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(1)

Chapter 2

R INGS

We intend to classify simple rings R using their (left) modules. For irreducible (left) R-modules M, J = ann

R

M is an ideal in R hence J = 0.

Moreover, Schur’s Lemma implies that ∆ = R

0

: = End

R

M is a division ring and M can be regarded as a left vector space over ∆. The embedding

ρ

: R ,→ End M then factors as

ρ

: R ,→ R

00

: = End

M ,→ End M,

with the left one being an embedding ι : R ,→ R

00

into ∆-linear maps.

To further investigate ι, the major tools we will develop is a kind of

“density theorems” or “approximation theorems” for ring homomorphisms into certain endomorphism rings. For this purpose we first study some re- lated and more flexible notions called (semi-)primitivity so that results on (completely reducible) modules in the previous chapter can be applied.

The major result is the classification theorem of (semi-)simple rings due to Wedderburn and Artin, under suitable finiteness assumption. More pre- cisely, simple artinian rings are nothing but the matrix rings M

n

( ) and semi-simple artinian rings are simply finite product of them.

Instead of giving the original proof, the proof via (semi-)primitivity and density presented here was due to Jacobson. The proof on Jacobson’s den- sity theorem was later simplified by Bourbaki via a version on density the- orem on completely reducible modules. This later density theorem turns out to be powerful in other applications.

1. Semi-primitivity and the radical

Definition 2.1. Let R be a ring and by R-modules we mean left R-modules.

(1) R is (left) primitive if it admits a faithful irreducible module M.

(2) R is (left) semi-primitive (s.p.) if every a ∈ R \ { 0 } acts non- trivially on some R-modules, i.e. ρ ( a ) 6= 0 for some ρ.

9

(2)

Clearly simple rings R are primitive: take M = R/I for a maximal left ideal I and notice that ann

R

M = 0. Also primitive implies semi-primitive.

Semi-primitivity is more flexible to work with due to the following fact:

Proposition 2.2. The following are equivalent:

(1) R is semi-primitive.

(2) R has a faithful completely reducible module M =

L

M

i

. (3) R is sub-direct product of primitive rings R

i

’s, namely

R

  ι //

!! !!

∏ R

i πi



R

i

P

ROOF

. (1) ⇒ (2): Take M =

La6=0

M

a

where ρ

a

( a ) 6= 0 on M

a

. (2) ⇒ (3): M is faithful means 0 = ann

R

M =

T

ann

R

M

i

, hence

R ,→ ∏ R/ann

R

M

i

= :R

i

.

R

i

is primitive since it is faithful on M

i

. Also R  R

i

clearly.

(3) ⇒ (1): Denote ρ

i

: R

i

,→ End M

i

. Let a ∈ R. If ρ

iπiι

( a ) = 0 for all i then a ∈

T

ker ρ

iπiι

=

T

ker π

iι

=

T

ker π

i

= { 0 } . 

Recall that for any left ideal I ⊂ R,

ann

R

R/I = ( I : R ) : = { a ∈ R | aR ⊂ I } ⊂ I,

which is the largest ideal contained in I. Also any irreducible R-module M is isomorphic to R/I for a maximal left ideal I. Hence

Corollary 2.3. (1) R is primitive ⇐⇒ ( I : R ) = 0 for some maximal left ideal I.

(2) R is s.p. ⇐⇒

TI

( I : R ) = 0 among all maximal left ideals.

Corollary 2.4. Let R be commutative. Then (1) R is primitive if and only it is a field, (2) R is s.p. if and inly if it is a sub-direct product of fields.

Example 2.5. (1) For a division ring∆, the matrix ring Mn()is simple.

(2) Primitive rings might not be simple. Let V be an infinite dimensional vector space over∆, L =EndV and F⊂ L be those linear maps f ∈ L with dimf(V) < ∞. Then F is a non-trivial proper ideal and hence L is not simple. But V is clearly irreducible as L-modules, hence L is primitive.

(3) Z is s.p. by Z,→pZ/(p)(sinceTp(p) =0 over all primes).

(3)

1. SEMI-PRIMITIVITY AND THE RADICAL 11

Definition 2.6. The Jacobson radical is the ideal J ( R ) ≡ rad R : =

\

I:max left

( I : R ) .

Proposition 2.7. For any maximal left ideal I, the ideal ( I : R ) is the intersection of all maximal left ideals I

0

with R/I

0

∼ = R/I. In particular,

J ( R ) =

\

I:max left

I.

P

ROOF

. By definition, for any R-module M, ann

R

M =

\

xM

ann

R

x.

Moreover, M is irreducible (M ∼ = R/I for a maximal left I) implies that M = Rx ∼ = R/I

x

for any 0 6= x ∈ M and I

x

: = ann

R

x is a maximal left ideal. Clearly ( I

x

: R ) = ann

R

M = ( I : R ) . This implies that

T

( I : R ) ⊃

T

I. Notice also that any f : R/I

0

∼ = R/I leads to I

0

= I

x

with x = f ( ¯1 ) .

The reverse direction

T

( I : R ) ⊂

T

I follows by definition.  Definition 2.8. An ideal P ⊂ R is primitive if ¯ R : = R/P is a primitive ring.

Corollary 2.9. P ⊂ R is a primitive ideal if and only if P = ( I : R ) for some maximal left ideal I. In particular J ( R ) =

T

P among primitive ideals.

P

ROOF

. If P = ( I : R ) = ann

R

M with M = R/I, then M is also a R : ¯ = R/P-module with ann

R¯

M = P/P = 0. Conversely, let M be a faithful irreducible ¯ R-module. By viewing M as a R-module via R  R, M is also ¯ irreducible with ann

R

M = P. But then M ∼ = R/I for some maximal left ideal I and hence P = ( I : R ) . The last statement on J ( R ) follows from

Proposition 2.7. 

Remark 2.10. In Proposition 2.7, one might wonder if ( I : R ) is indeed the intersection of all I

0

with ( I

0

: R ) = ( I : R ) . This is equivalent to ask if the primitive ring ¯ R : = R/ ( I : R ) has a unique faithful irreducible module M up to isomorphisms. This fails already for simple rings “without any finiteness assumptions”. Classification of all irreducible modules of a sim- ple ring is one the fundamental problems in ring theory. We will see shortly that the uniqueness does hold for simple (left) artinian rings.

