorphism of the group of invertible elementsof A into the multiplicative group K∗.

Proposition 11.1. Let A be a central simple K-algebra of dimension n2.

Then, for any x ∈ A, if Lx denotes the left multiplication by x in A, we have

detLx = (Nrd(x))n.

Proof. Let L be an extension of K such that L⊗A ∼= Mn(L). SincedetL1⊗x = detLx, we assume by replacing A by L⊗A that A is a ma-trix algebra over K. Thus, we have to show that if x ∈ Mn(K), and Lx

denotes the left multiplication by x in Mn(K), then detLx = (detx)n. LetMn(K) = S⊕ · · ·⊕S (n times), where S is the unique simple (left) Mn(K)-module. If lx denotes the endomorphism of S induced by x, then clearly, wehave detLx = (det lx)

n. But, with respect to a suitable K-basis for S, it iseasily seen that the matrix of lx is x itself. This proves the Proposition.

Theorem 11.1. Let A be a central simple algebra over a field K and let

e1, . . . , en2 be a basis of A over K. Then there exists a polynomial P (X1, . . . , Xn2)over K, homogeneous and of degree n, such that if x =

∑λiei, λi ∈ K, then

Nrd(x) = P (λ1, . . . , λn2).

For the proof of the theorem, we need the following lemmas.

Lemma 1. Let K be an infinite field and let P (X1, . . . Xn) be a polynomial

with coefficients in an extension field L of K such that P (k1, . . . , kn) ∈ Kfor all k1, . . . , kn ∈ K. Then the coefficients of P belong to K.

37

Proof. We proceed by induction on n.

Let n = 1. Let P =∑

0≤i≤m liXi ∈ L[X] be such that P (k) ∈ K for all

k ∈ K. Choose m + 1 distinct elements k0, k1, . . . , km ∈ K. This is possiblesince, by assumption, K is an infinite field. Then we have l0 + l1ki + · · · +lmk

mi = ai ∈ K for 0 ≤ i ≤ m. Since the Vandermonde matrix whose ith

row is (1, ki, . . . , kmi ) is non-singular, it follows from the above equations that

lj ∈ K for 0 ≤ j ≤ m.

Now assume by induction that the result holds for n − 1. Write P =∑0≤i≤mQi(X1, . . . , Xn−1)X

in where Qi are polynomials in X1, . . . , Xn−1 over

L. For any k1, . . . , kn−1 ∈ K,P (k1, . . . , kn−1, Xn) is a polynomial which hasthe property that for any k ∈ K,P (k1, . . . , kn−1, k) ∈ K. By the case dis-cussed above, we have that Qi(k1, . . . , kn−1) ∈ K. It now follows by inductionthat the coefficients of Qi are in K and hence the coefficients of P are in K.

Lemma 2. Let K be any field and X an indeterminate. Then, the only

elements of K(X) algebraic over K are those of K.

Proof. Let f/g ∈ K(X) be algebraic over K where f, g ∈ K[X] and (f, g) =1. Let f/g satisfy an equation

(f/g)n + a1(f/g)n−1 + · · · + an = 0

with ai ∈ K. From this equation, we get

(∗) fn + a1fn−1g + · · · + ang

n = 0

i.e. fn = −g(a1fn−1 + a2f

n−2g + · · · + angn−1)

which is not possible, since (f, g) = 1, unless g is a unit, i.e., g ∈ K∗. But inthat case (∗) is a non-trivial algebraic relation for X. This is a contradiction.

Proof of the theorem. Let L be a finite algebraic extension of K such thatwe have an isomorphism φ : L⊗A −→∼ Mn(L). Choose a K-basis e1, . . . , en2

of A. Let Eij be the n× n matrix which has the entry 1 in the (i, j)th placeand zero elsewhere. Then (Eij)1≤i,j≤n is an L-basis of Mn(L) and for each i,1 ≤ i ≤ n2, we can write

φ(1 ⊗ ei) =∑

1≤j,k≤n

aijkEjk


with aijk ∈ L. Let x =∑

i λiei be an element of A. Then

φ(1 ⊗ x) =∑

j,k

∑

1≤i≤n2

aijkλi

Ejk.

This shows that each entry of the matrix φ(1⊗x) is a linear homogeneouspolynomial in λ1, . . . , λn2 with coefficients in L. It therefore follows thatNrd(x) = detφ(1⊗x) is a homogeneous polynomial, say f , of degree n inλ1, . . . , λn2 with coefficients in L. But Nrd(x) ∈ K for every x ∈ A. Thus, ifK is infinite, Lemma 1 shows that the coefficients of f are actually in K.

Now, ifK is finite, we consider the central simpleK(X)-algebraK(X)⊗A.Then φ naturally extends to an isomorphism

L(X) ⊗K(X) (K(X)⊗A) −→∼ Mn(L(X))

which shows that for every x ∈ K(X)⊗A,Nrd(x) is again given by the samepolynomial f . Since K(X) is infinite, the coefficients of f belong to K(X).But the coefficients also belong to L and, by Lemma 2, K(X)∩L = K. Thiscompletes the proof of the theorem.

Chapter 4

Brauer Groups of some Fields

§ 12. The Field of Real Numbers.

Let R denote the field of real numbers. Let H denote the R-algebra ofdimension 4 with basis 1, i, j, k whose multiplication table is as follows:

1 i j k1 1 i j ki i -1 k -jj j -k -1 ik k j -i -1

It is easy to see that H is an associative algebra. We define a map : H → Hby

x = a+ ib+ jc+ kd 7→ x = a− ib− jc− kd

where a, b, c, d ∈ R. For x = a + ib + jc + kd ∈ H, we write N(x) = xx =a2 + b2 + c2 + d2. It is clear that N(x) = 0 if and only if x = 0, and forx 6= 0, x · (x/N(x)) = 1. Hence H is a division ring. In fact, H is a centraldivision algebra over R. For, if x = a + ib + jc + kd commutes with i, j, k,we have

xi = ix i.e. − b+ ai+ dj − ck = −b+ ai− dj + ck

which shows that c = d = 0, and x = a+ ib. The equation

xj = jx i.e. aj + bk = aj − bk

39

40 CHAPTER 4. BRAUER GROUPS OF SOME FIELDS

shows that b = 0, i.e., x ∈ R.

This algebra H is called the algebra of quaternions over R.

It is easy to check that x+ y = x + y and xy = yx for x, y ∈ H. Thismeans that x 7→ x defines an isomorphism of H onto its opposite H. Hencewe have an R-algebra isomorphism of H⊗RH onto H⊗RH, and the latter isisomorphic to M4(R) (see Corollary to Theorem 6.2). This shows that theclass of H in Br(R) is of order 2.

