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SK1

OF GRADED DIVISION ALGEBRAS

17

Proposition 5.4. For R = Z(T ) = K[Y ], there is a map Nrd : Div(T ) Div(R) such that the following diagram commutes:

D

//

Q

δT

//

Div(T )

NrdD

Z(D)

NrdQ

Nrd

(5.2)

NZ(D)/K

K

//

q(R)

δR

//

Div(R)

Moreover, Nrd is injective.

Proof. Let E = T [x 1] = D[x, x 1, σ], which with its grading by degree in x is a graded division r i n g w i t h E 0 = D a n d q ( E ) = Q . S i n c e i n d ( Q ) = i n d ( D ) [ Z ( D ) : K ] , b y ( 3 . 3 ) , f o r d D = E 0 ,

NrdQ(d) = NZ(D)/K

(NrdD(d)). This gives the commutativity of the left rectangle in the diagram.

For the right vertical map in diagram (5.2), note that there is a canonical map, call it N : Div(T ) Div(R) given by taking a T -module M (with ACC and DCC) and viewing it as an R-module; that is N(jhT (M)) = jhR(M). But, this is not the map Nrd : Div(T ) Div(R) we need here! (Consider N a norm map, while

our desired Nrd is a reduced norm map.) The map δR : q(R)Div(R) is defined the same way as δT

. Let ψ = δR NrdQ : QDiv(R). Then,

Q0 ker(ψ) as q(R)is abelian, and Dker(ψ) by the commutative left rectangle of (5.2). Prop. 5.3 thus yields ker(δT ) ker(ψ). Since δT is surjective, there is an induced homomorphism Nrd : Div(T ) Div(R) such that Nrd δT = ψ = δR NrdQ. This Nrd is the desired map.

We have a scalar extension map from R-modules to T -modules given by M T R M. This in- duces a map ρ: Div(R) Div(T ) given by ρ(jhR(M)) = jhT (T R M). For any r R, we have

T

R

( R / R r ) =

T / T r .

Th

u s f o r a n y g

T \ {0}

,

ρ(Nrd(δT (g))) = ρ(δR(NrdQ(g))) = ρ(jhR(R/RNrdQ(g))) = jhT (T/T NrdQ(g)) = δT (NrdQ(g)) = n δT (g),

using Remark 5.1(iv). This shows that ρ Nrd : Div(T ) Div(T ) is multiplication by n, which is an injection, as Div(T ) is a torsion-free abelian group. Hence Nrd must be injective.

Remark. Here is a description of how the maps Nrd : Div(T ) Div(R) and N : Div(T ) Div(R) and ρ: Div(R) Div(T ) are related, and a formula for Nrd on generators of Div(T ). Proofs are omitted. We

have

ρ Nrd = n idDiv(T )

;

(5.3)

and

N = n · Nrd.

(5.4)

Let S be any simple left T -module, and [S] the corresponding basic generator of Div(T ). Let M = annT (S), and let P = annR(S), which is a maximal ideal of R. Let k = matrix of size of T/M = dim(S), where

=

E n d T ( S )

, so

T /M

=

M k ( ) . T h

en,

Nrd([S]) = nS[R/P ],

where

nS

=

1 nk

dimR/P

(T/M) = ind(T/M).

(5.5)

We now consider an arbitrary graded division ring E. As usual, we assume throughout that torsion-free abelian group and [E : Z(E)] < .

E

is a

Lemma 5.5. Let E be a graded division ring, and let Q = q(E). Then, the canonical map SK1(E) SK1(Q) is injective.

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