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SK1

OF GRADED DIVISION ALGEBRAS

25

L e t c = N K / T c = N K / T ( a ) = N K / T (a)

=

NrdE(a).

Because

K

is

unramified

over

T

and

a

V K

,

( 1 ) = 1 , s o c 1 + M T . W e h a v e NT/F (c) = NT/F (NK/T (a)) = NK/F ,

(a) = NrdD(a) = 1 .

we

have

c

V T

and

By Hilbert 90, c = b/σ(b) for some b T . This equation still holds if we replace b in it by any F -

multiple of b. Thus, as

T

=

F

since T is unramified over F , we may assume that v(b) = 0. But

further, since T is unramified and cyclic Galois over F , its residue field T is cyclic Galois of degree p we have o v e r F al(T /F ) = hσi where σ is the automorphism of T induced by lies in the fixed field of σ in T , which is F , w i t h i s η V F 1 , w e c a n a s s u m e b = 1 , i . e . , In T there b 1 + M b/σ(b) with η on T . Hence, σ . = b/σ(b) = c = 1. Therefore, b T = b in T . By replacing b by .

S i n c e K i s u n r a m i fi e d a n d h e n c e t a m e o v e r T , P r o p . 4 . 6 s h o w s N K / T ( 1 + M K ) = 1 + M T .

So,

there is s 1 + MK with NK/F

(s) = b. Now, by Skolem–Noether, there is an inner automorphism ϕ

of D such that ϕ(T ) = T and ϕ|T = σ. Since E = CD(T ), we have ϕ is a (non-inner) automor- phism of E, and ϕ(K) is a maximal subfield of E (since K is a maximal subfield of E). We have

NrdE(ϕ(s)) = Nϕ(K)/ϕ(T )

(ϕ(s)) = ϕ NK/T (s)= σ(b). Thus, NrdE(s/ϕ(s)) = b/σ(b) = c.

Now, So a0

there is u a (mod

Dwith ϕ(s) = usu 1. D0). But further, a0 E

S o , ϕ ( s ) 1 + M D . ( 1 + M D ) = 1 + M E

Let (as

a0 = a/(s/ϕ(s)) = a , s , ϕ ( s ) ( 1 + M D a(sus 1u 1) ) E ). Also,

E.

NrdE(a0) = NrdE(a)NrdE(s/ϕ(s)) = NK/T (a)/c = 1.

Since [E : T ] < [D : F ] and E is inertially split over T (since it is split by its maximal subfield K which i s u n r a m i fi e d o v e r T ) , b y i n d u c t i o n o n i n d e x t h e t h e o r e m h o l d s f o r T o v e r E . H e n c e , a 0 E 0 . S i n c a a0 (mod D0), we thus have a D0, as desired. This completes the proof of Step 2. e

Step 3. Suppose D = P F Q, where gcd(ind(P ), ind(Q)) = 1, and suppose the theorem is true for L F Q and K F P for some maximal subfield L of P and K of Q. Then we show using Prop. B.3 below

that the theorem is true for D. L e t C = C D ( L ) . T h e n , C = C P F Q

(L

F F)

=

C L ( P )

F

CQ(F )

=

L

F

Q . A l

so,

L

FD

=

(L

F P)

L

(L

F

Q)

=

M ` ( L )

L

C

=

M ` ( C )

,

w h e r e ` = [ L : F ] = i n d ( P ) . T a k e a n y a ( 1 + M D ) D ( 1 )

. F o r 1 a L D = M ` ( C ) , P r o p . B . 3 s h o w s

that

there

is

c

1 + M C

with

ddet(a)

c

(mod

C 0

),

where

ddet

denotes

the

Dieudonne´

determinant.

Then,

H

ence, c

(1

+

M ` ( C ) M C ) 1 = NrdD(a) = Nrd C ( 1 ) w h i c h l i e s i n C 0 b y h y p o t h e s i s a s C = L (1 a) = NrdC (ddet(1 a)) = NrdC (c). F Q . T h a t i s , d d e t ( 1 a ) =

1

C / C 0

.

Hence, 1 a ker(ddet) = (L F D)0. Therefore, by Lemma B.2, a` D0. Likewise, by looking at

1 a K F D, we obtain ak D0 where k = [K : F ] = ind(Q). a D0, completing Step 3.

Since gcd(`, k) = 1, it follows that

Step 4. We now prove the theorem in full. Let F be a henselian field, and let D be a tame F -central division algebra. If char(F ) = 0, then D is strongly tame over F , so the theorem holds for D by Step 1.

I f c h a r

(F )

= p 6 = 0

we

h

ave

D

=

P

F

Q

w h e r e

Pi

s t h e p - p r i m a r y c o m p o n e n t o

fD

an

dQi

s t h e

tensor product of all the other primary components of D. So, gcd(ind(P ), ind(Q)) = 1. For any maximal s u b fi e l d L o f P , L F Q i s a d i v i s i o n a l g e b r a t a m e o v e r L w i t h i n d ( L F Q ) = i n d ( Q ) , w h i c h i s p r i m e t o p . S o , L F Q i s s t r o n g l y t a m e o v e r L , a n d t h e t h e o r e m h o l d s f o r L F Q b y S t e p 1 . O n t h e o t h e r h a n for any maximal subfield K of Q, we have K F P is tame over K and ind(K F P ) = ind(P ), which is a power of p; hence, K F P is inertially split, as noted in §2. Hence, by Step 2 the theorem holds for d ,

K F P . Thus, by Step 3 the theorem holds for D.

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