SK_{1 }

OF GRADED DIVISION ALGEBRAS

11

homogeneous (of degree v(a_{0})) with respect to the the grading on gr(F )[x] as in (2.3) with θ = λ. Also, f^{(λ) }has the same degree as fas a polynomial in x.

# The

λ-polynomials

are

useful

generalizations

of

polynomials

h

∈

V F

[x]

with

h ( 0 ) =6

0—these

are

the

0-polynomials.

# The

following

proposition

collects

some

basic

properties

of

λ-polynomials

over

henselian

fields, which are analogous to well-known properties for 0-polynomials, and have similar proofs.

See, e.g.,

[EP], Th. 4.1.3, pp. 87–88 for proofs for 0-polynomials, and [MW], Th. 1.9 for proofs for λ-polynomials.

# Proposition 4.2. Suppose the valuation v on F is henselian. Then,

(i) If f is a λ-polynomial and f = gh in F [x], then g and h are λ-polynomials and f^{(λ) }gr(F )[x]. So, if f^{(λ) }is irreducible in gr(F )[x], then f is irreducible in F [x].

= g^{(λ) }

h^{(λ) }

in

(ii) If f =

# P

n i=0

a_{i}x

i

is

an

irreducible

polynomial

in

# F [x]

with

a

n a 0 6 =

0,

then

f

is

a

λ-polynomial

for λ = (v(a_{0})

v(a_{n}))/n.

Furthermore, f^{(λ) }

= f a n h s f o r s o m e i r r e d u c i b l e m o n i c λ - h o m o g e n i z a b l e

polynomial h ∈ gr(F )[x].

( i i i ) I f f i s a λ - p o l y n o m i a l i n F [ x ] a n d i f f ( λ ) = g 0 h 0 i n g r ( F ) [ x ] w i t h g c d ( g 0 , h 0 ) = 1 , t h e n t h e r e e x i λ-polynomials g, h ∈ F [x] such that f = gh and g^{(λ) }= g^{0 }and h^{(λ) }= h^{0}. s t

(iv) If f is a λ-polynomial in F [x] and if f^{(λ) }in F with ea = b.

has a simple root b in gr(F ), then f has a simple root a

(v)

# Suppose

k

is

a

λ-homogenizable

polynomial

in

gr(F )[x]

with

k ( 0 ) =6

0,

and

suppose

f

∈ F [x]

with

e f = k. Then f is a λ-polynomial and f^{(λ) }= k.

Lemma 4.3. Let F ⊆ K be elds with [K : F ] < ∞. Let v be a henselian valuation on F such that K is defectless over F . Then, for every a ∈ K^{∗}, with ea its image in gr(K)^{∗},

^ N K / F ( a ) = N g r ( K ) / g r ( F ) ( e a ) .

Proof. Let n = [K : F ].

Note that [gr(K) : gr(F )] = n as K is defectless over F .

Let

f = x^{` }+ c_{` 1 }

x^{` 1 }

+

. . . + c

_{0 }∈ F [x] be the minimal polynomial of a over F . Then f is irreducible in

# F [x] and since v is henselian, f is a λ-polynomial, where λ = v(a) = v(c_{0})/n (see Prop. 4.2(ii)). Let f^{(λ) }

be the corresponding λ-homogenizable polynomial in gr(F )[x] as in (4.1). Then f^{(λ) }

(ea) = 0 in gr(K) (by

# Prop. 4.2(i) with g = x

a), and by Prop. 4.2(ii) f^{(λ) }

has only one monic irreducible factor in gr(F )[x],

say f^{(λ) }

= h^{s}, with deg(h) = `/s.

Since f^{(λ) }

(ea) = 0, h must be the minimal polynomial of ea over

gr(F ) and over is irreducible in

q(gr(F )). (Recall that since gr(F ) is integrally closed, a monic polynomial in gr(F )[x]

gr(F )[x]

iff

it

is

irreducible

in

q(gr(F ))[x].)

We

have

## N_{K/F }

(a)

=

(

1)

n c n / ` 0

.

Hence,

as

q ( g r

# (K))

∼ = g r

N (K) ⊗_{gr(F ) }q(gr(F )), g r ( K ) / g r ( F ) ( e a ) = N q ( g r ( K ) ) / q ( g r ( F ) ) ( e a ) = ( 1 ) n h ( 0 )

ns/`

= ( 1)^{n}(h(0)^{s})^{n/` }

= ( 1 ) n ( e c 0

^{n/`})

=

(

^

1 ) n c 0

n/`

^

= N_{K/F }

(a).

Remark. The preceding lemma is still valid if v on F is not assumed to be henselian, but merely assumed to have a unique and defectless extension to K. This can be proved by scalar extension to the henselization

F^{h }

o

f F . ( S i

n c e v e x t e n d s u n i q u e l y a n

d d e f e c t l e s s l y t o

K , K

⊗

F F h i

sa

fi e l d

, an

d

gr

(K

⊗

F F h )

∼

=

gr

gr

( K ) . )

Corollary 4.4. Let F be a eld with henselian valuation v, and let D be a tame F -central division algebra.

^ Then for every a ∈ D^{∗}, Nrd_{gr(D)}(ea) = Nrd_{D}(a).

Proof. Recall from §2 that the assumption D is tame over F means that [D : F ] = [gr(D) : gr(F )] and gr(F ) = Z(gr(D)). Take any maximal subfield L of D containing a. Then L/F is defectless as D/F is defectless, so [gr(L) : gr(F )] = [L : F ] = ind(D) = ind(gr(D)). Hence, using Lemma 4.3 and Prop. 3.2(ii),

we have,

^^ N r d D ( a ) = N L / F ( a ) = N g r ( L ) / g r ( F ) ( e a ) = N r d

_{gr(D)}(ea).