Exercise 2.1. Following Example 2.5 (2), show that the ring L/F is simple

which admits more than one irreducible modules.

(4)

Notice that the notion of primitivity of a ring depends on the left/right choices while the simplicity of a ring does not. The first example of a left primitive ring which is not right primitive was constructed by Bergman when he was an undergraduate (published in Proceedings AMS 1964).

On the contrary, we will now show that semi-primitivity on one side implies semi-simplicity on the other side. In fact we have

Theorem 2.11. The left and right Jacobson radicals coincide:

J

left

( R ) =

\

max left

I =

\

max right

I = J

right

( R ) . The proof is based on the notion of quasi-regularity:

Definition 2.12. Let R be a ring. An element a ∈ R is called (left, right) quasi-regular if 1 − a has a (left, right) inverse.

Nilpotent elements are clearly quasi-regular: if a

n

= 0 then w ( 1 − a ) = ( 1 − a ) w = 1 for w : = 1 + a + . . . + a

n1

.

Lemma 2.13. Let I be a left/right ideal such that every a ∈ I is left/right quasi- regular. Then all elements in I are quasi-regular.

P

ROOF

. We prove the left case. The right case is similar. If w ( 1 − a ) = 1, write w = 1 − a

0

then 1 − a − a

0

+ a

0

a = 1, hence

a

0

= a − a

0

a ∈ I.

By assumption we have ( 1 − a

00

)( 1 − a

0

) = 1 for some a

00

∈ R. This implies that 1 − a

00

= 1 − a and a

00

= a ∈ I. In particular 1 − a is invertible.  Definition 2.14. By Lemma 2.13, a left/right ideal consisting of (left/right) quasi-regular elements are referred as a quasi-regular left/right ideal.

Proposition 2.15. J ( R ) is the largest quasi-regular left ideal. In particular, J ( R ) = { a ∈ R | 1 − ba has a left inverse for all b ∈ R } .

P

ROOF

. We first prove that J ( R ) is a quasi-regular left ideal. If a ∈ J ( R ) is not left quasi-regular, then the left ideal R ( 1 − a ) 6= R and hence R ( 1 − a ) ⊂ I for some maximal left I. But a ∈ J ( R ) ⊂ I and 1 − a ∈ I then leads to the contradiction 1 ∈ I.

Next we prove that any quasi-regular left ideal Q is contained in J ( R ) .

If not, then Q 6⊂ I for some maximal left I. Then I + Q = R and we have

1 = a + q for some a ∈ I and q ∈ Q. But then a = 1 − q is invertible which

leads to the contradiction I = R.

(5)

2. DENSITY THEOREMS 13

Fro the last statement, a ∈ J ( R ) ⇒ ba ∈ Ra ⊂ J ( R ) which implies that a lies in the RHS. Conversely a lies in the RHS implies that Ra is a quasi-regular left ideal. Hence a ∈ Ra ⊂ J ( R ) .  P

ROOF OF

T

HEOREM

2.11. J

left

( R ) is the largest quasi-regular left ideal.

However from its original definition we know that J

left

( R ) is an ideal. Thus J

left

( R ) is also a quasi-regular right ideal. Proposition 2.15 then implies that J

left

( R ) ⊂ J

right

( R ) . Similarly we have J

right

( R ) ⊂ J

left

( R ) .  From now on, there is no need to distinguish left/right quasi-regular ideals as well as left/right J ( R ) .

Corollary 2.16. A ring R is left semi-primitive ⇐⇒ J ( R ) = 0 ⇐⇒ R is right semi-primitive.

2. Density theorems

Let R be s ring and M =

R

M be a R-module with ax =

ρ

( a )( x ) under

ρ

: R → End M. By definition R

0

: = End

R

M ⊂ End M consists of group endomorphisms f commuting with the R-action

f ( ax ) = a f ( x ) , a ∈ R, x ∈ M.

Then M is naturally a R

0

-module

R0

M.

Again let R

00

: = End

R0

M ⊂ End M be those group endomorphisms commuting with the R

0

-action. Then it is clear that

ρ

( R ) ⊂ R

00

and it is an immediate question to ask if they are equal. If we keep on defining R

000

: = End

R00

M ⊂ End M, then we have the trivial identity

R

000

= R

0

.

Indeed R

0

⊂ ( R

0

)

00

= R

000

. Also a ∈ R

000

means a commutes with the R

00

⊃ R action. Thus a ∈ R”.

Example 2.17. (1) For M=RM=LnR a free module of finite rank. By presenting M as row vectors we had seen in last chapter that R0 ∼=Mn(R)op(right multiplica- tion by matrices) and R00∼=R as scalar multiplications on the left. Thus ρ(R) =R00. The essential formula used is

f(vj) = f(vieij) = f(vi)eij ∈Rvj.

(2) The procedure also shows that (1) holds if the rank is infinite. Let M = LN

R. Then R0consists of row-finite infinite matrices.

(6)

Now let S⊂R0 ⊂M(R)opbe the subring consisting of aI+Af, a∈R,

where Af is a “finite n×n matrix” in M(R)for some n ∈ N. (If we put a = 0 then we get an ideal as in Example 2.5-(2).)

By viewing M = SM (matrix multiplication on right), the same procedure shows that S0 ∼= R as left scalar multiplications. Hence S00 ∼= R0 which is strictly larger than S.

Nevertheless, we do have the “finite approximation property” for the imbed- ding S,→S00: for any x1, . . . , xn∈ M and s00∈S00, there exists s∈S such that

sxi=s00xi 1≤i≤n.

(3) For R=Z, M :=Lp:primeZ/(p), we have R0 =homZ(M

pZ/(p),MqZ/(q)) ∼=

Z/(p)

since all cross terms vanish. Then R00 = EndR0M ∼= ∏ Z/(p)since the action on M by R0is component-wise. The embedding R,→ R00:

Z,→

Z/(p)

is not surjective. But the “finite approximation” as in (2) holds by the Chinese Remainder Theorem: for any x1, . . . , xn ∈ M and a00 ∈ R00, there are only a finite number of primes T such that a00xi 6= 0 for all i ∈ [1, n]. Over T we may solve a∈Z with a≡a00p (mod p)for p∈T. Then clearly

axi=a00xi, 1≤i≤n.

