Automorphisms and derivations of associative rings

Automorphisms and Derivations of Associative Rings Mathematics and Its Applications (Soviet Series) Managing Editor:...

Author: Kharchenko V.K.

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3.9.11. 'l( x)

=~

Definition. Let S

be

a subring

of R

.

The

form

a i x gi b i

is called an S -form, provided there is a nonzero ideal I of the ring R, such that 'l( a) E S for all a E I. For instance, by the construction of 3.4.1, any trace form is an RG-form. 3.9.12. Theorem. Let S be an properties BM and Sf, H = A(S). Then G and the principal H-trace form will be This theorem directly results from 3.9.13.

theorem.

there

Proposition. Under are

elements

a

E

(almost) intermediate ring obeying H is an M -subgroup of the group an S-form. the following proposition.

the

conditions

of

the

preceding

R, a '# 0, r i' Vi E R, S i E S,l ::; i::; m,

CHAPTER 3

185

such that for any x, y 'r H( yax)

where

E RF

the following equality holds:

=~(yr)s i'r(vi x), i

is the principal H-trace form, Let us first prove two lemmas.

'r H

is the principal G-trace form.

'r

3.9.14. Lemma. If the centralizer Z of an almost intermediate ring S, obeying condition BM, is generated by invertible elements, then the group A(S) is an M-group. Proof. As z is generated by invertible elements, then the algebra of the group A(S) is equal to z and, by lemma 3.9.4, is semisimple. It means that A(S) is an M-group. 3.9.15. Lemma. Under the conditions of theorem 3.9.12, for two principal forms of the groups G and H = A(S) the following relation is valid: (x) = 'r H ( dx) +

~I

.

L.}..

w. xv.) ~

h i

~

~

-\

+ ",,(w . xv .) .]]

gJ

(16)

]

~

for which the following conditions are met: (a)

tP~S) = 0 for all

j;

j

(b)

hj

(c)

Zd 11 ~ Zw i = 0;

E

H for all i; i

the

(d) ann(l)d =

z

left

annihilator

of

the

element d

in

is

Z

zero:

O.

Proof. Let us consider the algebra B as a right module over Z. Since Z is semisimple, we can find a direct decomposition B = Z e v. Let us choose a basis zl ,_., Z k of the algebra Z over C and extend it with elements v k +

1""

vn

E

V

up to the basis of the algebra B. Let

a dual basis, Le., (vi' w}= 0ij. where vi= zi at

WI ,..., W n

1 ~ i~ k

and the

brackets denote an associative bilinear nondegenerate form on B. k

Let A= ~ wiC, i= 1

D=

~ wiC, Then A= v.l and i> k

D=

be

z.l. As

186

AUTOMORPHISMS AND DERIVAnONS

the fonn is associative, we have (V, ZA)= (VZ, A) = 0, (Z, ZD) = (Z, D) = 0, i.e., A and D are left z-submodules of 1fJ, and, since they are not intersecting, then 1fJ = A $ D. Limiting the bilinear form to a pair of z-modules (Z, A), we get

z A== (Zz)*, i.e., z A==

z Z since a semisimple algebra is Frobenius. Thus,

we can find an element d E A, such that A = Zd and the left annihilator of d in Z is zero (d is an image of the unit at the isomorphism A == Z). In particular, WI = z'i d ,_., Wk = z*k d for some elements z'i ~-, z*k E z, these elements forming the basis of Z. On Z let us define a bilinear form [x, y] = (x, yd). This form is associative and nondegenerate since { z i ,_., zicl proves to be a dual basis to (z 1'-' z kl . Now we can construct a principal fonn which commutes with Z:

v H( x) =

z'i xZ1 + ... + zic xz /(

For the principal fonn which commutes with 1fJ we have: v( x)

= v H( dx) +

I

i> k

wi xVi'

in which case of importance for us is the fact that

Zd n

L

Zw i =

i> k

° and

that the left annihilator d in Z is zero. Let us choose a system of representatives of the right co sets of the group

over the

Gin H

subgroup

Gin'

which consists of the elements

I = hI' ~,_., ht, lying in H, and a system of representatives 1= gl' _., gs of the right cosets of the subgroup Gin H of the group G. Both these systems are finite, as G is a reduced-finite group. Now we have a presentation for the principal G-invariant fonn: ~

'l(x)= L"v(x)

hJ~

=

j, 1

+

I

I

j?li>k

i h.

(w.l xv.) ] l

I

t

(vi dx

=1

+

It should be remarked that

L

»

h

j

+

L L

j?li?11?2

cp~S) = 0

h.~

(w.l xv.)] l

at any

.

- 1

g = (h i g 1)

,

where

CHAPTER 3

187

i ~ 2. Indeed, in the opposite case there is, by property SI, an invertible

element

b

be 18,

g-le A(S) = H

such

that

and,

hence,

b- 1 sb= sg h j '11

for

e Gin H.

all

As

i.e.,

se S,

is

Gin

a

normal

subgroup, then g 1 e Gin H, which contradicts the choice of gl"'" g s' Thus, simplifying the notations, we find a correlation of the required type. The lemma is proved. 3.9.16. The proof of proposition notations of the preceding lemma, let us set

Preserving

3.9.13.

the

g

~ =

,,",.is - 1 .is I WI n I""'I g. (w. i j ] ]

j

I

/3

). Then, applying an element

e

~

to both

parts of equality (16), we get: -r

H« d . /3)x) =

Since such that

/3.

(17)

is a right ideal in

~

subbimodule in

"l( x) .

R F'

4S),

then

d.

is

~

a (S, R)-

By lemma 3.9.3, we can find an idempotent

fI!:: d~!:: fR F . Hence,

fez,

[(1- t)d]· ~ = (1- t)(d· ~)= 0, i.e.,

(1 - f) d e ~l .. Applying corollary 3.9.10. and using conditions (a), (c) and (d) for relation (16), we get f = 1. Now, due to proposition 3.7.5, one can find a finite set of elements such

VI ,_., V ned . ~,

y

that

for

suitable

a, r i' tie

R,

a

'#

0

and

any

e R F the following relation is valid: va

J'

= £... ~ -r( yr ~.)v.t ~ ~.. i

Let vi = d·

/3 i'

where

/3 i

e ~. Then by formula (17) we have

-r H( v it i x) = -r(t i x)/3 i' Herefrom we get -r H( yax) = -r H( =

L-r( yr i)v it i i

x) =

L -r( yr) -r H( vi t i x) = L'll: yr) [-r( t i x) . /3 i) . j,

i

Writing now the actions of /3 i in detail, we get an equality of the required type. The proposition is proved.

188

AUTOMORPmSMS AND DERIVATIONS

3.9.17. The validity of theorem 3.9.2 is now certain: if S is an intermediate subring, obeying conditions BM, RC and SI, then one can find a

nonzero ideal I of the ring R, such that

l' H(I) ~ S,

where

H = A(S) , but

being an essential two-sided ideal of the ring RH = S and, since S is rationally complete, we get S = s, which is the required proof. l' H(I),

3.9.18. Corollary. Let S be an (almost) intermediate ring obeying properties BM and Sf. In this case the ring S contains an essential ideal of the ring § = II R.,S).

Indeed,

for a nonzero ideal I of R and the set

l' H(I) !:; S

l' H(I)

is an essential ideal of §. 3.9.19. Finally, it should be remarked that the key statement of this paragraph (proposition 3.9.13:) remains to be valid under somewhat more general conditions as well (see remark 3.9.9. ). Namely, let R = R E9 _. E9 R, G = G n >- S n and let the almost intermediate ring S obey

condition BM, and let us assume A(S) = ( Hm >- S rrJ x G", where H < G, S m is a group of permutations of the first m components of the decomposition of R, while the group G" acts trivially on these components. Let us denote by e a unit of the first component and assume that j

> m and any g Let us set

Sn' Then m

L

l'G

E

( S)

ecP

g>-(lj)=O at all

G n.

n

l'G( x)

=

L fli) (x)

, where (li)

i=l

is a trace form for the group

f.~j) (x) is a trace form for the group

is a transvection from

G. Analogously,

l'A( s)

=

A(S).

j= 1

3.9.20.

exact, 8 1,_"

a 8 k E

E

Lemma. There is a nonzero element a E elements rl'-" r k' vl'-' v k E R

eR),

s, such that for any x,

y E

(to be more and elements

R

R F the following equality is valid:

k

1'A( S)( yax) =

L

1'-( yr )8 i1' -( Vi x). i= 1 G G

Proof. Using lemma 3.9.15, let us write the following equality:

189

CHAPTER 3

'Z'-( x) = G

(1 j)

m

J. = 1

~,

m

L, 'Z' H (dx) + L, L,(wixv)

+ £.J... W J• XV.) ~

J. = 1

-1

gJ

hi

>- (1 ])

.

+

~

+

where the elements d, wand v lie in the first component eR. Introducing renotations, we get a presentation ~

'Z' -( x) = 'Z'A(S)(dx) + L.,,(w. xv.) G

.~

h i~,

~

~

+ £.J.... WJ. xv.) J

g

-1 j

J

for which the following conditions are met: (a)

q,~S) = 0 for all "

j;

(b)

hiE /(S) for all i;

(c)

Zd

tl

L,ZWi = 0;

(d) the left annihilator of d in Ze is zero: ann~~d = 0 Now proposition

we are to repeat, nearly word 3.9.13. (see 3.9.16), replacing

per word, the proof of with R; 'Z' H' 'Z' with

R

'Z' fi.. S)' 'Z' G' respectively, paying a special attention to the fact that

dN

is

contained in the first component and, hence, the idempotent f will be equal to the unit e of the first component. Analogously, proposition 3.7.5 proves the existence of the elements a, r., t " a 7:- 0 from the first component for which the following equalities hold: ~

ya

~

= ~'Z'G( yr )vit i = ~'l'G( yr )vit i' ~

~

since e'Z' c;< x) = 'Z'G( x). The lemma is proved.

190

AUTOMORPIDSMS AND DERIVA nONS

3.10

Correspondence

Theorems

We can now summarize the data obtained in the preceding paragraph in a form traditional for the Galois theory. 3.10.1. Theorem. Let G be a regular group of automorphisms of a prime ring R. Then the mappings H ~ I( H), S ~ l( S) set a one-to-one correspondence between all regular subgroups of the group G and all intermediate subrings obeying conditions BM, RC and SI. The proof results immediately from theorems 3.5.2, 3.9.1 and 3.9.2. Since conditions BM and SI refer only to interactions of the elements of the algebra of the group G with intermediate subrings, then for outer groups we immediately get the following theorem. 3.10.2 Theorem. Let G be a finite group of outer automorphisms of a prime ring R. Then the mappings H ~ l/( H), S ~ l(S) set a one-toone correspondence between all the subrings of the group G and all intermediate rationally-complete sub rings of R. Another particular case, when the algebra of a group contains few idempotents and many invertible elements, arises when lB (G) is a sfield. This happens when R contains no zero divisors, i.e., it is a domain. Indeed, then R) also has no zero divisors and the algebra of any reduced-finite group is a sfield, and then we come to the next theorem.

ex

3.10.3. Theorem. Let G be a reduced-finite N -group of automorphisms of the domain R. Then the mappings H ~ l/ ( H), S ~ AI... S) set a one-to-one correspondence between all N-subgroups of the group G and all the intermediate rationally-complete subrings. It would be useful to remark here that the condition of rational completeness for intermediate rings in the case of domains is equivalent to a stronger elementary condition: if rs = sl for some 0"# s, sl E S, r E R, then rES (such subrings are called antiideals). This fact can, for instance, be deduced from theorem 3.10.3 and from the evident fact that a fixed ring of a domain is an antiideal: if rs

= sl'

then (rs

l = sIg,

i.e.,

r gs

= sl

and, hence, (r - rg)s = o. The next natural step is to consider groups H the algebras of which, lB ( H), are simple (see definition 3.1.7). This case is quite close to the general situation for M-groups, as studies of an arbitrary M-group can be reduced to such groups to the accuracy of matrix constructions. 3.10.4. Definition. A reduced-finite group of automorphisms, G, of a prime ring R is called an F -group if its algebra is simple.

191

CHAPTER 3

3.10.5. Let us present a general scheme of the above-mentioned reduction. Let B (G) = 11 e .. e B k be a decomposition of the algebra of an M-group G into G-simple components. Let us denote by e i a unit of the algebra Bi . Then Qi = ei Qei is an invariant subring and

QG=

Qfe ..

e Q ~, in which case the group G acts on Q in such a way that its algebra is isomorphic to B i' Le., it is G-simple. now consider the case of a G-simple algebra. B n be a decomposition into simple components. As decomposition is unique to the accuracy of permutation of the addends, for any g E G there is a permutation 1T: g of the numbers 1,..., n, such Let us

B (G)

= Bl e .. e

g

Let the then that

Since the algebra is G-simple, for every k there is an auto-

B k = B Irg { k)'

morphism gk E G, such that l1gk = Bk and, in particular, all the components of the decomposition are isomorphic. Let us denote by Gk a subgroup of all automorphisms for which 1T: g (k) = k. In other words, there are isomorphisms which leave the unit e k in its place. All groups Gk are certain to be mutually conjugated:

Gk = gj/G1g k ·

Let us consider a subring Q k = e kQe k' The action of the group Gk is naturally induced on it, and the algebra of the induced group is Bk

= e kB ( G). Let ~

E

and a certain g

Let us prove this fact. (Q

JJ F

E

G. If g = £ is an inner automorphism from

it is inner for ( Qk) F

and let us assume that

x~

= ~xg

for all Gk

x E Qk '

then

as well as the element e k b will correspond to it.

We then come to the conclusion that ~ and over a generalized centroid of the ring

ek b

are linearly dependent

Qk which is equal to

eke, i.e.,

~ E Bk .

If g is not an inner automorphism, then one can choose an element U E Qk in such a way that the following identity:

u~"#

0,

u~ E Qk

and for any

y E Q

we get

AUTOMORPIDSMS AND DERIVATIONS

192

i.e., by proposition 2.2.2, we get:

ue k ® e.rt< u;)

=0,

which is impossible,

as ue k = u, e.rt< u;) = u; . G

Let us , finally, show that Q k k == QG , the case k = 1 being sufficient for this purpose. Let us construct two mappings in the following way. If ljI(d)

a

E QG,

then we set

= d + d~ + .• + d gn .

Since

G

ef = e k

k

d E Ql

1

,

then we set

and the idempotents

are orthogonal, then qX ljI( d) ) = d. Hence,

The fact that these mappings preserve the operations is also trivial. Therefore, taking into consideration the fact that the ring Q can be presented as a ring of generalized matrices

studies of an M-group of automorphisms of a prime ring is, in a certain sense, reduced to studies of F-groups. Let us remark that a regular F -group is called quite regular (see 3.1.7).

3.10.6. Theorem. Let G be a regular group of automorphisms of a prime ring R. Then the mappings H ~ 11( H) and 8 ~ A(8) set a one-toone correspondence between all quite regular subgroups of the group G and all intermediate (prime) rings, obeying conditions BM and RC and having simple centralizers in B (G) . To prove it, it suffices to remark that by theorem 3.5.1 a fixed ring of an F -group has a simple centralizer (and is prime, by theorem 3.6.2), and to show that an intermediate subring under condition BM with a simple centralizer obeys condition SI. The first part of condition SI is fair, as any simple finite-dimensional algebra is generated by its invertible elements. The second part will be proved in a more general form. 3.10.7. Lemma. Let intermediate rings 8 and 81 have simple centralizers and obey condition BM. Then for any isomorphism g: 8 ~ 8 1 , ·h Wh IC

· IIy on RG,·h n,(S) . . ·ble acts I· d entlca elf er th e set 'V g contazns an znvertl element, or it equals zero. Proof. Let z be a centralizer of 8 in B(G), and ZI be that of

81 . The set

«p~S) is a

(Z, ZI) -bimodule. Let us view this set as a left z-

193

CHAPTER 3 (S)

.•

module. Let 0 '" t1> g = Ll + .• + L m be a decomposition of this module into irreducible components. Then L i == Ze i' where e i is a primitive idempotent from Z, Z = zel + Ze2 +.. + zen. Let us first assume that m ~ n. In this case the left z-module embeddible into t1>~S) and, hence, an

(S, R) -subbimodule

eR F ::! aR F ::! eI

in

for some I

E

is

t1>~S) contains an element a, the left

= as g

annihilator of which in Z is zero. As sa is

z

RF

and,

F, e

E

for all

by

z.

s E S,

then

aR F

lemma 3.9.3,

we

have

Since (1 - e) a

= 0,

then e = 1

and, hence, aR F ::! I. In particular, the left annihilator of a in the algebra is zero and, therefore, dim e /8 ( G) a = dime /8 (G), i.e., a invertible element.. /8 (G)

Let now m < n. Then

is an

t1>~S) is isomorphic to the left ideal of the

algebra Z and is, hence, a cyclic z-module

t1>~S) = Za. The right annihilator

of the element a in zl is equal to the right annihilator of t1>~S). The latter, however, is an ideal of zl and is, therefore, equal to zero. Allowing for the fact that (R, SI) -bimodule

suitable e

E

zl.

as g = sa

for all

s E S,

we see that

and, by lemma 3.9.3, we have But, as a(1 - e) = 0, we have

e

a

is an

RFe::! R F::! Ie

for a

= 1,

RF

Le., as above, a is

an invertible element of IB (G). The lemma is proved.

3.11

Extension

of Isomorphisms

A traditional problem of the Galois theory is that of finding criteria for an intermediate subring S to be a Galois extension over R G. The conditions for a general case being quite complex, we are not going to discuss them here. The considerations of this problem are commonly based on the theorem on the extension of isomorphisms which will be given below with two applications concerning the sought criteria for outer groups and domains. Let us begin with a simple example which shows the extension of isomorphisms over R G between intermediate subrings not to always be

AUTOMORPIllSMS AND DERIVATIONS

194

possible for arbitrary M-groupS. 3.11.1. Example (D.S.Passman). Let R be a ring of all matrices of the fourth order over a field F *- GF(2) and let G be a group of all (inner) automorphisms of this algebra. Let us set and let SI = (diag{a, a, b, b)1 S = (diag(a, a, a, b)1 a, bE F) a, bE F). Then

S == SI' in which case the corresponding isomorphism is

identical on R G= (diag (a, a, a, a)}. The both since their centralizers z, zl are generated by same time, the isomorphism between Sand SI automorphisms of R, as Z is not isomorphic to

rings are Galois subrings, invertible elements. At the cannot be extended to the ZI .

The situation is better under the assumption that subrings of F-groups.

S, SI

are Galois

3.11.2. Theorem. Let G be a regular group of automorphisms of a prime ring, S, SI be intermediate Galois subrings of quite regular groups.

Then any automorphism qJ: S -+ SI ' which is identical on

R G, can be

extended to the automorphism ip E G. This statement can be easily deduced from the correspondence theorem

for a ring R e R with a group cJ >.. ~ (see theorem 5.9.2). Here we shall give a direct proof of a somewhat more general statement which is going to be useful in a number of cases. Theorem 3.11.2 results from it by lemma 3.10.7. 3.11.3. Proposition. Let G be a regular group of automorphisms of a prime ring R, S, SI be intermediate subrings obeying condition HM.

If qr.

S -+ SI is an isomorphism identical on R G and if the ring S obeys condition Sl for a set of isomorphisms G n qJG, then qJ can be extended to the automorphism ip E G. Proof. Let us consider B (G) as a left Z-module, where Z = z{S). By lemma 3.9.4, the algebra Z is semisimple and, hence, B{G) is a completely reducible z-module. In particular, z can be singled out as a direct addend from B{G): B(G) = Z e v, where v is a left z-module. In Z let us choose a basis zl,M., Z k over C, in such a way that zl = 1. Let us extend this basis with elements from V to the basis b l ,..., b n of the

algebra B over C. Let b~ 'M" b*n be a dual basis.

195

CHAPTER 3

Let then

Let us

H = A(S) , H 1= J( Sl)'

also set

G(S) = Gin H ,

= Gin HI ,

where Gin is a subgroup of inner automorphisms from G. Let us choose a system of representatives of the right cosets of the group G( S) over the subgroup Gin which consist of elements 1 = hl'-" h m lying in H, and a system of representatives of the right cosets of the group G over the subgroup G(S) of the elements 1 = gl'-" g n' As G is a reducedfinite group, then both these systems are finite and, moreover, the set G(Sl)

forms a system of representatives of the right cosets of Gover

{h i g j}

Gin' Performing the same operations for the group

G(Sl) , we can find a

system of representatives (h ~ g' j) for Gin in G, where

h

~E

11.

For any

x let us set t H(X)=Lx h1 , tH(X)=Lx h :. Now, by lemma 3.4.1, we i I i

have two trace forms:

where v( x)

= L b t xb~. t

-'

Let us note that these forms are equal. We have

,

1(x)= LV(X) :i, j

h

g 1

Hence,

where

h ~,g~,= a h rr ('~,J,)g e~, (' J') ,

_h'

J= L(V(x)a)

I«l,J)

we

have

g' h'g' £(l,J)= LV(X) 1 J= "l(X)'

i, j

:i, j

Let us write the equality 'Z( x) 'I' = "1 (x) in detail:

qi. f3)

Let

be an arbitrary element from

L(Sl)

= L(S'I').

Then,

applying qi. f3) to both parts of the latter equality, we get (assuming, as usual, f3

f(x)'f3=Lsjt(rjx),

=L r

j

®

S

where

fP(f3)=Lr j ® sj,

j E L(S»:

~

1

[L.-tH(V(x)· gi (f3» i

i

gl

]

'I'

~

r

1

=L.-tH(V(x), lpgi (f3»

I

g'l

.

(18)

AUTOMORPHISMS AND DERIVATIONS

196

It should be remarked

that

4J{ S)

gil

=0

at

i

'* 1.

Indeed,

otherwise, by condition SI, one can find an invertible element such that

sa

= as"

belongs to

gi1a-1

affords

g-:-I

a E /B (G) ,

s E S. It means that the automorphisms

for all

gi 1 =(ag,J-IaE G(S),

and, hence,

A(S)

since

which

i = 1.

Let us assume that 4J{ S),-I rpgi can write

=0

for all i

gi

= 1).

Then we

(s) _lb., . . lng' ] ~,] ... i

( S)

4J

i

(including

'"

Ib.+ £.J4J ]

since the last two sums are zero. (It should be recalled that b m = Z m at m ~ k). By corollary 3.9.12, we can find a p E IL(S), such that

bel df 1 . P '* 0

and, simultaneously, b m '

at all j, 1 ~ j ~ k and all j.

Now,

according

to

i,* 1;

equality

P=0

b j ' qJ g'i 1 (

(18),

m> 1;

at

we

P ) =0

get

P) = 0

b j gi 1(

at all i

t H(V1(

x»

and all

= 0,

where

k

L

v 1( x) =

bmbel

m=1 '"

h.

xb~. Or, in more detail,

h.

h

h.

£.Jbm"bo"x i(b*m> ":::;:0.

i, m

As at

i

'* 1

the elements

hi do not lie in

Gin' then this identity, by

theorem 2.2.2, yields equality Lbmbo ® b*m = 0 in the ring

Q ® Q,

which

'* ( S) r1> rpg '* 0

is impossible, as (b*ml are linearly independent elements and bIb 0 = b O O. Thus, we can choose an element g = g

,- 1 i

E G, such that

.

By condition SI, for qJg one can find an invertible element a E /B (G), such rpg -1-1 -I -I _ that sa = as . Hence, sffJ :::;: (a g ) s gag , i.e., g- 1 b, where b = a g

-I

is the sought extension of qJ. The proposition is proved.

CHAPTER 3

197

3.11.4. Corollary. Let G be a reduced-finite group of automorphisms of a domain. Then any isomorphism over R G between intermediate subrings can be extended to an automorphism from A (R) . Indeed, it has already been noted that any intennediate subring of the domain obeys condition BM, and its centralizer in Q is a sfield. We are, therefore, to extend the group G to a quite regular one. In this case a fixed ring does not undergo changes and we can use proposition 3.11.3. 3.11.5. Corollary. Let G be a finite group of outer (for QJ automorphisms of a prime ring R. Then any isomorphism over R G between intermediate subrings can be extended to an automorphism from G. 3.11.6. Theorem. Let G be a finite group of outer (for QJ automorphisms of a prime ring R. An intermediate ring S will be a Galois extension of RG iff the group A(S) is normal in G, in which case a Galois group of extension SiR G is equal to G / A(S) .. Proof. Let A(S) be normal in G. Let us first show that CXS) is a

ex R G).

Galois extension of

ex R) Pi.. S)

and equal to ae exS),

By theorem 3.8.2, the subring

and, hence, it is invariant with respect to G: if h

he E(S), then (a g g

ge G,

exS) is Galois

-\

)g= a g , i.e.,

ageQ(R)A(S)=

= Q(S). Therefore, the action of the group G is induced on

exS) and the

induced group is isomorphic to G I A(S). This group will be outer: if a = g e G on exS), where a e exS), then for any s eRG we have

sa = s, i.e., a e

Z( R G)

= C.

It is also evident that a fixed ring of this

group is equal to ex R G) . The task now is to show that the induced automorphisms belong to A(S) (see 1.7). According to corollary 3.9.18 and propOSItIOn 1.4.13, we have A(S) = A(S), where S = IA(S) = Q(S) n R and, hence, it would be sufficient to find for every g e G ideals U1 ~

II'

I2

for ( I2

ui ~ S.

12 e

F,

By the definition that

such

U1 = II g

n Q( S»

n S g

we

exS»g

get

and

A(R)

Vi

from F(S-) , such that

there can be found ideals

12 n CXS) = = ri n Q(S) ~ R n exS) = S, and g g U1 !;;; I2 n Rn Q(S)= Ii n CXS)=

II ~ Ii. ~ R.

ui = (I 2 n

n S we have

of

U1

Herefrom

= U2 ' which is the required proof.

for

U2 =

198

AUTOMORPHISMS AND DERIVATIONS

Let us, inversely, assume S to be a Galois extension of RG with a group H!:: A(S). Let us show that the automorphisms from H have an extension up to those from the group G. If ~ E H, then there are U1' ~ E F (S),

such that

restriction of ~ on ~ +

cp

U1 ~

C1:

RG

can be extended to the automorphism

~ S.

According

to

corollary

3.11.5, a g E G.

As the restriction of g on 5 belongs to A(S) and g = ~ on ~ E F (5), then g = ~ on S as well. Let N be a group of all possible extensions of H. Then N ~ G, since G is a Galois group and N:2 A(S). Moreover, A(5) is a normal subgroup in N and N / A(S) == H. It should also be remarked that H is an outer group for s: if qX.s) = a- 1sa for a E O:S), s E s, then at s ERG we get

Let

0: R) N

~

0: R G) =

us

s

= a-

I sa, wherefrom we have

now

0: R/(S) = O:S) Q(R) G.

calculate and,

a

E

Z( R G)

0: R) N.

