X hits on this document

# OF GRADED DIVISION ALGEBRAS - page 18 / 29

113 views

0 shares

18 / 29

18

R. HAZRAT AND A. R. WADSWORTH

P r o o f . R e c a l l f r o m P r o p . 3 . 2 ( i ) t h a t N r d E ( a ) = N r d Q ( a ) f o r a l l a E , s o t h e i n c l u s i o n E Q y i e l d s a m a p S K 1 ( E ) = E ( 1 ) / E 0 Q ( 1 ) / Q 0 = S K 1 ( Q ) . A l s o r e c a l l t h e h o m o m o r p h i s m λ : Q of (2.6), which Q0, we have E m a p s Q 0 t o E 0 . S i n c e t h e c o m p o s i t i o n E Q λ E i s t h e i d e n t i t y m a p , f o r a n y a E ( 1 ) a = λ ( a ) E 0 . T h u s , t h e m a p S K 1 ( E ) S K 1 ( Q ) i s i n j e c t i v e .

Proposition 5.6. Let E be a graded division ring, and let Q = q(E). Then, Q ( 1 ) = ( Q ( 1 ) E 0 ) Q 0 .

Once this proposition is proved, it will quickly yield the main theorem of this section:

# Th

eorem

5 . 7 . L

et

E b e a g r a d e d d i v i s i o n r i n g .

Th

en,

SK1(E)

=

S K 1 ( q ( E ) )

.

Proof. Set Q = q(E). Since the reduced norm respects scalar extensions, Q(1) E 0 E ( . The image of 1 ) t h e m a p ξ : S K 1 ( E ) S K 1 ( Q ) i s E ( 1 ) Q 0 / Q 0 , w h i c h t h u s c o n t a i n s ( Q ( 1 ) E 0 ) Q 0 / Q 0 /Q0 = SK1(Q) = Q ( 1 )

(using Prop. 5.6).

# Thus ξ

is surjective, as well as being injective by Lemma 5.5, proving the theorem.

Proof of Prop. 5.6. Case I. Suppose

E We first treat the case where = Zn for some n N.

E

is finitely generated.

Let F = Z(E), a graded field, and let εi = (0, . . . , 0, 1, 0, . . . , 0) (1 in the i-th position), so

### E

=

LγZε1 . . . Zεn. For 1 i n, let ∆i = Zε1 . . . Zεi E ; a n d l e t S i = E i E γ , w h i = c h i s a i g r a d e d s u b - d i v i s i o n r i n g o f E . L e t Q i = q ( S i ) , t h e q u o t i e n t d i v i s i o n r i n g o f S i ; s o Q n = Q a s S n = E . S e t R 0 = Q 0 = E 0 . N o t e t h a t [ S i : ( S i F ) ] < , s o Q i i s o b t a i n a b l e f r o m S i b y i n v e r t i n g t h e n o n z e r o e l e m e n t s o f S i F . T h i s m a k e s i t c l e a r t h a t Q i Q i + 1 , f o r e a c h i .

F o r e a c h j , 1 j n , c h o o s e a n d fi x a n o n z e r o e l e m e n t x j E ε j . L e t ϕ j = i n t ( x j ) A u t ( E ) ( i . e . , ϕ j i s c o n j u g a t i o n b y x j ) . S i n c e ϕ j i s a d e g r e e - p r e s e r v i n g a u t o m o r p h i s m o f E , ϕ j m a p s e a c h S i t o i t s e l f .

Hence, ϕj extends uniquely a b e l i a n g r o u p , t h e r e i s ` j

to

an automorphism

N

jj such that ` ε

to F.

Qi, also denoted ϕj. Since each Then, if we choose any nonzero

E / z j F ` j F ε j is a torsion , we have

x ` j j E ` j ε j = E 0 z j . S o , x ` j j = c j z j f o r s o m e c j E 0 , a n d z j F = Z ( E ) . T h e n ϕ ` j j i n t ( c j z j ) = i n t ( c j ) . T h u s , ϕ ` j j | S i i s a n i n n e r a u t o m o r p h i s m o f S i f o r e a c h i , a s c j E 0 S i .

= int(xj

l j ) =

Now, fix i with 1 i n. We will prove: Q i Q ( 1 )

( Q i 1

Q(1)

) [ Q i , Q ] .

(5.6)

we

h ave Qi 1 W e h a v e S i = S i 1 [ x i , = x 1 ] = S i 1 [ i [ x i ] i] ( Q i 1 x i , ϕ t w i s t e d [xi, x i p o l y n o m i a l r i n g ) , w i t h ϕ ` i i 1, ϕi] (twisted Laurent polynomial ring). Likewise, within Qi an inner automorphism of Qi 1 . In

order to invoke Prop. 5.4, let

T = Qi 1

[ x i ] = Q i 1 [ x i , ϕ i ]

and let

R = Z(T ).

Since Si 1 automorphisms of Qi generated by ϕ of inner automorphisms of Qi. Since [xi] T Qi = q(Si 1 [ i+1 x i ] ) , w e h a v e q ( T ) = Q i . , . . . , ϕn, and let G = /( Skolem-Noether shows that

# ⊆ Aut(Qi) be the subgroup of

Inn(Qi)), where Inn(Qi) is the group Inn(Qi) is the kernel of the restriction

m a p A u t ( Q i ) A u t ( Z ( Q i ) ) , t h i s G m a p s i n j e c t i v e l y i n t o A u t ( Z ( Q i ) ) . F o r σ G , w e w r i t e σ | Z ( Q i )

for the

automorphism of Z(Qi) determined by σ. Note that G is a finite abelian group, since the images of the ϕ i h a v e fi n i t e o r d e r i n G a n d c o m m u t e p a i r w i s e . ( F o r , w e h a v e x j x k = c j k x k x j f o r s o m e c j k E 0 . H e n c e ϕ j ϕ k = i n t ( c j k ) ϕ k ϕ j a n d i n t ( c j k ) I n n ( Q i ) , a s c j k E 0 Q i ) . E v e r y e l e m e n t o f is an automorphism o f [xi] preseerving degree in xi, so an automorphism of T , since this is true of each ϕj. Therefore we S i 1

have a group action of

on T by ring automorphisms, and an induced action of

on Div(T ). Note that as

any ψ

permutes the maximal left ideals of T , the action of ψ on Div(T ) arises from an action on the

base of Div(T ) consisting of isomorphism classes of simple T -modules. That is, Div(T ) is a permutation

 Document views 113 Page views 113 Page last viewed Mon Jan 23 12:45:03 UTC 2017 Pages 29 Paragraphs 2607 Words 32418