This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
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.. 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_"'' 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
= 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
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
.. 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
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
ann0= v(o'- 0). Using
the
two-term
weak
As
v( 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
,
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
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
h
Or, in another expression,
If the Al
ideal
A
essential,
but
not
invariant,
i
'
V'i E
then
B ( G).
we
set
Al
= A.
h- I - I
= 11
A
i
9
i
,
while if A is an invariant left ideal, then we set
Let II is an ideal of I
is
LCiV'ixgi({Ji'
E IF ,
such that I
IF,
such that
h. g. ~ ~ l:: II'
ciV'i({JiII l:: R.
Then for any
We shall find an ideal df
I
r E IAI = A
we have
Let us show that A is an essential left ideal (if A is essential). To this end it is sufficient to show that an intersection A 11 A 9 is essential for -I
any g E G. Let v E R. Let us find 1m ideal I E
IF,
such that
Iv 9
l:: R
-I
and let J 9 l:: I, J
0* IJv g
«IJl v
E IF .
-I
11 A,
Then I Jv 9
which affords
11 A g) 11 A
* O.
Rv 11 (A gilA) and, hence,
Finally, 0 I E IF.
The
is a nonzero ideal in R. Therefore,
o* (IJ)gv 11 Agl:: latter
Iv 11 Agl:: R
intersection
A gilA, which affords that
* I( L 11 AI) l:: L 11 TAl
is
and, hence,
contained
in
~ are also essential.
for any nonzero left ideal Land
The lemma is proved.
6.3.11. It is now evident that the same statements are true for right ideals as well. 6.3.12. Lemma. Let G be an M-group of automorphisms of a semiprime ring R, A be an essentiaL one-sided ideaL of R. Then All R Gis an essentiaL one-sided ideaL of the ring R G • Proof. Let W be a left nonzero ideal of R G. Let us choose an arbitrary principal invariant form fe = 1" and by lemma 6.3.10 find an I E IF and essential left ideal A'l:: A, such that 't( A') l:: A. If 't(I) l:: R, Ie s;;; R, then 1(TW) = 1"(I)w l:: wand, hence, 't( A'II IW) l:: All W. If the latter intersection is zero, then, by lemma 5.6.8, we get 0 = e· (A'II IW) = Ii 11 (eI)W, i.e., eIW = 0 and e w=O. Since e is
300
AUTOMORPHISMS AND DERIVATIONS
an arbitrary homogenous idempotent, then
w= O.
The lemma is proved.
6.3.13. Proof of lemma 6.3.9. By lemma 6.3.8, the unit of a ring R G will be that of R. If A is an essential left ideal of the ring R, then, by lemma 6.3.12, the intersection A tl R G is an essential left ideal, i.e., by the condition it contains the unit and A= R, which is the required proof. The inverse statement can be proved in many ways in view of plentiful characterizations of semiprime artinian rings. We shall make use of the following characterization: if SZ = S and the semiprime ring S obeys the descending chains condition for Left ideaLs L, such that SL = L, then S is a semisimpLe artinian ring. So, let R be a semisimple artinian ring. By theorem 5.5.8, it suffices to consider the case when R is a simple artinian ring. Let l' be a principal invariant form. Then 'Z( R) is an essential twosided ideal of a semiprime ring RG (see lemma 3.8.1 and theorem 3.6.1). Using the above characterization, let us show that S = 'Z( R) is a semisimple artinian ring. If II:J 12:J M.:J In:J M' is a strictly decreasing chain of left ideals in S, such that SI n = In' then there is a number k, such that RI k = RI k + l'
Applying
to
this
equality
the
operator
l'
we
get
'Z( R)· I k = 'Z( R)' I k+ l' which is a contradiction. Then, by lemma 3.8.3,
and, applying l' to both parts, we get SZ = S. Thus, S is a semisimple artinian ring. Hence, S contains a unit, and this unit will be that in R G as well, since S is an essential ideal. Therefore, S = R G. The theorem is proved. R'Z( R)
= R,
REFERENCES M.Cohen [28]; M.Cohen, S.Montgomery [32]; V.K.Kharchenko [80]; D.S.Passman [135].
c. Primitive Rings. It should be recalled that a left pnmlllve ring is a ring with an exact irreducible left module. A primitive ring is certain to be prime if v is an irreducible exact module, I, J are nonzero ideals, then IJV = I(JV) = IV = V and, hence, IJ"# O. The task is to elucidate a possibility of transferring the property of
CHAPTER 6
301
primitivity from the ring R to R G and vice versa. As the prime dimension of a fixed ring is equal to the prime invariant dimension of the algebra of a group G (theorem 3.6.7), then it would be natural to consider the groups G with G-simple algebras B (G) . 6.3.14. Theorem. Let R be a left primitive ring. G be a reduced· finite group. with its algebra. B (G). being G-simple. Then the fixed ring RG is left primitive. Proof. Let V be an exact irreducible R-module. Let us show that it is finitely-generated as a left R G-module. By theorem 3.5.6 we can find a nonzero element a E R, such that n
Ra!::
I.
i= 1
that
av::f:.
RG a i 0,
for suitable
S
= {s
n
then
finitely-generated. Now we
a i E R. If v is an element from V, such
L
V = Rav!::
RG(a iV)
!:: V,
and,
therefore,
V
is
i= 1
can
find
a
maximal
ERG: sV!:: w} ::f:. O. Then
left
RG -module
W::f:. V.
Let
SR is a nonzero (R G_ R)-bimodule. In
this case its left annihilator in B (G) is an invariant ideal (as S!:: R G) and, is G-is simple, it equals zero. By corollary 3.7.4, we find since B(G) a two-sided nonzero ideal I!:; SR and get V = IV!:; SRV!:; SV!:; w, which is a contradiction. Thus, S = 0 and, hence, V / W is an exact invertible R G-module. The theorem is proved. In the case of an arbitrary M-group an analogous result is valid. It should be recalled that the ring R is called semiprimitive, if it has an exact completely reducible left module of a finite length. Let us begin with characterization of semiprimitive rings. 6.3.15. Proposition. The following conditions on a ring are equivalent: (1) R is a semlprzmltlVe ring; (2) R is a subdirect product of a finite number of primitive rings; (3) R has an essential two-sided ideal. decomposing into a direct sum of a finite number of primitive rings. Proof.
(1 ) -t (2). Let Vl'+.':~ V n be an exact left R-module,
1 ::; k::; n be its irreducible submodules. Let us denote by
the module V k' i.e.,
Nk = { r
module, in which case of primitive rings R/
n
n
Nk ,
= O.
V k'
a kernel of
E RI rV k = O}. Then V k is an exact R /
Nk
k= 1
Nk
N k-
It means that R is a subdirect product
k=l,2~.,
n.
AUTOMORPHISMS AND DERrvATIONS
302
(2)
Let R be a subdirect product of primitive rings
~(3).
Rl'-" Rn
and rc k: R ~ R k , 1 ::;; k::;; n be approximating epimorphisms. Let us assume that R is not approximated by a less number of primitive rings. Then Ik
= i;ek (1 N.;f. 0, ~ Ik n
then the sum
II
where
L
N.
~
~
= ker
rc.. Since ~
( .(1 N) (1 N k = ~;ck
j;ek
I. ]
+ _. +
In = I
0,
is direct.
Then, kerrc k n I k = 0 and, hence, I k == rc i I k) is an ideal of R k' which affords that I k are primitive rings. If now vI = 0, then, applying a projection
rc k' we get rc k( v) . rc k(I k) = 0, i.e.,
rc k( v) = 0 for all
k
and
v = O. Thus, I is the sought essential ideal. (3) ~(1). Let I = II ® _. ® In be a decomposition of an essential
ideal into a direct sum of primitive rings, and module for the ring
I k' 1 ::;; k::;; n. Then
R-module, and, on the other hand,
I kVk
I kVk = V k'
vk
is, on the one hand, a left i.e., V k
R-modules. Now we have to remark that the sum module. If and
rIkVk=O
be an exact irreducible are invertible left
v I .+ _. .+ v k
is an exact
for all k, then, as V k is exact, we have
rI = 0, which affords
rIk=O
r = O. The proposition is proved.
Let us recall an interaction ideals of a ring. If V is an invertible and, hence the module epimorphism r ~ rv, is determined.
between iirreducible modules and left module and O;f. v E v, then Rv = V I
[IeIeiI]V=O,
i> I
then
O:t:. VI E VI'
O. Therefore,
VI
n
I
ae 1VI == vi =
for
aVI
all
a
E I
and,
hence,
0 and, by symmetry, we get a direct
i> I
decomposition
v = VI
+ ... +
finitely-generated over submodule
Wi'
R G.
Let
instance, suggest that
vn .
In
S = (s E R
eI S
:t:.
This, in particular, implies that all every G
O. In
Vi
for
us
choose all
a
are
maximum
i).
Let
us,
the annihilator of
e1S
contains
: ':;V i k::: Wi
/B (G)
let
Vi
for
(1 - e1)/B (G) and is an invariant proper ideal, i.e., this annihilator is exactly equal to (1- eI)/B certain J
E
(G).
By theorem 3.7.3 we have
e I J k::: eISI
for a
IF. Therefore, we have
Hence,
which is a contradiction that shows that S =0. Thus, the direct sum
e i S = 0 for all i
and,
hence,
AU1OMORPHISMS AND DERIVA nONS
304
is the sought completely reducible exact module. Inversely, by proposition 6.3.15, a fixed ring has a finite prime dimension. By theorem 5.6.7, the ring R also has a finite prime dimension. Now theorem 5.5.10 and proposition 6.3.15 show that it would be enough to consider the case when R is prime. Let us prove a somewhat stronger statement which shows that for the ring R to be primitive, it is sufficient that one of the prime addends of the essential ideal of a fixed ring, the existence of which is stated by lemma 3.6.6., point (2), be primitive.
6.3.17. Lemma. Let G be an M-group of automorphisms of a prime ring R. Then if RG has a nonzero ideal which is a primitive ring, then R is primitive. Proof. Let 0 ~ T be an ideal of RG and V be an irreducible exact T-module. Then V = TV is an irreducible RG -module. Let A be a maximal modular left ideal of the ring R G corresponding to it.. It should be remarked that A contains no essential ideal of the ring R G: if A \;;;: A, then 0, where A= ann(l)v and, hence, AT = O. Then, by lemma 3.8.3, the left ideal RR G contains a nonzero ideal of the ring R, which will be denoted by I. ATV \;;;: ATv \;;;: Av =
Let
La =
RA +
A. Let us show that
If this is not the case, then, since
which affords
La:2
RR
G:2
I.
Let
La n
R G=
A.
A is maximal, we get
La:2
RG,
J
be a
r be a principal trace form,
nonzero ideal, such that 1( J) \;;;: R. Then
which, however, contradicts the choice of A, since 1(JI) is an essential ideal of the ring R G (see lemma 3.8.1). We now see that a set M of all the left ideals L of the ring R which are contained in R' = RR G+ R G and such that L n R G= A, is nonempty. It is evident that M is inductive by inclusion, i.e., by the Zorn lemma, there is a maximal element L EM. Let us prove that V = R' / L is an exact irreducible R-module. L1::d R
affords L1::d RRG + RG = R'. submodules. Then, the kernel
G
,
which
Consequently, V contains no proper of the module V is a two-sided ideal
K
contained in L, in which case A = L n R G:2 K G and, if K ~ 0, then KG is an essential ideal of the fixed ring, which contradicts the choice of A. Analogously,
L P R2, i.e.,
RV -::;:. O. The theorem is proved.
