12
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(Tr_{K/F }Tr_{K/F }(ea) = 0.
(a)) = v(a), then Tr_{K/F }
(ea) = Tr_{K/F }
(a), but if v(Tr_{K/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: Nrd_{D}(a) = N_{Z(D)/F }Nrd_{D}(a)^{δ},
(4.2)
for
any
a
∈
V D
\ M D
(thus,
Nrd_{D}(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 λpolynomials.
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 + M_{K }then se = 1 in gr(K). ^
So, as K is defectless over F , by Lemma 4.3
N_{K/F }
(s) = N_{gr(K)/gr(F }_{) }
(se) = 1 in gr(F ), i.e., N_{K/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}] =

K
:
_{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 [HwW_{1}], Prop. 2.2. Take any monic λpolynomial has the simple root b in gr(K) and the valuation on K is henselian, and g is λhomogenizable 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 λpolynomial (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)
+. . .+c_{1}x+tc_{0 }∈ 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)^{n}c_{0 }and N_{L/F }(d) = ( 1)^{n}c_{0}t. Thus, N_{L/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 ,
N_{K/F }
(1 + M_{K }) = N_{L/F }
N_{K/L }
(1 + M_{K })
= N_{L/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 + M_{K }, and Nrd_{D}(a) = N_{K/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 [HwW_{2}], 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
= N_{L/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
.