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## C OMPOSITIO M ATHEMATICA

### Compositio Mathematica, tome 5 (1938), p. 392-402

<http://www.numdam.org/item?id=CM_1938__5__392_0>

© Foundation Compositio Mathematica, 1938, tous droits réservés.

L’accès aux archives de la revue « Compositio Mathematica » (http:

//http://www.compositio.nl/) implique l’accord avec les conditions gé- nérales d’utilisation (http://www.numdam.org/conditions). Toute utili- sation commerciale ou impression systématique est constitutive d’une infraction pénale. Toute copie ou impression de ce fichier doit conte- nir la présente mention de copyright.

Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques

http://www.numdam.org/

(2)

### semi simple ring

by Jakob Levitzki

Jerusalem

Introduction.

In the

### present

note it is shown that each class of

### équivalent nilpotent

elements of a semi

### simple ring 1)

can be characterised

### by

certain characteristic

two

elements

### belong

to the same class if and

### only

if their characteristic numbers coincide. This is achieved

each

### nilpotent

element

to a certain normal form. As a conséquence we find that the number of classes is finite.

### Applying

our results to square matrices in a commutative or non commutative field we find that each

### nilpotent

matrix can be transformed into a Jordan form

### 2).

Essential use is made of the notion of the

which

an

### interpretation

of the rank of a matrix based on

the

### theory

of the ideals in a

### simple ring.

The convenience of this

is also shown

a few

in the

### appendix

to this paper.

I, ivotations and

remarks.

1. If

are r. h.

ideals of

a semi

then ive denote

### by (U,, U,, - - ., e,,,)

the

r.h. ideal SB which consists of all the éléments of the form

ri

If from

ri

ri

follows

### always

ri = 0,

i . = 1, ..., m then 38 is called the direct sum of the

### %;

and we

1) Two elements a and b of S are called équivalent if for a regular élément r

of S the relation r-lar = b holds.

a, 0 ...

2) The Jordan Form of a nilpotent matrix is

were each aA is

0 E 0... 0

### 00 E...O

a matrix of the form

### ...,

, e being the unit of the field.

(3)

write

or

If in

the

are

the

### ID3i

are different from zero and do not contain any subideals

and the zero

### ideal)

then the number m is called the

of

### length

of 38. Similar notations are used for I.h.

left

### hand)

ideals. If S8 is the zero

theii

0.

2. LEMMA

### 1. If aES, bES then | abS| laS/; Sab | Sb

Proof. This is an immédiate consequence of abS

aS and

Sab

Sb.

LEMMA 2.

b E S

.

Proof.

where

are

r.h.

i.e.

we obtain

abS =

...,

Each

is

either

a

### primitive

r.h. ideal or zero,

follow

### LEMMA 3. if

u E S and a2 = a then

### 1 Sal = 1 aS 1.

Proof. For a= 0 the lemma is trivial. For

let aS

where the

are

r.h.

i.e.

Let

a =

### SBmi =1 ai, ajeei,

then from a2 = a ive obtain

known argu-

the relations

if

### i =k

i = k and hence Sa =

which

=

..., m.

### Similarly

follows the relation

and hence

THEOREM 1.

a E S then

Proof.

S = aS

### -j- S8,

where % is a r.h. ideal. ’the unit e of S has a

### representation

of the form e = ab

where

eSS and

lemma

3 we have

and hence

lemma

.

and hence

NOTATION: If

then the common

### length

of aS and Sa is called the rank of a and is denoted

### by la 1

.

3. From 2 follows

and if

them also

=

In this case

we have the

### relation 1 l’,ai 1 = Z’, 1 ai 1

and wc say that the

ai form a direct sum.

### Conversely from 1 Y-m,a, 1 = Y-nll a,,’

it

follows that the r.h. ideals

the l.h. ideals

### Sai)

form a

direct sum. Thus the sum of

elements ei

...,

which form an

ei if

### is i = k)

direct.

(4)

4. From 2 and 3 it follows

that if

and ab =O

then

resp.

for each i =

### l,

..., in.

5. LEMMA 1. Let S be a semi

and

f

### primitive

r.h. resp. l.h. ideals.

1 ·

then the

Uf

S is a

Proof.

where

=

Then

either

or

### r2l2

is zero, since otherwise it would follow from

1 that

0. The

aSa

no divisors of

zero. From

follows also

similar

for

### rEe, lEI, rl =F

0 the relation rl

i.e. each

x =

l’el in

### Vf

has a solution which is

### unique,

since 38t has no divisors of zero.

each

### equation x -

rl = r’ l’ can be

### uniquely solved,

i.e. 7llil is a field.