CHAPTER 6
305
6.3.18. Theorem. A fixed ring of an M-group of automorphisms of a semiprime ring is semisimpie (in the sense of the Jacobson radical) iff the initial ring is semisimpie. Proof. Let R be semisimple. Since a set W of elements w of the n(a)
ring R, such that Rw!:;; L
R G a i forms an essential two-sided ideal of the
i= I
ring R, then W n R G is an essential ideal of the fixed ring (lemma 5.6.6). In particular, if R G is not semisimple, then we can find a fixed element w e W annihilating all the irreducible left modules of the ring R G • Let us consider a bimodule RG w R, and by 5.10.3 find an G
idempotent ee B(G), such that R wR;;;2 eI, ew= w, Ie F. Since Iw 0, then we can find an irreducible left R-module v, such that
*'
IwV
*' 0,
and, hence,
IV
= V,
wV
*' 0
and V
= IwV ~ LR G a i
finitely-generated left RG -module. We have a V = eIV $ (1 - e) IV and, hence, eIV is a nonzero (if
V
~
V is a
decomposition eIV = 0, then
wV= wIV= ewIV= we IV = 0) finitely-generated module over RG. If W is its maximal submodule, then eIV / W is irreducible, i.e., w;;;! weIV = wIV = wV, which affords R G wRV ~ W or eIV ~ w, which is a contradiction.
Inversely, let
J( R G) = O. It should be remarked that
J( R)
n
R G~
J( R G). Indeed, if a is a quasi-regular fixed element in R, then it will
I . . aIso be quasl-regu ar '10 R G : the equaI'lUes a e x df = ax + a + x = x e a afford ae xg=O and, hence, x= x.(a e xg)=(x e a)e x g= x g . Therefore, J( R) n R G= O. Let As was the case in lemma
form. 't( x)
~
= LJ(b i x)
T ~ I
g.
1
~ R
gi
. Let
T
1
gi
is isomorphic , as a ring, to a
IJ(R), in which case
~ I. It means that
ring R, i.e.,
be an arbitrary principal trace 6.3.10, let us present it as
be essential ideals of the ring R, such that
for all i. Then (IJ(R»
quasi-regular ideal g-l
T, I
'r
T(IJ( R»
9'1
T(IJ(R»gi«IJ(R»gi
Then TI(II J(R»
is a quasi-regular left ideal of the
II
be an ideal from
~ J(R)
and, hence,
b i II ~ I for all i. G G T 'r(IIJ( R» ~ J( R) n R = O. Now F, such that
, by lemma 5.6.6, we have 't( I I J(R » = O. Since affords
since
T(IJ( R»g i~ J( R).
Let now
5.6.8
=0
II J(R ) = 0 and, hence, R is
'r
is arbitrary, proposition
semisimple. The theorem is
AUTOMORPHISMS AND DERIVATIONS
306
proved.
REFERENCES
K.I.Beidar [13]; S.Montgomery, D.Passman [118]; J.-L.Pascaud [133].
D.Quite primitive rings. These rings have already been considered in relation with the Martindale theorem ( 1.14, 1.15). Since while transferring to a fixed ring the prime dimension can increase, then, as above, we should make this notion broader. A ring R will be called quite semiprimitive if it is a subdirect product of a finite number of quite primitive rings. Let us start with characterizing this class of rings.
6.3.19. Proposition. The following conditions on the ring Rare equivalent: (1) R is a quite semiprimitive ring. (2) R has an essential two-sided ideal decomposing into a direct sum of a finite number of quite primitive ring. (3) R has the least essential ideal which decomposes into a direct sum of a finite number of simple rings coinciding with their socles. (4) R has an exact completely reductive module of a finite length which is isomorphic to a left ideal of the ring R. Proof. (1) ~(2). One should repeat the proof of implication (2) ~(3) of proposition 6.3.16 substituting the word "primitive" with "quite primitive". (2) ~ (3). Let I = II EEl •. EEl In be a decomposition of an essential ideal into a direct sum of quite primitive rings and I k ' 1::;; k ::;; n. Then
(3)
~(4).
Let
H = HI H = HI
Hk
be a socle of
EEl •. EEl H n is the sought ideal.
EEl •. EEl Hn. In every ring
Hk
let us choose a
primitive idempotent e k' Then V k = H k e k is an irreducible exact H kmodule (lemma 1.14.2). The sum V = VI + .. + vn will be a left ideal of as HVk = V k' Besides,as left R -modules, the left ideals V k are irreducible. Therefore, V is a completely reducible R-module. If, now, rV=O, then (rH)V=O and, hence, (rHk>Vk=O, then rHk=O at all k. R,
Thus,
rH
(4)
=0
and r = O. Let VI'.' Vn be a set of minimal left ideals, the intersection
~(1).
of whose kernels (i.e., left annihilators in R) is zero. Let corresponding kernels. Then V k is an exact irreducible
NI ,•. , N n
be
RINk-module.
In
CHAPTER 6
particular,
307 R / Nk
is a prime ring, while R is a semiprime ring as a
subdirect product of prime ones: is
R =
n S
k= I
R / N k.
Well-known is the fact that every non-nilpotent minimal one-sided ideal generated by a primitive idempotent: if Vkv k *- 0, v k E v k , then
Vkv k =
vk
and
for
ekv k = v k
a
certain
ek E
vk'
in
which
case
(ifk - e k)v k = 0, but an intersection of the left annihilator of v k with v k is a left ideal and, hence, it equals zero; if now I is a nonzero left ideal in ekRe k'
Therefore,
then V k= RI and, e k R e k is a sfield.
Since
0c *- 0,
then
vk n
hence, Nk
=0
ekRe k
= ekVk = e kR e k I
and, hence,
ek +
Nk
~ I.
is the sought
primitive idempotent in the factor-ring R / N k. The proposition is proved. It should be remarked that the least essential ideal, arising in (3) is generated by all primitive idempotents of the ring R and, hence, it would be natural to call it a socle . When the ring R is arbitrary, a left socle is a sum of all minimal left ideals. If the ring is not semiprime, then the left socle is possible not to be generated by primitive idempotents (since minimal left ideal with zero multiplication can arise) and not to coincide with the right sockle. In the case of a semiprime ring, the left socle coincides with the right one and is generated by primitive idempotents. 6.3.20. Theorem. A fixed ring of an M-group of automorphisms of
a semiprime semiprimitive.
ring
is
quite
semiprimitive
iff the
initial
ring
is
quite
Proof. Let R be quite semlpnmltlVe. According to (3) of the preceding proposition, R has a finite prime dimension. Therefore, we can make use of theorem 5.5.10 and assume R to be quite primitive. Let H be the sockle of the ring R, r be the principal trace form. As H is the least ideal, then 'Z( H) ~ H, in which case 'l( H) is the least essential ideal of RG (lemma 3.8.1). If B = e l B + . .. en B .
is a decomposition of the algebra of the group into a direct sum of Gsimple components, then 'l( H)e i< R G and 'l(H) = r(H )el + ... + r(H )e n
is a decomposition of 'Z( H) into a direct sum of nonzero prime (theorem 3.6.7) rings. The task now is to show that each of the ideals 'Z( H)e minimal left ideal.
k
has a
AU1OMORPIDSMS AND DERIVAnONS
308
Let us assume that in 't(H)e l such an ideal cannot be chosen. Then we can find a sequence of elements sl"
~ ,M' S rn~M
from 't( H)e l' such that
It should be remarked that H'r(H) = H (lemma 3.8.3) and, hence, mUltiplying all the terms of the chain by H from the left. we get H sl
;;;;l
H
~ ;;;;l ••• ::?
H s rn
;;;2 •••
All the inclusions in this chain cannot be strict, since sl e Hand
H sl
is
a completely reducible module of a finite length. If H s k = H s k + l' then, applying 'r, we get 't(H)sk= 'r(H)sk+l' which is a contradiction. Inversely, by theorem 5.6.7, the ring R has a finite prime dimension. Theorem 5.5.10 and proposition 6.3.18 show that it would be sufficient to consider the case when R is a prime ring. Let e be a primitive idempotent from R G. Then eR G e is a sfield, in which case the group G is induced on the ring eR e and we have (eR e)G ~ eR G e ~ (eR e)G, i.e., (eR e)G is a sfield. If we now make use of theorem 6.3.9, we see that eRe is a semisimple artinian ring. In particular, a primitive idempotent f can be found in it, in which case fRf = feR ef is a sfield and f is a primitive idempotent of R. The fact that theorem 6.3.9 can be used here, results from the following lemma. 6.3.21. Lemma. Let G be an M-group of automorphisms of a prime ring Rand e be a fixed idempotent. Then a group G induced by the group G on the ring eR e will be an M-group, the algebra of which is
isomorphic to e/B (G). Proof. First, we should remark that eR e is a prime ring: if J, I < eR e, IJ = 0, then IRJ = IeR eJ = O. Second, it should be also remarked that G ~ A(eRe). If J ~ Ig~ R, then eJe ~ (eIe)g ~ eR e, in which case eJe, eIe are nonzero ideals in eR e. Third, e/B (G) (and even eQe) is naturally embedded into Q(eR e): if Ieqe ~ R, then eIe· eqe ~ eRe.
Then, if b is an invertible element from /B (G) and £ is an automorphism from G corresponding to it, then eb is invertible in e/B and e£ is induced by £, i.e., /B (G)::? e /B (G). Finally, if ~ e f/Jg,
g e G,
then
exe~ = ~exg e for all
x
e R. In
CHAPTER 6 eR e
Then
let us find a nonzero ideal
309 A,
such that ~A, ~ ~ eR e. Let
ax( ~a) = (a~ ) x g a is an identity on R.
a
E
A.
By theorem 2.2.2, we see
that g = £) is an inner automorphism and a ® ~a = a~- 1 ® ba, i.e., in the generalized centroid C one can find an element A, such that
(a~)b - 1. From here we get a relation in the ring type a· Ae . (b - 1e) = a~ or )i(~ - (Ae)( b- 1 e)) = ~ = Ab- 1 . e E 19 (G) e, which is the required proof. Aa =
C( eR e)
of the
and,
hence,
°
REFERENCES
M.Hacque [55]; N.Jaconson [65].
E. Goldie Rings. The conditions which determine this class of rings emerged in the famous works by A.Goldie on the orders of simple and semisimple artinian rings. As we have: no possibility of presenting here the basic notions of this theory, we shall limit ourselves with getting acquainted with the main definitions and facts. A left Goldie dimension (Gal·- dim v) of the left module v is the biggest number n, such that v contains a direct sum of n non-zero submodules. This dimension is also often called uniform. One can prove (it is not obvious ) that if a module does not contain a direct sum of an infinite number of nonzero submodules, then it has a finite Goldie dimension (see 6.3.55 below). A ring R is called a left Goldie ring if in R the condition of maximality is fulfilled for the left annihilator ideals, and R has no infinite direct sums of nonzero left ideals (i.e., as has been noted above, Gal-dim R< 00).
