III Algebras (Full)

Part III — Algebras

Based on lectures by C. J. B. Brookes

Notes taken by Dexter Chua

Lent 2017

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

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

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

The aim of the course is to give an introduction to algebras. The emphasis will be

on non-commutative examples that arise in representation theory (of groups and Lie

algebras) and the theory of algebraic D-modules, though you will learn something

ab out commutative algebras in passing.

Topics we discuss include:

–

Artinian algebras. Examples, group algebras of finite groups, crossed products.

Structure theory. Artin–Wedderburn theorem. Projective modules. Blocks.

–

No etherian algebras. Examples, quantum plane and quantum torus, differen-

tial operator algebras, enveloping algebras of finite dimensional Lie algebras.

Structure theory. Injective hulls, uniform dimension and Goldie’s theorem.

–

Hochschild chain and cochain complexes. Hochschild homology and cohomology.

Gerstenhab er algebras.

– Deformation of algebras.

– Coalgebras, bialgebras and Hopf algebras.

Pre-requisites

It will be assumed that you have attended a first course on ring theory, eg IB Groups,

Rings and Modules. Experience of other algebraic courses such as II Representation

Theory, Galois Theory or Number Fields, or III Lie algebras will be helpful but not

necessary.

Contents

0 Introduction

1 Artinian algebras

1.1 Artinian algebras

1.2 Artin–Wedderburn theorem

1.3 Crossed products

1.4 Projectives and blocks

1.5 K

2 Noetherian algebras

2.1 Noetherian algebras

2.2 More on A

(k) and U(g)

2.3 Injective modules and Goldie’s theorem

3 Hochschild homology and cohomology

3.1 Introduction

3.2 Cohomology

3.3 Star products

3.4 Gerstenhaber algebra

3.5 Hochschild homology

4 Coalgebras, bialgebras and Hopf algebras

0 Introduction

We start with the definition of an algebra. Throughout the course,

will be a

field.

Definition

(

-algebra)

A (unital) associative

-algebra is a

-vector space

together with a linear map

A ⊗ A → A

, called the product map, and linear

map u : k → A, called the unit map, such that

– The product induced by m is associative.

– u(1) is the identity of the multiplication.

In particular, we don’t require the product to be commutative. We usually

write m(x ⊗ y) as xy.

Example.

Let

K/k

be a finite field extension. Then

is a (commutative)

k-algebra.

Example.

The

n × n

matrices

(

) over

form a non-commutative

-algebra.

Example.

The quaternions

is an

-algebra, with an

-basis 1

, i, j, k

, and

multiplication given by

= j

= k

= 1, ij = k, ji = −k.

This is in fact a division algebra (or skew fields), i.e. the non-zero elements have

multiplicative inverse.

Example. Let G be a finite group. Then the group algebra

kG =

g : g ∈ G, λ

∈ k

with the obvious multiplication induced by the group operation is a k-algebra.

These are the associative algebras underlying the representation theory of

finite groups.

Most of the time, we will just care about algebras that are finite-dimensional

-vector spaces. However, often what we need for the proofs to work is not

that the algebra is finite-dimensional, but just that it is Artinian. These algebras

are defined by some finiteness condition on the ideals.

Definition

(Ideal)

A left ideal of

is a

-subspace of

such that if

x ∈ A

and

y ∈ I

, then

xy ∈ I

. A right ideal is one where we require

yx ∈ I

instead.

An ideal is something that is both a left ideal and a right ideal.

Since the multiplication is not necessarily commutative, we have to make the

distinction between left and right things. Most of the time, we just talk about

the left case, as the other case is entirely analogous.

The definition we want is the following:

Definition

(Artinian algebra)

An algebra

is left Artinian if it satisfies the

descending chain condition (DCC ) on left ideals, i.e. if we have a descending

chain of left ideals

≥ I

≥ · · · ,

then there is some N such that I

N+m

= I

for all m ≥ 0.

We say an algebra is Artinian if it is both left and right Artinian.

Example. Any finite-dimensional algebra is Artinian.

The main classification theorem for Artinian algebras we will prove is the

following result:

Theorem

(Artin–Wedderburn theorem)

Let

be a left-Artinian algebra such

that the intersection of the maximal left ideals is zero. Then

is the direct sum

of finitely many matrix algebras over division algebras.

When we actually get to the theorem, we will rewrite this in a way that

seems a bit more natural.

One familiar application of this theorem is in representation theory. The group

algebra of a finite group is finite-dimensional, and in particular Artinian. We

will later see that Maschke’s theorem is equivalent to saying that the hypothesis

of the theorem holds. So this puts a very strong constraint on how the group

algebra looks like.

After studying Artinian rings, we’ll talk about Noetherian algebras.

Definition

(Noetherian algebra)

An algebra is left Noetherian if it satisfies

the ascending chain condition (ACC ) on left ideals, i.e. if

≤ I

≤ · · ·

is an ascending chain of left ideals, then there is some

such that

N+m

for all m ≥ 0.

Similarly, we can define right Noetherian algebras, and say an algebra is

Noetherian if it is both left and right Noetherian.

We can again look at some examples.

Example. Again all finite-dimensional algebras are Noetherian.

Example.

In the commutative case, Hilbert’s basis theorem tells us a polynomial

algebra

[

, · · · , X

] in finitely many variables is Noetherian. Similarly, the

power series rings k[[X

, · · · , X

]] are Noetherian.

Example.

The universal enveloping algebra of a finite-dimensional Lie algebra

are the (associative!) algebras that underpin the representation theory of these

Lie algebras.

Example.