For a full endomorphism ring S = R

0

= End

R

M, we have seen that S = S

00

. This, together with Example 2.17, suggests the following notion:

Definition 2.18. A subring S ⊂ End

R

M is dense if for every f ∈ End

R

M and x

1

, . . . , x

n

∈ M there is a s ∈ S such that s ( x

i

) = f ( x

i

) for all i.

Remark 2.19. If R = ∆ is a division ring, M is then a vector space over ∆ and the condition is equivalent to: for every linearly independent v

1

, . . . , v

n

∈ M and any y

1

, . . . , y

n

∈ M, there is a s ∈ S such that s ( v

i

) = y

i

for all i.

Also if n = dim

M < ∞, then it is clear that the only dense subring S ⊂ End

M ∼ = M

n

( )

op

is the full matrix ring.

Here is a useful density theorem in the spirit of Example 2.17-(3):

Theorem 2.20 (Bourbaki). Let M =

R

M be completely reducible, R

0

: = End

R

M,

R

00

: = End

R0

M. Then R ,→ R

00

is dense.

(7)

2. DENSITY THEOREMS 15

P

ROOF

.

Step 1. Any submodule N

RM is a R00-submodule:

Indeed, the complete reducibility of M implies that M = N ⊕ P for some P ⊂

R

M. Let e ∈ R

0

be the corresponding projection onto N. Then for a

00

∈ R

00

,

a

00

N = a

00

e ( N ) = e ( a

00

M ) ⊂ e ( M ) = N.

Step 2. DenoteRMn

=

Lni=1 RM ei and let R0n :

=

End

(

RMn

)

. Then a00

R00determines a map a00(n)

End

(

R0nMn

)

:

Indeed any ` ∈ R

0n

is determined by `( u ) =

ni=1

`( u

i

e

i

) where

`( u

i

e

i

) =

n j=1

a

0ij

( u

i

) e

j

, a

0ij

∈ R

0

.

Now a

00

∈ R

00

gives a map on

R

M

n

by a

00(n)

u : =

ni=1

a

00

u

i

e

i

. Then

`( a

00(n)

u ) =

n i,j=1

a

0ij

( a

00

u

i

) e

j

=

n i,j=1

a

00

a

ij0

( u

i

) e

j

= a

00(n)

`( u )

for all ` ∈ R

0n

. This proves the claim.

Step 3. Proof of the theorem:

For n = 1, let N : = Rx

1

R

M, then Step 1 implies that N ⊂

R00

M.

Hence a

00

∈ R

00

⇒ a

00

x

1

∈ N = Rx

1

. That is, a

00

x

1

= ax

1

for some a ∈ R.

For general n ∈

N, Mn

is also completely reducible. By Step 2, we may apply the result for n = 1 to a

00(n)

and M

n

with x =

ni=1

x

i

e

i

∈ M

n

. Namely there exists a ∈ R such that ax = a

00(n)

x. This means ax

i

= a

00

x

i

for

1 ≤ i ≤ n as expected. 

As the first application, we derive “Jacobson’s density theorem”:

Theorem 2.21 (Jacobson). A ring R is primitive ⇐⇒ R is isomorphic to a dense subring of End

M where M is a vector space over a division ring ∆.

P

ROOF

. Given a faithful irreducible R-module ρ : R ,→ End

R

M, Schur’s Lemma implies that ∆ : = R

0

= End

R

M is a division ring and M is a ∆ vec- tor space. Theorem 2.20 then implies that R ⊂ R

00

= End

M is dense.

Conversely, if R ⊂ End

M is dense then for any x 6= 0 and y in M there exists a ∈ R such that ax = y. This implies that M =

R

M is irreducible and

thus R is primitive. 

If it happens that dim

M < ∞ then in fact R ∼ = End

M. This is the

starting point of the Wedderburn–Artin theorem on (semi-)simple (left) ar-

tinian rings discussed in the next section.

(8)

Exercise 2.2. Investigate how Bourbaki’s formal proof of the density theo- rem recovers the Chinese Remainder Theorem as in Example 2.17-(3).

It is clear that Theorem 2.20 does not take care of the phenomenon ex- plained in Example 2.17 (1) and (2) if R is not a division ring. This suggests the existence of a more general density theorem beyond the completely re- ducible case. A hint towards this generalization is to investigate the case for finitely generated modules over a p.i.d..

Now we impose the artinian property as our “finiteness condition”.

Theorem 2.22 (Wedderburn–Artin). The following are equivalent:

(1) R is simple and left artinian.

(2) R is (left) primitive and left artinian.

(3) R ∼ = End

M where ∆ is a division ring and dim

M < ∞.

P

ROOF

. (1) =⇒ (2) since simple implies primitive.

(2) =⇒ (3): let ρ : R ,→ End M be faithful irreducible, then ∆ : = R

0

= End

R

M is a division ring and ρ ( R ) ⊂ R

00

= End

M is dense by Theorem 2.20. (So far this is exactly the content of Jacobson’s density theorem.) It remain to prove dim

M < ∞ to conclude the equality.

Suppose the contrary and let x

i

∈ M, i ∈

N be linearly independent

over ∆. Then I

j

: = ann

R

x

j

is a left ideal of R and I

(n)

: =

Tnj=1

I

j

is the left ideal annihilating all x

j

, 1 ≤ j ≤ n for every n ∈

N. Now ρ

( R ) ⊂ End

M is dense implies that for 0 6= y ∈ M there exists a ∈ R with

ax

1

= 0, . . . , ax

n

= 0, ax

n+1

= y 6= 0.

Hence I

(1)

⊃ I

(2)

⊃ . . . is an infinite strictly decreasing chain which violates the left artinian assumption.

(3) =⇒ (1): End

M is anti-isomorphic to a full matrix ring M

n

( ) which is simple—since its ideals are of the form M

n

( I ) where I is an ideal of ∆ which must be 0. Also dim

M

n

( ) = n

2

, which shows that any left ideal chain has length at most n

2

. In particular R is left artinian (and also

left noetherian). 