The following statement presents the most important property of semiprime Goldie rings. 6.3.22. Proposition. Any left essential ideal of a semiprime left Goldie ring has a regular element. It should be recalled that an element r E R is called a regular element if sr"* 0, rs"* 0 for any s E R, s"* 0. 6.3.23. Definitions. Let R be a subring of a ring S. The ring is called a classical left ring of quotients of the ring R, and the ring R is called a left order in 5 iff the following conditions are met: (1) all regular element of the ling R are invertible in the ring 5; (2) all elements of the ring 5 have the form a- Ib, where 5
310
a, b
AU1OMORPlllSMS AND DERIVATIONS E
R and a is a regular element of the ring R. It is not any ring that has a classical left ring of quotients. Indeed, if
a, b are elements of R and b is a regular element, then ab- 1 = c- Id , where c, d E R and c is a regular element. Hence, ca = db and we come to the necessity of the following condition. The left Ore condition. For any elements a, b E R, where b is a regular element, one can find elements c, d, where c is a regular element, such that ca = db. This condition is known to be sufficient for the existence of a left classical ring of quotients (the Ore theorem). Moreover, the Ore condition guarantees the uniqueness of the left classical ring of quotients, this ring denoted by QciR). 6.3.24. Goldie theorem. A ring R is a left order in a semisimple artinian ring iff R is a semiprime left Goldie ring. Semiprime left Goldie rings have some interesting characterizations. Let us recall one of them which is required here. A ring R is called left nonsingular, if every essential left ideal has a zero right annihilator. 6.3.25. Johnson theorem. A semiprime ring R is a left Goldie ring iff it is nonsingular and contains no infinitive direct sums of nonzero left ideals. Now we go over to the basic topic. 6.3.26. Theorem. A ring of invariants of an M-group G of automorphisms of a semiprime ring R is a left Goldie ring iff R is a left Goldie ring. In this case automorphisms of the ring R are extended onto Qci R ) and QciR)G= QciRG).
Proof. A left Goldie ring is evident to have a finite prime dimension. Thus, theorem 5.5.10 and proposition 6.3.15 make it possible to limit ourselves with considering only a prime ring R (here one should pay attention to the fact that the Goldie conditions can be evidently transferred onto essential two-sided ideals and vice versa). Let R be a prime left Goldie ring. The condition of maximality for annihilators is preserved when going over to subrings and, hence, it is fulfilled in R G as well. If A, B are left ideals in RG and All B = 0, then for a principal invariant form -. we have 1(IA II IB) ~ r( I)A II -.( I)B = 0 ,
where the nonzero ideal I of the ring R is chosen in such a way that 1(I) ~ R. By lemma 3.8.1 we get IA II IB = O. This remark shows that if
CHAPTER 6
311
~
. +
~
. + ...
Ak
+ ...
is a direct sum of left ideals R G, then I~
. . + IA2 + ... + IA k + ...
is a direct sum of left ideals of R and, hence, Gol - dim R G< the required proof. The inverse statement is based on the following lemma.
00,
which is
6.3.27. Lemma. Let G be an M-group of automorphisms of a semiprime ring R. Then, if RG is a left Goldie ring, then the left RG_ module R is embeddible into a finite direct sum of copies of the regular module RG. Proof. Let us make use of theorem 5.10.4. By this theorem, a set W of elements a E R, such that R a is embeddible into a finite direct sum of copies of the module R G , contains an essential ideal. By lemma 6.3.12, an intersection W (") R G has a regular element a of the ring R G (see 6.3.22). The element a will be a regular element in R as well: if va = 0, then for any principal invariant form fe and an ideal I E F, such that fJ.I) I:: R, we get fJ.Iv)a = fJ.Iva) = 0, i.e., by lemma 5.6.8 we have Iev=O or ev=O and v=O. Now we have to remark that the mapping r -+ ra implements the isomorphism of the left R G -modules R == Ra. The lemma is proved. Let us go on proving the theorem. It should be remarked that the modules A, B have no infinite direct sums of nonzero submodules, then their direct sum A + B also obeys this property. If V = VI +.. + Vn +.. is an infinite direct sum in A + B, then we can group it into an infinite direct sum of infinite direct sums:
V=W1
. . +···+···
Wm == v ml
+ ... +
Vmn
+ ...
Only a
finite number of modules WI' •. ' Wm have nonzero intersections with A. Let, for instance, WI (") A= O. Then the module WI is embeddible into (A
-+
B) / A == B, which is impossible.
Now lemma 6.3.27 shows that R has no infinite direct sums of nonzero left R G -submodules and, moreover, nonzero left ideals. Let us check if R is nonsingular. If La = 0, where L is an essential left ide:u of the ring R, then, by lemma 6.3.12, an intersection
312
AUTOMORPmSMS AND DERIVATIONS
L ( l R G is an essential left ideal of R G. Proposition 6.3.22 results in a regular element tEL G. And now property 6.3.25 shows that R is a Goldie ring.
Let us prove, finally, that QCl(R)G= Qci R G). For this purpose let us consider a set T of regular elements in R G. We have already paid attention to the fact that nonzero divisors of the ring R G will be those in R
as well. Therefore, T consists of invertible elements of the ring - I
Let us show that form t
T
i.e., any element from
R = Qci R),
QciR).
QciR ) has the
- I
r, where t E T, r E R. Let a be a regular element from R. In this case the left ideal Ra will be essential: if v ( l Ra = 0, then the sum V + Va + Vcf- +.. is infinite and direct. By lemma 6.3.12, an intersection Ra ( l RG will be an essential Ra ( l T::F. 0. Let t = ra. Then ideal of the fixed ring, i.e., a- I -_ t- Ir E T- I R. If now a- l b 'IS an arb'Itrary eI ement from Qcl (R) '
then a - I b = t - I rb = T - I R . Now we have got an evident extension of the group G: (t - I r)g = t- I r g. Strictly speaking, the right-hand part of this equality is determined (by r) only on a certain essential ideal I of the ring R, but I contains a regular element to from T and, hence,
to R =
B'd eSI es, . It .IS necessary to chec k the correctness of the extension and the fact that it results in automorphisms. T- I I
~
T- I
- I
If t I
T - I R.
- I r I = t2 r 2 ,
G
then, by the Ore condition, we can find elements
such that
sl E R , s2 E T,
df
sltl = s2~ =
= Cls}rr - C l s2rj = C I (SIIl - S2L2)g = 0,
z
- t }L2)=O,
_} g
which implies correct-ness. Now, if t l l r l then - I(
then
preserving
s}rl
t3 r}
+
the
)g s2 r 2 = tl- 1 r}g
= s3 ~
and
-I g
We have tl r l - t2 r2 = since SIL} - S2L2 = t· (tilL} t.
t3 r [
is not obligatory
notations,
+
B'd eSI es,
- 1 g r 2 ·
~
= sf ~,
we
which
get I'f
affords
and, hence, -I -I g -I-I g -I-lgg -lg-lg. (t l r 1t 2 r 2) = (t 1 t3 s3 r 2 ) = tl t3 s3 r 2 = t} r l t2 r 2 'I.e.,
is an automorphism. Finally, an element G
Qc1 (R) = T
-1
R
G
t-
1r
will be fixed iff r
. The theorem is completely proved.
is
fixed,
g
i.e.,
313
CHAPTER 6
6.3.28. In the course of the proof we have established an interesting fact that when calculating a ring of quotients QclR) it is sufficient to make invertible only fixed elements
Q c1 (R) == T-
1 R.
REFERENCES
M.Cohen [28]; K.Faith [43]; I. Fisher, I.Ostenburg [48]; V.K.Kharchenko [69,80,81]; M.Lorentz, S.Montgomery, L.Small [99]; S.Montgomery [111]; A. Page [130]; D.Passman [135]; H.Tominaga [149]. The Goldie theory is presented in the following monographs: I.Herstein [57]; N.Iacobson [65]; I.Lambeck [91].
F. Noetherian Rings. It should be recalled that a module is Noetherian if there are no infinit~ ascending chains of submodule is generated by a finite number of elements. It is evident that a submodule of a Noetherian module will be Noetherian. The ring R is called
left Noetherian if such is the left regular module fact that a Noetherian.
finitely-generated
module
over
a
R
R. Well known is the
Noetherian
ring
is
also
6.3.29. Theorem. If a fixed ring of an M-group of automorphisms of a semiprime ring is left Noetherian. then such is the initial ring. Proof. Let us denote by W a set of all elements a E R, such that Ra
~
n
L.
RG a i
for some
i= 1
a1,•. an. Theorem 5.10.1, applied to a ring
antiisomorphic to R, shows W to be an essential ideal. A Noetherian ring is certain to be a Goldie ring. By theorem 6.3.26 and proposition 6.3.22 we can
find
Noetherian
in
W a
regular
RG -modules.
element
a.
Then
The theorem is proved.
R
==
Ra c-
n " RG a ~. L. i= 1
are
314
AU1OMORPJllSMS AND DERIVATIONS
6.3.30. Theorem. Let G s:: Aut( R) be a finite group of automorphisms of a left Noetherian ring R, which has an invertible in R order I GI- 1 E R. In this case the fixed ring Proof. Let
RG
is left Noetherian.
be a strictly ascending chain of left ideals in RG. Then, since R is a Noetherian ring, there can be found a number k, such that RI k = RI k + 1·
Let us apply the operator t = I GI - 1 '" £..g to both parts of this equality. As gE G
for any fixed element s, we get a contradiction I k = I k + 1. The theorem is proved. The latter theorem stops being valid for M-groups. Moreover, the t( s)
=
s
condition I GI- 1 E R cannot be changed with that of the absence of additive torsion. Let us give here an example by Chang and Lie which is based on a Nagarajan example.
6.3.31. Example. Let A=Z[a1,bl'a:z,b2 "_] be a ring of polynomials with integer coefficients. Let us denote by K the localization of A
relative
2A, i.e.,
K = (
~I
g E A, f E
2 A) , and let us consider a ring
R = K[ [x, y]] of the power series on variables x, y. Since K is a principal ideal ring, then R is Noetherian. Let us determine an automorphism g on R by the formulas:
x g =- x, yg= y, al=- a i +(a i + 1 x+ b i + 1 y)y, g bi
= b i + (a i + 1x + b i + 1 y) x. In this case the group G generated by g has the order of 2, and the ring R has no additive torsion. One can prove that RG is not a Noetherian ring. The proof is based on the result by Nagarajan, who claims that in the ring
(R /
2 R)
where
G
Pi =
there is a strictly ascending chain of ideals
a i + 1 x + b i+
1Y
+ 2 R.
315
CHAPTER 6
REFERENCES K.L.Chung, P.G.Lee [26]; M.Cohen, S.Montgomery [33]; D.Farkash, L.Snaider [44]; J.Fisher. J.Ostenburg [48]; S.Montgomery, L.Small [91]; S.Montgomery [111,113]; K.Nagarajan [123]; J.-L.Pascaud [133];
G.Prime and subdirectly undecomposible rings. The problem of invariant simple rings has been considered in 3.6, having showed that a ring of invariants of a finite group G is a finite direct sum of simple rings, provided there is no additive I GI-torsion in the initial ring. This result stops being valid when transferring to M-groups. Let us describe an example by Osterburg, which is based on an example by Zalesskii and Neroslavskii of a simple ring with a unit of a characteristic 2, having an outer automorphism of the order 2 with a non-simple ring of invariants. 6.3.32.
Example.