Some differential operator algebras are Noetherian. We assume

char k

= 0. Consider the polynomial ring

[

]. We have operators “multipli-

cation by

” and “differentiate with respect to

” on

[

]. We can form the

algebra

[

∂

∂x

] of differential operators on

[

], with multiplication given by

the composition of operators. This is called the Weyl algebra

. We will show

that this is a non-commutative Noetherian algebra.

Example.

Some group algebras are Noetherian. Clearly all group algebras of

finite groups are Noetherian, but the group algebras of certain infinite groups

are Noetherian. For example, we can take

G =











1 λ µ

0 1 ν

0 0 0





: λ, µ, ν ∈ Z







and this is Noetherian. However, we shall not prove this.

We will see that all left Artinian algebras are left Noetherian. While there is

a general theory of non-commutative Noetherian algebras, it is not as useful as

the theory of commutative Noetherian algebras.

In the commutative case, we often look at

Spec A

, the set of prime ideals of

. However, sometimes in the non-commutative there are few prime ideals, and

so Spec is not going to keep us busy.

Example. In the Weyl algebra A

, the only prime ideals are 0 and A

We will prove a theorem of Goldie:

Theorem

(Goldie’s theorem)

Let

be a right Noetherian algebra with no

non-zero ideals all of whose elements are nilpotent. Then

embeds in a finite

direct sum of matrix algebras over division algebras.

Some types of Noetherian algebras can be thought of as non-commutative

polynomial algebras and non-commutative power series, i.e. they are deformations

of the analogous commutative algebra. For example, we say

is a deformation

of the polynomial algebra

[

X, Y

], where instead of having

XY − Y X

= 0, we

have

XY − Y X

= 1. This also applies to enveloping algebras and Iwasawa

algebras. We will study when one can deform the multiplication so that it remains

associative, and this is bound up with the cohomology theory of associative

algebras — Hochschild cohomology. The Hochschild complex has rich algebraic

structure, and this will allow us to understand how we can deform the algebra.

At the end, we shall quickly talk about bialgebras and Hopf algebras. In a

bialgebra, one also has a comultiplication map

A → A⊗A

, which in representation

theory is crucial in saying how to regard a tensor product of two representations

as a representation.

1 Artinian algebras

1.1 Artinian algebras

We continue writing down some definitions. We already defined left and right

Artinian algebras in the introduction. Most examples we’ll meet are in fact

finite-dimensional vector spaces over

. However, there exists some more perverse

examples:

Example. Let

A =



r s

0 t



: r ∈ Q, s, t ∈ R



Then this is right Artinian but not left Artinian over

. To see it is not left

Artinian, note that there is an ideal

I =



0 s

0 0



: s ∈ R



∼

, and a matrix



r s

0 t



acts on this on the left by sending



0 s

0 0



. Since

is an infinite-dimensional

-vector space, one sees that there is an

infinite strictly descending chain of ideals contained in I.

The fact that it is right Artinian is a direct verification. Indeed, it is not

difficult to enumerate all the right ideals, which is left as an exercise for the

reader.

As in the case of commutative algebra, we can study the modules of an

algebra.

Definition

(Module)

Let

be an algebra. A left

-module is a

-vector space

M and a bilinear map

A ⊗ M M

a ⊗ m xm

such that (

)

(

) for all

a, b ∈ A

and

m ∈ M

. Right

-modules are

defined similarly.

-bimodule is a vector space

that is both a left

-module and a

right

-module, such that the two actions commute — for

a, b ∈ A

and

x ∈ M

we have

a(xb) = (ax)b.

Example.

The algebra

itself is a left

-module. We write this as

, and

call this the left regular representation. Similarly, the right action is denoted

These two actions are compatible by associativity, so A is an A-A-bimodule.

If we write

End

(

) for the

-linear maps

A → A

, then

End

is naturally a

algebra by composition, and we have a

-algebra homomorphism

A → End

(

)

that sends

a ∈ A

to multiplication by

on the left. However, if we want

to multiply on the right instead, it is no longer a

-algebra homomorphism

A → End

(A). Instead, it is a map A → End

(A)

, where

Definition

(Opposite algebra)

Let

be a

-algebra. We define the opposite

algebra

to be the algebra with the same underlying vector space, but with

multiplication given by

x · y = yx.

Here on the left we have the multiplication in

and on the right we have the

multiplication in A.

In general, a left A-module is a right A

-module.

As in the case of ring theory, we can talk about prime ideals. However, we

will adopt a slightly different definition:

Definition

(Prime ideal)

An ideal

is prime if it is a proper ideal, and if

and J are ideals with IJ ⊆ P , then either I ⊆ P or J ⊆ P .

It is an exercise to check that this coincides in the commutative case with

the definition using elements.

Definition

(Annihilator)

Let

be a left

-module and

m ∈ M

. We define

the annihilators to be

Ann(m) = {a ∈ A : am = 0}

Ann(M) = {a ∈ A : am = 0 for all m ∈ M} =

m∈M

Ann(m).

Note that

Ann

(

) is a left ideal of

, and is in fact the kernel of the

-module

homomorphism

A → M

given by

x 7→ xm

. We’ll denote the image of this map

by Am, a left submodule of M, and we have

Ann(m)

∼

Am.

On the other hand, it is easy to see that

Ann

(

) is an fact a (two-sided) ideal.

Definition

(Simple module)

A non-zero module

is simple or irreducible if

the only submodules of M are 0 and M.

It is easy to see that

Proposition.

Let

be an algebra and

a left ideal. Then

is a maximal left

ideal iff A/I is simple.

Example. Ann(m) is a maximal left ideal iff Am is irreducible.

Proposition.

Let

be an algebra and

a simple module. Then

∼

A/I

for some (maximal) left ideal I of A.