3. Semi-simple artinian rings and their modules

To extend the classification to the (not yet defined) “semi-simple” case,

the principal goal is to find the exact conditions on a ring to characterize it

as a finite direct sum of matrix rings.

(9)

3. SEMI-SIMPLE ARTINIAN RINGS AND THEIR MODULES 17

Definition 2.23. A left ideal I is nilpotent if I

k

= 0 for some k ∈ N. I is nil if every element a ∈ I is nilpotent: a

k

= 0 for some k = k

a

N.

It is clear that I is nilpotent ⇒ I is nil ⇒ I is quasi-regular. In general the reverse direction fails. However we have

Proposition 2.24. Let R be a left artinian ring. Then J ( R ) is nilpotent.

In particular J ( R ) = the largest nilpotent ideal = the largest nil ideal = the largest quasi-regular ideal.

P

ROOF

. Let J ( R ) = : J ⊃ J

2

⊃ . . . ⊃ J

k

= J

k+1

= . . . = : P. Then J is nilpotent if and only if P = 0. If P 6= 0 then we consider the set

S : = { I : a left ideal in P | PI 6= 0 } .

Then P ∈ S since PP = P

2

= J

2k

= P 6= 0. Let I be a minimal element in S.

Then PI 6= 0 implies that there is a b ∈ I with the left ideal Pb 6= 0. Since Pb ⊂ I and P ( Pb ) = P

2

b = Pb 6= 0, we must have I = Pb. Then

b = zb

for some z ∈ P and hence ( 1 − z ) b = 0. But z ∈ P ⊂ J implies that 1 − z is invertible. Hence we get b = 0 which is a contradiction.

The second statement follows from Proposition 2.15 that J ( R ) is the

largest quasi-regular ideal. 

Inspired by Proposition 2.2 on semi-primitivity, we introduce

Definition 2.25. A ring is semi-simple if it is a subdirect product of simple rings.

Lemma 2.26. Let M be a left artinian module which is a subdirect product of irreducible modules M

i

. Then M is a finite direct sum of some M

j

’s.

P

ROOF

. Let f : M ,→ ∏ M

i

be a subdirect product. For each component we have f

i

: M  M

i

. Let N

i

: = ker f

i

. Then

T

N

i

= ker f = 0. The artinian property implies that there is a minimal element among all finite intersections of N

i

’s. Denote this element by N = N

1

∩ . . . ∩ N

m

. Then N ⊂ N

i

for all i for otherwise N ∩ N

i

⊂ N will be smaller. This then implies that N ⊂

T

N

i

= 0 and we get an embeeding

M ,→

Mm

i=1

M

i

.

Since

Lmi=1

M

i

is completely reducible, we conclude that M is also com-

pletely reducible and equal to a sum

LjJ

M

j

for some J ⊂ { 1, . . . , m } . 

(10)

Proposition 2.27. Let R be a semi-simple ring whose ideals satisfy the DCC con- dition (i.e. artinian for two-sided ideals), then R =

Lni=1

R

i

where the ideals R

i

⊂ R are themselves simple rings.

P

ROOF

. In order to apply the above lemma, we need to introduce a mechanism to transform ideals into left modules.

Consider the ring M ( R ) = R × R

op

,→ End R

Z

which acts on R by ( a, b ) r : = arb.

Then submodules of

M(R)

R corresponds to ideals of R.

Now R is a semi-simple ring means that R ⊂

i

R

i

as a subdirect prod- uct of simple rings R

i

. The surjective map R  R

i

gives R

i

a left M ( R ) - module structure which is irreducible since N ⊂

M(R)

R

i

corresponds to an ideal of R

i

which must then be trivial. Lemma 2.26 then implies that R =

Lni=1

R

i

. Notice that R

i

⊂ R is an ideal (instead of a subring).  Remark 2.28. Later (in next section) we will define the enveloping algebra R

e

= R ⊗

F

R

op

for an F-algebra R to replace M ( R ) .

Now we are ready to prove the fundamental structure theorem for semi- simple left artinian rings:

Theorem 2.29 (Wedderburn–Artin, Jacobson). The following are equivalent:

(1) R is left artinian without non-trivial nilpotent ideals.

(2) R is left artinian and left semi-primitive.

(3) R is left artinian and semi-simple.

(4)

R

R is completely reducible (necessarily a finite direct sum).

(5) R ∼ =

Lni=1

R

i

with R

i

simple left artinian, i.e. R

i

∼ = End

i

M

i

.

P

ROOF

. (1) ⇔ (2): since they are both equivalent to left artinian and J ( R ) = 0 by Proposition 2.24.

(2) ⇔ (3): since the primitive/simple factors R

i

appeared in the sub- direct product are also left artinian (as indued from the surjective map R  R

i

), and then primitivity is equivalent to simplicity by Theorem 2.22.

(2) ⇔ (4): for, 0 = J ( R ) =

Tmax left

I implies R ,→

max left

R/I. Since R is left artinian, Lemma 2.26 then gives R =

Lni=1

R/I

i

.

For ⇐ , let

R

R =

L

M

i

with M

i

⊂ R being irreducible R-modules. This means that M

i

= I

i

is a minimal left ideal of R. Then

1

R

= e

1

+ . . . + e

k

, e

i

∈ I

i

.

(11)

3. SEMI-SIMPLE ARTINIAN RINGS AND THEIR MODULES 19

In particular a = ae

1

+ . . . + ae

k

and a 6= 0 implies ae

j

6= 0 for some j. So R is semi-primitive.

Moreover, R = I

1

+ . . . + I

k

and then R =

L

I

j

over a subset of { 1, . . . , k } . By reordering let this subset be { 1, . . . , m } . Then

R

R =

Mm

j=1

I

j

Mm1

j=1

I

j

⊃ . . . ⊃ I

1

⊃ 0

is a composition series for

R

R. Hence R is left artinian (and noetherian).

Since (1)–(4) are all equivalent, it suffices to show (3) ⇒ (5) ⇒ (2).