Let k
be a field of characteristic 2. Let us
consider an algebra Rl = k( y)[ x, x-I) over a field of rational functions of a variable y. Let g be an automorphism of this algebra which transfers x into xy and let G be a finite cyclic group generated by the automorphism g. Let us assume that ~ = Rl * G is a skew group ring and let us consider an automorphism h of the algebra ~, determined by the formulas h( x) = x- 1 and
h(g) = g- 1. One can show that
~ is a simple ring, h
is an outer automorphism, and the ideal t(~) = ( x + x~
Rt
X
E~)
of the
h:> is not simple. fixed ring contains no unit, i.e., Therefore, for characterizing the invariants of M-groups of simple algebras, we should somewhat extend the class of the rings under consideration. It should be remarked that if a simple ring R contains a unit, then
R = R F' If R
ex
has no unit, then
R
*" ex R) ,
but, however, R will be an
ideal in R), in which case this ideal will be contained in any other R), i.e., R is a heart of R) . nonzero ideal of the ring It should be recalled that the rings with hearts are called subdirectly undecomposable. It is known that any ring can be presented as a subdirect product of subdirectly undecomposable rings. In this case the hearts of cofactors, as the least ideals, either have zero multiplication or prove to be
ex
ex
316
AUTOMORPHISMS AND DERIVATIONS
simple rings. It is evident that the heart of a subdirectly undecomposable ring will be simple iff the ring is prime. It would be now natural to pay our attention to simple rings as hearts of subdirectly undecomposable prime rings. As when transferring to a ring of invariants the prime dimension can increase, we come to a natural generalization. 6.3.33. Lemma-definition. A ring R will be called almost simple if it obeys the following equivalent conditions. (1) R has the least essential ideal decomposing into a direct sum of a finite number of simple rings, which will be called a semi-heart. (2) the ring R is a subdirect product of a finite number of subdirectly undecomposable prime rings. Proof. ( 1 ) ~ (2). Let a = a 1 $ an be the semi-heart, and _0
ann R a k' Then
Tk =
and, hence,
R / Tk
a k == a / T k
II T k
=0
and
R
=
n 5
k= 1
R / Tk .
Let R be a subdirect product of rings
~ (l).
R / Tk
is a subdirectly undecomposable prime (since the heart is
prime) ring . It is evident that (2)
will be the heart of the ring
R 1,_. Rn
with
n k: R ~ R k' I S; k S; n be approximating simple hearts al'-' an and projections. Let us assume that n are the least possible numbers. Then df
Ik =
n
i* k
ker n i 1= O. The sum of ideals
I k II
IIi k i*k
Then, ker n k and, in particular,
.n
~*k
II I k =
I k
ker n i
contains an ideal
If
a n' is the semiheart of xa = 0, then 0 = n k(xa)
is
a nonzero
I d
I
ideal
of
a' k
+ _. + In will be direct, as
ker n k = 0
0 and, hence,
that aI' + .- + x = 0
II
II
I k == (Jk
=
n J!., I
k) is an ideal of
1rk 1( a k) == a k'
Rk '
Let us show
R.
=
n k(x)a k' i.e., and,
n k(x)
= 0
for all k
hence,
Ia k' = a. The lemma is proved ..
6.3.34. Theorem. A ring of invariants of an M-group of automorphisms of a semiprime ring is almost simple iff such is initial ring. In this case the semiheart of the ring R is generated as a one-sided ideal by a semiheart of R G . Proof. Let R be an almost simple ring. Then it has a finite prime dimension and, by theorem 5.5.10, it is sufficient to consider the case when
CHAPTER 6
317
is a prime ring with a heart 0" . Let -r be a principal trace fonn. Then 't( 0") 't( a") is a semiheart of the fixed ring.
R
If aE R
G
~ R.
Let us show that
and a't(O") =0, then 't(aO")=O and, hence,
aO"=O,
't( 0") is an essential ideal in R G • If I is an arbitrary essential ideal of the fixed ring, then
IO"
i.e., will
be an (R G- R) -bimodule with a zero left annihilator, i.e., this bimodule contains a nonzero ideal of the ring R (see 3.8.3) and, moreover, IO";;2 (J. Therefore, I;;2 I't( a) = -r(I0");2 't( 0") .
of R G
As R G has a finite prime dimension, then a certain essential ideal I is decomposed into a direct sum of prime rings I = II ® ® I no M.
We have 't( 0") = II 't( 0") EEl .M ® In r( (J). We now have to remark that are simple rings. If, for instance, J EEl
12 -r( 0") EEl
M'
0"# J
is
an ideal of
EEl In -r( a") is an essential ideal in
RG
I k 't( (J)
II-r( (J),
then
and, by the above
proved, it contains 't( 0"). Hence, J = II -r( a"). G
Inversely, let (J be a semi heart of the ring R . By theorem 5.6.7, the ring R has a finite prime dimension and, by theorem 5.5.10, we can limit ourselves with considering only the prime ring R. By lemma 3.8.3, the left ideal R(J contains a nonzero ideal I of the ring R. If J is an arbitrary nonzero ideal of the ring R, then J n RG is an essential ideal in RG (lemma 3.8.4) and, hence, J;;2 (J. This affords J;;2 R(J;;2 I, which means that I = R(J is a heart of the ring R. The theorem is proved. For simple rings with a unit we can now give a certain necessary and sufficient condition of decomposition of a fixed ring into a direct sum of simple rings. 6.3.35. M·group of its rings iff for a that 't(r) = 1. Proof. RO" = Rand
Corollary. Let R be a simple ring with a unit, G be an automorphisms. Then RG will be a direct sum of simple certain invariant form -r there is an element r E R, such If
't( R) 3 1 and
R G= r(R) = 't( R) 0"
0" is a semiheart of the ring
~
(J.
R G,
then
Inversely, if -r is a principal form,
then 't( R) is a semiheart of R G, i.e., 't( R) = R G3 1, which is the required proof. Returning to the case when G' is a finite group and R has no 1 G 1torsion, it should be remarked that theorem 3.6.9 is also valid for finite direct sums of simple rings (by theorem 5.5.10). As to the inverse statement,
318
AUTOMORPHISMS AND DERIVATIONS
one can easily give examples proving that it is not valid. At the same time, in this case the Bergman-Isaaks theorem makes it possible to calculate a semi heart more exactly. 6.3.36. Corollary. Let a fixed ring of a finite group of automorphisms of a semiprime ring R be a direct sum of a finite number of simple rings. Now, if there is no additive I GI-torsion, then a certain power of the ring is decomposed into a direct sum of simple rings. Proof. By theorem 6.3.4, the ring R is almost simple. Let a be its semiheart. Then a is an invariant relative AutR ideal, since it is equal to the intersection of all essential ideals. Moreover, in the factor-ring R / a there is no additive I GI-torsion : I GI a is an essential ideal in Rand, hence, I GI a= a; if I GI rE a, then I GI r=1 GI s, where SE a, i.e., r = SEa. Now the group G is induced onto R / a. Since the intersection a n R G is an essential ideal of the ring of invariants, then a ~ R G, i.e., there are no nonzero fixed elements in the factor-ring R / a. By the Bergman-Isaaks theorem we get R m= a, which is the required proof.
REFERENCES
V.K.Kharchenko [127]; I.Osterburg [128,129]; T.Sundstrom [147]; A.I.Zalessky. O.M.Neroslavsky [157].
H.Prime Ideals, Montgomery Equivalence. Here we shall consider relations between prime ideals of a given ring R and a ring of invariants R G for the case when G is a finite group of an invertible in R order. Since in this case fixed elements are determined by the equality x
=
t( x), where t
= I ~I ~g,
then we can go over to considering factor-
rings with respect to invariant ideals. Here we have (R / I) G= R G/ I G , where I is an invariant ideal. In particular, if p is a prime ideal of the ring
R,
then
gE G
is
an
invariant
ideal
and
we
can
go
over
to
considering a factor-ring R / n p g, which is a subdirect product of a finite number of prime rings. Let us start with presenting the well-known properties of such rings and general properties of prime ideals.
319
CHAPTER 6
6.3.37. Lemma. A semiprime ring is presented as a subdirect product of n prime rings iff its prime dimension ~ n (see 3.6.6). Proof.
Let
epimorphisms. Let
It
R
=
n SRi i= 1 In + 1
e .. e
and
TC i: R
--+
Ri
be
approximating
be a direct sum of ideals in
R.
Then
=0
and, hence, the kernel of TC k contains all the ideals Il'.' In + 1 with, possibly, only one excluded. It means that the intersection of all kernels of TC i contains at least one of the ideals TCk(It)·.·· TCiI n+ 1)
Il'." I n+ l' i.e., the prime dimension of R is not greater than n.
Inversely, by lemma 3.6.6, let us find in R an essential ideal I. which decomposes into a direct sum of prime rings I = II e .. e I m. m ~ n. Let
pk
be an annihilator of the ideal
Let us check if P k is a prime ideal. If
Ik
in R. Then (\ p k = O. k
aRb!:; P k '
then
I k aRb =
0 and,
moreover, I k a· I k' I kb = O. However, I k is a prime ring and, therefore. we have either I k a = 0 or I k b = 0, which is the required proof. 6.3.38. Lemma. Any prime ideal P contains a minimal prime ideal p' !:; P
Proof. It should be remarked that an intersection of any chain of prime ideals is a prime ideal: if AB!:; () P a' A~ () P a. then for a certain
f3
we have A~ P fJ and, hence. for all
B !:; P r and, hence. B!:; get the required proof.
r> f3
() P r = () P a. If
we get A~ P r C P fJ' i.e .•
we now use the Zorn lemma. we
6.3.39. Proposition. Let the prime dimension of a semiprime ring be equal 10 n. Then the following statements are valid. (a) Any irreducible presentation of a ring R as a subdirect product of prime rings contains exactLy n cofactors. (b) A presentation of R as a subdirect product of n prime cofactors R
is unique. i.e., if R =
n n, SR. = SR. i= 1 ~ i = 1 ~
I
then there is a permutation
(J,
such that ker TC'. = ker TC,J') , whae TC i' 'If i are the corresponding approximating projections. (c) The ring R has exactly n minimal prime ideaLs and their intersection equaLs zero. Proof. It should be recalled that a subdirect product is called irreducible if we cannot reduce a single cofactor without the rest projections to remain an approximating family. In terms of the kernels of projections it means that (\ ker TC. -:t; 0 for all k. Certainly, not any prime ring has an ~
i* k
V\~
~
320
AUTOMORPHISMS AND DERIVAnONS
invertible presentation, but if its prime dimension is finite, then the existence of such a presentation becomes obvious. So, let
R=
S
aE A
be an irreducible product,
Ra
approximating projections. Let I a =
na: R
~
Ra
be
ker n f3' Then
n f3~a
Ia
n
L.
L.
f3~a
I f3 ~
n
f3~a
ker n f3 nker n a = 0 .
Therefore, I a is a direct sum of nonzero ideals and, hence, the power of A does not exceed the prime dimension n. By lemma 6.3.37, n, on the contrary, does not exceed the power of A. If
R
=
n n SR. = S R~ i=I ~ i=I
n i'
and
are
n~
the
corresponding
projections, then for any i we have n
ker n I' . ker n '2 ..... ker n'n ~ n
j = I
Since
Ri
ker n']. = 0
~
ker n .. ~
is prime, it means that there is a number j, such that
ker n'j ~ ker n i' Analogously, by the number j one can find a number k, such that ker
nk ~
ker
n']" If
k
* i,
then
R=
S. Ra' a~
which contradicts
~
lemma 6.3.37. Therefore, ker n i = ker n'j and the mapping 0-: i -7 j implements the required permutation. Finally, by lemma 6.3.38. we can find n minimal prime ideals P 1,_., P n
then P;;;2 is proved.
the intersection of which is zero. If P is any other prime ideal, P k for a certain k, i.e., P cannot be minimal. The proposition
6.3.40. Lemma. If R is semiprime and has a finite prime dimension, then any not minimal prime ideal is essential. Proof. If Q is not minimal, and Q. u = 0, then for any minimal prime ideal P we have Q. u ~ P and, hence, u ~ P. However, the intersection of all minimal ideals is zero. The lemma is proved.
6.3.41. Spectrum. We have already encountered this notion when considering a commutative self-injective regular ring (see 1.9.3). For an arbitrary case on a set Spec R of all prime ideals of the ring R one can also set a topology using the same closure operation {P a }= {p
E
Spec RI P;;;2 n P a ).