Proof.

Pick an arbitrary element

m ∈ M

, and define the

-module homomor-

phism

A → M

(

) =

. Then the image is a non-trivial submodule,

and hence must be

. Then by the first isomorphism theorem, we have

∼

A/ ker ϕ.

Before we start doing anything, we note the following convenient lemma:

Lemma.

Let

be a finitely-generated

module. Then

has a maximal

proper submodule M

Proof.

Let

, · · · , m

∈ M

be a minimal generating set. Then in particular

, · · · , m

k−1

is a proper submodule of

. Moreover, a submodule

containing

is proper iff it does not contain

, and this property is

preserved under increasing unions. So by Zorn’s lemma, there is a maximal

proper submodule.

Definition

(Jacobson radical)

The

(

) of

is the intersection of all maximal

left ideals.

This is in fact an ideal, and not just a left one, because

J(A) =

{maximal left ideals} =

m∈M,M simple

Ann(m) =

M simple

Ann(M),

which we have established is an ideal. Yet, it is still not clear that this is

independent of us saying “left” instead of “right”. It, in fact, does not, and this

follows from the Nakayama lemma:

Lemma

(Nakayama lemma)

The following are equivalent for a left ideal

(i) I ≤ J(A).

(ii)

For any finitely-generated left

-module

, if

, then

= 0,

where

is the module generated by elements of the form

, with

a ∈ I

and m ∈ M.

(iii) G = {1 + a : a ∈ I} = 1 + I is a subgroup of the unit group of A.

In particular, this shows that the Jacobson radical is the largest ideal satisfying

(iii), which is something that does not depend on handedness.

Proof.

–

(i)

⇒

(ii): Suppose

I ≤ J

(

) and

M 6

= 0 is a finitely-generated

-module,

and we’ll see that IM  M.

Let

be a maximal submodule of

. Then

M/N

is a simple module,

so for any

¯m ∈ M/N

, we know

Ann

(

¯m

) is a maximal left ideal. So

J(A) ≤ Ann(M/N ). So IM ≤ J(A)M ≤ N  M.

–

(ii)

⇒

(iii): Assume (ii). We let

x ∈ I

and set

= 1 +

. Hence

1 =

y − x ∈ Ay

. Since

is a left ideal, we know

. In

other words, we know





Now using (ii) on the finitely-generated module

A/Ay

(it is in fact generated

by 1), we know that

A/Ay

= 0. So

. So there exists

z ∈ A

such

that 1 =

(1 +

). So (1 +

) has a left inverse, and this left inverse

lies in

, since we can write

= 1

− zx

. So

is a subgroup of the unit

group of A.

–

(iii)

⇒

(i): Suppose

is a maximal left ideal of

. Let

x ∈ I

. If

x 6∈ I

then

by maximality of

. So 1 =

for some

y ∈ I

and

z ∈ A

. So

= 1

− zx ∈ G

. So

is invertible. But

y ∈ I

. So

This is a contradiction. So we found that

I < I

, and this is true for all

maximal left ideals I

. Hence I ≤ J(A).

We now come to the important definition:

Definition (Semisimple algebra). An algebra is semisimple if J(A) = 0.

We will very soon see that for Artinian algebras, being semi-simple is equiva-

lent to a few other very nice properties, such as being completely reducible. For

now, we shall content ourselves with some examples.

Example. For any A, we know A/J(A) is always semisimple.

We can also define

Definition (Simple algebra). An algebra is simple if the only ideals are 0 and

It is trivially true that any simple algebra is semi-simple — the Jacobson

radical is an ideal, and it is not

. A particularly important example is the

following:

Example.

Consider

(

). We let

be the matrix with 1 in the (

i, i

)th

entry and zero everywhere else. This is idempotent, i.e.

. It is also

straightforward to check that

















0 · · · 0 a

0 · · · 0

0 · · · 0 a

0 · · · 0

0 · · · 0 a

0 · · · 0

















The non-zero column is, of course, the

th column. Similarly,

is the matrices

that are zero apart from in the

th row. These are in fact all left and all right

ideals respectively. So the only ideals are 0 and A.

As a left A-module, we can decompose A as

A =

i=1

which is a decomposition into simple modules.

Definition

(Completely reducible)

A module

is completely reducible

iff it is a sum of simple modules.

Here in the definition, we said “sum” instead of “direct sum”, but the

following proposition shows it doesn’t matter:

Proposition. Let M be an A-module. Then the following are equivalent:

(i) M is completely reducible.

(ii) M is the direct sum of simple modules.

(iii)

Every submodule of

has a complement, i.e. for any submodule

there is a complement N

such that M = N ⊕ N

Often, the last condition is the most useful in practice.

Proof.

– (i) ⇒ (ii): Let M be completely reducible, and consider the set

≤ M} : S

are simple,

is a direct sum

Notice this set is closed under increasing unions, since the property of

being a direct sum is only checked on finitely many elements. So by Zorn’s

lemma, it has a maximal element, and let N be the sum of the elements.

Suppose this were not all of

. Then there is some

S ≤ M

such that

S 6⊆ N

. Then

S ∩ N ( S

. By simplicity, they intersect trivially. So

is a direct sum, which is a contradiction. So we must have

, and

M is the direct sum of simple modules.

– (ii) ⇒ (i) is trivial.

– (i) ⇒ (iii): Let N ≤ M be a submodule, and consider

≤ M} : S

are simple, N +

is a direct sum

Again this set has a maximal element, and let

be the direct sum of those

. Again if

P ⊕ N

is not all of

, then pick an

S ≤ M

simple such that

is not contained in

P ⊕ N

. Then again

S ⊕ P ⊕ N

is a direct sum, which

is a contradiction.

