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OF GRADED DIVISION ALGEBRAS - page 20 / 29

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20

R. HAZRAT AND A. R. WADSWORTH

Hence, for

= δT (a) Div(T ), using the identification of G with al(Z(Qi)/Z(C)0) and the commutative

diagram (5.7), 0 = δ R ( N r d Q ( a ) ) = δ R N Z ( Q i ) / Z ( Q i )

G (Nrd

Q

i (a))

m

=

P

σG

σ δR(NrdQ

i ( a ) m )

=

N G

δR(Nrd

Q

i (a))

m

= m N G ( N r d ( δ T ( a ) ) ) = m N G ( N r d ( ) )

.

as we saw above, S i n c e D i v ( R ) i s t o r s i o n - f r Nrd(δT (a0)) = Nrd(δT (a)) e e , w e h a v e N G ( N r d ( ) ) = 0 , i . e . , N r d ( ) k e r ( N G ) = t h e r e i s c [ Q i , Q ] Q i w i t Nrd(δT (c)) = h N r d ( ) = N r d ( δ T ( c ) ) . L e t a Nrd( ) Nrd( 0 =

  • 0

    = IG( ). Therefore,

a / c Q ) = 0. i

.

Then,

B e c a u s e N r d : D i v ( T ) D i v ( R ) i s i n j e c t i v e ( s e e P r o p . 5 . 4 ) , i t f o l l o w s t h a t δ T ( a 0 ) = 0 i n D i v ( T ) . T h e r e - f o r e , a s T = Q i 1 [ x , ϕ i ] a n d q ( T ) = Q i , b y P r o p . 5 . 3 t h e r e i s a 0 0 Q i 1 w i t h a 0 0 a 0 ( m o d Q 0 i ) . S o , a 0 0 a

( m o d [ Q i , Q ] ) , a n d h e n c e N r d Q ( a 0 0 ) = N r d Q ( a ) = 1 , i . e . , a 0 0 Q proving (5.6). i 1

Q(1)

. T h u s , a ( Q i 1

Q(1)

) [ Q i , Q ] ,

The inclusion (5.6) shows that for any i, 1 i n and any a Q(1)

with b a (mod Q0).

Hence, by downward induction on i, for any q

Q i t h e r e i s b Q ( 1 ) Q ( 1 ) = Q(1) Qn there is Qi 1

d Q 0 Q ( 1 ) = E 0 Q ( 1 ) w i t h d q m o d Q 0 ) . S o , Q ( 1 ) completing the proof of Case I.

(Q(1)

E 0 ) Q 0 . T h e r e v e r s e i n c l u s i o n i s c l e a r ,

Case

II.

Suppose

E

is

not

a

nitely

generated

abelian

group.

The basic point is that E is a direct limit of sub-graded division algebras with finitely generated grade group, so we can reduce to Case I. But we need to be careful about the choice of the sub-division algebras to assure that they have the same index as E, so that the reduced norms are compatible.

Let F = Z(E). Since | E/ F | < , there is a finite subset, say {γ1, . . . , γk} of

E

whose images in

E/ generate this group. Let ∆0 be any finitely generated subgroup of E, and let ∆ be the subgroup F E generated by ∆0 and γ1, . . . , γk. Then, ∆ is also a finitely generated subgroup of of E, but with the

added property that ∆ +

F

=

E. Let

E

=

L E δ ,

δ

which is a graded sub-division ring of E, with E,0

= E 0 a n d

E

= ∆. Since ∆ +

F

=

E, we have

a EF = E. (For, take n d c F η . T h e n , E that γ any γ = d c E 0

E and write γ = δ + η with δ ∆ and η E F . ) B e c a u s e E F = E , w e h a v e Z ( E F , and any nonzero d E,δ ) = F E= FF . Note

[E: Z(E)] = [E,0 : F= [ E 0 : F 0 ] | E

: F ,0] | :( F | = [E : F ].

F ) | = [ E 0 : F 0 ] | (

+

F) :

F|

The graded homomorphism EZ(E )

F E is onto as EF = E, and is then also injective by di-

m e n s i o n c o u n t ( o r b y t h e g r a d e d s i m p l i c i t y o

f E

Z(E )

F)

.

Th

us,

E

Z(E )

F

=

E . I t f o l l o w s t h a t

q(E) q(Z(E ))

q

(F )

= q

( E ) . S

pec

i fi c a l l

y,

q(E) q(Z(E ))

q

(F )

=

( E

Z(E )

q(Z(E))) q(Z(E ))

q

(F )

=

E

Z(E )

q(F )

Therefore, for any a q(E), Nrdq(E ) = (EZ(E ) (a) = Nrdq(E) F) F (a). q

(F )

=

E

F q

(F )

(E) =q .

Now, if we take any a Q(1)

where Q = q(E), there is a subgroup ∆

E

with ∆ finitely generated

and ∆ +

F

a q(E)

(1)

(a) = NrdQ(a) = 1, we have, by Case I applied to E, and a E. Since Nrdq(E ) = E E 0 q ( E ) 0 ( Q ( 1 ) E 0 ) Q 0 , c o m p l e t i n g t h e p r o o f f o r C a s e I I .

Remark. (i) Prop. 5.6 for those E with

E

=

Z

was prove

di

n

[PY]

, a n d o u r p r o o f o f t h i s i s e s s e n t i a l l y

the same as theirs, expressed in a somewhat different language. Platonov and Yanchevski˘ı also in effect assert Prop. 5.6 for E with E finitely generated, expressed as a result for iterated quotient division rings

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