LEMMA 2.

A is a

or

set

elements

a semi

S then the

A SA is either

or it contains

a

### principal idempotent

element e and has a

A SA =

eSe

### +

N where eSe is a semi

### of

ASA.

Proof. Let 38 and 1 be a r.h. and a l.h.

subideal

of A S and SA

If the

SBt is

for each

38 and

### f,

then also A SA is

### nilpotent.

If now for certain 38 and 1

the

ID3f is not

### nilpotent,

then as in lemma 1 it follows that

7llil is a field since in this case

### le :A 0;

hence A SA contains

### idempotent

elements. Let now e be an

element of

### highest

rank which lies in

### ASA,

and suppose AS = eS

SA = Se

where 38

### and 1

are chosen such that ee =

= 0.

Then

in

### fact, suppose Ï ID3 =F 0

then there exists a

### pri-

mitive I.h. ideal f and a

r.h. ideal

such that

Hence

### (Lemma 1)

U f is a field and has a unit e’. Since e 1 = lUe = 0 we have ee’ = e’e = 0 and therefore

that e

is an

of

rank as e

is an

### idempotent

which lies in ASA and has the

From

follows that the

is a

ideal

in A SA which

with

the

REMARK. If ID3

### and 1

are different from zero and if e1 and e*

are units of 7lli and

then

0. We set

### e2=e* -ele*,

then e2 is also a unit

and ASA = eSe

where

e, el, e2 are

(5)

6. LEMMA 3.

and

0 then two

elements «,

can be

### found

such

that a == «a«, b =

### Pbfl.

Proof. Since aS and bS form a direct sum we may write S = aS

bS

### +

ID3 where ID5 is a r.h. ideal in S. If E is the unit of S we have E = el

e2

### +

e3 where ela = a;

=

and

if

### =kk. Similarly considering

the l.h. ideals Sa

ci i=k

0 i#k and

= a,

and

ei

### if i=-k*

2=k

From ab = ba = 0 follows further

=

= 0. If now e is

a

in the

aSa

case aSa is

we set e =

### 0),

then from aSab = basa = 0 and eS

aS Se

### C

Sa

follows eb = be = 0 and ele = e.

= e1-

we have

=

e 15

= a where

are

we find

such that

=

### é’e= 0, élei = él, aéi

= a.

If we set aSa = e.Se

eSê’

e’Se

e’Si5’ then

-ê’e’ =

### 0,

hence the element

is

### idempotent

and

«a == aoc::= «a«. Since now xb = b0153 == 0 if we substitute for x any of the

### elements e, e’, e’,

it follows that «b = =

### 0,

i.e.

the element b lies in the

which

the

as we may set

--- E - oc.

LEMMA 4.

and

### aja k=o for i -# k,

then there exists an

elements ai

...,

such that ai =

Proof. We prove

### by

induction. For ni = 2 the lemma is true

### according

to lemma 3. For m &#x3E; 2 we set b

then al and b

### satisfy

the condition of lemma

### 3, hence,

there exist two idem-

a,

such

oclalocl, b =

= 0.

For the

=

...,

of the semi

### simple ring f3Sf3

we may assume the

of the lemma

### (since

the number

of the elements in this

is less than

### m),

hence there exists

a

of

### orthogonal

elements «2, ..., OCm

### satisfying together

with oc, the relations in

### question.

II. Reduction to the normal

### form.

LEMMA 5. Let S be a

and

...,

an

matric-basis

S.

c =

- 1 ;

### CiCk - ° if Z .

Then the element c

### is equivalent to Xl £Î i c.

Cick

0 if k = i+l . en the element c is equival ent to LJÂ=l

### &#x3E;

Proof. The r.h. ideal

### Ci+lS

and the l.h. ideal

### Sei satisfy

the

conditions of lemma

there exists an

### idempotent ë’i+1,i+l

(6)

whieh is the unit of the field

and

ci;

Ci+i ,

rn - 1. From

follows further

=

i # k.

If

resp.

is an

### arbitrary

l.h. unit resp. r.h. unit of

then we set -C

+i = of

resp.

then we set

1 =

1--

+1 -

### i= M+I,M+l

and thus obtain an

### em+1, m-I-1 Em 2èi,iC*

,m , and thus obtain an

i =

...,

further

Ci,

=

then the

° ° ° ’ m - i

### -+-1 )

are different from zero as a consequence

and

### CiCi+1 *°

as in this case we have

=:

and therefore in

=

we

now the

k =

known

### argument)

to a matric basis

k =

...,

of S.