Since

hence,

0: R) N

= c.

A(S)

~

~

N,

then

O:S) H = Q(SH) =

As N is also a Galois group, then N = G and the theorem

is proved. 3.11. 7. Definition. Let G be a reduced-finite N -group of automorphisms of a domain R. Then a subgroup H of the group G is called almost normal in G if the least N-subgroup of the group G, which contains a normalizer

NG(H), coincides with G.

3.11.8. Theorem. Let G be a reducedjinite N -group of automorphisms of a domain R. The intermediate subring S is a Galois extension of R G iff the group A(5) is almost normal in G. In this case a

Galois group of extension 5 ~ RG is equal to NG(A(5» / 1(5). Proof. Let SH = R G for a group of automorphisms H ~ A(S). For of S, such that every hE H we can find nonzero ideals I, J I ~ Jh~ S. Now, by corollary 3.11.4, the automorphism h can be extended from J + R G to an automorphism g E G. One can easily check that g coincides with h on 5, since (sh_ sg)Jh=O for any sE 5. Let us denote by M a set of all extension of the elements of H. In order to show that A(S) is almost normal in G we have to check the validity of the following two statements: (a) the group A(5) is normal in M; (b) R M= RG.

CHAPTER 3

199

(a) Let me M, a e /(5). Let us choose a nonzero ideal I of s, such that I m s;;;; S. Then a restriction mam- 1 on I will be an identical automorphism. As an extension of automorphisms from the ideal on S (and even on C(5» is unique, then mam- 1 e /(5). (b) Let r e RM. Then r e RA(S)= §, since A(5) S;;;; M. By corollary 3.9.18, we can find a nonzero ideal I of 5 contained in 5. As H is a reduced-finite group of automorphisms for the domain S, then the ideal has a nonzero intersection with 5 H = R G. Let t be a G- fixed element, such that 5t ~ 5. Then rt e R M rI 5 = 5 H = R G , i . e . , (r g- r)t = (rt) g- rt = 0 for all g e G. Hence, r eRG, and the proof is completed. Inversely, let A(5) be almost normal in G, and let M be its normalizer. Let us, first, show that a restriction of m e M on 5 is an automorphism of S, Le., it belongs to A(S).

Since

and A(S) = A( 5). If now S

5

= R A(S)

m

E

M, a

E

1

have

a shared nonzero ideal, then A(S) , then ma = (mam - 1) m, in which -m

.1(S)

case mam- E ;(5). It means that S !:: Q of R, such find nonzero ideals I, J ( J rI 5-)m !:: Q .1( S) rI R

(JrlQ

A()

= S.

Besides,

•

-

As that

IriS!:; J

A(R), then we can I ~ J m !:: R. Hence,

mE

m

rI

-

S

= (J

rI

- m- 1 m

S

)

!::

.1(S)m-m

S)m=(JrlRrI Q

)

=(JrlS).

Since A(S) is a reduced-finite group, then J rI § and I rI § are nonzero ideals of § and, hence, a restriction of m on § belongs to A(S). The kernel of the mapping of the restriction M ~ A(S) is equal to A(S) and, hence, the induced group is isomorphic to M / A(5). The fixed ring of this group coincides with R M rI s. Now we have to remark that R M= R G, as when adding the inner automorphisms corresponding to the elements of the algebra of the group M, the fixed ring undergoes no changes. The theorem is proved. REFERENCES M.Artin [9]; L.N.Childs, F.R.De Meyer [27]; C.W.Curtis, I.Rainer [36]; J.Dieudonne [39]; I.M.Grousaud, I.L.Pascaud, I.Valette [50]; G.Hochschild [60,61]; N.Iacobson [62,64,65];

200

AUTOMORPIDSMS AND DERIVATIONS

T.Kanzaky [66]; G.Kartan [67]; V.K.Kharchenko [68,69,76,79,80,81]; H.Kreimer [89]; Y.Miyashita [l08]; S.Montgomery, D.S.Passman [118]; R.Moors [122]; T.Nakayama [124]; T.Nakayama, G.Azumaya [125]; E.Noether [126]; J.Osterburg [127,228,129]; A. Rosenberg, D.Zelinsky [141]; T.Sundstrom [147]; A.I.Shirshov [144]; H.Tominaga, T.Nakayama [150]; O.E.Villmayor, D.Zelinsky [152];

CHAPTER 4 THE GALOIS THEORY OF PRIME RINGS (THE CASE OF DERIV ATIONS) In this chapter we shall explore the problem of correlation between a given prime ring R and a subring of constants of a finite-dimensional Lie algebra L of its derivations. If the characteristic of a ring is equal to zero, then constants of finite-dimensional Lie algebras are related, but weakly, with the basic ring. Let, for instance, R = F ( x, y) be a free algebra with no unit. Let us consider a derivations J.1., such that yll = y, x Il = x. Then its derivation has the only constant, i.e., zero, in which case the Lie algebra generated by J.1. is one-dimensional. A reasonable restriction which can be imposed in the case of a zero characteristic is that the associative algebra generated by the considered derivations in a ring of endomorphisms of an additive group of the ring R is finite-dimensional. In this case, however, theorem 2.7.8 yields that L consists only of the derivations which are inner for the ring of quotients R), and the problems are reduced to those of centralizers of finite-dimensional subalgebras, which have already been considered in 3.3. Therefore, of primary interest is the case of a positive characteristic p, in which case the p-th power of any derivation is again a derivation, and all the constants of the initial derivation are also those of its p-th power. Hence, a restricted Lie algebra has to be viewed as L. It should be recalled that extend the situation a little under the assumption that derivations from L transfer R into C( R) but not necessarily to R, in which case, however, we set that for any J.1. E L there is a nonzero

ex

ideal I of R, such that I Il~ R. A set of all these derivations are designated by D( R) (see 1.8). This set forms a differential Lie C-algebra (see lemma 1.8.3), i.e., it is also a space over a generalized centroid C. Since when multiplied by a nonzero central element its constants undergo no changes, then it would be natural to view L as a Lie a-subalgebra in D( R) (see 1.2). Therefore, let L be a finite-dimensional restricted Lie a-algebra over C. If A is a subalgebra of derivations from L which are inner of Q, then A is an ideal in L and a subspace over c. Therefore, the action of a factor algebra E A( xll

= O)}

L

I A is induced on the ring of constants RA = (x

, in which case

RL

= (R A)L I

RI 'V J.1. E

A. This peculiarity, combined

201 V. K. Kharchenko, Automorphisms and Derivations of Associative Rings © Kluwer Academic Publishers 1991

E

202

AUTOMORPIDSMS AND DERIVATIONS

with the fact that RA is a centralizer of a finite-dimensional c-subalgebra from Q, makes it possible to restrict ourselves to considering Lie -algebras of outer derivations. Henceforth in this chapter we shall assume that L is a finite-

a

dimensional restricted lie a-algebra of outer for Q derivations from of a given prime ring R of a characteristic p > O.

4.1

Duality

for

Derivations

in Multiplication

D( R)

Algebra

Let s be a subring in a prime ring R. Let us denote by L oPes) a subring in the tensor product Q ® QOP (where, as usual, QOP is a ring antiisomorphic to Q with the former additive group), generated by elements of type S ® 1, 1 ® r, where S E S, r E R. (It should be recalled that the ring li..S) is generated by elements of the type 1 ® s, r ® 1 (see 3.9.8». a

If

a·

E

RF ,

f3 = Lvias i'

f3

If

~

= £.Jsi ® vi E L

~

is

a

op

(S),

mapping

then,

from

S

as

before,

to R,

then

we

set

we

set

f3~= LS~® vi' Let us denote by al. S a set of all f3 E LOP(S), such that a· f3 = 0, i.e., al. S = al. n L oP(S). If v!:: LOP( R) r;; = L(R), then we

vi- = [a E

should recall that

RF

I a· V

= O} .

4.1.1. Lemma. Let S be a subring of R, such that every nonzero (R, S) -subbimodule of R contains a nonzero ideal of the ring R. Then the following formula is valid:

m

La. Z +

i= i

where

~

gl = (

i

m (") a~ S n JB)l. ,

=1

~

a i are arbitrary elements from

from

R"

B is an arbitrary right ideal

L oPe S), Z is a centralizer of S in the ring of quotients RF .

Proof. The inclusion of the left part in the right one is obvious. Let us carry out the proof of the inverse inclusion by induction over m. If m = 1 and v is an element of the right part, then for any f3 E l/J the equality

q;: a l

.

P~

al . V·

f3 = 0 implies f3, where f3 runs

v .

f3 = O.

through l/J

Hence,

the

mapping

is correctly defined. We can

CHAPTER 4

203

choose a nonzero ideal

of

I

Iv!:: R, Ia l !:: R. If now

R, such that

f3

runs through 8 (1 ® I), then the domain of qJ values is contained in R, and the domain of the definition forms an (R, S) -subbimodule in R. If this subbimodule is zero, i.e.,

a l · 8 (1 ® I)

= 0,

then

al · 8

=0

and, hence,

v· 8 = 0, i.e., v E 81.. Therefore, the domain of qJ definition can be assumed to contain a nonzero ideal of the ring R. It is also evident that qJ is a homomorphism of the (R, S) -bimodules and, hence, there is an element ~

R F' such that (al

E

· f3)~ =

V·

f3. Substituting into this equality

f3(s®I), we get (al·f3)~s=(al·f3)s~, i.e., ~E (al ~ v

E

v) .

-

f3 = 0 indicates that

v

-

E

Now the relation

(8 (1 ® I) )1.= 8 1..

Therefore,

81.+ a l ~. Let

8 = 1

al ~

Z.

f3 with

the

k

( l a~s i= 1 ~

(I

lemma 8

be

proved

for

m= k

and

let

us

set

Then, according to the case when m = 1, we have

.

1. k+l a k + 1Z + 1B1 = ( (1

~=l

a1

S

1. 1B) •

(I

This is the required proof as, by the inductive supposition, we have 1

k

L

1B1" =

+ 1B

ai Z

1.

i=l

Lemma 4.1.1. is proved. 4.1.2. Lemma. Let S be a sub ring of the ring R, such that every nonzero (R, S) -subbimodule of R contains a nonzero ideal of the ring R. Let, then, z be a centralizer of S in R F. In this case, if and

a l ,·· arnE RF op

a l ~ ~z +

+ .• +

Cl 3Z

f3 ElL (S), such that alf3 :;eO, a2f3 =

a

l E

Proof.

(at s

(I

at

S .•

inclusion is equal to

4.1.3.

If (I

such

a-lnS )1.. ~ Z

a

f3

amz,

~f3 = •. =

does

then

there

is

a

a m f3 =0. not

exist,

then

By lemma 4.1.1, the right-hand part of this am z, The lemma is proved.

+ .. +

Lemma. Let the rings

Rand S obey the conditions of

204

AUTOMORPHISMS AND DERIVATIONS

lemma 4.1.2,

11 ,.., /lv

at ,•. , am

E

and let us assume that Z is a sfield. Let

RF

be derivations determined on S and assuming their values in

and let rt:t:. 0, r 2 ,•. , rv+ 1 be elements from fol/owing statements is valid:

[3

(a) there is an element

m

E

n

i=I

RF .

In this case one of the

aJ,s, such that ~

/1.

v

R,

L,r.[3'+r I·[3:t:.O ~ v+

(1)

i= 1

(b) there are elements

that for al/ s

and an element S the following equality is valid:

E

2 1,."

2

mE Z

r 1Z + r 2 Z + •. + at Z +

Proof. Let us consider a right Z-module

•. + am zV' Let m

such

t E R F'

d1 = r 1, ~ •• , d n be its ba..<;is. In this case we obviously have

m

n if: sen aJ, S - i = 1 ~ .

Let

Ir i' Id j ~ R for all

i, j, 1 ::;; i::;; v + 1, 1 ::;; j::;; n. Let us consider a set

i= 1 ~

fB =

nn

j= 2

be

I

a

nonzero

~.S. By lemma 4.1.2, the element

ideal

r1

]

of

I( r 1 . fB) is a nonzero (R, S) -subbimodule contains a nonzero two-sided ideal of the ring R.

us

fB 1 = fB (1 (9 I) the mapping

assume ~

that

for

any

f3

E

1

5

that

does not belong to

and, hence,

Let

such

R,

in n

~

R, Le., it .1 S

n fBI ~ ncr:, j=1

fB

~

where

fB, the left-hand part of the relation is zero. In this case

where [3 runs through /B t' is well-defined. This mapping is a homomorphism of the left R-modules,

205

CHAPTER 4

v

L

~[.r(r1/3)]=~[r1"/3(1®r)]= + rv+ 1/3(1 ® r)=

v

L

i= 1

ri"/3

Il

l(1®r)+

i=l Il

i)+ r(rv+1" /3)=

r(r 1 " /3

Since the domain of the ~ definition contains a nonzero ideal of the ring R, while the domain of its values lies in R, then there is an element t E R F , such that

(2) Let

S E

=

S, then

Lv

r i[ /3

Il

1(S ® 1) + /3 (s III ® 1)] +

i= 1

(3)

+rv + I "/3(s ®1)= =

Lv [(r "/3

i= 1

Il

~

Il

1)S + (r ~"/3) s

1] + (r v + 1 /3)s"

If we multiply relation (2) from the right by s and subtract it from (3), we get (r 1/3) [s, t] =

L V

(r i " /3)s

III

"

(4)

i E Z is a decomposition of

ri

i=l

If

/3 E /B 1 and

on the basis (d jl, then form

r i = rl

Z

i + ".,

Z

r i " /3 = ( r l /3 )z i

and equality (4) assumes the

(r l /3)([s, t]-Lzis 1l1 )=O"

By lemma 4.1.2, the element

rl

does not belong to

/B

f

and,

206

AU1OMORPIDSMS AND DERIVAnONS

hence, rIB I is a nonzero (R, S) -subbimodule in R, i.e., by the condition, it contains a nonzero two-sided ideal of the ring R. Therefore, the latter equality implies that statement (b) is fulfilled. The lemma is proved.

4.2

Transformation

4.2.1. the kind

Forms

A differential form

Definitions.

~aix~ibi'

of Differential

where ai,biE RF and

will be an expression of

~i

are words of derivations

~

from D( R). It should be recalled (corollary 2.3.2) that if a form turns to zero on a nonzero ideal of the ring R, then it will also be zero on R, and even on R F' In this respect the following agreement looks natural. 4.2.2.

Agreement. If T is a subset of

f( x) we shall write

D-polynomial

(or

f( Xl ,..., X d!;,; T

E

4.2.3.

f3

f3

=

f( x)

f( x)

* f3

form),

then

implies that there is a nonzero ideal

I

the of

presentation R, such that

il'-" inE I.

belongs

L r i f(s i x),

4.2.4.

T

Denotations. If f( x) is a differential form, and the element

= LSi ® r i

f( x) .

S;;

a polylinear differential

f(i l ,..., iJE I for all

then for the form

provided there is a nonzero ideal I of the T for all i E I. Analogously, if f( Xl ,_., x d is a

f( x)

ring R, such that f( i)

R F'

* f3

=

to

lLoP(S),

then,

as

before,

let

us

set

and

L f( xr) Lem rna.

!;,; TS for all f3 E lL The proof is obvious.

Si .

If

op

f( x) S;; T,

then

f( x) .

f3

S;; RT,

(S).

4.2.5. Let 111'-.,l1n be a set of derivations from D( R). Let us fix an order on this set: 111 < 11], < _. < 11 n and extend it in a standard way onto a set of all the words from I1t ,_., 11 n' i.e., we assume (as before) that a longer word is bigger than a shorter one, while the words of the same length are compared from the right, as in a dictionary. It should be recalled

207

CHAPTER 4

that the word 11 = p

~.•

om

is called correct if it contains no subwords of the

type JJ. i JJ. j : JJ. i'~ JJ.;, where

JJ. i > JJ. j and p is a characteristic of the given

ring. In other words, in a correct word we always have ~ S ~ S •. S and this chain should not contain p signs of equality in a row.

on ,

For two correct words, 111' 11 2 , let us denote by fl} 0 fl2 a correct word obtained from a combination of all the derivations included in Al and A . 'f fl } -- r} /1m,. •. rn /lmn A - /It I /ltn then A OA _ m,.+ tl •. JJ.nmn + tn , 2' I.e., I , °2 - r} .• rn o} °2 - JJ.}

mi + t

in which case, if one of the sums

i

proves to be greater then or

equal to p, then we assume that fl} fl2 = 0 is a zero word. A zero word should be told from an empty one. It is an empty word that the identical mapping jlJ = x corresponds to, which agrees with bringing a derivation to 0

a zero power, JJ.0 = 0. A zero word is corresponded to by a zero mapping 0: x ~ O. An empty word is, hence, a subword of any other word, while a zero word cannot be a subword of any other word. As it is common practice will be called a

in the theory of Lie algebras, a correct word fl}:f:. 0 subword of the correct word fl,

if fl = fltfl2 for a correct word A2 • Herefrom, when considering words we shall assume, unless otherwise stated, that they are not zero. However, sometimes it would be more convenient to carry out induction over the length, assuming a zero word to be as long as 1. 4.2.6. Let fl = JJ.~k •. JJ.:r:n be a correct word, mk > O. In this case well known is the fact (easily proved by induction over the length of the word 11) that for any x, y the following equality is valid: (xy)

Li

=

m' m' min C m k C m k+1 . . . C mn

"" £..

,,,

LioLi=Li

= xyLi+ mkx

J1.

k

k+ 1

k

A I

XLI

A "

_

yLl _

(5) -

yLi

+ ...

m'k m~ /I mk - 1 /I mk + I /I mn is where fl = JJ. k •• JJ.n ' the length of the word 2i = rk rk+}··rn less than that fl, and it is the least among the words different from fl, with which the element y enters the right-hand part. I

It affords that if

f3

E

L

op

(R), then

208

AUTOMORPlllSMS AND DERIVAnONS

(6)

(axL\b) = (axb)

L\

t. < Ii, in

~ b, where

where dots denote a sum of the terms of type which case t. is a subword of the word l1. Now the following formula is valid: '" L\ . +.L.(aixb) ",

(7)

i

where A i are correct subwords of the word A, with their length less than that of A. Indeed, if the length of the word l1 is one, then ax J.1. b = (axb)

a J.1. xb - axb J.1.. In a general case we get by induction

J.1. -

L\ ax b

= ax J.1. k X b = (ax J.1.k b) X + ... = [(axb) J.1. k

- axb

J.1. k

3

J + ... = (axb)

J.1. j

3

+ ...

- a

J.1.

k

xb-

L\

= (axb) + ...

Formula (7) can be modified for two cofactors:

X

L\

b = (xb)

L1

- m k( xb

J.1. k

)

S

(8)

+ ...

t. < Ii. Indeed, if the

where dots denote a sum of terms (xd )t., where length of the word we get by induction /\

x- b

is one, then

x

= X J.1. k Ii b = (x J.1. k b) Ii -

= (xb)

- (xb

l1

J.1. k

li

~ ;i )

- ( xb

+ _. =

J.1. k

J.1.

b = (xb)

(m k - 1) (x

2i

J.1.

J.1.

l1

J.1. k

)

Ii

J.1. k

k

b

J.1. k J.1. k

Ii

) - ( mk - 1 )( x b )

(xb) - m k( xb

J.1.

mk > 1, then

- xb . If

)

;i + ... =

-

+.- .

If mk = 1, then we get in the same way:

A

xL> b

= X J.1. k 3 b = (x J.1..

J{

b)

3

+ _. = (xb)

J.1. k

3

- (xb

Here we have made allowances for the fact that

J.1. k

)

3

+ _..

209

CHAPTER 4

By analogy with fonnula (6) we now see that (axb)

L1

* 13

== [ax(b .

13)] L1 -

[ax(m kb·

13

11k

)] X +._,

(9)

X

where dots denote a sum of the temlS (xad) , where ~ < 3, and ~ are subwords of the word d. Let us now assume that [J.1. ~_, Jl n} is a basis of a restricted Lie dalgebra over C. In this case the following equalities are valid: p_"''
(10)

where dots denote a linear combination of the words, the lengths of which are strictly less than the sum of the d

I

t

and d 2 lengths. In particular, if

d d 2 == 0, then the word dId 2 is presented as a linear combination of correct words of a strictly less length.

4.3

Universal

Constants

Let L be a finite-dimensional restricted Lie d-algebra of outer derivations. Let us fix a basis J.1. ,_., Jl n of the right vector space L over a generalized centroid C of a prime ring R. Let us assume that this basis is ordered:

JlI <

~

< .. < J1 n . Let d I'-" d

m

be a set of all correct words,

d I be an empty word. It is evident that m == P n. Let us consider a set D( L) of all D-polynomials with coefficients from Q which include derivations from L as operations.. Let us identify the elements from D( L) provided their difference is an identity on R. It is obvious that D( L) fonns a ring, in which case L acts on it as derivations. 4.3.1. Definition. A nonzero constant from D( L).

universal constant (for

L) is

an

arbitrary

210

AUTOMORPlllSMS AND DERIVATIONS

Lemma. If f( xl'""

4.3.2.

xrJ is a universal constant, then (11)

reduced

Proof. With no restrictions on generality one can consider f to be (see 2.1), i.e" to be presented as a polynomial from

A (x i j 11 :S i:S k, 1 :S j:S m), Let us choose a nonzero ideal R

IliI 1! : ;R' 1 < _

' m suc h a way th at

occurring

in

the

J.l

valid A

, an d f or a11 coe ff'IClents

~

presentation

of

the

f( x),

q

inclusions

are fulfilled, In this case for any integer 1> 1 and any

II q!:; R, qII!:; R

basic

reduced

'< _ n

of the ring

II

c

is

the

(I/) Ii-!:; II1 - I,

inclusion

d

IJ.( p -

Therefore,

I)n + 2

' we see that the values on I belong to R. Now we are to remark that , according to the definition of a

( II(p - 1) n+ 1)

f

j

I

-

1

an , hence, setting I =

universal constant, fli- = 0 and, hence, The lemma is proved, 4.3.3.

a-algebra

f(I)!:; Q L(] R

= R L,

Lemma-definition. For any finite-dimensional restricted Lie

L

there is a linear universal constant, f( x) = coefficients from C, which will be called a trace form, Proof. It would be sufficient to find an element f L

x

Cj

"j;

A

with

j

0 from

the

universal enveloping u for L (see 1.2.3), such that f , J.l i = 0, 1:S i:S n, where (J.l i) is the basis of Lover C, Indeed, presenting here f as a

Lx

D

A

linear combination of correct words j C j' we get j Cj is the universal constant sought. It should be remarked that u is a Frobenius algebra over the field of constants C L, Indeed, let us consider an arbitrary C L -linear epimorphism "': C --t C L,

and

determine

a

linear

correspondence to the linear combination

qr.

mapping

D

jC j

putting

U --t C

in

the coefficient.. at the word

h' case the I'mear f unctIon ' J.l Ip - 1_, J.l pn - 1 ' I ntIs

1

/l,

=

cp", trans f orms

' U In

C L and the kernel of this function contains no nonzero right and left ideals

as C

for

any

=
because in

U

U "j;

0

one

can

choose

a

ul

E

q>( ul u) ~ 0, In this case we have

U

in

such

A.( uU I c- 1)

a

= A( c-

way lUI u)

that

= 1,

the rule of commutation c.1 = .1 C +", acts, where dots denote

211

CHAPTER 4

a linear combination of words of a less length. Thus. U is a Frobenius algebra. Therefore. valid is the equality ann x< ann 1 u') = u'. where u' is the right ideal in U generated by elements ~ ._ .• Jl n. This ideal consists of linear combinations U

*"

D

u'. i.e..

of nonempty correct words. Hence. 1 e u' and. thus.

j Cj

*" O.

ann 1 U'

which is the required proof.

4.3.4. Let us now consider the identity (or. for short. R is a GI -ring). A restriction 11 of derivations onto -algebra Der Q in Der C. If R is a consists of only inner derivations (lemma intersect with L, Le., L == 11 ( L) ~ Der C.

a

4.3.5. Theorem. If RC

where

F

= (RC) L

®

consider a subring

is a prime

has a generalized

C is a homomorphism of the Lie GI -ring. then the kernel of 11 2.4.2) and it. therefore. does not

GI -ring. then

(12)

F C.

= C L.

Proof. Let

R

case when R

f( x) =

L x!1

j

c·] be

a

trace

RI = (RC)L C and show that

RI

form

== (RC

for

l

Let

L.

us

® C. For this

l

purpose it suffices to show that if ql..... q n are elements from (RC which are linearly independent over F. then they are also linearly independent over C. Let. on the contrary, C E

C. Applying Ll

j Cj

L q i A. i = O.

Then

L q i ).. i

C

=0

for an arbitrary

to both parts of the equality and carrying out

~. .q. f(A. .l.c) = O. Since f is a trace form and summation over j, we find £ ..l qi are linearly independent over F. then

f( A.ic) = 0 for all i.

Allowing

for the fact that C is an arbitrary element. we get an identity f( x) = 0 in the ring C. i.e .• by theorem 2.2.2. all coefficients C j are equal to zero, which is a contradiction. In order to show that ~ = R, let us fix a q E RC and consider a relation valid for all C E C: (13)

AUTOMORPHISMS AND DERIVAnONS

212

Through induction over the senior word let us show that a relation of the type (13), where 0 = .1} < _. < .1 k are correct words, e k following equality: q= Le'if(e"iq)

E

i

~

:;t

0, yields the

(14)

,

where the central elements e'., e". are independent of the choice of q. If the senior word in (13) is empty, then at e = 1 we ~

1

q= e 1f(q)

E

Let now

~

get

R1 .

be the senior word in (13). Let us substitute the product c' e" instead of e and write (13) using formula (5), setting x = c', y = e" q. In this case the senior term for e" q will have the .1

k

6

form e kc' (e" q) k. Therefore, if we substitute e = e" into relation (13), multiply it by c' and subtract f( c' e" q), then we get a relation of the previous type, where e" plays the part of e, while the senior word is less than ~

k'

Let us choose an element c' in such a way that not all the terms in the obtained difference were cancelled. For this purpose let us calculate the li k coefficient at (e" q) (here ~ k = Ilst" k' see formula (5». The word ~ k in decomposition (5) can be formed only for words ~, such that ~"0t" k =.1 at a certain 8'. As ~ occurs in presentation (13), then ~::;; ~ k' i.e., for 8' we have only two possibilities: 8' = 0; 8' = IlI' < Ils' Thus, let be the number of the word

Ilrt" k in presentation (\3),

number of the word Li k' Now the coefficient at (e" q) formula (5), to be equal to ( ') J.1s ek·mSe

~

li k

s(O)

s( r)

be the

is seen, by

( ') J.1r

£... cs(r)c r~O

By theorem 2.2.2, the element c' can be chosen in such a way that this coefficient be nonzero (in which case the choice of c' is q-independent). c' f(c" q)Therefore, applying the inductive supposition to f( c' c" q), we get equality (14), which implies theorem is proved.

