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24

R. HAZRAT AND A. R. WADSWORTH

Let n = ind(D), so char(F ) n. Take any s D, and let f = xk + ck 1xk 1 + . . . + c0 F [x] be the minimal polynomial of s over F . By applying the Wedderburn factorization theorem to f (see [ L ] , ( 1 6 . 9 ) , p p . 2 5 1 2 5 2 , o r A p p e n d i x A a b o v e ) , w e s e e t h a t t h e r e e x i s t d 1 , . . . , d k D w i t h ( 1 ) k c 0 =

( d 1 s d 1 1 ) . . . ( d k s d 1 k ) . H e n c e , a s D / D 0 i s a b e l i a n ,

NrdD(s)

=

[(

1 ) k c 0 ]

n/k

s

k

( d 1

s d 1

1

s

1 ) . . . ( d k s d 1 k s

1)

n/k

sn

(mod D0).

(B.1)

N o w , t a k e a n y a 1 + M D w i t h N r d D ( a ) = 1 . S i n c e c h a r ( n, Hensel’s Lemma applied over F (a) shows F ) t h a t t NrdD(s) = 1 NrdD(a) = Cor. h e r e i s s 1 + M F ( a ) 1 with sn = a. Then, by + M D M But, 4.7. 1 +m 1. + F (1 + m)n NrdD(sn) = =

If

m =6

0,

then

we

have

1

=

(1 + m)n

= 1 + n m + r with v(r) 2v(m), which would

imply

that

v(n m) = v(r) > v(m).

This cannot occur since char(F )

n; hence, m = 0.

Thus, by

(B.1)

a = sn NrdD(s) = 1 + m = 1 (mod D0), i.e., a D0. This completes Step 1.

Step 2. We prove the theorem if D is inertially split of prime power degree over F . This is a direct adaptation of Platonov’s argument in [P1] for discrete (rank 1) valuations. (When v is discrete, every tame division algebra is inertially split.)

Suppose ind(D) = pk, p prime and D is inertially split. Then, D has a maximal subfield K which is u n r a m i fi e d o v e r F ( c f . [ J W ] , L e m m a 5 . 1 , o r [ W 2 ] , T h . 3 . 4 ) T a k e a n y a ( 1 + M D ) D ( 1 ) . W e fi r s t p u s h a i n t o K . S i n c e K i s s e p a r a b l e o v e r F , t h e r e i s y K w i t h K = F ( y ) . C h o o s e a n y z V K w i t h z = y . K = F (z), by dimension count, as F (z) F (y). Note that az = z in D. If f is the minimal polynomial S o o f a z o v e r F , t h e n f V F [ x ] a s a z V D , a n d z = a z i s a r o o t o f t h e i m a g e f o f f i n F [ x ] . W e deg(f) = deg(f) = [F (az) : F ] [K : F ] = [F (z) : F ]. Hence, f is the minimal polynomial of z over F , so z is a simple root of f. By Hensel’s lemma applied over K, K contains a root b of f with b = z. Since b and az have the same minimal polynomial f over F , by Skolem–Noether there is t Dwith b = tazt 1. So az = t 1bt. Then, h a v e

We have bz

1

bz

1

1 + M D

,

as b = z. Therefore, a = t 1btz K, as b, z K, and bz we may 1 1 replace a by bz 1 , = (t 1btb 1)(bz 1). a ( m o d D 0 ) ; s o , N r d D ( b z so we may assume a K. 1) = NrdD(a) = 1, and

Let N be the maximal

the normal closure of K u n r a m i fi e d e x t e n s i o n F n r

over F , and let G = of F is Galois over F

al(N/F ). (cf. [EP],

Since K is unramified over F and Th. 5.2.7, Th. 5.2.9, pp. 124–126),

N

F n r

;

so

N

is

also

unramified

over

F.

Let

P

be

a

p-Sylow

subgroup

of

G

and

let

L

=

NP ,

the

fixed

field

of

P.

Thus,

[L : F ] = |G : P |,

which

is

prime

to

p,

and

N

is

Galois

over

L

with

al(N/L) = P .

Since

d [ L : F ] , i n d ( D ) 1 , D 1 d i v i s i o n L F Di r i n g a n d K 1 = L F K i s a fi e l d w i t h K 1 = L · K N . S o , K 1 i s u n gc r a m i fi e d o v e r F a sa n d h = = e n c e o , v e r L . W e h a v e N r d D 1 ( 1 a ) = N r d D ( a ) = 1 a n d 1 a 1 + M D

s o i f w e k n e w t h e r e s u l t f o r D 1 , w e w o u l d h a v e 1 a D 0 1 . B u t t h e n b y L e m m a B . 2 , a [ L : F ]

D0. But

we also have aind(D) D0, since SK1(D) is ind(D)-torsion (by [D], p. 157, Lemma 2 or Lemma B.2 above with L a maximal subfield of D). Since gcd [L : F ], ind(D)= 1, it would follow that a D0, as desired. Thus, it suffices to prove the result for D1.

T o s i m p l i f y n o t a t i o n , r e p l a c e D 1 b y D , K 1 b y K , 1 a b y a , a n d L b y F . B e c a u s e F K N w i t N Galois over F , any subfield T of K minimal over F corresponds to a maximal subgroup of al(N/F ) containing al(N/K). Since [N : F ] is a power of p, by p-group theory such a maximal subgroup is normal h

in

al(N/F ) and of index p. Thus, T is Galois over F and [T : F ] = p. So

al(T/F ) is a cyclic group ,

say al(T/F ) = hσi. Let E = CD(T ), the centralizer of T in D; so F T K E D. Note that K is a maximal subfield of E, since it is a maximal subfield of D.

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