### According

to a theorem. of Artin

there

exists a

### regular

element r which satisfies the relations

r-1 c2 k T -

k =

### 1,

..., n. Since a = M - 2+1 we obtain

I. =

k =

### 1 ,

..., n. Since a = i==l

we obtain

1’-1

l’ =a,

### q.e.d.

THEOREM 2. Let S be a

and

k

an

niatrie basis

S.

a =

=

ai =

and

0

Then a

to

+i

c

ent to k.Jit=l

+1

it +1 ...

.

1

### +

- ... - na-1-i-- q

Proof. The

follows

lem ma

4 and lemma 5.

THEOREM

S is a semi

and a is a

element

index m

### + 1 4)

then, there exists an

...,

each

Tank 1 such that

Proof. From

follows

if

since

### SaÂ+1 === Sa} implies SaÂ+Q === SaÂ

for each u, hence

### aG # 0

for each a which contradicts am+1 ==0. If

denotes the set

of all éléments

such that

and

then

is

### evidently

a 1,h, ideal which is différent from zéro in case

3 ) E. ARTiN, Zur Théorie der hyperkomplexen Zahlen [Hamb. Abh. 5 (1927), 251-260].

4 ) The index is m+1 if am+1 = 0 and ak #o for 1; in.

(7)

### Im-l

form a direct sum.

We set

and

### -,=-1+1,

then

duction we obtain the

### ( 2 ) SaÂ == 1;.. + 1;’, lÁ+1 ç: l;v’ 1; +1 1;’ ( À # i , ... , %l )

where each relation

We now set

-

...,

then

=

and we have

in

If we set

S = Sa

and E = e1

### + £\$Î- e§/ + e

where E is the unit of

and

then

it is well

we have

The element

is a unit of

and we have

and

1:, the

relations

If

### further eA

denotes any l.h. unit of

then from

follows also

= 0 since

and

### Suppose

now

Since it follows from

3 that

(8)

we obtain

since

there-

fore

=

=

From

we obtain

view of

=

and hence

From

follows

for Â = 1 and

### (3)

we further obtain

Not all the r.h. ideals which appear in the

side of

are

### necessarily

different from zero, since some of the

### e*

may

vanish. We denote

cl’..., Ct

### those e,*

which are different

from zero, and may assume that c1

since

### em* = e’ :A 0.

Then m1 = m, and we have

where we may assume m = m1 &#x3E; m2

### &#x3E; . ..

&#x3E; me. The ci

the relations

We now

each

### cÀ5

as a direct sum of

r.h.

ideals:

where the

### ciO’)

are chosen such that

cA Cf-O’

From

= 0 for

### À =F

À’ follows further

= 0 if

### À# À’,

and we obtain from

From

and the second

of

we now obtain

for

of

### simplification

we obtain

5) The last equality is a consequence of

### l;.aS ç I;.S and Il;.ar = Il;.1

(see

the definition of ’À)’

-

(9)

where mi =

### npl +P2+" ,Pi-l +Â’ Â = l,...,

Pú Po =: 0. Since

to

the

form an

### orthogonal system,

and since from the

of the

follows

the relations

and

the

of the theorem.

THEOREM

S is a semi

and a is a

element

index m

then a can be

in the

a =

ai where

a

and aiai’ = 0

i

i’.

each ai has the

ai =

where

Proof. From

### (15)

it follows that a can be

in the form

where

and

We

both sides of

hand

then: :

...

Since

and since the

### ideals akd6S

form a direct sum it follows in

that

and

= 0 for

a,

in case

we obtain

while

A ..

tting. o

(n.-k) a n. ).-

for each

we have in

to

### 1 ri,kl I

==1 the relations i’

DEFINITION. If S is a

then the

### representation ( 18 )

is called the normal

of the

element a

of

### S,

and the numbers nl, n2’ ...,

### nq

the characteristic of a.

Above definition is

the

THEOREM 5. Two

elements

a

are

valent

and

### only if

their characteristic numbers coincide.

Proof. The

follows

### easily

from theorems 2 and 4.