–

(iii)

⇒

(i): It suffices to show that if

N < M

is a proper submodule,

then there exists a simple module that intersects

trivially. Indeed, we

can take

to be the sum of all simple submodules of

, and this forces

N = M .

To do so, pick an

x 6∈ N

, and let

be submodule of

maximal among

those satisfying

P ∩ N

= 0 and

x 6∈ N ⊕ P

. Then

N ⊕ P

is a proper

submodule of M. Let S be a complement. We claim S is simple.

If not, we can find a proper submodule

. Let

be a complement of

N ⊕ P ⊕ S

. Then we can write

M = N ⊕ P ⊕ S

⊕ Q

x = n + p + s + q

By assumption,

and

are not both zero. We wlog assume

is non-

zero. Then

P ⊕ Q

is a larger submodule satisfying (

P ⊕ Q

)

∩ N

= 0 and

x 6∈ N ⊕

(

P ⊕ Q

). This is a contradiction. So

is simple, and we are

done.

Using these different characterizations, we can prove that completely reducible

modules are closed under the familiar ex operations.

Proposition.

Sums, submodules and quotients of completely reducible modules

are completely reducible.

Proof.

It is clear by definition that sums of completely reducible modules are

completely reducible.

To see that submodules of completely reducible modules are completely

reducible, let

be completely reducible, and

N ≤ M

. Then for each

x ∈ N

there is some simple submodule

S ≤ M

containing

. Since

S ∩ N ≤ S

and

contains x, it must be S, i.e. S ⊆ N. So N is the sum of simple modules.

Finally, to see quotients are completely reducible, if

is completely reducible

and N is a submodule, then we can write

M = N ⊕ P

for some P . Then M/N

∼

P , and P is completely reducible.

We will show that every left Artinian algebra is completely reducible over

itself iff it is semi-simple. We can in fact prove a more general fact for

-modules.

To do so, we need a generalization of the Jacobson radical.

Definition

(Radical)

For a module

, we write

Rad

(

) for the intersection

of maximal submodules of M, and call it the radical of M.

Thus, we have Rad(

A) = J(A) = Rad(A

Proposition.

Let

be an

-module satisfying the descending chain condition

on submodules. Then M is completely reducible iff Rad(M) = 0.

Proof.

It is clear that if

is completely reducible, then

Rad

(

) = 0. Indeed,

we can write

M =

α∈A

where each S

is simple. Then

J(A) ≤

α∈A





β∈A\{α}





= {0}.

Conversely, if

Rad

(

) = 0, we note that since

satisfies the descending

chain condition on submodules, there must be a finite collection

, · · · , M

maximal submodules whose intersection vanish. Then consider the map

i=1

x (x + M

, x + M

, · · · , x + M

)

The kernel of this map is the intersection of the

, which is trivial. So this

embeds

as a submodule of

. But each

is simple, so

is a submodule

of a completely reducible module, hence completely reducible.

Corollary.

is a semi-simple left Artinian algebra, then

is completely

reducible.

Corollary.

is a semi-simple left Artinian algebra, then every left

-module

is completely reducible.

Proof.

Every

-module

is a quotient of sums of

. Explicitly, we have a

map

m∈M

A M

)

Then this map is clearly surjective, and thus M is a quotient of

is not semi-simple, then it turns out it is rather easy to figure out radical

of M , at least if M is finitely-generated.

Lemma.

Let

be left Artinian, and

a finitely generated left

-module, then

J(A)M = Rad(M ).

Proof.

Let

be a maximal submodule of

. Then

M/M

is simple, and is in

fact A/I for some maximal left ideal I. Then we have

J(A)





= 0,

since J(A) < I. Therefore J(A)M ≤ M

. So J(A)M ≤ Rad(M).

Conversely, we know

J(A)M

is an

A/J

(

)-module, and is hence completely

reducible as

A/J

(

) is semi-simple (and left Artinian). Since an

-submodule

J(A)M

is the same as an

A/J

(

)-submodule, it follows that it is completely

reducible as an A-module as well. So

Rad



J(A)M



= 0,

and hence Rad(M ) ≤ J(A)M.

Proposition. Let A be left Artinian. Then

(i) J(A) is nilpotent, i.e. there exists some r such that J(A)

= 0.

(ii)

is a finitely-generated left

-module, then it is both left Artinian and

left Noetherian.

(iii) A is left Noetherian.

Proof.

(i)

Since

is left-Artinian, and

(

)

r ∈ N}

is a descending chain of

ideals, it must eventually be constant. So

(

)

(

)

r+1

for some

. If

this is non-zero, then again using the descending chain condition, we see

there is a left ideal

with

(

)

I 6

= 0 that is minimal with this property

(one such ideal exists, say J(A) itself).

Now pick

x ∈ I

with

(

)

x 6

= 0. Since

(

)

(

)

, it follows

that

(

)

(

)

= 0. So by minimality,

(

)

x ≥ I

. But the other

inclusion clearly holds. So they are equal. So there exists some

a ∈ J

(

)

with x = ax. So

(1 − a)x = 0.

But 1 − a is a unit. So x = 0. This is a contradiction. So J(A)

= 0.

(ii)

Let

(

)

. Then

i+1

is annihilated by

(

), and hence

completely reducible (it is a module over semi-simple

A/J

(

)). Since

is a finitely generated left

-module for a left Artinian algebra, it satisfies

the descending chain condition for submodules (exercise), and hence so

does M

i+1

So we know

i+1

is a finite sum of simple modules, and therefore

satisfies the ascending chain condition. So

i+1

is left Noetherian,

and hence M is (exercise).

