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22

R. HAZRAT AND A. R. WADSWORTH

the Wedderburn factorization theorem. We now carry over this theorem to the setting of graded division algebra. (This is used in proving Th. 3.3).

Theorem A.1 (Wedderburn Factorization Theorem). Let E be a graded division ring with center T (with

E

torsion-free abelian). Let a be a homogenous element of E which is algebraic over T with minimal

polynomial ha T [x]. of a such that ha = (x

Then, ha splits completely in E. Furthermore, there exist n conjugates a1, . . . , an

an)(x

an

1) . . . (x

a1) in E[x].

Proof. The proof is similar to Wedderburn’s original proof for a division ring ([We], see also [L] for a nice a c c o u n t o f t h e p r o o f ) . W e s k e t c h t h e p r o o f f o r t h e c o n v e n i e n c e o f t h e r e a d e r . F o r f = P c i x i E [ x ] a n d a E , o u r c o n v e n t i o n i s t h a t f ( a ) m e a n s P c i a i . S i n c e E i s t o r s i o n - f r e e , w e h a v e E = E h \ { 0 } .

factorization f

= gk

some conjugate

a0 of a.

I:

Let

f

E[x] with

f (a)

=

g(a0)k(a), for

then

E satisfies k(a) T · E, any ring with T Z(E).)

in E[x].

If a

(Here E

could

be

P i b i x i . T h e n , f = P P bik(a)ai. But, k(a) = te, where t T and e E. b i k x Proof. Let g = P i , s o f ( a ) = P P 1 1)ite = g(eae 1)k(a). b Thus, f(a) = bi(eae i t e a b i e a te = = i e

II: Let f E[x] be a non-zero polynomial. Then r E is a root of f if and only if x

r is a right

divisor of f in E[x]. (Here, E could be any ring.)

Proof. We have xi

ri = (xi

1

  • +

    xi

2r + . . . + ri 1

)(x

r) for any i 1. Hence,

f

f(r) = g · (x

r)

(A.1)

for some g E[x]. So, if f(r) = 0, then f = g · (x

r). Conversely, if x

r is a right divisor of f, then

equation (A.1) shows that x that f(r) = 0.

r is a right divisor of the constant f(r). Since x

r is monic, this implies

III: If a non-zero monic polynomial f E[x] vanishes identically on the conjugacy class A of a (i.e., f(b) = 0 for all b A), then deg(f) deg(ha).

P r o o f . C o n s i d e r f = x m + d 1 x m + . . . + dm E[x] such that f(A) = 0 and m < deg(ha) with m as small 1 a s p o s s i b l e . S u p p o s e a E γ , s o A E γ , a s t h e u n i t s o f w e m a y a s s u m e t h a t e a c h d i E i γ T [ x ] , s o m e d i of f(b) is 0 for each b A, T . m i n i m a l w i t h d j T , E are all homogeneous. Since the E-component a n d s o m e e E s u c h E, write c0 := ece Because f For any c Choose j 1. Thus t h =6 a t e d j . e. d j m1 d 0 j 6 = d j d 0 ` d E[x]. Now, for all b A, we have ` f 0 for ` d 0 but 1 + . . . + d 0 m xm Let j. + < x = = f 0 ( 1 b 0 ) = [ f ( b ) ] 0 = 0 0 = 0 . S i n c e e A e = A , t h i s s h o w s t h a t f 0 ( A ) = 0 . L e t g = f f 0 , w h i c h h a s

d e g r e e j < m w i t h l e a d i n g c o e ffi c i e n t d j

d 0 j . T h e n , g ( A ) = 0 . B u t , d j

d 0 j E j γ

\ {0} ⊆ E. Thus,

( d j

d 0 j )

1g is monic of degree j < m in E[x], and it vanishes on A. This contradicts the choice of f;

hence, m deg(ha).

We now prove the theorem. Since ha(a) = 0, by (II), ha E[x] · (x

a). Take a factorization

ha = g · (x

ar) . . . (x

a1) ,

where g E[x], a1, . . . , ar A and r is as large as possible. Let k = (x

ar) . . . (x

a1) E[x]. We claim

that

k(A) = 0,

where

A

is

the

conjugacy

class

of

a.

For,

suppose

there

exists

bA

such

that

k ( b ) =6

0.

Since k(b) is homogenous, we have k(b) E. (b) = 0, (I) implies as b A; But, ha = gk, and h hence, a t h a t g ( b 0 ) = 0 f o r s o m e c o n j u g a t e b 0 o f b . W e c a n t h e n w r i t e g = g 1 · ( x b0), by (II). So ha has a right

factor (x

b0)k = (x

b0)(x

ar) . . . (x

a1), contradicting our choice of r.

Thus k(A) = 0, and using (III),

we have r deg(ha), which says that ha = (x

ar) . . . (x

a1).

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