4.3.6.

Remark.

RC = (RC )L@ F

It should be noted that in

the

C. The

proof of the

213

CHAPTER 4

preceding theorem use has been made only of the fact that the homomorphism 1J: L --+ Der C is an embedding, so that in the formulation of theorem 4.3.5 the condition "R is a GI -ring" can be replaced with the condition "L= 1J(L)".

4.3.7. Definition. A subring S of the ring R will be called almost intermediate if f( x) !:; S for a certain trace form f( x) . The subring S is called intermediate if R:2 S:::> RL 4.3.8. Theorem. If S is an almost intermediate subring, a, bare nonzero elements from Q, then aSb:t: O. In particular, R L is a prime ring. Proof. Let, on the contrary, aSb = O. Then for the trace form we have a ®

C j

af( x) b

=a

D.

~ x J C j b = 0, i.e., by theorem 2.2.2, we b = 0, which yields C j = 0 at all j, which is a contradiction.

have the relation

4.3.9. Theorem. If a polynomial identity is fulfilled on it is also fulfilled in R. Proof. Let

g(

Z

i) = 0 be an identity in

R L,

then

R L. Then, by lemma 4.3.2,

we can find a nonzero ideal on which aD-identity

D. J = 0 holds. g( '" ~c j xi) J

By theorem 2.4.1, on the ring

RF

found. If now c,J ;t: 0, then we set o

at ];t:]o. We thus get an identity proved.

4.4

Shirshov

the identity

z"~J = c-.J 1Z g(

•

~

g(~CjZi}=O can be

at

z) = 0 on

j

=

jo' and

R F . The

Z ';J'

•

=0

theorem is

Finiteness

If R = C is a field, then one can show (see below corollary 4.4.4) that it has a finite dimension over C L. In a general case R is not a finitely generated right (or left) R L -module.

4.4.1. Example. Let C ( x, y) be a free associative algebra over a field of characteristic 2, and let J1 be its derivation, such that f.J{ x) = 0, J1( y) = y. Then Jl = J1 and, hence, L= J1C is a one-dimensional restricted Lie d -algebra, The algebra R L is generated by monomials which

214

AUTOMORPIDSMS AND DERIVATIONS

have an even degree over y. Any monomial not belonging to presented as

where max (k) + 1

xk y. w,

wE RL.

"k

RL

can be

If now (xkl YW'i) is a finite set of L

L

monomials, then x y e: £.. x 1 Y R + R and, hence, R cannot be generated as a module over R L by a finite number of elements. 4.4.2. Theorem. In a prime ring R there is a nonzero ideal locally-finite over R L (see definition 3.1.9). This theorem results from the following proposition. 4.4.3. Proposition. For any nonzero element u of the ring R there

are elements a"# 0, rl'-" r t

ax

=

t

L

r i u'f

i= 1

i

E

R, such that at all

x

E

R

(15)

x) ,

where 'fi are mappings from R to R L, which have the form 'fi = f( six) for suitable S i E R and a trace form f. Let us show the way theorem 4.4.2 is deduced from this proposition. Let us arbitrarily choose an element u. Let us then consider a set I of elements a for which there exist elements r l ,_., r m(a) E R, such that at all x E R equality (15) holds. One can easily see that IR is a Shirshov locallyfinite ideal over R L. The fact that I"# 0 follows directly from proposition 4.4.3. Proof of proposition 4.4.3. Let f( x) be a trace form for L. Let us consider a relation

(16)

where,

Vi,-·

as

usual,

V k , r i' t i

are

are elements of the ring Q, in which case

obvious that in the ring fulfilled, where

correct

f( x) =

R

L

Cj

V k "#

words, O. It is

the relation of the type " £..A, C j u) x 6 ]. = uf( x) is x

6 j.

Through induction over the senior word A k let us show that if a relation of type (16) holds, then the proposition is valid. If A k = 0, then relation (16) assumes the form

215

CHAPTER 4

v1x =Lriuf(tix). i

(17)

Let us choose a nonzero ideal I f(I) ~ R, IVI ~ R, Ir i !;; R, It is; I

the product

IVl:?

such a way that

of the ring R in such a way that for all i. Then, since R is prime,

is nonzero. Let us choose elements a

= 80 Vj

80'

aI'

~

E I,

in

a 1 a 2 :;t: O. Substituting into (17) the expression

a 1 a 2 x instead of x and multiplying by 80 from the left, we see that relation (15) is valid for the elements a, r i = 80 r i and mappings fi(X)

. f3 f3 E

= t(t i

a l )a2 x).

Let us carry out an induction step. If we apply an arbitrary operator to both parts of equality (16) and make use of equality (6), then, if

V±,

we again get a relation of type (16) with a smaller senior word.

Now we are to show that one of the coefficients V'I' V'2 ,.. of the newly obtained relation does not turn into zero. Let us calculate the coefficient at X Ilk , where a k = Jls3. k (see words a, such that S'

Li k in decomposition (6) can be formed only for 03. k = a at a certain a". Since a is encountered in

presentation (16), then

a ~ a k'

formula (5». The word

Le., for S' only the following relations are

possible: S' = 0; a" = Jlr < Jls' Let

8( r)

Jlr3. k' 8(0) be the number of the word

3.

be the number of the word k'

By formula (6) we can now

II

find a new coefficient at x k: (18)

Let

us

appl y

lemma

Jls' Jl s - I'·' JlI' assuming that

(b) cannot be fulfilled, as Jl i

C=

4.1. 3 to the ring R and a sequence s = R, r l = msVk , n = 1, a l = v k . Condition Z and the linear combination of derivations

over C lies in L. Therefore, one can choose an element

such a way that v ~(O) 4.4.4.

:;t: 0,

f3

E

V~ in

which completes the proof of proposition 4.4.3.

Corollary ( M. Weisfeld). Let

T

be a sfield of a

216

AUTOMORPmSMS AND DERIVATIONS

characteristic p > 0, and let L be a finite-dimensional restricted Lie dalgebra of derivations of the sfie/d T (possibly, the one containing inner derivations). Then the right and left dimensions of T over the sjield of constants, T L, are finite. Since the conditions are symmetric, it suffices to show that the right dimension of T over T L is finite. Let us carry out induction over the dimension of L. If L= 0, then there is nothing to prove. If A is an ideal of inner derivations, then T L= (T A)L/ A, in which case the center Z of the sfield T A can strictly contain C, but, nonetheless, dim z «L / A)· Z):S; dime ( L / A) and, hence, using inductive supposition, it is sufficient to consider the cases when L= A or A = O. In the former case the associative algebra B( L) generated over C by the elements corresponding to derivations from L is finite-dimensional and is, hence, a sfield. Therefore, I T: T.r, = I T: C( B) I :s; dime B, where C( B) is a centralizer of B( L) in the sfield T (and, certainly, C( B) = T L). If A = 0, then, by theorem 4.4.2, we immediately have I T: T.r, < The corollary is proved.

00.

4.4.5. Corollary. Let S be an almost intermediate subring. Then any nonzero (R, S) -subbimodule and any nonzero (S, R) -subbimodule of R contains a nonzero two-sided ideal of the ring R. Proof. Let U be an (R, S) -subbimodule and let u be its nonzero element. By proposition 4.4.3, for suitable elements valid is equality (13), the right-hand part of which belongs to u at all x from an appropriate nonzero ideal I. Therefore, aI k U and, taking into account the fact that U is the left ideal, we get RaI k u. If U is an (S, R) -subbimodule, then, going over to a ring which is antiisomorphic to R, we immediately get the required statement. The corollary is proved.

4_5

The

4.5.1.

Correspondence Theorem_

Lemma. The ring of constants

R L is rationally complete in

R.

Proof. If moreover, RHR L r the required proof.

Hr

k

L R ,

then for any

J1

E L

we have

~= 0, i.e., by theorem 4.3.8, we get

r

~

Hr =

0 and,

~= 0, which is

CHAPTER 4

217

4.5.2. Theorem (Correspondence). Let R be a prime ring of a characteristic p > 0, and L be a finite-dimensional restricted Lie -algebra

a

of outer derivations from D( R). Then the mappings A --+ R A; S --+ D f- R) set a one-to-one correspondence between all restricted Lie -subalgebras in L and all intermediate rationally-complete subrings in R. The proof will be divided into a number of statements.

a

4.5.3. Lemma. The centralizer of an almost intermediate ring S in RF

is equal to C. Proof. Let [S, a]

= O.

If

f( x)

= LC j

t:.

x J is a trace form,

such

that .t( x)!:; s, then by formula (11) we have an identity [f(x), a] = 0 on a nonzero ideal of the ring R. By theorem 2.2.2, we get relations Cj

® a-

C j

a ® 1 = O. Since one of the coefficients is not equal to zero,

then 1 ® a = a ® 1. i.e., a

E

C. The lemma is proved.

Theorem. Let

4.5.4.

).L be a derivation from

D( R), such that

S Jl.= 0 for an almost intermediate ring S. Then ).L E L .. Proof. Let ).L1'.").L n be a basis of Lover C. Let us first suppose t:. that ).L, J.lt p., ).L n is a strongly independent set. Let f( x) = L C j x J!:; S. In R the relation

holds, i.e., Lcj/JJl.+Lcj/J!::(O). By

.t(x)Jl.!:;(O)

theorem 2.2.2, we have

Cj

® 1 = 0, which implies that all

Cj

are zero.

Therefore, the elements ).L,).Ll p.,).Ln cannot be strongly independent over C, i.e., there are elements c, c p .' c n E C, such that ).Lc + ).L1 c 1 + .•. is an inner derivation for Q. As L contains no inner for Q

+ ).Lnc n = ad~

derivations,

then

condition f( x) Jl. i

C :F.

~

4.5.3, the element ).L = - L).L i C i

C-

0

and,

hence,

{O}, it means that

D( R)

R

Ds< R)

+ L).L i C i c- 1 = ad~c- 1. By the

[~c- 1,

~c- 1 lies in C. Hence,

1 E L.

~ {O},

i.e., by lemma

ad~c- 1 = 0

and, therefore,

f(x)]

The theorem is proved.

4.5.5. Definitions and intermediate ring S and denote by D s(R)

).L

notations. Let us fix an almost D S(R ) a set of all derivations from

which map S into zero. Then, by theorem 4.5.4, we get that is a Lie -su balgebra in L. By § let us denote a subring

a

•

It is obvious that

S;:?

Sand § is an intermediate ring.

218

AUTOMORPIllSMS AND DERIVATIONS

4.5.6.

S.

Theorem. The ring S contains a nonzero ideal of the ring

Let us choose a basis J.lI ,•. , J.l k of the algebra Ds(R ) and extend it to the basis ~ ,•. , J.l n of the algebra L. Let us assume this basis to be ordered, J.lI > ~ > •. > J.ln. Any correct word in this basis has the fonn d j

=d

(1) (2) j d j ,

where d

(2) . j IS

a correct word from

~ ... ,

(1)

J.l k' and d j

is a

correct word from J.l k + 1"" Jl n' Theorem 4.5.6 will now result from the following proposition. Proposition. There

4.5.7.

are

a I ,•. , a k E R not all of s, such that at all

elements

which are zero, and there are elements

sl'."

St E

x E R the following equality is valid: pk

(2)

'L. " (xa .) Ll

. 1 J=

where 'ri (x)

'r i

are

= f( xr i ) '

t

j

(19)

L'riX)Si'

]

i=I

mappings whl'le

A

(2,)

ti ]

from

R

into

which

R L,

are correct wards firom

~ ,•. ,

have

the

form

J.l k'

4.5.8. Let us first show how theorem 4.5.6 results from it. The left hand- part, v( x), of equality (19) does not turn identically to zero on R, since otherwise, using corollary 2.5.8, we would immediately get all a j equal to zero. Equality (19) shows that v(r)

is contained in S (2)

suitable nonzero ideal I of R. On the other hand, the words d,]

for a act on

in a trivial way and, hence, V(r):2 V(SI) = SV(r), i.e., v(I) generates a left ideal A of the ring S in S. Going over to a ring which is antiisomorphic to R, we can also find a right nonzero ideal of the ring S, contained in S. Now ASB is a two-sided ideal of the ring S, contained in s, where ASB:t= 0 by theorem 4.3.8. S

4.5.9. Lemma. Let S be an almost intermediate ring. Then there is an almost intermediate ring derivations

~,and

Proof.

51

Let

=

SI ~

s, such that

J.l., S 1 '~ R

for

all basic

5,

us

consider

a

subring

SI={SESlsJ.l.iER,

219

CHAPTER 4

1 ~ i ~ n). As D S (R ) I

= D s(R ),

elements I

EF

SI:2 S

R L,

(1

then

then there is nothing to prove. Let, therefore, there be

S, such that J.I. in such a way that s i I

J.L

is an almost intermediate ring. If

SI

s

E L, S E

s:

Jl.

:f:.

JI

0, S 1 =

R

J.I.

(11) to the ring I). In this case (sf(J»

o.

Let us choose an ideal

for all i, 1 ~ i ~ n. Let, then, J f(J) s: S and f(J) s: I (applying formula

be a nonzero ideal, such that

Therefore, (sf( J» J.I. =

o.

1

s:

R

and, hence,

sf(J)

s:

SI.

This affords s J.I. f( J) = 0, i.e., by theorem 2.2.2,

s J.I.= 0 is a contradiction. Proof of proposition 4.5.7. By lemma 4.5.9, we can assume S

J.I.

1s:

R

L

Let f( x) = x rewrite the constant as

6

j C

1

1)

L( L(xr ..)

.

~

]

.

~]

!: S.

j

Using formulas

(7)

and (8) we

can

(2)

J ).11

= t( x)

(20)

Let us replace the right-hand part of equality (20) with

Lq -rJx ) s q

and, starting from it, prove equality (19). Let us choose among words /::,. .

10

.

(2)

braces, a semor word /::,. v

~ xrvj)

6

j

,

for

which

,

(2) i

'

for which correct words can occur (1)

and let /::"W

rvw :f:. o.

If such

be a senior word in the sum a

choice

is

impossible

to

]

implement, then all words /::,. (~) are empty and equality (20) turns into (19). This peculiarity is a foundation for carrying out induction over /::,.~) . An induction step will be carried out by induction over /::,.~). In other words, we shall transform equality (20) to the same form but with a (1) .

smaller word, /::"W

10

(2).

braces at /::,. v , and

10

such a way that for the words

/::,.(~) > /::,. ~) only empty words could be encountered in braces. ~

Let

f3 = L r A. ®

sA. E

r;~!: l. op(S). Let us apply the operator

* f3

to both parts of (20). Allowing for the fact that the words /::,.(~) act trivially

220

AUTOMORPlllSMS AND DERIVATIONS

on S, we see that only empty words will still be encountered in braces for (2)

(2)

the words d i > d v the words d rvw .

.

Let us, usmg identities (9), transform the braces for

.

(2) (2) i ::; d v . Then the coefficient under

f3 = 0,

(1)

.

dw

WIll be equal to

and we can make use of the induction supposition, having

,

showed that not all coefficients zero.

r vj of the newly obtained expression are

Let us consider a coefficient

for the word

d

(1) j(O)

(1 )

= :1 w

,

where, as usual,

Since all the words

d

(1) j

do no exceed

(1)

dw ,

then the

word

(1) J1t:1 w,

decomposition (9) can appear only for the words of type

J1t

either have

= 0,

or

Jlt::; J1l' Let

_(1) J1t d w =

(1) d j( t)'

(1)

:1 w

in

where

Then, by formula (9), we

Allowing for corollary 4.4.5, let us apply lemma 4.1.3 to the ring Rand the subring S, assuming that r l = m1 r vw, n = 1, a l = r vw' In this case statement (a) yields

r' vj(O) :t:. 0, which is the sought result.

Let us show that statement (b) cannot be fulfilled. Indeed, it means that ;

E

J1 =

for

a derivation

RF ,

such that [s,

;1 = 0,

[f(x),

S J.1= O.

i.e.,

Therefore,

;1=

I

J1 jet) c t

sJ.1 for all

there

can

SE S.

be

Setting

by theorem 2.2.2, the element

J1

E D s(r) =

found

J11 c + ._ + J1 k C .

On

an

element

s= f(i), we get

;

lies in the

other

c

and hand,

J1 belongs to the subspace J1 k + 1 C + .- + J1 n C and, as all J1 i the basis of L, then J1 = O. This fact contradicts the linear

however, form

independence of the elements ( J1 1' J1 j(t)}' The proposition is proved.

4.5.10. From theorems 4.5.4, 4.5.6 and lemma 4.5.1 one can now easily deduce the correspondence theorem formulated above.

CHAPTER 4

221

Indeed, by lemma 4.5.1, for the Lie a-subalgebra AS;; L the ring of constants RA is an intermediate rationally-complete subring in R. Inversely, if S is an intermediate rationally-complete subring in R, then, by theorem 4.5.6, one cam find a nonzero ideal H of the ring S, contained in S, and, hence, for any S E S there is an inclusion Hs S;; H S;; S, i.e., due to rational completeness, s E S. Therefore, S = S is a ring of constants for the Lie a-algebra Ds(R). Finally, by theorem 4.5.4, for any restricted Lie a-algebra A in L we have the equality D iR) = A, which fact R

completes the proof of the correspondence theorem.

4.6

Extension

of Derivations

Let us now consider the problem in what cases the extension S ~ R L is a a-extension, where S is a rationally-complete subring. 4.6.1. Definitions and notations. Let A be a Lie a-subalgebra in L. Let us consider an idealizer I( A) = = ( J.l. ELI [A, J.l.] ~ A) of this algebra. One can easily see that I( A) is a differential Lie C A -algebra, but can be not a space over C. Let us say that A is a quasi-ideal in L provided I( A) C = L. Obviously, if L acts trivially on C, then any quasiideal is an ideal and vice versa. 4.6.2. Theorem. An intermediate ring S, R ~ S ~ R L will be a of the ring of constants R L iff the corresponding to it Lie subalgebra DS
a-extension

a-

This theorem will be deduced from the following one. (Extension). Any derivation of an almost Theorem 4.6.3. . intermediate ring S in R, mappmg R Ln S to zero, can be extended to a derivation from L. Proof Let us assume that J.1:. S -7 R be a differential mapping, such that (R

Ln S) Jl=

0, but J.l. cannot be extended to a derivation from L.

L

t::.

Let us consider a trace form f( x) = C j x J ~ S. By theorem 2.2.2, this form does not tum to zero on R. For any

222

S E

AUTOMORPfUSMS AND DERIVAnONS

S

the following equality can be written: ~

[""",x

~j

c.s] ]

J1. ~ ~ J1. -(""",x jc.s )=0 ]

(21)

which is valid for all x running through a nonzero ideal of the ring R. One can assume that an element s can be chosen in such a way that the expression in braces does not turn to zero on R. Indeed, if such an element cannot be chosen, then we get t( x)S J1.= 0, which yields t( x) . R L. S J1.= 0, which contradicts theorem 4.3.8. It would be more convenient to reduce to a contradiction a somewhat more general relation (which holds by formula (7» df

If'(x) = [

~

.""'" (xb.) ] ] E J

..1 j

]

J1. -(

~

""",(xrv )

~v

(22)

)=0.

vE V

Let us assume that in this relation the expression in square brackets assumes the values in S for any x from a suitable ideal of the ring R. Let us also assume that all the elements b j' r v are nonzero elements from R C, V:F. 0 . Let us consider a set U of the words (22), such that

flv:F.~~),

~v

occurring in presentation

i.e., with their first letters not lying in

Ds(R).

By induction over a senior word in the set u u ( J1.!l j' j E J), and assuming that J1.I > ~ > M. > J1. n > J1., let us reduce equality (22) to a contradiction. The basis for the induction is the case when

U

u (J1.!l j' j

E

J)

= 0. In this case the expression in braces disappears and we see a contradiction to corollary 2.5.8. Let us carry out an induction step. Let us first assume that the senior word fl is encountered in braces. Then all the words ~ v' v E V which are greater than fl and encountered in braces will be the words from J1.I"'· J1. k' i.e., they will act trivially on S. Therefore, for any

f3

E

n

vE V

IS

LV

where Vo is a set of those indices v

E

!:: L

op (S), at v E

"0

we get:

v, for which the word

~v

is

greater than ~ . Therefore, the application of operator * f3 to the expression in braces results in the disappearance of all the terms containing in their presentation

223

CHAPTER 4

words greater than !J., and new terms do not appear from them. The term containing the word !J. = !J. v(O) is also obvious to disappear, but new terms will arise from it. Let us consider the terms arising when applying the operator * f3 to relation (22) and containing in their presentation the word A , where, as usual, !J. = J.l. s A = J.l.~s .•• (see formulas (5), (9». The word A can appear in decomposition (9) only from the words of type

J.l.tA, where

J.l.t ~ J.l.s. Let

J.l.tA = !J.v(t); J.l.A = J1ll j(O). Then, making allowances for the fact that [A] Il

* f3

= [A *

f3]

Il_ (A

* f3

Ill, '1'( x) *

we can write the term of the expression presentation A: x( mSrv(s)·

f3

Ils

""

+ L.t rv(t) . t

f3

Il t

-

r j(O)·

f3

containing in its

f3 Il),

in which case the last two addends may be absent (provided the words !J. v( t) in braces and the word A in square brackets, respectively, do not participate in (22». Now we choose

f3

e (l

v

rt. S

in such a way that the written term

does not turn to zero. Indeed, in the opposite case, by lemma 4.1.3, the

L

linear combination J.l.s + J.l.tCt + J.l.Co acts trivially on S, in which case the corresponding element commutes with the elements of an almost intermediate ring R Ln s. By lemma 4.5.3, we get

If in this relation

of derivation

Co *" 0, then - J.l.sc 1_ L J.l.tctcQ 1 is an extension

o

J.l. on R. If c O= 0, then we have a contradiction with the

linear independence of the derivations (J.l.s' J.l.t) • Therefore, choosing f3 in the above mentioned way, we see that the inductive proposition can be applied to the expression '1'( x) * f3. Let us now assume that the senior word has the form J1ll = J.1.ll j(O). Then all the words in braces with their length greater than !J. will be the

224

AU1OMORPHISMS AND DERIVAnONS

words from

J.; ,_. J.l k and, hence, choosing

PE

(J b.i~ fi (J r~S, j ] v

we see

that the senior word for '¥( x) * P gets smaller and, second, the term in braces containing the word Ll will have the form x( b j(O) .

Jl:

P ).

By lemma 4.1.3, one can choose P in such a way that this term does not turn to zero. If we now use the inductive proposition, then the theorem is proved. 4.6.4. Theorem (Rigidity). Any monomorphism of an almost intermediate ring S to Rover R L is identical. Proof. Let J1: S ~ R be an monomorphism acting identically on L

S fi R • Then for any trace form

~

we have [f(x)] - (f( x) } = 0, and we get a relation of type (22). Let us deduce this relation to a contradiction by induction over the length of the senior word occurring in braces, assuming that the expression in square brackets lies in S and does not turn to zero on R. If the expression in braces is not present, then the contradiction results from the fact that the kernel of J.l is zero. . an aImost mterme . d'late nng. . As S ~;;;;? R L fi S , then S ~ IS If

PE

Po~ = p.

that

(J

v

xi s,

f( x)

then we can choose an element

f30

E

lL

op

(S), such

.

In thIS case we have

'¥( x)

* P = [ T( x) * Pol ~ - ( Tl (x) * P} .

I

is obvious that the senior word in braces for '¥ ( x) * P has a less length and it WOUld, therefore, suffice to choose P in such a way that the expression in square brackets does not turn to zero on R. Allowing for the fact that the senior term,

T( x)

* PO'

is equal to (x (b j(O) . Po»

ll.

j(O)

,where

Ll j(O) is a senior word, we come to the conclusion that it suffices to choose

P

in such a way that b· f30 *- 0 for a given nonzero b = b j(O) . Let us suppose that it is impossible, i.e., that for ~

m .is POE [;lb i

Jl

we have b· Po=O. Without any violation of generality

any

225

CHAPTER 4

one can assume that (b J are linearly independent over C elements, while the

number

J.L dE B 0 =

m

b}

. I Sll

(1.t5""J.

~=

/30 E BO

~:

is the least of all possible,

m

and,

2

and

~ -+

by lemma 4.1.2,

t:= 0,

b}/3

b· /30' where

then

Po

i.e.,

J.L bl·B 0*0'

Po = ° and,

b·

runs through

b· Bo

* 0,

where

In this case, if

hence,

the

mapping

B O' is well-defined. One can

easily see that ~ is a homomorphism of left R-modules. the domain of whose definition is an (R, 5 I) -bimodule and, hence, by corollary 4.4.5, it contains a nonzero ideal of the ring R. Therefore, by the definition of a ring of quotients, one can find a nonzero (as such

that

/31 = /3o(s ® 1)

for E

So

Herefrom we have

bB 0

all

* 0)

element

If

S E

t E RF '

5,

then

and, hence,

J.L J.L b l /3 o(s t - ts)

=°

and,

as

blBO

contains,

by

corollary 4.4.5, a nonzero ideal of the ring R, then sJ.Lt = ts for all s E 5. In particular, choosing s from R L and using lemma 4.5.3, we see that t is a nonzero central element, i.e., the latter equality can be cancelled by t. Thus, s J.L= s for all s E 5. The theorem is proved. Proof of theorem 4.6.2. Let us assume that a quasi-ideal D s(R) corresponds to the subring 5. It means that in the idealizer I( Ds(R» one can choose a basis of the space Lover C. Let us, therefore, assume the basic derivations J.l.l ,..., J.l. n to lie in the idealizer. Let us consider a restriction j1 of an arbitrary derivation J.l. E I( DS
I

of 5, such that

one can find a subring 51 r;;;. 5, such that

I J.Lr;;;. 5. By lemma 4.5.9, J.L -51 r;;;. Rand 51 = 5. By theorem

4.5.6, one can find an ideal A of the ring

f.l.t E

Ds(R),

then

Jl... J.L [ IJ, J.L] J.L J.L (51) 1r;;;.5 1 1+(5}1) =0,

can be chosen as I.