321
CHAPTER 6
It is the obtained topological space that is called spectrum (or a prime spectrum) of the ring R. The above formulated lemmas are quite obvious from the viewpoint of this topological space. For instance, a decomposition of a semiprime ring into a subdirect product of prime rings means that a set of kernels of the corresponding projections is dense in Spec R, and the prime dimension is the least power of a dense subset, and so on. If a certain group G acts on the ring R, the action of this group is induced on the space Spec R. In this case there arises a space of orbits Spec R / G which is determined as a factor-space Spec R relative the equivalence relation P '" Q H 3 g
E G: P
g = Q.
6.3.42. Theorem (Montgomery). Let G be a finite group of automorphisms of the ring R, G!;;;; Aut R with its order invertible in R. In this case the following statements are valid. (a) If P is a prime ideal of the ring R , then P
n R G= PI n ._ n P n' where
ring
RG
n:S;
I GI, PI ~_, Pn are prime ideals of the
uniquely determined by the condition that
Pi /
P
n
are
RG
minimal prime ideals in R G/ P n R G . (b) If P, Q are prime ideals of Rand P n R G= Q n R G, then Q = P g for a certain g E G. (c) If P is a prime ideal of the ring R G , then there is a prime ideaL P of the ring R, such that P is minimal over P n R G, i. e., P / P n R G is a minimal prime ideal of the ring R G/ P n R G. In this case if Q is another prime ideaL of the ring R with the same property, then Q = P g for a certain g E G. If I is an ideal of the ring R, such that I
n
R
G
p, then Proof. !;;;;
I!;;;; P
g
for a certain g
E G.
(a) Let us consider an invariant ideal P
n
R G= J
J =
(l P g. gE G
We
have
n R G. Going over to a factor-ring R / J and allowing for the
fact that (R / J)G= R G/ I n R G , we can assume that J=O. By theorems 3.6.7 and 5.5.10, the ring RG has a finite prime dimension :s; I GI and we are now to use point (c) of proposition 6.3.39. (b) If P n R G= Q n R G, then in the factor-ring Ii = R / n P g we have Q n Ii G = O. Hence, the invariant ideal (l Q g of a semiprime ring G
Ii
has no fixed elements. This does not contradict the Bergman-Isaaks theorem -g g g only in the case when n Q = 0, i.e., n Q !;;;; n P . Analogously, n P g!;;;; n Q g. Now the factor-ring ii has two presentations in the form of
322
AUTOMORPI-llSMS AND DERIVA nONS
a subdirect product of prime rings. Cancelling spare cofactors from every presentation and making use of proposition 6.3.39. (a,b), we get p g= Qh -\
for certain h, g E G, i.e., Q = p gh , which is the required proof. (c) Let us consider a set M of all ideals I of a ring R, such that I G ~ I II R G ~ P. This set is nonempty and inductive. By the Zorn lemma, there are maximal elements in M. Let us show that all maximal elements of M are prime ideals and lie in one orbit of the spectrum. If p is a maximal element, and A· B ~ P, in which case A:::J P,
B:::J P,
then
A Gt;,
P and
B Gt;,
p. Since the ideal P is prime, we (AB )G~ P G~
get A G. B G t;, P and, moreover, (AB) G t;, p. However, which is a contradiction. Let Q be another maximal element from M. invariant ideals I = II P g, J = II Q g. We have
i.e., S E
Let
us
p,
consider
I + J E M and, by the Zorn lemma, we can find a maximal element M containing I + J. We have II P g ~ S, i.e., since S is prime, we
can find
agE G, - 1
such
that
P
g~ Sand -\
P
~ S
-
g
\
,
and, gh-
due
to
I
maximality, P = S g . Analogously, Q= Sh Now Q= P , which is the required proof. Let now I = II P g be an intersection of all maximal elements of M. Then I G= P II R G, and I is an invariant ideal. Going over to a factor ring R / I and allowing for the fact that (R / I) G = R G/ I G, we can assume that I = O. We are now to show that in this case p is a minimal prime ideal. If p is not minimal, then by lemma 6.3.9 it will be essential. By lemma 5.10.5, the left ideal Rp contains a two-sided essential ideal W of the ring R. We have
This means that WE M . By the Zorn lemma, let us find a maximal element ph of the set M containing \N, Now we come to a contradiction with the fact that W is an essential ideal:
323
CHAPTER 6
Let Q be another prime ideal, such that p is minimal over Q G • Then Q E M and we can imbed Q into a maximal element Q s:; ph. Let us consider a factor-ring Ii. = R Iii Q g. If P h* Q, then phis not a minimal prime ideal of Ii.. By lemma 6.3.40, it means that
phis an
essential ideal of ji. An intersection P hli jiG= P will be also essential, but this is impossible since p is a minimal prime ideal in ji G= R GI QG. The theorem is proved. This theorem proves that every prime ideal p of a ring of invariants uniquely determines a semiprime ideal p., such that p is minimal over p. ' and .l!. has the form p G for a certain prime ideal P of the ring R. 6.3.43. Definitions.
p, JI. of a fixed ring
Prime ideals
called Montgomery equivalent, p ~ JI., if .l!. =..Y. Each class of equivalence contains not more than n = Theorem 6.3.41, shows the mapping f, which puts the class of minimal over P G prime ideals into correspondence p E Spec RIG, to be one-to-one. Moreover, by point (b), will be uniquely determined by both the invariant ideal intersection
P Ii R G= P G.
and on the space Spec R I M
~
M df
Jl.1-
¢::} p~
are
I G I elements. of equivalence to the orbit the orbit P
and
(1 p g
gE G
the
It enables one to determine on the orbit space G M
and pl-
RG
the order relation
_ _ df
P~Q¢::}IiP
g
S:;IiQ
g
JI..
6.3.44. Theorem. The mapping topological spaces preserving the order:
f
is a homeomorphism of the
G M f: Spec RIG == Spec R / - .
Proof. We should check the fact that if the prime ideal Q contains an intersection (1 a
A= (1 Q g g
t:J.
P
(1 P a, g
~,where
Pa
are prime ideals, then
~ = B. Then in the factor-ring
Q::1 (1 P
Ii. = R I
B * 0 and, by the Bergman-Isaaks theorem, we get
(1 P
a
~ = B Gt;;,
a, g A
~. Let
we have
B G* 0,
i. e. ,
A, which is a contradiction. The theorem is proved.
In relation with the notion of Montgomery equivalence there arises a number of interesting problems which can be generally formulated in the following way: to study the properties of the points of the spectrum of the fixed ring, which are stable relative the Montgomery equivalence. We have
AUTOMORPIllSMS AND DERIVAnONS
324
already encountered one of these properties, i.e., primitivity (it should be recalled that an ideal P of the ring R is called primitive if the factor-ring R / P is primitive). Indeed, if P is a pnmtt1ve ideal of the fixed ring, and P is its corresponding prime ideal of the ring R, then, going over to a factor-ring R = R / (\ P g, we can assume that P is minimal and, hence, not essential. Its annihilator will be a nonzero ideal and a primitive ring. By lemma 6.3.17 and theorem 5.5.10 , the ring R is semiprimitive, and by theorem 6.3.16, RG is also semiprimitive. Since the minimal decomposition into a subdirect sum of prime rings is unique, we say that all the ideals which are Montgomery equivalent to the ideal P, will be primitive. Another example of a Montgomery-stable characteristic is given by the monotonicity of the mapping f. It should be recalled that the height of the ideal P is the maximum length of a strictly descending chain of prime ideals P ::> P2 ::> ". ::> Pn' The Montgomery-equivalent points of the spectrum are evident to have the same height. One of Montgomery"unstable (or, to be more exact, not always stable) characteristics is the depth of an ideal, i.e., the maximal length of a strictly ascending chain of prime ideals P c P2 c ". c Pn. 6.3.45. Example. Let R be a ring of all linear transformations of a countable-dimension space v over a field F with a zero characteristic. This ring is naturally identified with a ring of infinite finite-row matrices (1.14.4). Let us consider a conjugation g by a diagonal matrix diag( - 1, 1, 1 ~"). In this case the ring of invariants of a group G = (g, I } is isomorphic to F $ R, i.e., we have two Montgomery-equivalent ideals PI
=F
$ 0,
P2
=0 $
R,
one of them being maximal, another not maximal.
The same example shows the condition "R / P is a Goldie ring" to be Montgomery"unstable, in spite of the fact that it is transferred onto rings of invariants and vice versa. 6.3.46. Morita contexts. Montgomery-equivalence proves to be closely related with attenuated Morita"equivalence. Let us consider the properties of the points of a spectrum, which are determined by factor-rings. Setting such a property is equivalent to singling out a class R of prime rings. In this case Montgomery-equivalence is set by the relation between
prime rings:
M
R -
S
iff there exists a prime ring A and a group
invertible order, such that
R
==
AG /
P, S ==
AG/
PI
G
of an
for certain minimal
prime ideals P, PI of a fixed ring AG. It should be recalled that a Morita context is a sequence (R, v, w, S), where R, 5 are rings; v is an (R, S) -bimodule; W is an
CHAPTER 6
325
(S, R)-bimodule,
and determined are the multiplications V ® s W --+ R, W ® R V --+ s, such that a set of all the matrices of the type
(=;)
forms an associative ring relative the matrix operations of
multiplication and addition. The rings R nd S with unity are called Moritaequivalent if there is a Morita context (R, v, w, S), where VW = R, WV = S. This notion has been studied in detail and its essence is that the categories of left modules over R and S are equivalent iff Rand S re Morita-equivalent. The Morita context (R, v, w, S) is called prime if the corresponding ring ( contain
t,
=~) is prime.
If in this case the rings Rand S
then, as in example 6.3.45, we can consider a conjugation
onto the element
(10-10). For the
group
G=
(1, g)
we have
M G= R $
g
s,
i.e., the rings R and S are Montgomery-equivalent . The following theorem shows that the inverse statement is also , to a certain extent, valid. Let us consider two conditions on the class of prime rings R. Inv. If R is a prime ring and G !:: Aut R is a group of its automorphisms of an invertible order, such that R G is prime, then R GE R iff RER. Mar. If the nonzero prime rings Rand S are related by a prime Morita context and R E R, then S E R . 6.3.47. Theorem.
If a class of prime rings
R
obeys conditions
Inv and Mar, then it is stable under Montgomery equivalence. Proof. Let us first remark that condition Mar shows that the class R together with the ring R contains all its two-sided ideals. Indeed, if I
is an ideal, then we have a prime Morita context (R, I, I, I). Let R be a prime ring, G a group of its automorphisms of an invertible order. Let us decompose the algebra 1B (G) of the group G into a direct sum of G-simple components
We can find a nonzero ideal set
I = J2.
Then
J
of R, such that
e i Ie j ~ R for any
are fixed idempotents, then the group which case
G
ei
J,
Je i ~ R.
i, j,1 ::;; i, j::;; n. Since
Let us e i' e j
acts on all components e i Ie j ' in
AUTOMORPmSMS AND DERIVATIONS
326
The next step will be a remark that (e i Ie i)G that the algebra of an induced group is equal to 6.3.20). Therefore, in the ring of invariants IG ei I
is a prime ring, so e i B (G)
(see lemma
RG we have an essential ideal
decomposing into a direct sum of prime nonzero rings (e i Ie)G = G
.
Let P be a minimal prime ideal of a ftxed ring. Then p is not an essential ideal and, hence, it has cannot have nonzero intersections with all the ideals eiI G . Let, for example,
pn e1IG=0. Then
=e1I G+ pi p
e1I G
is an ideal of RGI p, i.e., RGI pER iff e l IG E R. Therefore, we are to show that if e 1 I
GE
R, then all the components
e i IG also lie in R. Since (e l Ie I)G = e l I G, then, by property
Inv we get
e l Ie l
E
R .