(3) ⇒ (5): left artinian implies artinian (DCC on ideals), Proposition 2.27 then gives R =

Lsi=1

R

i

with the simple ring R

i

⊂ R being an ideal. Now R

i

R

j

⊂ R

i

∩ R

j

= 0 for i 6= j, hence any left ideal of R

i

is also a left ideal of R and this shows that R

i

is left artinian. The statement R

i

∼ = End

δ

M

i

follows from the case for simple artinian rings in Theorem 2.22.

(5) ⇒ (2): simple implies primitive so R =

Lni=1

R

i

is semi-primitive.

Also R

i

is left artinian for 1 ≤ i ≤ n implies R is left artinian.  Remark 2.30. The above decomposition R =

Lni=1

R

i

into simple compo- nents (ideals) is unique. In fact we only need to require that R

i

are inde- composable ideals to get the uniqueness.

Next we study the structure of modules over semi-simple artinian rings.

Lemma 2.31. Let R be simple (left) artinian, then R has a unique irreducible module up to isomorphism. A representative is given by a minimal left ideal. In particular all minimal left ideals are isomorphic.

P

ROOF

. Minimal left ideals 0 6= I ⊂ R exist by the artinian property.

By definition I is also an irreducible R-module. For any other irreducible R-module M, R is simple implies that ann

R

M = 0. In particular Ix 6= 0 for some x ∈ M. Schur’s Lemma then implies I −→

x

M is an isomorphism. The argument holds for all choices of M, hence the uniqueness. 

Example 2.32. We recall that a maximal left ideal is a maximal element among all proper left ideals I ( R while a minimal left ideal, which is the dual notion, is a minimal element among all non-zero left ideals 0(I.

(1) For a division ring∆, the maximal left is 0, the minimal left is ∆.

(2) One may see Lemma 2.31 directly by noticing that for R = Mn(), ∆ a division ring, 1 = ni=1eii with eiibeing orthogonal idempotents. Hence RR = Ln

i=1Reiiand left ideals I ⊂R (submodule ofRR) are precisely partial sums: J ⊂ {1, . . . , n},

IJ=M

j∈JMn()ejj

(12)

as a column spaces over∆ with indices from J.

Thus IJ is left maximal⇐⇒ |J| = n−1 and I is left minimal ⇐⇒ |J| = 1.

Notice that R = IJ⊕IJc and R/IJ ∼= IJc. It is unusual that a quotient module can also be understood as a submodule. Nevertheless this does happen for simple (left) artinian rings.

We will see below that this holds for all semi-simple artinian rings by showing that all their modules are completely reducible.

(3) Consider the group algebra R = F[G]for a finite group G over a field F such that p=char F divides|G|. R is artinian. Let z= g∈Gg then hz = z=zh for all h∈G, hence Fz⊂R is an ideal. Also z2=z∑ g= |G|z=0 in R, hence Fz is a nilpotent ideal and R is not semi-simple. We will see later that F[G]is indeed semi-simple if p6 | |G|.

Exercise 2.3. Show that the Weyl algebra W =

C

[ x, ∂ ] of polynomial dif- ferential operators is simple but not left artinian. Show that there are more than one, in fact infinitely many, irreducible left W-modules.

Theorem 2.33. If g : R

1

R

2

is a ring isomorphism between simple left artinian rings R

i

= End

i

M

i

, then there is an isomorphism s : M

1

→ M

2

such that

g ( A ) = sAs

1

. for A ∈ R

1

.

Moreover s is semi-linear in the sense that there is an isomorphism on the coeffi- cients σ :

1

∼ =

2

such that s ( ax ) =

σ

( a ) s ( x ) for a ∈

1

, x ∈ M

1

.

P

ROOF

. M

1

is an irreducible left R

1

-module. M

2

can also be viewed as an irreducible left R

1

-module via the map g : R

1

→ R

2

. That is, for A ∈ R

1

, y ∈ M

2

, Ay : = g ( A ) y. Lemma 2.31 then implies that there is an R

1

-module isomorphism s : M

1

→ M

2

. This means

s ( Ax ) = g ( A ) s ( x ) , ∀ x ∈ M

1

,

hence g ( A ) = sAs

1

. Now s induces a group isomorphism ˜g : End M

1

→ End M

2

via B 7→ ˜g ( B ) : = sBs

1

which restricts to g on R

1

. Also ∆

i

can be identified as C

End Mi

( R

i

) with left scalar multiplications, hence ˜g restricts to

σ

: ∆

1

∼ =

2

. The above formula for A = a ∈

1

gives the result.  Corollary 2.34. Denote by G the group of semi-linear transformations on a finite dimensional vector space M over ∆, then there is a short exact sequence

1 →

×

→ G →

I

Aut ( End

M ) → 1,

where I ( s )( A ) : = sAs

1

.

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4. FINITE DIMENSIONAL CENTRAL SIMPLE ALGEBRAS 21

P

ROOF

. If s ∈ ker I then sA = As for all A ∈ End

M. But then s is an

invertible scalar multiplication as expected. 

Theorem 2.35. Let R be a left artinian ring.

(1) If R is semi-simple then any left/right R-module is completely reducible.

(2) There is an one to one correspondence between the isomorphism classes of irreducible left/right R-modules and the simple components of R/J ( R ) P

ROOF

. First we assume that R is semi-simple. By Theorem 2.29 R is a finite direct sum of matrix rings over division rings. Hence the left case implies the right case by taking R

op

.

For (1), Theorem 2.29 implies that R =

iΛ

I

i

where each I

i

is a min- imal left ideal (contained in some simple component of R). Let M be a R-module and 0 6= x ∈ M. Then x ∈ ∑ I

i

x with I

i

x = 0 or I

i

x ∼ = I

i

by Schur’s Lemma. Hence M =

xM

Rx =

iΛ, xM

I

i

x as a sum of irre- ducible modules. Hence M is completely reducible.

(2) follows from Lemma 2.31 since I

i

is a minimal left ideal in a simple component R

i

of R.

