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6

R. HAZRAT AND A. R. WADSWORTH

Remark 3.1. The reduced norm for an Azumaya algebra is defined using a splitting ring, and in general splitting rings can be difficult to find. But for a graded division algebra E we observe that, analogously to the case of ungraded division rings, any maximal graded subfield L of E splits E. For, the centralizer C = CE(L) is a graded subring of E containing L, and for any homogeneous c C, L[c] is a graded subfield of E containing L. Hence, C = L, showing that L is a maximal commutative subring of E. Thus,

b y L

e m m a 5 . 1 . 1 3 ( 1 )

, p.

1 4 1 o f [ K ]

, as

E i s A

zumaya,

E

TL

=

E n d L ( E )

=

M n ( L ) . T h

us, we can compute

reduced norms for elements of E by passage to E T L.

We have other tools as well for computing NrdE and TrdE:

Proposition 3.2. Let E be a graded division ring with center T . Let q(T ) be the quotient eld of T , and let q(E) = E T q(T ), which is the quotient division ring of E. We view E q(E). Let n = ind(E) =

ind(q(E)). Then for any a E, (i) charE(x, a) = charq(E)(x, a), so

NrdE(a) = Nrdq(E)

(a)

and

TrdE(a) = Trdq(E)

(a).

(3.2)

(ii) If K is any graded sub eld of E containing T and a K, then

NrdE(a) = NK/T

(a)n/[K :T ]

and

TrdE(a) =

n [K :T ]

TrK/T

(a).

(iii) For γ

E , i f a E γ t h e n N r d E ( a ) E n γ

a n d T r d ( a ) E γ . I n p a r t i c u l a r , E ( 1 )

E 0 .

(iv)

Set δ = ind(E)

i n d ( E 0

) [ Z ( E 0

) : T 0

]

.

I f a E 0

, then,

N r d E ( a ) = N Z ( E 0 ) / T 0

Nrd

E 0 ( a )

δ

T 0

and

Trd

E

(a) = δ Tr

Z ( E 0 ) / T 0 T r d

E 0 ( a ) T 0 .

(3.3)

Proof. (i) The construction of reduced characteristic polynonials described above is clearly compatible with

scalar extension of the ground ring. Hence, charE(x, a) = charq(E) a 1 in E T q(T ) ). The formulas in (3.2) follow immediately.

(x, a) (as we are identifying a E with

(ii) Let ha = xm + tm 1xm 1 + . . . + t0 q(T )[x] be the minimal polynomial of a over q(T ). As noted in [HwW1], Prop. 2.2, since the integral domain T is integrally closed and E is integral over T , we have ha T [x]. Let `a = xk +sk 1xk 1 +. . .+s0 T [x] be the characteristic polynomial of the T -linear function

on the free T -module K given by c 7→ ac. By definition, NK/T

(a) = ( 1)ks0 and TrK/T

(a) =

sk 1

. Since

q(K) = K T q(T ), we have [q(K) : q(T )] = [K : T ] = k and `a is also the characteristic polynomial for

the q(T )-linear transformation of q(K) given by q 7→ aq. So, `a = h

k/m a

. Since charq(E)

(x, a) = h

n/m a

(see

[R], Ex. 1, p. 124), we have charq(E)

( x , a ) = ` n / k a

. Therefore, using (i),

NrdE(a) = Nrdq(E)

(a) =

( 1 ) k s 0

n/k

= N K / T ( a )

n/k

.

The formula for TrdE(a) in (ii) follows analogously.

n/m a (iii) From the equalities charE(x, a) = char (x, a) = h noted in proving (i) and (ii), we have q(E) m1 NrdE(a) = [( 1)mt0]n/m and TrdE(a) = . A s n o t e d i n [ H w W 1 ] , P r o p . 2 . 2 , i f a E γ , t h e n i t n m t s m1 minimal polynomial ha is γ-homogenizable in T [x] as in (2.3) above. Hence, t0 Eand t E γ .

Therefore, NrdE(a) E

a n d T r d ( a ) E γ . I f a E ( 1 )

then a is homogeneous, since it is a unit of E,

and since 1 = NrdE(a) En deg(a)

, necessarily deg(a) = 0.

( i v ) S u p p o s e a E 0 . T h e n , h a i s 0 - h o m o g e n i z a b l e i n T [ x ] , i . e . , h a T 0 [ x ] . H e n c e , h a i s t h e m i n i m a l p o l y n o m i a l o f a o v e r t h e fi e l d T 0 . T h e r e f o r e , i f L i s a n y m a x i m a l s u b fi e l d o f E 0 c o n t a i n i n g a , w e h a v e N L / T 0 ( a ) = [ ( 1 ) m t 0 ] [ L : T 0 ] / m . N o w ,

n / m = δ i n d ( E 0 ) [ Z ( E 0 ) : T 0 ] m = δ [ L : T 0 ] / m .

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