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R. HAZRAT AND A. R. WADSWORTH

Remarks 4.5. (i) Again, we do not need that v be henselian for Cor. 4.4. It suffices that the valuation v on F extends to D and D is tame over F .

(ii) Analogous results hold for the trace and reduced trace, with analogous proof. In the setting of ^

Lemma 4.3, we have: if v(TrK/F TrK/F (ea) = 0.

(a)) = v(a), then TrK/F

(ea) = TrK/F

(a), but if v(TrK/F

(a)) > v(a), then

(iii) By combining Cor. 4.4 with equation (3.3), for a tame valued division algebra D over henselian

field F , we can relate the reduced norm of D with the reduced norm of D as follows: NrdD(a) = NZ(D)/F NrdD(a)δ,

(4.2)

for

any

a

V D

\ M D

(thus,

NrdD(a)

V F

\ M F

)

and

δ

=

ind(D)

ind(D) [Z(D)

:

F]

(cf.

[E],

Cor.

2).

The next proposition will be used several times below. It was proved by Ershov in [E], Prop. 2, who refers to Yanchevski˘ı [Y] for part of the argument. We give a proof here for the convenience of the reader, and also to illustrate the utility of λ-polynomials.

Proposition 4.6. Let F K be elds with henselian valuations v such that [K : F ] < and K is tamely r a m i e d o v e r F . T h e n N K / F ( 1 + M K ) = 1 + M F .

Proof. If s 1 + MK then se = 1 in gr(K). ^

So, as K is defectless over F , by Lemma 4.3

NK/F

(s) = Ngr(K)/gr(F )

(se) = 1 in gr(F ), i.e., NK/F

( s ) 1 + M F . T h u s N K / F

( 1 + M K ) 1 + M F . T o p r o v e

that this inclusion is an equality, we can assume [K : F ] > 1. We have [gr(K) : gr(F )] = [K : F ] > 1, since tamely ramified extensions are defectless. Also, the tame ramification implies that q(gr(K)) is separable over q(gr(F )). For, q(gr(F )) · gr(K)0 is separable over q(gr(F )) since gr(K)0 = K and K is separable over gr(F )0 = F . But also, q(gr(K)) is separable over q(gr(F ))·gr(K)0 because [q(gr(K)) : q(gr(F )) · gr(K)0] =

|

:

F |, which is not a multiple of char(F ). Now, take any homogenous element b ∈ gr(K), b ∈6 gr(F ),

and let g be the minimal polynomial of b over q(gr(F )). Then g gr(F )[x], b is a simple root of g,

λ = deg(b), by [HwW1], Prop. 2.2. Take any monic λ-polynomial has the simple root b in gr(K) and the valuation on K is henselian, and g is λ-homogenizable where f F [x] with f(λ) = g. Since f(λ)

by Prop. 4.2(iv) there is a K such that a is a simple root of f and ea = b. Let L = F (a) K. Write

h is a λ-polynomial (because f is) and h(λ) e f = x n + c n 1 x n 1 + . . . + c 0 . T a k e a n y t 1 + M F , a n d l e t h = x n + c n 1 x = f ( λ ) = g in gr(F )[x]. Since h n1 (λ)

+. . .+c1x+tc0 F [x]. Then has the simple root b in gr(L),

h h a s a s i m p l e r o o t d i n L w i t h d = b = e a b y P r o p . 4 . 2 ( i v ) . S o , d a 1 1 + M L . T h e p o l y n o m i a l s f a n d are irreducible in F [x] by Prop. 4.2(i), as g is irreducible in gr(F )[x]. Since f (resp. h) is the minimal poly- h

1) = t, n o m i a l o f a ( r e s p . d ) o v e r F , w e h a v e N (a) = ( 1)nc0 and NL/F (d) = ( 1)nc0t. Thus, NL/F (da L / F s h o w i n g t h a t N L / F ( 1 + M L ) = 1 + M F . I f L = K , w e a r e d o n e . I f n o t , w e h a v e [ K : L ] < [ K : F ] , a n d K i s t a m e l y r a m i fi e d o v e r L . S o , b y i n d u c t i o n o n [ K : F ] , w e h a v e N K / L ( 1 + M K ) = 1 + M L . H e n c e ,

NK/F

(1 + MK ) = NL/F

NK/L

(1 + MK )

= NL/F

( 1 + M L ) = 1 + M F .

Corollary 4.7. Let F be a eld with henselian valuation v, and let D be an F -central division algebra w h i c h i s t a m e w i t h r e s p e c t t o v . T h e n , N r d D ( 1 + M D ) = 1 + M F .

Proof.

T a k e a n y a 1 + M D

and any maximal subfield K

of D with a K.

Then, K

is defectless over F ,

since D is defectless over F . So, a 1 + MK , and NrdD(a) = NK/F ( a ) 1 + M F b y t h e fi r s t p a r t o f t h e p r o o f o f P r o p . 4 . 6 , w h i c h r e q u i r e d o n l y d e f e c t l e s s n e s s , n o t t a m e n e s s . T h u s , N r d D ( 1 + M D ) 1 + M F . F the reverse inclusion, recall from [HwW2], Prop. 4.3 that as D is tame over F , it has a maximal subfield L with L tamely ramified over F . Then by Prop. 4.6, o r

1 + M F

= NL/F

( 1 + M L ) = N r d D ( 1 + M L ) N r d D ( 1 + M D ) 1 + M F ,

so equality holds throughout

.

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