[J.l.,

f.l.t1

ED;' R)

and, hence, 51J.Lr;;;. 5-

S lying in and,

51'

If now

hence,

and, thus, the ideal

A

2

226

AUTOMORPlllSMS AND DERIVATIONS

Therefore, the mapping J.L -+

fi is a homomorphism from

I( DS
on a restricted Lie subalgebra L of D(S), in which case S L = R L, as I( Ds(R » contains the basis of L. It is also evident that the kernel of the latter homomorphism is equal to D S(R) and, hence,

As far as the homomorphism of the restncUon is C L -linear and L is finitedimensional over C L, then L is also finite-dimensional over C L. Let now, vice versa, S L=

L

fi I

for a finite dimensional Lie a-algebra

RL

!;;;;; D(S)

E

jl

Let us now show that L cannot contain inner derivations. Let L. One can find a nonzero ideal I of the ring S, such that

~

certain

fi coincides with a

S !;;;;; R and, hence, by the theorem of extension,

J.L E L on

well. Indeed, if s

I

+

R L.

It means that

i E I, then

E S,

(si) Jl.

fi coincides with

= (si) jl

and

J.L on S

= si jl,

si Jl.

as

i.e.,

by theorem 4.3.8, we get fi = J.L on S. Let us assume that ~s - s~ = sJl. for a certain ~ E C(S) and all s E S. Having chosen an arbitrary element a from the ideal I of S, such that I~!;;;;; S, ~I!;;;;; S, we get the equality b 1sa - asb2

= as Jl. a,

where b 1 = a~,

basic derivation, not lying in where L\

b.l = ~a.

D S
Let us assume that J.L is a ~

a trace form f( x) = Lx

are some words from basic derivations of

j

j C

j'

Ds(R ) , not turn to

zero on R, and assume the values in S at x E J, where J is a nonzero ideal of R. By theorem 4.4.6, we can find a nonzero ideal A of the ring S lying in S. We have: Af( x) !;;;;; S, and, hence, we get the following identity for all d E A: b1(d· f(x»a- a(d· f(x»b2

= a{df(x»

Jl.

a.

Using theorem 2.2.2, we see that a(d ® C j) a = 0, and since a is an arbitrary element from I, while the element d is also an arbitrary element from A, then IA = 0, which contradicts theorem 4.3.8. Let ~ be a set of all derivations from L whose restrictions onto S

belong to

coincides

L. Then the kernel of the natural homomorphism

with

D S(R),

and

the

image

is

equal

to

L} -+ L

L, i.e.,

CHAPTER 4 D L(S) R

= -1 T~ /

227

Ds(R), as by theorem 4.5.4, we have

L

= D R L(S).

Then, the Lie d-subalgebra Ll C of L has only elements from R L as its constants and, hence, by the correspondence theorem, Ll C = L. The task now is to show that

[Jl.t' J1.]= [ .lIt, .al = 0, Jl.t E I( Ds(R» and (S

J.l.1 J.l. [ J.l.j , J.l.] ) !;;;; S

L} = I( D s(R» .

.a = ° and,

as

J1.

E

JL.. J.l.j )

+ (S·

If

hence,

and, hence,

S

[J1.,

Jl.t]

lJ1.l' J1.] = 0,

then

Ds(R),

J1. E DS(R),

J.l.1

!;;;; R

Ds< R)

Jl.t

E Ll ,

then

E D

J R).

If

which -

= S,

so that, using

theorem 4.5.6, one can find the ideal A of the ring S, such that or, in other words,

.al E

D R L( S) =

proved.

REFERENCES R.Baer [11 l;

A.Berkson [22]; N.Jacobson [63]; V.K.Kharchenko [75,771; M.Weisfeld [153].

L, i.e.,

Jl.t

E ~.

affords

The

A

J.l.1

!;;;;

theorem

S

is

CHAPTER 5 THE GALOIS THEORY OF SEMIPRIME RINGS In this chapter we shall, using the metatheorem, transfer the basic results of the Galois theory for automorphisms and derivations onto semlpnme rings. The task is, in essence, purely syntaxical, i.e., to reformulate the results in the form of truthfulness of Horn formulas. One can easily see that the basic theorems cannot be formulated in an elementary language, which makes us analyze not the theorems proper, but the key statements in their proofs. Let us begin with considering the simplest case of Lie a-algebras of outer derivations where the situation is much simpler since derivations (unlike automorphisms) act trivially on central idempotents. Considerations of the inner part will be postponed until automorphisms have been studied. Let us fix notations for a semiprime ring R, its left and symmetrical Martindale rings of quotients R F , Q and for a generalized centroid C. By we shall denote a set of all idempotents of the ring C.

5.1

Essential

Trace

E

Forms

5.1.1. Definitions and notations. Let L be a finitely-generated C-submodule of derivations from D( R), closed under the operations of a restricted Lie a-algebra, i.e., [ )1, )11] E L, )1P E L, if )1, ~ E L, where

o< p is a characteristic of the ring R. Let us also assume that L contains no nonzero inner for Q derivations. By lemma 1.6.21, we can expand L into a direct sum of cyclic submodules L = ~ C <37 ._ <37 )1 n C. In this case ~ ,_.

)1 n

will be strongly independent: if L)1 i C i

= ad~,

as L contains no other inner derivations and, hence, Let us also fix a Lie

a-algebra

The linear form

L

m

f =

)11'"' )1n; C j E

)1 i C i =

= 0,

0, 1 :;::; i:;::; n.

and its C-basis ()11 ~-, )1 n) .

L x '1

L'l. j

C "

J=

words from

then L)1 i C i

]

where .11 < _. < .1 m are correct

C will be called an essential trace form for L,

228 V. K. Kharchenko, Automorphisms and Derivations of Associative Rings © Kluwer Academic Publishers 1991

229

CHAPTER 5

L /' J

f( x) Jl. = (

if

C

j) Jl. = 0

sup [e (c j) e(L1 j») = 1 .

and

It

should

be

]

recalled

that

x. e( P-t .. Jl k) = e(

P-t) •.

a

is

e( x)

support

of

element

the

e( Jl k)'

5.1.2. Proposition. The algebra L has an essential trace form. Proof. As the number of the correct words is finite (it equals m = pn) and any word can be presented as their linear combination using the

Jl1 Jl = Jl~ + [ Jl1• Jl]. JlP = ¢ p].

substitutions L1

j

Jl i =

LL1 cli 1 and. k k

then

we

have

hence.

~ ~ Jl.i ~ (k) ( L"L1 . C .) Jl . = L" L1 . c· + L" L1 k C . C . . • ]

.

]

]

~

]

.

]

]

k,'

i.e.. the equality f( x) Jl. = 0 will relations for the coefficients C j :

Fk

•

~) = .( C

~

]

]

~J

result from

~ k) -- 0 • c kJl.1 + L"C .c( .. . ] ~J

the

following

system

of

k = 1 ••.• m. i= 1 •.•• n.Now the

]

condition of the existence of the essential trace form is written as a Hom formula:

m 3 ~ ..... cmV' x & p(c .)& & j= 1

]

k, i

Fk,

.(2)=0&

~

&[(&j xe·c .=O)~ x=o] ] ] •

where

ej

= e (L1 }.

while by P we denote a predicate which singles out a

generalized centroid p( x) = T <=> X E C. Since C is a closed subset in Q. then -it is a strictly sheaf predicate (see 1.11.9). in which case. by lemma 1.12.6. P( x) = T on a stalk r Q) iff x belongs to the generalized centroid of the stalk.

sJ

-

_

~-

As the relations L1 j J.l i = L"L1

_(k) k C ij

are fulfilled for the derivations

induced on the stalk, then, by lemma 4.3.3, the above formula is also valid on all the stalks and, by the metatheorem, it is also valid on the ring Q of the global sections. The proposition is proved.

AUTOMORPIDSMS AND DERIVATIONS

230

5.1.3.

Corollary.

Let

a, b

R F' and let I be an essential ideal

E

of the ring R. If aI L b = O. then e(a) e( b) = O.

Proof.

/:1

f( x) = Lx J c·]

Let

be

an

essential

trace

form.

As

D(I) = D(R) (lemma 1.8.4), then one can find an essential ideal J, such

that

J

n/:1

~

for all basis derivations. Therefore, ( J p)

(;;;; I

J (;;;; I

n

and, hence,

the differential identity af( x) . b = 0 holds on J P . By theorem 2.2.2, we get

which

e(c)·e(L1)a0b=0,

e(a)·e(b)sup(e(c.)e(~

j

]

5.1.4.

Theorem.

]

affords

e(a)·e(b)=

.))=0. which is the required proof.

Let

J.l be a derivation from

D( R). such that

(R~~=O. Then J.lE L.

Proof. Let us consider a c-module L + J1.c. By lemma 1.6.2, the submodule L is obtained from it as a direct addend L + J.lC = L Ef1 J.loC =

11. C Ef1 .- Ef1 J1. n C Ef1 J.loC. 11. < ~ < _. < J1.n < J.lo. n

L

If

i= 0

J1..c. = ~ ~

Let

ad~

us

bring

then

,

the

in

order

the

~

element

basis

derivations

commutes with all

constants for L. Let I be an essential ideal, such that for the essential trace form f the inclusion f(I) (;;;; R holds. Then on I we have the identity ~f( x) -

f(x)~ =

e(L1 .) = 0

for

J1. i c i = 0, 0

s: i s:

]

Thus, I

O.

i.e.,

every

by

Therefore.

j.

~

E

we

get

(~x - x~)e

C

and

ad~ =0,

(c j) i.e.,

n, since the sum of the basic submodules is direct.

Jl.o, J1.t '-' J1.n

is a strongly independent set of derivations and on

we have a reduced identity

~/:1

2.2.2, for every j

~

~/:1~

O=LJx J c .D+LJX ]

theorem

2.2.2,

theorem

we get

e( J.lo) = O. The theorem is proved.

.

J. ]

j

Dc

]

.=

e(c}e(L1}e(J.lo)=O,

f(x)

Jl.o . By

and, hence,

231

CHAPTER 5

S.2

Intermediate Subrings

Our nearest aim is to transfer theorem 4.5.6 into a semlpnme case. Preserving the notations of the preceding paragraph, let us fix an intermediate S, i.e., R:::::> S:::::> RL. Let § = RA, where A= Ds(R) = ring (JL E D(R)I JL(S) = 0). By theorem, 5.1.4. the inclusion

AS;;; L is valid.

S.2.1. Lemma. A is a finitely-generated submodule of the C-module L. Proof. Let us note that A is a closed submodule of the complete

module

Po=limJL a and

L. If

any a and any

s

E

S

JLa E A, then

and, hence,

eaPo(s) = eaJLa(s)=O

Po(s) = 0

and

Po

E

at

A. By lemma

1.6.16, the module A is injective and, therefore, it is obtained as a direct addent from the finitely-generated module L and is, hence, finitely-generated itself. The lemma is proved. 5.2.2. Theorem. The ring S contains a two-sided ideal of the ring the annihilators of which (both left and right) in R F are zero. For the metatheorem to be used, it is necessary to single out elements from the set S. For this purpose it is natural to introduce a unary predicate determined by this set (see 1.11.9). This predicate, however, will not be strictly sheaf if S is not orthogonally complete. Let us consider a closure S,

A

SE

and a strictly sheaf predicate

case the value of 5.2.3. (a)

If

(b)

PS

on the stalk

PS

r

A

determined by the set

SE.

p is determined by the set

Pp(

In this A

SE ).

Lemma. JL E D( R) and

P p(

A

SE)

j.J(S) = O. then

A

j.J( SE ) =

o.

is an almost intermediate subring on the stalk for the

d-algebra of the induced derivations L = p ,J.L). (c) The set of derivations of the stalk

r

p

over

A

p p( SE) is equal

to A = Pp(A). (d) If SESE. then one can find an ideal sI

u

Is

~

S and annihilators of I in

I

of

R L. such that

R F are zero.

Proof. (a) As all derivations act trivially on the central idempotents (2.5.2) and are continuous (1.6.3), then the equality j.J(S) = 0 implies j.J(SE) = 0

232

AUTOMORPIDSMS AND DERIVATIONS 1\

and JlI.. SE )=0. (b) Let f( x) be an essential trace form. Then f(I)!:: S for a certain ideal I E IF. It affords f( IE) !;;;;; SE and, since f is continuous, we have

1\

1\

1\

P p( IE) is a two-sided

f( IE ) !:: SE. Now we are to recall that 1\

ideal of the stalk Pp( RE) (see 1.12.5), while the induced form a trace form for L in the stalk. (c) As L is a restricted Lie

a-algebra acting on the stalk, while

1\

P p( SE) is an almost intermediate ring, then any derivation

p p(

which acts trivially on

1\

SE)

I will be

L, i.e.,

belongs to

jl

jl of the stalk

= P p( J1),

J1

E

L

(see 4.5.4). Now the truthfulness of the formula 'V x( p S<x) -+ jl( x) = 0) on the stalk implies that it is also true on the ring of sections over a certain neighborhood of the point p, i.e., there is a central idempotent e ~ p, such 1\

that eJ1( SE ) = 0, i.e., eJ1 E A and jl = P p( eJ1)

_

EA. 1\

(d) Let U be a set of all elements s from SE satisfying condition (d). Let us show that it is a closed set containing SE. lf s

= LSi e i

and Je i!;;;;; R, then by corollary 5.1.3, an ideal J L

has zero annihilators in

RF

and

J

L

ei

~ S.

Therefore, if

S

i E S,

then

SJLU JsL~ S.

Let now

S = lim sand

a choose an essential ideal 5.1.3, the ideal

J

L

L

of

(eaq)I aJa=O

I

in

U

I a s a ~ S. For every

J a, such that

ea J a

has zero annihilators and

SaeaJ~IaVeaJ~IaSa!;;;;;S annihilators

saIa

are

RF

for all

a

R.

if,

and, hence,

for

let us

Then, by corollary

L L eaJ a~ R .

I=IeaJ~Ia.

Let zero:

~

a

instance,

Therefore, Then

the

qI = 0,

then

q=sup (e a ) q=O. a

Moreover,

SI!;;;;;ISaeaJ~Ia!;;;;; S, and analogously, Is~ S. The lemma is proved. a Proof of theorem 5.2.2. Let us expand the module A into a direct sum of cyclic submodules A = J11 C $ ._ $ J1 k C. Let us recall that the proof of theorem 4.5.6 was based on proposition 4.5.7, which can be written on the stalk parameter t:

r

lP as the formula c!>( t)

which depends on a natural

233

CHAPTER 5 t St' r 1 , ... , rt'if x & Ps(s)& i=1

3 a1, .•• , a k, sl"'" p

¥

&

L1(~)

L

(xa)

j= 1

t

L

i= 1

(2) f(xr i)s i & sup ( e(a )e (.1 j )} = 1, j

(2)

where A j are correct words from J1I'-" J1 k' while f( x) is an essential trace form which can be considered as a strictly sheaf operation. On every stalk

r

p

a formula ct>( t(

p»

is true, which results from

prOpOSItIOn 4.5.7 and lemma 5.2.3. By corollary 1.11.21, we see that formula cI>( t) is true for a certain /\

on the ring RE. Now the proof is completed, as was the case in theorem 4.5.6. Let us set t

pk

F( x) =

L I (xa ] .)

(2)

L1j

.

J=

and choose an essential ideal

I

besides, A be an ideal of the ring

in such a way that RL

f(Ir i) !:,;; R.

with zero annihilators in

Let,

R F' such

that S i A!:,;; S (see lemma 5.2.3). Then V = F(I) A is a left ideal of the ring S contained in S. Its left annihilators is equal to zero: if qF(I) A = 0, then qF( x)a = 0 is a differential identity on I and, by theorem 2.2.2, for all j valid is e(q)· e(a) e(A )e (a) = 0, which affords e(q)a = 0 and, since a is an arbitrary element from A, then e(q) A= 0, i.e., q = O. In an analogous way we shall find the right ideal w of the ring contained in S, the right annihilator of which in

RF

S

is zero. Now VR LW

is the sought ideal: if, for instance, qV R LW = 0, then by lemma 5.1.3, for any VE V, WE W we have e(qv)e(w)=O, which yields we(qv)=O or We(qv)=O, i.e., qv=O and, hence. qV=O and, therefore, q=O. The theorem is proved.

234

AUTOMORPHISMS AND DERIVATIONS

5.3 The Correspondence Theorem for

Derivations

a

-algebra of outer 5.3.1. Theorem. Let L be a restricted Lie derivations from D( R). Let us assume that L is finitely-generated as a module over c. Then the mappings A ~ R A, S ~

a

D

f. R)

set a one-to-one

-subalgebras of L, which are correspondence between restricted Lie finitely-generated over c, and intermediate rationally complete subrings. In this case an extension s!:: R L will be differential iff D s(R) is a quasiideal of L. It should be recalled that the subring S is called rationally complete if the inclusion Hr!:: S implies rES for an essential ideal H of S and an element r E R. If A is a Lie a-subalgebra of L, then it is called a quasi-ideal if its idealizer I( A) = (1 E LI[l, AJ !:: A) contains a basis of L, Le., I( A)c = L. This theorem is deduced by standard considerations from theorems 5.1.4 and 5.2.2, as well as from the following extension theorem. 5.3.2. Theorem. Let S be an intermediate ring, Jl. S ~ R be a differential mapping acting trivially on R L. Then J.l is extended to a derivation from L. Proof. Let us first extend J.l to a differential mapping from S E into

Let

R E .

us

set

J1{ a) = L J.(s

pe

for

a

linear

combination

u=Lsie i , si E S, eiE E. This is a correct definition: if LSiei=O,

then for any constant and, hence, L

j E R L, such that

J.l(s)eij=L

E R,

we get

ei j

J1{ s i(ei j » = = J.(LsieiJ)=O,

corollary 5.1.3, we get L J.l( Si)e i = differential E-linear mapping.

"

ei j

o.

s =

Le., by

Direct calculations show J.l to be a

Let us extend J.l to the differential E-linear mapping from

RE. Let

E R L~ S

"

SE

into

lim sa and let J1. values on sa be determined. Then, since

aE A

" is complete, there is a lim J.(s a)' which will be henceforth the module RE A

assumed t =

to

be

J.(s).

Let,

simultaneously,

s =

IW

lim 'J.( t p). Then (J.l(s)- t)eae p = J.l(sa)eae p - J.l(tp)eae p = J.(saeaep- tpe a )= J.l(seae p - sepea)=O

t

P and we set

235

CHAPTER 5

aE

for all

A,

f3

As

E B.

sup AX B

e a e p = 1, then

and, hence, the

J1(s) = t

definition of "'s) is correct. The differentiality of Jl is directly checked: if s= l~ sa' t= li: t p ' st =

lim sat

AX B

then

[Jl(st)-

- "'s)t- s",t)]eaep=O,

(.I'

f'

The next step will be an extension of

A

as

Jl to the E-linear

A

mapping

A

jl: RE -+ RE. For this purpose let us consider

SE as an E-module

(see

A

remark 1.6.9). Since SE is complete, then it is an injective E-module (see 1.6.16) and, by the definition of injectivity (1.6.15), the homomorphism of E-modules

A

j1: SE

A

-+ RE is extended to a homomorphism

A

A

A

Jl: RE -+ RE. A

Let us now consider {l as a unary operation on RE. Due to Elinearity, this operation is strictly sheaf, its action on the stalks being of such A

a character that the restriction of (l on p p( SE) (see 5.2.3, (b» will be a derivational mapping acting trivially on the constants of the almost intermediate ring

A

p p( SE ). By theorem 4.6.3, the restriction of

A

P p( SE) will be extended to a derivation from the following formula is valid: 3 cl'-"

C

nV x

-+ Jl-( x ) =

where

C1 x

n

&

p c( c)& (p

i=l

111 +

_. +

L=LJliC, Pc(x) = T<=>

/l

on

Pp(L), i.e., on every stalk

s< x) -+

Cn x

I1n)

XE

c, Ps(X)= I<=>

A

SE.

xE

By the metatheorem, this formula is also true on is proved.

A

RE. The theorem

5.3.3. Theorem. Any monomorphism Jl of an intermediate ring S into Rover R L is identical. Proof. By analogy with the above theorem, let us extend Jl to an A

E-linear monomorphism A

A

(l: RE -+ RE.

Then

j1: SE

on

A

-+ RE, and then to an E-linear

every

stalk

a

Horn

formula

(l(x) = x) will be true. The validity of this formula on required statement.

A

mapping

V x( P s(x) -+

RE will give the

236

AUTOMORPIDSMS AND DERIVAnONS

5.4 Basic Notions of the Galois Theory of Semiprime Rings (the Case of Automorphisms) 5.4.1. Let us start with a simple example, vividly illustrating vanauons in some notions of the Galois theory when going over to a semi prime case.

II

F a be a direct product of isomorphic fields Fa == F ae A and H be a finite group of automorphisms of the field F. Let us determine the action of H on the product R in a componentwise manner,

Let R =

fa~_)h = = C."' ~~_). On the other hand, on

fh(a) = Cf(a»h, i.e., (.",

one can naturally determine the action of the direct product G = the

groups

(.", fa~-)

Ha

(_., ga"")

isomorphic

= = C."'

H:

to

II

ae

Ha

R

of

A

fgca)=f(a)g(a),

i.e.,

ga fa ,_.). In this case the subrings of fixed elements

for the group H and for the group direct product of copies of F H .

G

coincide and are isomorphic to the

5.4.2. Reduced finiteness. The most natural and direct transfer of this notion is as follows: a module B (G) is finitely generated over C and a factor-group G / Gin is finite with respect to a subgroup of inner for Q automorphisms. Such a definition, however, excludes from consideration in the above example the group G, which is a Galois closure of a finite group H. By the same reason we have to transfer to a "local" variation of the same notion. It should be recalled that for an automorphism g e A (R) of a semiprime ring R we have determined a conjugation module, 4>g=(ae

d"t xe

R xa= ax g ),

4>g = qJ gC, the support of which,

which e( qJg),

action of the automorphism g on i(g)Q the invertible (in i(g) Q) element qJg. automorphism g will be inner on e Q, Ge =(g e GI i(g) ~ eJ is a subgroup of 5.4.3.

is

a

cyclic

has been denoted by

2:

ge G

module and sup( ell G: GcJ <

00

J = I,

i(g). The

coincides with the conjugation by Moreover, for any e ~ i(g) the which, in particular, implies that the group G (for a fixed e).

Definition. A group of automorphisms

reduced-finite if its algebra B(G) =

c-submodule

4>g

G ~ A (R)

is called

is finitely-generated as a

C-

CHAPTER 5

237

5.4.4. Closure of a group. Returning to example 5.4.1, we should remark that the action of every g E G on cofactor F a coincides with that of a certain h E H, which is a -dependent. It is this peculiarity that results in a coincidence of the invariants of G and those of H. Under general conditions we, by analogy, come to the notion of a local belonginness to the group: the automorphism g locally belongs to the group H, provided there is a dense family of idempotents (e a E Cl a E A), such that the action of g on eaQ coincides with that of a certain ha E H, i.e., sup(e: gleQ

E H1eQ )

= 1.

5.4.5. Lemma. If the automorphism g locally belongs to the group H, then gE Al!(H), i.e., g acts identically on QH. Proof. Let g action on eaQ coincide with that of ha E H. For g

H

any

g

g

r E Q we have ear = (e a r) = ( ear) · . d hence, d con Iuon we have e ha an, a -- e ag -1

ea(r - r g

)=0. Therefore,

the required proof.

ha

h

= e a a r, but, by the g( r g - r ) -- 0 , ea

-1

-1

O=sup ea'(r- r g )= r - r g a

. I.e.,

, which is

5.4.6. Definition. A group Q is called closed if any automorphism locally belonging to the group G lies in G. We immediately get the following corollary. 5.4.7. Corollary. Any Galois group is closed. 5.4.8. Let us consider in detail the notion of closure for the case when R has a finite prime dimension, i.e., Q = Q1 ED _. ED Qn is a direct sum of a finite number of prime rings (see 3.6.6). Let us start with an example somewhat more complex than that in 5.4.1. Let

't> ED _. ED 't>,

Q= l

T

n

where

't>

is a prime ring. If a group H acts

J

on 'b, then on Q we have, first, a product H n= H x _. x H acting in a componentwise manner and, second a group of permutations Sn' rearranging n the addends (ql ED .- ED q n> = q _ 1 ED _. ED q _ 1 • Therefore, the action n

(1)

n

(n)

of a semidirect product G = H n>. S n is determined on Q. It is obvious that G is a closed group, the inverse statement being also valid. 5.4.9.

Lemma. Let a closed group

G act transitively on the

238

AUTOMORPHISMS AND DERIYAnONS

components of Q (which is equivalent to the fact that Q is G-prime). In this case rings Qi are mutually isomorphic and they can be identified in such a way that G = H n>.. S n' Proof. Because of transitivity, we can find elements g i E G, such that

g

i

Q)

= Qi

+ l' 1 $; i

$;

n- 1

(in

particular,

(Ji i

= 1.

isomorphisms

Let us identify addends (i.e.,

(Ji j

elements

Qi

OJ)

at

with respect to the system of and

q i E Qi

(Ji i

will

q i) E Q j

considered to be the same). In this case the transvection (J ji~-'

mutually

j> i and (Jij =

isomorphic). Let us set (Jij = gig i + 1-· g j _ 1 at j < i,

are

Qi

be

(1~_, (Jij' 1~.,

1,

1) permuting the the i-th and j-th components locally belongs to G

and, hence, G contains the whole group of permutations S n. Let

be a subgroup of all automorphisms from G acting trivially

HI

on the first component

QI. To every automorphism

hI

E

HI

let us put in

correspondence an automorphism h, the action of which on QI coincides with that of hI' while on the other components it is identical. This automorphism belongs to G since the group is closed. Let H be a group of all such automorphisms. Components Qi are acted upon by conjugated groups

<1} H(J 1i; Hn

=

and, due to the closure, the group G contains a direct product

II CJ)} H(JIi·

Since we have identified

1::::; i::::; n

system of isomorphisms (Jij' then the groups

with respect to the

Qi

(JiiH(Jli

are also identified.