Next let D be any finite dimensional division algebra over R. The centerK of D, being a finite algebraic extension of R, is isomorphic to either C orR. In case K = C, we have D = C, since C is algebraically closed. SupposeK = R. Then any maximal subfield of D being a proper extension of R isisomorphic to C. Hence, by Corollary 3 to Proposition 8.1, dimR D = 4. Wefix a maximal subfield L of D and choose i ∈ L such that i2 = −1. ThenL = R(i). By the Skolem-Noether Theorem, the automorphism z 7→ z of L,given a+ ib 7→ a− ib, can be extended to an inner automorphism of D. Thismeans that there exists an element u ∈ D such that uzu−1 = z for all z ∈ L.Since z 7→ z is an automorphism of order 2, we have u2z = zu2 for all z ∈ L.Since L is a maximal subfield of D, we have u2 ∈ L. We claim that u2 isactually in R. For, if u2 6∈ R, then R(u2) = L and uzu−1 = z for all z ∈ L.Contradiction. Hence u2 ∈ R. We assert that u2 < 0. Indeed, if u2 = a > 0,then u = ±√

a. This is impossible. Hence u2 = −a for some a ∈ R, a > 0.We put j = u/

√a and ij = k. Then we have j2 = −1. Since j−1ij = −i,

we see that k = ij = −ji and k2 = ijij = −i2j2 = −1. Moreover k 6∈ L,since k ∈ L implies j ∈ L. Since dimLD = 2, 1, j is a basis for D as a vectorspace over L. Hence 1, i, j, k is a basis for D over R. It is now easy to verifythe multiplication table given in the beginning of the section. Thus we haveshown that any finite dimensional division algebra over R is isomorphic toC,R or H. Thus the only central division algebras over R are R and H andwe have the following

Theorem 12.1. The Brauer group of R is cyclic of order two and is gener-

ated by the class of H.

41

§ 13. Finite Fields.

In this section we show that Br(K) = 1 for any finite field K, i.e., there areno non-commutative finite division rings.

In what follows, by a form f over a field K, we mean a homogeneouspolynomial of degree ≥ 1 in one or more variables with coefficients in K. AfieldK is said to be Ci if every form f(X1, . . . , Xn) in n variables and of degreed, with n > di, has a non-trivial zero in K, i.e., there exist a1, . . . , an ∈ K,not all zero such that f(a1, . . . , an) = 0.

The field R is not Ci for any i. An algebraically closed field is C0.

Proposition 13.1. Let K be a C1 field. Then Br(K) = 1.Proof. We show that if D is any central division algebra over K, thenD = K. Let [D : K] = n2. We fix a basis w1, . . . , wn2 of D over K.We know, by Theorem 11.1, that there is a form f(X1, . . . , Xn2) of degree

n such that if x =∑n2

i=1 aiwi, ai ∈ K, then Nrd(x) = f(a1, . . . an2). Ifn > 1, then n2 > n and since K is C1, there exist a1, . . . , an2 ∈ K, not all

zero, such that f(a1, . . . , an2) = 0. The element x =∑n2

i=1 aiwi 6= 0 andNrd(x) = f(a1, . . . , an2) = 0. This is a contradiction since x is a unit. Hencen = 1, i.e., D = K.

Theorem 13.1 (Chevalley). A finite field is C1.

Proof. Let K be a finite field of characteristic p and let q = pr be thenumber of elements in K. For a polynomial f ∈ K[X1, . . . , Xn], let Z(f)denote the number of zeros of f in Kn. Our theorem is a consequence of thefollowing stronger

Theorem 13.2. Let f be a polynomial of degree d in n variables over K. If

n > d, then Z(f) ≡ 0 (mod p).

Proof. We may assume that f is a non-constant polynomial. In the field K,we have the identity

Z(f) · 1 =∑

(x1,...,xn)∈Kn

(1 − f q−1(x1, . . . , xn)

).

The polynomial F (X1, . . . , Xn) = 1− f q−1(X1, . . . , Xn) is of degree d(q− 1).


We write

Z(f) · 1 =∑

(x1,...,xn)∈Kn

F (x1, . . . , xn).

Let

F =∑

i1+···+in≤d(q−1)

ai1...inXi11 · · ·X in

n , ai1...in ∈ K.

The above formula can be written as

Z(f) · 1 =∑

i1+···+in≤d(q−1)

ai1...in

∑

(x1,...,xn)∈Kn

xi11 · · ·xin

n .

We have

∑

(x1,...,xn)∈Kn

xi11 · · ·xin

n =

(∑

x∈K

xi1

)(∑

x∈K

xi2

)· · ·(∑

x∈K

xin

).

Since i1 + · · · + in ≤ d(q − 1) < n(q − 1), we have ik < q − 1 for some k.Consider

∑x∈K xik , ik < q − 1. If ik = 0, then

∑x∈K xik = q = 0. Suppose

ik > 0. Since K is finite, K∗ is cyclic. Let θ ∈ K∗ generate K∗. We thenhave ∑

x∈K

xik =∑

x∈K∗

xik =∑

0≤m≤q−2

θmik =∑

0≤m≤q−2

(θik)m.

Now,

(θik − 1)∑

0≤m≤q−2

(θik)m = (θik)q−1 − 1 = 0.

Since ik < q − 1, θik 6= 1. Hence

∑

0≤m≤q−2

(θik)m = 0.

Thus∑

x∈K xik = 0. This shows that Z(f) · 1 = 0, i.e., Z(f) ≡ 0 (mod p).

Corollary. If K is a finite field, then Br(K) = 1.Proof. This follows immediately from the above theorem in view of Propo-sition 13.1.

43

§ 14. Tsen’s Theorem.

In this section we show that a function field in one variable over an alge-braically closed fieldK (i.e., a finite extension of the field of rational functionsin one variable over K) is C1.

Let L/K be a finite algebraic extension. We fix an algebraic closure Ω ofL. Let f =

∑ai1...inX

i11 · · ·X in

n . Clearly, σ(f) is a form over σ(L) of degreed. We define the norm of f , N(f) by

N(f) =

(t∏

i=1

σi(f)

)pe

where σ1, . . . , σt denote all the K-monomorphisms of L into Ω and pe is thedegree of inseparability of L/K. Clearly, N(f) is a form over K of degreetped = [L : K]d. Moreover, if (x1, . . . , xn) ∈ Kn is a zero of σi(f), then it isalso a zero of f (apply σ−1

i to the equation σi(f)(x1, . . . , xn) = 0).

Proposition 14.1. Let K be a Ci field and L be a finite algebraic extension

of K. Then L is Ci.