If R is only left artinian, then ¯ R : = R/J ( R ) is left semi-primitive and left artinian since J ( R ¯ ) =

Tmax left

¯I = J ( R ) = 0. Then Theorem 2.29 shows that ¯ R is also semi-simple. Now (2) follows from the semi-simple case since any irreducible R-module M satisfies J ( R ) M = 0 and hence M is also an

irreducible ¯ R-module. 

Of course (1) fails if R is not semi-simple by Theorem 2.29. Hence in general there are interesting non-completely reducible modules for left ar- tinian rings which is not semi-simple.

4. Finite dimensional central simple algebras

In order to classify simple artinian rings R = M

n

( ) , it is equivalent to classify division rings ∆. Notice that the center F = C ( R ) of a simple ring is necessarily a field: for c ∈ F \ { 0 } , Rc = cR ⊂ R is an ideal hence Rc = R and bc = cb = 1 for some b ∈ R. Thus R and ∆ are vector spaces over F.

In general a ring R is called an algebra over a commutative ring S (an S-algebra) if S ⊂ C ( A ) . R is central over S if S = C ( A ) . Thus every simple ring R is a central simple algebra (CSA) over its center field F.

Definition 2.36. Let R be a CSA/F. R is centrally finite if dim

F

R < ∞. In this

case R is called a finite dimensional CSA. Otherwise it is centrally infinite.

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Centrally finite implies artinian trivially, but not the converse: if R is artinian, R ∼ = M

n

( ) for a division F-algebra ∆. Hence R is centrally finite

⇐⇒ dim

F

∆ < ∞.

Thus it is a good idea to start with some classical results on division F-algebras A with n = dim

F

A < ∞. For a ∈ A, 1, a, a

2

, . . . , a

n

are linearly dependent over F, hence there is a polynomial relation f ( a ) = 0 for some f ( x ) ∈ F [ x ] . The monic minimal polynomial f

a

( x ) ∈ F [ x ] exists since F [ x ] is a p.i.d., also f

a

( x ) is irreducible over F since A is a division ring.

Example 2.37. (1) If F = F (algebraically closed, e.g. ¯¯ Q, C), then fa(x) is linear which implies a∈F. Hence A=F.

(2) Next we consider the case F=R:

Theorem 2.38(Frobenius). A finite dimensional divisionR-algebra A is isomorphic to R, C or H.

PROOF. If F =R, then R[√

−1]is algebraically closed implies fa(x) =x−a (a∈R) or fa(x) =x2+px+q∈R[x]with p2−4q<0 (a6∈R).

In the second case we set b=a+p/26∈R and get b2=r∈R<0, where r=p2/4−q.

Inspired by Hamilton’s quaternionH=R⊕ (RiRjRk), we define A0:= {b∈ A|b2R≤0}.

Claim 2.39. A0⊂ A is vector subspace overR. Hence A=R⊕A0.

It is equivalent to showing that if a, b∈ A0are linearly independent then a+ b∈A0. Let a2=u∈R, b2=v∈R, then∃p, q, r, s∈R such that

(a+b)2=u+ (ab+ba) +v= p(a+b) +q, (a−b)2=u− (ab+ba) +v=r(a−b) +s, which leads to 2(u+v) = (p+r)a+ (p−r)b+ (q+s).

But anyR-relation a=tb+t0gives a2=t2b2+2tt0b+t02. b6∈R implies t=0 (a = t0) or t0 = 0 (a = tb). In both cases we get contradictions. Hence we must have p=r=0 and then(a+b)2=q∈R. Again a+b6∈R implies q<0. Hence a+b∈ A0as claimed.

Now A=R⊕A0and we have a positive definite quadratic form on A0defined by Q(a) = −a2for a∈A0. The associated symmetric bilinear form

B(a, b):=Q(a, b) −Q(a) −Q(b) = −(ab+ba) is then an inner product on A0.

If A)R, pick i∈ A0with i2= −1. Then A⊃RRi=C.

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4. FINITE DIMENSIONAL CENTRAL SIMPLE ALGEBRAS 23

If A)C, pick j∈ (Ri)in A0with j2= −1. Then 0=B(j, i) = −(ji+ij), i.e. ij= −ji.

Let k := ij, then k2 = ijij = −i2j2 = −1 ⇒k ∈ A0. Also k ⊥ i, j since B(k, i) =

−(iji+iij) =0 and(B, j) =0. So A⊃RRiRjRk=H.

If A ) H, then there exists` ∈ A0,`2 = −1 and` ⊥ i, j, k. However,`ij =

−i`j=ij` ⇒ `k=k`, which contradicts to`k= −k`, so A=H.  (3) For finite division rings the situation is even simpler:

Theorem 2.40(Wedderburn). Any finite division ring A is a finite field.

PROOF. Let F = C(A)be a finite field with|F| = q and let n = dimFA. If n=1 we are done. If n>1 we will derive a contradiction.

Consider the conjugation group action of G := A× on A×. The class equa- tion reads as qn−1 = q−1+i[A× : C(xi)]where xiare representatives of the conjugacy classes not from F.

Notice that Ai := {g ∈ A | gxi = xig} is a division subring( A with A×i = C(xi). Let ni := dimFAi < n. Since A is also a vector space over Ai, n= [A : Ai][Ai: F]implies ni |n. Hence we have

qn−1=q−1+

i(qn−1)/(qni−1)

with each summand being inZ[q]. Moreover, xm−1=d|m`d(x)where`d(x) ∈ Z[x]is the irreducible cyclotomic polynomial for xd=1. Hence`n(q)divides the sum as well as qn−1, and then`n(q) | (q−1).

However, since`n(q) =w(q−w)where w’s are primitive n-th roots of 1, we have|q−w| > |q−1|for all w. This leads to a contradiction. 

Historically the first centrally infinite division ring was discovered by Hilbert it in his study of independence of geometry axioms!

Exercise 2.4. Let F be a field and σ ∈ Aut F, and ∆ = F (( x, σ )) be the ring of σ-twisted formal Laurent series

i=n

a

i

x

i

with n ∈

Z and ai

∈ F, with the non-commutative product being defined by xa =

σ

( a ) x.