The groups H nand S n generate a semidirect product H n>.. S n' Let now g be an arbitrary automorphism from G. This automorphism in some way permutes components of Q, i.e., there is a .

permutatIOn

1!,

such

Qg _ i - Q n(i).

that

invariantly on all the addends which on Qi

Qi. Let

The

gibe

automorphism

- I

g1!

acts

an automorphism the action of

coincides with that of glC 1, while on the other components

they are identical. Then

g i E (Jli 1 H(JI i

and

g

= gr-

g n 1C E H n >.. S n'

The

lemma is proved.

ring

5.4.10. Note. It is obvious that in the preceding lemma the fixed QG is isomorphic to Q~, where the isomorphism carries out a

diagonal mapping

q

~

q(JII

+ ._ +

q (Jln

239

CHAPTER 5

5.4.11. Let us now consider the structure of an arbitrary closed group of automorphisms of a ring R which has a finite prime dimension. Let c:x R) = QI Ea ._ $ Q n and G be a closed group. Then every element g E G rearranges, in this or that way, components Qi' i.e., G acts on a set of indices (1 ~_, n}. This set falls into orbits II U 12 U _. U I k' Let

Qa =

i

L

E

Ia

Qi' 1 ~ a:S k. Then we can apply lemma 5.4.9 to the ring

Qa

Qa has the form Ha ),. Sn. where na is a number of elements of the orbit Ia. Now, since G is closed, we immediately get and to the group G, i.e., the restriction of G on na

Assuming, for simplicity, a

E

I a at 1 ~ a:S k, we also get

5.4.12. Noether groups (N-groups). This unchanged: a group G!;;;; A(R) is called an N-group invertible

element

of

its

algebra

18(G) =

2.

4>g

notion remains provided every determines

an

gE G

automorphism lying in G. It is easy to see now that any Galois group is a closed N-group. 5.4.13. Maschke groups (M-groups).The definition is preserved: a reduced-finite group G is called an M-group provided its algebra is semiprime. Let B be an arbitrary non-semiprime algebra with a unit over a regular self-injective commutative ring c. Let us assume that the module c B is projective and generated by a finite number of invertible elements. It is obvious that the algebra of any reduced-finite group of automorphisms obeys these conditions (non-semiprimeness, possibly, excluded). 5.4.14. Proposition. There exists a semiprime ring R with a generalized centroid C and a reduced-finite group of automorphisms G, the algebra of which is isomorphic to B, while a fixed ring R G is not semiprime.

240

AUTOMORPIDSMS AND DERIVAnONS

Proof. Since the module

C

B

contains a free submodule, then it is a

projective generating module in the category of modules over C. It means that the rings C and R = Endc B are Morita-equivalent ([36], theorem 4.2.9). In particular, R is a regular and self-injective ring. Therefore, it coincides with its complete left ring of quotients ([91], remark on p.152) and, moreover, R = R F' The center of the ring R is isomorphic to C ([36], corollary 4.36), so that C is also a generalized centroid of R. Let us define the embedding of B into R using the operators of right multiplication rb(x) = xb. For every invertible element bE B let us determine the inner automorphism of the ring R corresponding to the element r b . Let G be a group of such automorphisms. As B is generated by invertible elements, then IB (G) = r B == B, in which case the fixed ring of this group coincides with the centralizer of r B in the ring R. Since B has a unit, then this centralizer coincides with the ring of left multiplications 1 B' The latter ring is anti isomorphic to B and, hence, it cannot be semi prime. The proposition is proved. This proposition shows that if, in the process of going over to fixed rings, we wish to remain within the class of semiprime rings with confidence, then we should consider the groups the algebras of which are semiprime. The condition of semiprimeness of an algebra does not greatly restrict the class of the groups considered. If, for instance, in R we have no nonzero nilpotent elements, then the ring of quotients Q does not have them either, i.e., the algebra of any reduced-finite group is semiprime. Another important class of M-groups gives an analog of the Maschke theorem. 5.4.15. Theorem. If G is a finite group of automorphisms of a semiprime ring R having no I G I-torsion, then the algebra of the group G is semiprime. Proof. Let us assume that in b = Lq> gC g'

such that

!B (G)

bIB (G)b = O.

there exists a nonzero element

Let us consider a finite

Boolean

G

algebra generated

by

idempotents

~b),

~q>gCg).

If e

is

a

minimal

idempotent of this algebra, then for eb we have a presentation Leq> gC g' where g runs only through the automorphisms for which moreover, algebra of By generalized

e(q>gC g ) ~ e

and,

e( q> g) ~ e, i.e., g E Ge . This implies that eb E !B (G e ) and the the group Ge is not semlpnme. proposition 5.4.14 we can find a semiprime ring Rl with a centroid eC and a group of inner automorphisms G1, such that

241

CHAPTER 5

= eB(Ge ),

in corollary 1.3.7 from of the group G and, torsion and, besides, B(G 1)

G

which case 1) I is not semlpnme. This fact contradicts the Bergman-Isaacs theorem. Indeed, Ge is a subgroup hence, eC and, likewise, 1), have no additive I Gelconjugations on the elements elPg determine the Ge

action on R1, in which case 1)Ge = 1)G1 • The theorem is proved.

5.4.16. Regular groups. If an M-group H is given, then we can extend it to an N-group by adding all the inner automorphisms corresponding to invertible elements from B ( H). The obtained group will have the same algebra B (G) = B (H) and will, therefore, be reduced-finite, the fixed ring II

Q H remaining unchanged. We can now extend G to a closed group G by adding all the automorphisms locally belonging to G. In this case we also QG=

A

II

have QG = Q Hand /B (G ) = B ( a) . The latter equality needs proof and suggests the reader the idea of doing this easy task himself. Below (5.5.6) we shall see that the group 5.5.6).

II

G

can be not reduced-finite (see corollary

5.4.17. Definition. An N-group G will be called regular if it is a closure of an M-group. Below (5.5.9) we shall see that any closed N-subgroup of a regular group is a closure of a certain reduced-finite group with the same algebra.

5.5 Stalks of an Invariant Sheaf for a Regular Group. Homogenous Idempotents Let R be a semiprime ring, G be a reduced-finite group of its automorphisms. The elements of the group G will be viewed as unary operations. In this case the invariant sheaf r is a correct stalk (1.11.4). In this paragraph we shall see the way the stalks of an invariant sheaf are arranged in nearly all the orbits of the spectrum.

5.5.1. Definition. A nonzero central idempotent e is called homogenous if Ge = Gf for any nonzero f ~ e and every orbit p has not more than one point from U(e). Here Ge = [g E GI i(g) 2: e) (see 5.4.2). 5.5.2. Lemma. A set of all homogenous idempotents is dense in E. For proof it suffices to establish that any idempotent fEE, such

242

AUTOMORPIDSMS AND DERIVATIONS

I

that G: G~ < 00 (see 5.4.3) has a homogenous subidempotent shall start from the following remark.

e

~ f.

We

5.5.3. Remark. If a point P E U(f) and g E Gf , then P g= p. Proof. Let a E p. Then, in line with the von Neumann regularity of a centroid C, we have e( a) E p. Allowing for the fact that g acts identically (by conjugation) on all the idempotents less than i(g) ~ f, we get p3e(a)f=(e(a)f)g=e(a)gf=e(a g )f. Since

f~p and the ideal

p

is simple, then e( a g) and, hence, a g lie in p, i.e., p g = p, which is the required proof. Therefore, a subgroup Gf of a finite index acts trivially on all the points from U(f). This implies that every orbit generated by a point from U(f) has a finite number of elements (not greater than the index). Let P be one of the points the orbit of which has the greatest possible number of -

g2

gl

gk

elements P= (p = P, P ,... , P ) . Using the fact that the spectrum is Hausdorff (see 1.9.13), let us find a neighborhood U( e 1) of the point . pomts

p 92 ~_, p gk

oth er th an

p contained in U( f) and containing no

p. Let us consider a Boolean algebra

generated by the sets U(e Ig ) = U(eI)g, gE G. Let G= g= h· gi

and

hE G f ,

then

U(er) = U(eI)g

U Gfg ~..

i= 1

is the least subset of the algebra /B

containing the point

different neighborhoods

not intersect.

then

U(e gi), 1 ~ i

~

U( e) k

U(e)g g

= U (e)

do

and,

hence,

Then, if

and, hence, the Boolean

algebra /B is finitely generated and, therefore, finite. Let u( e) be the least set of this algebra containing

g PI "* PI'

]lJ)

IV

If

in

p. Then

p

PI'

g.

pt

every

u( e g)

In particular, E u(

e)

and

neighborhood

we can find two points of the orbit of

PI

which

contradicts the maximality of k. Thus, the neighborhood U(e) has not more than one point of every orbit. Since Gf;;2 Ge at f:S; e and the index I G: GJ is finite, then, reducing, if necessary, e, we get that G f = Ge at 0"* f ~ e. The lemma is proved.

5.5.4. Proposition. There exists a system of mutually orthogonal homogenous idempotents [e a , a E AJ, such that

243

CHAPTER 5

II ea

A

RE =

A

RE ,

aE A

where ea = sup ( idempotents.

eg,

g

E

G)

is

a system of fixed

mutually orthogonal

L

Proof. Let us denote through a set of all sets of mutually nonintersecting domains of type U(e), where e is a homogenous idempotent, e = sup( e g, g E G). In this case is a non-empty directed set. By the

L

Zorn Lemma, in Let us

show

that

L

we can find a maximal element U

aE A

U(e a)

is

(J

= (U( ea)'

a

E

A).

a dense set in the spectrum, i.e.,

sup ea = 1. If this is not the case, then e'· e a = 0 for a certain nonzero a idempotent e'. Let us find a homogenous idempotent e < e' (lemma 5.5.2). Then e . ea = 0, since all e a are fixed and, hence, (J

U( u( e) )

L , which

E

contradicts the maximality of

Let us consider a homomorphism

~: r ~

(J.

II ea r

from Q into the a direct product. As sup e a = I, then this homomorphism is an embedding. If raE ea Q, then, by the definition, of a sheaf there is a global section r E Q, such that re a = r a. This implies that ~ is an epimorphism. The proposition is proved. Allowing for the fact that a closed group of automorphisms of a direct product of invariant components falls into a direct product of induced groups, let us concentrate our attention on homogenous idempotents and rings of sections over the domains detennined by them.

Proposition. Let e be a homogenous idempotent. Then there

5.5.5.

can be found automorphisms

gl =

1,

92 ,•. ,

g kEG.

such that the ring of

A

global sections RE is presented as a direct sum A

e RE EEl e

where are

e=

g A g 1\ A 2 RE EEl ••• e k RE EEl (l - e) RE.

(1)

k

Le gi =sup(e g , gE G). i= 1

identified through

the

If the components

automorphisms

gi 1 g j'

egiQ

then

the

and following

244

AUTOMORPHISMS AND DERIVATIONS

inclusions are valid:

(2) 1\

where H

a closure of the group G;

G is

= (g

HI

is a projection of the stabilizer

G, e g = e) on e Q which is identically extended onto

E

(1 - e) Q;

a group of permutations rearranging the components eg1Q; G' is a group acting identically on eQ. Proof. As e is a homogenous idempotent, then a subgroup Ge has a finite index in G, in which case H = ( g E G, e g = e) ;;;l Ge , i.e., the Sk is

stabilizer H of the idempotent e has a finite index in G. Let be

a

(e g ,

decomposition (e gi ,

gE G)=

of

the

1~ i

~

group k).

G into

Besides,

a

e=

union

i= I

of cosets.

( e g) =

sup

k

U

G=

Hg, ~

Then

L e gi

is

a

i

gE G

fixed idempotent, since, according to homogeneity, the idempotents e gi are mutually orthogonal. Now it is the decomposition of the unit, 1 = e + er;J;. + ._ + e gk + (1 - e), that results in decomposition (1). Let us connect to every automorphism h E H an automorphism hI' the action of which on eQ coincides with that of h, while it is identical on the remaining addends of sum (1). Then we have

1\

hI

E

1\

G, since

G is

closed. Now the group

HI of all such automorphisms is easily identified with

a projection of H onto

eQ,

gi I HI g i

product

On the addends

e gi Q

the conjugated subgroups

are acting and, due to the closure, the group HI

x

with respect -I g i HI g i

g"i. I HI g2

to

the

x ._ x gk system

I HI g k .

Ii

contains a direct

If we identify the addends

of isomorphisms

gi I g j'

then

e gt Q

the

groups

are also identified and the direct product will have the form

Moreover, due to the closure, the group (1.-, gilg j'-' g-/g i"" 1) G:::>_ Hk),. S

1

and,

hence,

1\

G

k

HI'

will contain all the transvections

the group of permutations. Thus,

k·

So, if g

E

g', the action

G, then let us determine an automorphism

of which on (1 - e) Q coincides with that of g- I, while on eQ it is identical. Let G' be a subgroup of all such automorphisms. For every

i, 1 $; i

$;

k

we can find a

j, 1 $; j $; k, such that

e

gjg-

- e

gj

,

, I.e., g

245

CHAPTER 5 eQ in decomposition (1). If

rearranges, in some way, addends of

n is a

corresponding permutation component, then (gg') n- 1 acts invariantly on all the

components

e g1 Q,

i.e.,

(gg')n- 1 E Hlk

and,

hence,

g E (H 1k >.. S k) x G' , which is the required proof.

5.5.6. Corollary. If G is a closed and reduced-finite group, then the ring Q has a decomposition

into invariant components, which is corresponded to by a decomposition of the group

in which case the group

G1

is a group of inner automorphisms of the

component QI' while the rings Q a are prime. Proof. By formula (2) and proposition 5.5.4, it is sufficient to show that the homogenous idempotent e, for which Ge "# G, is minimal. Let g EO Ge and 0"# f < e. Let us denote by gl an automorphism coinciding with g on

fQ

and identical on (1 -

f) Q;

automorphism coinciding with g on (1 -

while by

92

we shall denote an

and identical on

f) Q

fQ.

Then

gl E Ge _ f = Ge = Gf 3 g2 is a contradiction. This corollary shows that a closure of a reduced-finite group is, as a rule, not reduced-finite. If, for instance, the spectrum contains no isolated points, then only the groups of inner automorphisms will be reduced-finite and closed.

g == gl g2' but

5.5.7. homogenous

Theorem. Let G be a reduced-finite group, e idempotent, H = (g E G, e g == e). Then there

automorphisms

g] == 1,

P

E

92,-., g k from the group

be a are

G, such that for any point

U( e) the following statements are valid: (a) The orbit

P is equal to (p

(b) The stalk

r

gl

.-, P

gk

).

p (Q) is decomposed into a direct sum of mutually

isomorphic stalks of a canonical sheaf

246

AUTOMORPHISMS AND DERIVATIONS

r

T-p(Q) =

p

r

g\(Q)$ ..• $

p

(3)

gk(Q).

(c) If the stalks of r(Q) at the points of the orbits are identified with the help of isomorphisms induced by automorphisms gl'-" g k' then the

Gp will have the form

closure of the group

group induced by the group (d) The

group of the

projections

H

Hp

If

p.

is reduced-finite and its algebra equals the

algebra

of

the

B( Hp) = P p(B (G». (e) If G is a Maschke group, then (f)

r

on the stalk

G is a Noether group, then

group

on

G

the

stalk:

Hp is also a Maschke group. Hp

is also a Noether group, in

which case

Proof.

Let

gI'-" g k

be automorphisms, the existence of which is

claimed by proposition 5.5.5. Then each of the neighborhoods U(e gi) contains exactly one point of the orbit p, which proves (a). (b) If 3 is an element of the stalk determined by the section s over an invariant neighborhood W of the point p, then W will be a

r"

neighborhood of each of the points

and, hence, s

p gi, 1 ::;; i::;; k

will

determine elements 31,,,,, 3 k of the stalks r lP g\ ._, r I' g k, respectively. Let us show that the mapping f: 3 ~ 31 + ._ + 3 k sets the isomorphism

r- p == r As the mappings homomorphism. Let all .

of the pomts

3i p, p

P

g\

s

$ ... $

~

3i

r

g k'

P are homomorphisms, then

will also be a

f

be zero. It implies that there are neighborhoods VI ,.., g2

,..., p

gk

Vk

,such that the restrictions of s on each of them k

are zero. One can also assume that

VI!;;;; u( e).

Let us set

Then, using structure (2) of the group G, we get WI

=

U

gE G

V

Vo

=n

g=

-\

V~i

i= 1 ~

247

CHAPTER 5 k

g k VOl!: U Vi' i= 1 i= 1

=U

As

WI

is an invariant neighborhood of the point

p

and the restriction of 8 on this neighborhood is zero, then 8 = 0 in the stalk of an invariant sheaf. Let now 81,." 8 k be arbitrary elements of stalks at the points of the orbit determined by sections 8 1", 8 k over the neighborhoods VI ,.., V k of the • points of the orbit. Now we can assume that vi!: U ( e gl) ,t.e.,

In this case a section

81+

=

8

•. +

over

8k

U( e)

8i E

e gl Q.

determines the element 8

of the stalk T- 1I , such that £(8) = 81 + .• + 8 k' (c) Now the isomorphisms

gi: r

r

11 ~

eQ ~ e g1 Q induce isomorphisms

g i:

If we identify the stalks of the points of the orbit p with respect to the system of these automorphisms, then the groups 5 k acting on

eQ

JP gl

r

on the stalk

r

p g i'

of h

r p'

will be induced on the stalk

(and, by identification,

p

Besides, the group gil Hg i

since any neighborhood of the point

E H

to that of the point

H

is induced

is induced on the stalk

p goes over, under the action

Gp acts transitively on

p). As the group

the components of decomposition (3), then, by 5.4.9, its closure has the form

H~).. 5 k' where ~ is a group induced by a stabilizer (g E G I P g= p), which, since the idempotent e is homogenous, is equal to H. (d) Let us show that B( Hp) = pp(B (G». Let g If

pp(qJ) :t:.0,

However,

g E

u( i(G».

In

homomorphism

then

(lemma

1.9.18),

Gi ( g), and, hence, particular,

Pp

xip = ip;i, where

to

x

g

p g= p,

the identity

e(qJ) ~ p,

acts i.e.,

E

G and qJ

i.e.,

E lP g .

p E U(i(g».

identically on all points of g E H.

xqJ = qJx g .

In

Let

us

this

case

apply we

a get

is any element of the stalk and g is an automorphism

induced by g. Therefore, ip E lP g . In order to prove the inverse inclusion, let us consider a ring R together with the action with the group H and an H-invariant sheaf r'. It should be remarked that the point p coincides with its orbit and the stalk of from

Hp

,

the sheaf r neighborhood U(e)

(J

V).

on the orbit (p ) is equal to the stalk r p (since any V of the point p contains an H-invariant neighborhood

248

AUTOMORPHISMS AND DERIVAnONS

can find elements

o =t:

and come to the conclusion that the predicate

qJb

=

v

rp

E

r

is true on the stalk

xE r p

qJ E ex r p)' We 0 =t: aqJ = d E r p ,

xqJ for a, b

Let qJx h =

all

and for a nonzero

of the stalk

r p'

such that

r'

This is a strictly sheaf predicate for the sheaf

p'

i.e., there are preimages, such that dx h b = axv is an identity on Q, in which case e(d), e( b) e: p. Let us transform this identity to a reduced form By

d(1- J(h))xhb+ dep'hxephb= axb.

d(l- i( h)) ® b= 0, i.e.,

and,

l-e(h)Ep

conjugation by

e(d)·

hence,

0-

2.2.2,

theorem

e( h))· e( b) = 0,

e(h)e:kerp p '

i.e.,

wherefrom

get

we

get

Since

Pp(qJJ=t:O.

a

P p(ep J determines the same automorphism, h, on the stalk

that a conjugation by ep does, then ep = cip h' , where generalized centroid of the stalk (equal to

ep E P p(lB

we

c

is an element of a

P p( C) (see 1.1.26)).

Thus,

(G) L

Let us show that finitely-generated

module

dimensional over the field

has a finite-reduced order. As

Hp

C, then

over

is

lB ( H p) = P p(lB (G))

Cp' Then, all the automorphisms from

conjugations in the ring e Q and

Ge

S;;; H,

since

P

E

finite-

Ge act by

u( e), and hence, the

induced automorphisms P p( G~ will be inner for the stalk: Hence, , Hp:( Hp) in'::S;' H: GJ <

is a

lB (G)

P p( G~

S;;; ( H p)

in'

00.

(e) Let us check if lB ( H p) is a semi prime algebra. Let us first remark that lB (G) is an injective c-module (since it is finitely-generated) and, hence, lB (G) is a closed set in the topology determined by idempotents. Therefore, a unary

predicate,

such

that

p(s) = T H

S E

strictly sheaf, in which case on a stalk we shall have P p(lB (G)) = lB (H p)'

If

s lB (H p )s = 0,

now

predicate is true on the stalk:

v

x

P(x) ~

sxs

=

o.

then

the

lB (G)

pes) =

will T H

following

be

sE sheaf

CHAPTER 5

249

Consequently,

there

is

an

e1 ~ p

idempotent

and

a

preimage

s e e 1 B ( G), such that s B ( G) s = 0 and, hence, s = O. (f) Let

and elements

iii e

and ip. ip' = 1. We can find an idempotent

Hp

cp, cp' e B (G) , such that

qxp' = cp' cp = e 1, P p( cp) = ip.

(1- e 1 + cp)(l- e l + qJ)= =(1- e l + cp')(l- e l + cp)=l,

automorphism

b, where

is an automorphism from

+ cp belongs to

b = 1 - el

e1 e p

i.e.,

Then

an

inner

PpCb)

H. Obviously,

coinciding with the conjugation on ip.

Hp

1\

Let ve G. By statement (c), it suffices to show a projection

Gp' i.e.,

belong to the closure of the group

vp

to

to show that its action on

every addend of (3) is induced by a certain automorphism from G. Let us, for instance, consider the first addend. For every g e G let us denote by Eg a set of all central idempotents £ ~ e, such that v == g(modGe ) on £Q. Since G is a Noether group, and B(Ge ) = e/3(G) is an orthogonally complete set, then Eg is also an orthogonally complete set. This implies that in idempotent £( g) . Let

v- = ( bg .)p ~ p

U( £( g

there is the biggest

G = U Geg.. Since v locally belongs to the group i= 1

~

G, then sup( £(g i») = e. Therefore, the point neighborhoods

Eg

n

i» .

As

in

this

p belongs to one of the

neighborhood

v= bg.e G, ~

then

on the first addend of decomposition (3). The theorem is

completely proved. 5.5.S. Corollary. Let G be a reduced-finite Noether group, and e be a homogenous idempotent. If v locally belongs to G, then for a fixed idempotent e there is a decomposition into an orthogonal finite sum

e=

e1 + M.

+ em in such a way that that of a certain g i E G, 1 ~ i < m.

Oll

e i Q the action of v coincides with

Proof. It is the fact that we have established when proving point (f) of the preceding theorem.

5.5.9. Proposition. Let

G

be a reduced-finite N-group. Then any 1\

closed N-subgroup F of the group G is a closure of a reduced-finite subgroup H k F, such that /3 ( H) == B (F). Proof. By proposition 5.5.4, we have a decomposition

Q

==

IT eaQ. a

250

AUTOMORPHISMS AND DERIVATIONS

Since F is a closed group, then the decomposition F =

II F a

corresponds to a it.. If in each cofactor F a we can find a dense reduced-finite subgroup Ha' such that

II Ha

will be the sought one. a Therefore, with no violation of generality, one can assume that we have 1 = e g1 + ._ + e It containing

then

B ( Ha) = B ( Fa)'

is

H=

for a certain homogenous idempotent e and

gk

sufficient

lB (F) = Fin'

for

us

to

find

r

p

reduced-finite

i

E

G.

group

such that If::= F-p for a dense set of points p

the space of orbits. Indeed, in this case for every v on the stalk

a

g

the formula \::f x

XV

E F

there is an

H !:;;; F

p

of

h E H, such that

= xh is valid. Viewing v and h as

strictly sheaf operations for a G-invariant sheaf (see corollary 5.5.8) and applying the metatheorem, we see that v coincides with h in a certain neighborhood of the point p. As p is arbitrary from a dense set, then 1\

vE H.

Let us fix a nonzero idempotente1 :::; e and show that there exists a nonempty neighborhood U(f),

f:::; e 1 and a reduced-finite group

such that H(f)p = Fp for all

p E U (f).

H(f)!:;;; F,

Let us choose a point p

in such

a way that the index I F P :( Fp) inl had the biggest value (by theorem 5.5.7, point

(f ),

it

is

not

greater

than

representatives of cosets of (Fp) in in

k! . I G: GJ).

Let

ipl ,-., ip n

be

F-. Let us fix preimages of these p

automorphisms, q>l'-" q> n E F and consider them to be strictly sheaf operations for a G-invariant sheaf (see corollary 5.5.8). On the stalk the following Horn formulas hold:

3 b i j \::f

x

\::fl:{(\::fx

cp cp.

b ij( x i) bx

cP

i

]

= xCPm b ij&

e( b ij)

=1

= X cpo] b) ----7 b = 0] ,

where in the former series of formulas i, j run through all the values from 1 to n, while in the latter one i:t j. By the metatheorem, these formulas set sheaf predicates, i.e., by definition 1.11.11, one can find a neighborhood U(f) of the point p, over which these formulas are valid. As F is a closed group, then, violating no generality, one can assume that outside u( f) the operations q>l'-" q> n act identically. Let H( f) be a group

251

CHAPTER 5

generated by

'PI"'" 'P n and by inner automorphisms corresponding to the

elements from i B (F) (extended onto (1 - i) Q by an identical action). In this case the validity of the above mentioned formulas indicates that I H(f): H(f) inl = n, i.e., H(f) is reduced-finite. By corollary 1.11.18, these formulas are also true on all the stalks

rq

under

fie

U(f), i.e., the

. I = n has the biggest possible value, which fact implies index I H( f)-q : (F-) q ~n

that H(f)q

= Fq .

Now, as was the case in proposition 5.5.4, from a dense set of fixed idempotents (i lone can choose a dense class (i a 1 of mutually orthogonal

II i a Q

II

idempotents. In this case Q =

and the group H = H(f a) is the a a sought one. The proposition is proved. By way of concluding this paragraph, let us consider Maschke groups for rings of a finite prime dimension. In this case the set E of central idempotents of Q is finite and, hence, the topology determined by them is discrete. By lemma 3.6.6, the ring Q is decomposable into a direct sum of ideals which are prime rings

where

Qi

= Qe i'

ei

are minimal idempotents. The neighborhood

contains only a point Pi = (1 - e i) C and the stalk

r ,,}Q)

U( e i)

is equal to Qi

and, hence, r ,JR) = Rei' Each of the idempotents e i is homogenous and the ring Q can be presented as a direct sum of stalks of the invariant sheaf

(4) Let us consider these stalks in detail. By theorem 5.5.7, we have a decomposition of, for instance, the first addend

in which case on the prime ring closure of a restriction a~

G

on

e 1 Q an M-group

eQ has the form

e1 a,

a ~ a + a g2 + ... + a gn

H}

Ht>.. S n'

acts, so that the Now the mappings (5)

,

252

AUTOMORPHISMS AND DERIVAnONS

set an isomorphism

Returning to the initial ring. we get the following result.