(iii) Follows from (ii) since A is a finitely-generated left A-module.

1.2 Artin–Wedderburn theorem

We are going to state the Artin–Wedderburn theorem for right (as opposed to

left) things, because this makes the notation easier for us.

Theorem

(Artin–Wedderburn theorem)

Let

be a semisimple right Artinian

algebra. Then

A =

i=1

for some division algebra D

, and these factors are uniquely determined.

A has exactly r isomorphism classes of simple (right) modules S

, and

End

) = {A-module homomorphisms S

→ S

}

∼

and

dim

) = n

If A is simple, then r = 1.

If we had the left version instead, then we need to insert op’s somewhere.

Artin–Wedderburn is an easy consequence of two trivial lemma. The key idea

that leads to Artin–Wedderburn is the observation that the map

→ End

(

)

sending

to left-multiplication by

is an isomorphism of algebras. So we need

to first understand endomorphism algebras, starting with Schur’s lemma.

Lemma

(Schur’s lemma)

Let

, M

be simple right

-modules. Then either

∼

, or

Hom

(

, M

) = 0. If

is a simple

-module, then

End

(

)

is a division algebra.

Proof.

A non-zero

-module homomorphism

→ M

must be injective, as

the kernel is submodule. Similarly, the image has to be the whole thing since

the image is a submodule. So this must be an isomorphism, and in particular

has an inverse. So the last part follows as well.

As mentioned, we are going to exploit the isomorphism

∼

End

(

This is easy to see directly, but we can prove a slightly more general result, for

the sake of it:

Lemma.

(i)

is a right

-module and

is an idempotent in

, i.e.

, then

∼

Hom

(eA, M ).

(ii) We have

eAe

∼

End

(eA).

In particular, we can take e = 1, and recover End

)

∼

Proof.

(i) We define maps

me (ex 7→ mex)

Me Hom(eA, M )

α(e) α

We note that

(

) =

(

) =

(

)

e ∈ Me

. So this is well-defined. By

inspection, these maps are inverse to each other. So we are done.

Note that we might worry that we have to pick representatives

and

for the map

, but in fact we can also write it as

(

)(

) =

, since

idempotent. So we are safe.

(ii) Immediate from above by putting M = eA.

Lemma. Let M be a completely reducible right A-module. We write

M =

where

}

are distinct simple

-modules. Write

End

(

), which we

already know is a division algebra. Then

End

)

∼

and

End

(M) =

)

Proof.

The result for

End

(

) is just the familiar fact that a homomorphism

→ S

is given by an

m × n

matrix of maps

S → S

(in the case of vector

spaces over a field

, we have

End

(

)

∼

, so they are matrices with entries in

k). Then by Schur’s lemma, we have

End

(M) =

End

)

∼

We now prove Artin–Wedderburn.

Proof of Artin–Wedderburn.

is semi-simple, then it is completely reducible

as a right A-module. So we have

∼

End(A

)

∼

We now decompose each

(

) into a sum of simple modules. We know each

(

) is a non-trivial

(

) module in the usual way, and the action of

the other summands is trivial. We can simply decompose each

(

) as the

sum of submodules of the form

















0 0 · · · 0 0

· · · a

−1

0 0 · · · 0 0

















and there are

components. We immediately see that if we write

for this

submodule, then we have

dim

) = n

Finally, we have to show that every simple module S of A is one of the S

. We

simply have to note that if

is a simple

-module, then there is a non-trivial

map

A → S

(say by picking

x ∈ S

and defining

(

) =

). Then in the

decomposition of

into a direct sum of simple submodules, there must be one

factor

such that

is non-trivial. Then by Schur’s lemma, this is in fact an

isomorphism S

∼

This was for semi-simple algebras. For a general right Artinian algebra, we

know that

A/J

(

) is semi-simple and inherits the Artinian property. Then

Artin–Wedderburn applies to A/J(A).

Some of us might be scared of division algebras. Sometimes, we can get away

with not talking about them. If

is not just Artinian, but finite-dimensional,

then so are the D

Now pick an arbitrary

x ∈ D

. Then the sub-algebra of

generated by

be commutative. So it is in fact a subfield, and finite dimensionality means it is

algebraic over

. Now if we assume that

is algebraically closed, then

must

live in k. So we’ve shown that these D

must be k itself. Thus we get

Corollary.

is algebraically closed and

is a finite-dimensional semi-simple

k-algebra, then

∼

(k).

This is true, for example, when k = C.

We shall end this section by applying our results to group algebras. Recall

the following definition:

Definition

(Group algebra)

Let

be a group and

a field. The group algebra

of G over k is

kG =

g : g ∈ G, λ

∈ k

This has a bilinear multiplication given by the obvious formula

(λ

g)(µ

h) = λ

(gh).

The first thing to note is that group algebras are almost always semi-simple.

Theorem

(Maschke’s theorem)

Let

be a finite group and

p - |G|

, where

p = char k, so that |G| is invertible in k, then kG is semi-simple.

Proof.

We show that any submodule

of a

-module

has a complement.

Let π : U → V be any k-vector space projection, and define a new map

|G|

g∈G

gπg

−1

: U → V.

It is easy to see that this is a

-module homomorphism

U → V

, and is a

projection. So we have

U = V ⊕ ker π

and this gives a kG-module complement.

There is a converse to Maschke’s theorem:

Theorem. Let G be finite and kG semi-simple. Then char k - |G|.

Proof.

We note that there is a simple

-module

, given by the trivial module.

This is a one-dimensional k vector space. We have

D = End

(S) = k.

Now suppose

is semi-simple. Then by Artin–Wedderburn, there must be

only one summand of S in kG.