Proof. Let φ be a form over L in n variables and of degree d with n > di.Starting with φ0 = φ, we define inductively forms φi, i = 0, 1, . . . as follows:Suppose φk−1 is defined and is a form in Nk−1 variables. Introduce nNk−1

variables x11, . . . , x1n, . . . , xNk−11, . . . , xNk−1n and put

φk = φk−1

(φ(x11, . . . , x1n), . . . , φ(xNk−11, . . . , xNk−1n)

).

It is easy to see that φk is a form in Nk = nk+1 variables of degree Dk = dk+1.Since n > di, we have Nk/D

ik = (n/di)k+1 → ∞ as k → ∞. We choose k

so large that Nk > miDik, where m = [L : K]. Write φk = f(X1, . . . , XNk

).Then N(φk) is a form over K in Nk variables of degree mDk. Since Nk >miDi

k and K is Ci, it follows that N(φk) has a nontrivial zero in K. Ourproposition now follows from the following

Lemma. Suppose φ = φ0 has no non-trivial zero in L. Then φk has no

non-trivial zero for all k.

Proof. We proceed by induction on k, starting with k = 0. Suppose k > 0and φk−1 has no non-trivial zero. Let

φk(x) = φk−1

(φ(x11, . . . , x1n), . . . , φ(xNk−11, . . . , xNk−1n)

)= 0


with xij ∈ L. Since φk−1 has no non-trivial zero, we have

φ(x11, . . . , x1n) = · · · = φ(xNk−11, . . . , xNk−1n) = 0.

Since φ has no non-trivial zero, it follows that xij = 0, 1 ≤ i ≤ Nk−1, 1 ≤ j ≤n.

Theorem 14.1. Let K be an algebraically closed field and let L be a finitely

generated field extension of K of transcendence degree r. Then L is Cr.

Proof. Since L is a finite algebraic extension of a field of rational functionsin r variables over K, by Proposition 14.1, we may (and do) assume thatL = K(t1, . . . , tr), the ti being algebraically independent over K.

We now proceed by induction on r and prove the following stronger

Proposition 14.2. Let K be an algebraically closed field and let f1, . . . , fm

be m forms over L = K(t1, . . . , tr) in n variables and of the same degree d.If n > mdr, then f1, . . . , fm have a non-trivial common zero in L.

Proof. For r = 0, L = K is algebraically closed, and the above assertion isproved in Appendix I. Suppose r ≥ 1 and assume that the assertion is truefor F = K(t1, . . . , tr−1). We put tr = t so that L = F (t). The coefficientsof f1, . . . , fm are rational functions in t, so that multiplying f1, . . . , fm by asuitable polynomial in t, we may assume that the coefficients of f1, . . . , fm areactually in the polynomial ring F [t]. We introduce variables xij1 ≤ i ≤ n, 0 ≤j ≤ s (s to be chosen later). We write hi(f) =

∑s

j=0 xijtj. It is easy to see

that we can write fk(h1(t), . . . , hn(t)) =∑ds+l

j=o φkjtj, where l is the largest of

the degrees of the coefficients of f1, . . . , fm and φkj are forms over F of degreed in n(s+ 1) variables xij. It is clear that if φkj, 0 ≤ j ≤ ds+ l, 1 ≤ k ≤ m,have a common non-trivial zero in F , then f1, . . . , fm have a common non-trivial zero in L. Now the φkj are m(ds+ l + 1) in number and are forms ofdegree d in n(s+ 1) variables. Hence, by induction hypothesis, the φkj havea common non-trivial zero if

n(s+ 1) > m(ds+ l + 1)dr−1.

Since n > mdr, this inequality holds if

s >m(l + 1)dr−1 − n

n−mdr.

We choose s this large and get a common non-trivial zero for f1, . . . , fm.

45

Corollary 1. (Tsen’s Theorem). A function field of one variable over an

algebraically closed field is C1.

Corollary 2. There are no finite dimensional non-commutative division

algebras over a function field of one variable over an algebraically closed

field, i.e., its Brauer group is trivial.

Remark. Theorem 14.1 is a particular case of the following:

Let K be a Ci field and L a finitely generated field over K of transcendence

degree r. Then L is Ci+r.

(For a proof, see Nagata, “Note on a paper of Lang concerning quasialgebraic closure”, Memoirs of the college of Science, University of Kyoto,Series A, Vol. XXX, Mathematics No. 3, 1957.)


Chapter 5

Representations of FiniteGroups

§ 15. The Group Algebra.

Let K be a field and G, a finite group of order g. Let KG denote the free K-module on the setG = x1, . . . , xg. The group operation inG induces a mul-tiplication in KG, defined as follows: for a, b ∈ KG, we have a =

∑g

i=1 λixi

and b =∑g

i=1 µixi with λi, µj ∈ K; we define ab =∑

1≤i,j≤g λiµjxixj. Withthis multiplication KG becomes an associative K-algebra. The identity el-ement of the group G is the identity element of KG. This algebra KG iscalled the group algebra of G over K. The group G is canonically identifiedwith a subgroup of the group of all units in KG.

Exercises.

1. KG is simple ring if and only if g = 1.

2. KG is semi-simple ring implies “g 6= 0 in K”.

3. If U(A) denotes the group of units in aK-algebraA, then Homgroups(G,U(A))is canonically bijective with HomK−alg.(KG,A).

We now determine the center of KG.

Proposition 15.1. Let C1, . . . , Cr be the conjugacy classes of the elements

of G and let zi =∑

x∈Cix. Then the elements z1, . . . , zr form a basis (over

K) of the center of the ring KG.

47

48 CHAPTER 5. REPRESENTATIONS OF FINITE GROUPS

Proof. We have xzi = zix for every x ∈ G and so the elements z1, . . . , zr arein the center of KG. Suppose now that a =

∑g

i=1 λixi, λi ∈ K,xi ∈ G; is anelement of the center of KG. We show that λi = λj whenever xi and xj areconjugates. This is clear because we have xi = xxjx

−1 for some x ∈ G anda = xax−1. Thus a is a linear combination of the elements z1, . . . , zr. But theelements z1, . . . , zr are linearly independent over K and so the proposition isproved.

Remark. It is easily seen that zizj =∑

1≤k≤r dijkzk, where dijk ∈ Z.

§ 16. Representations.

By a representation of a finite group G over a field K, we mean a pair (V, ρ),where V is a vector space over K and ρ : G → AutK(V ), a homomorphismof groups.

Suppose (V, ρ) is a representation of G over K. Then ρ : G → AutK(V )can be uniquely extended to a homomorphism of the K-algebras ρ : KG →EndK(V ) by defining

ρ

(g∑

i=1

λixi

)=

g∑

i=1

λiρ(xi).