Then, we have a prime Morita context (e1 Ie l' e l Ie i' e i Ie I , e i Ie) and, hence, e i Ie i E R. By property Inv we get e i I G= (e i Ie) GE R. The theorem is proved. The results obtained in sections A -G and in the above theorem imply that Montgomery-stable are the classes of semisimple prime rings for radicals considered in section A. the class of prime subdirectly nondecomposable rings, the class of primitive rings and that of quite primitive rings. Montgomery- unstable is the class of simple artinian rings, Goldie rings, Noetherian and simple rings. Here one essential remark should be made. When considering Montgomery-equivalence not on the whole union of prime rings but on its certain parts, unstable properties can be come stable. For instance, the Gel'fand-Kirillov dimension becomes a stable characteristic in the class of PI -rings (S.Montgomery, L.Small). The classical Crull dimension is stable in the class of affine PI -rings (J.Alev), as well as in the class of Noetherian PI -rings (S.Montgomery, L.Small). REFERENCES J.Alev [1]; M.Artin, W.Shelter [10]; K.I.Beidar [13]; M.Lorenz, S.Montgomery, L.Small [99]; S.Montgomery [110,114]; S.Montgomery, D.Passman [116]; S.Montgomery, L.Small [119,120,121].
327
CHAPTER 6
I. Modular Lattices. Many properties studied in the theory of rings can be set in tenns of the lattice of left (right) or two-sided ideals (for instance, Noetherity, artinity, Crull dimension, Goldie dimension, etc.). If a ring of invariants R G obeys such a property, then this fact gives an information only on certain invariants (left or two-sided) ideals and there naturally arises a problem of an interrelation of the lattice of all the ideals with that of invariant ideals. As the lattice of ideals is modular (Le., I + (J ( l K) = (I + J) ( l K if I!:: K), then we come to the problem of studying fixed points of finite groups acting on modular lattices. The following two theorems proved by J.Fisher [45] and P .Grzesc-zuk, E.p .Puchilovsky [33] give us a bulky infonnation on relations between Land LG.
It should be recalled that a lattice is an algebraic system L with two binary commutative associative operations /\ and v, interrelated by the equalities: P 1 (x v y) /\ P 2 (x /\ y) V P3
x /\
X
=X
X
=X
X
=
X
, X V X
= x.
On the lattice one can introduce a relation of a /\ b = a. From the point of view of this order the exact lover boundary, while the operation v boundary of the two elements. The mapping of called strictly monotonous if the inequality
a partial order a:S; b ~ the operation /\ calculates calculates the exact upper the lattices f: L -+ Ll is
a < b implies ~
.t( a) < f(b). ~
A
strictly monotonous mapping can not always preserve the exact lower and upper boundaries of incomparable elements. The lattice L is called modular if for all a:S; c from L the following equality holds: Ml.
av(b /\ c)=(av b)/\ c.
Since a:S; a v b, a:S; a v c, then the above conditional identity yields the following identities: M2.av(c/\(a v b»=(av c)/\(av b)=
a v (b /\ (a v
c».
These two formulas result in the rule of substitution:
328
AUTOMORPIDSMS AND DERIVATIONS
if
a /\ b = 0 and c /\ (a v b) = 0, then
where 0 is the least element of the lattice. Indeed, if c /\ (a v b) = 0, av(b/\(avc»,
i.e.,
b /\ (a v c) = 0,
then
a = a v (c /\ (a v
and,
b/\(avc):Sa
b»
=
hence,
b/\(av c):S b/\ a=O.
In the theory of modular lattices the latter rule plays approximately the same role as the lemma on substitution in the theory of linear spaces which results in the theorem of the existence of a basis. The following condition which readily follows from M 1, plays the same role in the theory of modular lattices as the theorem of homomorphisms. M3. If a:S x:S a v b, then a v (x /\ b) = x.
This immediately affords that the mapping
f b(x) = x /\ b
sets an
isomorphism [a, a v b) == [a /\ b, b]. It should be recalled that for elements x:S y the interval (z I x:S z:S y) is denoted by [x, y]. A direct product of the lattices is determined in a natural way: ( a 1~-, a rJ v ( bl'-" b rJ = (a 1 v b p _, a n ( a p _, a rJ
/\ ( bl'-"
b
rJ =
V
b rJ ;
(a 1 /\ b p _, an /\ b rJ .
6.3.48. Theorem (J .Fisher). Let G be a finite group of automorphisms of a modular lattice L. Then there is a strictly monotonous mapping f: L -+ L G x _. x LG of the lattice L into a direct product of a finite number of copies of a sub lattice LG of fixed elements. Proof. Let us denote through F a set of all unary terms in a signature (G, /\ , V ). The elements of F will be called G-lattice polynomials.
Examples of such polynomials can be the terms
v
SE
S
(
/\
tE
T
xi:)s,
where
S, T are subsets of G. Each of such polynomials can be viewed as a mapping f: M -+ M. By induction over the construction of the term f one can easily prove that the corresponding to it mapping will be monotonous (but, possibly, not strictly monotonous): a:S b -+ t(a):S feb). A G-Iattice polynomial f is called invariant if all its values lie in L G (for instance, /\ G x g, V G x g are G-invariant). In order to prove the Fisher theorem, it is sufficient to construct a finite set of invariant G-Iattice polynomials f} ,_., f k' such that for a :;::; b the equalities f i a) = f i(b ),
1 :S i:S k imply a = b. Therefore, we shall say that f
trivializes the couple
CHAPTER 6
329
(a. b) if a
band t(a)
~
= f(b).
6.3.49. Lemma. Let S be a subset of G, and g E G\ S. coupLe (a, b) is triviaLized by each of the foLLowing poLynomiaLs xt ,
/\
SU (g)
tE
(1)
- 1
V
{(
/\ SE
xS) t i t
S
S
E
(2)
(g) ).
U
/\
then the couple (a, b) is trivialized by the polynomial
SE
Proof. By the condition we have: a g /\ V
(
/\ SE
S
S
(
/\
S
tE
It
/\ SE
1
/\ SE
as)t
S
should
as) t-
as=b g /\
SE
If the
-1
v (
b S•
S
S
remarked
~ a and, hence.
(3)
as) g
/\ SE
be
v
S
-1
=
(4)
that
tE
xS
S
(
for /\
SE
S
every
we
t E S
have
~ a, i.e., applying
as/ -1
to
equality (4) an automorphism g and neglecting all the terms in square brackets in the right-hand part, we get an inequality a gv (
/\ SE
a s) ~
S
/\ SE
and, moreover. bgv (
/\ SE
S
S
as) ~
b
(5)
S
/\ SE
S
b S.
Let us now make use of the fact that
/\
the lattice is modular. we get: /\ SE
S
b S=
/\ SE
S
bS/\(bgv
/\ SE
S
a~=(
a s~
S
SE
/\ SE
S
/\
S
b S= (
/\ SE
S
as /\ a g) v
/\ SE
S
a s=
/\ SE
S
b S and that
bS/\bg)v
By equality (3) we find SE
/\
SE
S
as
330
AUTOMORPI-llSMS AND DERIVATIONS
The lemma is proved. Let us arbitrarily order the elements of the group 1 = gl ,~., g no Let I
1 = i(1)
1 and let I) be an independent set.. In this case a 1\ b i:t: 0, 1 ~ i ~ n + 1 and, hence we can suppose that b i ~ a. Let, then, c = ~ v .. van. Then, (bl~· b n+
by the inductive proposition, the Goldie dimension of the lattice [0, c] is equal to n - 1. In particular, not more than n - 1 of elements from (b i) can have a nonzero "intersection" with c. Let b i bi
1\ C
=
° at
k
f i = ( b n + 1 Vb)
< i 1\ C
n + 1. Let us set
~
at
suffices to show that (f1 ~., f If 1 ~ i
~ k,
then
f i
1\
=bi
c:t:
1\ C
° at at
1 $ i $ k,
i::; k
and
k < i ~ n. In order to get a contradiction, it
nl
is an independent set.
333
CHAPTER 6
If
f
k
i
< i:S; n, then, accounting for the modularity, we have:
/ \ ( f 1 V ... V
f
i _ 1v
f
i + 1 v ... v
frJ:S;
:s; (b n + 1 v b i) /\ (b 1 v ... v b i - I v b i + 1 v ... v b n v b n + 1) = = b n + 1 v [ (b n + 1 v b i) /\ (b 1 v ... v b i - I v b i + 1 v .• •. v brJ = b n+ 1 .
Since f i:S; c and b n + 1 /\ We
now
C
= 0, we get the required equality.
have to show
f i = ( b n + 1 v b i) /\ c = O.
that
f i :t:.
0
at
b i + I /\ b i = 0,
Since
i, k < i:S; n.
all then,
by
the
Let
rule
of
substitution, we get b n + 1 /\ (c vb) = O. From this fact and from formulas M2 we have
hence,
c= cV(bn+I/\(cVb)= (biv c)/\ /\(bn+lV c), and,
[(biv c)/\ al)/\[(bn+lV c)/\ al):S;
c/\ al=O.
Since
al
is
a
uniform element, this is possible only if (b i v c) /\ a l = 0 or (b n+ I v c) /\ /\ a 1 = O. As far as
b i /\
C
= b n+ I
1\ C
= 0, then the rule of substitution
results in a contradiction with the fact that the element c v a l is essential. The lemma is proved. 6.3.57.
Lemma.
For any
Gol- dim L:S; Gol- dim[O, a)
aE
the following inequality holds:
L
+ Gol- dim[ a, 1).
Proof. Let us first note that if (a, Xl ,..,
in L, then
( a v x I -., a v
X
n}
X
n} is an independent set
is an independent set in
[a , 1).
Indeed,
using the modularity, we see that for any i, 1:S; i:S; n, we have: (a v x) /\ (a v
Xl
v ... v a v
X
i-
1 v a v x i+ I v ... ) =
= a v «a v Xi) /\ (xl v ... v x i _ l V x i + I v ...
»= a
Then, the inequality in the formulation of the lemma gets trivial provided one of the dimensions on the right is infinite. Let us, therefore, assume that they are both finite. Let us choose the biggest by the number of elements family all
(x I _. x n}, for which
xi will be uniform,
{ a,
Xl ,•. , X n}
n:S; Gol - dim[ a, 1) and
are independent. Then
a v
Xl
v .. v
Xn
will be
334
AUTOMORPHISMS AND DERrvATIONS
d = Gol - dim[O, a]
= q>x J) -----7 q> = 0,
'V qX 'V x
(i
* J)
(10)
Since these formulas set sheaf predicates, we can find a neighborhood u( f) of the point p, such that the form 'rf = He has a presentation
'rf
=
n
m
~=
J=
L L (b . xb*.)] . 1 . 1 ]
h.
",
the coefficients of which, as elements of the ring fQ, and the automorphisms hl'-' hn' as automorphisms of the ring fQ, obey conditions (9), (10) and (11). Besides, the ring
copies
f
fQ = fQ $ ... $
Q decomposes into a direct sum of isomorphic
fQ,
342
AU1OMORPmSMS AND DERIVATIONS
and the form ff is f f $ .. $ ff" Let us present the form f f as
Let I be an ideal from ( fI) g k r:;, R. If
F, such that ( Ib j) g k ~ R,
(b j I) g k ~ R,
x, a are arbitrary elements from I, then
(12) It should be remarked that we can find an element that lP(a 3 b*} ":t see that f ~ITf) =
° for a certain
= c1bi + .. +
° and,
T
E
F
so that
S;
e
S; e( ({i..a 1a 2
gk
I
E
and
W
df
gk
Wjk are zero. Let us set R. Besides,
a
gk
*
jk = a 4 ({i..a 3 (b j)
= af
gk
)
E
a4
E
W
jk
=f
I, such that
R, in which case not all elements
a= a4 a 3 . Then equality (12) yields
and, hence,
which
».