COROLLARY. The number

the

classes

elements

a semi

is

REMARK. It is

### easily

seen how lemma 5 in

theorem

### 2,

the above definition and theorem 5 should be modified in case S is semi

(10)

III.

to rnatrices.

a1...a1

### Let .":..’: al ... an

be a matrix in an

field F. If for

1

s, : ... .

m, and

where

at least for one

the relations

then the r rows

...,

### r )

are called left hand

short:

### 1,h. ) dependent.

Otherwise these

rows are called I.h.

the

hand

short:

### r.h.) dependence

of the rows, as well as the I.h. and the

r.h.

and

### independence

of the columns is defined.

DEFINITION. The maximal number of I.h.

### independent

rows

of a matrix is called the 1.h, rank of the rows.

### Similarly

the r.h.

rank of the rows as well as the I.h. rank and the r.h. rank of the columns is defined.

In

### general,

the r.h. rank of the rows

### (columns)

is different

from the 1,h, rank of the rows

the

theorem can be

r.h. i-ank

the rows

each matrix

second order is

Io the l.h. rank

the rows

then the

### field

F is commutative.

As to the I.h. rank of thc rows

### (columns)

and the r.h. rank of

the columns

### (rows )

it will be shown in

that

are

to each other.

### «i...«1

We now consider the set S of all square matrices a

«1 ... etn

of order n

### (a

shorter notation : a =

sum and

as follows

WC

as it is well

a

of

n.

THEOREM 6.

then the

Sa is

to the l.h.

rank

the rows

a and the

aS is

to the r.h. rank

the columns

a.

Proof. We denote

...,

the

the

### Âth roW

of which coïncides with the

### a.th

row of a while all the other elements are zeros. We assume

### (for

the convenience of

that the r l.h.

### independent

rows are the r first ones. From the l.h.

of any r

### +

1 rows follows for each r

### + À,

Â. = 1, ...,

n - r the existence of r elements

=

...,

such that

(11)

=

k =

### 1,

..., n. If further

=

...,

### n )

denotes the matrix the

row of which is

to

,

0

=

,

and

### hence,

the l.h. ideal Sa is a subset of

...,

On the other

### hand,

if ek denotes the matrix ail the elements of which are zeros,

### except

the élément in the

row and

column which is

### equal

to the unit of the field

### F,

then eka = ak and hence

therefore

Sa.

If further

### =1 siai == 0; Si E S,

it follows from the l.h.

### independence

of the r first rows that all the elements of the

column of

### s).

are zéros, and hence siai = 0

...

in other

the

...,

### r)

form a direct sum: Sa . =

Since

the I.h. ideals

...,

are

the first

### part

of the theorem is

The

second

follows

### By 1,

2 and theorem 6 w e now obtain:

THEOREM 7. The l.h. rank

the rows

matrix is

r.h. rank

### of

the corumns.

As to the r.h. rank of the rows and the l.h. rank of the columns of a matrix a, their

follows

the

of a and

the theorem

### just proved.

It is clear how the results of II should be

to the

matrices of the

S : Each

matrix can be

to a Jordan

the structure

which is

theorems

4 and 2.

I v :

It

### might

be of some interest to show the usefulness of the

### interpretation

of the rank stated in theorem 6 also

the

### following examples.

As in section III we denote

5 the

### simple ring

of all matrices of

order in an

field F.

1. If a e s

=

### 1 SI,

then a is called a

### regular

matrix.

In this case we have Sa = aS = S. This

### implies

for each b E S

the relations Sab =

baS =

and

2.

;

In

### fact,

a is an immediate consequence of

To

we write

### a == a + b + (- e) b,

where e is the unit

(12)

of

then

oc and 1 we

Sab

Sb.

3. If

c E S

In

since

### bS bcS,

we may write bS = bcS

where

we obtain abS =

and

,

Remark.

the above

### inequality

we obtain the "law

,

4. If

and s &#x3E;

then a

for each natural

number t. Since

### aSDa 2S D ... ;2 aSS

and since ais D ai+lS

...,

would involve

to the

### assumption,

it follows that for a certain i

### (1 i C s )

the relation aiS == ai+1S must

which

successive

=

for each

and hence

### at =1=-

0 for

each t.

5. The theorem in 4 may be considered as a

case of the

### and s &#x3E; 1 al l

then there exists an element c of ,S and a naturel number t such that c = cl C2 ... et t where

### the ci

are elements

of the set al, a2, ... , as and Ck

0 for each k

### 20th,

1937.)

6) For matrices in a commutative field this was proved by FROSErrm’s, LUber den Rang einer Matrix [Sitzungsber. d. Akad. Berlin 1911, 20-29].

7) This follows easily from § 2 of the paper: J. LEVITZKI, Über nilpotente Unterringe [Math. Ann. 105 (1931), 620-627].

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