Now V becomes a KG module via ρ. On the other hand given a KG-moduleM , we have (i) M is canonically a vector space over K and (ii) a K-algebrahomomorphism ρ : KG→ EndK(M), given by the KG-module structure onM . By restricting ρ to G, we get a group homomorphism ρ : G→ AutK(M)and hence a representation (M,ρ) ofG overK. Through this correspondence,we identify representations of G over K with the KG-modules (and from nowon by a representation of G over K, we shall mean a KG-module).

Further we shall confine ourselves to the study of representations whichare finite dimensional over K.

Remarks.

1. Suppose (V, ρ) and (V ′, ρ′) are two representations of G over K. Thena K-linear map f : V → V ′ is KG-linear if f ρ(x) = ρ′(x) f forevery x ∈ G.

49

2. Given a K-linear map f : V → V ′, we can construct a KG-linear mapf : V → V ′ by setting

f =

g∑

i=1

ρ′(x−1i ) f ρ(xi).

In fact for every x ∈ G we have

f ρ(x) =

g∑

i=1

ρ′(x−1i ) f ρ(xi) ρ(x)

=

g∑

i=1

ρ′(x) ρ′(x−1x−1i ) f ρ(xix)

= ρ′(x)

g∑

i=1

ρ′(x−1x−1i ) f ρ(xix)

= ρ′(x) f

which proves the assertion.

Examples.

1. The K-vector space K can be made into a KG-module by setting

(g∑

i=1

λixi

)λ =

(g∑

i=1

λi

)λ for λi, λ ∈ K,xi ∈ G.

This is called the trivial representation of G over K.

2. The (left) module structure of KG over itself is called the (left) regular

representation of G over K.

Theorem 16.1. (Maschke). Let G be a finite group of order g and K a

field. Then KG is a semi-simple ring if and only if “g 6= 0 in K” (i.e., if pis the characteristic of K, then either p = 0 or p 6 | g).Proof. KG is semi-simple implies “g 6= 0 in K” by exercise 2 in § 15.

Conversely, suppose “g 6= 0 in K”. We shall prove that every left ideal ofKG is a direct summand of KG.


Let a be a left ideal of KG. We have a K-linear map f : KG → a

satisfying (i) f(KG) = a and (ii) f |a = identity.

We know, by Remark 2 above, that f : KG→ a is KG-linear and hencef = 1

gf is also KG-linear. If a ∈ a, we have

f(a) =1

gf(a) =

1

g

g∑

i=1

x−1i f(xia)

=1

g

g∑

i=1

x−1i (xia) (Since xia ∈ a and f |a = identity.)

= a.

This shows that f : KG → a is a KG-linear map and satisfies (i)f(KG) = a and (ii) f |a = identity and hence a is a direct summand ofKG. This proves that KG is a semi-simple ring.

Note. From now on K will denote an algebraically closed field and G a finitegroup of order g such that “g 6= 0 in K”.

§ 17. Simple Modules over KG.

Theorem 17.1.

(i) The number of non-isomorphic simple KG-modules is equal to the num-

ber r, of conjugacy classes in G.

(ii) If n1, . . . , nr are the dimensions (over K) of these simple modules, then

n21 + · · · + n2

r = g.

Proof. By Maschke’s theorem, we know thatKG is a semi-simpleK-algebra.Since K is algebraically closed, we have KG = A1×· · ·×As, where each Ai isa matrix algebra over K. But then the center of KG is precisely K×· · ·×K(s times) and hence we get that

s = dimension of the center of KG (over K)= r, the number of conjugacy classes in G (Proposition 15.1)

But s is also the number of non-isomorphic simple KG-modules and sothe statement (i) is proved.

51

Now let the dimension of Ai over K be n2i . A simple KG-module is

isomorphic to a minimal left ideal of Ai for some i and has dimension ni overK. It follows therefore that n2

1 + · · · + n2r = g.

Corollary 1. If G is abelian, then a simple KG-module is one dimensional

over K.

Proof. Since G is abelian, we have r = g and the equation n21 + · · ·+n2

g = ggives that each ni = 1.

Corollary 2. If V is a KG-module and x ∈ G, then the action of x on Vcan be represented by a diagonal matrix with respect to a suitable basis of V .

Proof. Let H denote the subgroup of G generated by x. Then KH is a sub-algebra of KG and V can be regarded as a KH-module. Now V decomposes(over KH) as a direct sum of simple KH-submodules, say V1, . . . , Vm. SinceH is abelian, by the corollary 1 above, each Vi is one dimensional over K,say Vi = Kvi. Then v1, . . . , vm form a K-basis of V with respect to whichthe action of x on V is obviously represented by a diagonal matrix.

Remark. The above theorem and the corollaries are false if the field is notalgebraically closed. Example? (Look at the field of rational numbers and acyclic group of odd prime order.)

§ 18. Characters and the Orthogonality Rela-

tions.

Let (V, ρ) be a representation of G over K. The function : G → K, givenby, x 7→ trace (ρ(x)) is called the character of the representation (V, ρ) andis denoted by χρ or χV or simply χ.

Choosing a basis of V over K, each ρ(x), x ∈ G, can be represented bya matrix (ρij(x)), with ρij(x) ∈ K. Then by the definition of trace, we haveχρ(x) = trace (ρ(x)) =

∑i ρii(x).

Remark. It is seen easily that

(i) the character is a class function on G, i.e., χ takes the same value onthe elements of a given conjugacy class in G and;

(ii) isomorphic KG-modules have the same character.


Examples.

1. Let (K, ρ) be the trivial representation. Then χK(x) = 1 for everyx ∈ G.

2. Let (KG, ρ) be the regular representation. Then χKG(x) = 0 if x 6= 1.

The character of a simple KG-module is called an irreducible character.

Proposition 18.1.

(i) If a KG-module V is the direct sum of two submodules V1 and V2, then

χV = χV1+ χV2

.

(ii) The character χ of any representation can be written as a sum of irre-

ducible characters (called the irreducible components of χ).

Proof. (i) is obvious and (ii) follows from (i) since any KG-module is adirect sum of simple KG-submodules.

Example. Let χKG be the character of the regular representation. Letχ1, . . . , χr be the characters and n1, . . . , nr, the dimensions (over K) of thenon-isomorphic simple KG-modules. Then

χKG =∑

1≤i≤r

niχi.

Remark: A consequence of Schur’s Lemma. Let V be a simple KG-module. Then EndKG(V ) = K.