Consider the following two ideals of kG: we let

g ∈ kG :

= 0

This is in fact a two-sided ideal of

. We also have the center of the algebra,

given by

g ∈ kG : λ ∈ k

Now if char k | |G|, then I

⊆ I

. So we can write

kG =

⊕ I

⊕ · · · .

But we know

acts trivially on

and

, and they both have dimension 1.

This gives a contradiction. So we must have char k - |G|.

We can do a bit more of representation theory. Recall that when

algebraically closed and has characteristic zero, then the number of simple

modules is the number of conjugacy classes of

. There is a more general result

for a general characteristic p field:

Theorem.

Let

be algebraically closed of characteristic

, and

be finite.

Then the number of simple

modules (up to isomorphism) is equal to the

number of conjugacy classes of elements of order not divisible by

. These are

known as the p-regular elements.

We immediately deduce that

Corollary.

|G|

for some

and

is prime, then the trivial module is the

only simple kG module, when char k = p.

Note that we can prove this directly rather than using the theorem, by

showing that

ker

(

kG → k

) is a nilpotent ideal, and annihilates all simple

modules.

Proof sketch of theorem.

The number of simple

modules is just the number

of simple

kG/J

(

) module, as

(

) acts trivially on every simple module.

There is a useful trick to figure out the number of simple

-modules for a given

semi-simple A. Suppose we have a decomposition

∼

i=1

(k).

Then we know

is the number of simple

-modules. We now consider [

A, A

the k-subspace generated by elements of the form xy − yx. Then we see that

[A, A]

∼

i=1

(k)

(k), M

(k)]

Now by linear algebra, we know [

(

)

, M

(

)] is the trace zero matrices, and

so we know

dim

(k)

(k), M

(k)]

= 1.

Hence we know

dim

[A, A]

= r.

Thus we need to compute

dim

kG/J(kG)

[kG/J(kG), kG/J(kG)]

We then note the following facts:

(i) For a general algebra A, we have

A/J(A)

[A/J(A), A/J(A)]

∼

[A, A] + J(A)

(ii) Let g

, · · · , g

be conjugacy class representatives of G. Then

+ [kG, kG]}

forms a k-vector space basis of kG/[kG, kG].

(iii)

, · · · , g

is a set of representatives of

-regular conjugacy classes, then



[kG, kG] + J(kG)

o

form a basis of kG/([kG, kG] + J(kG)).

Hence the result follows.

One may find it useful to note that [

kG, kG

] +

(

) consists of the elements

in kG such that x

∈ [kG, kG] for some s.

In this proof, we look at

[

A, A

]. However, in the usual proof of the result

in the characteristic zero looks at the center

(

). The relation between these

two objects is that the first is the 0th Hochschild homology group of

, while

the second is the 0th Hochschild cohomology group of A.

1.3 Crossed products

Number theorists are often interested in representations of Galois groups and

-modules where

is an algebraic number field, e.g.

. In this case, the

’s

appearing in Artin–Wedderburn may be non-commutative.

We have already met one case of a non-commutative division ring, namely

the quaternions H. This is in fact an example of a general construction.

Definition

(Crossed product)

The crossed product of a

-algebra

and a

group G is specified by the following data:

– A group homomorphism φ : G → Aut

(B), written

(λ) = λ

;

– A function

Ψ(g, h) : G × G → B.

The crossed product algebra has underlying set

g : λ

∈ B.

with operation defined by

λg · µh = λµ

Ψ(g, h)(gh).

The function Ψ is required to be such that the resulting product is associative.

We should think of the

as specifying what happens when we conjugate

pass µ, and then Ψ(g, h)(gh) is the product of g and h in the crossed product.

Usually, we take

, a Galois extension of

, and

Gal

(

K/k

). Then

the action

is the natural action of

on the elements of

, and we restrict to

maps Ψ : G × G → K

only.

Example.

Consider

, and

. Then

Gal

(

C/R

)

∼

{e, g}, where g is complex conjugation. The elements of H are of the form

e + λ

where λ

, λ

∈ C, and we will write

1 · g = g, i · g = k, 1 · e = 1, i · e = i.

Now we want to impose

−1 = j

= 1g · 1g = ψ(g, g)e.

So we set Ψ(

g, g

) =

−

1. We can similarly work out what we want the other

values of Ψ to be.

Note that in general, crossed products need not be division algebras.

The crossed product is not just a

-algebra. It has a natural structure of a

G-graded algebra, in the sense that we can write it as a direct sum

BG =

g∈G

Bg,

and we have Bg

· Bg

⊆ Bg

Focusing on the case where

K/k

is a Galois extension, we use the notation

(

K, G,

Ψ), where Ψ :

G × G → K

. Associativity of these crossed products is

equivalent to a 2-cocycle condition Ψ, which you will be asked to make precise

on the first example sheet.

Two crossed products (

K, G,

) and (

K, G,

) are isomorphic iff the map

G × G K

(g, h) Ψ

(g, h)(Ψ

(g, h))

−1

satisfies a 2-coboundary condition, which is again left for the first example sheet.

Therefore the second (group) cohomology

{2-cocycles : G × G → K

}

{2-coboundaries : G × G → K

}

determines the isomorphism classes of (associative) crossed products (K, G, Ψ).

Definition

(Central simple algebra)

A central simple

-algebra is a finite-

dimensional k-algebra which is a simple algebra, and with a center Z(A) = k.

Note that any simple algebra is a division algebra, say by Schur’s lemma.

So the center must be a field. Hence any simple

-algebra can be made into a

central simple algebra simply by enlarging the base field.

Example. M

(k) is a central simple algebra.

