SK_{1 }

OF GRADED DIVISION ALGEBRAS

25

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

=

Nrd_{E}(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 N_{T/F }(c) = N_{T/F }(N_{K/T }(a)) = N_{K/F },

(a) = Nrd_{D}(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 bη .

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 + M_{K }with N_{K/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 = C_{D}(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

Nrd_{E}(ϕ(s)) = N_{ϕ(K)/ϕ(T }_{) }

(ϕ(s)) = ϕ N_{K/T }(s)^{ }= σ(b). Thus, Nrd_{E}(s/ϕ(s)) = b/σ(b) = c.

Now, So a^{0 }

there is u ≡ a (mod

∈ D^{∗ }with ϕ(s) = usu ^{1}. D^{0}). But further, a^{0 }∈ E ∩

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

Let (as

a^{0 }= a/(s/ϕ(s)) = a , s , ϕ ( s ) ∈ ( 1 + M D a^{}(sus ^{1}u ^{1}) ∈ ) ∩ E ). Also,

E.

Nrd_{E}(a^{0}) = Nrd_{E}(a)^{ }Nrd_{E}(s/ϕ(s)) = N_{K/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 ≡ a^{0 }(mod D^{0}), we thus have a ∈ D^{0}, 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

C_{Q}(F )

=

# L

⊗

## F

Q . A l

so,

L

⊗

_{F}D

∼ =

(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 = Nrd_{D}(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) = Nrd_{C }(ddet(1 ⊗ a)) = Nrd_{C }(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^{` }∈ D^{0}. Likewise, by looking at

1 ⊗ a ∈ K ⊗_{F }D, we obtain a^{k }∈ D^{0 }where k = [K : F ] = ind(Q). a ∈ D^{0}, 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.