For, by Schur’s lemma, we have D = EndKG(V ) is a division algebraover K. Since V is finite dimensional over K, so is D over K. But K isalgebraically closed and hence D = K.

Proposition 18.2. Let (V, ρ) and (V ′, ρ′) be two simple KG-modules and

f : V → V ′, a K-linear map. Let f = 1gf (see Remark 2 in §16).

(i) If V is not isomorphic to V ′, then f = 0.

(ii) If V = V ′ and ρ = ρ′, then f is a scalar multiplication in V and

trace (f) = trace (f).

53

Proof. (i) follows from Schur’s lemma since f is KG-linear.(ii) That f is a scalar multiplication follows from above consequence of

Schur’s lemma. Also,

trace (f) =1

g

g∑

i=1

trace (ρ(x−1i ) f ρ(xi)) = trace (f).

Corollary. If V is a simple KG-module, then the characteristic of K does

not divide the dimension of V over K.

Proof. Apply (ii) of the above Proposition to an f of trace 1.

We introduce a scalar product <,>, in the K-vector space of all K-valuedfunctions on G. If φ and ψ are two such functions, we define

< φ,ψ >=1

g

∑

1≤i≤g

φ(x−1i ) · ψ(xi).

Now let (V, ρ) and (V ′, ρ′) be two simple KG-modules and χ and χ′, theircharacters. Let f : V → V ′ be a K-linear map. Choosing K-bases forV and V ′, we can identify ρ(x), ρ′(x) (for x ∈ G) and f with matrices(ρij(x)), (ρ

′kl(x)) and (fli) respectively. Then f is identified with the ma-

trix (fkj) where

fkj =1

g

∑

x,l,i

ρ′kl(x−1) fli ρij(x).

If V is not isomorphic to V ′, by Proposition 18.2. (i), we have f = 0, i.e.,fkj = 0 for all k, j and all fli ∈ K. It follows that

1

g

∑

x∈G

ρ′kl(x−1)ρij(x) = 0 for all k, l, i, j.

From this we get that

< χ′, χ > =1

g

∑

x∈G

χ′(x−1)χ(x) =1

g

∑

x∈G

(∑

k

ρ′kk(x−1)

)·(∑

i

ρii(x)

)

=1

g

∑

k,i

(∑

x∈G

ρ′kk(x−1) · ρii(x)

)= 0.


If V = V ′ and ρ = ρ′, by Proposition 18.2 (ii), we get that

fkj = 0 if k 6= j and

fkk =α

n

where n is the dimension of V over K and α =∑

i fii. By making suitablechoices for the fii, it is seen easily that

1

g

∑

x∈G

ρkl(x−1)ρij(x) = 0 if k 6= j and l 6= i and

1

g

∑

x∈G

ρki(x−1)ρik(x) =

1

n.

This gives

< χ, χ > =1

g

∑

x∈G

χ(x−1) · χ(x)

=1

g

∑

x∈G

(∑

k

ρkk(x−1)

)(∑

i

ρii(x)

)

=∑

k

1

g

∑

x∈G

ρkk(x−1)ρkk(x)

= 1.

From the above discussion, we get the following

Theorem 18.1 Let χ1, . . . , χr denote the characters of a complete set of

mutually non-isomorphic simple KG-modules. Then < χi, χj >= δij, the

Kronecker symbol.

Corollary 1. If K has characteristic 0, then a KG-module is determined by

its character.

Proof. Let S1, . . . , Sr be a complete set of mutually non-isomorphic simpleKG-modules and χ1, . . . , χr be their characters. Let V be a KG-module andwrite a decomposition of V = V1⊕ · · ·⊕Vs, where the Vj are simple KG-modules. Let χ be the character of V and ni be the number of the Vj whichare isomorphic to Si. Then we have χ = n1χ1 + · · · + nrχr. By the above

55

theorem, we have ni =< χ, χi > which shows that V is determined by itscharacter.

Corollary 2. If K has characteristic 0, then a KG-module is simple if and

only if its character χ satisfies < χ, χ >= 1.

Proof. Let V be a KG-module with χ its character. As in the proof of theabove corollary, we have χ =

∑i niχi. Then we get that < χ, χ >=

∑i n

2i .

If < χ, χ >= 1, we find that ni = 1 for some i and hence the other nj arezero which shows that V is simple.

§ 19. Integrality of Characters.

In the rest of this chapter, the ground field K will be assumed to be C.

A complex number α is called an algebraic integer if it is integral over Z,i.e., α satisfies a relation of the type αn + a1α

n−1 + · · ·+ an = 0 with ai ∈ Z.We know, by Appendix I, that the algebraic integers form a subring of Ccontaining Z.

Proposition 19.1 If a rational number is an algebraic integer then it must

be a rational integer.

Proof. Let b/c be a rational number (where b, c ∈ Z and c 6= 0) satisfyingan equation

(b/c)n + a1(b/c)n−1 + · · · + an = 0

with ai ∈ Z. We can assume that (b, c) = 1. From the above equation weget bn = −c(a1b

n−1 + · · ·+ ancn−1) which shows that c divides bn in Z. Since

(b, c) = 1, it follows that c must be an invertible element in Z and henceb/c ∈ Z.

Proposition 19.2 Let G be a finite group and (V, ρ) a representation of Gwith χ its character. Then

(i) For any x ∈ G,χ(x) is an algebraic integer.

(ii) If V is simple with its dimension over C being n and the conjugacy class

containing x has h elements, then (hχ(x))/n is an algebraic integer.

Proof. (i) With respect to a suitable basis of V, ρ(x) becomes a diagonalmatrix, say with (ζ1, . . . , ζn) as its principal diagonal. If x has order m, we


have ζmi = 1. Thus the ζi are roots of unity and hence χ(x) = ζ1 + · · · + ζn

is an algebraic integer.

(ii) Let C1, . . . , Cr be the conjugacy classes of elements of G and letzi =

∑y∈Ci

y. We know that z1, . . . , zr form a basis of the center of thegroup algebra CG and that they satisfy equations

zizj =∑

1≤k≤r

dijkzk (See the Remark in §15)

where dijk ∈ Z. This shows that z1Z⊕ · · ·⊕zrZ is a commutative subringof CG. Since it is finitely generated Z-module, it is integral over Z (byAppendix I). Let ρ : CG → EndC(V ) be the homomorphism of C-algebrasobtained from ρ. Let x ∈ Ci. Since zi is integral over Z, ρ(zi) is integral overZ. But zi being in the center of CG, ρ(zi) is CG-linear and since V is simple,(by consequence of Schur’s lemma) ρ(zi) is a scalar multiplication by someα ∈ C. It follows that α is integral over Z, i.e., it is an algebraic integer.Now hχ(x) = trace (ρ(zi)) = nα which shows that (hχ(x))/n is an algebraicinteger.