The point of talking about these is the following result:

Fact.

Any central simple

-algebra is of the form

(

) for some division

algebra

which is also a central simple

-algebra, and is a crossed product

(K, G, Ψ).

Note that when

and

, then the second cohomology group has

2 elements, and we get that the only central simple

-algebras are

(

) or

(H).

For amusement, we also note the following theorem:

Fact (Wedderburn). Every finite division algebra is a field.

1.4 Projectives and blocks

In general, if

is not semi-simple, then it is not possible to decompose

as a

direct sum of simple modules. However, what we can do is to decompose it as a

direct sum of indecomposable projectives.

We begin with the definition of a projective module.

Definition

(Projective module)

-module is projective

if given modules

M and M

and maps

M 0

then there exists a map

P → M

such that the following diagram commutes:

M 0

Equivalently, if we have a short exact sequence

0 N M

M 0,

then the sequence

0 Hom(P, N) Hom(P, M

) Hom(P, M) 0

is exact.

Note that we get exactness at

Hom

(

P, N

) and

Hom

(

P, M

) for any

at all.

Projective means it is also exact at Hom(P, M ).

Example. Free modules are always projective.

In general, projective modules are “like” free modules. We all know that free

modules are nice, and most of the time, when we want to prove things about

free modules, we are just using the property that they are projective. It is also

possible to understand projective modules in an algebro-geometric way — they

are “locally free” modules.

It is convenient to characterize projective modules as follows:

Lemma. The following are equivalent:

(i) P is projective.

(ii) Every surjective map φ : M → P splits, i.e.

∼

ker φ ⊕ N

where N

∼

P .

(iii) P is a direct summand of a free module.

Proof.

– (i) ⇒ (ii): Consider the following lifting problem:

M P 0

The lifting gives an embedding of

into

that complements

ker φ

(by

the splitting lemma, or by checking it directly).

–

(ii)

⇒

(iii): Every module admits a surjection from a free module (e.g. the

free module generated by the elements of P )

–

(iii)

⇒

(i): It suffices to show that direct summands of projectives are

projective. Suppose P is is projective, and

∼

A ⊕ B.

Then any diagram

M 0

can be extended to a diagram

A ⊕ B

M 0

˜α

by sending

to 0. Then since

A ⊕ B

∼

is projective, we obtain a lifting

A ⊕ B → M

, and restricting to A gives the desired lifting.

Our objective is to understand the direct summands of a general Artinian

-algebra

, not necessarily semi-simple. Since

is itself a free

-module, we

know these direct summands are always projective.

Since

is not necessarily semi-simple, it is in general impossible to decompose

it as a direct sum of simples. What we can do, though, is to decompose it as a

direct sum of indecomposable modules.

Definition

(Indecomposable)

A non-zero module

is indecomposable if

cannot be expressed as the direct sum of two non-zero submodules.

Note that since

is (left) Artinian, it can always be decomposed as a finite

sum of indecomposable (left) submodules. Sometimes, we are also interested in

decomposing A as a sum of (two-sided) ideals. These are called blocks.

Definition

(Block)

The blocks are the direct summands of

that are inde-

composable as ideals.

Example. If A is semi-simple Artinian, then Artin-Wedderburn tells us

A =

and the M

) are the blocks.

We already know that every Artinian module can be decomposed as a direct

sum of indecomposables. The first question to ask is whether this is unique. We

note the following definitions:

Definition

(Local algebra)

An algebra is local if it has a unique maximal left

ideal, which is J(A), which is the unique maximal right ideal.

If so, then

A/J

(

) is a division algebra. This name, of course, comes from

algebraic geometry (cf. local rings).

Definition

(Unique decomposition property)

A module

has the unique

decomposition property if

is a finite direct sum of indecomposable modules,

and if

M =

i=1

then n = m, and, after reordering, M

= M

We want to prove that

as an

-module always has the unique decomposi-

tion property. The first step is the following criterion for determining unique

decomposition property.

Theorem

(Krull–Schmidt theorem)

Suppose

is a finite sum of indecom-

posable

-modules

, with each

End

(

) local. Then

has the unique

decomposition property.

Proof. Let

M =

i=1

We prove by induction on

. If

= 1, then

is indecomposable. Then we

must have n = 1 as well, and the result follows.

For m > 1, we consider the maps

: M

M M

: M

M M

We observe that

i=1

◦ β

: M

→ M

Since

End

(

) is local, we know some

◦ β

must be invertible, i.e. a unit, as

they cannot all lie in the Jacobson radical. We may wlog assume

◦ β

is a

unit. If this is the case, then both

and

have to be invertible. So

∼

Consider the map id −θ = φ, where

θ : M M

i=2

−1

Then φ(M

) = M

. So φ|

looks like α

. Also

i=2

φ|

i=2

looks like the identity map. So in particular, we see that

surjective. However, if φ(x) = 0, this says x = θ(x), So

x ∈

i=2

But then

(

) = 0. Thus

= 0. Thus

is an automorphism of

with

φ(M

) = φ(M

). So this gives an isomorphism between

i=2

∼

i=2

and so we are done by induction.

Now it remains to prove that the endomorphism rings are local. Recall the

following result from linear algebra.

Lemma

(Fitting)

Suppose

is a module with both the ACC and DCC on

submodules, and let f ∈ End

(M). Then for large enough n, we have

M = im f

⊕ ker f

Proof. By ACC and DCC, we may choose n large enough so that

: f

(M) → f

(M)

is an isomorphism, as if we keep iterating

, the image is a descending chain and

the kernel is an ascending chain, and these have to terminate.