Then, by lemma 6.3.70. we can find an element a4
Tc i ~ R, we get
ITf({f..a 1a 2 ) = ({i..ITa 1 a 2 f) ~ IP( ITf) = 0,
hence,
such
j. In the opposite case by formula (9) we
cnb* n' Choosing
contradicts the choice of f
~ E If,
gk
W
({i..fJ,Ia»~
jk'
If ({i.. f J,Ia» = 0, then (12) turns into an identity with automorphisms. Since formula (11) is true, theorem 2.3.1 yields the following relations
at every k. Formula (10) shows that such relations are possible only when f gk W Ok ]
=f
gk ({i..a gk(
b*]
l k) =° for
all
j, k
(see lemma 1.6.24). This is
a contradiction that proves the lemma. 6.3.74. Proposition. Let R be a semiprime ring. G be its Mgroup of automorphismso If ~ is a dense left ideal in RG. then ~ is a dense left ideal in R. Proof. Let ((i..~) = 0, but IP be a nonzero homomorphism of the left R-modules IP: R ~ E(R). By the preceding lemma, we can find an
CHAPTER 6
element a
343 E R,
such that 'Z(a)
Rand 0
E
-:t
qi." 'Z(a»
E R
for a trace fonn
-r. Since!!. is dense, the right annihilator of the left ideal !!.l = !!.-r(a) RG
is zero. If
p is the right annihilator of this ideal in
principal
arbitrary
trace
form,
!!.1-r1(pI) = '(!!.lPI) = 0, i.e., arbitrary (see 5.6.8). Thus,
I
and
,( pI) = 0
R,
and, hence,
E
p = 0,
- 1
in
-r1 is an IF ,
then
as
-r1 is
This is a contradiction, which proves the proposition. 6.3.75. Proof of theorem 6.3.69 is based on proposItIOns 6.3.74 and 6.3.72, as the proof of theorem 3.8.2 was based on lemmas 3.8.3 and 3.8.4, and is carried out by the same scheme. Let
us
'4nax ( R). Let ~
first E
show
that
Chmx ( R G)
HaIr( A, R G), where
h
naturally
imbedded
into
A is a dense left ideal of the fixed
ring. Let us detennine the correspondence ~
Let us show that ~
is
h: R A
~
R
by the fonnula
is a mapping. Let
V= (Lra(a a~) I Lraaa= 0). a a
Then V is also a left ideal and for any fixed form essential ideal I, such that 'Z(I) ~ R, we have T
-r and for an
Lir a(aaS» = L'Z(ir a)(a as) = [L'Z(ir a)aa] S =
[-r(Liraaa)]s =0,
where i is an arbitrary element from we get
V =
O. It means that ~
h
I.
Therefore, by propoSItIOn 5.6.8,
is a homomorphism of the left R-modules.
Its domain of definition is a dense left ideal of the ring determines a certain element from
~
h.
and, hence,
Sh
be also denoted by
Sh
is an imbedding of
It is now obvious that the mapping
'4nax ( R G) into '4nax ( R) .
R
Chmx ( R), which will h: is
~
344
AU1OMORPfllSMS AND DERIVATIONS 1\
Now we are to show that the image of If
~ E '4nax ( R G) ,
Therefore,
x
= L r a a a'
IA[~ h_ (~ ~g]
=0
where
coincides with '4nax ( R)G
h
-\
raE I,
I g
.
!:: R, then
and, hence, the image h is contained in
1\
G
'4nax (R) . 1\
Inversely, let
e Qnax ( R). Then I-l
f(DO)S -
D( R)
Dq>!: R
has
for a certain
I e F. Let us find a dense ideal
a
unique
D
e G. Let
DO I:: D, such
Then we can detennine q>1-l e Hom R( IDO' R) by the formula Now we can easily see that q>~q>1-l is the sought
dq>I-l={dq»I-l_dl-lq>.
extension. Its uniqueness results from the fonnula (dq» I-l = dq>1-l+ d I-l q>. which we used to set the homomorphism q>1-l. The lemma is proved.
The extensions
6.4.20.Lemma.
fl1.-.• fl n
of basic derivations onto
Qnax (R) will be strongly independent.
Proof.
Let
"" £.Jc i x Ii·"= ax - xa
x e Qnax ( R). Let us find a dense ideal
for
C i'
and
a e Qmax (R)
D. such that
all
Da I:: R.
Dc i I:: R.
Then we get an identity lld1ci)(xd)l-li=(d1a)xd-d1x(da).
~. de D.
By
theorem
e( D)
= 1.
2.3.1
we can find
then we get
Ci
e( Jl i)
remark that e(Jl) = e(fl): if The lemma is proved.
=0
e( Jl i) . dIe i ® d = 0 at all i.
and e( C i)e ( Jl)
Rl-le=O.
then
= O.
Since
We now have to
iqlie=(iq)l-le- il-leq=O.
6.4.21. Theorem (I. Piers Dos Santos). quotients the following formula is valid:
Let us prove some more auxiliary lemmas.
For a maximal ring of
352
AUlOMORPIDSMS AND DERIVATIONS
Let
121
= Al < A2 < "' < A m
be correct words of basic derivations.
1 = el'"" em be supports of the operators A 1'"" A m' Let us fix a denotation
F, such that
I for an ideal
= e j I +(1 E( R) = E(I) = E(e . IH· ]
decomposition hull
I
Ie j ~ R,
I
il
j
~ R,
1~ j
~
m, A direct
e j)I induces decomposition of the injective E«1 - e .) I), This enables one to determine ]
multiplications of the elements of the injective hull by the central idempotents where is the corresponding decomposition of correspondence
e j E( R) , Mk
=
w
an
under
into a direct sum, To every word A j abelian the
group
A j e j E( R)
correspondence
which
Aj wH
let us put into
is
w.
isomorphic Let
us
to set
k
L$
A j e j E( R), and introduce on M k the structure of the left R-
j=l
module:
rAw =
D' (r il" w).
tl' • il" = il
By the Leibnitz formula we have:
=
so that the determined action indeed turns M k into a left module. In each of k
these modules let us consider submodules N k =
L
$ A j e j R.
j=l
is an essential submodule in Mk' Proof. Let us carry out induction over k. If k = 1, then
6.4.22.
Lemma.
Nk
f'1.
can
be naturally identified with E( R), in which case Nl is identified with R and now we have make use of the fact that R is an essential I-submodule in the injective hull E( R) . Let N k _ 1 be an essential submodule in Mk _ l' Let us consider a submodule
Wk
= IA k e k
and note that
Wk t1 M k- 1
= O.
Indeed, if
353
CHAPTER 6
O:;t
SE Ie k ,
~A' tt then sAke k = £..P kS ks sAks(mod M k _ 1). Now, by the
inductive supposition, Wk
+-
N k _ 1 is an essential submodule in
Wk
+-
M k _ 1.
Since Nk:::J. Wk:" N k _ l' we now have to prove that Wk +- M k-l is an essential submodule in Mk • For this purpose let us make use of an evident fact: if U eVe W is a chain of modules, and v / U is an essential submodule, then v is an essential submodule of W. Under our conditions, k
Mk / M k _ 1
This mapping transforms
L Ai
e k E( R) at the mapping
is isomorphic to
into
Wk
ek I
and, hence,
+-
Wk
Wi
i= 1
~
wk'
M k _ 1 / M k _ 1 is
an essential submodule in e k E( R). The lemma is proved.
6.4.23. Lemma. If W is a nonzero element of the injective hull then there is an element a E R, such that O:;t t(a) we R, where f( x) is an essential trace form and f( a) E R. E( R),
Proof. Let ideal from above,
t( x)
F, such that
=L x I
.
e· = e (A .). SInce ]
such that
]
eke(ck)w:#
!J.J
!J.
J C j'
and by I let us denote an 1 S; j S; m, where, as
,Ic j ' Ie j !:: R,
m
sup e. e( c .) = 1, then there is a number .
J=
1
]
]
L
O. By corollary 6.4.6, we get
i.e., we can find a constant t
E I L,
k,
RI· eke(ck)w:;t 0,
such that (c kt) e kW :# 0 (it should be e( c k) = c k' c' k ).
recalled that by the regularity of the center C, we have In M k let us consider an element h =
LA j(
Cj
t)e j w. By the construction,
the k-th component of this sum is nonzero and, hence, h is a nonzero element in Mm' By lemma 4.6.22, we can find an element S E I, such that 0:# sh E N m.
Let us interrelate every element ~: I ~ E( R) A
sg (x)
= ~(xs)
acting
by
at any
the
g =
formula
LA j wi E
!J.
~(x) = Lx
x, s E I. Besides, if
with a mapping
Mn j
wi'
E Nm
and ~ is a zero mapping, then
Now we can find an element
have
g E N m' then the image of
~ will be contained in R. And, finally, the main point:
shows that if g of N m .
We
theorem 2.3.1
g is a zero element
s1 E I, such that
A
sh(s1):;t0, i.e.,
354
AU1OMORPHISMS AND DERIVATIONS
A
h(Sls);t:O,
however,
at
XE I
A
we
have
A
h(x)= f(xt)w.
Setting
"
a= sIst, we get O;t: h(sls)= f(a)w= Sh(sl)E R. The lemma is proved. L
6.4.24. Proposition. If P is a dense left ideal in R , then Rp is also a dense left ideal in R. Proof. Let Rpw = 0 for a certain nonzero WE E( R). By the preceding lemma, we can find an element a E R, such that O;t: f( a) w E R and f(a)
E R L.
PI = Pf( a) - 1 in
Since P is dense, the right annihilator of the left RL
R L -ideal
and, hence, it is zero in R. Now we have
O;t: PI' [f(a)wl =
[plf(a)lw~ pw=O.
This is a contradiction which proves the proposition.
6.4.25. Now the proof of theorem 6.4.21 can be obtained by nearly word per word repetition of theorem 6.3.58 , using lemma 6.4.18 and proposition 6.4.24. 6.4.26. Corollary. Martindale ring of quotients:
The
following
equalities
are
valid for
a
The proof immediately results from theorem 6.4.21 by applying corollary 6.4.6 and an imbedding R F ~ Clnax .
6.4.27. Corollary. If
R
is a left Goldie ring, then
Indeed, in this case the maximal ring of quotients coincides with the classical one.
RL
6.4.28. Corollary. If R is a regular left self-injective ring, then is also regular and left self-injective. REFERENCES: V.K.Kharchenko [70,77]; J.Piers Dos Santos [105]; A.Z.Popov [137].
355
CHAPI'ER 6
6.S Hopr Algebras In this paragraph we shall consider a general concept of the action of Hopf algebras embracing both the case of automorphisms and that of derivations. Let us recall some definitions pertaining to the notion of Hopf algebras. 6.S.1. The definition of a co-algebra is obtained from the category definition of an algebra by reversing all the arrows. Namely, comultiplication on a linear space H is a linear mapping A: H ~ H ® c H, where C is a basic field. We have already observed earlier that the use of the functor Hom( - ,C) results in reversing of arrows (3.3). This observation prompts that if on a finite-dimension space H a multiplication is given, then on a dual space If a co-multiplication is induced, and vice versa, if on H a comultiplication is set, then a multiplication is induced on If (i .e., If transforms into an algebra which is not, generally speaking, associative). Now there arises a natural desire to present this multiplication and co-multiplication in a common or alike form. This can be achieved in terms of structural constants. Let a multiplication U and co-multiplication A which are not related, be given on H. Let us fix a certain basis (h i' i E I) of H. In this case the multiplication is uniquely set by a system of structural constants ( c(~ ~ E CI ~]
i, j, k
E
I) which determines the products of basic elements
' " (k) h . U(h.,h.) df = h.·h.=£.Jc" k ~]
~]
k~]
(1)
Analogously, the co-multiplication on the basic elements can be set by structural constants ( us:(k) .. ~]
E
CI"~, ], k
E I
)• .