Corollary. The dimension (over C) of a simple CG-module divides the order

of the group.

Proof. Let n be the dimension of the simple CG-module and χ, its character.Let C1, . . . , Cr be the conjugacy classes with hi, the number of elements inCi. Choose xi ∈ Ci. Since < χ, χ >= 1, we get that

g =∑

x∈G

χ(x−1)χ(x) =∑

1≤i≤r

χ(x−1i )χ(xi).

This givesg

n=∑

1≤i≤r

χ(x−1i )

hiχ(xi)

n

and hence g/n must be an algebraic integer by Proposition 19.2. But g/nbeing a rational number must be in Z, i.e., n|g.

§ 20. Burnside’s Theorem.

Lemma. Let G be a finite group, (V, ρ) a simple CG-module of dimension

n over C and x an element of G such that the conjugacy class containing x

57

has h elements with (h, n) = 1. Then either χρ(x) = 0 or ρ(x) is a scalar

multiplication in V .

Proof. Choosing a suitable basis of V we may assume that ρ(x) is a diagonalmatrix, say with (ζ1, . . . , ζn) as its principal diagonal. If x has order m, thenthe ζi are mth roots of unity and we can write ζi = ζai where ζ is a primitivemth root of unity and 0 ≤ ai ≤ m− 1. If all the ζi are equal, then ρ(x) is ascalar matrix. Suppose then that not all the ζi are equal. Since (h, n) = 1,there exist integers b and c such that bh+ nc = 1. So

χρ(x)

n= b

hχρ(x)

n+ cχρ(x)

is an algebraic integer by Proposition 19.2. Since ζa1 , . . . , ζan are not equal,we have ∣∣∣∣

χρ(x)

n

∣∣∣∣ =|ζa1 + · · · + ζan|

n< 1.

The conjugates of χρ(x)/n are again algebraic integers. Also they are ofmodulus less than 1 since a conjugate of χρ(x)/n is of the form (ζja1 + · · ·+ζjan)/n (for some j coprime to m) and ζja1 , . . . , ζjan are not equal. Theproduct of all these conjugates is a rational number of modulus less than 1which is an algebraic integer, thus by Proposition 19.1 it is 0. It follows thatχρ(x)/n = 0.

Theorem 20.1. (Burnside). A group of order paqb is solvable, p and qbeing primes.

Proof. Let G be a group of order paqb. We shall prove by induction on theorder of the group. A group of prime order being solvable (see Appendix II),we can assume a, b > 0.

In view of the induction hypothesis it suffices to show that G has a propernon-trivial normal subgroup. For if G1 is such a normal subgroup of G, thenthe orders of G1 and G/G1 are of the form pcqd and are strictly less thanthe order of G. Therefore by induction hypothesis G1 and G/G1 are solvableand hence G is solvable. Let us now show that such a G1 exists. Let H bea Sylow p-subgroup of G and x 6= 1 be an element in the center of H. Weshall distinguish two cases.

Case (i). x is in the center of G:

Then (x) is a normal subgroup of G and we can take G1 = (x).


Case (ii). x is not in the center of G. (In particular [G,G] 6= 1).

Then the centralizer of x in G is a proper subgroup of G containing Hand therefore has index qc in G for some c > 0. Thus the conjugacy classcontaining x has qc elements.

Let χ1, . . . , χr be all the distinct irreducible components of the charac-ter of the regular representation of G (χ1 being the character of the trivialrepresentation) and let 1 = n1, . . . , nr be their respective degrees. Then thecharacter χ of the regular representation satisfies χ = n1χ1 + · · · + nrχr.Since x 6= 1, we have

0 = χ(x) = 1 + n2χ2(x) + · · · + nrχr(x).

It is not possible that χi(x) = 0 for all i ≥ 2 for which q 6 | ni. For,otherwise, we would get an equation of the type 1 + qα = 0 with α analgebraic integer. But that would mean that −1/q is an algebraic integercontradicting Proposition 19.1.

Thus there exists an i ≥ 2 for which q 6 | ni and χi(x) 6= 0. If (V, ρ) is thecorresponding representation, then we know, by the above lemma, that ρ(x)is a scalar multiplication.

If ni = 1, then [G,G] is a proper normal subgroup of G since χi is notthe trivial character. So in this case we can take G1 = [G,G].

If ni > 1, then, since V is simple, ρ(G) can not consist of scalar multipli-cations alone. In this case we can take

G1 = y ∈ G : ρ(y) is a scalar multiplication in V .

This completes the proof of the theorem.

Appendix I

All the rings considered below will be assumed to be commutative.

§ 1. Integral Extensions.

Let A be a subring of a ring B. We say that x ∈ B is integral over A if itsatisfies an equation

xn + a1xn−1 + · · · + an = 0

with ai ∈ A. We say that B is integral over A if every x ∈ B is integral overA.

Proposition 1.1. Let B = A[x1, . . . , xn] be a finitely generated ring over A.

The following statements are equivalent:

1. B is integral over A.

2. The elements x1, . . . , xn are integral over A.

3. B is finitely generated A-module.

Proof. 1.⇒ 2.: Follows from definition.2.⇒ 3.: For each i, 1 ≤ i ≤ n, we have a relation

xni

i + ai1xn1−1i + · · · + aini

= 0, aij ∈ A.

By induction on∑αi it is clear that the monomial xα1

1 · · ·xαn

n is in the A-submodule of B generated by the monomials xβ1

1 · · ·xβn

n , 0 ≤ βi ≤ ni. HenceB is a finitely generated A-module.

59


3. ⇒ 1. Let B be generated as an A-module by y1, . . . , ym and let x beany element of B. Then

xyi =∑

1≤j≤m

aijyj aij ∈ A,

that is ∑

1≤j≤m

(aij − xδij

)yj = 0, 1 ≤ i ≤ m.

Multiplying the ith equation by the (i, k)th cofactor of the matrix(aij −xδij

)

and summing over i, we get

det(aij − xδij

)yk = 0 1 ≤ k ≤ m.

Since 1 ∈ B, this implies that

det(aij − xδij

)= 0.

Expanding this we get a monic equation in x with coefficients in A.

Corollary. Let A be a subring of a ring B. The elements of B which are

integral over A form a subring of B containing A.

Proof. Let x1, x2 ∈ B be integral over A. Then A[x1, x2] is integral over A,so that x1 + x2 and x1x2 are integral over A.

Proposition 1.2. Let A be a subring of B, with B integral over A. If B is

a field , then so is A.