If m ∈ M, then we can write

(m) = f

)

for some m

. Then

m = f

) + (m − f

)) ∈ im f

+ ker f

and also

im f

∩ ker f

= ker(f

: f

(M) → f

(M)) = 0.

So done.

Lemma.

Suppose

is an indecomposable module satisfying ACC and DCC

on submodules. Then B = End

(M) is local.

Proof.

Choose a maximal left ideal of

, say

. It’s enough to show that if

x 6∈ I

then x is left invertible. By maximality of I, we know B = Bx + I. We write

1 = λx + y,

for some

λ ∈ B

and

y ∈ I

. Since

y ∈ I

, it has no left inverse. So it is not an

isomorphism. By Fitting’s lemma and the indecomposability of

, we see that

= 0 for some m. Thus

(1 + y + y

+ · · · + y

m−1

)λx = (1 + y + · · · + y

m−1

)(1 − y) = 1.

So x is left invertible.

Corollary.

Let

be a left Artinian algebra. Then

has the unique decompo-

sition property.

Proof.

We know

satisfies the ACC and DCC condition. So

is a finite

direct sum of indecomposables.

So if

is an Artinian algebra, we know

can be uniquely decomposed as a

direct sum of indecomposable projectives,

A =

For convenience, we will work with right Artinian algebras and right modules

instead of left ones. It turns out that instead of studying projectives in

, we

can study idempotent elements instead.

Recall that

End

(

)

∼

. The projection onto

is achieved by left

multiplication by an idempotent e

= e

The fact that the

decomposes as a direct sum of the

translates to the

condition

= 1, e

= 0

for i 6= j.

Definition

(Orthogonal idempotent)

A collection of idempotents

}

is or-

thogonal if e

= 0 for i 6= j.

The indecomposability of P

is equivalent to e

being primitive:

Definition

(Primitive idempotent)

An idempotent is primitive if it cannot be

expressed as a sum

e = e

+ e

where e

, e

are orthogonal idempotents, both non-zero.

We see that giving a direct sum decomposition of

is equivalent to finding

an orthogonal collection of primitive idempotents that sum to 1. This is rather

useful, because idempotents are easier to move around that projectives.

Our current plan is as follows — given an Artinian algebra

, we can quotient

out by

(

), and obtain a semi-simple algebra

A/J

(

). By Artin–Wedderburn,

we know how we can decompose

A/J

(

), and we hope to be able to lift this

decomposition to one of

. The point of talking about idempotents instead is

that we know what it means to lift elements.

Proposition.

Let

be a nilpotent ideal in

, and let

be an idempotent of

A/N ≡

A. Then there is an idempotent e ∈ A with f = ¯e.

In particular, we know

(

) is nilpotent, and this proposition applies. The

proof involves a bit of magic.

Proof.

We consider the quotients

A/N

for

i ≥

1. We will lift the idempotents

successively as we increase

, and since

is nilpotent, repeating this process

will eventually land us in A.

Suppose we have found an idempotent

i−1

∈ A/N

i−1

with

i−1

. We

want to find f

∈ A/N

such that

= f.

For

i >

1, we let

be an element of

A/N

with image

i−1

A/N

i−1

Then since

− x

vansishes in

A/N

i−1

, we know

− x ∈ N

i−1

. Then in

particular,

− x)

= 0 ∈ A/N

. (†)

We let

= 3x

− 2x

Then by a direct computation using (

†

), we find

, and

has image

i−1

−

i−1

A/N

i−1

(alternatively, in characteristic

, we can use

= x

). Since N

= 0 for some k, this process gives us what we want.

Just being able to lift idempotents is not good enough. We want to lift

decompositions as projective indecomposables. So we need to do better.

Corollary. Let N be a nilpotent ideal of A. Let

1 = f

+ · · · + f

with {f

} orthogonal primitive idempotents in A/N. Then we can write

1 = e

+ · · · + e

with {e

} orthogonal primitive idempotents in A, and ¯e

= f

Proof. We define a sequence e

∈ A inductively. We set

= 1.

Then for each

i >

1, we pick

a lift of

· · ·

∈ e

i−1

, since by

inductive hypothesis we know that f

+ · · · + f

∈ e

i−1

/N . Then

i+1

= e

i+1

= e

i+1

We let

= e

− e

i+1

Then

¯e

= f

Also, if j > i, then

= e

i+1

and so

= (e

− e

i+1

= 0.

Similarly e

= 0.

We now apply this lifting of idempotents to

(

), which we know is

nilpotent. We know

A/N

is the direct sum of simple modules, and thus the

decomposition corresponds to

1 = f

+ · · · + f

∈ A/J(A),

and these

are orthogonal primitive idempotents. Idempotent lifting then gives

1 = e

+ · · · + e

∈ A,

and these are orthogonal primitive idempotents. So we can write

A =

where

are indecomposable projectives, and

(

) =

is simple.

By Krull–Schmidt, any indecomposable projective isomorphic to one of these

The final piece of the picture is to figure out when two indecomposable

projectives lie in the same block. Recall that if

is a right

-module and

idempotent, then

∼

Hom

(eA, M ).

In particular, if M = fA for some idempotent f, then

Hom(eA, f A)

∼

fAe.

However, if e and f are in different blocks, say B

and B

, then

fAe ∈ B

∩ B

= 0,

since B

and B

are (two-sided!) ideals. So we know

Hom(eA, f A) = 0.

So if

Hom

(

eA, f A

)

= 0, then they are in the same block. The existence of a

homomorphism can alternatively be expressed in terms of composition factors.

We have seen that each indecomposable projective P has a simple “top”

P/P J(A)

∼

Definition

(Composition factor)

A simple module

is a composition factor

of a module M if there are submodules M

≤ M