(2)
It would be natural to suggest that a co-multiplication r! and a multiplication .1'" is set on If by the same sets of structure constants
356
AUlOMORPHISMS AND DERIVATIONS
and, respectively,
= L. . C(~~ h*. ® ~J ~
u*(h* \
k'
.1,
(3)
h*. J
J
A*( * *) ~ ~(k) * hi' Ii j = £.Jo ij Ii k .
(4)
Ll
k
where, as usual, by h~ ,M" h*n we denote the dual basis of a dual space. This proposition proves to be correct, and thus, fonnulas (1 )-( 4) show that in principle the notion of a co-algebra is as natural as that of an algebra. For the case of an infinite dimension, fonnulas (3) and (4) remain valid, but in this case there arise some question to be answered. First, the set
Il does not form the basis of a dual space. Nevertheless, any functional a E Hom( H, C) is set by its action on basic elements (h i 1 and, (h*il i E
hence a product
can be identified with an element
IT Clfi ,
and inversely, any element
detennines the functional:
(II a .If.)( h J.) = ~
~
II
ie I
a(h) lfi
IT a i lfi
of the direct
of the direct product
a " Therefore, any element from J
If is uniquely presented as an infinite linear combination of elements h*i' It would be now natural to detennine rf and tJ,* on infinite combinations with
respect to linearity
rf( LA k h*k ) = LA ku*(h k) L1*(LAi h*i' L J1. j h*j)
=L
(5)
(6)
A iJ1. jL1*(h*i ' h*j)
and here arises the second question: how to reduce the similar terms in the right-hand parts of fonnulas (5) and (6). Here of some help is the fact that in fonnulas (1) and (2) the sums are finite, which implies that for fixed i, j
there is only a finite number k, such that
c< .k~ ::t. 0, ~J
and for every
fixed k there is only a finite number of pairs (i, J), such that
o~;::t. O.
Therefore, the right-hand parts of formulas (5) contain only finite sets of similar terms. And, finally, the last and most tragic for formula (3) peculiarity is that in the right-hand part of (5) there is an infinite direct sum of basic tensors, i.e., it is an element from (H ® H)*. This element can be not a finite sum of basic tensors of infinite linear combinations of functionals (h *i 1, i.e., it can lie not in
If ® If and, as a result, rf will not be a
357
CHAPTER 6
* co-multiplication in H. Therefore, for an infinite-dimension case we can only claim that if H is a co-algebra, then H* is an algebra, which is sufficient for putting all the notions of the theory of algebras into correspondence with dual notions of the theory of co-algebras. For example, co-commutativity, co-associativity and the existence of co-unity in H mean commutativity, associativity and the existence of unity in H*, respectively. Let us write these notions in terms of comultiplication.
Co-commutativity. If t:J... h) = L h(~) ® h(~), ~ ~
then
""
L.J
h(~) ® h(~) = ~ ~
t:J... h).
Co-associativity. L
h(~) ® L1(h(~») ~
~
If t:J... h) = L
h(~)
in the tensor product
Co-unity is a functional "" (1) (2) L.J£( hi) hi
~
H
®
h(~), ~
then
LL1(h (~») ® ~
h(~) = ~
® H ® H.
£: H4 H,
such
that
Lh(~)£(h(~»)= ~
~
"" (1) (2) L.J hi ® h i '
= h, where, as above, t:J... h) = A bialgebra is an associative algebra with unity,
H, on which set are a co-associative multiplication L1: H 4 H ® H and a co-unity £: H 4 H, which are homomorphisms of C-algebras. An antipode of a bialgebra H is an antihomomorphism of algebras, s: H 4 H, such that
AI ~ " (1) h were '-'\ h, = " L.J hi
10. '. Each of the elements s E G acts on L and. hence. uniquely extends to an automorphism of R. In the space y( s) let us fix a certain
basis
J.lt ,.. .•
J.L n~· and determine the action J.L i J.L j) = 8ij J.L j' where 8ij is a
Kroneker delta. and J.L i r(h» = 0 at h;t:. s. Since R is a free algebra. then such an action can be extended to s-derivations of R. Therefore. r is implemented by skew derivations. Let us relate a comb r to a Hopf algebra C < r >. which will be called a free hull of the comb G. As has been noted above. the group G acts on a free algebra C
. where
L=
L y( s).
Hence. we can detennine
a crossed product C < r > C;; C < L> * G with a trivial system of factors (or. in other words. a skew group algebra). The basis of this product consists of all possible words sJ.lt .• J.Ln. where s E G. J.Li are the elements of fixed bases of teeth. A product of such words is detennined by a multiple use of the formula J.Ls = s J.Ls. Let us define co-multiplications on C < G>. For the words of a unit length let us set N..J.L)= J.L® 1+ s® J.L. (J.LE r(s». Ll( s) = s ® s. (s e G).
CHAPTER 6
363
A is extended onto words of greater length and on linear combinations in such a way that co-multiplication becomes a homomorphism of algebras. Therefore, C < r > turns into a bialgebra. This bialgebra has a co-unity E: C < r > -+ C, which is zero on all the words including basic derivations, and is equal to unity on the elements from G. Let us determine an antipode S on C < r >. Let us set S( Jl.) = - h- 1 Jl., S( h) = h- 1, where Jl. E y( h) • Let us extend it to an anti-automorphism s: c < r > -+ C < r >. The fact that E is a co-unity and S is an antipode is easily checked by induction over the length of a word. Thus, C < r > turns into a Hopf algebra. It is absolutely obvious that the action of this Hopf algebra on the algebra R is equivalent to the implementation (possibly, not exact, Le., with a kernel) of the comb r with skew derivations. Inversely, if r is a subcomb of a comb of all skew derivations of the algebra R, then on R one can determine the action of a free hull C < r > by viewing the word .. s llr Iln s- III Iln sPt.-Jl.n as a superpoSItiOn x =«(x) ) •. ) .
The algebraic structure of a set of skew derivations can be characterized by the notion of a relatively primitive element of a free hull. An element wee < r > is called relatively primitive if there can be found an element S E G, such that
LX w) = w ® I +
S
® w.
Let us assume that G is a comb of all skew derivations of a certain algebra R. If w = w(s 1,.., S n' Pt ,.., Jl. ~ is a certain relatively primitive element, then, by the definition, its action on R will be a skew derivation, Le., there is a Jl. E L s ' such that (x E R).
This implies that w(s 1,...,
S
n'
Pt ,.. Jl.rJ =
w
determines a certain partial operation on G:
Jl..
Let us consider some examples. Let Jl., v be skew derivations corresponding to automorphisms S and h, respectively. If Jl.h= Jl., v s= v, then the commutator Jl. v - vJl. is a relatively primitive element. If Jl. h= - Jl., then is a relatively primitive element. If s = hand Ji2 = Jl. + v,
Jil
2 Jl. v - 2v Jl. - v 2 is a relatively primitive element. In more general terms, let us consider a certain tooth y(s) in an arbitrary comb. v s= v, then
Then y( s ) s = r( s - 1 s s) = r( s) , Le., s acts on y( s ) as a non-degenerate linear transformation. Let us assume that in y(s) a basis Pt ,.., Jl. n is chosen
364
AUTOMORPIllSMS AND DERIVATIONS
in such a way that the matrix of transformations is an elementary Jordan matrix with
an eigen
number
a, i.e.,
/1f= a/11 + /12'
/1;= a
1'2 + J1:,,-
/1~=
a*" ± 1, then a subspace )(s)2 stretched in < r > onto the products of pairs of elements of the tooth )(s) has no nonzero relatively primitive elements. If a = - 1, then the dimension of the a /1 n. One can show that if C
subspace of relatively primitive elements in
)(s)2 is equal to
[n; 1].
If
a = 1 and the characteristic of the main field is zero, then this dimension is n equal to [2]. 6.5.6. Algebraic dependences between skew derivations. Here in the spirit of Chapter 2 we shall consider the problem of dependences between skew derivations. This problem is equivalent to studying generalized identities with automorphisms and skew derivations. Let R be a prime ring. Any its skew derivation has a unique R), so that, as extension up to the skew derivation of a ring of quotients usual, we shall assume that the objects under investigation are determined on R) as well. Moreover, it would be reasonable to consider not only the skew derivations transferring R in R, but also s-derivations from R in R) transferring a certain nonzero ideal of the ring R in R. Let
ex
ex ex
and
sEA (R)
let
us
set
!L s = ( /1
is
an
s -derivation
of
QI3 I < R, I*" 0, I J.1 . The solution of the problem of describing algebraic dependences implies studying, to a certain extent, relatively primitive elements. In the simplest case, when an automorphism commutes with skew derivations (i.e., basic automorphisms act trivially on all the teeth of a comb), relatively primitive elements are described easily: their linear combinations form a (restricted, provided the characteristic is positive) Lie algebra generated by the elements of the comb. Therefore, the situation differs but slightly from the case of common derivations. 6.5.9. Theorem. Let a prime ring the type F( x ~i ~ k)
=
O. where
F(
z(~·
k)
R
obey a polynomial identity of
is a generalized polynomial with
AU'IOMORPIDSMS AND DERIVATIONS
366
the coefficients from R F ; L\ 1 ~-, L\ n are mutually different correct words from a reduced set of skew derivations commuting with all the corresponding automorphisms, and hI".' h m are mutually outer automorphisms. In this case the identity F( z =
an g n 0 g n'
or L a ·a . g. 0 g ] . = La. g. ® g ~.. . . ~J~ .~~
2., ]
°
~
ci1
Hence, we have a i a j = at i c# j and = a i. It is possible only one of the coefficients a i is equal to unity, while the other are zero, i.e.,
367
CHAPTER 6
the initial dependence assumes the fonn g = g, which is the required proof. Therefore, the linear space generated by G is a group algebra of the group G. In this case the structure of the Hopf algebra induced from H to C[ G], is the same as on the group algebra C[ G] as a Hopf algebra. Thus, any Hopf algebra contains the biggest Hopf subalgebra which is a group algebra. Analogously, in H we can find "Lie" elements which always act as derivations (in the theory of Hopf algebras such elements are called primitive). Let L= (1 E HI ~1) = 1 ® 1 + 1 ® 1) We can easily see that L is a linear space, and if 11' Lz E L, then 1112 - 1211 E L, i.e., L fonns a Lie subalgebra in H. If the basic field has a positive characteristic p, then
~1 p) = (1 ® 1 + 1 ® 1) p = 1 P® 1 + 1 ® 1 P and, hence, L fonns a restricted Lie algebra. Let us show that the sub algebra generated by L in H will be isomorphic to the universal enveloping of L if p = 0, and to the universal p-enveloping if p> O. To this end it would suffice to show that different
correct words from a certain fixed basis ( 11 p_) of the space L will be linearly independent. Let us carry out induction over the length of a correct word. Let all the words less than v are linearly independent and let us
v La
assume that = i Vi' where Vi are less words. Let us apply to both parts of the latter equality a co-multiplication L1(V)= V'oV"
or, cancelling the sums right, we get
V'oV" =
v'. v"
v
"* (2)
=v
v' ® v" =
La i Vi ® 1 + Ll ® ai V
v' ® v" =
L
. '0" .l,ViVi=Vi ,
n
