24

R. HAZRAT AND A. R. WADSWORTH

Let n = ind(D), so char(F ) n. Take any s ∈ D^{∗}, and let f = x^{k }+ c_{k 1}x^{k 1 }+ . . . + c_{0 }∈ 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 ,

Nrd_{D}(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

≡ s^{n }

(mod D^{0}).

(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 Nrd_{D}(s) = 1 Nrd_{D}(a) = Cor. h e r e i s s ∈ 1 + M F ( a ) 1 with s^{n }= a. Then, by + M D M But, 4.7. 1 ⊆ +m 1. + ∈ F (1 + m)^{n }Nrd_{D}(s^{n}) = =

# 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 = s^{n }≡ Nrd_{D}(s) = 1 + m = 1 (mod D^{0}), i.e., a ∈ D^{0}. 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 [P_{1}] for discrete (rank 1) valuations. (When v is discrete, every tame division algebra is inertially split.)

Suppose ind(D) = p^{k}, 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 ∈ D^{∗ }with b = tazt ^{1}. So az = t ^{1}bt. Then, h a v e

# We have bz

1

bz

1

∈ 1 + M D

,

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

=

N^{P },

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 ]

∈ D^{0}. But

we also have a^{ind(D) }∈ D^{0}, since SK_{1}(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 ∈ D^{0}, as desired. Thus, it suffices to prove the result for D_{1}.

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 = C_{D}(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.