Proof. Let x 6= 0 be an element of A. Then 1/x ∈ B and therefore satisfiesa relation (

1

x

)n

+ a1

(1

x

)n−1

+ · · · + an = 0

with ai ∈ A. Multiplying this equation by xn, we get

x(a1 + a2x+ · · · + anxn−1) = −1.

This shows that 1/x ∈ A and hence A is a field.

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§ 2. The Hilbert Nullstellensatz.

Proposition 2.1. Let K be a field and L = K[x1, . . . , xn] be a finitely

generated ring over K. If L is a field, then L is algebraic over K.

Proof. We proceed by induction on n, the case n = 1 being trivial. Byinduction hypothesis L = K(x1)[x2, . . . , xn] is algebraic over K(x1). Weshow that x1 is algebraic over K. Indeed, suppose that x1 is transcendentalover K. We have equations

fi0xni

i + fi1xn1−1i + · · · + fini

= 0

with fij ∈ K[x1] and i = 2, . . . , n. Let φ = f20f30 · · · fn0. Then x2, . . . , xn

are integral over the ring K[x1, φ−1]. By Proposition 1.1, L = K[x1, . . . , xn]

is integral over K[x1, φ−1]. Hence by Proposition 1.2, K[x1, φ

−1] is a field.This is a contradiction since x1 is transcendental over K.

Corollary. Let K be an algebraically closed field. Then any maximal ideal

in K[x1, . . . , xn] is generated by x1 − a1, . . . , xn − an for some ai ∈ K.

Proof. Let m be a maximal ideal of K[x1, . . . , xn]. Then the ring L =K[x1, . . . , xn]/m is a finitely generated K-algebra which is a field. Hence Lis algebraic over K and therefore L = K. Hence there exist a1, . . . , an ∈ Ksuch that xi − ai ∈ m. Clearly x1 − a1, . . . , xn − an generate m.

LetK be an algebraically closed field and let a be an ideal inK[x1, . . . , xn].We say that (a1, . . . , an) ∈ Kn is a zero of a if f(a1, . . . , an) = 0 for allf(x1, . . . , xn) ∈ a. Clearly (a1, . . . , an) is a zero if and only if a is containedin the ideal generated by x1−a1, . . . , xn−an. Let a be generated by (fα)α∈I .Then (a1, . . . , an) is a zero of a if and only if fα(a1, . . . , an) = 0 for all α ∈ I.For an ideal a of K[x1, . . . , xn] we denote by V (a) the set of zeros of a.

Theorem 2.1. (Hilbert Nullstellensatz.) Let K be an algebraically closed

field and let a be an ideal of K[x1, . . . , xn]. Let f ∈ K[x1, . . . , xn] be such

that f(a1, . . . , an) = 0 for all (an, . . . , an) ∈ V (a). Then fm ∈ a for some

positive integer m.

Proof. We consider the polynomial ring K[x1, . . . , xn, z] in n + 1 variables.Let b be an ideal generated by a and 1 − zf in K[x1, . . . , xn, z]. We observethat b is not contained in any maximal ideal of K[x1, . . . , xn, z]. Indeed, let m

be any maximal ideal containing b. Since, by the corollary to Proposition 2.1,m is generated by x1−a1, . . . , xn−an, z−an+1 with ai ∈ K, (a1, . . . , an, an+1)


is a zero of the ideal b. Then (a1, . . . , an) ∈ V (a) and hence f(a1, . . . , an) = 0.But 1 − an+1f(a1, . . . , an) = 1, i.e., (a1, . . . , an+1) is not a zero of 1 − zf .Contradiction. So we have b = K[x1, . . . , xn, z], that is,

h1f1 + · · · + hrfr + hr+1(1 − zf) = 1

with fi ∈ a and hi ∈ K[x1, . . . , xn, z]. Putting z = 1/f in the above poly-nomial identity and multiplying by a suitable power fm of f , we see thatfm ∈ a.

Corollary. Let f1, . . . , fm be forms in n > m variables over K. Then

f1, . . . , fm have a common non-trivial zero in K.

Proof. Suppose that (0, . . . , 0) is the only common zero of f1, . . . , fm inKn. Let a be the ideal generated by f1, . . . , fm. Then V (a) = (0, . . . , 0). ByHilbert Nullstellensatz we can find a positive integer N such that

xNi =

∑

1≤j≤m

hijfj, hij ∈ K[x1, . . . , xn].

We may assume that the hij are polynomials of degree < N . We assertthat the monomials xα1

1 · · ·xαn

n , 0 ≤ αi < N , generate K[x1, . . . , xn] as aK[f1, . . . , fm]-module. Take a monomial xβ1

1 · · ·xβn

n . If βi ≥ N , then

xβi

i = xβi−Ni

∑

1≤j≤m

hijfj

so that xβ1

1 · · ·xβn

n is a linear combination of monomials xγ1

1 · · ·xγn

n ,∑γi <∑

βi, with coefficients in K[f1, . . . , fm]. Now the assertion is immediate byinduction on

∑βi.

Hence, by Proposition 1.1, K[x1, . . . , xn] is integral over K[f1, . . . , fm].Since m < n, this is impossible by transcendence degree consideration.

Appendix II

§ 1. Group acting on a set.

Let G be a group and E be a set. An action of G on E is a map φ : G×E → Esuch that (denoting φ(x, a) by xa), for all x1, x2 ∈ G and a ∈ E,

1. x1(x2a) = (x1x2)a and

2. 1a = a.

We also say that G acts on E (via the action φ).

For an x ∈ G, the map φx : E → E such that a 7→ xa is obviously atransformation of the set E (i.e., a bijective map of E onto itself). If S(E)denotes the group of all transformations of E, it is easily seen that the mapφ : G → S(E) such that x 7→ φx is a homomorphism of groups. Conversely,given any homomorphism of groups ψ : G → S(E), we obtain an action ψof G on E, in a natural way, by setting ψ(x, a) = (ψ(x))(a). Thus giving anaction of G on E is equivalent to giving such a homomorphism.

We say that G acts transitively on E if for every a, b ∈ E, there exists anx ∈ G such that xa = b.

Given an action of G on E, we have an equivalence relation on E namelyfor a, b ∈ E, a ∼ b if there exists an x ∈ G such that xa = b. The equivalenceclasses (under this equivalence relation) are called the orbits of E. For an a ∈E, the orbit containing a is also called the orbit of a (or the orbit through a)and is denoted by E(a). We say that an orbit is trivial if it contains onlyone element. It is clear that G acts transitively on each orbit in an obviousmanner.

