SK_{1 }

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^{∗ }

δ_{T }

//

Div(T )

### Nrd_{D }

# Z(D)^{∗ }

Nrd_{Q }

## Nrd

(5.2)

N_{Z(D)/K }

K^{∗ }

//

q(R)^{∗ }

δ_{R }

//

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 ,

Nrd_{Q}(d) = N_{Z(D)/K }

(Nrd_{D}(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(jh_{T }(M)) = jh_{R}(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 δ_{R }: q(R)^{∗ }→ Div(R) is defined the same way as δ_{T }

. Let ψ = δ_{R }◦ Nrd_{Q }: Q^{∗ }→ Div(R). Then,

Q^{0 }⊆ ker(ψ) as q(R)^{∗ }is abelian, and D^{∗ }⊆ ker(ψ) by the commutative left rectangle of (5.2). Prop. 5.3 thus yields ker(δ_{T }) ⊆ ker(ψ). Since δ_{T }is surjective, there is an induced homomorphism Nrd : Div(T ) → Div(R) such that Nrd ◦ δ_{T }= ψ = δ_{R }◦ Nrd_{Q}. 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(jh_{R}(M)) = jh_{T }(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(δ_{T }(g))) = ρ(δ_{R}(Nrd_{Q}(g))) = ρ(jh_{R}(R/RNrd_{Q}(g))) = jh_{T }(T/T Nrd_{Q}(g)) = δ_{T }(Nrd_{Q}(g)) = n δ_{T }(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 id_{Div(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 = ann_{T }(S), and let P = ann_{R}(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]) = n_{S}[R/P ],

where

n_{S }

=

1 nk

dim_{R/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 SK_{1}(E) → SK_{1}(Q) is injective.