5.5.10. Theorem. Let a ring R have a finite prime dimension, G be an arbitrary M-group of its automorphisms. Then a certain sub ring $ containing an essential ideal K of R is decomposed into a direct sum of prime rings

on each of which a Maschke group

where all components

$1 ..... $ n

among the components

Hi

acts, in which case

occur (to the accuracy of an isomorphism) while the number t

is equal to the

invariant prime dimension of the ring R. Proof. Let I be an essential ideal of R. such that

IE!;; R. Let us

$ k .... $ k' I

t

choose an essential ideal J in such a way that J gl!:; I for all automorphisms g i E G participating in the construction of all isomorphisms (5). Let $

g2

= Je 1 + (Je1) + ... + Je ~ + (Je k2)

where the idempotents e1• e k

2

."

g;

+ ...

are determined by decomposition (4). while

the automorphisms g2..... g;... carry out the identification of stalks in theorem 5.5.7 (point (c». thus participating in the definition of isomorphisms (5). We come to the conclusion that $!:; R and mappings (5) set the isomorphism

$

G _

= (Je 1)

HI

EE> ... EE>(Je k ) t

Ht

The task now is to prove if $ contains an essential ideal. By the definition of the group A(R) and using the equality A(R) = A(J). one can

CHAPTER 5

253

find an essential ideal K, such that K

S ;;;;? Ke l

+

~

Ke 1

+ ... +

~

Ke.~

for all

J gj

+

Ke

"2g;+ ... -

g i' Then we have

K.

The theorem is proved.

5.6

Principal Trace Forms

In this paragraph we shall denote through e a homogenous idempotent, while for groups of automorphisms, idempotents, etc. we preserve the notations from proposition 5.5 ..5. Using the metatheorem, we shall construct the "principal" trace forms rJx) and fJx) for the groups Hand G, respectively. For this purpose it is sufficient to formulate our problems as Hom formulas, the validity of which for prime rings has been established. Let us start enumerating strictly sheaf and sheaf predicates and functions which will be used (not only in the present paragraph). 5.6.1. Let us put the unary operations of projections

e

1C ,1C B' 1C g

into

correspondence with each of the sets C,!B (G) , (J> g' In more general terms, let s be a closed E-linear set in Q. Then, considering S as a module over E (see 3.6.19), we see that it is an injective module. Consequently, it is singled out as a direct addend. In particular, the E-linear projection 1C s: Q ~ S is determined. This projection is a strictly sheaf operation. But, it goes without saying, that the projection 1C S is not uniquely determined by the set S. In particular, if S is a Csubmodule, then 1C s can be chosen as a homomorphism of C-modules. For us of major importance is the very existence of this projection, but it is going to be used for eliminating excessive implications: for instance, V b P J b) --1 If/(b) can be replaced with V x If/( 1C J x) ). 5.6.2. The unary operation of support sheaf (see 1.1.10).

e( x), as we know, is strictly

5.6.3. Let us fix a system of coset representatives hI = 1, hz,-., h n for a subgroup Hin of inner for eO automorphisms in H (see 5.5.5). This system is finite, since Hin;;;;? Ge . These automorphisms can be viewed as unary strictly sheaf operations for H-invariant and G-invariant sheaves. In

254

AUTOMORPHISMS AND DERIVAnONS

this case one has to remark that the restriction of an H-invariant sheaf on a closed domain U( e) does not differ from that of a canonical sheaf, as the automorphisms from H do not replace the points of U(e) and do not change the idempotents less than e 5.6.4..

Let

Permutations from

be

Sk Sk

a

subgroup

determined

formula

by

(as well as all automorphisms from

(2).

/I

G) can be

viewed as strictly sheaf operations for a G-invariant sheaf, transvections (1 i) being, in particular, such operations. 5.6.5. Let us get down to constructing principal trace forms. Let us seek the form 're as a sum n

m

h.

L {La"xJb" __ I

'rJx) =

~J

j = 1 •

(6)

~J

Let us write down the required properties of the form as formulas. (1) The invariance: n &

I {::} 'Ii x 'Ii y

'rJx) n B( y) =

i= 1

m = TCB(y)'rJx)&

& j

=1

'rJx)

h j

=

'rJ x) ,

where, for simplicity, instead of the right part of (6), where are replaced with left one.

n B( a ij) and

a i j and

b ij

TC B(b ij)' respectively, we have written the

(2) The non-degeneracy: e UJ rx» = e( r) . e

r{::}'lir'3x

r' {::} 'Ii Now

the

formula

r'3 x

e('rJ xr» = e( r)· e.

'3 a .. '3 b . . I& r& r' ~J

~J

expresses

all

the

properties

required. The validity of this formula for prime rings has been established (see lemma 3.4.3 and remark 3.4.5) and, by the metatheorem, we can assume the existence of the required form 're to have been proved, provided

CHAPTER 5

255

G is a Maschke group. The form 'fe can now be considered as a strictly sheaf operation for a canonical sheaf over the domain U( e). In this case its values l'p at the

P E U( e) will be principal trace forms on stalks r p . Using decomposition (1) and identification, one can present the form fe as a sum of copies 'fe e ... e 'fe Or, in more exact terms,

points

Therefore, fe can be also viewed as a strictly sheaf operation for a Ginvariant stalk. As an illustration of this strictly stalk operation, let us prove the following useful statement.

5.6.6. Lemma. Let G be a Maschke group of automorphisms of a semiprime ring R. If I is an essential ideal of the ring R. then the left (right) annihilator of IG in R F is zero and. in particular. IG is an essential ideal of the ring R G . Proof. Let a· I G= O. Then for the principal trace form fe one can find an essential ideal J, such that fJ.. J) !;;;; I and, hence, afJ..x) = 0 at x E J. By corollary 2.3.2, this identity is fulfilled on Q as well. Let us

project this identity on the stalk a'f H( x) = 0

a B (H p) = 0

in

the

prime

and, hence,

ring

a =O.

r P' r As

P E u(e), thus getting an identity p'

By the

theorem point

p

2.2.2, has

we

have

been chosen

arbitrarily, then ea = 0, and we are now to make use of the fact that a set of homogenous idempotents is dense. The lemma is proved.

5.6.7. Theorem. If a semiprime ring R has an infinite prime dimension and G is an M-group of its automorphisms. then RG also has an infinite prime dimension. Proof. Let us assume that R G has a finite prime dimension. If e is a homogenous idempotent and e l "., en"' are pairwise orthogonal nonzero idempotents less than e, then the fixed idempotents el , e2 "" en," are also pairwise orthogonal. Let us consider the principal trace forms For each of them let us find an ideal In E F, such that fen(I rJ we have an infinite direct sum of ideals in R G

fe 1"., fe n ". ~ R.

Then

AUTOMORPHISMS AND DERIVAnONS

256

All the ideal in this chain are nonzero: if, for instance, rei (I I) = 0, then, by corollary 2.3.2, we have fe (Q) = 0 and, projecting this equality on any stalk 1

r

p,

P

E

U( e l ), we get a contradiction.

Therefore, the idempotent e has only a finite number of less idempotents.Analogously, in Q there are no infinite sets of pairwise orthogonal idempotents ea , where e a is homogenous, and we are to use now proposition 5.5.4. The theorem is proved. Let us formulate the principal property of trace forms in a somewhat different way.

5.6.8. Proposition. Let L be a nonzero one-sided ideal of a ring if fJL) = 0, then eL = O. Proof. Let L be a left ideal, VEL. Then, by corollary 2.3.2, the identity e' fJxv) = 0 holds on the ring R F . The formula r which defines the form re shows that e( v)e = 0, i.e., ve = 0 and Le = 0, which is the required proof. By way of concluding this paragraph, let us consider the problem of uniqueness of principal trace forms. R. Then

5.6.9. Theorem. The principal trace form fe is uniquely determined to the accuracy of the replacing variable XI = bx, b being an invertible element from

e iB ( G).

Proof. By lemma 3.4.3 and remark 3.4.5, the projection (re> p will be a principal trace form on the stalk. By theorem 3.4.9, this projection is uniquely determined to the accuracy of the substitution Xl = bx. If now i~) is another trace form, then, viewing it as a strictly sheaf operation, we see the validity of the following formula on the stalk

By

the

i~ )(x) =

3 b i ,bV X ij\x) = rJbx)& bIb = bb i = 1 & P!8(b)& PI/J(b j ).

=

metatheorem,

on

this

rJbx) and, hence,

formula

ti

1)(

x) =

is

true

elB

as

well,

i.e. ,

fJ bx), which is the required proof.

257

CHAPTER 5

5.7

Galois

Groups

5.7.1. Theorem. Let G be an M-group of automorphisms of a semiprime ring R. Then the centralizer of a fixed subring R G in a ring of quotients R F is equal to the algebra lB (G) of the group G. Proof. Let us fix a homogenous idempotent e and let a be an element which commutes with fixed elements of the ring R. It suffices to show that p p ( a) sheaf predicate

E

lB ( H p) for all points

x = 1t'B(x) is true at

P

U( e),

E

since in this case the

x = p p( a) on the stalks at nearly all

points of the spectrum and, by the metatheorem, a = 1t'B( a), Le., a E lB (G) . Let I be an essential ideal of the ring R, such that f"J. I) ~ R. Then, making allowances for decomposition (1), we get a'CJ.x) - 'CJ.x) a = 0 for all

x

E

I. By corollary 2.3.2, this identity is also valid on

R F . Going

a'Cp(x) = 'C p(X) a.

As, by the

over to a stalk, we find on it an identity construction,

the

form

'C p

is

principal,

then

'C per p)

intermediate ring of the stalk, i.e., by theorem 3.5.1,

aE

is

an

almost

lB ( H p)' which is

the required proof.

5.7.2. Theorem. An automorphism h belongs to the Galois closure of an MN-group G

iff h locally belongs to

G, i.e.,

A( R G) =

G.

Proof. Let h E A( R G). Let us first show that h affects identically the G-fixed idempotents (which may not belong to R). Let, on the contrary, for such an idempotent. Let us choose an essential ideal I in such a

fh:;: f

~ R. Then fIG ~ R G and, hence, (f - f~I G= 0, Le., lemma 5.6.6 yields the equality f = f h. Now we can consider h as a strictly sheaf unary operation for a Ginvariant stalk. It is sufficient to show that for every point P E u( e), where e is a homogenous idempotent, there can be found an automorphism way that

1\

fI

g E (G)-p, the action of which on the stalk

r

p

coincides with that of h

on this stalk. Indeed, in this case the formula V x xh = x g which is true on the stalk by the metatheorem, will also be true in a certain invariant neighborhood of the point p, i.e., h will locally belong to the group G. It should be remarked that the idempotents

pp( e gi )

in decomposition

(3) are minimal central idempotents of the stalk. Therefore, the automorphism h

rearranges them in some way, Le., one can find a permutation

5

E

Sk

258

AUTOMORPHISMS AND DERIVAnONS

(see 5.5.7), such that the stalk

r

p

hO affects all the components of decomposition (3) of

in a fixed way. Now we are to show that the restrictions of

hO on this components lie in

Hp .

We have f,ix)h = f,i x) for all x from a suitable essential ideal I. This equality holds on

RF

as well (see 2.3.2). If we project it on the first

component of the stalk, then we get an identity stalk

r

peR F). Since the form

'r p

'rp(x)hO = 't"p(x)

is principal, then

't"peR F)

on the

is an almost

intermediate ring and, hence, by theorem 3.5.2, the restriction hO on the first component belongs to H p. Analogously, the restriction hO on the other (identified) components also belongs to hence,

HP .

Therefore,

hO

E H

~

and,

h E H; >.. S k. The theorem is proved.

5.8 Galois Subrings for Regular Groups Let G be a regular group of automorphisms of a semiprime ring R, S be an intermediate ring, R:d S:d RG By theorem 5.7.1, a centralizer of S in the ring RF is contained in the algebra lB (G) of the group G. This centralizer will be denoted by z. The set z coincides with the intersection of the kernels of all mappings x ~ sx - xs, S E S and, hence, z is a closed set in the topology determined by the central idempotents. Hence, we have a strictly sheaf unary operation of projection 7r z (see 5.6.1). Let us recall the conditions on an intermediate ring arising when considering the prime case. BM (Bimodule condition). Let e be an idempotent from /B (G), such that se = ese for any s E S. Then there is an (idempotent) f E Z , such that ef = f, fe = e. SI (Sufficiency of invertible elements). A c-algebra Z is generated by its invertible elements and if for an automorphism g E G there is an element b E lB (G), such that sb = bs g for all s E S, then there is an invertible element in e( b)Q with the same property. RC (Rational completeness). If A is an essential ideal of S, and Ar s;:;; S for a certain r E R, then rES. When going over to the semi prime case the formulation of condition

259

CHAPTER 5

SI has been somewhat changed. 5.8.1. Theorem. Any intermediate Galois subring of an M-subgroup of the group G obeys conditions BM, SI and Re. This theorem will result from the following somewhat more general statement. 5.8.2. Proposition •• Let G be an M-group of automorphisms of a semiprime ring R. Then (a) if a s E RG

is an element from

R F'

such that

sa

= asa

then ap = p, pa = a for an idempotent p E B(G);

(b) if g E A(R) and b is an element. such that

for all

= bs g

for all s E RG. then there can be found an invertible element in e( b)Q with the same property; (c) the fixed ring RG is rationally complete in R. In order to deduce the theorem from this proposition, It IS sufficient to remark that, by theorem 5.7.1, a centralizer z of the fixed ring RH of an M-subgroup H::;; G coincides with the algebra B( H) of this subgroup (and, in particular, is generated by invertible elements). sb

Proof of proposition 5.8.2. (a) Let e be a homogenous idempotent. Then by the condition, fJ..x) a = afJ.. x)a for all x of an essential ideal of the ring R. By corollary 2.3.2, this equality is also valid for all x E R F' If P E U(e), then

when

projecting

'fp(x) a = a'f p( x)a-.

As

this

equality A

'fp( RE)

on

the

stalk

r

p'

we

get

is an almost intermediate ring of the

stalk, then, b y 3.7.6., we see that the following formula is true on the stalk (it should be recalled that the algebra of the group

Hp

equals

p p(B )

(see 5.5.7»:

3 f

af= f &

fa =

a & n IB (£) = f.

Since the homogenous idempotent e and the point P E U(e) have been arbitrarily chosen, then, by the metatheorem, the formula under consideration (with a replaced by a) is true on the ring of global sections, which is the required proof. (b) For simplicity, let call an element qJ almost invertible if it is invertible in the ring e( qJ)Q. Let T be a set of all almost invertible

260

AUTOMORPIDSMS AND DERIVATIONS

elements for which the condition of statement (b) is fulfilled. It is obvious that T is a a closed set and, hence, the set E( T) of the supports of elements from T will be also closed. Our task is to show that e( b) E E( T), for which purpose it is sufficient to find a dense family { f} of the idempotents for which e( b)f E E(T). Let e be an arbitrary homogenous idempotent, e ~ e( b). Then, as was the case in the preceding point of discussion, we can find an identity fJ..x)b

= bfJ..x)g

on the ring

R F . If this is a reduced identity, then, by

theorem 2.2.2, we have fJ..x)b = brJ.. x) = 0, which is impossible. Nonreducibility of the identity implies that e( b)cJ> hg :t= 0 for a certain h E G (naturally, under the condition that fJ..x) is reduced). Therefore, we can find an almost invertible nonzero element x E RF ,

in which case

e(q» ~ e(b).

q>,

such

that

If we assume

xq> = q>x x= S E R

hg G

for

all

, then we

shall get q> E T. As the set of all homogenous idempotents is dense and e is arbitrary among them, then we have e( b) E E( T), which is the required proof. (c) Let A be an essential ideal of the ring RG. Let us show that its left annihilator in Q is zero. If Ar = 0, then for any principal trace form fe and any ideal I we have AfJ..rI) = f( ArI) = O. Choosing I in such a way that fe values lie in R, we get fJ..rI) = 0, which implies that rJ..rx) is a zero form. However,

(see 5.6.7), proof.

e( r) . e

=0

re is a principal trace form and, hence,

and, consequently, e( r)

= 0,

which is the required

now Ar k R G , then A( r - r g) = 0 at all g and the proposition is proved. Let us now get down to proving the inverse statement.

If

E

G.

Hence,

r = r g,

5.8.3. Theorem. Let G be an MN-group. Then any intermediate ring obeying conditions BM, SI and RC is a Galois subring of a regular subgroup of the group G. This theorem will obviously result from the following theorem. 5.8.4. Theorem. Let G be an MN-group, S be an intermediate ring obeying conditions BM and SI. Then A(S) is a regular group and the ring S contains an ideal which in Q are zero.

w of the ring 5

= II

A.. S), the annihilators of

5.8.5. Let us start with describing the situation arising on nearly all stalks of canonical and invariant sheaves.

CHAPTER 5

261

By proposition 5.5.9, the group A(S) is a closure of a reduced-finite N-group V, the algebra of which is equal to that of the group A(S). Since Z is generated by its own invertible elements, then B ( A(S» = Z and B(V)

= z.

It is obvious that the set of idem po tents homogenous for both G and v is dense. Let us fix one of such idempotents, e, and let p E U(e). By 1\ k theorem 5.5.7, the projection (G)-p has the fonn Hp),. Sk' where Hp is a regular (i.e., an MN-) group of automorphisms of the stalk A(S)-p

is

a

A(S)-p = (F

closed

m),.

subgroup

in

1\

then,

(G)-p,

by

Since

Tp'

5.4.11,

we

get

Sm> x Gil. Carrying out, if necessary, a rearrangement, one

can assume that S m acts on the first while Gil acts on them identically.

components of decomposition (3),

m

1\

Lemma. Let us denote by

5.8.6.

S· EGa closure of S· E G in 1\

the topology determined by G-fixed idempotents, and let Then

SI is

an

almost

intermediate

ring

of the stalk

conditions BM and SI, in which case the group

!>p( S . E G).

SI =

r

p'

obeying

A(SI) coincides with the

group A(S)-p.

Proof. If I

is

an

essential

ideal II

then 1\

fJ..p

» ~ SI'

p( I· E G

G

fee I· E ) r;,;;

of the ring

R,

such

that

1\

s· E G

and,

hence,

i.e., SI is an almost intermediate subring. 1\

Let us denote by 7r S a strictly sheaf projection n s: Q -+ S . E G (for an invariant sheaf). Then conditions BM and SI will be presented as implications:

[ 'v' x ~3f

[ 'v' x ) ns(yb

ef=f&fe=e&f=nif),

n s(x)b = b7r '

I

" s< x)h 1----7 I 3 b ,b 'v' y

h

JII

= b7rs(Y) & bl:>

'"

= bb = e(b),

262

AUTOMORPHISMS AND DERIVATIONS

where 7r B,7r Z are strictly sheaf projections (it should be recalled that B, Z x = 7r z( x)

are closed). As

of the su bring

ZI

in

SI

V y 7r f y) x = X7r s( y), then a centralizer

iff Q(

r p)

coincides with the image of

7r z'

i. e . ,

= pp(Z).

Allowing for the fact that for any h, b, e the above mentioned implications are valid on all rings of sections, we come to the ZI

conclusion (proposition 1.11.18) that they are also valid on pp( Q), i.e.,

SI

obeys conditions BM and SI. Since A(SI)

the

~

Gp.

stalk

Gp " is a regular group, then it is a Galois group and, hence, Let

r- p .

g

E

and the formula V x 7r f x)g = 7r f x) be valid on

G,

We

can

find

a

fixed

idempotent

such

f ~ p,

that

A

7r s (Xf)g = 7r s (xf)

for all

xE R· EG.

Let us determine

xh= xgf+

A

(1 - f)x. Then g =

h

E

8

h=

for all

8

8 E

S' E

G,

i.e.,

h

E

A(S)

and, hence,

A(Sl) ~ l( S)p' The inverse inclusion is obvious as

A(S)-p, i.e.,

A

A( S . E G) = A(S). The lemma is proved.

5.8.7. Lemma. The algebra of the group A(S) is semiprime. Proof. Since, by the condition, Z is generated by its invertible elements, then the algebra of the group A(S) coincides with Z. Let us consider the formula

By lemma 3.9.4, this formula is true on the stalk

r

p'

As this is a Horn

formula, and the point p is arbitrary, then this formula is also true on the ring of global sections. The lemma is proved. 5.8.8.

A

Proposition. In a ring of global sections

RE

there are

k E S . EG.

such that

A

elements

a, r l .-, r k' v I .-, v k

e(a) = e. and for any

and elements

8 1,_"

8

x, y the following equality holds k

'l'1i. s)( yax)

=

.L fJ yr)8 ifJ Vi x),

~

=1

(7)

263

CHAPTER 5

r AI.. S) is a principal V-trace form determined by the homogenous

where

idempotent

e, and V is a reducedjinite group,

such that

1\

A(S) = v,

B(A(S» = B(V).

Proof. Let us consider the forms operations

for

the

invariant

sheaf.

r AI.. S) and fe as strictly sheaf The

predicate

P s(P

f.. x) =

1\

will also be strictly sheaf. Since the statement of the theorem is set by a Horn formula "'k' then, by corollary 1.11.20, it is T

XES· E G)

¢:::>

sufficient to establish that it is true on the stalk

r p'

It should be recalled that we are under the conditions of lemma

cp~S» .. (1])

3.9.22, and it is only necessary to check if the projection of the first component of the decomposition of f

E

l' p

equals

on

zero at any

k

H P' m + 1 $ j $ k.

Let us assume that this is not the case. Then, by condition SI, we

r

can find an invertible in sb= bs

for all

s

hI X ...

SI' where

E

element b, such that

p

x hk

SI =

)..

(1 ])

/\

Pp

(S . E G). Let us consider an automorphism

g = (b- h - . 1 x 1 x ._ x 1 x h . b-- 1 x 1 x _. xl) ).. (1 ]). ]

]

Let s = sl EF> _. EF> s n

From

here

we

get

E

-

sf

SI' In this case

=

h,

S j]

A(S)p = (F

g G!: (F

m)..

S rrJ x Gil, as

proposition is proved.

j> m.

and, rn)..

hence,

Sm> x G"

s

g

=

s.

(see 5.8.5).

Therefore, However,

This is a contradiction and, hence, the

264

AUTOMORPmSMS AND DERIVATIONS

Lemma. For any finite set

5.8.9.

of elements from

( ~1'M' ~ k)

A

there is an ideal

S . EG

i,

such that

IE JF ( R).

l~i~k.

for

~ ifJ I} ~ S

all

=

Proof. It suffices to show that the set V (v E QI 3 Iv E JF (R), vfJ I v}!:: S) is closed in the topology determined by fixed central

idempotents, and contains S. E G. Let v = sl e + .M +

S

n en' where

S

i E S, e i E E

G

. Let us find an

ideal J E JF, such that e i J!:: R. Then let us choose an way that

fJI}!:: J.

We

get

IE JF

eifJI}!:: QG(J R= RG!:: S

in such a

and,

hence,

vfJ I} !:: S.

Let,

then,

va E V, e a E E G, sup(e a) = 1. Let vafJJa} !:: S, eaJa !:: R. Then

where

ve a = vae a ,

i.e .,

I a , J a be ideals from

JF, such that

I = LeaIaJa is an essential ideal of the

ring R. In this case we get

vfJI}!:: LvfJe aIaJa} = LveafJI aJa} =

a

a

=LVafJI a(eaJa »!:: LVafJI a)!:: a

a

S.

The lemma is proved. 5.8.10. The proof of lemma 5.8.4 now becomes evident. Let a sum of all ideals assume that 1W = O. homogenous for both By lemma 5.8.9, we

be

of the ring 5 = J/ li.,S) contained in S, and let us Let us choose an idempotent e S; e(l), which is groups, G and V, and make use of proposition 5.8.8. can find an ideal IE JF, such that at y, x E I the

rightMhand part of relation (7) will lie in S. Therefore, and, hence,

W

l'rI(S)(IaI)=O. Let us recall that

'rI( S)(

IaI)!:: W

m 'rl(s)(x)=Lte1j)(x) and, 1

hence, 'rI(S)(X) = 'rl(slex). Therefore, the form

l'rI(S)(x)

turns to zero

on IaI + (1 - e) I. Reducing, if necessary, I, we can assume the last sum to lie in R and to form, therefore, an ideal from JF. Thus (see theorem 2.2.2),

HI(S)(x) is a zero form. Consequently,

form. Going now over to the stalk

r

p' we get

HJx) will also be a zero

p p( 1) = 0, as

're

is a

265

CHAPTER 5

principal trace form. Allowing for the fact that p is an arbitrary point from U(e), we get e· e(l)=O, while our choice was e~ e(l). Thus, this is a contradiction and the theorem is proved.

5.9 Correspondence and Extension Theorems Now we can summarize the results obtained above as a correspondence theorem. We will also show that a certain form of the extension theorem can be easily deduced from it. 5.9.1. Correspondence theorem. Let G be a regular group of automorphisms of a semiprime ring R. In this case mappings H -+ 1I( H), s -+ 1(S) set a one-to-one correspondence among all regular subgroups of the group G and all intermediate subrings obeying conditions BM, RC and SI. Proof. It immediately results from theorems 5.7.2, 5.8.1, 5.8.3 and corollary 5.4.7. 5.9.2. Extension theorem. Let G be a regular group of automorphisms of a semiprime ring Rand S S" be intermediate Galois I,

subrings of M-subgroups. If q>: S , -+ S " is an identical on RG isomorphism, and the ring s' obeys condition SI for mappings from q>G, then q> is extended to an isomorphism from G. Proof. Let us consider a ring R = R $ R with a group

G=

cl-). ~.

Then

S = ( s Et> s rp,

S E

S/}

will be an intermediate subring.

Its centralizer in C( R) = Q(R) $ C( R) is equal to a direct sum of centralizers of the rings S' and S". Therefore, S obeys condition BM and the first part of SI. The second part of SI results from the fact that S' obeys this condition for the mapping from q>G. Finally, condition RC can also be easily verified: if I is an essential ideal of S' and r 1 (9 r 2 E R, in

which

case

[i Et> q>( i) 1·

. (r 1 Et>

Ir1 !:: S', qX.,I) r 2 !:: S" and, hence, r2

= qX.,s 1)

for

qi..ir 1) = q>( is l ),

a i.e.,

certain

r1E

for

all

i E I,

then

S', r 2 E S" and, in particular,

sl E S';

I( r 1 - sl) = 0

r 1 $ r 2 = sl Et> qX.,sl) E S.

r 2) E S

and,

as

qX.,ir 1)

hence,

= q>( i)r 2'

r 1 = sl'

then

Therefore,

AUTOMORPlllSMS AND DERIVATIONS

266

G2 . Indeed, since

Let us now remark that A(S) is not contained in in S

the

=I

opposite

J( S)

=R

Let sl

e

({i...s 1)

H 1

case, $ R

H 2,

due

to

= qJ( sl)

A(S) = HI

x

H2

and,

hence,

which is impossible, since qJ is an isomorphism.