(1) Show that ∆ is a division ring. (2) Moreover, let k = F

σ

be the fixed field of σ. Then C ( ) = k if σ has infinite order. In this case ∆ is centrally infinite. Give an explicit example of such a ( F, σ ) . (3) Show that C ( ) = k (( x

r

)) when σ has order r < ∞ and in this case ∆ is centrally finite.

From now on we focus on the basic structural theorems on finite di-

mensional CSA/F. It is based on tensor product of F-algebras.

(16)

Let A, B be F-algebras with vector space base x

i

and y

j

respectively.

Then the tensor product over F is simply

D : = A ⊗

F

B = (

M

Fx

i

) ⊗

F

(

M

Fy

j

) ∼ =

M

F x

i

⊗ y

j

,

with the product structure between A ⊂ D and B ⊂ D, under the standard embedding a 7→ a ⊗ 1 and b 7→ 1 ⊗ b, defined by

ab = ( a ⊗ 1 ) · ( 1 ⊗ b ) = a ⊗ b = ( 1 ⊗ b ) · ( a ⊗ 1 ) = ba.

So D = AB = BA. The following simple characterization is useful:

Proposition 2.41. Let A, B ⊂ D be sub F-algebras. Then D ∼ = A ⊗

F

B if (1) ab = ba for all a ∈ A, b ∈ B.

(2) There is a F-base x

i

of A such that any z ∈ D can be written uniquely as z = ∑ b

i

x

i

with b

i

∈ B.

If dim

F

D < ∞, then (2) is satisfied by D = AB and [ D : F ] = [ A : F ][ B : F ] .

P

ROOF

. Define the F-bilinear map A × B → D by ( a, b ) 7→ ab. This then induces A ⊗

F

B → D, which is a ring homomorphism by (1). The map

is indeed a bijection by (2). 

Corollary 2.42. (1) For any F-algebra B, M

n

( F ) ⊗

F

B ∼ = M

n

( B ) . (2) M

m

( F ) ⊗

F

M

n

( F ) ∼ = M

mn

( F ) .

P

ROOF

. For (1), take D = M

n

( B ) with A = M

n

( F ) ⊂ D and B ,→ D under the embedding b 7→ diag ( b, . . . , b ) ∈ D. Then both conditions in Proposition 2.41 are satisfied. Hence A ⊗

F

B = D.

For (2), simply apply (1) to B = M

n

( F ) and then use the isomorphism of block decompositions of matrices M

m

( M

n

( F )) ∼ = M

mn

( F ) .  Definition 2.43. The algebra A

e

: = A ⊗

F

A

op

is called the enveloping algebra of A. If A ,→ B, then B =

Ae

B via ( ∑ a

i

⊗ b

i

) y = ∑ a

i

yb

i

.

For B = A, this shows that submodules of

Ae

A are precisely ideals of A. Hence A is simple ⇐⇒

Ae

A is irreducible. Also End

Ae

A = C ( A ) .

For the last statement, let f ∈ End

Ae

A, then

f ( acb ) = f (( a ⊗ b ) c ) = ( a ⊗ b ) f ( c ) = a f ( c ) b.

Hence f ( a ) = a f ( 1 ) = f ( 1 ) a and we may identify f with f ( 1 ) ∈ C ( A ) .

The following two theorems are fundamental:

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4. FINITE DIMENSIONAL CENTRAL SIMPLE ALGEBRAS 25

Theorem 2.44. Let A be a finite dimensional CSA/F with n = dim

F

A. Then A

e

: = A

F

A

op

∼ = M

n

( F ) .

P

ROOF

. Let R = A

e

, then

R

A is irreducible and R

0

: = End

R

A = C ( A ) = F. Since n = dim

F

A < ∞, the density theorem (Theorem 2.20) for F- algebras implies that for all a

00

∈ R

00

: = End

R0

A = End

F

A, there exists a ∈ R = A

e

with a 7→ a

00

. That is, A

e

 End

F

A ∼ = M

n

( F ) . Since both A

e

and M

n

( F ) have the same dimension n

2

, the map is an isomorphism.  Theorem 2.45. Let A be a f.d. CSA/F, A ⊂ B as a F-subalgebra. Then

(1) B ∼ = A ⊗

F

C with C = C

B

( A ) .

(2) There is a bijection between ideals of C and ideals of B via I ⊂ C 7→ AI ⊂ B.

(3) C ( C ) = C ( B ) .

P

ROOF

. (1): Theorem 2.44 ⇒ A

e

is simple ⇒

Ae

A is irreducible ⇒ B =

M

Ac

i

as A

e

-modules (by Theorem 2.35). Since ( a ⊗ 1 ) 1

A

= a = ( 1 ⊗ a ) 1

A

and Ac

i

∼ = A as A

e

-modules, by choosing c

i

such that c

i

7→ 1

A

we then have ac

i

= ( a ⊗ 1 ) c

i

= ( 1 ⊗ a ) c

i

= c

i

a. Moreover ac

i

= 0 ⇒ a = 0.

If c ∈ C : = C

B

( A ) , by writing c = ∑ a

i

c

i

with a

i

∈ A, the condition ca = ac for all a ∈ A is equivalent to aa

i

= a

i

a for all i, that is a

i

∈ C ( A ) = F.

This shows that C =

L

Fc

i

. Proposition 2.41 ⇒ B ∼ = A ⊗

F

C = AC.

(2) I ⊂ C ⇒ AI ⊂ AC = B. We claim that AI ∩ C = I: if A has a F-base 1 = x

i

, . . . , x

n

, then B =

L

x

i

C as a vector space over F. If d = ∑ x

i

d

i

∈ AI ∩ C, d

i

∈ I, then by (1) the expression is unique even allowing d

i

∈ C.

Since d = x

1

d is such an expression, we thus have d = d

1

∈ I.

To show that every I

0

ideal in B is of the form AI, we simple take I : = I

0

∩ C. Since I

0

= ∑ Ad

i

with d

i

∈ C ∩ I

0

= I, we get I

0

= AI.

(3) C ( B ) ⊂ C

B

( A ) = C ⇒ C ( B ) ⊂ C ( C ) . Also c ∈ C ( C ) commutes with all elements in B = AC. Hence C ( C ) ⊂ C ( B ) as well.  The condition on A is necessary: for B = A ⊗

F

C, C

B

( A ) contains C and C ( A ) . Hence C ( A ) must be the ground field to conclude C

B

( A ) = C.

