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R. HAZRAT AND A. R. WADSWORTH

• (iii)

For all q, s Q, δ(sqs 1) = δ(q). This is clear, as δ is a homomorphism into an abelian group.

• (iv)

For all q Q, δ(NrdQ(q)) = n δ(q), where n = ind(Q). This follows from (iii), since Wedderburn’s

factorization theorem applied to the minimal polynomial of q over Z(Q) shows that NrdQ(q) =

1 n i=1 Q for some si Q. s i q s i ( v ) I f N r d Q ( q ) = 1 , t h e n δ ( q ) = 0 . T h i s i s i m m e d i a t e f r o m ( i v ) , a s D i v ( T ) i s t o r s i o n - f r e e .

# L

emma

5 . 2 . T a k e a n y f , g

T \ {0}

w i t h T / T f

=

T / T g , s o d e g

(f )

= d e g ( g ) . I f d

eg

(f )

1 , t h e r e e x i s t

s, t T \ {0} with deg(s) = deg(t) < deg(f) such that fs = tg.

: T/T f

# (1 + T f) = s + T g. By the division algorithm, s can be chosen

with deg(s) < deg(g). We have

fs + T g = f(s + T g) = f (1 + T f) =

(f + T f) =

(0) = 0

in T/T g.

Hence, fs = tg for some t T . Since deg(f) = deg(g), we have

deg(t) = deg(s) < deg(g) = deg(f).

Proposition 5.3. Consider the group homomorphism δ : QDiv(T ) de ned in Remark 5.1(ii) above. Then ker(δ) = DQ0.

Proof. (cf. [PY], proof of Lemma 5) Clearly, Dker(δ) and Q0 ker(δ), so DQ0 ker(δ). For the reverse inclusion take h ker(δ) and write h = fg 1 with f, g T \ {0}. (As Q is a central localization of T , g may be chosen in Z(T ), but we do not need this.) Since δ(fg 1) = 0, we have δ(f) = δ(g), so d e g ( f ) = d e g ( g ) . I f d e g ( f ) = 0 , t h e n h D , a n d w e r e d o n e . S o , a s s u m e d e g ( f ) > 1 . W r i t e f = p f 1 w i t h p i r r e d u c i b l e i n T . T h e n , T / T p i s o n e o f t h e s i m p l e c o m p o s i t i o n f a c t o r s o f T / T f . I f g = q 1 q 2 . . . q k w i t h e a c h q i i r r e d u c i b l e i n T , t h e n t h e c o m p o s i t i o n f a c t o r s o f T / T g a r e ( u p t o i s o m o r p h i s m ) T / T q 1 , . . . , T / T q k . δ δ(f) ( g ) , jh(T /T f ) i . e . j h ( T / T g ) , w e B m u s t h a v e T / T p = T / T q j f o r s o ecause m e j . W r i t e g = g 1 q g = = 2

w h e r e q = q j . B y L e m m a 5 . 2 , t h e r e e x i s t s , t T \ { 0 } w i t h d e g ( s ) = d e g ( t ) < d e g ( p ) = d e g ( q ) a n d p s = t q .

# Then, working modulo Q0, we have

h = f g 1 = ( p f 1 ) ( g 1 q g 2 )

1

f 1 ( p q

1 ) ( g 1 g 2 )

1

f 1 ( t s

1 ) ( g 1 g 2 )

1

( f 1 t ) ( g 1 g 2 s )

1

.

L e t h 0 = ( f 1 t ) ( g 1 g 2 s ) 1 . S i n c e h 0 h ( m o d Q 0 ) , w e h a v e δ ( h 0 ) = δ ( h ) = 0 , w h i l e d e g ( f 1 t ) < d e g ( f iterating this process we can repeatedly lower the degree of numerator and denominator to obtain h00 Dwith h00 h0 h (mod Q0). Hence, h DQ0, as desired. ) . B y

R e m a r k . S i n c e K 1 ( Q ) = Q / Q 0 , P r o p . 5 . 3 c a n b e s t a t e d a s s a y i n g t h a t t h e r e i s a n e x a c t s e q u e n c e

K1(D)

K1(Q)

δ D i v ( T )

0.

(5.1)

This can be viewed as part of an exact localization sequence in K-Theory. We prefer the explicit description of Div(T ) and δ given here, as it helps to understand the maps associated with Div(T ).

Let R = Z(T ) = K[y]. So, q(R) = Z(Q). We define Div(R) just as we defined Div(T ) above. Note that this Div(R) coincides canonically with the usual divisor group of fractional ideals of the PID R, since for a R \ {0}, the simple composition factors of R/Ra are the simple modules R/P as P ranges over the prime ideal factors of the ideal Ra.

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