(gl x g2) >. (l, 2) E E( S). g2

closure,

gl

$ sl

Then

for

' which shows that

any

sl E S

I

we

have

gi is the sought extension.

The theorem is proved.

5.10

Shirshov

Finiteness.

Structure

of Bimodules

The use of an invariant sheaf when transferring the results of the Galois theory from prime rings onto semiprime ones, enables one to ignore some facts on invariants of M-groupS. Nonetheless, these facts are of their own interest and can also be obtained using sheaves. In this paragraph we prove the theorem on finiteness in the Shirshov sense (see 3.1.9) and describe the structure of (R, R G) -subbimodules of R F • 5.10.1. Theorem. Let G be a Maschke group of automorphisms of a semiprime ring R. Then R has an essential ideal which is locally-finite (in the Shirshov sense) over a fixed ring RG. Proof. For simplicity, let us call an element a E R finite if the right ideal aR is contained in a certain finitely-generated RG -submodule of the ring R: aR~

i= I

It is obvious that a set of all finite elements forms a two-sided ideal W of the ring R. Our task is to prove that its annihilator in R is zero. For this purpose it is sufficient to show that We"#. 0 for any homogenous idempotent

e. Let e be a homogenous idempotent, and let us consider a sequence of Hom formulas: cP n ¢:::> 3 a, t l , ... , tn, r l , ... , r n

(8)

n

& 'if x

ax=

L

i=1

ea"#. 0 &

eti'fJrix) ,

267

CHAPTER 5

where fe is a principal trace form viewed as a strictly stalk operation (see 5.6). Since by the construction efe= 're' then, by proposition 3.7.2 (where V

= (1 ),

holds

at

on every stalk p e U( e) .

r. p

of the invariant sheaf one of formulas

According

to

metatheorem

1.11.13

and

tP n

corollary

1\

1.11.20, one of formulas tP n holds on the ring of sections eRE. Let us choose an essential ideal I in such a way that fJ.r i I) !:: R, Iet i!: R, Ia!: R, and then we get IaI !:W, as if u, ve I, then n

(uav)x

= ua(vx) = L

(uet ~.)fJ. r ~.vx) .

i= 1

We have eW~ eIaI = IeaI:t:. O. The theorem is proved. 5.10.2. Theorem. Let G be a Maschke group of automorphisms of

a semiprime ring R. Then for any (R, R G) -submodule an idempotent Ie~

e e B (G)

and an

essential

ideal

v

of R F

of R,

I

there is

such

that

V= Ve.

Proof. Let us show that the right annihilator of V in the algebra B (G) is generated by an idempotent. Let us write this statement as a Hom formula 3 f

f2 =

& 'V d( 'V w

f& nB(f) = f& V t

nv(t)f= 0&

nv< w)n B( d) =: 0 ---7 fn B ( d)

= n B( d». 1\

Here n B' nv are strictly sheaf projections on Band VE, respectively (see 1\

1.11.9). As the annihilators of V and VE coincide, then this formula really gives the required statement. If e is a homogenous idempotent and p e U( e), then the stalk r. p has a finite prime dimension and , hence, the algebra B(G)p is semisimple, i.e., its anyone-sided ideal is generated by its idempotent. Now we are to use the metatheorem. Let e = 1 - f, where

ann<;)v =

f B( G). Let us consider an element

a e R to be finite with respect to v, if there is a finite set vI ,_., v n e such that

V,

268

AUTOMORPHISMS AND DERIYATIONS

n

aRe!: L RViR G. i= 1

It is obvious that the set W of all finite with respect to v elements forms a two-sided ideal of the ring R, and the task now is to show that the annihilator of this set is zero. Let e be an arbitrary homogenous idempotent. Let us consider a sequence of Horn formulas: cP n <=> 3 a, vI" .. v n, r l , ..., r n, t l , ... , t n T;f

X

n x

&

.

]=

I

7r.jv.)= v.& axe = ]

]

n

= L et i vi 'fee r i x) & ea:t:. O. i= I

As ere

= -re ,

Fp,

then by proposition 3.7.2, on every stalk

P

E U(e),

one of formulas cP n holds. By the metatheorem and corollary 1.11.20, we A

eRE.

see that one of these formulas holds on A

Let us show that for any W E VE there is an ideal I Iw c v. For this purpose it is sufficient to show contains

(WE RFI3 IE F, Iw!: v)

where ViE Iei w

C

and

R,

= aE lim A

we

we can find an ideal

suitable and

Ia E F ,

E

such that

F,

then

I E F ,

set

w=Lvie i •

Jae a !: R. In this case

= 1) and Iw k L I a Jaw k L I a Jaw a k V. Therefore, we can find an ideal v,

I

the

Finally,

for JaE F

that

and is closed. If

get

wa ' wa Ia!: V

I=LIaJaea, where

sup e a

then

V, e i E C,

VE

F, such that

E

if

we

set

IE F

(as

such

that

Now IaI kW and, hence, e W;;J Iae I :t:. O. Since a set of one-sided homogenous idempotents is dense, then W is an essential ideal. The theorem is proved. In a symmetrical way we get the following theorem. IEt iV i!::

5.10.3.

Ia k R, 'fee r i I) k R.

Theorem. Let

G

be an M-group

semiprime ring R. Then for any (R be found an idempotent e

E

of automorphisms

G. R) -subbimodule

w of

RF

of a

there can

!B (G) and an essential ideal I of R, such that

269

CHAPTER 5 eI!:: W= eW.

Theorem 5.10.1 has a left analog as well. Let us formulate it in a somewhat different way, taking into account the fact that the coefficients fJ,.r i x) in formula (8) are linear from the right with respect to x over RG.

5.10.4. Theorem. Let G be a Maschke group of automorphisms of a semiprime ring R. Then a set W of elements a of the ring R, such that the left R G -module Ra is embeddible into a finite direct sum of copies of a regular module R G, contains the essential ideal of the ring R. Proof. Let e be a homogenous idempotent. Let us denote by W e a set of all elements a E R for which the left analog of formula (8) holds. In this case W e is a nonzero ideal of the ring R and e W e~ 0 . Let a EWe and I be an essential ideal of the ring R , such that fJ,. Ir JJ !:: R, 1 :$; k

:$;

n. Then formula (8) shows that at

xia ~ fJ,.xir})

. out an em bedd'mg carnes

LI

R"~a

i

E

I

the mapping

+... +fJxir ~ mto

~R G

+ .• + T

R G"

There fore, I W""'W ean d

n

W e is the sought ideal. The theorem is proved.

e

Theorems 5.10.2 and 5.10.3 make it possible to obtain a statement which is, in a certain sense, inverse to lemma 5.6.6. Corollary. Let G be an M-group of automorphisms of a p be an essential ideal of a fixed ring. Then the left the right ideal PRJ contains the essential ideal of the ring R. It suffices to show that the right (left) annihilator p in If pa = 0, then for any principal trace form f one can find I of R, such that f(aI) !:: R. We have ideal pf(aI) = f(paI) = 0, i.e., f(aI) = 0 and, according to the basic property of principal trace forms, we get a = 0 (see 5.6.8).

5.10.5. semiprime ring ideal Rp (and Proof. /1J (G) is zero. an essential

R,

REFERENCES J.M.Goursaud, J.-L.Pascaud, J.Valette [50]; I.M.Coursaud, I.Osterburg, J.-L.Pascaud, I.Valette [51]; V.K.Kharchenko [70,80]; J.-L.Pascaud [132,333]; A.I.Shirshov [144]; A.V.Yakovlev [155].

CHAPTER 6 APPLICA TIONS This concluding chapter is of a somewhat mosaic character and is mainly intended for the advanced reader. Here some problems are considered, which are not interrelated but they are in this or that way associated with the same basic topics: free algebras, non-commutative invariants, radicals, Goldie rings, rings of quotients, Noetherian rings, etc. It is clearly impossible to try to give a closed presentation of such a wide range of material within a monograph (let alone a single chapter). Therefore, use without proof published elsewhere will be made of classical theorems of the ring theory, such as the Cohn criteria on freedom of subalgebras, Goldie theorems, the lemma on composition (the Grabner basis), etc .. The last section explores the general concept of the action of Hopf algebras embracing both the cases of automorphisms and derivations, and presents certain approaches to studying skew derivations from this point of view.

6.1

Free

Algebras

Let F ( x) be a free associative (but non-commutative) algebra over a field F with a set of generators x. On F ( x) one can define a positive-degree function v, by setting arbitrarily the value of v on the generators and expanding v naturally, first on the monomials and then on the whole of the algebra. In this case we assume the degree of a polynomial to be equal to the maximum of the degrees of the monomials occurring in it with nonzero coefficients. In a free algebra the weak algorithm with respect to the degree function introduced in such a way ([34], proposition 4.2,) holds. Let us fix a function v throughout the section. An automorphism of a free algebra will be called homogenous if it transforms homogenous elements into homogenous ones with their degree preserved. For instance, if v is a usual x-degree, i.e., v( x) = 1, where {x

J

= x, then the homogeneity of an automorphism g implies that it acts

on the generators linearly, i.e.,

xl = La i j xj' ai j E

F.

6.1.1. Theorem. Let G be a finite group automorphisms of a free algebra. Then the mappings

270

V. K. Kharchenko, Automorphisms and Derivations of Associative Rings © Kluwer Academic Publishers 1991

of

homogenous

271

CHAPTER 6

S

-4

A(S)

H -4 I( H)

= {g

= {s

E

AutF ( X) I Tj S

E

S(S g

E F ( X) I Tj h E H(S h

=

= S)) ,

S))

set up a one-to-one correspondence between all subgroups of the group G and all intermediate free algebras. A subalgebra S:2 I(G) will be a Galois extension of I( G) iff it is invariant relative G. In this case a subgroup A(S) is normal in G and the Galois group of the extension I( G) of S is isomorphic to G / A(S). To prove this theorem, it will be sufficient to check the validity of the following statements: (1) The algebra of invariants of any finite group of homogenous automorphisms is free. (2) Any finite group of homogenous automorphisms is a Galois group, i.e., a group of all automorphisms of the algebra F ( x) over its algebra of invariants. (3) Any free subalgebra which contains the algebra of invariants of a finite group of homogenous automorphisms is an algebra of invariants of a certain group. (4) Any automorphism of an intermediate algebra which leaves the elements of the algebra of invariants of a given group of homogenous automorphisms fixed, can be extended to an automorphism of the algebra F ( x).

We shall show that statement (1) is valid for any (not necessarily finite) group of homogenous automorphisms, while statements (2) and (3), as well as (4) hold for a finite group of any (not necessarily homogenous) automorphisms. In this case it should be noted that any embedding of an intermediate subalgebra into F ( x) which leaves the elements of the algebra of invariants of a finite group fixed, can be extended to an automorphism of the algebra

F ( x).

6.1.2. Proposition. The algebra of invariants of any group of homogenous automorphisms of a free associative algebra is free. Proof. Using Cohn's criterion on freedom of a subalgebra of a free algebra ([50], proposition 7.2), it is sufficient to show that the subalgebra I( G) satisfies the weak algorithm relative the degree function vo

Let from

aI' a 2 ,-, a n be a right

I( G), and let v( a 1) ~ v( a 2 ) ~ _0 ~ v( a

are v-dependent in the algebra F

<X

v-dependent set of nonzero

)

°

J

°

Since the elements

I( G), they will also be

elements al'_o, an

v-dependent

Let us choose the least number k, such that aI' a 2 ,_, a k

in

are still

v-dependent in F ( X ). Since in a free algebra the weak algorithm holds, one of the elements (it will be a k> since k is minimal) will be right vdependent on the preceding ones in

F ( x), i.e.,

272

AUTOMORPHISMS AND DERIVATIONS

v(a k -

L

i< k

ai b i )< v( a 0·

(1)

In this formula under the sum sign we can eliminate the terms with their degree strictly less than v(a i b)

=

v(a)

v(a k)'

Therefore,

one

can

assume

that

either

+ v( b) = v(a k) or b i = O. Additionally, inequality (1) is

not violated if every

bi

is replaced with its highest degree homogenous

component, i.e., one can assume that b i are homogenous elements. Now let g be an arbitrary automorphism from the group G. As g is homogenous, we have

v(a k -

L i
aib i )= v([ak-Laibilg)= v(a k -

aib;).

L i
Allowing for the fact that the degree of the difference of any two elements is not greater than the maximum of the degrees of these elements, we get V(" a .(b ~. - b~)) = v(fa k- "a .b'!J- [a k - "a.b L..~ ~ L..~~ L..~~.J) < (2)

If

now

for

a

i

certain

we

g

have g

because the elements b i' b i are homogenous and, hence, relation (2) implies the right v-dependence of the set a 1,_· a k _ l' which contradicts the choice of k. v( b i - b) = v( b)

Therefore, dependent on proposition.

bi

E

I( G), and relations (1) show

aI' a 2 .-, a k _ 1

in

the

algebra

ak

I( G) .

to be right vThis

proves

the

6.1.3. Lemma. Let R satisfy a two-term weak algorithm. Then, either C( R) = R or R is not a ring of skew polynomials, where C( R) is a two-sided Martindale ring of quotients. Proof. It should be remarked that in C( R) there are no zero divisors. If qJlfI = 0, qJ -:t= 0, lfI-:t= 0, then for some p, q E R we have o -:t= pqJ E R, 0 -:t= lfI q E R and we come to a contradiction with the fact that

= O. Let us assume that

( pqJ) ( lfIq) = p( qJlfI ) q

C( R) -:t= R.

Among the elements

qJ E C( R)

not

273

CHAPTER 6

lying in R, let us choose one in such a way that the set Rcp (") R has a nonzero element pcp of the least possible degree. Then for a certain nonzero q E R we have qJq E R. In the ring R let us consider the equality (pcp) q = p(cpq). Since the two-term weak algorithm holds in this ring, we have two options [34]: (1) pcp = PA + s, where A, SE R, v(s)< v(p) (if v(pcp)~ v(p»; where A, s E R, v(s) < v( pcp) (if (2) p= (pcp)A + s, v( pcp)

< v( p». In

v(

A cp - A»

the

former

case

we

have

A cp -

A) = s,

= v(s) < v( pcp). Given the choice of the element

i. e . , cp we get

A E R, i.e., cp E R, which is a contradiction. In the second case we get A cpA - 1) = - s, i.e., as above, cpA - 1 E R. It implies cpA E R. The equality A cpA) = p - s, however, gives

qJ -

v( qJA) = O. Since in a ring with a two-term weak algorithm a set of elements of a zero degree forms a sfield, then cpr = 1 for a certain r E R and, as

ex R)

has no zero divisors,

rcp = 1. Let us now show that if the degree of

r is positive, then R is a ring of skew polynomials. Let x be an element of the least positive

degree from R. Let O:t:. I be an ideal of the ring R such that cpI ~ R. Let us arbitrarily choose a nonzero element i E I. Then we have r( cpxi) = xi. Therefore, due to the two-term weak algorithm, r = xj1 + 8, j1, 8 E R, v( 8) = O. If now y is another element of the least positive degree, then r = yJl + 8', or xj1- yJl= 8'- 0, in which case max(v(Xj1),v(yj1')} >0= v(o'- 0). Using

the

two-term

weak

As

v(<\)= v(82 )=0.

algorithm,

we

get

v(8x) = v(8) + v(x)= vex)

y = x~

for

+ all

~,

where 8 E.1=

(al v(a)} = 0, then 8x = x8 h + 8(1. [t is clear that h is a monomorphism of the sfield .1, while (J is its h-derivation. The task now is to show that x and .1 generate the ring R. Let us carry out induction over the degrees. If s E Rand v(s) > vCr), then the equality r(cpsi)= si yields s= rJl'+ s', in which case v(s')< v(s), v(j1") = v(s)- v(r)< v(s) and one can make use of the supposition of induction. If

then the equality r( cpsi)= si affords where v(s') < v(s). We, therefore, get sJl' - xj1 = 8 - s' and, hence, since the algorithm is two-term and weak, s= xJl"+ s", in which case v(Ji")= v(s)- v(x) < v(s), v(s")< v(s), i.e., one again make use of the induction proposition. As the condition of the existence of an n-term weak algorithm is left-right symmetric [34], then h is an automorphism. This is the proof of the lemma. v(s)

~

v(r),

r = sJl' + s' = xj1 + 8,

274

AUTOMORPIDSMS AND DERIYATIONS

6.1.4. Proposition. Any finite group of automorphisms of a free associative algebra is a Galois group.

ex F

Proof. Since the algebra of a group is contained in then, by the previous lemma, it is also contained in

F (

X).

(

x) ),

All

the

invertible elements of the algebra F < X ), however, are exhausted by the elements of the field F. Therefore, the algebra of the group coincides with F and the only inner automorphism, which is identical, belongs to any group. Now if we, for instance, make use of theorem 3.2.2, the proposition will be proved. 6.1.5. Proposition. Any free subalgebra of a free assoczatlve algebra which contains the algebra of invariants of a finite group, is itself an algebra of invariants of a certain group. Proof.

Let

F (

Y)

be a free subalgebra, G be a finite group of

automorphisms and I( G) ~ F ( Y). Let, then, IA(F ( X ) ) = S. While proving the previous proposition we have seen the algebra of the group G to coincide with F. Therefore, for the group G and ring F ( Y) the conditions of corollary 3.9.20 hold (as, incidentally, for any finite group of automorphisms of a ring with no zero divisors). Thus, by this theorem, the ring it

s.

contains a nonzero two-sided ideal A of the algebra

F ( Y)

is

evident

that

A

is

also

an

ideal

in

F ( Y).

Then

Therefore,

ex F ( Y) ) = ex A)= Q(S). As F ( Y) is a free algebra, it is a ring of skew polynomials only when its rank is equal to one, i.e., I YI = 1. In particular, in this case the algebra is commutative and, hence, the following identity with automorphisms holds: Lx g )( gE G

Lyg)=( gE G

As the group G contains reduced

identity.

By

no

theorem

non-trivial

=

ex F

6.1.6. the algebra

< Y) ) ~ 5, i.e., Proposition.

F (

inner

2.3.1, in

xy = yx, but from the very beginning non-commutative (i.e., I xl > 1 ). Therefore, one can use F ( Y)

gE G

gE G

F ( Y)

F ( X)

F (

X)

lemma

=

automorphisms, we

find

this an

is

a

identity

has been assumed to be

6.1.3.

We

have

5, which is the required proof.

Let G be a finite group of automorphisms of

x) and 5 be a subalgebra containing

embedding of 5 into the algebra

F (

I(G).

x), which is identical on

Then

any

I( G), can

be extended to an automorphism of F ( X ) . This proposition results immediately from corollary 3.11.4. One has.

275

CHAPTER 6

however, to remark that, by lemma 6.1.3, the group A(F ( X») coincides with the group of automorphisms of F ( x). Theorem 6.1.1 is deduced from the propositions proved above using standard considerations. It should be remarked that in this theorem there is no need to assume that I xl > 1, as it is evident that the theorem is also valid when I xl = 1. Let us now go over to a free associative algebra over a field of a positive characteristic, p. 6.1.7. Definition. A derivation J1. of the algebra F ( x) is called homogenous if it transforms homogenous elements into either homogenous ones with the their degree preserved, or to zero. For instance, if v is a usual degree, i.e., v( x) = 1, x = {x i}' then homogeneity implies that acts linearly on the generators, i.e.,

J1.

xl-! = Lx ]·a . '. ~

~J

6.1.8. Theorem. Let L be a finite-dimensional restricted Lie algebra of homogenous derivations of the free algebra F ( x). In this case A

the mappings A ~ F ( x) , s ~ Der S F ( x) sets up a one-to-one correspondence between the restricted Lie algebras of L and intermediate free L

subalgebras. Then the differential extensions of F ( x) correspond to the ideals of L and vice versa. Proof. By theorem 4.5.2 (on correspondence) and by theorem 4.6.2 (on extension), it is sufficient to show that any free sub algebra is rationallycomplete in F ( x), that a set of the constants of a finite-dimensional restricted Lie algebra of homogenous derivations forms a free subalgebra, and also that a homogenous nonzero derivation cannot be inner (for Q). The rational completeness of free subalgebras results from lemma 6.1.3. The same lemma also affords that the derivations which are inner for

ex F

also be inner for F ( x), however, the equality for x EX, f E F ( x) \ F results immediately in a contradiction and, hence, L consists of outer derivations. The following proposition remains also valid for infinite-dimensional Lie algebras over a field of an arbitrary characteristic. v( x)

(

x»)

= v(xf -

will

fx)

6.1.9. Proposition. A set of constants of any Lie algebra of homogenous derivations forms a free subalgebra. Proof. It coincides with the proof of proposition 6.1.2 nearly to the word. One only has to replace the word "automorphism" with the word

"derivation" and to consider a "derivative" b·b~. ~ ~

b~ instead of the difference ~

276

AU1OMORPIDSMS AND DERIVATIONS

6.1.10. Well known is the fact that any finite group is presented by linear transformations of a certain finite-dimensional space V. If we consider an arbitrary basis of V as x, then we see that any finite group is presented by homogenous automorphisms of a free algebra F ( x). Therefore, theorem 6.1.1 shows, in particular, that the lattice of free subalgebras which contain a fixed free subalgebra can be antiisomorphic to the lattice of subgroups of an arbitrary finite group. An analogous problem of presentation arises for an arbitrary restricted Lie algebra as well. It is still more interesting , since, according to the Baer theorem, a restricted Lie a-algebra of derivations of a finite extension of fields K:;? k is generated over K by one element and, in particular, can be arbitrary in no way. The following proposition shows that any restricted differential Lie algebra over a field C is presented by homogenous derivations of a certain ring C ( X ). In this case if the initial algebra acts trivially on C (i.e., it is a usual Lie p-algebra), then we have it presented by derivations of a free algebra. 6.1.11. Proposition. Any restricted differential Lie C-algebra can be presented as an algebra of outer derivations of a prime ring with a generalized centroid c. Proof. Let us first show that any restricted Lie -algebra D is embedded into its universal enveloping UD. Let M be a certain quite ordered basis of the space D over a field C, and let (c i) be the basis of C over a field of constants, P. Let us

a

consider a free associative algebra

Fa = P <

generated by the symbols (J Ji' la, where /1 E factor-algebra of Fa determined by the relations

where

.

/1[P]=La./1.c., /1,/1.E M, a.;E P; ~

~

~

~

(J Ji'

M,

la >,

which

is

freely

a E (c J. Let U be the

CHAPTER 6

where

277

c 1c 2 =L.a i Ci

, cl' c 2 ' CiE (e i ), aiE P.

In the set of symbols order in such a way that the set of words from

( 0" J1' 1 a) let us introduce a certain complete

0" J1. > 0" P ¢::>

( 0" J1.' 1 a)

11 > p and

0" J1.

< 1 a for all 11, a. On

the order > > will be determined in the

following way. Let us assume that the word f is bigger than the word g (f> > g), if the word f M obtained from f by crossing out all the symbols from (la) is bigger than the word q M under a standard order of words. If f M = g M' then bigger is the word which is bigger under a standard order (with respect to the order ». Now we can easily check that the set of the written relations of the algebra Fa is closed relative the compositions (see [23] and [24]). This implies that the images of the words not containing subwords of the type O"J1.O"p at

11> P; O"~; lcO"J1.; lCI lez, form a basis of the algebra Fa (the

Grabner basis). Let us now denote by

{l,

La the images of the symbols 0" J1.' 1 a in

u, respectively. In this case the mapping ~: L. a ill i Ci ~ L. a i (l i LC i determines a certain embedding of the space D into u. It is evident that ~(D) generates U as an associative p-algebra with a unit. Then, U turns into a C-module when identifying Lc with an element e E C, since

( L.a i Lc) == C. In this case c{l = Lc {l = {lLc + L.a i LC i = {lc + cJ1., where c J1. = L. a i C i. Therefore, the task now is to check if ~ is a homomorphism of restricted Lie rings. Using identities (4) (1.1.2), we get

~ ( [ III c1' 112 S]) = ~ ([ III ' p:~] c 1 c 2 + III S Ci2 + ~ c 1 C;I) =

Ci2

c;l

= [ {l1' ~]c 1 c 2 + {l1 S + ~ c1 = [ {l1 c 1' ~ c2l ~« III + 112 } pl) = ~(I1~ pl + ~pl + w( 111 ,112)) = = {l1

p

p

,

p

+ {l2 + W( (l1' ~) = ( ~ +~) ,

~« I1C)[ pl) = ~ul plcp +

j1T(C)) = {l PcP + (IT(C) = ({lc) p.

278

AUTOMORPIDSMS AND DERryA TrONS

Therefore, U is a universal enveloping and ~u is an embedding. Let us add to the set of generators of the algebra FO one more symbol, y, which is less then the former ones, and let us add to the set of relations relations of the type ie Y = Yie' C E (c i) . In this case the obtained set of relations is again closed with respect to the compositions. Hence, D does not intersect with the center of U ( P), where P is the image of y under a natural homomorphism. In particular, D is a restricted Lie a-subalgebra of the factor-algebra U ( P) / z, where z is the center

P).

of U (

Thus, in order to prove the proposition, it is sufficient to show that the factor-algebra over the center D / Z of any restricted a-algebra D is presentable in an appropriate way. Let (J 11'

of

C ( M) be a free associative c-algebra generated by the set

/1 EM. On the generating elements the

elements

from

D

by

C ( M) let us determine the action

the

formula

L

(J~=L(JI1Ci' o

where

[/1, v] = /1 i C i' Now the formula of derivation (xy) 11 = xl1 Y + xyl1 makes it possible to propagate the action of D onto c( M). One can easily check that

Thus, q;:D~

proved.

we

get

a

homomorphism

of

the

restricted

DerC ( M), the kernel of which is equal to

z.

Lie

a-algebras,

The proposition is

6.1.12. By way of concluding the section, let us remark that it is still unknown if we can be free from the condition of homogeneity in theorems 6.1.1 and 6.1.8. We have seen that to this end one should show that the algebras of invariants of a finite group of automorphisms is free (and the algebra of constants of a finite-dimensional restricted Lie algebra is also free). These problems have been considered elsewhere. Thus, for instance, P. Cohn ("On the Automorphism Group of the Free Algebra of Rank 2", Preprint, 1978) has showed any finite group of automorphisms of an invertible order of a free algebra of rank 2 to be conjugated with a certain linear (i.e., affecting the generators linearly) group. Hence, the algebra of its invariants is free. The Cohn's approach is based on the Cherniakievich-Makar-Limanov theorem stating that every automorphism of a free algebra of rank 2 is tame.