Corollary 2.46. Let A be a f.d. CSA/F and C be an arbitrary F-algebra. Then B = A ⊗

F

C is central/simple if C is central/simple.

Hence if A

1

, . . . , A

r

are f.d. CSA/F then A

1

F

. . . ⊗

F

A

r

is also a f.d. CSA/F.

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The second statement in Corollary 2.46 allows us to introduce

Definition 2.47 (Brauer group Br ( F ) ). Let A, B be f.d. CSA/F. We say that A ∼ B if there are m, n ∈ N such that M

m

( A ) ∼ = M

n

( B ) . Since A ∼ = M

p

( ) , B ∼ = M

q

(

0

) , this is equivalent to M

mp

( ) ∼ = M

nq

(

0

) , hence equivalent to

∆ ∼ =

0

since divisions are local rings (as shown in last chapter).

It is clear that ∼ is an equivalence relation, namely A ∼ B ⇒ B ∼ A and A ∼ B ∼ C ⇒ A ∼ C. Hence we may define the abelian group

Br ( F ) : = ({ isomorphism classes of f.d. CSF/F } , ⊗

F

) / ∼ ,

with 1 = [ F ] = [ M

n

( F )] for all n ∈ N and [ A ]

1

= [ A

op

] (by Theorem 2.44).

Hence Example 2.37 implies Br ( F ¯ ) = { 1 } = Br (

Fq

) and Br (

R

) = h[

H

]i ∼ = {± 1 } . Notice that

C is not R-central simple since C

(

C

) =

C.

Remark 2.48. Later we will show that for any p-adic number field F = Q

p

, Br ( Q

p

) −→(

Q/Z, +) . This is part of the “local class field theory”.

For “global class field theory”, there is a Hasse exact sequence 0 → Br ( Q ) →

M

ν

Br ( Q

ν

) → Q/Z → 0,

where ν runs through all finite places ν = p (primes) as well as the infinite place ν = ∞ (Q

: = R). Here [ A ] 7→ [ A

ν

] : = [ A ⊗

Q

Q

ν

] is non-trivial only for a finite set of places. Also Br ( R ) is identified as

12

Z/Z.

Similar statements hold for a finite extension field K/Q. After we de- fined the Galois cohomology, we will interprete Hilbert’s theorem 90 in terms of H

1

and show that the Brauer group is exactly H

2

.

The first statement in Corollary 2.46 leads to

Definition 2.49. Let A be a f.d. CSA/F and E/F an extension field. Then A

E

: = A ⊗

F

E is f.d. CSA/E. E is called a splitting field of A If A

E

∼ = M

n

( E ) . Example 2.50. Consider the standard embedding H ,→ M

2

( C ) of rings

a + bi + cj + dk = z + wj 7→ a + bi c + di

− c + d

i

a − bi

!

= z w

− w ¯ ¯z

! .

It follows that H

R

C ∼ = M

2

( C ) .Notice that C is a maximal subfield of H.

Proposition 2.51. Let A, B be CSA/F and E an extension field of F. Then (1) If E splits A and B then it splits A

op

and A ⊗

F

B.

(2) If A = M

n

( B ) then E splits A if and only if E splits B.

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4. FINITE DIMENSIONAL CENTRAL SIMPLE ALGEBRAS 27

P

ROOF

. (1) follows from the definitions easily. For (2) we only need to prove the direction ⇒ : we have

A

E

∼ = M

n

( B

E

) ∼ = M

n

( M

s

( )) ∼ = M

ns

( ) where ∆ is a central division algebra over E.

If E splits A then A

E

∼ = M

m

( E ) . Hence M

ns

( ) ∼ = M

m

( E ) which implies that m = ns and ∆ = E since ∆ and E are local rings.  In particular, E splits A = M

n

( ) if and only if E splits the division F- algebra ∆. From this we see that splitting fields always exist. Indeed E = F ¯ is a splitting field since the only division algebra over ¯ F is ¯ F itself. Hence

A

F¯

= M

n

(

F¯

) ∼ = M

n

( M

s

( F ¯ )) ∼ = M

ns

( F ¯ ) .

However ¯ F is generally an infinite extension. Much better results hold:

Theorem 2.52. Let ∆ be a finite dimensional division F-algebra. A finite dimen- sional extension field E of F splits ∆ if and only if E is a subfield of A : = M

r

( ) for some r such that C

A

( E ) = E.

In particular, taking r = 1 shows that any maximal subfield E of ∆ is a split- ting field for ∆. (There could be infinitely many of them, e.g. C ,→ H.)

We only give the proof that a maximal subfield E of ∆ splits ∆. The general case is similar and is left to the readers.

First we noticed that ∆ = End

op

∆ consists of F-linear maps commute with ∆

op

. Consider R : =

opE

=

op

F

E which acts on ∆ as a faithful irreducible R-module. R

0

: = End

R

∆ ⊂ End

F

∆ consists of F-linear maps which commute with R. Thus R

0

= C

( E ) = E since E is a maximal sub- field. This implies R

00

: = End

R0

∆ = End

E

∆ ∼ = M

n

( E ) where n = dim

E

∆.

The density theorem for F-algebras then implies that

op

F

E = R ∼ = R

00

= M

n

( E ) .

Remark 2.53. In particular, [ ∆ : F ][ E : F ] = n

2

[ E : F ] ⇒ [ ∆ : F ] = n

2

. Since n = [ ∆ : E ] , then [ E : F ] = [ ∆ : F ] / [ ∆ : E ] = n

2

/n = n as well. Thus the phenomenon occurs in the special case ∆ = H is really a general one.

There are other interesting basic results on CSA. All of them are derived from consideration of A

e

-modules and the density theorems as above. We mention only two of them and leave the proofs as exercises:

Exercise 2.5. (1) (Skolem–Noether Theorem) show that any automorphism

of a f.d. CSA is inner. (2) Let A ⊂ B be a semi-simple sub-algebra of a

f.d. CSA B, then C

B

( C

B

( A )) = A.

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