63


For a ∈ E, the set Ga := x ∈ G : xa = a, which is a subgroup of G, iscalled the isotropy subgroup at a. An element a ∈ E is called a fixed pointof E (for the given action of G) if Ga = G, i.e., the orbit of a is trivial. IfGa = 1 for every a ∈ E, we say that G acts freely on E.

For any subgroup H of G, we denote by G/H, the set of left cosetsdetermined by H.

Proposition 1.1. Let G be a group acting on a set E. Then for any element

a ∈ E, there exists a natural bijection of the set G/Ga onto E(a).

Proof. Consider the canonical map : G → E(a) defined such that x 7→ xa.We have (xh)a = x(ha) = xa for every x ∈ G and h ∈ Ga which shows thatthis map passes down to a map η : ZG/Ga → E(a), namely; xGa 7→ xa, forx ∈ G. Clearly η is surjective and moreover

η(x1Ga) = η(x2Ga)

⇒ x1a = x2a

⇒ x−12 x1 ∈ Ga

⇒ x1Ga = x2Ga

Which shows that η is bijective.

Examples.

1. Given a group G and a set E, there is always a trivial action: (x, a) 7→ afor all x ∈ G and a ∈ E. Here the orbits are obviously trivial.

2. Let H be a subgroup of G. Then H acts on G by left translation:(h, a) 7→ ha for h ∈ H and a ∈ G. Here the orbits are the right cosetsof H and the isotropy subgroups of H are trivial, i.e., H acts freely onG.

In particular, the action ofG on itself by left translation gives a monomor-phism of the groups G→ S(G). Hence

Caley’s Theorem. A finite group is a group of permutations.

3. A group G acts on itself by inner conjugation: (x, a) 7→ xax−1 forx, a ∈ G. Here

(a) the orbits are the conjugacy classes of elements of G

65

(b) the isotropy group at a ∈ G is the centraliser of a in G and

(c) the fixed points of G are the elements of its center.

If G is finite, by Proposition 1.1, we get that the number of conjugatesof an element of G is equal to the index of its centraliser in G.

Recall that a finite group whose order is a power of a prime p is called ap-group.

Proposition 1.2. Let G be a p-group acting on a finite set E. Then the

number of fixed points of E is congruent to the number of elements in Emodulo p.

Proof. Suppose E(a) is nontrivial for some a ∈ E. By Proposition 1.1, wehave the index of Ga in G = card (E(a)) > 1 and hence (G being a p-group)p divides card (E(a)).

Let E0 be the set of all fixed points of E and a1, . . . , ak ∈ E be a completesystem of representatives of the non-trivial orbits of E. Then we have

card (E) = card (E0) +k∑

i=1

card (E(ai)).

This proves the proposition since p divides each card (E(ai)).

Corollary 1. If p does not divide the card (E), then there exists at least one

fixed point of E.

Corollary 2. The center of a p-group is non-trivial.

Take E = G and make G act on itself by inner conjugation.

§ 2. Sylow’s theorem.

Let G be a finite group of order n. Let p be a prime dividing n. Suppose pm

is the maximal power of p dividing n. Then a subgroup of G of order pm, ifit exists, is called a Sylow p-subgroup of G. We now prove the existence of aSylow p-subgroup of G.

Theorem 2.1 (Sylow). Let p be a prime dividing the order of a finite group

G. Then G contains a Sylow p-subgroup.


Proof. Let n be the order of G. Write n = pmq with (p, q) = 1. We firstprove a

Lemma.(

n

pm

)≡ q (mod p) (in particular p ∤

(n

pm

)).

For, we have, over the field Z/(p), a polynomial identity

(1 + x)n =((1 + x)pm)q

= (1 + xpm

)q.

Compairing the coefficients of xpm

, we have the lemma.

To complete the proof of the theorem, we take E, the set of all subsetsof G each containing pm elements and make G act on E by left translation.Clearly the cardinality of E is

(n

pm

). Since p ∤

(n

pm

)), there is an A ∈ E such

that the cardinality of the orbit E(A) is not divisible by p. Now consider theisotropy group GA at A. By Proposition 1.1 and the choice of A, p does notdivide the index of GA in G. Hence pm divides the order of GA. In particularpm ≤ order (GA).

On the other hand, since hA = A for every h ∈ GA, we have GAa ⊆ Afor every a ∈ A. This shows that

order (GA) = card (GAa) ≤ card (A) = pm

and hence pm = card (GA). Thus GA is a Sylow p-subgroup of G.

Corollary.

1. For every power pα, of a prime p, dividing the order of a group G, there

exists a subgroup of order pα. (In particular, the converse of Lagrange’s

theorem is true for p-groups).

2. Further, every such subgroup is contained in a Sylow p-subgroup of G.

Proof. Not a difficult exercise!

§ 3. Solvable groups.

A group G is said to be solvable if there exists a sequence of subgroups

G = G0 ⊃ G1 ⊃ · · · ⊃ Gn = 1

67

such that Gi+1 is a normal subgroup of Gi and Gi+1/Gi is abelian (0 ≤ i ≤n− 1). Such a sequence is called a solvable series of G.

Remark. Every abelian group is solvable.

Proposition 3.1.

1. Any subgroup and any quotient group of a solvable group are solvable.

2. Conversely, if there exists a normal subgroup H of a group G such that

H and G/H are solvable, then so is G.

Proof. 1. Let G be a solvable group having a sovable series

G = G0 ⊃ G1 ⊃ · · · ⊃ Gn = 1.

Suppose H is a subgroup of G. Then

H = G0 ∩H ⊃ G1 ∩H ⊃ · · · ⊃ Gn ∩H = 1

is easily seen to be a solvable series of H. Further if H is a normal subgroupof G and η : G→ G/H is the canonical homomorphism, then

G/H = η(G0) ⊃ η(G1) ⊃ · · · ⊃ η(Gn) = 1

is again seen to be a solvable series of G/H.

2. Conversely, suppose H is a normal subgroup of G such that H andG/H are solvable. Let η : G → G/H be the canonical homomorphism.Suppose

H = H0 ⊃ H1 ⊃ · · · ⊃ Hn = 1

andG/H = G′

0 ⊃ G′1 ⊃ · · · ⊃ G′

m = 1

are solvable series of H and G/H respectively. Then it follows that

G = η−1(G′0) ⊃ · · · ⊃ η−1(G′

m) ⊃ H1 ⊃ · · · ⊃ Hn = 1

is a solvable series for G (noting of course that η−1(G′m) = H etc.).

Proposition 3.2. Any p-group is solvable.

Proof. Exercise. (Look at the center of the group and proceed by inductionon the order of the group.)

Exercise. A group of order pq, where p and q are primes, is solvable.

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