279

CHAPTER 6

Since so far no examples of not tame automorphisms of free algebras have been known, our problem is not less interesting for a group of tame automorphisms as well. It should also be added that so far no examples of even infinite groups with non-free algebras of invariants have been known. For derivations the situation is somewhat different. G.Bergman gives an example with a non-free kernel. In a free algebra of rank 3 let us consider the derivation J1, determined by the formulas xJ.1= xyx+ x, yJ.1=_ yxy, zJ.1=_ x.

One can show that the kernel of this derivation is generated by the elements

+ x + z +1

qJ = xyz

r= xy

These elements are Cohn's theory, one not free. If the basic generated by J1 is By way of explicit form.

+1 = zyx + x + z q= yx

S

related by qJq = IS, with the help of which, using the can easily show that the subalgebra generated by them is field has the characteristic p> 0, then the Lie p-algebra infinite-dimensional. conclusion let us formulate the problems arising in their

6.1.13. Will a free algebra of invariants of an arbitrary group of automorphisms of a free algebra be free? The same problem is also open for discussion for tame, finite, and finite groups of an invertible order. 6.1.14. Will an algebra of constants of a finite-dimensional restricted Lie algebra of derivations of a free algebra over a field of positive characteristic be free?

REFERENCES G.Bergman [19]; G.Bergman, P.Cohn [20]; Bokut' [23, 24]; P.Cohn [34,35]; A.Czerniakiewich [37]; V.K.Kharchenko [83]; A.T.Kolotov [87]; D.R.Lane [92]; L.G.Makar-Limanov [101]; W.S .. Martindale, S.Montgomery [107]; A.I.Shirshov [143].

280

AUTOMORPIllSMS AND DERIVATIONS

6.2

Noncommutative

Invariants

In this section we shall dwell on the problem of finite generality of algebras of noncommutative invariants of linear groups. The Koryukin's theorems presented here show finite generality not to exist, as a rule, and explain the reason of this phenomenon, which can be attributed to the fact that symmetrical groups act on homogenous components. This peculiarity taken into account, we come to an analog of the Hilbert-Nagata theorem. 6.2.1. Definitions. Let v be a finite-dimensional space over a field > I, G be a certain group of its linear transformations. Let us

F, dim V

denote by F < v) the tensor algebra of this space.

< v> = F+ v;:

F

If we fix a basis

X =

V

®

V

-+

V

®

V

® v;: ...

(x p _' xn), then the algebra

F

< v) can be viewed

as a free associative algebra F ( X ). The action of the group G is uniquely extended onto F ( v). The noncommutative invariants or simply invariants are fixed elements of the algebra F ( v). A set of all invariants will be denoted by InvG. Definition. The support space of a subset

6.2.2.

the least, by inclusion, subspace

W b:

v, such that

A b: F (

v)

is

Ab: F ( w).

6.2.3. Lemma. Any subset has a support space. If a subset

A

is

stable relative G, i.e., A g = A, g E G, then its support space is also stable relative the action of G. Proof. It is obvious that for subspaces u and W the equality F ( U ) ( l F ( w) = F ( U ( l w) is valid. Therefore, if Ab: F ( U ) , then

A b: F ( W),

support

space.

A= A g b: F (

A b: F ( U

If

wg ) ( l F ( Wg ::1 wand,

space, then The lemma is proved.

A

w)

g

(l

w),

which implies

= A, g E G

= F ( wg

(l

and

the existence of a A b: F (

w) and, hence, if

W

w),

then

is a support

since the space W is finite-dimensional,

wg = w.

6.2.4. Theorem. The algebra of noncommutative invariants InvG is finitely generated iff the group G acts on the support space W of the algebra InvG as a finite cyclic group of scalar transformations.

CHAPTER 6

281

In one respect the theorem is obvious: if the restriction of G on W is generated by the transformation g: W --7 w, such that gw = aw, am = 1 , then the algebra InvG is equal to F ( rJ9 m) and, hence, it is finitely generated (here m is the order of g, i.e., a k *- 1 at 0 < k < m). The proof of the inverse statement will be divided into a number of steps.

6.2.5. It should be recalled that on the n-th tensor degree of the space v, besides the linear group GL(V) , a group of permutations 5 n acts, rearranging (in a similar way) the factors in all tensor sums:

In this case the action of GL(V) and 5 n commute and, hence, homogenous component of the degree n of the algebra Inv G is invariant relative the action of Sn' i.e., for every invariant r of the degree n we have n! new invariants (not necessarily different),

r Tr ,

nE

5 n'

6.2.6. Let x = (x p _' x~ be a basis of the space v, with its elements called letters. Every element f of the tensor algebra is uniquely presented as a linear combination

I;a w w w

of different words from

X.

If

a w *- 0, then we shall say that the word W is used in the presentation of f. If A is a certain subset of a tensor algebra, then we shall say that the letter x occurs in A if x occurs in at least one of the words used in the presentation of elements of A. And, finally, a sequence of letters Yl' Y2'·' Ym.- will be considered agreed with A, if at least one of the words of type Yl Y2-' Ym is used in the presentation of a certain element f

E

A.

6.2.7. Lemma. If a multiplicative closure of a finite nonempty set of words M is closed relative the action of symmetric groups, then any infinite sequence of letters Yl' Y2-.' Y m•• occurring in M agrees with M. Proof. Let m be the maximum of the lengths of the words from M. Let us consider a word w = WI'. W m, where wi is a word from M wit h the letter Y i included. Let us rearrange the letters in the word w in such a way that the new word had the form wTr = Yl Y2·· Ym z. By the condition, this word occurs in the multiplicative closure of the set M, i.e., wTr = v 1-. v s, where

v i lies in M. Since the length of viis not greater

282

AUTOMORPI-llSMS AND DERIVATIONS

than m, then, comparing both presentations of the word conclusion that v I = Yr- Y k EM. The lemma is proved.

wlr:

we come to the

6.2.8. Lemma. If the algebra (A) generated by a finite set A of the elements of a tensor algebra is homogenous and closed relative the action of symmetric groups, then any infinite sequence of the letters occurring in A agrees with A. In order to prove the lemma, it is sufficient to choose as M a set of all the words used in the presentation of A and to make use of the previous lemma, taking into account that the fact that the word ml occurs in the presentation of

al

E

(A), and the word

occurs in the presentation of

~

~ E (A) implies, due to homogeneity, that the word presentation of a l a2 .

ml m2

occurs in the

6.2.9. Lemma. Let g E GL(V) and a field F be algebraically closed. Then, if g is not a scalar transformation, then there is a basis X of the space V and an infinite sequence of letters (from X) which does flat agree with In v( g) . Proof. Let us first assume that in a certain basis X a transformation

Xl

is set by a diagonal matrix = a i xi' a i E F. Let also, for definiteness, a l -:;:. ~. Let us construct, by induction, a sequence of letters g

YI ,_. Y i"-' If

a l -:;:. I,

then we set

if

YI = xl;

a l = 1, then

we assume

YI = ~. If YI ,_., Y i - I are constructed and the element Yr- Y i - I Xl is not fixed relative g, then we set Yi = xl; in the opposite case we set Yi

= x2·

By the construction, all the words

YI'- Y m are not fixed relative

g. However, in the case under discussion the algebra Inv( g) is a linear enveloping of fixed words. This fact implies that none of the words Yr" Y m is used in the presentation of the

elemenl~

f E Inv( g), i.e., the sequence

constructed does not agree with Inv( g). In a general case let us choose a basis g

has

a Jordan

xi = ax2 ,

normal

X

X{ = aXI + X],

form:

where the transformation and

xi = aX2 + x 3'

or

in which case the subspace "".3 F + _. + Xn F is invariant relative g. Let us consider a homogenous invariant f of the automorphism g and present it as f

where

f3,

= f3

rEF,

m

Xl

+

r

m- I

Xl

X

2+

f

I'

and the presentation of

fl

does not include the words

283

CHAPTER 6 xlm' xlm- I X 2' Then we have

in which case in the presentation of f2 no use is made of the monomials xlm' xlm-l x 2 · A s

g f- f ,we get

Allowing for the fact that the eigenvalue a of the non-degenerate transformation g is other than zero, we get f3 = O. This implies that the sequence of letters xl' xl ~_, xl'" does not agree with proof of the lemma.

Inv( g). This is the

6.2.10. Lemma. Let K be an extension of a field F; G be a group of linear transformations of the space V over F. In this case the algebra of

invariants of the group G in F ( v) is finitely generated invariants of the group G in F ( The proof is obvious.

V

iff the algebra of

® K) is finitely generated over K.

6.2.11. Let us go over to proving theorem 6.2.4. Let InvG be finitely generated. By lemma 6.2.10, the field F can be considered algebraically closed. Let W be a support space of the algebra of invariants. By lemma 6.2.3, this space is invariant and, hence, by restricting the action of G on W one can assume, with no generality violated, that V = w. If g is not a scalar automorphism from G, then, by lemma 6.2.9, there is a basis of the space V = W and a sequence of letters from this basis which does not agree with Inv( g). However, InvG S;;; Inv( g), and, hence, such a sequence of letters does not agree with InvG either and, by lemma 6.2.8, the algebra Inv G cannot be finitely generated. Thus, G is a group of scalar transformations. If it is infinite, then it has no nonzero invariants at all and, hence, 0 = V = W is a contradiction. Therefore, G is finite. Since G is embeddible into F (to every scalar transformation we put into correspondence its eigenvalue), then it is, as a finite subgroup of the multiplicative group of the field, cyclic. This is the proof of the theorem.

Corollary. Let G be an almost special group of matrices (i.e., the kernel of a mapping g -+ det g has a finite index in G). If the algebra InvG is finitely generated, then G is a finite cyclic group of scalar matrices. 6.2.12.

284

AUTOMORPIDSMS AND DERIVATIONS

Proof. It is sufficient to show that the support space of the algebra of invariants coincides with the whole space v_ Let us construct a standard polynomial

where summation is carried out by all the permutations easily see that

1r _

One then can

sg = det g - G _ Since det G is a finite group in the field,

we can find a number m, such that (det g) m = 1 for all g E G _ Therefore, we have S mE Inv G _ Let w be a support space of the element Sm _ Let us choose its basis v I ,__ , v k and extend it to the basis v I ,__ , v n of the space v_ Let H be a linear transformation v i ~ xiIn this case S( xl'--' x~ = S(vl'--' v~

sax VI ,--, v ~

V =

h

E F( v I ,..- v k)'

=

det h - S(vp _' v~

and,

which is impossible as k < n _ Therefore,

hence, r = n,

w, and the corollary is proved_

6.2.13. Corollary. The algebra of noncommutative invariants of an irreducible group is either trivial or finitely generated. The proof is obvious. 6.2.14. The proof of theorem 6.2.4 shows that the main obstacle for finite-generatedness of an algebra of invariants is the action of symmetric groups on homogenous components. Allowing for the fact that the construction of the invariant fTr by an invariant f and a permutation 1r is of no calculation difficulty, it is natural to study the algebra of invariants as an associative algebra, on the homogenous components of which symmetrical groups are acting. Now the question arises: if there exists a finite number of invariants, such that all the rest are expressed through them by using the operations of algebra and the actions of symmetrical groups on homogenous components? The answer to this question is positive for reductive groups (it should be recalled that a reductive group is a linear group with all its rational representations completely reducible). 6.2.15. Definition. The subalgebra A ~ F(v) is called an ssubalgebra if it is homogenous and closed relative the actions of symmetric groups on homogenous components. The ideal I of A is called an S-ideal if it is an s-subalgebra. 6.2.16. Theorem. The algebra of noncommutative invariants of a reductive group is finitely generated as an S-algebra. A general scheme of this theorem is the same as that in the NagataHilbert theorem.. Let us start with an analog of the Hilbert theorem on the

CHAPTER 6

285

basis.

6.2.17. Theorem. Any strictly ascending chain of s-ideals in is finite. The proof of this theorem is based on the Higman lemma.

F(V)

6.2.18. Lemma. For any infinite sequence of words of a finite alphabet one can choose an infinite subsequence, every word of which can be obtained from any subsequent word of the subsequence by crossing out the letters. Proof. On the set of all the words let us introduce a relation of a partial order u ~ v iff u can be obtained from v by crossing out letters. Let us call the set of words p Higman if for any infinite subsequence of words from P we can choose an increasing subsequence. Our task is to show a set of all the words from

X::::

(x 1._, x n) is Higman.

For the word W let us set wi:::: (vi W f::. v). Since x~ is a set of all the words not including the letter xi, we can assume that all the sets ~

x-] are Higman ones. By induction over the length of the word

W

let us

show that all the sets wi are also Higman. Let w:::: xv, where x is a letter, v is a word for which vl- is a Higman set. An arbitrary word u E 1Ir1- which does not belong to x~ can be presented as axb, where a E x~, bE vl-. To this end it is sufficient to find the first letter x occurring in the presentation (since, if b> v, then axb> xv).

Let us consider an arbitrary sequence of words from

a.

where

~

E

x~, b.

~

E

v~, in which the cofactors

xb ~.;

wl- :

could be absent.

Since x~ is a Higman set, one can eliminate a part of the terms from this sequence in such a way that in the remaining infinite sequence the cofactors a i will be increasing. As v~ is also a Higman set, from the sequence obtained one can eliminate a part of the terms as well, so that in the remaining infinite subsequence the right cofactors b i will be also increasing, which is the required proof. Therefore, all the sets wi are Higman. Let wI' W2 W n-' be a sequence of words from which one cannot choose an increasing subsequence. In this case not all the words w2 ,_., W n.- lie in wi, i.e., one can find a p_

pair Wk

WI

<

<

Wm

Wk'

Analogously, not all the words wk + 1'-' wn '_· lie in

0,

i.e.,

at a certain m > k, and so forth. This is the proof of the lemma ..

286

AUTOMORPIDSMS AND DERIVATIONS

6.2.19. The proof of lemma 6.2.17. Let II

<

12 < ...

I m<

...

be an infinite strictly ascending chain of S-ideals in F(V) .Let us fix a basis

x = {x p_' x nl of a space v. From the elements of the set

let us choose a homogenous element Z m with a minimal senior word W m in the lexicographical ordering of the words. By the Higman lemma, one can find numbers i < j, such that Wj

= YI . ..

Ym

wi = Y r 1 Y r 2

•••

(

I m\ I m _ I

Yk EX, 1 :s; k :s; m) (1 :s; r l < r 2 < ... < r s:S; m)"

Y rs

Let us denote through h the word obtained from W j by crossing out the letters standing in the positions r I ,.., r s. Let wi h = Yn(I Yn( m). Then n

r-

is a permutation, in which case

= ( wi h) a, where

Wj

Let us write an element

Zi

as

Wi

+

(J

= 1e I .

Lw. aw' w

where summation is

carried out over monomials W which are less than wi and aware from the basic field. (With no generality violated, one can assume that the senior coefficient is equal to the unit). We have

It should be remarked that the word (wi h) a is the senior word in

the presentation of (z i h) a. If

W

<

wi'

then, by crossing out the letters

occurring in the positions r I ,.., r s in the monomials (w i h) a and (wh) a, we get the same word, h. Therefore, the first from the left difference between the words (wi h) a and (wh) a is attained at one of the positions r I ,.., r s , i.e., where the words

Since

a

the

element

is less than w j. However, this element belongs to I which contradicts the choice of Z j. The theorem is proved.

j \ I j _ I,

Z

j - ( Z i h) a

( wi h) = w j'

Wi and w are located, respectively.

then

the

senior

word

of

287

CHAPTER 6

6.2.20. Let h be an arbitrary element of F ( v), G be a reductive group of linear transformations of a space v. Let Mh be a space generated in F ( v) by the orbit of the element h, while N h be a subspace generated by all the differences h g - h, where g E G. Since automorphisms from G do not change the degrees of elements, then the space M h is finitely-dimensional. Its invariant subspace N h has a co-dimensionality not greater than one, in which case the group G acts identically on the factor-space Mh / N h . Moreover, the presentation G -+ GL(MJ will be rational if in

Mh

we choose a basis consisting of the elements of the orbit

h. Therefore, if if fixed element h* E

Mh

Mh

* Nh ,

then, due to reductivity, one can find a

not lying in

Nh .

In any case we get a presentation (3)

h= h*+ h',

where

h* is a fixed element from

6.2.21.

Let

Lemma.

A

Mh , while

h belongs to

Nh .

be a certain S-suba/gebra

in

F (

v);

f 1,_., f m be its homogenous elements. Then any element v from an S-ideal generated by the elements f 1,_., f m has a presentation

(4) where

h i j are either homogenous elements from

A,

or elements of the

field, while (J j are suitable permutations. The proof results immediately from the fact that any element of the type uf iV, where u, v are words, is transferred through a permutation into fi

uv.

6.2.22. Lemma. An intersection of an S -ideal generated in F ( v) by homogenous invariants f 1,.., f m of a reductive group G, with the algebra of invariants InvG coinciding with an S-ideal generated by these elements in InvG. Proof. Let us denote by I an S-ideal generated by elements f 1,•. , f m in F ( v), and by II an S-ideal generated by these elements in InvG.

The inclusion

II

~

I

(I

InvG

is obvious.

Let

v

E

I

(I

InvG.

288

AUTOMORPIDSMS AND DERlVAnONS

Then by lemma 6.2.21, we get presentation (2). Let us consider a finite set ( (i, j) I h i j *- O}. Let us prove that v E II by induction over the number k of the elements of this set At k = 0 we have nothing to prove. In decomposition (4) let us single out one of the terms v= (fh)

0'

+ y,

(5)

where y is a sum of all k - 1 addends of the type (fh) 0', i.e., ~, O'j y= £.J.f.h .. ) , B

in

which

~

case

(6)

~J

B

contains

elements.

k- I

Let

gE G.

As

[(fh)O'] g - (fh)O' = (f( h g - h»O', we have

0=

V

g

V

-

= (f( h

g

- h»

O'~,

g

+ £.J.f .(h .. - h . .) B

~

~J

O'j

~J

(7)

.

For the element h let us consider a decomposition formula

where

(3),

equality (7) by

some

a g and summing them up over all , 0'

Hi j

for

G

0= (fh)

where

H = L(h g - h)a g

+ L(f. h B

~

'

.. )

O'j

~J

h = h*+ h'

a g E F.

by

Multiplying

g from G, we get

(8)

,

are certain homogenous elements from

According to (3) and (5), the invariant

F (

v).

v - ( fh*) 0'

is

equal

to

(fh') 0' + y. Writing y by formula (6), and substituting (fh') 0' by formula (8), we get

*

v-(fh)

0'

~,

'

=£.J.f.(h .. -h ..» B

~

~J

~J

0' j

.

Now we apply the inductive proposition to the right-hand part of the above equality, i.e., v - (fh*) 0' 6.2.23.

E

Lemma.

II' but then v E II' which is the required proof.

Let

A+

be an ideal of an

S-subalgebra

289

CHAPTER 6 AS;;; F

Then.

< V) which consists of all the elements of

if

A with a zero free term. S -ideal in A by homogenous elements is generated as an S-subalgebra (with a unit) by the same

A+ is generated as an

f 1,_., f m. then A elements. The proof is

h omogenous element

V E

carried

out

by

,,', A+. If v = "fJ.f

induction .

~

over

h h , ,)CT j ,were ~J

the

degree

h iJ'

E

of

a

A, then the

degree of h i j is strictly less than that of v and the induction is obvious. 6.2.24. From the lemmas proved above one can easily deduce now theorem 6.2.16. Let us consider an S-ideal

in

J

F ( V)

generated by a

set (InvG)+ of invariants with no free terms. By theorem 6.2.17, one can find a finite set of homogenous invariants fl'-" f m generating this S-ideal. By lemma 6.2.22, the intersection J n InvG is generated by the same elements as the S-ideal in InvG. Since this intersection is equal to (Inv G) +, then, by lemma 6.2.23, the algebra of invariants is generated by the set f 1,_" f m as an s-algebra, which is the required proof.

6.2.25. Remark. As was the case in the Nagata-Hilgebt theorem, reducibility of a group is necessary only in decomposition (3), the latter demanding a somewhat weaker condition, i.e., splitting of sequences of homomorphisms of finite-dimensional G-modules:

o----7

N ----7 M ----7 M / N ----7

0,

where G acts rationally on M and trivially on M / N. This condition, for instance, works for some rational representations of groups GL(h, F) and SL( n, F) in GL(V), if F is an algebraically closed field of a zero characteristic, as well as for rational representations of any semi -simple algebraic group r s;;; GL(n, k) over a field of a zero characteristic. 6.2.26. Hilbert series. The problem of describing groups G s;;; GL(V), the algebras of non-commutative invariants of which are finitely generated, was posed by S .Montgomery at the conference "Noether Rings and Rings with Polynomial Identities" at Durham University in 1979, as well as by the authors of the "Dnestrovskaya Notebook" (problem N 2.88). E.Formanek and W.Dicks have chosen a way to solve this problem which is different from the one presented here. Their approach is based on studying Hilbert series of the algebras of invariants. It should be recalled that a Hilbert series of a graded F-algebra

290

AUTOMORPHISMS AND DERrv A110NS

such that the dimensions of the components integer coefficients

Ri

are finite, is a series with

H( R) = 1 + L(dim RrJtn. n<: 1

This series can be viewed as an element of a ring of the power series Z r[t]], and as a function, provided t is considered to be running through a set of either real or complex numbers. R = InvG is a Let now G !;;;; CL(V). Then, as we know, homogenous free subalgebra in F homogenous subspace U!;;;; InvG, determine an auxiliary series

(

v). such

In particular, one can find a Inv G:=: F ( u). Let us that

p(U)= Ldim(U rlRrJtn. n~

1

From the obvious decomposition of graded spaces F< U>:=:Fc:B(u0F< U»

we get H( F( U) ) = 1 + P( U) H( F ( U) ). Hence, in the ring of series we have the relation H(InvG)

- 1

= 1 - p(u).

In particular, this affords that

InvG

1 I _ f{ t)' where f

series has the form

is finitely generated iff its Hilbert is a polynomial.

We know (theorem 6.2.4) that, as a rule, a Hilbert series has no form of the kind. However, calculations of a Hilbert series in an explicit form gives an essential positive information on the algebra of invariants (unlike the very fact of being not finitely generated). Hilbert series for the algebras of commutative invariants were calculated as far back as the last century. For instance, well known is the Molien's theorem published in 1897: if a field F has a zero characteristic and a group G is finite, then the Hilbert series of the algebra the form:

H(t) =

1

TGT

~

L...

gE G

1

det(1 _ gt)

F[ vl G

has

291

CHAPTER 6

An analogous result is also valid for the algebras of non-commutative invariants (see [2]). 6.2.27. Theorem. Let F be a field of a zero characteristic, G be a finite subgroup in GL(V). Then we have

H(t) =

1 "" TGT L..l_

1 tr{g)t '

where tr is a trace of a linear transformation. In the works by G.Almkwuist, W.Dicks and E.Formaneck [3], this formula was transferred to the case of compact groups in GL(V) for a field of complex numbers. The authors called their result "The non-commutative Molien-Weyl compact subgroup in GLn(C). In this case H(InvG) =

J G

theorem".

Let

dJl

1- t tr(g) ,

where dJl is a normalized Haar measure on

G

1 and I tl < n'

REFERENCES G.Almkvist [2]; G.Almkvist, W.Dicks, E.Formanek [3]; I.Dieudonne, I.Carrol [40]; Dnestrovskaya Notebook [41]; G.Higman [58]; V.K.Kharchenko [82,83]; A.T.Kolotov [87]; A.N.Koryukin [88]; T.Molien [109]; H.Weil [151]; M.C.Wolf [154]; An analogous range of problems for reduced-free algebras: R.Guaralnick [54]; V.K.Kharchenko [82]. I.V.L'vov, V.K.Kharchenko [90]; S.Montgomery [112]; I. Fisher, S.Montgomery [122];

G

be a

292

AUTOMORPHISMS AND DERIVATIONS

6.3 Relations of a Ring with Fixed Rings The trend we are going to discuss in this quite a bulky section was initiated by the Bergman-Isaaks theorem (theorem 1.3.6) and is associated with the fact that when studying finite groups of automorphisms of arbitrary rings, the most important notions of the classical Galois theory, such as finite dimension, separability, normality lose their primary sense and the problem of relations of a fixed ring R G with the basic ring R gains apriority. The Bergman-Isaaks theorem shows that when in R there is no additive I G1torsion, the nilpotency of R G implies that of R, which fact opens perspectives for a positive solution of the basic problem of the trend arising here, Le., the problem of transferring certain properties from a fixed ring to the basic one, and vice versa. At the same time, in the latest investigations the theorem itself has had a limited application since studies of M- grou ps have gained in importance, as, on the one hand, they embrace the case of an arbitrary finite group of automorphisms of a domain (or of a ring without nilpotent elements), while on the other hand, it embraces the case of infinite groups to the extent which is necessary for considering Galois closures of finite groups. The reader will see that the key moment in proving nearly all the results presented in this section is the use (in different variations) of the Shirshov local finiteness of a certain essential ideal over a fixed ring (see 5.10). When considering finite groups, the most natural restrictions (i.e., the absence of an additive I G I-torsion in the ring) can be somewhat attenuated (in literature one can find limitations of the kind: the existence of an element Y with a unit trace t(y) = 1; non-degenerality of the form of a trace; Rt( R) 31, etc.) or even eliminated at all, as, for instance, when studying domains or rings without nilpotent elements. We are going to consider theorems under most natural, in our opinion, limitations: when viewing the problems referring to subclasses of prime rings the limitation is the semiprimity of the algebra of a group; when considering the problems pertaining to all the rings, it is the absence of additive I GI-torsion or inversability of the order of a group. These limitations do not always correspond to maximum generality, and sometimes somewhat weaker limitations prove to be essentially more cumbersome, obscuring the essence. The reader interested in corresponding generalizations will easily find them in the references. It is obvious that for every property of a ring one can pose a problem on transference of this property from a fixed ring onto a ring and vice versa. Therefore, perspective might prove a general approach to the problem: to describe in some terms the properties for which the problem posed has a positive solution. A step in this direction is consideration of radical problems.

CHAPTER 6

293

A. Radicals of algebras. It should be recalled that in the class of associative algebras over a field F, a radical is a mapping p setting every algebra R into correspondence to its ideal p( R) and obeying the following properties: